Conditions for spacetime to have flat spatial slices [Archive]

PeterDonis

Nov10-10, 12:19 PM

In another thread,

http://www.physicsforums.com/showpost.php?p=2973770&postcount=45,

some questions came up about what the conditions are for a spacetime to admit flat spatial slices, and for a spacetime to have a time-independent "scale factor" (see definition below). These questions seemed interesting enough to warrant a new thread.

One key definition:

(A) The term "scale factor" is here extended from what I believe is its normal usage, which refers to the coefficient a(t) in front of the spatial part of the FRW metric in at least one of its "standard" forms. Physically, a(t) tracks the "comoving distance" between nearby geodesics that are "at rest", meaning that they are the worldlines of observers who see the universe as homogeneous and isotropic. By analogy, we can extend the term to refer to the "comoving distance" between nearby worldlines "at rest" in other spacetimes (with some caveats to the definition of "comoving", since the "nearby worldlines at rest" won't always be geodesics), and specifically to the fact that, in the other spacetimes we will be considering, there is no time-dependence in the "scale factor", meaning, roughly speaking, that the "size" of a given region of space does not change with time.

The reason this came up in the other thread was my use of the term "spatially flat" to describe the FRW spacetime with k = 0. Passionflower pointed out that, since the scale factor in this spacetime is time-dependent, even though each hypersurface of constant FRW coordinate time is flat, the "size" of a given region of space changes from one spatial slice to another, so, for example, a geometric figure "drawn" using geodesics in one spatial slice would have "expanded" in a later slice. This raised the question of what would be required of a spacetime to ensure that this didn't happen.

Two proposed conditions, based on the discussion in the other thread, are:

(1) For a spacetime to admit a metric with a time-independent scale factor, it must be stationary. The "canonical" example here is the Kerr spacetime, which is stationary but not static, and which does not admit any metric with flat spatial slices. It does, however, have the property that none of the metric coefficients depend on the time (in any of the standard coordinate charts--this is, of course, a restatement of the fact that the spacetime is stationary), which means that the "scale factor" is constant in time; contrast this with, for example, any of the FRW spacetimes, which have a scale factor that varies with time (see above).

(2) For a spacetime to admit a metric with flat spatial slices *and* a time-independent scale factor, it must be static. The "canonical" example here is, of course, Schwarzschild spacetime, which admits a coordinate chart (the Painleve chart) with flat spatial slices, and which also shares the time-independent scale factor property with the Kerr metric (of which it is a special case). (Of course there are other charts for this spacetime as well, which do not have flat spatial slices, but that doesn't matter for our purposes here as long as there is *some* chart that does.)

Condition (1) seems straightforward: a time-independent scale factor requires a time-independent metric.

I'm not sure about condition (2), though, because I'm not sure the interior portion of Schwarzschild spacetime--with r < 2M--qualifies as "static". (This portion is covered by the Painleve chart, so we can't finesse the issue by only considering the exterior portion.) The definitions in Wald are, briefly, that "stationary" means there is a timelike Killing vector field, and "static" means that field is hypersurface orthogonal. In the exterior of Schwarzschild spacetime, \partial_{t} meets both these conditions, but it doesn't in the interior since it's no longer timelike. As far as I know, there is no other Killing vector field that *is* timelike in the interior (let alone hypersurface orthogonal).

There is one additional condition that we didn't propose an answer for in the other thread:

(3) What is required for a spacetime to admit a metric with flat spatial slices (but not necessarily a time-independent scale factor)? The "canonical" example here would be the FRW spacetime with k = 0, which has flat spatial slices but a scale factor that varies with time.

The key property that enables the construction of the metric for FRW spacetime in simple form is that it is isotropic; however, that condition can't be the right one for admitting flat spatial slices because it is too strong--Schwarzschild spacetime is not isotropic (it can be represented in so-called "isotropic coordinates", but that's not the same thing--the radial direction is still fundamentally different from the other two spatial directions). So maybe "spherically symmetric" is the right condition? I believe FRW spacetimes meet the textbook definition for that, even though they're not usually thought of in that way.

Can any of the experts here shed any more light on conditions (1), (2), and (3)?

Passionflower

Nov10-10, 12:58 PM

I'm not sure about condition (2), though, because I'm not sure the interior portion of Schwarzschild spacetime--with r < 2M--qualifies as "static". (This portion is covered by the Painleve chart, so we can't finesse the issue by only considering the exterior portion.)
People often confuse (I am not saying you do) things here:

There are two Schwarzschild solutions one is a vacuum solution and then there is an interior solution which is obviously not a vacuum solution. But the confusion comes because often the region 0 < r < 2M of the vacuum solution is also called interior.

The interior solution is static, however the vacuum solution is only static for 2M < r < infinity.

PAllen

Nov10-10, 01:17 PM

For the purposes of this discussion, I have another question about what you want 'spatially flat' to mean. From discussion in another thread I initiated, it is clear that you can have valid coordinate systems with no timelike coordinate at all (and this is possible even in Minkowski spacetime of SR). So, if we're going to talk about something like time, and 'slices of constant time', we need to agree on a coordinate independent definition. Perhaps we need to say that a spatial slice is locally Euclidean, as bcrowell suggested in my other thread. I'm sure others can suggest better definitions, but just a coordinate t being constant means nothing.

PeterDonis

Nov10-10, 01:28 PM

People often confuse (I am not saying you do) things here:

There are two Schwarzschild solutions one is a vacuum solution and then there is an interior solution which is obviously not a vacuum solution. But the confusion comes because often the region 0 < r < 2M of the vacuum solution is also called interior.

The interior solution is static, however the vacuum solution is only static for 2M < r < infinity.

Good point; I should clarify that here by "Schwarzschild interior solution" I mean specifically the interior *vacuum* solution, i.e., the vacuum solution for r < 2M.

PeterDonis

Nov10-10, 01:33 PM

So, if we're going to talk about something like time, and 'slices of constant time', we need to agree on a coordinate independent definition. Perhaps we need to say that a spatial slice is locally Euclidean, as bcrowell suggested in my other thread. I'm sure others can suggest better definitions, but just a coordinate t being constant means nothing.

Roughly speaking, by "time" I mean "a coordinate that's timelike". :wink: I guess a somewhat more precise way of saying that would be that the vector field generated by the partial derivative with respect to the coordinate must be a timelike vector field. I think (but am not positive) that's enough to guarantee that any slice of the spacetime with a constant value of such a coordinate must be a spacelike hypersurface.

pervect

Nov10-10, 02:00 PM

PAllen

Nov10-10, 02:20 PM

I'm not sure if this will help any, but here goes:

"Flat" is a little ambiguous, but usually people mean that the Riemann tensor vanishes everywhere.

Occasionally one will see the term used loosely in some other sense.

I'm not sure if saying that "flat" means that the opposite sides of a parallelogram are always equal in length is equivalent to the above definition or not, though I suspect it's very close even if it turns out to be slightly "off". Of course, you'll need some notion of "parallel" to make this work.

If you take a spatial slice (setting a timelike coordinate constant is one way of slicing space-time, it doesn't really matter how you do it as long as you manage it somehow), a metric is induced on the spatial slice by your space-time metric. Basically you can use the space-time metric to compute the lengths of curves that lie entirely in the spatial slice, and this is sufficient to determine the distances between points on that space-like slice and hence the metric on the space-like surface.

It's not clear to me how to tell if a given metric allows a flat space-like slice to be taken, however.

Agreeing with the ambiguity referenced here, it is easy to have a purely spacelike hypersurface on which the normal notions of spatial flatness fail (e.g. the triangle inequality is not true, even locally, everywhere; for example, I believe this is true for 'natural' spacelike slices of an accelerated frame in SR; this is because a little bit of mixed signature metric components are mixed into the induced 3-metric). So there must be some agreed on definition of a flat spatial slice. Pervect's parallellogram definition is a good one. For my purposes, I was more interested in triangle inequality (which amounts, I think, to positive definite spatial metric). (The triangle inequality guarantees that a spatial geodesic is a true local minimum).

PeterDonis

Nov10-10, 02:53 PM

If you take a spatial slice (setting a timelike coordinate constant is one way of slicing space-time, it doesn't really matter how you do it as long as you manage it somehow), a metric is induced on the spatial slice by your space-time metric. Basically you can use the space-time metric to compute the lengths of curves that lie entirely in the spatial slice, and this is sufficient to determine the distances between points on that space-like slice and hence the metric on the space-like surface.

This is the procedure for finding a "flat spatial slice" that I was thinking of. For example, if we take the Painleve metric and set dt = 0 (where t is the Painleve t-coordinate), the induced 3-metric on the resulting t = constant hypersurface is the metric for Euclidean 3-space in spherical coordinates. I think we're safe in saying that Euclidean 3-space is flat. :wink:

"Flat" is a little ambiguous, but usually people mean that the Riemann tensor vanishes everywhere.

Using the procedure above, this would mean, I take it, "the Riemann tensor of the induced 3-metric on the spacelike slice vanishes everywhere." We also have to add the proviso that PAllen mentioned, that the induced 3-metric can't be pseudo-Riemannian, it has to be Riemannian (i.e., signature +++, which I think captures the idea that it needs to be a "purely spatial" metric--I use the term "Riemannian" instead of "positive definite" because of course we can still have zero distances if we calculate the metric from a point to itself). I think that will work as the definition of a "flat spatial slice" for this discussion.

It's not clear to me how to tell if a given metric allows a flat space-like slice to be taken, however.

Darn it!

bcrowell

Nov10-10, 04:40 PM

The reason this came up in the other thread was my use of the term "spatially flat" to describe the FRW spacetime with k = 0. Passionflower pointed out that, since the scale factor in this spacetime is time-dependent, even though each hypersurface of constant FRW coordinate time is flat, the "size" of a given region of space changes from one spatial slice to another, so, for example, a geometric figure "drawn" using geodesics in one spatial slice would have "expanded" in a later slice. This raised the question of what would be required of a spacetime to ensure that this didn't happen.

This sounds like what the volume expansion \theta and expansion tensor \theta_{ab} are designed for. I'd suggest using those tools rather than reinventing the wheel. The relevant definitions are given in Wald and in Hawking and Ellis.

I'm not really seeing how this relates to the question of flat spatial slices.

Mentz114

Nov10-10, 05:55 PM

It's not clear to me how to tell if a given metric allows a flat space-like slice to be taken, however.

I think, one has to find a coordinate transformation so that the metric written in the new coordinates has a spatial part which is E3.

It's possible to do that in the Schwarzschild and FLRW spacetimes.

This coordinate transformation of the FRW ( in Cartesian coords)
[tex]
\begin{align*}
dt'&=dt\\
dx'&=dx-\frac{2\,x}{3\,t}dt\\
dy'&=dy-\frac{2\,y}{3\,t}dt\\
dz'&=dx-\frac{2\,z}{3\,t}dt\\
\end{align*}
[/itex]

transforms the FLRW metric into -

[tex]
\left[ \begin{array}{cccc}
\frac{4\,{z}^{2}+4\,{y}^{2}+4\,{x}^{2}-9\,{t}^{2}}{9\,{t}^{2}} & -\frac{2\,x}{3\,t} & -\frac{2\,y}{3\,t} & -\frac{2\,z}{3\,t} \\\
-\frac{2\,x}{3\,t} & 1 & 0 & 0 \\\
-\frac{2\,y}{3\,t} & 0 & 1 & 0 \\\
-\frac{2\,z}{3\,t} & 0 & 0 & 1}\end{array} \right]
[/itex]

which has the required E3 spatial slices and clearly the frame-field defined by the differentials represents a comoving observer of some sort.

PeterDonis

Nov10-10, 06:29 PM

This sounds like what the volume expansion \theta and expansion tensor \theta_{ab} are designed for.

<sound of me hitting myself over the head and saying "D'oh!">

Yes, you're right, I need to refresh my memory on the definitions of those and the required conditions for them to vanish.

I'm not really seeing how this relates to the question of flat spatial slices.

Not flat spatial slices per se, but the question of whether the term "spatially flat" is appropriate when the scale factor is not time-independent, or whether that term has connotations of a constant (in time) scale factor (which would mean a vanishing expansion) as well as no spatial curvature on each individual spatial slice.

PeterDonis

Nov10-10, 06:31 PM

I think, one has to find a coordinate transformation so that the metric written in the new coordinates has a spatial part which is E3.

It's possible to do that in the Schwarzschild and FLRW spacetimes.

I was looking for some more general set of conditions that determine when this is possible for a spacetime, something like the ones I proposed in the OP (for example, is spherical symmetry necessary? is it sufficient?).

Mentz114

Nov10-10, 06:39 PM

I was looking for some more general set of conditions that determine when this is possible for a spacetime, something like the ones I proposed in the OP (for example, is spherical symmetry necessary? is it sufficient?).

I think the actual symmetry is irrelevant, but obviuosly the more there is the easier to analyse. I was editing my post while you were posting, sorry if this causes confusion.

[edit]

It seems clear that a transformation of the type

\begin{align*}
dt'&=dt\\
dx'&=dx-f_1(x,y,z,t)dt\\
dy'&=dy-f_2(x,y,z,t)dt\\
dz'&=dx-f_3(x,y,z,t)dt\\
\end{align*}

where f_i(x,y,z,t) are chosen to reflect the symmetry of the metric. The tensor product of that coframe basis will always give a spatial part diag(1,1,1).

I doubt if this is general enough to cover all cases, though.

Mentz114

Nov10-10, 07:31 PM

(3) What is required for a spacetime to admit a metric with flat spatial slices

At the risk of becoming a bore. To find a transformation that gives a flat spatial part of the metric in a more general case, where the metric is diag( -g00, g11, g22, g33), a transformation of this form is required, ( I'm choosing Cartesian coords and abusing notation)

\begin{align*}
dt'&=f_0(x,y,z,t)dt\\
dx'&=dx-f_1(x,y,z,t)dt\\
dy'&=dy-f_2(x,y,z,t)dt\\
dz'&=dx-f_3(x,y,z,t)dt\\
\end{align*}

with the additional constraint that g_{00}=-f_0^2+f_1^2+f_2^2+f_3^2 which recovers the original g00.

Given a (diagonal) metric this becomes an algebraic problem.

PAllen

Nov10-10, 08:36 PM

At the risk of becoming a bore. To find a transformation that gives a flat spatial part of the metric in a more general case, where the metric is diag( -g00, g11, g22, g33), a transformation of this form is required, ( I'm choosing Cartesian coords and abusing notation)

\begin{align*}
dt'&=f_0(x,y,z,t)dt\\
dx'&=dx-f_1(x,y,z,t)dt\\
dy'&=dy-f_2(x,y,z,t)dt\\
dz'&=dx-f_3(x,y,z,t)dt\\
\end{align*}

with the additional constraint that g_{00}=-f_0^2+f_1^2+f_2^2+f_3^2 which recovers the original g00.

Given a (diagonal) metric this becomes an algebraic problem.

I was aware that you could diagonalize any metric at a point, but not globally. Then your argument shows you can construct a local euclidean slice given the local diagonalization, which is obvious. It seems to provide a recipe for a global flat spacial slice only if global diagonalization is possible. Isnt that a fairly specialized condition for a metric to have?

Mentz114

Nov10-10, 09:29 PM

I was aware that you could diagonalize any metric at a point, but not globally. Then your argument shows you can construct a local euclidean slice given the local diagonalization, which is obvious. It seems to provide a recipe for a global flat spacial slice only if global diagonalization is possible. Isn't that a fairly specialized condition for a metric to have?

You could be right.

We can always find a Minkowski frame locally by using the coordinate coframe but that isn't the same thing as finding a transformation that transforms the spatial part of the metric to diag(1,1,1) or the equivalent. I don't really know what the latter means. I think the best way to look for such a transformation is to use the frame-field method, and that's what I've been on about ( rather feebly).

The transformation dxu -> sqrt(guu)dxu means we can rewrite the metric as diag(-1,1,1,1) immediately, but this is interpreted as a local frame.

This is probably not helping to establish which spacetimes can admit E3 spatial slices, so I'm going to bed.

JDoolin

Nov11-10, 06:53 AM

Roughly speaking, by "time" I mean "a coordinate that's timelike". :wink: I guess a somewhat more precise way of saying that would be that the vector field generated by the partial derivative with respect to the coordinate must be a timelike vector field. I think (but am not positive) that's enough to guarantee that any slice of the spacetime with a constant value of such a coordinate must be a spacelike hypersurface.

I was reviewing a book I read some time ago "Relativity Visualized" and came to wonder what this means: "a coordinate that's timelike"

In this book, Epstein makes all of his space-time diagrams as "proper-time vs. space" Thus, the equation we're all familiar with

d\tau^2=dt^2-dx^2

is turned on its head:

dt^2=d\tau^2+dx^2

Within this context, the (horizontal) x-component represents the position, the (vertical) \tau component represents the age of the particle, and the lengths of arcs represent the actual amount of time passed for a particle from the frame of an inertial observer.

More importantly, the effects of a "stretching" of space-proper-time becomes somewhat visualizable, using one's knowledge of euclidian geometry.

I just thought I would bring this up here, since you said "a coordinate which is timelike," these two "time-like" coordinates, \tau and t, perform differently.

Another question I would ask, though is whether both of them are actually coordinates. The value of the proper-time, \tau actually represents a property of an object passing through the space, while the value of t represents an actual coordinate value in an inertial reference frame. (In tensor lingo, would this be saying that t is contravariant, while \tau is covariant?)

When Epstein works with geometries in (x,\tau) coordinates, he seems to be mixing coordinates of different types; a location in space, and a property of a particle at that location. Under this context, he stretches the space-"proper-time" and is able to make quite a few interesting predictions. However, at no point does he actually stretch the coordinate time, t. There's a rule in the book that the length of the paths is equal to the actual time passed, and it seems like he sticks with it throughout.

Is it possible that all of this stretching of space-time in general relativity is actually stretching of (coordinate space, proper time), while it leaves unstretched (coordinate space, coordinate time?)

If one is weightless inside a windowless room, he cannot tell whether he is in flat space, where his clock is reading coordinate time, or if he is in a gravitational field, where his clock is reading proper-time. But when one measures the motion of the planets with respect to the positions of distant galaxies, he should be able to determine the speed of coordinate time; even if his local clocks are measuring proper-time; and he can measure the speed of his own clocks relative to this universal coordinate time.

PAllen

Nov11-10, 08:39 AM

Passionflower

Nov11-10, 10:00 AM

Also, if on the scale of the room, there area deviations from Euclidean spactial geometry, these are directly measurable using, e.g. pieces of string pulled tight between different points and measuring angles and lengths.
Could you demonstrate this by showing us the calculations for the deviations from Euclidean spatial geometry?

If you have not done this I urge you to do this as you then will realize that the effect is beyond anything we can directly measure with state of the art measuring apparatus let alone with pieces of strings.

PAllen

Nov11-10, 10:15 AM

Could you demonstrate this by showing us the calculations for the deviations from Euclidean spatial geometry?

If you have not done this I urge you to do this as you then will realize that the effect is beyond anything we can directly measure with state of the art measuring apparatus let alone with pieces of strings.

Read my hypothesis:

"Also, if on the scale of the room, there area deviations from Euclidean spacial geometry,"

Either the deviations are or are not signficant at lab scales. I am well aware that we are never likely to have a lab at a location where this is true.

Passionflower

Nov11-10, 10:16 AM

Read my hypothesis:

"Also, if on the scale of the room, there area deviations from Euclidean spactial geometry,"

Either the deviations are or are not signficant at lab scales. I am well aware that we are never likely to have a lab at a location where this is true.
So can you show me your calculations?

PAllen

Nov11-10, 10:32 AM

So can you show me your calculations?

What's to calculate? Do the internal angle of a triangle add up to 180 detrees? Are there deviations from the pythagorean theorem at 1000 meter scales compared to 10 meter scales?

Passionflower

Nov11-10, 10:56 AM

What's to calculate?
Let's say the Euclidean deviation for an elevator 3 meters high on the surface of a non rotating planet with surface area X and Mass Y? Are you going to use Schwarzschild coordinates or do you think these will be meaningless, if so, how will you approach the calculation otherwise?

PAllen

Nov11-10, 11:14 AM

Let's say the Euclidean deviation for a 3 meter heigh elevator on the surface of a non rotating planet with surface area X and Mass Y? Are you going to use Schwarzschild coordinates or do you think these will be meaningless, if so, how will you approach the calculation otherwise?

Two separate questions here: real world measurments and modeled measurements. In the real world of a lab scale, I wouldn't worry about coordinates or metric at all. I would, for example, set up a 1 km triangle of laser pointers and measure angles with protractor.

For modeled measurements, attempt to compute only invariants that modeled some physical situation. Then, I wouldn't care about the coordinates. For the example above, the trickiest question, which I don't completely know how to solve is computing an observed spacial angle between a pair of null geodesics.

Passionflower

Nov11-10, 11:36 AM

For modeled measurements, attempt to compute only invariants that modeled some physical situation. Then, I wouldn't care about the coordinates. For the example above, the trickiest question, which I don't completely know how to solve is computing an observed spacial angle between a pair of null geodesics.
You would not care about coordinates? Ok. So how would you approach it without using coordinates? Say we got three points arranged in a triangle, how are you going to model that?

You disagree we need coordinates to solve just about any problem in GR?

I am not totally sure but it seems you have the wrong ideas about using coordinates. Coordinates without a metric are not very useful but together with the metric they are very useful. While it is true that a particular coordinate value does not necessarily translate into a directly measurable quantity, we can use these values, or their differences, together with the metric, to calculate physically meaningful results.

PAllen

Nov11-10, 12:22 PM

You would not care about coordinates? Ok. So how would you approach it without using coordinates? Say we got three points arranged in a triangle, how are you going to model that?

You disagree we need coordinates to solve just about any problem in GR?

I am not totally sure but it seems you have the wrong ideas about using coordinates. Coordinates without a metric are not very useful but together with the metric they are very useful. While it is true that a particular coordinate value does not necessarily translate into a directly measurable quantity, we can use these values, or their differences, together with the metric, to calculate physically meaningful results.

I agree completely with your last quote. The key is that when I was calculationg things, I never attached meaning to coordinate values or differences, by themselves. I only attach meaning to calculations of observable from the coordinates in combination with the metric. The question I ask myself about spatial angle between intersecting null geodesics is to me non-trivial. It will depend on the world line of observer chosen. Given some timelike world line, at some point of which two null geodesics intersect, I would want to define a spacelike slice 'perpendicular' to the world line, project the metric and null geodesic paths to this slice, and figure the angle by dot product of the induced paths using the induced metric. I don't know how to do all of this, but this is the problem I would pose for myself to attach meaning a given coordinates and metric.

Even more non-trivial, is to compare measurements of angles at locations 1 km apart, I would want to define some specific simultaneity convention that is defined physically, not in terms of coordinates. Then I would attempt to compute predictions using some such definition using coordinates and metric. In all cases, I would only want to attach meaning to measurable quantities computed using coordinates and metric.

PeterDonis

Nov11-10, 01:07 PM

Another question I would ask, though is whether both of them are actually coordinates. The value of the proper-time, \tau actually represents a property of an object passing through the space, while the value of t represents an actual coordinate value in an inertial reference frame.

Correct, they are not both coordinates. The proper time \tau along a given worldline is an invariant; it's the same regardless of what coordinate system we use to calculate it. Invariant quantities have a direct physical meaning--in this case, the time actually experienced by an observer following the worldline. The coordinate time t is just a number; we can set things up so this number has some relationship to quantities like \tau that have direct physical meaning, but there's nothing in relativity that *requires* us to do so.

(In tensor lingo, would this be saying that t is contravariant, while \tau is covariant?)

No; contravariant and covariant are different concepts, and refer to how we represent geometric quantities in particular coordinate systems; the terms don't apply to individual coordinates or invariant quantities.

When Epstein works with geometries in (x,\tau) coordinates, he seems to be mixing coordinates of different types; a location in space, and a property of a particle at that location. Under this context, he stretches the space-"proper-time" and is able to make quite a few interesting predictions. However, at no point does he actually stretch the coordinate time, t. There's a rule in the book that the length of the paths is equal to the actual time passed, and it seems like he sticks with it throughout.

That last rule seems OK as far as it goes--it just amounts to choosing coordinates so that the "time" coordinate directly represents proper time along a worldline. However, there's a catch to this: in general, you can only do it *locally*--i.e., within a certain (usually small) distance in space--and time!--from a given event. You can't, in general, construct a *global* coordinate system where the "time" always directly represents actual time passed along every timelike curve, or even every "interesting" timelike curve.

Because of this, I'm not sure about Epstein's approach: it sounds like he's encouraging the reader to confuse the very different concepts of coordinate values and invariant quantities. (For example, see my next comment below.) I haven't seen his book so I can't really comment on any specifics, but that's the way it looks to me based on what you've said.

Is it possible that all of this stretching of space-time in general relativity is actually stretching of (coordinate space, proper time), while it leaves unstretched (coordinate space, coordinate time?)

If one is weightless inside a windowless room, he cannot tell whether he is in flat space, where his clock is reading coordinate time, or if he is in a gravitational field, where his clock is reading proper-time. But when one measures the motion of the planets with respect to the positions of distant galaxies, he should be able to determine the speed of coordinate time; even if his local clocks are measuring proper-time; and he can measure the speed of his own clocks relative to this universal coordinate time.

This is an example of the confusion I just described. There is no such thing as "the speed of coordinate time" in general; if you have a set of coordinates where the "time" seems to have a direct physical meaning, that just means you've been lucky enough to find coordinates that, at least in some region of spacetime, correlate closely to invariant quantities like the proper time along particular worldlines. There are also no such things as "clocks which read coordinate time", if by "clocks" you mean "actual, physical clocks". Actual, physical clocks *always* read the *proper time* along their worldlines, and nothing else. If you're lucky, you can find coordinates such that the "coordinate time", at least along some particular worldlines you're interested in, correlates with the proper time along those worldlines; but the physically meaningful quantity is the proper time.

In my experience, to make progress in understanding relativity, you have to focus on the physically meaningful quantities, the invariants; trying to think about things in terms of assigning any meaning to coordinates themselves just leads down an endless rabbit hole. That means giving up any thoughts about whether the coordinates are "stretched", "unstretched", etc. and focusing on questions that have direct physical meaning, like: "How much proper time would an observer following a given worldline experience between event A and event B?" If to answer that question you find it helpful to use a certain system of coordinates for easier calculation, that's fine; but the only thing that has actual physical meaning is the answer to a question like that, itself.

For example, take this simple comparison you give: someone floating weightless in flat spacetime, versus someone freely falling in a gravitational field. You say the former observer's clock reads coordinate time; that's wrong. Both observers' clocks read their proper time. In flat spacetime, it happens to be possible to set up a global coordinate system such that, for a given observer, coordinate time directly represents that observer's proper time. (But put in a second observer in relative motion to the first, and that coordinate system won't directly represent the second observer's proper time.) When gravity is present, so spacetime is curved, you can't in general set up such a global coordinate system; but it turns out that you *can* set up a coordinate system such that, for the one particular freely falling observer you're interested in, coordinate time directly represents *his* proper time. (For the idealized case of an observer freely falling "from rest at infinity", these coordinates are called Painleve coordinates.)

The other case you mention, measuring the motion of the planets with respect to distant galaxies, confuses me, because I'm not sure what you think we would be measuring in this way. If I am sitting at rest on the surface of a planet (let's say one with much stronger gravity than Earth's, so the time dilation factor is more pronounced), and I observe the motions of distant planets orbiting the same star as mine, against the background of the rest of the universe, the times I record will be according to my proper time, so they will be affected by my position in the gravitational field (i.e., they would be *different* times--in general, smaller time lapses--than would be recorded by someone much higher up). I would have to have some way of figuring out my radial coordinate r, relative to the mass M of my planet, to get a handle on how much my proper time is dilated relative to observers far away.

JDoolin

Nov11-10, 03:03 PM

I would have thought one of the few things universally agreed is that a clock measures proper time in its own world line, always.

Sure, but in a flat spacetime, the proper time of a stationary clock is the same as the coordinate time. My point was that the proper time is easy to find; it's what the clock measures.

What about a larger clock? You take the motions of the planets around the solar system, and mark them against the background of distant stars. Or you take the motions of stars in the galaxy and mark them against the background of distant galaxies. I would think that as the clock becomes larger, as you take into account more and more of universe, it's "proper time" gets closer to the coordinate time.

The proper time of individual "small" clocks are measured in reference to this larger celestial clock.

Note, proper time is not covariant, it is invariant: all observers and coordinates produce the same value for it beyond purely unit conventions (that is the same value given some local defintion of second).

Also, if on the scale of the room, there area deviations from Euclidean spactial geometry, these are directly measurable using, e.g. pieces of string pulled tight between different points and measuring angles and lengths.

Okay, so are there ever area deviations from Euclidean spatial geometry? Yes, I guess, under Lorentz Transformation, you can have length contraction and thus the lengths and oblique angles change. Are there also such deviations due to gravity?

I have read that a string passed through the center of the earth, stretched tight must be longer than a string passed through empty space across the same distance. Unfortunately, I wasn't convinced... (The argument didn't seem to be there to support the claim.)

However, there were two different claims. One was that somehow the graph of proper time vs. space was drawn on paper where the proper time was stretched out in the region of gravity. The second claim was that the space was stretched relative to the space around it.

It just occurred to me, that perhaps somehow, the basis of the claim rests on the existence of a "proper space" analog of "proper time."

Just as we had dt^2=d\tau^2 + dx^2 we can have dx^2=ds^2 + dt^2. The first equation has a fairly elegant description in "Relativity Visualized." I'm not sure about the second one. Is it too strange to describe in any way the human mind can understand?

But just to start to make the attempt, I'm imagining pulling the rope down through the center of the earth. It is very easy down to the halfway point, because the rope is falling into the hole. When I get to the center, I am pulling at a certain rate, but because the time is running slower here, it should seem at the surface to be entering the hole slower than I am pulling. I am sorry; this gedanken experiment didn't go anywhere, as far as I can tell. But maybe I am missing something.

PAllen

Nov11-10, 04:02 PM

Sure, but in a flat spacetime, the proper time of a stationary clock is the same as the coordinate time. My point was that the proper time is easy to find; it's what the clock measures.

What about a larger clock? You take the motions of the planets around the solar system, and mark them against the background of distant stars. Or you take the motions of stars in the galaxy and mark them against the background of distant galaxies. I would think that as the clock becomes larger, as you take into account more and more of universe, it's "proper time" gets closer to the coordinate time.

The proper time of individual "small" clocks are measured in reference to this larger celestial clock.

I wouldn't think so. I would define local clock time in terms of a physical phenomenon (atomic oscillation, radioactive decay, etc.

I have read that a string passed through the center of the earth, stretched tight must be longer than a string passed through empty space across the same distance. Unfortunately, I wasn't convinced... (The argument didn't seem to be there to support the claim.)

.

I was avoiding such complexity by proposing such a method only at lab scales. That IF there were euclidean deviations at lab scales, they would be easy to directly measure. In no way was I worrying about issues like the tension needed to stretch a string over large distances.

pervect

Nov11-10, 05:02 PM

The issue isn't tension, I think. The "string stretched tight" is just to give a notion of the path along which the distance is measured. If we ignore the rotation of the Earth, it's fairly well known that the spatial slice of constant Schwarzschild time has a non-Euclidean spatial geometry.

I have no idea of what arguments the OP read, or where but it's more or less a standard textbook result.

The simplest case is when we not only ignore the rotation of the Earth, but we assume it has a constant density, independent of depth. It's standard to consider the spatial slice through the equator, with theta = 0. Then you have the metric (MTW, pg 614)

dr^2 / (1-2m(r)/r) + r^2 d phi^2

And they have a nice embedding diagram in figure 23.1 on that page. It should be obvious by inspection that this isn't Euclidean, looking at the embedding diagram makes it more obvious.

m(r) is given earlier by eq 23.19 and is just the integral of 4 pi r^2 rho dr. Thus when rho is constant m(r) = 4/3 pi r^3 rho, r<R, and m(r) = M for r>= R, R being the radius of the spherical non-rotating planet.

As an aside, 4 pi r^2 dr is NOT a volume element, even though it may look like one at first glance.

PeterDonis

Nov11-10, 07:19 PM

What about a larger clock? You take the motions of the planets around the solar system, and mark them against the background of distant stars. Or you take the motions of stars in the galaxy and mark them against the background of distant galaxies. I would think that as the clock becomes larger, as you take into account more and more of universe, it's "proper time" gets closer to the coordinate time.

The proper time of individual "small" clocks are measured in reference to this larger celestial clock.

I'm sorry to keep harping on this, but I really think this is a very misleading way of looking at things. First of all, as I noted in my last post, the "time" I measure by observing the motion of stars and galaxies against a background of more distant objects will be affected by my particular position in a gravitational field. But more important, "proper time" is a property of specific worldlines--more precisely, it's an invariant quantity associated with a specific segment of a specific worldline between two specific events. That means that in order to talk about proper time, you need to first specify the worldline and the events. There is no "proper time of the universe" or "proper time of the Milky Way galaxy" per se; you have to pick out a specific worldline (say, the worldline of the center of mass of the Milky Way galaxy) and specify events on it (say, the event of two successive "meridian passages" of the Solar System, as observed at the center of mass of the Milky Way) before you can talk about proper time. (The same goes for "proper distance", which is the spacelike analogue of proper time, of course; you have to first specify a particular spacelike curve, and two events on it, before you can talk about proper distance.)

When you talk about "coordinate time", I suspect what you are really thinking about (though you may not realize it at first) is the proper time as experienced by a particular observer traveling on a particular worldline that has some special property you're interested in. I think it would make for a much clearer discussion, and probably help you to conceptualize what's going on, if you would explicitly state, in every case, who those particular observers are and what particular worldlines they're traveling on, instead of using terms like "coordinate time" as though they automatically have a well-defined physical meaning.

JDoolin

Nov12-10, 10:53 AM

I'm sorry to keep harping on this, but I really think this is a very misleading way of looking at things. First of all, as I noted in my last post, the "time" I measure by observing the motion of stars and galaxies against a background of more distant objects will be affected by my particular position in a gravitational field. But more important, "proper time" is a property of specific worldlines--more precisely, it's an invariant quantity associated with a specific segment of a specific worldline between two specific events. That means that in order to talk about proper time, you need to first specify the worldline and the events. There is no "proper time of the universe" or "proper time of the Milky Way galaxy" per se; you have to pick out a specific worldline (say, the worldline of the center of mass of the Milky Way galaxy) and specify events on it (say, the event of two successive "meridian passages" of the Solar System, as observed at the center of mass of the Milky Way) before you can talk about proper time. (The same goes for "proper distance", which is the spacelike analogue of proper time, of course; you have to first specify a particular spacelike curve, and two events on it, before you can talk about proper distance.)

When you talk about "coordinate time", I suspect what you are really thinking about (though you may not realize it at first) is the proper time as experienced by a particular observer traveling on a particular worldline that has some special property you're interested in. I think it would make for a much clearer discussion, and probably help you to conceptualize what's going on, if you would explicitly state, in every case, who those particular observers are and what particular worldlines they're traveling on, instead of using terms like "coordinate time" as though they automatically have a well-defined physical meaning.

You're saying that I'm talking about the proper time as experienced by a particular observer traveling on a particular worldline, and you are absolutely right.

You can visualize as the moon goes round the earth, the earth goes around the sun, the sun goes around the milky way, and the milky-way undulates with other galaxies in a local cluster. By doing so you are invoking a reference frame of a particular observer traveling on a particular worldline.

What I'm saying is that any hypothetical observer who can back up and see all of these things at the same time is imagining a Minkowski reference frame of the size and scope of whatever events he visualizes.

But to attempt for a much clearer discussion:

We imagine that colonies are erected on several planets and moons in the solar system, and we decide to share an interplanetary time standard.

The solar body with the straightest worldline is the sun, so we imagine a hypothetical observer at the center of the sun, facing the star Polaris. And each time earth completes a complete circle around the sun, from the perspective of this observer, we call it a solar year.

Colonies on Earth, Ganymede, and Mars all share the same time standard, but the intersolar time committee has its work cut out. None of them can actually see the solar system from the perspective of their hypothetical observer at the center of the sun, so they have to do a few tricky calculations. But they are able to do it, and when they come on the news in the morning, they are able to report the exact time as would be measured by that hypothetical observer.

I think what we have here is a spacetime that is "locally Minkowski" where the locality is quite large, but we can make it larger by taking more care in defining our solar observer:

One problem that occurs over the next few hundred years is this: over time, Polaris moves across the sky relative to the background galaxies. This means that the coordinate system as defined is very slowly rotating, to maintain a facing toward Polaris. So instead of using Polaris to define our constant reference direction, we use a background of distant galaxies.

By switching to a more distant reference point, we keep our East, North, and Up more constant, so we've increased the size of our "Local Minkowski-ness" Errors that would have crept up in dozens of years, now will take hundreds of years.

There are also slight errors that creep up over time due to the fact that the sun's worldline is not quite straight, the sun is actually under constant acceleration toward the center of the galaxy.

By switching to a reference frame based on the worldline of the center of the galaxy, we improve things; errors that would have crept up in hundreds of years, will now take thousands or millions of years, perhaps.

Ideally, we would use a worldline of an observer which never accelerates at all, and construct a coordinate system from that observer's reference frame. I think this is theoretically possible construction.

Not only is it a theoretically possible construction, but it's the construction of spacetime that comes most naturally to mind whenever any person first realizes that the it is actually the earth that goes around the sun; not vice versa.

What we automatically imagine, is the view "from afar (http://video.google.com/videoplay?docid=8842256077873416888#)" and as larger and larger regions come into our view, the space becomes more and more Minkowski. Though perhaps there are regions right around gravitational bodies where the space is stretched, from the distance, you see the birds-eye view, where hills and valleys can be ignored, and placed into a large Euclidian space with a "global time" (meaning--everyone within the system can reference a single hypothetical gravity-immune observer's positon and time mapping, for every event in the system.)

PeterDonis

Nov12-10, 11:28 AM

What we automatically imagine, is the view "from afar (http://video.google.com/videoplay?docid=8842256077873416888#)" and as larger and larger regions come into our view, the space becomes more and more Minkowski. Though perhaps there are regions right around gravitational bodies where the space is stretched, from the distance, you see the birds-eye view, where hills and valleys can be ignored, and placed into a large Euclidian space with a "global time" (meaning--everyone within the system can reference a single hypothetical gravity-immune observer's positon and time mapping, for every event in the system.)

Thanks, this does make it a lot clearer. Something like this is done today with what is called the "Earth-centered inertial" (ECI) frame, which is constructed as the local inertial frame of someone moving along the geodesic worldline of the Earth's center of mass, but with (I believe this is correct) the time coordinate scaled so that it matches the proper time rate at the Earth's surface (on the "geoid", which is the equipotential surface "at sea level", more or less). If you took this frame and rescaled the time so that it matched the proper time rate of an observer at Earth's position in the Sun's gravity well, but far away from the Earth (for example, orbiting the Sun in the Earth's orbit but on the opposite side of the Earth), you would have something closer to what you are talking about, at least with reference to the Earth. Then, as you say, you could continue to "correct" the time coordinate by rescaling to match the proper time rate far outside the Sun's gravity well, the galaxy's gravity well, the gravity well of the Local Group of galaxies, etc.

There's a catch to this, though: the Universe is expanding. Suppose we make all these corrections and arrive at a local inertial frame which tracks the proper time of an observer at Earth's spatial location "now" who is moving exactly with the cosmological "Hubble flow", i.e., this observer's only "motion" is due to the expansion of the Universe. The way to distinguish such an observer, experimentally, is that only this observer (and the family of observers like him, at different spatial locations, e.g. at some point in the Andromeda galaxy) sees the Universe as isotropic--it looks the same in all directions. The most sensitive measure of this that we know of currently is the CMBR, which does *not* look isotropic to us because the Earth is not "at rest" with respect to the expansion of the Universe as a whole; but we can certainly construct what the local inertial frame of an observer that was just passing Earth now but that *did* see the CMBR as isotropic would look like.

So we have what I'll call the "Earth-centered cosmological" (ECC) reference frame, which is the local inertial frame of this observer just passing Earth now that sees the CMBR as isotropic, so his only "motion" is due to the expansion of the Universe. Now consider a similar observer a million parsecs from Earth (distance "now" as measured in the ECC frame). The current estimate of the Hubble constant is about 70 km/sec per million parsecs, so this second observer, in the ECC frame, would appear to be moving away from Earth at 70 km/sec. Pick the event on the second observer's worldline that has the time coordinate "now" according to the ECC frame, and call that event E; the origin of the ECC frame will be the event of Earth "now", which we'll call event O.

Now consider the following: the time since the Big Bang at event O, according to the ECC frame, is 13.7 billion years. How much proper time has the second observer experienced since the Big Bang at event E? The problem is that the "obvious" answer in the ECC frame is *wrong*. The obvious answer is that, since the second observer is moving in the ECC frame, time dilation will make his proper time since the Big Bang at event E *less* than the Earth's proper time since the Big Bang at event O. However, the actual general relativistic cosmological models that best match the data say that the second observer's proper time since the Big Bang at event E will be the *same* as the Earth's proper time since the Big Bang at event O.

The resolution of this "paradox" is, of course, that the expansion of the Universe means spacetime, as a whole, is not flat, even though each spatial slice (hypersurface of constant "cosmological time") in the spacetime of the Universe as a whole is flat (i.e., Euclidean), at least according to our current best-fit model. Because spacetime is not flat, the local inertial frames at different events on the "ECC" worldline (e.g., the observer "at rest" at the Earth's position in the Universe as a whole now, vs. the observer "at rest" at the Earth's position in the Universe as a whole a billion years ago) do not "line up" with each other; the frame of the observer in the past will appear to be moving, relative to the frame of the observer now, just as the frame of an observer at a different spatial position now will appear to be moving. So there is no such thing as a "global Minkowski" frame for the Universe that works everywhere and at all times; the best you can do is to set up one that works "locally" around a particular event of interest (such as the Earth now), and since the constant-time slice of the Universe is Euclidean, you *can* in principle use this frame to cover the entire Universe (or at least everything we can see), as long as you remember that you can't draw correct deductions over extended time ranges using this frame (e.g.., how much proper time since the Big Bang for an observer a million parsecs away). (And if it turns out that we have to change our best-fit model with future data, such that the constant-time slices of the Universe turn out not to be Euclidean, then a frame like the "ECC" frame would be limited in the spatial range it could cover, as well as the time range.)

JDoolin

Nov13-10, 08:58 AM

JDoolin

Nov13-10, 10:29 AM

Now consider the following: the time since the Big Bang at event O, according to the ECC frame, is 13.7 billion years. How much proper time has the second observer experienced since the Big Bang at event E? The problem is that the "obvious" answer in the ECC frame is *wrong*. The obvious answer is that, since the second observer is moving in the ECC frame, time dilation will make his proper time since the Big Bang at event E *less* than the Earth's proper time since the Big Bang at event O. However, the actual general relativistic cosmological models that best match the data say that the second observer's proper time since the Big Bang at event E will be the *same* as the Earth's proper time since the Big Bang at event O.

I can't vouch for the data, but lets see whether we agree about what the "obvious" answer is. I'll give you my "obvious" answer, then you can see if you agree.

Let's take a galaxy, for instance, at 400 megaparsecs (approximately 1.3 billion light years away). According to Hubble's Law, this galay should be moving away from us at 70 X 400 = 28,000 km/second=.093c. So the time-dilation factor at 1.3 billion light years away would be:

\frac{1}{\sqrt{1-(v/c)^2}}=1.004

So where our galaxy aged 13.7 billion years, the distant galaxy would age 13.65 billion years. Now we also need to account for the travel-time of the light, so we can subtract about 1.3 billion years from this figure. The actual age of the galaxy we see should be about 12.35 billion years old.

Take notice that at 1.3 billion light years, the effect of 70 km/s per megaparsec is quite small. The effect of the delay due to the speed of light is huge by comparison.

I've shown that the "obvious" answer (at 1.3 billion light years distance) is that the second observer ages 13.65 billion years, while the earth ages 13.7 billion years. 13.65 billion and 13.7 billion are very close. This means that any observer within 1.3 billion light years should have approximately the *same* amount of proper time since the Big Bang event.

So now I need some clarification on what you've said above. "the second observer's proper time since the Big Bang at event E will be the *same* as the Earth's proper time since the Big Bang at event O" When you say this, are you already taking into account the speed of light delay? For instance, the image of a galaxy 1.3 billion light years away should not look 13.7 billion years old, but should look 12.4 billion years old, because the light originated from an event 1.3 billion years ago.

PeterDonis

Nov13-10, 12:56 PM

I've shown that the "obvious" answer (at 1.3 billion light years distance) is that the second observer ages 13.65 billion years, while the earth ages 13.7 billion years. 13.65 billion and 13.7 billion are very close. This means that any observer within 1.3 billion light years should have approximately the *same* amount of proper time since the Big Bang event.

According to the "naive" calculation in the ECC frame, yes. But the difference will get larger as you go out to larger distances in the ECC frame. In the "cosmological" frame, the frame in which the FRW metric is written, *all* observers at the same cosmological time t have experienced *exactly* the same proper time since the Big Bang. See next comment.

So now I need some clarification on what you've said above. "the second observer's proper time since the Big Bang at event E will be the *same* as the Earth's proper time since the Big Bang at event O" When you say this, are you already taking into account the speed of light delay? For instance, the image of a galaxy 1.3 billion light years away should not look 13.7 billion years old, but should look 12.4 billion years old, because the light originated from an event 1.3 billion years ago.

Remember that I said event E and event O are, by definition (or perhaps by construction) *simultaneous* in the ECC frame. That means event E is the event on the second observer's worldline that would be assigned the same time coordinate in the ECC frame as event O; in other words, event E is the event at which the Earth observer's "surface of simultaneity" at event O intersects the second observer's worldline. So the "speed of light delay" is already taken into account (at least, I think that's how you're using the term).

However, the way the ECC frame is constructed, its "surface of simultaneity" at event O exactly matches the (Euclidean) surface of constant cosmological time t = 13.7 billion years in the FRW frame, the (curved) spacetime metric in which cosmology is normally done. So according to the FRW frame, the second observer's proper time since the big bang at event E will be *exactly* 13.7 billion years, *not* 13.65 billion years.

JDoolin

Nov13-10, 01:36 PM

So we have what I'll call the "Earth-centered cosmological" (ECC) reference frame, which is the local inertial frame of this observer just passing Earth now that sees the CMBR as isotropic, so his only "motion" is due to the expansion of the Universe. Now consider a similar observer a million parsecs from Earth (distance "now" as measured in the ECC frame). The current estimate of the Hubble constant is about 70 km/sec per million parsecs, so this second observer, in the ECC frame, would appear to be moving away from Earth at 70 km/sec. Pick the event on the second observer's worldline that has the time coordinate "now" according to the ECC frame, and call that event E; the origin of the ECC frame will be the event of Earth "now", which we'll call event O.

Now consider the following: the time since the Big Bang at event O, according to the ECC frame, is 13.7 billion years. How much proper time has the second observer experienced since the Big Bang at event E? The problem is that the "obvious" answer in the ECC frame is *wrong*. The obvious answer is that, since the second observer is moving in the ECC frame, time dilation will make his proper time since the Big Bang at event E *less* than the Earth's proper time since the Big Bang at event O. However, the actual general relativistic cosmological models that best match the data say that the second observer's proper time since the Big Bang at event E will be the *same* as the Earth's proper time since the Big Bang at event O.

You've double-defined event O. Saying that both the big bang is "O" and earth now is "O." I want to define four events. You've defined "E" as the "the event on the second observer's worldline that has the time coordinate "now" according to the ECC frame" That one is good. The other events I'm defining are:

O: The Big Bang.
N: The event of here and now. Earth Circa 2010, 13.7 billion years after O.
P: The event that we are currently seeing, looking at the galaxy that is moving along a path from O to E.

Now, in my previous post, I only considered what to expect out to about 1.3 billion light years. And I only gave lip-service to the speed-of-light delay. In this post, I want to make "The obvious answer" a little more complete. Specifically, I think the "obvious answer" out to 1.3 billion light years is pretty close to what we see. It is out beyond 6 billion light years where the "obvious answer" is clearly wrong.

Here is the problem with the "obvious answer". If you are observing galaxies out beyond 7 billion light years (which we do), (Event P at 7 billion light years--Event E at 14 billion light years, it means that the velocity of those stars is greater than the speed of light. If this "obvious answer" were correct, we should not see any stars beyond 7 billion light years distance.

However, we see here:

http://www.astro.ucla.edu/~wright/sne_cosmology.html

We have supernovae all the way out to 12 giga-parsecs (39 billion light years.) That means the "obvious" answer cannot be right. It should be noticed though, that this is under the assumption that the hubble-constant is a global parameter. A careful analysis of the graph indicates that hubble's constant is 70 km/s per megaparsec in a region nearby, out to about 6 billion light years, but seems to be much smaller as you go out beyond that region.

Mentz114

Nov13-10, 01:41 PM

The big bang didn't happen 'somewhere'. It was 'everywhere'. So you're free to choose any point as the 'center', and choosing the point you're at is most convenient. ( sorry about all the 'quotes', but it's a subtle distinction).

JDoolin

Nov13-10, 01:43 PM

Remember that I said event E and event O are, by definition (or perhaps by construction) *simultaneous* in the ECC frame. That means event E is the event on the second observer's worldline that would be assigned the same time coordinate in the ECC frame as event O; in other words, event E is the event at which the Earth observer's "surface of simultaneity" at event O intersects the second observer's worldline. So the "speed of light delay" is already taken into account (at least, I think that's how you're using the term).

I posted, and then saw you had posted. I'm still not quite sure how the event E takes into account the speed-of-light delay. Hopefully the diagram I drew, which includes point P will show what I mean by taking into account the speed of light delay.

The event E which intersects the earth observer's "surface of simultaneity" will not be seen on earth for a long, long time. However, the event P which intersects the observer's "past light cone" is what is seen on earth NOW.

JDoolin

Nov13-10, 02:34 PM

The big bang didn't happen 'somewhere'. It was 'everywhere'. So you're free to choose any point as the 'center', and choosing point you're at is most convenient. ( sorry about all the 'quotes', but it's a subtle distinction).

PeterDonis brought up the idea of an ECC, Earth Centered Cosmological frame, in which the anisotropy of the CMBR disappeared. It appears that if we adjusted our velocity by about 600 km/second, in the appropriate direction, then we would be able to make this anisotropy disappear.

I would agre that once we did this, then a "point" in our space-ship is an actual center of the universe. Since we're free to set off that space-ship at any time we want, and any place we want, we can choose any event to be the center of the universe. But if we choose a point, we have to choose a specific velocity as well. A "point" in space, is a worldline in space-time.

Now, the interpretation of General Relativistic Cosmology is that all of these cosmologically centered worldlines are parallel (or at least they don't cross). The divergence of those cosmologically centered worldlines is not due to real motion of the particles. It is due, instead to an increasing scale factor. In the past, that scale factor goes right down to zero. However, the unscaled distance remains more-or-less the same throughout time.

This is how I have come to understand the idea, though I realize that perhaps my understanding is not complete.

But how can I reconcile this with my idea that the big bang happened "somewhere." Honestly, I don't think I can. If you trace back a set of straight lines to where they intersect, you get a "somewhere" But if you trace back a set of lines that curve, and do not cross, then you get an "everywhere". My diagram in post 37 assumes that the two lines do cross somewhere in the past. Is there any chance that we could even use this as an approximation? That we could maybe just "pretend" that two galaxies traveling at velocity away from each other were once, far in the past at the same place, at the same time? I guess we wouldn't have to call it the big bang; just the meeting event. And maybe they never met, but just assume that the majority of the time they were moving away from each other, they acted as though they met.

Mentz114

Nov13-10, 03:24 PM

JDoolin

Nov13-10, 04:23 PM

I think you are on the right track. If the big-bang started at a singularity, then all the constituents that make up the universe were once in same place ( the only place) at t=0.

I don't have time to reply to this at length, but this document says it much better, and has loads of good diagrams.

http://ls.poly.edu/~jbain/philrel/philrellectures/12.RelativisticCosmology.pdf

The attached diagrams came from the referenced article. They both are graphs of space vs. time. The Robertson Walker diagram has elements that go to infinite speed. The "conformal" version has all elements stationary. Are these two diagrams supposed to be different diagrams of the same thing, or are they different geometries in the same coordinates?

Is the time parameter in the Robertson Walker diagram supposed to be proper time of particles, or does it represent coordinate time for a hypothetical observer?

Mentz114

Nov13-10, 04:53 PM

The attached diagrams came from the referenced article. They both are graphs of space vs. time. The Robertson Walker diagram has elements that go to infinite speed. The "conformal" version has all elements stationary. Are these two diagrams supposed to be different diagrams of the same thing, or are they different geometries in the same coordinates?
Same geometry, different coords.

Is the time parameter in the Robertson Walker diagram supposed to be proper time of particles, or does it represent coordinate time for a hypothetical observer?

Those diagrams illustrate the horizon that exists in an expanding FRW universe. The first diagram uses scaled coordinates, so everything starts in the same place and remains the same distance apart. The straight red lines represent horizons. As time progresses, each worldline loses touch with the others as they move outside its future light cones.

In the second diagram, comoving coordinates are used, so we can pick any worldline as our reference frame. In these coords all the other matter appears to be receding. The horizon lines now become the red curves. I think the point is that the physics is the same whichever of these coords we use.

I haven't got time to do it right now but I'll get the actual transformations and post them.

Is the time parameter in the Robertson Walker diagram supposed to be proper time of particles, or does it represent coordinate time for a hypothetical observer?

In both diagrams, the vertical axis is coordinate time.

PeterDonis

Nov13-10, 06:55 PM

Is the time parameter in the Robertson Walker diagram supposed to be proper time of particles, or does it represent coordinate time for a hypothetical observer?

I'm not sure what "coordinate time for a hypothetical observer" means. The Robertson-Walker coordinates are constructed in such a way that coordinate time directly represents the proper time of observers who are "at rest" in those coordinates--in other words, observers whose values of the spatial coordinates remain constant (and who see the universe as isotropic for all time). As you noted, the actual proper distance between any two such observers increases with time because of the increase of the scale factor; this is a more precise way of saying that the universe is "expanding".

The "conformal" coordinates in the other diagram can be constructed from the Robertson-Walker coordinates in two steps, which I'll describe here in highly "heuristic" fashion: (1) take the initial singularity, which is a point (a single event), and "stretch it out" into a line; (2) take the Robertson-Walker time coordinate and "stretch" it by an amount that increases as the singularity is approached, so that after the stretching is done, all light rays travel on 45 degree lines. This makes it easy to see the causal structure of the spacetime.

PeterDonis

Nov13-10, 07:02 PM

You've double-defined event O. Saying that both the big bang is "O" and earth now is "O."

I didn't mean to double-define O; I meant O to always refer to the Earth now. Sorry if I wasn't clear.

Here is the problem with the "obvious answer". If you are observing galaxies out beyond 7 billion light years (which we do), (Event P at 7 billion light years--Event E at 14 billion light years, it means that the velocity of those stars is greater than the speed of light. If this "obvious answer" were correct, we should not see any stars beyond 7 billion light years distance.

However, we see here:

http://www.astro.ucla.edu/~wright/sne_cosmology.html

We have supernovae all the way out to 12 giga-parsecs (39 billion light years.) That means the "obvious" answer cannot be right. It should be noticed though, that this is under the assumption that the hubble-constant is a global parameter. A careful analysis of the graph indicates that hubble's constant is 70 km/s per megaparsec in a region nearby, out to about 6 billion light years, but seems to be much smaller as you go out beyond that region.

This is what I was referring to as the curvature of the FRW spacetime: as you go back in time, the local inertial frames on a given worldline (say, that of the cosmological observer who is at Earth's spatial location now) don't "line up" with each other. The changing slope of the graph is evidence of that.

PeterDonis

Nov13-10, 07:09 PM

Hopefully the diagram I drew, which includes point P will show what I mean by taking into account the speed of light delay.

Yes, it's clear, and it matches what I understood you to be saying. Event E is spacelike separated from the event you've labeled N, but that doesn't prevent us from calculating the proper time between the big bang (which you've labeled O) and event E, along the "straight" worldline connecting them. As we've seen, that calculation gives an answer which can't be right.

JDoolin

Nov13-10, 07:10 PM

In the conformal map (see post 42, above), the light rays are always parallel to the light cones. In the Robertson Walker coordinates, the light rays don't follow the light cones. Is there a mistake in one of the diagrams?

The light rays come to a full stop in the Robertson Walker diagram, stopping and turning around and coming the opposite direction. Is that expected?

(Now I see, Mentz post #43, the lines are not light rays but horizon lines? I'm not quite clear on what that means.)

PeterDonis

Nov13-10, 07:22 PM

In the conformal map, the light rays are always parallel to the light cones. In the Robertson Walker coordinates, the light rays don't follow the light cones. Is there a mistake in one of the diagrams?

The light rays come to a full stop in the Robertson Walker diagram, stopping and turning around and coming the opposite direction. Is that expected?

Yes, it's expected, and no, there's no mistake in the diagrams. The light rays actually do follow the light cones in the Robertson Walker coordinates, but it's hard to see because the bottom of the diagram is crunched together (that's one of the reasons conformal coordinates are useful). The two opposite "sides" of the past light cone actually originate from the *same* event, the initial singularity (since everything originates there), but they "come out" in different directions that aren't quite opposite--there's a little bit of "tilt" in each ray towards the other one. As the expansion of the universe decelerates, and the light cones tilt inward (until they're vertical at "now"), the two light rays are bent back together to meet at "here and now".

Part 3 of Ned Wright's cosmology tutorial at http://www.astro.ucla.edu/~wright/cosmo_03.htm gives more details about these diagrams.

JDoolin

Nov13-10, 08:00 PM

Yes, it's clear, and it matches what I understood you to be saying. Event E is spacelike separated from the event you've labeled N, but that doesn't prevent us from calculating the proper time between the big bang (which you've labeled O) and event E, along the "straight" worldline connecting them. As we've seen, that calculation gives an answer which can't be right.

So does it seem to you that if we go out to about one or two billion light years that we have what appears to fit a Minkowski approximation, and then only once you get out past 6 billion light years does it seem that the space is noticeably stretching? Because if you agree with me that this is essentially the case, then I can begin to discuss the invisible elephant in the room, which is acceleration.

The "obvious" calculation we're doing makes the assumption of a uniform, peaceful expansion at the beginning of the universe. This is highly unlikely, because one should expect nuclear explosions, or matter-anti-matter explosions, of great force (or possibly even greater unknown particle-interactions) in the instants immediately after the big bang.

I don't know how to treat these interactions in General Relativity, but I do have some idea how to treat them in Minkowski spacetime. You've seen that the calculation without acceleration cannot be right. Would you humor me long enough to see that the calculation with acceleration actually could be right?

PeterDonis

Nov13-10, 09:24 PM

So does it seem to you that if we go out to about one or two billion light years that we have what appears to fit a Minkowski approximation, and then only once you get out past 6 billion light years does it seem that the space is noticeably stretching?

I don't know that I disagree, but I don't know that I would put it this way either. The "stretching" doesn't happen in space; it happens in time, as the scale factor changes. Obviously you can kind of convert time to space because looking farther away means looking at things the way they were a longer time ago; but I'm not sure I would characterize what we see as we look out that far as "stretching". Maybe my further comments below will help to clarify what I'm getting at.

Because if you agree with me that this is essentially the case, then I can begin to discuss the invisible elephant in the room, which is acceleration.

We have to be careful with terminology here as well, because the cosmological observers we've been discussing (the ones at rest in the Robertson-Walker coordinates) are not accelerating; that is, they don't feel any acceleration. And again, I'm not sure that the fact that the local inertial frames of cosmological observers at different times don't "line up" is most usefully viewed as an "acceleration", although that term is often used (as in, the expansion of the universe is now accelerating, but was decelerating earlier in its history).

The "obvious" calculation we're doing makes the assumption of a uniform, peaceful expansion at the beginning of the universe. This is highly unlikely, because one should expect nuclear explosions, or matter-anti-matter explosions, of great force (or possibly even greater unknown particle-interactions) in the instants immediately after the big bang.

I don't know how to treat these interactions in General Relativity, but I do have some idea how to treat them in Minkowski spacetime. You've seen that the calculation without acceleration cannot be right. Would you humor me long enough to see that the calculation with acceleration actually could be right?

The calculations we've been doing are kinematic; they don't get into the detailed dynamics of what's going on with the matter-energy in the universe. The Robertson-Walker models abstract all that out by treating the matter-energy in the universe as a perfect fluid, which can take one of three simple forms characterized by different equations of state relating the pressure, p, to the energy density, rho:

(1) "Matter dominated": a fluid with zero pressure, p = 0. This is a good approximation at the cosmological level for non-relativistic matter (meaning, the average speed of the individual "particles" in the fluid, which are basically galaxies or galaxy clusters, is much less than the speed of light).

(2) "Radiation dominated": a fluid with equation of state p = 1/3 rho. This is the equation of state for a "fluid" made of pure radiation (for example, the CMBR).

(3) "Vacuum dominated": a fluid with equation of state p = - rho. This is the equation of state for a "fluid" which is due to a cosmological constant (other terms used are "vacuum energy" or "dark energy").

The current best fit model for the evolution of the universe is: first the "inflation" phase, in which the equation of state was vacuum dominated with an extremely large effective energy density rho, meaning that the universe "expanded" exponentially; then, after the phase transition that ended inflation, a radiation dominated phase, which lasted roughly until the time of "recombination" (electrons and nuclei combining into atoms, which made the universe basically transparent to photons) when the universe was about 100,000 years old (all these times are very approximate, I'm going from memory here); then a matter dominated phase, which lasted until a few billion years ago (I believe); and finally, another vacuum dominated phase but with a very, very small effective energy density, causing the expansion of the universe to start "accelerating" again (it had been decelerating during the radiation and matter dominated phases).

The reason I go into all this is to illustrate that nowhere in any of this did I have to specify what, exactly, was going on *within* the cosmological fluid, in terms of nuclear reactions, explosions, whatever. The only thing that matters for the overall dynamics of the universe is the equation of state, and the effective equation of state of the cosmological fluid, on an overall level, can remain the same while the fluid undergoes violent internal changes. (Part of how this can work is that the identity of the individual "particles" that compose the fluid changes over time: in the early universe, they were elementary particles like quarks, electrons, and photons; then they were atoms of hydrogen, helium, and a few other elements; and for the past few billion years, at least, they've been galaxies and galaxy clusters. But on a gross cosmological level, all of these different "fluids" can be described by the same simple equations of state I gave above.)

So as far as the overall dynamics of the universe is concerned, we can get away with *not* modeling all the details you mention. (We do have to model them, at least to some extent, in order to predict finer details like the ratios of abundances of different elements that we should expect to see in intergalactic space.) Another way of saying this is that, in fact, the Robertson-Walker models do *not* make the assumption of a "peaceful" expansion of the early universe; they only make the assumption that, whatever might be going on at a detailed level, it can be adequately modeled at an overall level by a fluid with one of the above equations of state. That assumption appears to be working pretty well so far.

I'm not sure if you want to get into the actual derivation of the Robertson-Walker metric; the Wikipedia page, http://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric, has a decent (if brief) discussion. Also, technically, these models do not go all the way back to the initial singularity (the word "singularity" means the equations break down and the model can't make predictions). Currently the actual models, as I understand it, more or less assume that the inflation phase started with a state the size of the Planck length or thereabouts (not a zero-size initial singularity), which began expanding exponentially; how that state came to be is not known, though there are various proposals with no real way of testing any of them experimentally at this time.

JDoolin

Nov14-10, 01:10 PM

Yes, it's expected, and no, there's no mistake in the diagrams. The light rays actually do follow the light cones in the Robertson Walker coordinates, but it's hard to see because the bottom of the diagram is crunched together (that's one of the reasons conformal coordinates are useful). The two opposite "sides" of the past light cone actually originate from the *same* event, the initial singularity (since everything originates there), but they "come out" in different directions that aren't quite opposite--there's a little bit of "tilt" in each ray towards the other one. As the expansion of the universe decelerates, and the light cones tilt inward (until they're vertical at "now"), the two light rays are bent back together to meet at "here and now".

Part 3 of Ned Wright's cosmology tutorial at http://www.astro.ucla.edu/~wright/cosmo_03.htm gives more details about these diagrams.

Mea Culpa. I did not notice that on the lower triangles both the left side and the right side are going in the same direction.

Thank you for the link to Ned Wright's page, but that leads me to further questions. Earlier, I was saying that I did not know how to handle acceleration in general relativiy. Ned Wright handles a change-of-reference frames in the GR by doing a Galilean Transformation (attached)
Note that this is not a Lorentz transformation, and that these coordinates are not the special relativistic coordinates for which a Lorentz transformation applies. The Galilean transformation which could be done by skewing cards in this way required that the edge of the deck remain straight, and in any case the Lorentz transformation can not be done on cards in this way because there is no absolute time. But in cosmological models we do have cosmic time, which is the proper time since the Big Bang measured by comoving observers, and it can be used to set up a deck of cards. The presence of gravity in this model leads to a curved spacetime that can not be plotted on a flat space-time diagram without distortion. If every coordinate system is a distorted representation of the Universe, we may as well use a convenient coordinate system and just keep track of the distortion by following the lightcones.

I'm not sure what he's saying, but he seems to be saying that since Lorentz Transformation does not apply to these coordinates, that Galilean Transformation actually DOES apply. Also, this suggests that the vertical component of the graph is NOT the coordinate time of the central observer, but rather it is the "cosmological time" which is the "proper time since the Big Bang measured by comoving observers."

So effectively, you're doing a velocity transformation, but it is unusable as acceleration, because the two reference frames are "comoving."

So my question is, if you want to handle a real acceleration; how is this handled? I presume it is not done by either galilean transformation or by lorentz transformation.

We have to be careful with terminology here as well, because the cosmological observers we've been discussing (the ones at rest in the Robertson-Walker coordinates) are not accelerating; that is, they don't feel any acceleration. And again, I'm not sure that the fact that the local inertial frames of cosmological observers at different times don't "line up" is most usefully viewed as an "acceleration", although that term is often used (as in, the expansion of the universe is now accelerating, but was decelerating earlier in its history).

I know RW coordinates are not accelerating. The question is whether they can handle acceleration. And I'm not talking about gravitational acceleration. I'm talking about acceleration due to the pressure caused by primordial particles decaying, creating pressure, reaching critical mass, exploding, etc. One can handle these accelerations in Minkowski space-time rather easily. But in R-W coordinates, it appears that Lorentz Transformation will not do the job. So will Galilean Transformation do it?

JDoolin

Nov14-10, 02:05 PM

I don't know that I disagree, but I don't know that I would put it this way either. The "stretching" doesn't happen in space; it happens in time, as the scale factor changes. Obviously you can kind of convert time to space because looking farther away means looking at things the way they were a longer time ago; but I'm not sure I would characterize what we see as we look out that far as "stretching". Maybe my further comments below will help to clarify what I'm getting at.

We have to be careful with terminology here as well, because the cosmological observers we've been discussing (the ones at rest in the Robertson-Walker coordinates) are not accelerating; that is, they don't feel any acceleration. And again, I'm not sure that the fact that the local inertial frames of cosmological observers at different times don't "line up" is most usefully viewed as an "acceleration", although that term is often used (as in, the expansion of the universe is now accelerating, but was decelerating earlier in its history).

Actually, I'm not quite sure what you mean by observer's lining up.

The calculations we've been doing are kinematic; they don't get into the detailed dynamics of what's going on with the matter-energy in the universe. The Robertson-Walker models abstract all that out by treating the matter-energy in the universe as a perfect fluid, which can take one of three simple forms characterized by different equations of state relating the pressure, p, to the energy density, rho:

(1) "Matter dominated": a fluid with zero pressure, p = 0. This is a good approximation at the cosmological level for non-relativistic matter (meaning, the average speed of the individual "particles" in the fluid, which are basically galaxies or galaxy clusters, is much less than the speed of light).

(2) "Radiation dominated": a fluid with equation of state p = 1/3 rho. This is the equation of state for a "fluid" made of pure radiation (for example, the CMBR).

(3) "Vacuum dominated": a fluid with equation of state p = - rho. This is the equation of state for a "fluid" which is due to a cosmological constant (other terms used are "vacuum energy" or "dark energy").

I don't understand the idea of a pressure less than zero. Also, what is rho?

The current best fit model for the evolution of the universe is: first the "inflation" phase, in which the equation of state was vacuum dominated with an extremely large effective energy density rho, meaning that the universe "expanded" exponentially; then, after the phase transition that ended inflation, a radiation dominated phase, which lasted roughly until the time of "recombination" (electrons and nuclei combining into atoms, which made the universe basically transparent to photons) when the universe was about 100,000 years old (all these times are very approximate, I'm going from memory here); then a matter dominated phase, which lasted until a few billion years ago (I believe); and finally, another vacuum dominated phase but with a very, very small effective energy density, causing the expansion of the universe to start "accelerating" again (it had been decelerating during the radiation and matter dominated phases).

To me, it seems like there should be a radioactive decay dominated era. You have a matter-dominated era, where the particles are not, on average, old enough to have decayed. Then particle start decaying, and the pressure builds, and you have cascading reactions. This would have, on average, a pressure far beyond the pressure in the radiation dominated era. Also, the net effect of these interactions would be to mash the matter in the universe together, providing the seeds for stars and galaxies.

In the attachment, I took the liberty of modifying the RW diagram to show worldlines of particles flying out of a "secondary bang" imagining some region where a large number of particles reached critical density at the same time, producing an enormous reaction.

If we don't have a good explanation already for our 600 km/s peculiar velocity with the CMBR, this could explain it. Also, this would explain why nearby, we have a hubble's constant of 70 km/s per megaparsec, and further out we have a much smaller hubble constant.

PeterDonis

Nov14-10, 02:21 PM

I know RW coordinates are not accelerating. The question is whether they can handle acceleration. And I'm not talking about gravitational acceleration. I'm talking about acceleration due to the pressure caused by primordial particles decaying, creating pressure, reaching critical mass, exploding, etc. One can handle these accelerations in Minkowski space-time rather easily. But in R-W coordinates, it appears that neither Lorentz Transformation will not do the job. So will Galilean Transformation do it?

Wright calls the transformation he uses a "Galilean transformation" because, loosely speaking, it doesn't "change the time direction" the way a Lorentz transformation does; it leaves the surfaces of constant cosmological time invariant, and just transforms the space coordinates within each surface. In General Relativity, this kind of transformation is perfectly legitimate, as long as you transform *everything* accordingly. It just so happens that this particular transformation, in addition to leaving the surfaces of constant cosmological time invariant, also preserves the isotropy of space--the fact that everything looks the same in all directions. So it actually leaves the entire Robertson-Walker metric invariant; the metric looks exactly the same in the transformed coordinates as it does in the original coordinates.

This means that, for example, the cosmological observers--the ones "at rest" in the Robertson-Walker coordinates--are still not accelerating in the transformed coordinates; all that's changed is which particular set of constant space coordinates each observer is at. So I think the simple answer to your question is that Wright's "Galilean transformation" is not intended to "handle accelerations".

However, as I said in my previous post, I don't think a model of the universe as a whole is meant to handle accelerations in the sense you mean, because it simply doesn't model things to that level of detail. See my next post responding to your next post for more on this, since it fits in better in response to what you said in that post.

PeterDonis

Nov14-10, 03:18 PM

Actually, I'm not quite sure what you mean by observer's lining up.

I mean that the local inertial frames at, say, the event you labeled N (the Earth now), and another event P, a point on the worldline of the cosmological observer that passes through N, but ten billion years to the past of N (according to that observer's proper time) are different inertial frames, the same way that the inertial frames of observers moving at different velocities are different inertial frames. If we extended both local inertial frames far enough in time that they overlapped, their coordinate lines would not line up, just as the coordinate lines of observers moving at different velocities do not line up.

I don't understand the idea of a pressure less than zero. Also, what is rho?

Ned Wright has a page that explains briefly how the negative pressure of the vacuum works here: http://www.astro.ucla.edu/~wright/cosmo_constant.html.

Rho is the energy density of the cosmological fluid, as seen by an observer at rest in the cosmological coordinates. Some people write \rho to mean the mass density, so the energy density would then be \rho c^{2}; I like to work in units where c = 1, so energy density and mass density are the same thing.

To me, it seems like there should be a radioactive decay dominated era. You have a matter-dominated era, where the particles are not, on average, old enough to have decayed. Then particle start decaying, and the pressure builds, and you have cascading reactions. This would have, on average, a pressure far beyond the pressure in the radiation dominated era. Also, the net effect of these interactions would be to mash the matter in the universe together, providing the seeds for stars and galaxies.

What you say about pressure here isn't right. The pressure in the radiation dominated equation of state is actually a limiting case of the pressure due to highly relativistic particles, which is what would be produced by reactions such as you describe. Basically, as the particles in the fluid become more relativistic, the pressure builds from zero (the non-relativistic limiting case) to 1/3 rho (the relativistic limiting case).

I believe you're correct, though, that internal reactions in the cosmological fluid would have the effect of producing "clumps" of higher density, which would then contract gravitationally to form stars and galaxies. I don't know offhand how large this effect is believed to be compared with the effect of "primordial density fluctuations", which are basically quantum fluctuations in the initial state at the beginning of the inflation phase, magnified by many orders of magnitude during the inflation.

In the attachment, I took the liberty of modifying the RW diagram to show worldlines of particles flying out of a "secondary bang" imagining some region where a large number of particles reached critical density at the same time, producing an enormous reaction.

Again, I believe this is basically correct, but I don't know how large an effect it would be compared to others. The only caveat I can see is that, because the universe is expanding, there may be limits on how high the density and pressure can get in such a localized "bang" compared to what they were in the actual initial "bang".

If we don't have a good explanation already for our 600 km/s peculiar velocity with the CMBR, this could explain it. Also, this would explain why nearby, we have a hubble's constant of 70 km/s per megaparsec, and further out we have a much smaller hubble constant.

I agree your suggestion could in principle explain the peculiar velocity of a particular system of matter now (e.g., the solar system). The variation in the Hubble constant, however, is due to the change in the expansion rate of the universe, and is already accounted for in the basic Robertson-Walker models. (Different models predict different specific "curvatures" in the diagram, but they all predict *some* curvature.)

Mentz114

Nov14-10, 03:42 PM

JDoolin, you should be aware that the FRW model is an large scale approximation whose symmetries are isotropy and homogeneity. On a smaller scale than cosmic we know that clusters of galaxies have peculiar ( ie not Hubble flow) motions and galaxies within clusters also have this kind of motion, superimposed on the flow. PeterDonis makes this point also.

GR allows us to write models of the whole universe and also to work out how this universe would look to different observers, which is where the coordinate transformations come in. A great benefit is that the essential physics and symmetries of the model will not be lost when we do coordinate transforms. You've seen two examples in FRW spacetime. The one that corresponds best to our experience is the comoving frame which sees all other matter as receding with cosmological red-shift.
It's also possible to define a frame where clock rates change with time so everything looks static.

I think my point is - some of your questions, good though might be, are outside the range of what GR can tell us.

On the question of Gallilean vs Lorentz transformations - this depends on what geometry you choose for your 3D hyperslices, when making a transformations. As you can see in the lecture notes, there are Euclidean ( Galliean relativity) or hyperbolic ( SR) as possibilities. No doubt the coords Ned Wright was using had Euclidean spatial hyperslices (Painleve coords, probably).

PeterDonis

Nov14-10, 03:55 PM

[B]No doubt the coords Ned Wright was using had Euclidean spatial hyperslices (Painleve coords, probably).

No, Painleve coordinates apply to the Schwarzschild spacetime. Wright was using the Robertson-Walker coordinates for the critical density case, where the spatial slices are Euclidean. Wright doesn't specifically write down the metric in the coordinates he's using, but I believe it would look like this:

d\tau^{2} = dt^{2} - a(t) \left[ dr^{2} + r^{2} \left( d\theta^{2} + sin^{2} \theta d\phi^{2} \right) \right]

Every "cosmological observer" has constant spatial coordinates r, theta, phi in this coordinate system; and it should be obvious from the metric above that the time coordinate t directly reflects the proper time experienced by these observers. All the time variation (the "expansion" of the universe), and hence all the curvature of spacetime, is reflected in the dynamics of the scale factor a(t).

The spatial coordinate in Wright's "Galilean transformation" diagram would be r in the metric above; his transformation is basically just a spatial translation to put a different cosmological observer at the origin r = 0.

Mentz114

Nov14-10, 04:01 PM

No, Painleve coordinates apply to the Schwarzschild spacetime.
Yes, there is a Painleve chart for FRW dust. Do you want me to post it ?

Every "cosmological observer" has constant spatial coordinates r, theta, phi in this coordinate system; and it should be obvious from the metric above that the time coordinate t directly reflects the proper time experienced by these observers.

Phew - how can say that when you've just seen two different coordinate transformations of FRW representing observer frames with completely different proper times ?

PeterDonis

Nov14-10, 04:06 PM

Yes, there is a Painleve chart for FRW dust. Do you want me to post it ?

Yes, please.

Phew - how can say that when you've just seen two different coordinate transformations of FRW represnting observer frames with completely different proper times ?

Just read it off the metric I wrote: for an observer at rest in the spatial coordinates, dr = dtheta = dphi = 0. So the metric reads d\tau^{2} = dt^{2}. Any transformation that leaves the metric in the same form (which, as I understand it, Wright's "Galilean transformation" does) will preserve that property. But of course, a transformation into some other coordinates where the metric looks different (such as Wright's "special relativistic" coordinates) may not.

Mentz114

Nov14-10, 04:15 PM

Just read it off the metric I wrote: for an observer at rest in the spatial coordinates, dr = dtheta = dphi = 0. So the metric reads d\tau^{2} = dt^{2}. Any transformation that leaves the metric in the same form (which, as I understand it, Wright's "Galilean transformation" does) will preserve that property. But of course, a transformation into some other coordinates where the metric looks different (such as Wright's "special relativistic" coordinates) may not.

You're right about the raw coordinates, no transformation is needed because it the spatial slice is already conformally flat. I didn't read your post properly. Anyhow, I did put a 'probably' after my Painleve speculation :smile:

I'll look up the Painleve chart as soon as my coffee is brewed and consumed.

Mentz114

Nov14-10, 05:02 PM

I had a problem editing my previous post so I made a new one.

\begin{align*}
dt_p&=dt\\
dx_p&=dx-\frac{2\,x}{3\,t}dt\\
dy_p&=dy-\frac{2\,y}{3\,t}dt\\
dz_p&=dz-\frac{2\,z}{3\,t}dt
\end{align*}

which gives a metric,

\left[ \begin{array}{cccc}
\frac{4\,{z}^{2}+4\,{y}^{2}+4\,{x}^{2}-9\,{t}^{2}}{9\,{t}^{2}} & -\frac{2\,x}{3\,t} & -\frac{2\,y}{3\,t} & -\frac{2\,z}{3\,t}\\\
-\frac{2\,x}{3\,t} & 1 & 0 & 0\\\
-\frac{2\,y}{3\,t} & 0 & 1 & 0\\\
-\frac{2\,z}{3\,t} & 0 & 0 & 1
\end{array} \right]

The spatial part is static now.

The Einstein tensor transformed by this frame field has only one non-zero component,
G_{00}=\frac{4}{3\,{t}^{2}}
which up to a factor is the SET of static non-interacting dust whose density varies with time. The Riemann scalar is 80/27t^4 which shows there is an unremoveable curvature singularity at t=0.

JDoolin

Nov14-10, 05:05 PM

I'm a little troubled that the Robertson Walker chart is either mapping coordinate time or proper time, depending on who gives me an answer. If it is supposed to be proper time, then I understand what you mean by events "not lining up" The problem is that the intersection of two world-lines on a space versus proper time graph do not indicate co-location in space (a collision).

If the diagram represents space vs. proper time, also, this casts into doubt the presence of the light-cones *anywhere* in the diagram. Light does not age, and therefore it should not move forward in proper time. All the light rays in a space vs. proper time diagram should be horizontal. And the intersection of light rays and world-lines on the space versus proper time diagram are all but meaningless. (since as mentioned before, they don't "line up")

PeterDonis told me that Lewis Epstein was encouraging the confusion of proper time and coordinate time, but it appears to me that Robertson Walker have already successfully muddled the two. Epstein, in fact, made great strides in clarifying the difference.

Mentz114

Nov14-10, 05:47 PM

I'm a little troubled that the Robertson Walker chart is either mapping coordinate time or proper time, depending on who gives me an answer.
I'm not sure what you mean by 'chart' or 'mapping'. Care to elaborate ?

The problem is that the intersection of two world-lines on a space versus proper time graph do not indicate co-location in space (a collision).
The coincidence of worldlines always means collision. Same place, same time. No transformation can change that.

Coordinate time is a parameter in the model, proper time is the integral of the Lorentzian interval along a worldline.

JDoolin

Nov14-10, 06:20 PM

I'm not sure what you mean by 'chart' or 'mapping'. Care to elaborate ?

The coincidence of worldlines always means collision. Same place, same time. No transformation can change that.

Coordinate time is a parameter in the model, proper time is the integral of the Lorentzian interval along a worldline.

It's my question in post #42. You're sometimes saying that the vertical coordinate in the Robertson-Walker diagram represents the proper time of particles. Other times, you're acting like it is the actual time passed by the central observer. You can either have one or the other. Not both.

The only place where those two definitions can be shared is along the single line representing the worldline of the "stationary" particle.

As I've pointed out before, the parameters of \tau and t are very different. In an x vs t diagram, you are correct. The coincidence of worldlines always mean a collision, and no legitimate transformation can change that...EXCEPT FOR a transformation into an x vs \tau coordinate system, which is totally NOT a legitimate transformation, because \tau is not a coordinate. It is a property of matter.

And when two pieces of matter cross the same coordinate space, when they reach the same PROPER AGE, that does NOT mean that they are colliding. It means they just happened to reach that same spot in space when they happened to be the same proper age. That does NOT mean they reached the same spot at the same TIME.

So, I repeat, (more emphatically, this time,) that an intersection on a space vs proper time graph, is all but physically meaningless.

Mentz114

Nov14-10, 07:51 PM

JDoolin

Nov14-10, 08:14 PM

OK.

On a spacetime diagram, the vetical axis is t, coordinate time. For the vertical worldline we can calculate the proper time on the clock,

d\tau^2=dt^2-dx^2=dt^2 \Rightarrow d\tau=dt

because dx=0. For the tilted worldlines we get a different proper time. So every observer thinks his clock is showing coordinate time. Can you see why people say that the vertical axis shows proper time ? Because for the stationary observer it does. None of this is of the least significance. Assume that in a spacetime diagram it is t vs x.

This is why I'm saying I get different answers depending on who answers. Ned Wright Says (http://www.astro.ucla.edu/~wright/cosmo_03.htm) "But in cosmological models we do have cosmic time, which is the proper time since the Big Bang measured by comoving observers, and it can be used to set up a deck of cards."

It has to be proper time that he's talking about, because, he then proceeds to do a Galilean Transformation on the diagram. The reason that it's valid to do this transformation is that the horizontal lines in the diagram represent lines of constant PROPER time. (Specifically, proper time for particular observers, who are following particular worldlines.)

While I have no objection to graphing proper time vs. position, this way, I think we deserve a certain amount of clarity about which is being graphed, because it can be either one or the other, but not both. If we have a space vs. proper-time graph, the speed of light should be represented as horizontal lines; not light cones. If we have a space vs TIME graph, then it is not valid to do a galilean transformation, because the relativity of simultaneity has to come into play.

(I don't know what you do, when the Lorentz Transformations are only valid "locally" but you need to have some kind of transformation that looks like the Lorentz Tranformations locally, and whatever they do with the rest of the stretching space, I really cannot say.)

Mentz114

Nov14-10, 09:02 PM

I can see why you're confused. The FRW model is special in this way - because of the homogenity and isotropy, every worldline is the same.

While this diagram is drawn from our point-of-view, the Universe is homogeneous so the diagram drawn from the point-of-view of any of the galaxies on the diagram would be identical.

From my previous post you can see that for the stationary observer proper time and coordinate time coincide. Now combine that with the quote above, and one concludes that all the observers' clocks are showing coordinate time.

Ned Wright says 'we do not have absolute time', which I find puzzling, but we do have 'cosmic time', which coincides with any observer if we choose them to be at rest ( vertical wl). But coordinate time also has the property that it coincides with the stationary clock time. Therefore cosmic time = coordinate time, as far as I can see.

Looking at the spacetime diagram that shows all the worldines as straight and radiating from t=0, r=0. In a Minkowski diagram, those lines should all be curving away from the central stationary observer, because they are accelerating. They are straight because there's been some serious deformation of the axes, so this is not a Minkowski ST diagram. Therefore it's not odd that to change frames we use Gallilean transformations. If we plotted the worldlines in a Minkowski diagram, the LT would connect instants on the worldlines.

There are lots of coordinate transformations going on in those notes that are not explicitly stated. But he does say how the conformal coordinates are found

Sometimes it is convenient to "divide out" the expansion of the Universe, and the space-time diagram shows the result of dividing the spatial coordinate by a(t). Now the worldlines of galaxies are all vertical lines.(my emphasis)

PeterDonis

Nov14-10, 09:17 PM

This is why I'm saying I get different answers depending on who answers. Ned Wright Says (http://www.astro.ucla.edu/~wright/cosmo_03.htm) "But in cosmological models we do have cosmic time, which is the proper time since the Big Bang measured by comoving observers, and it can be used to set up a deck of cards."

It has to be proper time that he's talking about, because, he then proceeds to do a Galilean Transformation on the diagram. The reason that it's valid to do this transformation is that the horizontal lines in the diagram represent lines of constant PROPER time. (Specifically, proper time for particular observers, who are following particular worldlines.)

While I have no objection to graphing proper time vs. position, this way, I think we deserve a certain amount of clarity about which is being graphed, because it can be either one or the other, but not both.

I agree that some clarification of terms is in order:

(1) The FRW *spacetime* is a geometric object; it can be described using a number of different coordinate systems or charts (or metrics--see next item), but it's the same geometric object no matter what chart/metric is used to describe it. So statements about invariant quantities, like the proper time experienced by a given observer with a given worldline between two events on that worldline (e.g., the big bang and the Earth "now", events O and N in your terminology from an earlier post), are independent of the specific coordinate chart/metric being used.

(2) What I've been calling the Robertson-Walker *metric* is a specific coordinate system used to describe the FRW spacetime, along with the expression for the metric using that coordinate system. (The expression "coordinate chart" is also sometimes used to refer to a coordinate system.) This coordinate system is useful because it has the property I described, that in this particular coordinate system, the "time" coordinate t happens to directly represent the proper time experienced by "comoving observers", observers who remain at the same spatial coordinates (r, theta, phi) for all time. But this is a property of the particular coordinate system; a different coordinate system may not have it (see next item).

(3) What Mentz114 is calling the Painleve chart for the FRW spacetime is a different coordinate system used to describe that spacetime, in which the metric looks quite different than it does in the Robertson-Walker coordinate system. In this coordinate system, the "time" coordinate t does *not* directly represent the proper time of "comoving" observers (at least, I don't think it does based on looking at the metric--Mentz114, please correct me if I'm wrong). But again, this is a property of the specific coordinate system.

(4) Finally, a note on "coordinate time" vs. "proper time". Coordinate time is just what it says: in a coordinate system, there will be one coordinate which is "timelike" and three which are "spacelike" (unless you're using null coordinates, which I won't go into here). The one timelike coordinate is "coordinate time". It's nice if you can set up the coordinate system so the timelike coordinate has at least some kind of actual physical meaning, but there's nothing that requires you to. Proper time, on the other hand, is an invariant quantity, and as I noted above, it must be the same regardless of what coordinate system is used to calculate it. But, as I've also noted, in order to even talk about proper time you have to specify a worldline and events on that worldline (as in the example I gave, the proper time experienced by a "comoving observer" between events O and N).

With the above clarification of terms, hopefully the meaning of what Ned Wright was saying is clearer. Because he explicitly says that he is using the proper time of comoving observers to set up his "deck of cards", I deduce (though he doesn't say so explicitly) that he is using the Robertson-Walker coordinate system. In that coordinate system, the "Galilean transformation" he does is just a spatial translation, moving the origin of the spatial coordinates to a different comoving observer, without changing anything else.

If we have a space vs. proper-time graph, the speed of light should be represented as horizontal lines; not light cones. If we have a space vs TIME graph, then it is not valid to do a galilean transformation, because the relativity of simultaneity has to come into play.

I don't understand either of these statements. What is the difference between a "space vs. proper time" and "space vs. time" graph? In either case, you have the horizontal dimension representing space and the vertical dimension representing time (the only difference is that in the first case, you're specifying that the time coordinate you're using in the vertical dimension directly represents proper time for observers who stay at constant values of all the space coordinates, as with, for example, the Robertson-Walker coordinates as I defined them above). And in either case, light can't possibly travel on horizontal lines, because light doesn't travel in lines of constant time; it travels on null worldlines, and the lines of constant time (the horizontal lines in the diagrams) are spacelike lines. Also, the "Galilean transformation" Ned Wright does leaves the surfaces of constant time invariant; as I noted above, it's just a spatial translation, without changing anything else. Pure spatial translations don't bring in any of the issues involved with relativity of simultaneity. (This is true even in the standard Minkowski coordinates of special relativity; I can always move the spatial origin to a new spatial location without affecting anything else except the specific space coordinates I assign to specific events.)

JDoolin

Nov15-10, 06:51 AM

I agree that some clarification of terms is in order:

(1) The FRW *spacetime* is a geometric object; it can be described using a number of different coordinate systems or charts (or metrics--see next item), but it's the same geometric object no matter what chart/metric is used to describe it. So statements about invariant quantities, like the proper time experienced by a given observer with a given worldline between two events on that worldline (e.g., the big bang and the Earth "now", events O and N in your terminology from an earlier post), are independent of the specific coordinate chart/metric being used.

2) What I've been calling the Robertson-Walker *metric* is a specific coordinate system used to describe the FRW spacetime, along with the expression for the metric using that coordinate system. (The expression "coordinate chart" is also sometimes used to refer to a coordinate system.) This coordinate system is useful because it has the property I described, that in this particular coordinate system, the "time" coordinate t happens to directly represent the proper time experienced by "comoving observers", observers who remain at the same spatial coordinates (r, theta, phi) for all time. But this is a property of the particular coordinate system; a different coordinate system may not have it (see next item).

(3) What Mentz114 is calling the Painleve chart for the FRW spacetime is a different coordinate system used to describe that spacetime, in which the metric looks quite different than it does in the Robertson-Walker coordinate system. In this coordinate system, the "time" coordinate t does *not* directly represent the proper time of "comoving" observers (at least, I don't think it does based on looking at the metric--Mentz114, please correct me if I'm wrong). But again, this is a property of the specific coordinate system.

(4) Finally, a note on "coordinate time" vs. "proper time". Coordinate time is just what it says: in a coordinate system, there will be one coordinate which is "timelike" and three which are "spacelike" (unless you're using null coordinates, which I won't go into here). The one timelike coordinate is "coordinate time". It's nice if you can set up the coordinate system so the timelike coordinate has at least some kind of actual physical meaning, but there's nothing that requires you to. Proper time, on the other hand, is an invariant quantity, and as I noted above, it must be the same regardless of what coordinate system is used to calculate it.

But, as I've also noted, in order to even talk about proper time you have to specify a worldline and events on that worldline (as in the example I gave, the proper time experienced by a "comoving observer" between events O and N).

Yes, proper time (OF A WORLDLINE) is an invariant quantity, but proper time is NOT A COORDINATE. Coordinates are contravariant; not invariant. Proper time is a property of a worldline. By itself, proper time does not tell you when the event occurs, unless you also know the world-line-path that the particle has followed.

A good coordinate time will tell you when the event occurred, even if you don't know the world-line-path taken to arrive at that event.

With the above clarification of terms, hopefully the meaning of what Ned Wright was saying is clearer. Because he explicitly says that he is using the proper time of comoving observers to set up his "deck of cards", I deduce (though he doesn't say so explicitly) that he is using the Robertson-Walker coordinate system. In that coordinate system, the "Galilean transformation" he does is just a spatial translation, moving the origin of the spatial coordinates to a different comoving observer, without changing anything else.

I don't understand either of these statements. What is the difference between a "space vs. proper time" and "space vs. time" graph? In either case, you have the horizontal dimension representing space and the vertical dimension representing time (the only difference is that in the first case, you're specifying that the time coordinate you're using in the vertical dimension directly represents proper time for observers who stay at constant values of all the space coordinates, as with, for example, the Robertson-Walker coordinates as I defined them above). And in either case, light can't possibly travel on horizontal lines, because light doesn't travel in lines of constant time; it travels on null worldlines, and the lines of constant time (the horizontal lines in the diagrams) are spacelike lines. Also, the "Galilean transformation" Ned Wright does leaves the surfaces of constant time invariant; as I noted above, it's just a spatial translation, without changing anything else. Pure spatial translations don't bring in any of the issues involved with relativity of simultaneity. (This is true even in the standard Minkowski coordinates of special relativity; I can always move the spatial origin to a new spatial location without affecting anything else except the specific space coordinates I assign to specific events.)

In Minkowski Spacetime, a spatial translation is done by taking the paper your graph is drawn on, and moving it, to the left or right. A velocity change is done by Lorentz Transformation.

But in Friedmann-Walker, (Let me see if I've got this right) a pure spatial translation is represented by a Galilean Transformation, and an actual velocity change is simply beyond the scope of General Relativity.

What makes up a geometric object? Are you sure you can graph an invariant vs. a contravariant quantity, and that qualifies as a geometric object? It seems to me that the thing that Robertson Walker did was just set the contravariant coordinate-time equal to the invariant proper time of a bunch of particles that don't even exist, except by statistical average.

I'd like to know why; what was their point? What made this necessary?

I'd also like to know whether, once you set your coordinate time and proper time equal, is it even possible to determine the effect of a large acceleration?

And if I take\tau=t and plug it into d\tau^2=dt^2-dx^2 I get dx=0, suggesting that nothing can ever change its position. (correction: This is to be expected since the \tau represents properties of particles for whom dx is equal to zero. However, I still think that saying \tau is "timelike" is an overgeneralization. \tau and t are really quantities of a fundamentally different sort.)

PeterDonis

Nov15-10, 10:30 AM

Yes, proper time (OF A WORLDLINE) is an invariant quantity, but proper time is NOT A COORDINATE. Coordinates are contravariant; not invariant. Proper time is a property of a worldline. By itself, proper time does not tell you when the event occurs, unless you also know the world-line-path that the particle has followed.

I think some more clarification of terminology is in order; I should have clarified this before since we've been using the term "proper time" in more than one sense.

(1) Proper time *along a particular worldline between two particular events* is an invariant; geometrically, it's the analogue in spacetime of the invariant "length" of a particular line segment in a Euclidean space.

(2) Proper time *along a particular worldline*, without specifying events on the worldline, is a *parameter*: a range of real numbers you can use to label events on the worldline, by arbitrarily assigning some particular event the value 0 and then labeling every other event by its invariant proper time along the worldline from the event with the value 0 (with earlier events having negative proper time and later events having positive proper time).

(3) Once you have a labeling of events on a worldline by the proper time parameter, you can then look for a coordinate system that uses that same event labeling as its time coordinate. If you're really lucky, you can find a coordinate system that does this, not just for one worldline, but for a whole family of worldlines that are picked out by some symmetry property of the spacetime. This is what is meant by "coordinate time directly represents proper time" for a particular family of observers (in the case I've been discussing, the "comoving" observers).

(Actually, technically *any* definition of a coordinate system with a time coordinate implicitly specifies a "family of observers", in a sense I'll define below; but that family may not make much sense physically, depending on how the coordinate system is defined. See next comment.)

A good coordinate time will tell you when the event occurred, even if you don't know the world-line-path taken to arrive at that event.

But when the event occurred is frame-dependent (i.e., coordinate system dependent); it's not an invariant. That means that specifying when an event occurs requires specifying a coordinate system; and specifying a coordinate system requires specifying a time coordinate; and specifying a time coordinate implicitly specifies a family of worldlines, the integral curves of the vector field defined by that time coordinate (more precisely, defined by the partial derivative operator with respect to that time coordinate).

Depending on the particular coordinate system, this family of curves may or may not make much sense physically, interpreted as the worldlines of a family of observers; but in spacetimes with particular symmetries, those symmetries will pick out certain families of curves, and therefore certain coordinate systems that have those curves as the integral curves of their time coordinates; and those curves will (at least in the cases we're discussing) make sense as the worldlines of a family of observers. The time coordinate of the coordinate system then directly represents the proper time of those observers, in the sense I described above; and therefore specifying when events occur according to this time coordinate *is* specifying the proper time of those events with reference to that family of observers, without having to know the specific worldline path taken to arrive at the event.

In all these cases, though, when events occur is still frame-dependent, and as I think I've said before, I think it's a mistake to insist on thinking about relativistic physics in terms of frame-dependent quantities. Coordinate systems can be a calculational convenience, but they are not *necessary* for doing physics; you can write all the actual physics, if necessary, solely in terms of invariant quantities, without ever specifying a coordinate system.

In Minkowski Spacetime, a spatial translation is done by taking the paper your graph is drawn on, and moving it, to the left or right. A velocity change is done by Lorentz Transformation.

But in Friedmann-Walker, (Let me see if I've got this right) a pure spatial translation is represented by a Galilean Transformation, and an actual velocity change is simply beyond the scope of General Relativity.

You could do a "velocity change" in the FRW spacetime, but the resulting coordinate system would no longer respect the symmetries of the spacetime--the metric would look different (because space would no longer look isotropic in the "moving" frame--Earth itself is an example of such a "moving frame", since the CMBR does not look isotropic to us). That makes it a different case from Minkowski spacetime, where a Lorentz transformation leaves the metric looking the same in the transformed coordinates as it does in the original coordinates.

What makes up a geometric object? Are you sure you can graph an invariant vs. a contravariant quantity, and that qualifies as a geometric object? It seems to me that the thing that Robertson Walker did was just set the contravariant coordinate-time equal to the invariant proper time of a bunch of particles that don't even exist, except by statistical average.

I'd like to know why; what was their point? What made this necessary?

The FRW metric is determined by the condition that the universe is homogeneous and isotropic (it looks the same everywhere and in all directions). We know this condition isn't exactly fulfilled by our universe, but it's close, and the condition makes the mathematics tractable for expressing solutions in closed form. More detailed models take the FRW solution as a starting point and do an expansion about it in powers of small perturbations from exact isotropy, which gives more precise answers but requires numerical solutions.

The existence of a family of observers whose proper time is directly represented by the time coordinate in the FRW metric is something that *appears* in the solution, not something that is put in at the start. Basically, it amounts to looking at the integral curves of the FRW time coordinate, as I described above.

I still think that saying \tau is "timelike" is an overgeneralization. \tau and t are really quantities of a fundamentally different sort.)

As you note in your correction, saying that coordinate time directly represents proper time is not the same as setting tau equal to t in the metric to begin with; you first impose a condition on the metric (such as your dx = 0, which I stated in an earlier post as dr = dtheta = dphi = 0), and then see what the metric looks like with the condition imposed. But that also means that even though tau and t *are* different kinds of quantities, there can still be a relationship between them under certain conditions. For example, we impose the condition that all the space coordinates are constant on the FRW metric and obtain d\tau^{2} = dt^{2}. Since there are no other coefficients on either side, that means that tau and t must be measured in the same units, so it makes sense to talk about t "representing" tau for the particular family of observers that meets the condition we imposed. Also, since the signs are the same on both sides, tau must represent a timelike interval (positive squared length, using the sign convention I've been using), so it makes sense to call tau "timelike". (And in fact, this latter property holds in general: for *any* observer moving on a timelike worldline, tau along that worldline will have a positive square.)

PeterDonis

Nov15-10, 01:49 PM

What makes up a geometric object? Are you sure you can graph an invariant vs. a contravariant quantity, and that qualifies as a geometric object?

I realized on reading over my last post that I didn't fully respond to this point. A geometric object is, by definition, invariant; the object itself remains the same regardless of what coordinate system you use to label points in it. You don't "make" a geometric object by graphing anything; you draw graphs to illustrate how a particular coordinate system represents points, curves, etc. in a geometric object (or a portion of one). What you graph are always coordinates, but sometimes, as I noted in my last post, a particular coordinate happens to represent something with direct physical meaning, such as coordinate time representing proper time (in the sense I described in my last post).

JDoolin

Nov15-10, 03:03 PM

I think some more clarification of terminology is in order; I should have clarified this before since we've been using the term "proper time" in more than one sense.

(1) Proper time *along a particular worldline between two particular events* is an invariant; geometrically, it's the analogue in spacetime of the invariant "length" of a particular line segment in a Euclidean space.

(2) Proper time *along a particular worldline*, without specifying events on the worldline, is a *parameter*: a range of real numbers you can use to label events on the worldline, by arbitrarily assigning some particular event the value 0 and then labeling every other event by its invariant proper time along the worldline from the event with the value 0 (with earlier events having negative proper time and later events having positive proper time).

(3) Once you have a labeling of events on a worldline by the proper time parameter, you can then look for a coordinate system that uses that same event labeling as its time coordinate. If you're really lucky, you can find a coordinate system that does this, not just for one worldline, but for a whole family of worldlines that are picked out by some symmetry property of the spacetime. This is what is meant by "coordinate time directly represents proper time" for a particular family of observers (in the case I've been discussing, the "comoving" observers).

When things are proven mathematically, there is a certain inevitability of the next step in the process. You recognize your axioms and state them clearly, and then those axioms lead inevitably to certain conclusions. Even then, you acknowledge that if your assumptions are false, then your conclusions would also be false.

I find Special Relativity to be an axiomatically sound theory. Namely because the Lorentz Transformations leave the speed of light constant, but they allow for acceleration. But it seems to me that you've kind of nailed it here with General Relativity. "If you're really lucky, you can find a coordinate system that does this"

We start by making an assumption; I don't know what it is--there's no axiom behind General Relativity. If you ask "Are you assuming that the density is the same throughout," the answer is no. If you ask, "Are you assuming that the proper time is some universal parameter," the answer is no. There's no starting point.

You just say, let's assume the coordinate time is equal to the proper time, and then you run with it. "If you're lucky" you find a coordinate system that does this. And hey, you got lucky:

http://www.astro.ucla.edu/~wright/photons_outrun.html

You just need the coordinate system to stretch over time, and you need to have the particles to be appearing to move apart, but it's just an illusion formed by the stretching of space. And then, voila, you've created a system where Special Relativity no longer works. Yay!

So when I ask, why do Robertson Walker think they can set proper time to be coordinate time, I'm also asking, is there anything axiomatic that FORCES them to throw away the results of Special Relativity? Is there some assumption that they made that made the proper-time = coordinate time assumption inevitable?

I don't care how LUCKY they got in coming up with a system that throws away Special Relativity theory. I want to know the assumption they made that requires them to throw away the Special Theory of Relativity.

PAllen

Nov15-10, 03:31 PM

I don't care how LUCKY they got in coming up with a system that throws away Special Relativity theory. I want to know the assumption they made that requires them to throw away the Special Theory of Relativity.

Not sure if this is really what you're asking, but gravity is what killed Special Relativity. Before, and after General Relativity, there were researchers attempting to come up with consistent gravity+SR, without accepting GR (or other theories with spacetime curvature). Mathematically consistent theories were set up, but they all conflicted with experiment. There are proofs of impossiblity of an SR based theory matching known experiments, but, as you note, proofs always assume something you might find a way to remove.

Anyway since a pretty common view among physiscists is that GR is 'next theory to fall', and given that an SR based gravity would provide a great fit for Quantum Gravity, you could be a hero if you were to produce one.

PeterDonis

Nov15-10, 03:50 PM

So when I ask, why do Robertson Walker think they can set proper time to be coordinate time, I'm also asking, is there anything axiomatic that FORCES them to throw away the results of Special Relativity? Is there some assumption that they made that made the proper-time = coordinate time assumption inevitable?

I don't care how LUCKY they got in coming up with a system that throws away Special Relativity theory. I want to know the assumption they made that requires them to throw away the Special Theory of Relativity.

I think you're misunderstanding the process of arriving at a model in general relativity. First of all, the FRW solution (like any solution of the Einstein Field Equation) does not "throw away special relativity". At any event in the spacetime, you can set up a local inertial frame in which the laws of SR hold locally. You can't set up a *global* Minkowski coordinate system in which the laws of SR hold that covers the entire spacetime, because the spacetime is curved, and SR assumes a flat spacetime. This is not something unique to the FRW solution; it's true of any solution in GR (except, of course, the trivial solution of Minkowski spacetime itself, a spacetime that's globally flat, zero curvature everywhere).

Second, as I said before, the FRW model does not *assume* that proper time is equal to coordinate time. Let me lay out the steps in the process more explicitly:

(1) We are looking for a solution to the Einstein Field Equation that describes (to some appropriate level of approximation) the universe as a whole.

(2) We observe that, to some appropriate level of approximation (and after correcting for our own peculiar velocity), the universe appears homogeneous and isotropic. Therefore, in looking for a solution to the EFE, we decide to try imposing the condition that the universe be homoegenous and isotropic (meaning, spatially).

(3) The condition of homogeneity and isotropy picks out a certain particular subset of solutions to the EFE. After some mathematical work, we find that that subset of solutions has the property that the metric, by appropriate choice of coordinates, can always be written in the following form:

d\tau^{2} = dt^{2} - a(t) \left[\frac{1}{1 - k r^{2}} dr^{2} + r^{2} \left( d\theta^{2} + sin^{2} \theta d\phi^{2} \right) \right]

where t, r, \theta, and \phi are the coordinates, and k is a constant that can take one of three values: +1, 0, or -1. (The case I've been discussing up to now, where the spatial slices are flat, is the case k = 0, which makes the metric look like the one I wrote in an earlier post.)

(4) We notice that the metric as written above has the property that, for any worldline with constant values for all the space coordinates, d\tau^{2} = dt^{2}. That means three things: first, those worldlines are also integral curves of the t coordinate (since the dt^{2} term is the only term on the RHS); second, the t coordinate directly represents the proper time parameter along those worldlines (since there are no other coefficients on either side of the equation); and third, since the worldlines are timelike, we can define a family of observers moving along those worldlines, who we call "comoving" observers, and we can they say that the t coordinate directly represents their proper time.

So we never *assumed* that coordinate time was equal to proper time; we *discovered* that, if we look for solutions to the EFE that are spatially homogeneous and isotropic, those solutions can be described by a coordinate system in which coordinate time directly represents proper time for "comoving" observers. (That last qualifier, by the way, is key: some of your comments seem to imply that you think the FRW coordinate system somehow has coordinate time equal to proper time period, for *all* observers, which is false. A non-comoving observer--one whose spatial coordinates in the FRW coordinate system change with time--will find that their proper time is *not* represented directly by coordinate time.) Of course this coordinate system is *not* a Minkowski coordinate system such as we would use in SR; it can't be, because it's a global coordinate system covering the entire spacetime, and the spacetime it is describing is curved, and Minkowski coordinates can't globally represent a curved spacetime. That's not "throwing away SR"; it's recognizing that SR has a limited domain of applicability.

A final note on my use of the word "lucky": the "luck" comes in finding a particular symmetry property that allows a particular solution (or set of solutions) of the EFE to be written in a form that looks simple. In the general case, this is not possible. It just so happens that a number of spacetimes of interest in physics *do* happen to have particular symmetry properties that allow us to write them in a form that looks simple.

PeterDonis

Nov15-10, 04:20 PM

I find Special Relativity to be an axiomatically sound theory. Namely because the Lorentz Transformations leave the speed of light constant, but they allow for acceleration. But it seems to me that you've kind of nailed it here with General Relativity. "If you're really lucky, you can find a coordinate system that does this"

We start by making an assumption; I don't know what it is--there's no axiom behind General Relativity. If you ask "Are you assuming that the density is the same throughout," the answer is no. If you ask, "Are you assuming that the proper time is some universal parameter," the answer is no. There's no starting point.

Roughly, the general assumptions of GR are

(1) That we describe physics by means of invariant geometric objects in spacetime, which is itself a geometric object (a 4-dimensional pseudo-Riemannian manifold with a locally Minkowskian metric);

(2) That if we can express a given physical law in terms of invariant geometric objects in flat spacetime (i.e., in SR terms), that law will continue to hold if we allow the spacetime to be curved (which it must be in the presence of "gravity"), as long as we replace ordinary derivatives by covariant derivatives, which account for the curvature of spacetime.

These general assumptions are enough to get pretty close to the Einstein Field Equation. As far as I know, the shortest route the rest of the way is the one Hilbert discovered, about the same time Einstein was finishing his 1915 paper that announced GR:

(3) We assume that the dynamics of spacetime (or spacetime coupled to matter-energy, if the latter is present) is determined by a principle of least action.

(4) We assume that the appropriate action for spacetime is the unique action (which Hilbert found) that depends only on the metric and its first and second derivatives.

(5) We assume that, if matter-energy is present, we are given its action as well. I'm not sure if there are any general conditions on the matter action other than it being in appropriate Lagrangian form.

This is enough to get us to the Einstein Field Equation.

These are general assumptions, though, not assumptions particular to any specific physical problem. For a particular problem, of course, you will need additional specifications, such as the specific form of the matter action, or equivalently its stress-energy tensor (e.g., a perfect fluid with a particular equation of state--or the specification that you're looking for a vacuum solution, with zero stress-energy tensor), and any particular properties the solution must have (e.g., isotropy or spherical symmetry).

PeterDonis

Nov15-10, 06:45 PM

And when two pieces of matter cross the same coordinate space, when they reach the same PROPER AGE, that does NOT mean that they are colliding. It means they just happened to reach that same spot in space when they happened to be the same proper age. That does NOT mean they reached the same spot at the same TIME.

So, I repeat, (more emphatically, this time,) that an intersection on a space vs proper time graph, is all but physically meaningless.

Reading back through the thread, I saw this post, which helped me understand better what you meant by "a graph of space vs. proper time". This is definitely *not* what is being shown in any of the spacetime diagrams we have been discussing; the "time" is always *coordinate* time, which in some cases happens to also represent the proper time of a particular family of observers, as I've described.

However, your statement here does bring up a further issue, which is: suppose we have a coordinate system where coordinate time directly represents the proper time of some family of observers, and two of those observers happen to pass through the same spatial point? Wouldn't this raise the question you raise above, about how the coordinate system can possibly represent events accurately, if it's possible for two observers to be at the same point in space at the same *proper* time, but *not* necessarily at the same "time"?

I believe (but see one caveat at the end of this post) the answer is that no such question can ever arise, because whenever you have a coordinate system where coordinate time directly represents the proper time of a family of observers, each such observer has a unique worldline that can never intersect the worldline of any other such observer. This follows from the way the family of observers is picked out: their worldlines are the set of integral curves of the time coordinate. None of those integral curves can ever intersect: each event in the spacetime lies on one and only one such integral curve. So if, for some reason, the spatial coordinates were set up such that two different observers' worldlines both passed through the same spatial point (i.e., the same set of values for the spatial coordinates), they would *have* to do so at different proper times.

Going further along this line, note that we can use these integral curves to label the spatial points, such that each observer in the family of observers, for whom coordinate time directly represents proper time, has their own unique space point (meaning, label for their particular integral curve) at which they remain for all time. If there are three spatial dimensions in the spacetime, then the labels for the integral curves will need to contain three numbers to uniquely specify each curve. This amounts to finding a set of spatial coordinates that "matches up" with the time coordinate, in the sense that each set of unique values for the spatial coordinates is paired up one-to-one with a unique worldline in the family of integral curves of the time coordinate (and thus with a unique observer in the family of observers). So we can always find a coordinate system that not only has coordinate time directly representing proper time for a particular family of observers, but also has each observer "at rest" at his own unique spatial point for all time. The Robertson-Walker coordinates we've been discussing are such a coordinate system, with respect to the "comoving" observers.

Note: The one caveat I referred to above is that there may be singular "events" at which integral curves of the time coordinate can "intersect". For example, the initial singularity (the Big Bang) in FRW spacetime is such an event--all timelike worldlines originate there. However, the word "singularity" is key: technically, the initial singularity is not actually "part of the spacetime", because the curvature is infinite there (also, the spatial part of the metric collapses to zero, meaning that "space" at the initial singularity is no longer three-dimensional, but a single point). Physically, this means GR breaks down at the singularity; we need new physics (e.g., a quantum theory of gravity) to understand what's going on there. But leaving out singularities, I believe what I said above is correct.

PAllen

Nov15-10, 08:30 PM

Note: The one caveat I referred to above is that there may be singular "events" at which integral curves of the time coordinate can "intersect". For example, the initial singularity (the Big Bang) in FRW spacetime is such an event--all timelike worldlines originate there. However, the word "singularity" is key: technically, the initial singularity is not actually "part of the spacetime", because the curvature is infinite there (also, the spatial part of the metric collapses to zero, meaning that "space" at the initial singularity is no longer three-dimensional, but a single point). Physically, this means GR breaks down at the singularity; we need new physics (e.g., a quantum theory of gravity) to understand what's going on there. But leaving out singularities, I believe what I said above is correct.

Curious about this. Are you saying if you remove singular points, you can construct a single coordinate patch (not just any, but a nice one!) covering any solution of GR? That there are no GR solutions analagous to a sphere, which cannot be covered in one patch but has no singularity of any kind? Or were you implicitly referring to some class of cosmologic solutions that have this property?

PeterDonis

Nov15-10, 09:30 PM

Curious about this. Are you saying if you remove singular points, you can construct a single coordinate patch (not just any, but a nice one!) covering any solution of GR? That there are no GR solutions analagous to a sphere, which cannot be covered in one patch but has no singularity of any kind?

No, I wasn't saying that. I was only saying that within a given coordinate patch, leaving out singularities, what I said was (as far as I know) true. However, on thinking it over, I realize that there are some subtleties, which are worth mentioning, if I may be permitted to "think out loud" for a little.

The specific example I used, that of FRW spacetime, *does* have the property that a single coordinate patch can be used to cover the entire spacetime, and what I said in my previous post *is* true for FRW spacetime, without doubt (I pretty much gave the explicit construction).

Consider another simple example, Schwarzschild spacetime. The "obvious" coordinate patch in which what I said in my last post is true is the Schwarzschild exterior coordinates, for which the Schwarzschild time coordinate fulfills the requirements I gave. However, this patch does not cover the horizon or the region inside the horizon. A coordinate patch which does cover the interior region is the Schwarzschild interior coordinates; in this coordinate patch, the t coordinate is spacelike (so it can't be used as a "time" coordinate there) and the r coordinate is timelike. I think, but am not certain, that the r coordinate meets the requirements in the interior region; to be certain, I'd have to think some more about what the surfaces of constant r look like and whether a set of integral curves orthogonal to those surfaces can be used to uniquely label spatial points in the surfaces. I think they can, but I'm not certain.

There are, of course, other coordinate systems that can be used to describe Schwarzschild spacetime. The Painleve coordinates cover both the exterior and the "future interior" region, and the Painleve time coordinate satisfies the conditions (as long as we include the bit about leaving out singularities, since each integral curve of Painleve time ends at the central singularity at r = 0--this is also true for the r coordinate of the Schwarzschild interior coordinates). If we restrict ourselves to talking about black holes that form as a result of the collapse of a massive object (such as a star), then the two regions I just mentioned cover the entire spacetime in question.

However, if we allow ourselves to consider mathematical solutions that may or may not be "acceptable" physically, there is a "maximal analytic extension" of Schwarzschild spacetime which includes two more regions, the "past interior" region and the "second exterior" region. This maximal analytic extension is described by Kruskal coordinates, and one version of these (the version that does not use null coordinates) has a timelike coordinate that I believe meets the requirements I gave, although again I'm not certain. Assuming it does, the integral curves of this time coordinate all have finite endpoints both past and future, on the past and future singularities, so again the bit about leaving out singularities is needed.

Assuming that what I've said so far is correct, I believe similar comments would apply to the other "black hole" spacetimes, the most general of which is the Kerr-Newman spacetime (which includes all other "black hole" spacetimes, including Schwarzschild, as special cases).

However, I realize that there are a *lot* of other possible spacetimes that are solutions to the EFE. From the examples above, I think there may be at least two other conditions that would need to be satisfied for what I said to be true: first, the spacetime would need to be time orientable (no closed timelike curves, so something like the Godel solution, which has CTCs, would not work); second, I think the spacetime would need to be simply connected (no "wormholes" or other topological anomalies).

PeterDonis

Nov15-10, 10:34 PM

The Painleve coordinates cover both the exterior and the "future interior" region, and the Painleve time coordinate satisfies the conditions (as long as we include the bit about leaving out singularities, since each integral curve of Painleve time ends at the central singularity at r = 0--this is also true for the r coordinate of the Schwarzschild interior coordinates).

On further consideration, there's another subtlety here. The integral curves of the Painleve time coordinate are non-intersecting, so that's all right; but they do *not* have constant values of the Painleve radial coordinate r. So if we wanted to use these curves to uniquely label "spatial points", we would have to use some other labeling that didn't involve the Painleve r coordinate. There are such labelings: the simplest one I can think of is to label each integral curve of Painleve time by the event at which that curve crosses the horizon at r = 2M; we can use, for example, the Kruskal spatial coordinate X to uniquely label each such event.

So I do have to add another qualification to what I said: the spatial coordinates that "match up" with the time coordinate, so that the integral curves of the time coordinate have constant values of the spatial coordinates for all time, may *not* be the same as the spatial coordinates of the coordinate system that gave rise to the time coordinate!

PAllen

Nov15-10, 11:06 PM

Consider another simple example, Schwarzschild spacetime. The "obvious" coordinate patch in which what I said in my last post is true is the Schwarzschild exterior coordinates, for which the Schwarzschild time coordinate fulfills the requirements I gave. However, this patch does not cover the horizon or the region inside the horizon.

Of course, coordinate time in these coordinates is not actual proper time for any observer except at r=infinity. You used an interesting phrasing earlier:

"suppose we have a coordinate system where coordinate time directly represents the proper time of some family of observers"

Do you mean something other than equals for "directly represents"? Or are you thinking of some simple transform of the standard Schwarzschild coordinates that normalized t to equal tau ?

George Jones

Nov16-10, 04:01 AM

Let T and R be Painleve coordinates.
On further consideration, there's another subtlety here. The integral curves of the Painleve time coordinate are non-intersecting, so that's all right; but they do *not* have constant values of the Painleve radial coordinate r.

Other subtleties:

1) even though R = r, \partial / \partial R \neq \partial / \partial r, so their integral curves are quite different;

2) even though T \neq t, \partial / \partial T = \partial / \partial t, so their integral curves are the same;

3) the worldline of an observer who falls freely from infinity is not an integral curve of \partial / \partial T;

4) the integral curves of \partial / \partial R intersect orthognally the worldline of an observer who falls freely from infinity.

Here, t and r are standard Schwarzschild coordinates.

1) is an example of what Penrose calls Woodhouse's Second Fundamental Confusion of Calculus. Even though R = r, \partial / \partial R is not the same as \partial / \partial r because lines of constant (T,\theta,\phi) are not the same as lines of constant (t,\theta,\phi). For lines of constant (R = r,\theta,\phi), T and t differ by a constant, and hence 2). Because \tau = T on the worldline, 3) might seem a little odd. However, g \left( \partial / \partial T , \partial / \partial T \right) = 1 - 2M/R \neq 1, so integral curves of \partial / \partial T cannot be worldlines of observers. The 4-velocity of an observer who falls freely from infinity,

\bf{u} = \frac{\partial}{\partial T} - \sqrt{\frac{2M}{R}} \frac{\partial}{\partial R},

is used to derive 4).

If anyone wants, I can elaborate mathematically on the above.

JDoolin

Nov16-10, 06:56 AM

I think you're misunderstanding the process of arriving at a model in general relativity. First of all, the FRW solution (like any solution of the Einstein Field Equation) does not "throw away special relativity". At any event in the spacetime, you can set up a local inertial frame in which the laws of SR hold locally. You can't set up a *global* Minkowski coordinate system in which the laws of SR hold that covers the entire spacetime, because the spacetime is curved, and SR assumes a flat spacetime. This is not something unique to the FRW solution; it's true of any solution in GR (except, of course, the trivial solution of Minkowski spacetime itself, a spacetime that's globally flat, zero curvature everywhere).

This is a major part of what I can't understand. The Effect of the Lorentz Transformations are essentially directly proportianal to distance in space and time. i.e. If you Lorentz Transform an event that is 2 light years away, the effect will be roughly twice as much as if you Lorentz Transform an event that is 1 light-year away.

If you do a Lorentz Transform on an event that is a billion light years away, the effect is roughly a billion times as much as if you do an LT on an event that is 1 light year away. Maybe I'm misinterpreting what you're saying when you say "the laws of SR hold locally." The way I'm taking your meaning is that you can Lorentz Transform events within a certain radius in spacetime, but events beyond that radius are not Lorentz Transformed.

Maybe you mean something different by "the laws of SR hold locally."

Second, as I said before, the FRW model does not *assume* that proper time is equal to coordinate time. Let me lay out the steps in the process more explicitly:

(1) We are looking for a solution to the Einstein Field Equation that describes (to some appropriate level of approximation) the universe as a whole.

(2) We observe that, to some appropriate level of approximation (and after correcting for our own peculiar velocity), the universe appears homogeneous and isotropic. Therefore, in looking for a solution to the EFE, we decide to try imposing the condition that the universe be homoegenous and isotropic (meaning, spatially).

But you reject the possibility that maybe space ISN'T stretching, and you reject the possibility that perhaps there was an era of non-uniform acceleration, and you reject the possibility that the galaxies might actually be moving apart. And you reject the possibility that Lorentz Transformations might actually work on the long range.

(3) The condition of homogeneity and isotropy picks out a certain particular subset of solutions to the EFE. After some mathematical work, we find that that subset of solutions has the property that the metric, by appropriate choice of coordinates, can always be written in the following form:

d\tau^{2} = dt^{2} - a(t) \left[\frac{1}{1 - k r^{2}} dr^{2} + r^{2} \left( d\theta^{2} + sin^{2} \theta d\phi^{2} \right) \right]

where t, r, \theta, and \phi are the coordinates, and k is a constant that can take one of three values: +1, 0, or -1. (The case I've been discussing up to now, where the spatial slices are flat, is the case k = 0, which makes the metric look like the one I wrote in an earlier post.)

So it is an available option to just set a(t)=1 and k=0, right? So Minkowski spacetime actually is a possible solution to the Einstein Field Equations? And then we don't have to throw away Special Relativity.

(4) We notice that the metric as written above has the property that, for any worldline with constant values for all the space coordinates, d\tau^{2} = dt^{2}. That means three things: first, those worldlines are also integral curves of the t coordinate (since the dt^{2} term is the only term on the RHS); second, the t coordinate directly represents the proper time parameter along those worldlines (since there are no other coefficients on either side of the equation); and third, since the worldlines are timelike, we can define a family of observers moving along those worldlines, who we call "comoving" observers, and we can they say that the t coordinate directly represents their proper time.

Ah, I see. Yes, of course, if we have comoving observers, then of course, they would all share the same proper time. But it is one thing to define a family of observers moving along those parallel worldlines. It is quite another to claim that the galaxies in the real universe are a family of observers moving along those worldlines.

So we never *assumed* that coordinate time was equal to proper time; we *discovered* that, if we look for solutions to the EFE that are spatially homogeneous and isotropic, those solutions can be described by a coordinate system in which coordinate time directly represents proper time for "comoving" observers. (That last qualifier, by the way, is key: some of your comments seem to imply that you think the FRW coordinate system somehow has coordinate time equal to proper time period, for *all* observers, which is false. A non-comoving observer--one whose spatial coordinates in the FRW coordinate system change with time--will find that their proper time is *not* represented directly by coordinate time.) Of course this coordinate system is *not* a Minkowski coordinate system such as we would use in SR; it can't be, because it's a global coordinate system covering the entire spacetime, and the spacetime it is describing is curved, and Minkowski coordinates can't globally represent a curved spacetime. That's not "throwing away SR"; it's recognizing that SR has a limited domain of applicability.

I'm sorry. Unintentionally, I've been switching back and forth between two ideas, and now there are three. The third idea is what you are explaining, that a set of comoving observers share a proper time, and that proper time is the same as their coordinate time. That's valid. The first idea is that those comoving observers are the galaxies in the real universe, who only appear to be moving apart because of the stretching of space. (That's wierd, but not where my argument was coming from.) What I was thinking about were galaxies moving apart from each other, with real recessional velocities, whose proper times were all different. In this environment, it would be ridiculous to simply set proper time equal to coordinate time, because the galaxies wouldn't be comoving.

A "limited domain" of applicability for SR seems to me, the same as "throwing it away." If I told you that "rotation" had a limited domain of applicability, it would mean that if you turn to the left or right, only nearby objects respond. Things in your room might change positions relative to your facing, but distant stars would not cooperate; they would remain in the same place; stubbornly remaining in front of you as you spin around, because rotation is "only valid locally".

If you say "SR is valid only locally" you're saying that only nearby objects are affected by the Lorentz Transformations. It is absurd. Either SR is valid or it's not.

A final note on my use of the word "lucky": the "luck" comes in finding a particular symmetry property that allows a particular solution (or set of solutions) of the EFE to be written in a form that looks simple. In the general case, this is not possible. It just so happens that a number of spacetimes of interest in physics *do* happen to have particular symmetry properties that allow us to write them in a form that looks simple.

We have discussed General Relativity on the small scale, (say 0 to 1000 Astronomical Units across) and the large scale (say, beyond 1 billion light years) and recognized that there is a curvature associated with each one. The curvature at the local level is caused by gravitating bodies such as the earth, moon, sun, etc. The curvature at the global level is caused by making the mathematical generalization that

d\tau^{2} = dt^{2} - a(t) \left[\frac{1}{1 - k r^{2}} dr^{2} + r^{2} \left( d\theta^{2} + sin^{2} \theta d\phi^{2} \right) \right]

where t, r, \theta, and \phi are the coordinates, and k is a constant that can take one of three values: +1, 0, or -1.

One thing that I think we have established is that the large scale curvature, if it occurs at all, occurs at a level that is almost imperceptible up to a scale of at least a billion light years, and a billion years. Yet no one will entertain the idea that the large scale curvature actually is null; that a(t)=1 and k=0; that is, that the universe actually is, on the large scale, Minkowski.

It seems to me like this should be a starting point. That we should be willing to explore this simplest of possible options, and see what the actual expectations would be.

JDoolin

Nov16-10, 07:08 AM

You could do a "velocity change" in the FRW spacetime, but the resulting coordinate system would no longer respect the symmetries of the spacetime--the metric would look different (because space would no longer look isotropic in the "moving" frame--Earth itself is an example of such a "moving frame", since the CMBR does not look isotropic to us). That makes it a different case from Minkowski spacetime, where a Lorentz transformation leaves the metric looking the same in the transformed coordinates as it does in the original coordinates.

If you consider an observer that changes velocity at an event after t=0, he will not see a metric looking the same. He will se a dipole anisotropy. Possibly you are considering the Lorentz Transformation without thinking about the path the accelerating observer takes through it.

PeterDonis

Nov16-10, 11:08 AM

Or are you thinking of some simple transform of the standard Schwarzschild coordinates that normalized t to equal tau ?

I was thinking of this, but you're correct, this is another subtlety. (I see George Jones has pointed out further subtleties as well.) Probably I need to step back and re-think what I was saying and come up with a better formulation.

PeterDonis

Nov16-10, 11:12 AM

1) is an example of what Penrose calls Woodhouse's Second Fundamental Confusion of Calculus.

Heh, good phrase. Can you give a reference? I've read a fair amount of Penrose's writing (at least his writing for the lay reader) and I haven't come across this one.

If anyone wants, I can elaborate mathematically on the above.

I can follow what you've written, but I'll have to digest it some more; I may have questions after I do.

PeterDonis

Nov16-10, 11:14 AM

If you consider an observer that changes velocity at an event after t=0, he will not see a metric looking the same. He will se a dipole anisotropy.

Yes, this is what I was saying.

PeterDonis

Nov16-10, 12:41 PM

This is a major part of what I can't understand. The Effect of the Lorentz Transformations are essentially directly proportianal to distance in space and time. i.e. If you Lorentz Transform an event that is 2 light years away, the effect will be roughly twice as much as if you Lorentz Transform an event that is 1 light-year away.

...Maybe I'm misinterpreting what you're saying when you say "the laws of SR hold locally." The way I'm taking your meaning is that you can Lorentz Transform events within a certain radius in spacetime, but events beyond that radius are not Lorentz Transformed.

That's not what I mean; see below.

A "limited domain" of applicability for SR seems to me, the same as "throwing it away." If I told you that "rotation" had a limited domain of applicability, it would mean that if you turn to the left or right, only nearby objects respond. Things in your room might change positions relative to your facing, but distant stars would not cooperate; they would remain in the same place; stubbornly remaining in front of you as you spin around, because rotation is "only valid locally".

If you say "SR is valid only locally" you're saying that only nearby objects are affected by the Lorentz Transformations. It is absurd. Either SR is valid or it's not.

There are several issues conflated here, which I'll try to disentangle; hopefully this will also clarify some of the terms (e.g., "domain of applicability") I'm using.

First, the general issue of the "validity" of theories: If you absolutely must have a "go/no go" decision on SR, so to speak, then SR is *not* valid, just as Newtonian mechanics is not valid. Both are approximate theories that have known limitations. Newtonian mechanics can't handle objects moving at speeds large enough relative to the speed of light. SR can't handle situations where gravity must be taken into account.

GR is also an approximate theory with known limitations, but its domain of applicability is wider than both Newtonian mechanics and SR, since it includes both as special cases. If we specialize to weak gravity and slow speeds, we get Newtonian mechanics; if we specialize to negligible gravity (but allow relativistic speeds) we get SR.

GR's known limitations are: (1) It predicts spacetime singularities in certain situations, which basically amounts to saying that it admits it can't cover those particular situations and new physics is needed; (2) It isn't a quantum theory, and the general belief is that a quantum theory of gravity is needed (for example, to cover those situations where GR predicts singularities).

Second, there's the issue of what, given the above, it means to say that SR holds "locally". In the standard interpretation of GR (where gravity = spacetime curvature), SR holds "locally" in a curved spacetime in the same sense that Euclidean geometry holds "locally" on a curved surface, such as the surface of the Earth. The Earth's surface is not Euclidean, but I don't have to worry about its curvature when I'm measuring the square footage of my house; the curvature is too small to matter. But if I try to measure the area of the state of Alaska, for example, I'd better take the Earth's curvature into account or I'll get the wrong answer; in other words, Euclidean geometry does *not* hold on the Earth's surface when you get to that large a scale.

Does that mean that, for example, if my house is in the middle of the state of Alaska, and I stand in the middle of my house and spin around, my house spins but the state of Alaska as a whole doesn't? Of course not. But it does mean, for example, that my spinning around doesn't change the area of the state of Alaska; it's still different than it would be if the Earth's surface was flat. So whatever coordinate transformation is being induced on the entire surface of the Earth by my rotation, it must preserve the non-Euclidean geometry of that surface. If that means that such a transformation is different in some way than a "standard" rotational transformation in flat Euclidean space, then okay, it's different. But locally (within my house), I can still treat the transformation as a standard rotation in flat Euclidean space, as long as I remember that I can only make that approximation over a small enough distance.

Similarly, if I'm in a curved spacetime and I change my velocity, locally (i.e., over a small enough patch of spacetime that the effects of curvature are negligible--same basic criterion as I used above for my house vs. Alaska) I can model this by a standard Lorentz transformation, provided I set up local Minkowski coordinates around the event of the velocity change (just as I can set up local Euclidean coordinates inside my house, even though they don't do a good job of representing the entire state of Alaska). It may well be that the transformation induced on distant parts of spacetime will *not* be a standard Lorentz transformation, because it will have to preserve the global curvature (i.e., non-Minkowskian geometry) of the spacetime. But certainly *some* transformation will be induced; the entire universe will look different after the velocity change, not just a local patch, just as it's not only my house that spins around me when I spin.

Third, there's the issue of "interpretations" of GR. I said above that gravity = spacetime curvature is the standard interpretation. However, it is true that it is not the *only* interpretation. (One good discussion of this is Kip Thorne's, in his book Black Holes and Time Warps: Einstein's Outrageous Legacy, which I highly recommend, and not just for this specific issue but as a generally very good presentation of relativity for the lay reader.) Another way to interpret GR is by treating the metric as a field on a background spacetime that is flat--i.e., Minkowski. Basically, you start by writing the metric as

g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}

where \eta_{\mu \nu} is the standard Minkowski metric and h_{\mu \nu} is the extra field that accounts for the effects of gravity. Then you try to figure out what h_{\mu \nu} is by doing an expansion in powers of some parameter; when this method was first investigated in the 1950's and 1960's, by Feynman among many others, the motivation was to look for a quantum theory of gravity, so h_{\mu \nu} was taken to be (sorry for the bit of jargon here) a massless spin-two quantum field, the "graviton", on the background spacetime, and the expansion was just a standard perturbation expansion in powers of the graviton's quantum coupling constant (which is related to, but not necessarily the same as, Newton's gravitational constant), adding more and more different Feynman diagrams for different possible virtual graviton exchanges, similar to the methods that had worked so well for quantum electrodynamics. The end result of this process would be an expression for the action of spacetime, to some level of approximation anyway, which could be used, in the classical limit (i.e., letting Planck's constant go to zero), to derive a field equation by the same route that Hilbert had used in 1915 (which I referred to in an earlier post).

Of course there are an infinite number of terms in the perturbation expansion, but remarkably, in the case of gravity, it turned out that there was a way to calculate the sum of all of them, which converged to a finite answer, which, remarkably, turned out to be the *same* action that Hilbert had calculated in 1915! So basically, the theory of a massless spin-two field on flat Minkowski spacetime turns out to be GR, at least in the classical limit. But there are two key points about this:

(1) The flat "background" spacetime is unobservable: the actual, physical spacetime (meaning, the actual physical metric that determines proper times and proper distances) is curved, just as in the standard interpretation of GR. This is why this "field on flat spacetime" model is called an "interpretation" of GR instead of a different theory: it makes exactly the same predictions for all experiments as the "curved spacetime" model.

(2) The assumption of a flat background spacetime restricts the possible solutions in a way that the standard curved spacetime model of GR does not. For example, asymptotically flat solutions, such as the Schwarzschild spacetime, are allowed. But it does not, as I understand it, allow solutions such as the FRW spacetime, at least in the k = 1 and k = -1 cases (I'm not sure whether the k = 0 case would be allowed--it does have flat spatial slices, but conformally it doesn't look the same as Minkowski spacetime). So even though the field equation is the same, the underlying assumptions of the flat spacetime model are more restrictive and exclude solutions which are certainly relevant in physics, and to our discussion here.

In this thread I've been talking entirely within the curved spacetime model, since that's the standard interpretation, and since the flat spacetime interpretation ends up making the same predictions anyway for experiments. But see below for some further comments specifically about Minkowski spacetime as a solution of the EFE.

But you reject the possibility that maybe space ISN'T stretching, and you reject the possibility that perhaps there was an era of non-uniform acceleration, and you reject the possibility that the galaxies might actually be moving apart. And you reject the possibility that Lorentz Transformations might actually work on the long range.

The comoving observers (not galaxies, necessarily--see next comment) *are* moving apart, in the sense that the proper distance between them is increasing with time. I'm not sure what you mean by "an era of non-uniform acceleration"; if you mean that the rate of expansion of the universe (the rate of change of the scale factor a(t) with t) may have changed in the past, it has--we know that by the curvature in the Hubble diagram that you mentioned in an earlier post. If you mean that various individual pieces of matter may have accelerated non-uniformly, against the background of FRW spacetime overall, I agree that certainly may have happened, but as I said before, these details are averaged out in the overall FRW models we've been discussing (though they are treated numerically in more detailed models). I talked about how transformations would work long range above.

So it is an available option to just set a(t)=1 and k=0, right? So Minkowski spacetime actually is a possible solution to the Einstein Field Equations? And then we don't have to throw away Special Relativity.

Minkowski spacetime *is* a solution to the EFE, but only if there is no matter-energy present--i.e., the stress-energy tensor is zero identically. That isn't true of the universe, and the FRW solutions to the EFE are valid in the presence of matter-energy (non-zero stress-energy tensor). In the presence of matter, we can't set a(t) = 1 and k = 0 by fiat; we have to work out the dynamics and see. When we do that, we find that a(t) must change with time, and that there are three possible values for k, and which one actually holds for our universe is something we have to determine by measuring the overall density of matter-energy in the universe, the curvature of the Hubble diagram, etc.

Ah, I see. Yes, of course, if we have comoving observers, then of course, they would all share the same proper time. But it is one thing to define a family of observers moving along those parallel worldlines. It is quite another to claim that the galaxies in the real universe are a family of observers moving along those worldlines.

I agree, and I don't think I've claimed the latter, only the former. Individual galaxies, galaxy clusters, etc. may be moving with respect to the cosmological coordinates. The assumption of a perfect fluid on a cosmological scale allows that, as long as the motions average out to zero, just as with the molecules in an ordinary fluid.

I'm sorry. Unintentionally, I've been switching back and forth between two ideas, and now there are three. The third idea is what you are explaining, that a set of comoving observers share a proper time, and that proper time is the same as their coordinate time. That's valid. The first idea is that those comoving observers are the galaxies in the real universe, who only appear to be moving apart because of the stretching of space. (That's wierd, but not where my argument was coming from.) What I was thinking about were galaxies moving apart from each other, with real recessional velocities, whose proper times were all different. In this environment, it would be ridiculous to simply set proper time equal to coordinate time, because the galaxies wouldn't be comoving.

If individual galaxies are not "comoving" (if they are changing their spatial "location" in cosmological coordinates with time), then their proper times will *not* be directly related to coordinate time. That's quite true. The "comoving" observers are abstractions, and there may not be any actual observers in the actual universe who are exactly "comoving" in this sense. However, the condition for determining whether an observer is "comoving" (do they see the universe, for example the CMBR, as isotropic) is quite clear and physically realizable.

One thing that I think we have established is that the large scale curvature, if it occurs at all, occurs at a level that is almost imperceptible up to a scale of at least a billion light years, and a billion years. Yet no one will entertain the idea that the large scale curvature actually is null; that a(t)=1 and k=0; that is, that the universe actually is, on the large scale, Minkowski.

It seems to me like this should be a starting point. That we should be willing to explore this simplest of possible options, and see what the actual expectations would be.

As I noted above, Minkowski spacetime is only a solution of the EFE if there is no matter-energy present--if the stress-energy tensor is zero. So the actual universe *cannot* be a Minkowski spacetime. That's why we are forced to consider models that are more complicated than Minkowski spacetime. As Einstein said, "Make everything as simple as possible--but not simpler."

PeterDonis

Nov16-10, 03:03 PM

Other subtleties:

Just to check that I've got this right, after some digestion, here's my grasp of these four items, slightly out of order:

2) The transformation between t and T doesn't "change the direction" of the integral curves; it just reparametrizes them. This means that, in both coordinate systems, we can use the integral curves of "time" to uniquely label spatial points (each curve has constant values of r = R, \theta, and \varphi).

For the rest of this, I'll leave out the angular coordinates (assume them held constant) and only talk about "time" and "radius".

1) The lines of constant T, which are integral curves of \partial_{R}, "cut at a different angle" from the lines of constant t, which are integral curves of \partial_{r}. So even though the integral curves of "time" stay the same, the "spatial slices" cut through them can be different if they're cut at different angles.

3) It looks to me like the 4-velocity you gave, which gives us 4), also gives us 3), since it makes it obvious that the 4-velocity is *not* just \partial_{T}, so the integral curves of the 4-velocity can't be the same as the integral curves of \partial_{T}. The integral curves of the 4-velocity are "tilted inward", while the integral curves of \partial_{T} (and thus, of course, \partial_{t}) are "vertical".

4) Since the 4-velocity "tilts inward", the integral curves of \partial_{R}, to be orthogonal to them, must "tilt downward" relative to the integral curves of \partial_{r}, which are "horizontal".

PeterDonis

Nov16-10, 05:30 PM

4) Since the 4-velocity "tilts inward", the integral curves of \partial_{R}, to be orthogonal to them, must "tilt downward" relative to the integral curves of \partial_{r}, which are "horizontal".

Oops, I think this should be "tilt upward", since this is spacetime, not space, so "orthogonal" works differently.

JDoolin

Nov17-10, 07:14 AM

(1) The flat "background" spacetime is unobservable: the actual, physical spacetime (meaning, the actual physical metric that determines proper times and proper distances) is curved, just as in the standard interpretation of GR. This is why this "field on flat spacetime" model is called an "interpretation" of GR instead of a different theory: it makes exactly the same predictions for all experiments as the "curved spacetime" model.

(2) The assumption of a flat background spacetime restricts the possible solutions in a way that the standard curved spacetime model of GR does not. For example, asymptotically flat solutions, such as the Schwarzschild spacetime, are allowed. But it does not, as I understand it, allow solutions such as the FRW spacetime, at least in the k = 1 and k = -1 cases (I'm not sure whether the k = 0 case would be allowed--it does have flat spatial slices, but conformally it doesn't look the same as Minkowski spacetime). So even though the field equation is the same, the underlying assumptions of the flat spacetime model are more restrictive and exclude solutions which are certainly relevant in physics, and to our discussion here.

In this thread I've been talking entirely within the curved spacetime model, since that's the standard interpretation, and since the flat spacetime interpretation ends up making the same predictions anyway for experiments. But see below for some further comments specifically about Minkowski spacetime as a solution of the EFE.

If you are applying rotation to surface of a planet, you can do it in steps. Take whatever coordinates you have and map them, one-to-one into \mathbb{R}^2, the mapping that you CAN apply the rotation. Then do the rotation, and convert back.

It should be the same with Lorentz Transformation; simply map whatever coordinates you have into \mathbb{R}^4, apply the Lorentz Transformation, and then convert back.

When you say " the actual, physical spacetime (meaning, the actual physical metric that determines proper times and proper distances) is curved." It is curved with respect to what? The answer is, that it is curved with respect to the Minkowski coordinates. Even if we don't know what they are, can't we at least say the Minkowski Coordinates exist?

The answer is really simple. Don't apply the metric. Just use the original unmodified event coordinates, and you'll have Minkowski spacetime. Maybe it's not possible to go backwards, because we've got the transformed coordinates, but we don't know what the transformation actually was.

But those Minkowski coordinates within which the Lorentz Transformations still exist, whether or not we know how to map to them.

The comoving observers (not galaxies, necessarily--see next comment) *are* moving apart, in the sense that the proper distance between them is increasing with time. I'm not sure what you mean by "an era of non-uniform acceleration"; if you mean that the rate of expansion of the universe (the rate of change of the scale factor a(t) with t) may have changed in the past, it has--we know that by the curvature in the Hubble diagram that you mentioned in an earlier post. If you mean that various individual pieces of matter may have accelerated non-uniformly, against the background of FRW spacetime overall, I agree that certainly may have happened, but as I said before, these details are averaged out in the overall FRW models we've been discussing (though they are treated numerically in more detailed models). I talked about how transformations would work long range above.

Minkowski spacetime *is* a solution to the EFE, but only if there is no matter-energy present--i.e., the stress-energy tensor is zero identically. That isn't true of the universe, and the FRW solutions to the EFE are valid in the presence of matter-energy (non-zero stress-energy tensor). In the presence of matter, we can't set a(t) = 1 and k = 0 by fiat; we have to work out the dynamics and see. When we do that, we find that a(t) must change with time, and that there are three possible values for k, and which one actually holds for our universe is something we have to determine by measuring the overall density of matter-energy in the universe, the curvature of the Hubble diagram, etc.

If a(t) is not equal to 1, then I must ask you for an extraordinary level of clarity in what t defines. Is t the proper time at x,y,z, or is t the proper time at 0,0,0, or are these two assumed to be the same thing? Are you assuming that t at x,y,z is the age of particles that have traveled from here to x,y,z over the age of the universe, or are you assuming that the particles at x,y,z have always been at x,y,z?

If the universe is Minkowski, then I would say t represents the proper time at 0,0,0, and a(t)=0. But FRW says that the universe is stretching over time, which means a(t)~t and t represents the proper time of the actual galaxies at x,y,z, but somehow that proper time of the galaxies at x,y,z is the same as the proper time at 0,0,0. Meaning, they've taken as a-priori that the galaxies at x,y,z are comoving with the galaxies at 0,0,0.

If somehow, it can be shown that the stress energy tensor causes the scale of space to stretch over time, that's the sort of rigour that I'm looking for. I'm certainly no expert, but my impression has been that the stress energy tensor operates on velocity and proper time; i.e. properties of matter; not properties of space.

I agree, and I don't think I've claimed the latter, only the former. Individual galaxies, galaxy clusters, etc. may be moving with respect to the cosmological coordinates. The assumption of a perfect fluid on a cosmological scale allows that, as long as the motions average out to zero, just as with the molecules in an ordinary fluid.

If individual galaxies are not "comoving" (if they are changing their spatial "location" in cosmological coordinates with time), then their proper times will *not* be directly related to coordinate time. That's quite true. The "comoving" observers are abstractions, and there may not be any actual observers in the actual universe who are exactly "comoving" in this sense. However, the condition for determining whether an observer is "comoving" (do they see the universe, for example the CMBR, as isotropic) is quite clear and physically realizable.

As I noted above, Minkowski spacetime is only a solution of the EFE if there is no matter-energy present--if the stress-energy tensor is zero. So the actual universe *cannot* be a Minkowski spacetime. That's why we are forced to consider models that are more complicated than Minkowski spacetime. As Einstein said, "Make everything as simple as possible--but not simpler."

Let's investigate these results, which say, "Minkowski solution" -> "No Matter, No Energy." Let's see whether or not, they make any assumptions that the universe is made up of (approximately) comoving particles, or whether they assume that the amount of matter in the observable universe is finite. If they make either of these assumptions, or equivalent ones, they've already precluded the Minkowski solution.

PeterDonis

Nov17-10, 10:47 AM

When you say " the actual, physical spacetime (meaning, the actual physical metric that determines proper times and proper distances) is curved." It is curved with respect to what? The answer is, that it is curved with respect to the Minkowski coordinates.

No, the answer is that curvature is, essentially, an invariant; whether or not a spacetime is curved does not depend on what coordinates we use to describe it. It's a real, physical property that corresponds to real, physical measurements. I say "essentially" because curvature is not quantified by a single invariant quantity; there are a number of them, and often, in solving problems in GR, we don't express curvature solely in terms of those invariants, but in terms of geometric objects like the Riemann curvature tensor that do transform when we change coordinates. That's just for calculational convenience, and doesn't change the physics.

Let me give an example. Suppose we have two objects freely falling towards the Earth. Object 1 is at radius R (from the Earth's center), and object 2 is at radius R + r (slightly further away). Both objects start out at rest with respect to the Earth at time t = 0; here "time" refers to coordinate time in a Schwarzschild coordinate system with the Earth as the central mass. How will these objects move? We know from the physics of tidal gravity that the radial separation between them, which starts out as r, will increase with time, as they both fall toward the Earth.

What does this mean geometrically? Look at the problem in the t-r plane of the Schwarzschild coordinate system. We have two geodesics (two curves) in this plane which are initially parallel: at time t = 0, dr/dt is 0 for both curves. However, as time passes, the curves separate; the distance between them increases. That is a manifestation, geometrically, of curvature (initially parallel geodesics changing separation--that can't happen in a flat Euclidean space or a flat Minkowski spacetime), and it's a physical effect, independent of the coordinates. For example, we could measure it by running a string between the two objects (making sure that it is light enough and elastic enough that it will not affect their motion) and measuring how it stretches as time passes.

Now suppose we decide to adopt the "flat spacetime" model I described in a previous post, and decree that we are going to use Minkowski coordinates come hell or high water. What will we find? We will find that we can't use the simple metric corresponding to those coordinates to determine actual physical distances and times; we will need to add this extra field, h_{uv}, to the metric we actually use to calculate distances and times. For example, the Minkowski coordinate system would assign a constant coordinate separation between the two falling objects I just described (since they are both freely falling and their initial velocities are equal), but their *physical* separation, as we've seen, increases with time, so to obtain a metric that accurately represents distances throughout the spacetime, we'll need to modify the Minkowski metric. (Again, this is another way of saying that spacetime is curved, as as *physical* effect, independent of coordinates.)

Even if we don't know what they are, can't we at least say the Minkowski Coordinates exist?

As I just described, we can certainly decide to *adopt* "Minkowski coordinates", because GR allows us to use any coordinate system we want. But as we've just seen, since the actual spacetime is curved, physically, the Minkowski metric cannot accurately represent it. In that sense, no, the Minkowski coordinates do not "exist".

The answer is really simple. Don't apply the metric. Just use the original unmodified event coordinates, and you'll have Minkowski spacetime.

What does "don't apply the metric" mean? Without the metric, you have no way to translate the coordinates of events, which are just numbers, into actual physical distances and times. You can't dictate what the spacetime geometry is just by assigning coordinates to events. Suppose I decree that everyone must use Euclidean coordinates to label locations on the Earth's surface. Does that make the Earth's surface flat?

If a(t) is not equal to 1, then I must ask you for an extraordinary level of clarity in what t defines. Is t the proper time at x,y,z, or is t the proper time at 0,0,0, or are these two assumed to be the same thing? Are you assuming that t at x,y,z is the age of particles that have traveled from here to x,y,z over the age of the universe, or are you assuming that the particles at x,y,z have always been at x,y,z?

If the universe is Minkowski, then I would say t represents the proper time at 0,0,0, and a(t)=0. But FRW says that the universe is stretching over time, which means a(t)~t and t represents the proper time of the actual galaxies at x,y,z, but somehow that proper time of the galaxies at x,y,z is the same as the proper time at 0,0,0. Meaning, they've taken as a-priori that the galaxies at x,y,z are comoving with the galaxies at 0,0,0.

You keep on saying "assume" and "define" and so on. None of the things you are saying here are assumed or defined. They are all *discovered* as aspects of the solution to the EFE that we obtain when we impose the condition that the universe is isotropic. I described how the process works in a previous post, but I'll revisit one item here: how we determine what the coordinate "t" defines. Here, again, is the process:

(1) We look at the metric we obtain as a solution of the EFE under the condition that the universe is isotropic, and discover that it can be written in the form I gave.

(2) We look at the coordinate "t" as it appears in this metric (which, so far, is just an arbitrary coordinate, we haven't assumed anything about its physical meaning) and discover that, if we consider a curve with all the spatial coordinates held constant (which is an integral curve of \partial / \partial t), the coordinate t is "the same" as the actual lapse of proper time \tau along that curve (as given by the metric). (I elaborated on what "the same" means in my previous post.)

(3) We therefore have *discovered* that, *if* an observer were to move along such a curve (which is a timelike curve, so it can be the worldline of an observer), that observer's proper time would be the same as coordinate time. This is true for any such curve, so there is a whole family of curves, each of which can be the worldline of its own unique "comoving" observer.

(4) Thus, we have *discovered* that the coordinate t "directly represents the proper time of comoving observers" in the sense just given.

We didn't have to *assume* anything here; this is all just logical deduction from the metric, which is a solution of the EFE given the condition of isotropy. But let's go on and consider the scale factor a(t) (which, note, we did *not* have to consider in any of the above--all of the above is true independently of what the scale factor is or what it physically means):

(5) We note that the spatial part of the metric is multiplied by a factor a(t), and ask what this factor means. Since the spatial part of the metric that a(t) multiplies is a "standard" metric for one of three known geometric surfaces (hypersphere for k = 1, Euclidean 3-space for k = 0, or "hyperbolic space" for k = -1), the effect of a(t) is to determine the distance scale of space (because it multiplies all the terms in the spatial metric equally), with "space" being the appropriate geometric object for a given value of k. Since a(t) can vary with time, this means the distance scale of space can vary with time, meaning coordinate time. But since coordinate time directly represents proper time for comoving observers, if a(t) varies with coordinate time, comoving observers will see the same variation, which will appear to them physically as a change in their physical separation with time.

Again, we didn't have to assume anything; we deduced the physical meaning of a(t) by looking at the metric and applying known geometric facts and what we deduced above about time and comoving observers.

If somehow, it can be shown that the stress energy tensor causes the scale of space to stretch over time, that's the sort of rigour that I'm looking for. I'm certainly no expert, but my impression has been that the stress energy tensor operates on velocity and proper time; i.e. properties of matter; not properties of space.

The stress-energy tensor does describe properties of matter, not space, but the Einstein Field Equation tells us that the stress-energy tensor can *affect* the properties of space. So if you believe the EFE, then you ipso facto believe that the stress-energy tensor can affect the properties of space.

Let's investigate these results, which say, "Minkowski solution" -> "No Matter, No Energy." Let's see whether or not, they make any assumptions that the universe is made up of (approximately) comoving particles, or whether they assume that the amount of matter in the observable universe is finite. If they make either of these assumptions, or equivalent ones, they've already precluded the Minkowski solution.

The fact that Minkowski spacetime is only a solution of the EFE if the stress-energy tensor is identically zero is mathematically proven; it doesn't involve any assumptions beyond those required to derive the EFE itself. The fact that the stress-energy tensor can only be identically zero if there is no matter or energy in the universe is part of the physical definition of the stress-energy tensor. There's no wiggle room to "investigate these results" as you suggest; it would be like investigating whether the derivative of x^2 is 2x.

PeterDonis

Nov17-10, 01:37 PM

PeterDonis

Nov17-10, 05:49 PM

If you are applying rotation to surface of a planet, you can do it in steps. Take whatever coordinates you have and map them, one-to-one into \mathbb{R}^2, the mapping that you CAN apply the rotation. Then do the rotation, and convert back.

It should be the same with Lorentz Transformation; simply map whatever coordinates you have into \mathbb{R}^4, apply the Lorentz Transformation, and then convert back.

There's one other issue involved with this that I didn't mention. If the surface of the planet is a sphere, you can't map it one to one into \mathbb{R}^2, because the topology is different. You have to include a "point at infinity" as well. Similar remarks might apply when trying to map a curved spacetime into \mathbb{R}^4; for example, I don't know if you can do it with any of the FRW spacetimes without running into this type of difficulty. (This is all in addition to the point I already made, that whatever the final transformation turns out to be, if you *can* do the "map-transform-map back" procedure, it must preserve the intrinsic geometry of the original spacetime; you can't make the surface of the Earth flat by applying a rotation in this way, even assuming you find a way to deal with the "point at infinity" issue.)

JDoolin

Nov17-10, 06:42 PM

There's one other issue involved with this that I didn't mention. If the surface of the planet is a sphere, you can't map it one to one into \mathbb{R}^2, because the topology is different. You have to include a "point at infinity" as well. Similar remarks might apply when trying to map a curved spacetime into \mathbb{R}^4; for example, I don't know if you can do it with any of the FRW spacetimes without running into this type of difficulty. (This is all in addition to the point I already made, that whatever the final transformation turns out to be, if you *can* do the "map-transform-map back" procedure, it must preserve the intrinsic geometry of the original spacetime; you can't make the surface of the Earth flat by applying a rotation in this way, even assuming you find a way to deal with the "point at infinity" issue.)

You can map all the points on the surface of the earth into \mathbb{R}^2, but there are two points on the earth mapped to each point. Sorry, I didn't make that clear. I'm not talking about a one-to-one mapping. I'm just saying that if you are rotating, you can draw a line through the center of the earth, and each point on the earth has an r and theta with respect to that line around which you are rotating.

No, you're right, once you map all the points this way, you had better have kept track of the third coordinate which will tell you whether the point is on your hemisphere, or the opposite hemisphere.

Every point on the earth's surface has a coordinate relative to your axis of rotation, but every coordinate relative to your axis of rotation is associated with an infinite number of points on the earth.

I may add more if I remember what my point actually was. :wink:

Yes, you're right. I said you could convert to \mathbb{R}^2 and convert back. That is not quite true. You can convert fairly easily into cylindrical coordinates; though. Not sure if that was quite my point either.

What it comes down to is that I wouldn't be entirely satisfied with a rotation, unless we had the whole business defined in \mathbb{R}^4. If you take the earth speeding by at 99% of the speed of light and rotate around the coordinates around the point speeding by, you get quite a different result than if you rotate the earth around a point sitting on the earth.

I'm running into difficulty, though, because even in \mathbb{R}^4 there is some difficulty. From the observer's perspective at an instant, we regard the earth as a set of simultaneous events. If there were no such thing as slowing of time in a gravity well, then it would be a simple matter to claim that we have a set of simultaneous events. But with gravity, we cannot merely take a clock far enough out into space and say, it measures the "real time" for that observer.

Yet, I feel that the "real time" does exist for that observer. What I mean is that there is an objective meaning of universal simultaneity, which is NOT related to the proper time of a particle following a geodesic. At each instant, every observer has a momentarily comoving universal reference frame, even though, in general, there is no other particle in that reference frame in the entire universe.

I feel like you're saying "Ah, we have no way of determining a universal reference frame because everything is moving." And then, in frustration, you say "Let's just use all the moving stuff as our universal reference frame."

What I'm hearing is "even light bends in gravity, and therefore there is no such thing as a straight line, therefore, define geodesics as straight lines." To me, these things simply don't follow one another.

To me, it seems like, yes, light bends in gravity, but the vast majority of light that reaches us is coming straight. And we can use those straight lines to define what we mean by straight. I honestly don't even think I can begin to fathom how anyone could think a geodesic is a straight line. I honestly don't even think I can begin to fathom how anyone could see a "great circle" on a sphere as a straight line. To me, these ideas are very very distinct.

In regards to a(t) the scale factor which is a function of cosmological time, I do not think it is appropriate for the cosmos to have an age independent of the matter that is passing through it. I certainly think that the universe is less dense now than it was before, and that means things are moving apart. But that is not a matter of scale. That is a matter of movement.

If all of what I am saying means that I don't "believe" the Einstein Field Equations, I really don't know. I do know that these graduate texts on Riemannian Geometry are beyond me, and I ordered some undergraduate texts from the library. But I think there might be some more fundamental disagreement if the Einstein Field Equations make the assumption a priori that distant galaxies are comoving. And I'm afraid that might be the case, given that a large chunk of "Relativity, Gravitation, and World Structure" were devoted to criticizing Eddington's flawed assumptions about homogeneity, and I do, definitely agree with Milne.

PeterDonis

Nov17-10, 10:55 PM

But with gravity, we cannot merely take a clock far enough out into space and say, it measures the "real time" for that observer.

Yet, I feel that the "real time" does exist for that observer. What I mean is that there is an objective meaning of universal simultaneity, which is NOT related to the proper time of a particle following a geodesic. At each instant, every observer has a momentarily comoving universal reference frame, even though, in general, there is no other particle in that reference frame in the entire universe.

There's nothing stopping you from setting up such a reference frame "at each instant", but to have a complete description of the spacetime, you have to have one at *every* instant, and they need to match up with each other smoothly and continuously. Depending on your state of motion, the rest of the universe may look simple in such a description, or it may not. As long as your frame meets the continuity requirement above, and as long as it allows you to calculate the invariant quantities of the spacetime correctly, your frame is just as "valid" as any other. Whether or not you can assign a reasonable physical meaning to all the frame-dependent quantities in such a frame is a separate question.

I feel like you're saying "Ah, we have no way of determining a universal reference frame because everything is moving." And then, in frustration, you say "Let's just use all the moving stuff as our universal reference frame."

As *one possible* universal reference frame--the one that happens to match up well with the symmetry we imposed as a condition on the solution, the symmetry of isotropy, and which therefore makes the universe look simple to any observer that happens to be moving in a way that matches up with that symmetry. Once again, if you're in a different state of motion, there's nothing stopping you from setting up your own personal reference frame and assigning coordinates to every event in the entire spacetime in that frame; but in such a frame, the universe probably won't look as simple (it won't look isotropic, for example, because of your state of motion).

What I'm hearing is "even light bends in gravity, and therefore there is no such thing as a straight line, therefore, define geodesics as straight lines." To me, these things simply don't follow one another.

To me, it seems like, yes, light bends in gravity, but the vast majority of light that reaches us is coming straight. And we can use those straight lines to define what we mean by straight. I honestly don't even think I can begin to fathom how anyone could think a geodesic is a straight line.

But these "straight" lines that light travels on *are* geodesics! And geodesics like these *are* what we use to define what we mean by "straight". But you have to remember that this is "straight" in *spacetime*, not just in space. The paths of light rays grazing the Sun look bent in *space*, but in *spacetime* they are as straight as it's possible for a line to be. In regions of spacetime that are flat to within some accuracy of measurement, the paths of light rays will be "straight" in the sense you're thinking (the Minkowski frame sense of "straight") to that same accuracy. The "vast majority of light that reaches us" is coming from such regions of spacetime, so those paths do appear "straight" to us in the everyday sense. But they are still geodesics of spacetime.

I honestly don't even think I can begin to fathom how anyone could see a "great circle" on a sphere as a straight line. To me, these ideas are very very distinct.

You're right, they are. And the question is, which idea is the better idea for use in modeling a particular geometry that we're interested in? If the geometry you want to model is a Euclidean plane (or Euclidean 3-space), then obviously you need to use the Euclidean notion of "straight line". But if the geometry is that of a 2-sphere (not a 2-sphere embedded in Euclidean 3-space, but the intrinsic geometry of the 2-sphere itself), then you need to use great circles as your "straight lines" if you want those straight lines to meet the axioms and postulates of geometry. The only exception is the parallel postulate, of course, but the whole point of studying the intrinsic geometry of the 2-sphere in the first place is that in that geometry the parallel postulate doesn't hold--on a 2-sphere there simply are no "straight lines" in the Euclidean sense.

Similarly, in spacetime, if the geometry you are studying is the actual flat Minkowski spacetime, then of course you want to use Minkowski straight lines as your "straight lines". But if the spacetime geometry you are studying is not flat, then you need to find the right notion of "straight line" for that geometry if you want your straight lines to satisfy the appropriate axioms and postulates. If you really don't like calling such objects "straight lines", well, that's why the word "geodesic" was invented. :wink:

In regards to a(t) the scale factor which is a function of cosmological time, I do not think it is appropriate for the cosmos to have an age independent of the matter that is passing through it.

I'm not sure what you mean by this; the FRW solution to the EFE is certainly not "independent of the matter"--the actual dynamics of the scale factor a(t) depend crucially on the specific equation of state for the matter-energy that is present.

I certainly think that the universe is less dense now than it was before, and that means things are moving apart. But that is not a matter of scale. That is a matter of movement.

I would say that these are two different ways of saying the same thing--or maybe two different ways of describing the same physical reality. It's a matter of "scale" if you look at it one way, and it's a matter of "movement" if you look at it another way. The two viewpoints are not mutually exclusive; they're like looking at the same object from different vantage points.

If all of what I am saying means that I don't "believe" the Einstein Field Equations, I really don't know. I do know that these graduate texts on Riemannian Geometry are beyond me, and I ordered some undergraduate texts from the library.

I can't remember if this has been linked to before, but you might try John Baez' article on "The Meaning of Einstein's Equation" here:

http://math.ucr.edu/home/baez/einstein/

He goes into the sorts of things we've been talking about at a pretty basic level--he mentions differential geometry but only in passing, more or less. I think this page does a pretty good job of describing the basic physical content of the EFE without requiring you to dig into the heavy mathematical machinery.

You might also try the pages linked to at Baez' relativity tutorial index page here:

http://math.ucr.edu/home/baez/gr/

Also, as I mentioned before, I recommend Kip Thorne's Black Holes and Time Warps if you want a good non-technical book on relativity (as well as giving a lot of interesting historical background). Thorne also does a good job of getting across the physical content without requiring you to dig into heavy math.

But I think there might be some more fundamental disagreement if the Einstein Field Equations make the assumption a priori that distant galaxies are comoving. And I'm afraid that might be the case, given that a large chunk of "Relativity, Gravitation, and World Structure" were devoted to criticizing Eddington's flawed assumptions about homogeneity, and I do, definitely agree with Milne.

I haven't read the book you refer to, but there appears to be a free ebook download available so I'll look through it. One thing to bear in mind, though, is that the EFE itself is separate from particular solutions such as those used as cosmological models; obviously the latter will be dependent on the reasonableness of the conditions imposed (such as homogeneity and isotropy), but that doesn't invalidate the EFE itself, it just means we need to investigate other possible conditions and see how the various models, with various different conditions imposed, match up with experimental data. The FRW models are the front runners right now because they appear to do the best job at that (more precisely, the more detailed models that include perturbations about the FRW "baseline" do the best job to date).

One should also bear in mind that what look like different solutions to the EFE may actually be describing the same spacetime geometry, just in different coordinates. The Milne model, for example (on Wikipedia here: http://en.wikipedia.org/wiki/Milne_model), looks at first glance like a distinct solution that might address some of the potential issues with the FRW models, but it turns out to be a special case of the FRW models.

JDoolin

Nov18-10, 12:12 PM

I can't remember if this has been linked to before, but you might try John Baez' article on "The Meaning of Einstein's Equation" here:

http://math.ucr.edu/home/baez/einstein/

He goes into the sorts of things we've been talking about at a pretty basic level--he mentions differential geometry but only in passing, more or less. I think this page does a pretty good job of describing the basic physical content of the EFE without requiring you to dig into the heavy mathematical machinery.

This is helpful, and no, I had not seen it before. I wonder if you could verify what Baez says, that the following is a good representation of the Einstein Field Equations in Plain English

We promised to state Einstein's equation in plain English, but have not done so yet. Here it is:

Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is proportional to its volume times: the energy density at the center of the ball, plus the pressure in the x direction at that point, plus the pressure in the y direction, plus the pressure in the z direction.

I find this to be comforting, because it means I do not have to conflict with the Einstein Field Equations.

The key is that to apply the EFE, we begin by assuming we have a small ball of freely falling particles initially at rest with respect to one another. I am starting with a small ball of freely falling particles that have relative velocity with one another of v= d/t, where t is the time since the big bang. When we consider any group of particles which follow geodesics from the big bang, there can be no such "small ball" of comoving particles.

So the case I'm considering does not "conflict" with the EFE's, but it does lie outside the scope of the EFE's. In other words, I think I do "believe" the EFE's are true, but I'm pretty sure they don't apply in this case.

We have to start from scratch, considering a ball of particles with not v=0, but v=r/t.

JDoolin

Nov18-10, 12:31 PM

One should also bear in mind that what look like different solutions to the EFE may actually be describing the same spacetime geometry, just in different coordinates. The Milne model, for example (on Wikipedia here: http://en.wikipedia.org/wiki/Milne_model), looks at first glance like a distinct solution that might address some of the potential issues with the FRW models, but it turns out to be a special case of the FRW models.

Check out the discussion page for the Milne Model, because there are some things there that came from the actual book. When I tried to put actual quotes from Milne in the main article, they were removed.

Wikipedia's policy is to use secondary; not primary sources. You must go back to "Relativity, Gravitation, and World Structure" to actually get any idea of what Milne actually wrote.

PeterDonis

Nov18-10, 12:49 PM

The key is that to apply the EFE, we begin by assuming we have a small ball of freely falling particles initially at rest with respect to one another. I am starting with a small ball of freely falling particles that have relative velocity with one another of v= d/t, where t is the time since the big bang. When we consider any group of particles which follow geodesics from the big bang, there can be no such "small ball" of comoving particles.

So the case I'm considering does not "conflict" with the EFE's, but it does lie outside the scope of the EFE's. In other words, I think I do "believe" the EFE's are true, but I'm pretty sure they don't apply in this case.

We have to start from scratch, considering a ball of particles with not v=0, but v=r/t.

And if you read further to Baez' page on the Big Bang, in the same series of pages, you'll find that he covers this case, which *is* covered by the EFE.

PeterDonis

Nov18-10, 01:08 PM

Check out the discussion page for the Milne Model, because there are some things there that came from the actual book. When I tried to put actual quotes from Milne in the main article, they were removed.

I'm shocked, yes, shocked that such a thing could possibly happen on Wikipedia! :surprised

You're right, one shouldn't offer Wikipedia as an authoritative source. But see below.

You must go back to "Relativity, Gravitation, and World Structure" to actually get any idea of what Milne actually wrote.

I've downloaded the ebook and am working my way through it. From what I've read so far, I see some justice on both sides of the argument on the discussion page you referred to. However, a key point that I didn't really see brought up in that discussion is that we've learned a *lot* about both the theoretical aspects of relativity and the experimental facts about cosmology (not just the discovery of the CMBR, which is mentioned in the discussion) since Milne wrote his book. For example, as the last paragraph of the actual Wiki article notes, we have a lot of evidence now that we didn't have in the 1930's concerning how exactly the conditions of homogeneity and isotropy actually apply in the universe (i.e, pretty exactly--to one part in a few hundred thousand in the CMBR, for example). I agree with the position you took in the discussion that the Wiki article should fairly represent what Milne actually wrote at the time, not what secondary sources say; but it looks to me like Milne's model itself (apart from the more general remarks he makes about what constitutes the actual physical observations we make) is not a good fit to the data based on our current knowledge.

JDoolin

Nov18-10, 06:43 PM

And if you read further to Baez' page on the Big Bang, in the same series of pages, you'll find that he covers this case, which *is* covered by the EFE.

Are you talking about this page?

http://math.ucr.edu/home/baez/einstein/node7.html

The case I'm describing is definitely not there. What do you think you see that sounds like it has anything to do with what I am talking about?

I said that the particles were moving apart and d=v*t; or likewise, v=d/t. No comoving particles ANYWHERE.

JDoolin

Nov18-10, 06:54 PM

I'm shocked, yes, shocked that such a thing could possibly happen on Wikipedia! :surprised

You're right, one shouldn't offer Wikipedia as an authoritative source. But see below.

I've downloaded the ebook and am working my way through it. From what I've read so far, I see some justice on both sides of the argument on the discussion page you referred to. However, a key point that I didn't really see brought up in that discussion is that we've learned a *lot* about both the theoretical aspects of relativity and the experimental facts about cosmology (not just the discovery of the CMBR, which is mentioned in the discussion) since Milne wrote his book. For example, as the last paragraph of the actual Wiki article notes, we have a lot of evidence now that we didn't have in the 1930's concerning how exactly the conditions of homogeneity and isotropy actually apply in the universe (i.e, pretty exactly--to one part in a few hundred thousand in the CMBR, for example). I agree with the position you took in the discussion that the Wiki article should fairly represent what Milne actually wrote at the time, not what secondary sources say; but it looks to me like Milne's model itself (apart from the more general remarks he makes about what constitutes the actual physical observations we make) is not a good fit to the data based on our current knowledge.

Pardon me, but does the current model really "FIT" that well? We have no real explanation for inflation. We have no dark energy. We have no dark matter. We have a theory that is inconsistent with quantum mechanics. But we have an equation that matches up really well.

The thing is, I can show you a conformal mapping between the Milne model and the Robertson Walker Metric, which preserves the speed of light and has exactly the same events, if you're interested. All the same variables are there. All the same events are there, except the singularity is transformed into a lightcone in Milne's version, and it's turned into a plane in the "comoving matter" version.

The only difference is that in Milne's version, the transformation makes sense, because you're converting from proper time into coordinate time, whereas in the "comoving matter" version, your converting from proper time to meaningless arbitrary cosmological time coordinates, chosen arbitrarily to make it "look like" all the particles are comoving.

JDoolin

Nov18-10, 07:28 PM

The thing is, I can show you a conformal mapping between the Milne model and the Robertson Walker Metric, which preserves the speed of light and has exactly the same events, if you're interested. All the same variables are there. All the same events are there, except the singularity is transformed into a lightcone in Milne's version, and it's turned into a plane in the "comoving matter" version.

I have all the images I need, so I will go ahead and do it.

This image is correct as a mapping of proper time vs. distance, or more meaningfully, proper time vs. rapidity.

http://www.physicsforums.com/attachment.php?attachmentid=29832&d=1289686770

What goal do you have, then? Do you want to map it into a coordinate system where all of the worldlines are vertical? Then it should look like this:

http://www.physicsforums.com/attachment.php?attachmentid=29831&d=1289686770

As for why anyone should want to do such a thing, that is a matter of preference and opinion, only! It was an a priori decision made by Einstein which has only been confirmed by circular reasoning.

On the other hand, would you like to map it into space vs. coordinate time?

Then it should look like the lower diagram below:

http://www.wiu.edu/users/jdd109/stuff/img/milnemetric.jpg

This diagram has been on my blog since September. It represents two different conformal mappings of the Robertson-Walker Diagram.

All the difference is in how you parameterize your variables.

(*Correct Variables:
r=rapidity, relative to central particle;
t=proper time of particles on unaccelerated paths from big bang.;
(both are "invariant" properties of matter.)

Incorrect variables:
r=distance in "real universe coordinates";
t=time in "real universe coordinates";
(both are contravariant properties of space.)
*)

t=1;
e0 = Table[{r, 0}, {r, -10, 10}];
e1 = Table[{r, t}, {r, -10, 10}];
comovingWorldLines = Transpose[{e0, e1}];
ListLinePlot[comovingWorldLines]
e0 = Table[{0 Sinh[r], 0 Cosh[r]}, {r, -1.5, 1.5, .1}];
e1 = Table[{t Sinh[r], t Cosh[r]}, {r, -1.5, 1.5, .1}];
milneWorldLines = Transpose[{e0, e1}];

JDoolin

Nov18-10, 08:30 PM

I didn't have the light-cone in the earlier diagram.

I put it in, and realized there is also a subtle mistake in the "comoving particle" conformal mapping that doesn't happen in this milne mappping.

In the Friedmann-Walker diagram, the light "from the big bang" crosses every single worldline. But in the "comoving particles" diagram, the light just passes a finite number of worldlines.

In the milne diagram, you have that subtle error fixed, and the light "from the big bang" crosses every worldline, just as it is in the Friedmann Walker diagram.

PeterDonis

Nov18-10, 08:48 PM

The case I'm describing is definitely not there. What do you think you see that sounds like it has anything to do with what I am talking about?

I said that the particles were moving apart and d=v*t; or likewise, v=d/t. No comoving particles ANYWHERE.

The page I linked to describes a ball B which is "expanding at t = 0". That ball corresponds to your particles moving apart with v = d/t. Baez says that we can't directly apply his equation (2) (which is his statement of the EFE) to ball B because its particles aren't at rest relative to each other at time t = 0. He then defines a second ball, B', which *does* have all its particles at rest relative to each other at time t = 0, and which has the same radius and the same acceleration as ball B at time t = 0. This allows him to apply his equation (2) to ball B', and then show that the equation he derives for the radius r of ball B' vs. time, *also* holds for the radius R of ball B vs. time. So he's showing that the EFE *does* apply to the type of ball you defined, where the particles are moving apart at time t = 0.

PeterDonis

Nov18-10, 08:59 PM

Pardon me, but does the current model really "FIT" that well? We have no real explanation for inflation. We have no dark energy. We have no dark matter. We have a theory that is inconsistent with quantum mechanics. But we have an equation that matches up really well.

And the Milne model, as far as I can see, suffers from the same problems. See below.

The thing is, I can show you a conformal mapping between the Milne model and the Robertson Walker Metric, which preserves the speed of light and has exactly the same events, if you're interested. All the same variables are there. All the same events are there, except the singularity is transformed into a lightcone in Milne's version, and it's turned into a plane in the "comoving matter" version.

I already know about this; it's what I was referring to before when I said that the Milne model is describing the same spacetime geometry as the FRW models, just in different coordinates. See next comment.

The only difference is that in Milne's version, the transformation makes sense, because you're converting from proper time into coordinate time, whereas in the "comoving matter" version, your converting from proper time to meaningless arbitrary cosmological time coordinates, chosen arbitrarily to make it "look like" all the particles are comoving.

What you are saying amounts to this: you like the coordinate system in Milne's model better than you like the FRW coordinate system. That's fine; particular coordinate systems don't matter. What matters is the spacetime geometry. That's the same either way. It's like saying that you prefer to use a Mercator projection rather than a stereographic projection to map the surface of the Earth.

You apparently don't believe this; you appear to think that the Milne model and the FRW models are describing fundamentally different objects. From what I've read so far, I would have to disagree; it looks to me, so far, like what I said above is valid--both models are describing the same geometry, just in different coordinates. The diagrams you posted give me no reason to change that conclusion; for example, your statement that "All the differences is in how you parametrize your variables" indicates to me that what you're illustrating are simple coordinate transformations that don't change the geometry, and hence, don't change the physics.

PeterDonis

Nov18-10, 09:13 PM

In the Friedmann-Walker diagram, the light "from the big bang" crosses every single worldline. But in the "comoving particles" diagram, the light just passes a finite number of worldlines.

In the milne diagram, you have that subtle error fixed, and the light "from the big bang" crosses every worldline, just as it is in the Friedmann Walker diagram.

I'm not sure that what you're saying about the Friedmann-Walker diagram is correct. Physically, in a spatially infinite universe (which is what's required for an "infinite number of worldlines" in the sense you're using the term), I would *not* expect light reaching us now from the Big Bang event to have crossed *all* of that infinite number of worldlines, because only a finite amount of proper time has elapsed, and light can only cover a finite distance in a finite time--what you're suggesting would require the light to cover an infinite distance in a finite time. The "comoving particles" diagram (a more standard name for it would be a "conformal" diagram, as it is called on the Ned Wright page I linked to earlier) has the advantage that it makes the reasoning I just gave obvious.

As far as I can tell, your "Milne" diagram corresponds to the diagram on the Ned Wright page here...

http://www.astro.ucla.edu/~wright/cosmo_02.htm

...in what Wright calls "special relativistic" coordinates. This means to me that you have raised an interesting question about the light cones, since it certainly appears in your "Milne" diagram and Wright's "special relativistic" diagram that the light from the "Big Bang" *does* cross all the "infinite number" of worldlines. I'll have to think about this one some more.

George Jones

Nov18-10, 09:20 PM

JDoolin

Nov19-10, 07:57 AM

I'm not sure that what you're saying about the Friedmann-Walker diagram is correct. Physically, in a spatially infinite universe (which is what's required for an "infinite number of worldlines" in the sense you're using the term), I would *not* expect light reaching us now from the Big Bang event to have crossed *all* of that infinite number of worldlines, because only a finite amount of proper time has elapsed, and light can only cover a finite distance in a finite time--what you're suggesting would require the light to cover an infinite distance in a finite time. The "comoving particles" diagram (a more standard name for it would be a "conformal" diagram, as it is called on the Ned Wright page I linked to earlier) has the advantage that it makes the reasoning I just gave obvious.

As far as I can tell, your "Milne" diagram corresponds to the diagram on the Ned Wright page here...

http://www.astro.ucla.edu/~wright/cosmo_02.htm

...in what Wright calls "special relativistic" coordinates. This means to me that you have raised an interesting question about the light cones, since it certainly appears in your "Milne" diagram and Wright's "special relativistic" diagram that the light from the "Big Bang" *does* cross all the "infinite number" of worldlines. I'll have to think about this one some more.

At least on this page, I can see that I'm talking about the same thing as Ned Wright. He correctly notes that the observable universe is the entire universe. He also correctly notes that our past lightcone passes the most distant galaxies at x=c t_0/2

(He does not mention that the proper age of those asymptotically most distant galaxies is zero, so that past light-cone does cross the singularity. I imagine the era of hydrogen recombination to be a big hyperbolic arc in the diagram, and that represents the furthest distance we can actually see.)

Technically, the Friedmann Walker Diagram is ambiguous at the singularity. The question is what is the slope of the past light-cone at the point (0,0). If that slope is zero, then the light should cross all of the past light-cones. If the slope is nonzero, then it must cross only a finite number of light-cones.

(Sorry--in Wright's diagram (at http://www.astro.ucla.edu/~wright/cosmo_02.htm ), I looked again, and the speed of light, as drawn, is definitely a non-zero slope at the singularity. That means that it is not a conformal mapping with the image below it, because it does not cross every worldline.) As for me, I think that the speed of light at that singularity point should be faster than the speed of light for any particle at that point. So I think the line should be horizontal there.

JDoolin

Nov19-10, 08:18 AM

JDoolin

Nov19-10, 08:28 AM

The Milne universe is an FRW universe that is empty of matter and energy. See

http://www.physicsforums.com/showthread.php?p=1757634&highlight=milne#post1757634

http://www.physicsforums.com/showthread.php?p=2872975#post2872975.

You are linking to links of you asserting that this is true, but as I told PeterDonis, this is one occasion, where you should really go back to the original text, because the secondary sources do not justly represent Milne's ideas.

PeterDonis

Nov19-10, 09:05 AM

We're not suggesting that the light has to cover an infinite distance, but that an infinite number of particles lie within a finite distance.

A finite *coordinate* distance, yes. But have you integrated the metric over that finite coordinate distance to confirm that it covers a finite *proper* distance? The fact that the model indicates density going to infinity as you approach that finite coordinate distance is a big red flag to me that there's actually an infinite proper distance (meaning the actual physical density remains finite), and that the apparent infinite density is an artifact of the peculiar coordinate system you are using. Actual infinite physical density is not physically reasonable.

JDoolin

Nov19-10, 09:26 AM

A finite *coordinate* distance, yes. But have you integrated the metric over that finite coordinate distance to confirm that it covers a finite *proper* distance? The fact that the model indicates density going to infinity as you approach that finite coordinate distance is a big red flag to me that there's actually an infinite proper distance (meaning the actual physical density remains finite), and that the apparent infinite density is an artifact of the peculiar coordinate system you are using. Actual infinite physical density is not physically reasonable.

Infinite physical density is physically reasonable, I think. For an infinite physical density, what is required is that several particles occupy the same place at the same time.

Certainly, you can't have any two particles sharing the same quantum numbers. But these particles have different momenta, and hence different quantum numbers. This situation can only exist for an instant.

However, since the proper time experienced by a particle traveling asymptotically approaching the speed of light is zero, that "instant" is effectively, the whole lightcone; it lasts forever in the frame of the central observer. Infinite physical density for an instant (a single event) is physically reasonable, and due to time dilation, that instant is effectively forever.

As for integrating the proper time, I hope you'll forgive me. As I mentioned before, the Einstein Field Equations need to be done over from scratch assuming this case where particles are diverging as v=d/t. So, no, I have not completed this task. That, I think, is something for better minds than mine. I just want it to be realized that the Einstein Field Equations were derived under the assumption of a finite number of comoving particles, and it does not cover the case of an infinite number of non-comoving particles. Then perhaps someone; (maybe you!) could re-do the EFE's from scratch, develop the theory and see what comes out of it.

PeterDonis

Nov19-10, 11:21 AM

Infinite physical density is physically reasonable, I think. For an infinite physical density, what is required is that several particles occupy the same place at the same time.

To be precise, what's required is that an *infinite* number of particles occupy the same place at the same time. But I think you realize that--see below.

Certainly, you can't have any two particles sharing the same quantum numbers. But these particles have different momenta, and hence different quantum numbers. This situation can only exist for an instant.

However, since the proper time experienced by a particle traveling asymptotically approaching the speed of light is zero, that "instant" is effectively, the whole lightcone; it lasts forever in the frame of the central observer. Infinite physical density for an instant (a single event) is physically reasonable, and due to time dilation, that instant is effectively forever.

I can't be sure here whether or not you're committing a common confusion in relativity. An instant (event) is a single point. A null worldline--the path of a light ray--has "zero lapse of proper time", but it is *not* an instant--it is *not* a single point. It consists of a series of distinct events. Your language in the quote above is ambiguous; you need to be clear and explicit about whether the infinite density in your model occurs at a single point, or along a whole null worldline.

I suspect that what you mean is the former (the infinite density only occurs at a single point), in which case I would strongly discourage using the language "due to time dilation, that instant is effectively forever", since that language promotes exactly the confusion that I described. The instant of infinite density in your model, I suspect, is supposed to be a single point; it's only the particular coordinates you've chosen that distort that single point into what appears to be a null cone. Physically, it's a single point.

However, saying that the infinite density is only for a single point does *not* make it physically reasonable. It only means that your model "predicts its own downfall", just as GR does for spacetimes in which a singularity occurs. Your model is telling you that it can't cover that point--you need some new physics to tell you what happens there. The FRW models in GR handle that, as has been mentioned before, by *not* claiming to cover events all the way to the initial singularity; they only cover things back to (roughly) the beginning of the inflationary phase, where the universe was very small, hot, and dense, but not infinitely so. In your model, that would correspond (as you note) to drawing a hyperbola very *close* to the "null cone" boundary, corresponding to the (very small but not zero) proper time of the beginning of inflation (let's say), and saying that that's the actual physical limit of what your model covers; before that, new physics is needed to say what happened and how the initial state in your model came to be, just as with the FRW models.

As for integrating the proper time, I hope you'll forgive me. As I mentioned before, the Einstein Field Equations need to be done over from scratch assuming this case where particles are diverging as v=d/t.

No, the EFE doesn't need to be reworked, as I noted in a previous post. (Small nitpick: the usual term is "Einstein Field Equation", singular, even though it does have multiple tensor components. The tensors on each side of the EFE are considered single geometric objects, just as we normally write vector equations as relating single objects, even though vectors have multiple components.)

But that's actually not required anyway for what I was asking for. I said "integrate the metric", and you already have an expression for the metric; how that metric arises from the EFE is a separate issue. Integrate that expression over the appropriate range of coordinate values, and see what you come up with.

JDoolin

Nov19-10, 12:59 PM

To be precise, what's required is that an *infinite* number of particles occupy the same place at the same time. But I think you realize that--see below.

I can't be sure here whether or not you're committing a common confusion in relativity. An instant (event) is a single point. A null worldline--the path of a light ray--has "zero lapse of proper time", but it is *not* an instant--it is *not* a single point. It consists of a series of distinct events. Your language in the quote above is ambiguous; you need to be clear and explicit about whether the infinite density in your model occurs at a single point, or along a whole null worldline.

See if I can make it a little less ambiguous without going into the math right now. If I take a "null world line" that comes straight from the big bang, then that world-line has a proper time of zero. You go out to that distance, and take away one meter, you're going to be in a region of incredibly high density. You go to 1 millimeter away from the null world-line, and you'll get many orders of magnitude higher of density. You go 1 micrometer, it will be may orders higher density still; you go to 1 nanometer, many more orders of magnitude. I could go on forever, but I hope you get the picture. The density is just a function that asymptotically approaches infinity as you approach the light ray. You can't say precisely what the density is AT the singularity.

I suspect that what you mean is the former (the infinite density only occurs at a single point), in which case I would strongly discourage using the language "due to time dilation, that instant is effectively forever", since that language promotes exactly the confusion that I described. The instant of infinite density in your model, I suspect, is supposed to be a single point; it's only the particular coordinates you've chosen that distort that single point into what appears to be a null cone. Physically, it's a single point.

I meant both. Why is it appropriate to map the singularity to a straight line, as is done in the "comoving" conformal mapping, but it is not appropriate to map the singularity to a light cone? The infinite density occurs at a single point in the Friedmann Walker Diagram (one single event), but in both the standard model, and in the Milne model, that single event is stretched out to infinity.

However, saying that the infinite density is only for a single point does *not* make it physically reasonable. It only means that your model "predicts its own downfall", just as GR does for spacetimes in which a singularity occurs. Your model is telling you that it can't cover that point--you need some new physics to tell you what happens there. The FRW models in GR handle that, as has been mentioned before, by *not* claiming to cover events all the way to the initial singularity; they only cover things back to (roughly) the beginning of the inflationary phase, where the universe was very small, hot, and dense, but not infinitely so. In your model, that would correspond (as you note) to drawing a hyperbola very *close* to the "null cone" boundary, corresponding to the (very small but not zero) proper time of the beginning of inflation (let's say), and saying that that's the actual physical limit of what your model covers; before that, new physics is needed to say what happened and how the initial state in your model came to be, just as with the FRW models.

I'm not sure if you're being entirely fair. The model I'm describing leaves out a singularity. The model you're describing leaves out a singularity plus some additional time including the inflationary stage.

You want me to stop at some finite time after the big bang. But I don't want to stop there. You name a time after the big bang, and I will name an earlier time after the big bang. No matter what time I choose, the universe has a density at that time. If you choose ZERO, then, I'm stuck. You're right. That is outside the scope of the theory. But it is also outside the scope of General Relativity, so what is the difference?

No, the EFE doesn't need to be reworked, as I noted in a previous post. (Small nitpick: the usual term is "Einstein Field Equation", singular, even though it does have multiple tensor components. The tensors on each side of the EFE are considered single geometric objects, just as we normally write vector equations as relating single objects, even though vectors have multiple components.)

But that's actually not required anyway for what I was asking for. I said "integrate the metric", and you already have an expression for the metric; how that metric arises from the EFE is a separate issue. Integrate that expression over the appropriate range of coordinate values, and see what you come up with.

You asserted that the EFE doesn't need to be reworked. I asserted that the EFE does need to be reworked, and backed up that statement, by pointing out that the EFE's assume a finite number of comoving particles. You asserted that the case I'm talking about has been tried, but I told you I couldn't find it anywhere in whatever you linked to. (By the way, I've been, off and on, looking since 2001 to try to find one "respectable" text on General Relativity that considered this case, and asking for it, and the only place I have found it is in Milne. So it is not a surprise to me if you can't find it on the Baez site.)

So, actually I don't have the metric. I would guess that the metric is {{-1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}} because of symmetry. And yes, it's quite easy to show that the proper distance in this case is finite.

PeterDonis

Nov19-10, 01:36 PM

I meant both. Why is it appropriate to map the singularity to a straight line, as is done in the "comoving" conformal mapping, but it is not appropriate to map the singularity to a light cone? The infinite density occurs at a single point in the Friedmann Walker Diagram (one single event), but in both the standard model, and in the Milne model, that single event is stretched out to infinity.

In a Mercator projection of the Earth's surface, the North and South Poles are mapped to infinitely long lines. But physically, they're not lines; they're points. The "comoving" conformal mapping does something similar to the initial singularity: it takes what is physically a single point and maps it into an infinitely long line. As I understand it, your model does something similar, except that instead of taking what is physically a single point (like the Earth's North Pole) and mapping it into a line, it maps it into a null cone. I'm just trying to understand if you agree that that's what your model is doing, or if you intend it to be doing something else.

You want me to stop at some finite time after the big bang. But I don't want to stop there. You name a time after the big bang, and I will name an earlier time after the big bang. No matter what time I choose, the universe has a density at that time. If you choose ZERO, then, I'm stuck. You're right. That is outside the scope of the theory. But it is also outside the scope of General Relativity, so what is the difference?

Again, I'm just trying to get clear about whether or not your model actually claims that infinite density is physically reasonable. You were saying earlier that it was; now your comments here indicate that it is not--that your model can get arbitrarily close to the infinite density point (but at any such arbitrarily close point, the density will still be finite), but does not actually cover the infinite density point. As long as the latter is the case, I have no issue. (How close you actually want to get to the initial singularity in your model based on the data you have is a separate issue--the FRW models *can* be extended back arbitrarily close to the singularity, just as yours can; the reason they usually aren't is only because we estimate that the start of the inflation phase is where the new physics we need will actually end up coming in. We won't know for sure until we know what that new physics is.)

You asserted that the case I'm talking about has been tried, but I told you I couldn't find it anywhere in whatever you linked to.

Read my post #103 again. I described there how the page I linked to, in which you claimed you couldn't find anything relevant, *is* relevant. If you can't respond to the specifics I gave there, other than just saying "I can't find anything that applies", I don't think we can get any further on this particular point.

So, actually I don't have the metric. I would guess that the metric is {{-1,0,0,0},{0,1,0,0},{0,0,1,0},{0,0,0,1}} because of symmetry. And yes, it's quite easy to show that the proper distance in this case is finite.

The metric you just gave is a representation in matrix form of the line element:

ds^{2} = - dt^{2} + dx^{2} + dy^{2} + dz^{2}

This is the standard Minkowski metric, which covers a range of minus infinity to plus infinity for all four coordinates (t, x, y, z). That doesn't correspond to the diagram you drew, which only covers a limited range of coordinate values; so the above metric can't be the one that applies to your diagram.

JDoolin

Nov20-10, 09:21 AM

Regarding Density:

You're asking whether infinite density is physically reasonable. Well, density is defined as a number of particles per unit area. Even though the local density at any point is finite by my description, If you take the average density of the universe,

\frac{\int \rho dV}{V}

at any given time, yes, the top will be infinite, and the bottom will be finite. Yes, an infinite density of the universe is reasonable. The average density of the universe is always infinite, because you always have an infinite number of particles within a finite distance.

So there are different meanings of an infinite density. You can have

(1) an infinite number of particles in a finite volume, (All the time in the Milne model)

or (2) a finite number of particles in a zero volume (never happens in the Milne model),

or (3) an infinite number of particles in a zero volume (approached asymptotically at r=c*t)

Regarding Mapping the Singularity to a Straight Line vs a Null Light Cone.

I erred, I think, which is leading to some confusion.

We have (1 FWD) the Friedmann Walker Diagram, (2 CPD) the comoving particle conformal mapping Diagram, and (3 MMD) the Milne/Minkowski conformal map.

Earlier I said http://www.astro.ucla.edu/~wright/cosmo_02.htm[/url] ), I looked again, and the speed of light, as drawn, is definitely a non-zero slope at the singularity. That means that it is not a conformal mapping with the image below it, because it does not cross every worldline.) As for me, I think that the speed of light at that singularity point should be faster than the speed of light for any particle at that point. So I think the line should be horizontal there.
"

My error was in thinking that ONCE that line was horizontal at the origin, that it can turn. In fact, the mapping is not from the singularity to the light cone, but from this horizontal line to the lightcone.

You might well ask why this light does not bend back toward the infinite density behind it, which is a very good question, so my answer is going to seem glib. That lightcone surface is not causally connected to anything within the lightcone. If I send a beam of light to chase after a beam of light, it will never reach it.

You can receive light from any matter that is asymptotically close to the null light-cone, but the light-cone, itself is not "covered" by the model.
In other words, there is no straight line speed-of-light path from the big bang to any event inside the lightcone. Only events immediately following the big-bang can ever be observed within the lightcone.

I think this is actually a pretty reasonable conclusion, though it is contrary to the standard model. We shouldn't be able to see the Big Bang itself, since it already happened in our reference frame. But we can see events immediately after the Big Bang.

Regarding whether Baez considered this topic:

[QUOTE=PeterDonis;2993062]The page I linked to describes a ball B which is "expanding at t = 0". That ball corresponds to your particles moving apart with v = d/t. Baez says that we can't directly apply his equation (2) (which is his statement of the EFE) to ball B because its particles aren't at rest relative to each other at time t = 0. He then defines a second ball, B', which *does* have all its particles at rest relative to each other at time t = 0, and which has the same radius and the same acceleration as ball B at time t = 0. This allows him to apply his equation (2) to ball B', and then show that the equation he derives for the radius r of ball B' vs. time, *also* holds for the radius R of ball B vs. time. So he's showing that the EFE *does* apply to the type of ball you defined, where the particles are moving apart at time t = 0.

I'm sorry. I still don't see it on the page where I think you were saying it was (http://math.ucr.edu/home/baez/einstein/node7.html). Maybe there's something in the equations that I'm missing. I kind of skimmed over them, looking for something in the text. If you see it is there in the math, or I am looking on the wrong page, I need more information.

PeterDonis

Nov20-10, 10:48 AM

Well, density is defined as a number of particles per unit area. Even though the local density at any point is finite by my description, If you take the average density of the universe,

\frac{\int \rho dV}{V}

at any given time, yes, the top will be infinite, and the bottom will be finite.

I assume you mean "number of particles per unit volume" (not "area"), and that V is a spatial volume. How is the corresponding volume element dV related to a product of coordinate differentials (dx, dy, dz, or whatever spatial coordinates you are using) in your diagram? This question is related to my question about coordinate distance vs. proper distance and what metric you are using (since the metric and the volume element are related). Remember my earlier comment that the metric that applies to your diagram can't be the ordinary Minkowski metric, since whatever coordinates the diagram is drawn in do not have infinite ranges.

My error was in thinking that ONCE that line was horizontal at the origin, that it can turn. In fact, the mapping is not from the singularity to the light cone, but from this horizontal line to the lightcone.

This sounds basically the same as what I said, just with an extra mapping step, so to speak: the mapping from FWD to CPD maps a point (the initial singularity) into a horizontal line, and then you do a mapping from CPD to MMD that maps the horizontal line into a null cone.

Regarding whether Baez considered this topic:

Read the first two sentences from my post (that you quoted) again: The page I linked to describes a ball B which is "expanding at t = 0". That ball corresponds to your particles moving apart with v = d/t. Do you see how ball B here covers the case you were considering?

Now read these sentences again: He then defines a second ball, B', which *does* have all its particles at rest relative to each other at time t = 0, and which has the same radius and the same acceleration as ball B at time t = 0. This allows him to apply his equation (2) to ball B', and then show that the equation he derives for the radius r of ball B' vs. time, *also* holds for the radius R of ball B vs. time. Do you see how this shows that the EFE covers the case you were considering?

JDoolin

Nov21-10, 10:43 AM

I assume you mean "number of particles per unit volume" (not "area"), and that V is a spatial volume. How is the corresponding volume element dV related to a product of coordinate differentials (dx, dy, dz, or whatever spatial coordinates you are using) in your diagram? This question is related to my question about coordinate distance vs. proper distance and what metric you are using (since the metric and the volume element are related). Remember my earlier comment that the metric that applies to your diagram can't be the ordinary Minkowski metric, since whatever coordinates the diagram is drawn in do not have infinite ranges.

From http://casa.colorado.edu/~ajsh/sr/wheel.html we have the lovely animation:

http://casa.colorado.edu/~ajsh/sr/lorwheel.gif

Andrew Hamilton has a nice diagram of the effect of Lorentz Transformation around the event at the origin.

This diagram is pure Minkowski diagram.

What you see in this diagram is events from the top light-cone being moved to other places in the top lightcone. What you never see is an event passing down from the future lightcone past the null light cone.

The transformation keeps on going and going and going forever, with each of an infinite number of worldlines blissfully thinking that it is at the center of that lightcone.

Now, there may or may not be anything outside that lightcone but if the model is correct, then it does not matter, because in order to get intothe lightcone from outside the lightcone, one would need to push past an infinite number of particles.

Still, this is the "ordinary Minkowski Metric." It's just that the entire observable universe lies within a single lightcone of the Minkowski Spacetime. The universe that we know only occupies a fraction of the Minkowski spacetime.

This sounds basically the same as what I said, just with an extra mapping step, so to speak: the mapping from FWD to CPD maps a point (the initial singularity) into a horizontal line, and then you do a mapping from CPD to MMD that maps the horizontal line into a null cone.

But there are differences in how that initial singularity is mapped.
Namely: The CPD maps the initial singularity (0,0) to a horizontal line, while the MMD maps the horizontal line through the singularity (0,0) to a light-cone. That represents the surface of \tau^2=t^2-r^2=0 The whole lightcone represents a singularity in the \tau variable. In the MMD representation, the (0,0) point remains a single event.

If I'm not mistaken, the CPD simply does not recognize the horizontal line in the FWD. But how would the horizontal line be mapped, if it were? Would it become a vertical line "at infinity?"

Read the first two sentences from my post (that you quoted) again: The page I linked to describes a ball B which is "expanding at t = 0". That ball corresponds to your particles moving apart with v = d/t. Do you see how ball B here covers the case you were considering?

Now read these sentences again: He then defines a second ball, B', which *does* have all its particles at rest relative to each other at time t = 0, and which has the same radius and the same acceleration as ball B at time t = 0. This allows him to apply his equation (2) to ball B', and then show that the equation he derives for the radius r of ball B' vs. time, *also* holds for the radius R of ball B vs. time. Do you see how this shows that the EFE covers the case you were considering?

Thank you for clarifying what you saw.

Suppose that, at some time t=0, she identifies a small ball B of test particles centered on her. Suppose this ball expands with the universe, remaining spherical as time passes because the universe is isotropic.

Okay, I see this now, but can you understand why I didn't recognize it as the same case? This example says the ball "expands with the universe." When I read that, I interpreted it to mean "It is a ball which stretches as the universe stretches." I see now that Baez actually meant to talk about something else, but I can't see how you get v = d/t out of this. He literally says later there is "nothing special" about time, t=0. If you have v=d/t, then you should have infinite density at t=0.

Let R(t) stand for the radius of this ball as a function of time. The Einstein equation will give us an equation of motion for R(t). In other words, it will say how the expansion rate of the universe changes with time.

It is tempting to apply equation (2) to the ball , but we must take care.

See, he's already attempting to apply the EFE before he even gives the simple answer. He never says, R(t)= v_{max} t + R_0, so I would say he is not really considering the idea at all; not in any clear mathematical sense.

This equation applies to a ball of particles that are initially at rest relative to one another -- that is, one whose radius is not changing at . However, the ball B is expanding at t=0. Thus, to apply our formulation of Einstein's equation, we must introduce a second small ball of test particles that are at rest relative to each other at t=0.

What has he done? He has said we can't simply apply the Einstein Field Equations. We have to define our coordinate system, THEN apply the Einstein Field Equations.

To me, nothing seems to have changed. He is still using the EFE's, but he gave a tiny bit of lip-service to the idea of not using them. He sure didn't start over from scratch, assuming d=v/t.

PeterDonis

Nov21-10, 12:55 PM

Andrew Hamilton has a nice diagram of the effect of Lorentz Transformation around the event at the origin.

Yes, I understand how all this works. (I find all of the diagrams on Hamilton's pages to be very helpful, btw.) But the description you give based on this diagram...

It's just that the entire observable universe lies within a single lightcone of the Minkowski Spacetime. The universe that we know only occupies a fraction of the Minkowski spacetime.

...is *not* compatible with the description you give a little later concerning how the initial singularity is mapped:

Namely: The CPD maps the initial singularity (0,0) to a horizontal line, while the MMD maps the horizontal line through the singularity (0,0) to a light-cone. That represents the surface of \tau^2=t^2-r^2=0 The whole lightcone represents a singularity in the \tau variable. In the MMD representation, the (0,0) point remains a single event.

If the observable universe is only a fraction of Minkowski spacetime, as you say above, then when you draw a diagram showing a light cone emerging from the initial singularity (the "big bang" event), the light cone is not a "mapping" of the big bang event, or anything else; it's just showing the limiting case of light rays going outward from the big bang event in opposite directions. In other words, the light cone is the boundary of the portion of Minkowski spacetime that is causally connected to the big bang event, but that certainly doesn't mean the light cone "is" the big bang event, or is a "mapping" of the big bang event into some other set of coordinates. You're using Minkowski coordinates throughout, and you're saying that we, now, on the Earth, lie on some timelike worldline in a Minkowski spacetime that, in the far past, met up with all the other timelike worldlines in the "universe" we observe, at an event called the "big bang". All those worldlines must lie within the future light cone of that big bang event, so you've just drawn in the future light cone to mark the spacetime boundary of "our universe", as opposed to the rest of the infinite Minkowski spacetime.

It is true that all points on the light cone have zero spacetime interval from the big bang event, but as I said in an earlier post, that does *not* make the entire light cone a single event! There are other ways of assigning coordinates in Minkowski spacetime that make it clear that there are distinct events on light cones. This is in contrast to the "mapping" that is done in, for example, the "CPD"--see next comment.

Also, in your model, there is nothing preventing other "big bang-like" events from happening *outside* the "big bang" light cone (after all, there is an infinite Minkowski spacetime for events to happen in), and sending signals (light rays or timelike worldlines) *into* the "big bang" light cone that represents our universe. As far as I can tell, your model doesn't account for that at all.

If I'm not mistaken, the CPD simply does not recognize the horizontal line in the FWD. But how would the horizontal line be mapped, if it were? Would it become a vertical line "at infinity?"

There actually isn't a "horizontal line" at the bottom of the FWD (I assume that's what you were asking about). There are lines going left and right that approach the horizontal as close as you like, but there are none that are exactly horizontal. In the CPD, the lines approaching the horizontal are mapped to vertical lines on the left and right sides, further and further out from the center. In the "special relativistic" diagram on Wright's page, those lines are mapped to "timelike" lines that get closer and closer to the bounding light cone, without ever reaching it.

The point about all three of these mappings is that the "singularities"--the single point at the "bottom" of the FWD, the line at the bottom of the CPD, and the bounding light cone of the "SR" diagram--are true *boundaries* of the entire spacetime: there is *no* spacetime at all outside of these boundaries. These diagrams are not embedded in any larger diagrams (for example, there is no "larger" Minkowski spacetime in which the "SR" diagram is embedded); they show the *entire* spacetime, *everything* that physically exists in the model.

Okay, I see this now, but can you understand why I didn't recognize it as the same case? This example says the ball "expands with the universe." When I read that, I interpreted it to mean "It is a ball which stretches as the universe stretches." I see now that Baez actually meant to talk about something else, but I can't see how you get v = d/t out of this. He literally says later there is "nothing special" about time, t=0. If you have v=d/t, then you should have infinite density at t=0.

First, a small clarification: by "t = 0" Baez did not mean the instant of the big bang, but just some arbitrarily defined "origin" of a time coordinate. He says "at some time, t = 0", not "at the time of the big bang, t = 0".

That said, you are correct that, at the actual "t = 0" of the big bang, the density (and hence the spacetime curvature) is infinite, which is why, as I've said before, General Relativity "predicts its own downfall" whenever a spacetime singularity--a point of infinite density, infinite curvature, etc.--occurs. The EFE becomes mathematically singular at those points, so we can't use it to predict what happens. But we can get as close to the singularity as we like, and everything will still be finite, so the EFE, mathematically, works just fine. (Whether it still gives predictions that are accurate physically is a separate question--we don't know at this point because we can't make any observations of regimes where the density, curvature, etc. gets large enough to be comparable to what the EFE says it would have to have been, for example, before the inflation phase started.)

See, he's already attempting to apply the EFE before he even gives the simple answer. He never says, R(t)= v_{max} t + R_0, so I would say he is not really considering the idea at all; not in any clear mathematical sense.

Yes, he is; you're missing the point of his argument. He is saying that (provided we are not at a spacetime singularity, as I noted above) *the initial velocities of the particles in the ball don't matter*, because the EFE relates the ratio of *acceleration* and *radius* of the ball to the density and pressure at the center--velocity doesn't enter into it at all. So we can apply the EFE to a ball with *any* distribution of initial velocities, using the method he describes.

JDoolin

Nov22-10, 03:52 PM

PeterDonis

Nov22-10, 05:09 PM

When I said "A singularity in the tau variable," I think you over-interpreted my meaning. I meant that the value of tau is zero everywhere on the light-cone, but I did not mean that having the same tau represented the same event. \tau^2=t^2-r^2=0 means every event where t=r. That includes more than the big bang event.

Ok, good, that makes it clear what the lightcone in your model means--except that this...

The light cone is the set of events for whom the space-time-interval between them and the big bang is zero. It IS a mapping: the HORIZONTAL PLANE THROUGH (0,0) in the FWD is mapped to the LIGHT CONE in the MMD. It doesn't map an EVENT to the lightcone, it maps the WHOLE PLANE to the lightcone.

...is *not* correct if what you said above is true. There is no "horizontal plane through (0, 0)" in the FWD. See my comments further below.

However, in the CPD (Comoving Particle Diagram), according to what (I think) you've been saying, the big bang itself is mapped to the horizontal line, and the horizontal line from the FWD is not mapped anywhere!

This is correct, because, as I just noted, there is *no* "horizontal line" in the FWD (I said this in a previous post as well). The FWD includes lines that get as close to "horizontal" as you like, but none that are exactly horizontal. Yes, that means that this...

In the CPD, you have a SINGLE EVENT from the FWD which is mapped to MULTIPLE LOCATIONS in the Comoving Particle Diagram.

...is true; that's usually how "conformal" diagrams work. (A Mercator projection of the Earth's surface, for example, is "conformal" in this sense--it maps the North and South poles to horizontal lines, not points.) But that does *not* mean that this...

That means that you'll see light from the same event at multiple times.

is true. Consider the Mercator projection again: the lines of longitude (great circles through the poles) are mapped to vertical lines, which *appear* to meet the poles at "different places". But that's an artifact of the "infinite distortion" that the projection makes at the poles. In the same way, the *apparent* "multiple light rays" from the initial singularity in the CPD are an artifact of the "infinite distortion" that this diagram makes at the initial singularity.

How do we deal with this? The only really consistent way is to accept that these "conformal" diagrams *cannot* actually represent the singularities (just as we don't actually use the Mercator projection at the poles). What the "multiple light rays" in the conformal diagram are actually indicating is that, at very short times after the initial singularity, worldlines which emerged from that singularity "in different directions" will be causally disconnected; the more "different" the initial directions are, the longer it will take for the worldlines to become causally connected again. The light that is reaching is now from "close to the big bang" is coming from worldlines that emerged from the big bang in a direction that was "more different" from ours than light that reached us from close to the big bang some time ago.

I realize the above is a somewhat vague description; when I have more time I can try to make it more precise if needed.

In your model, the Big Bang wasn't a single event; it was a huge number of events which all occurred simultaneously throughout the universe.

Strictly speaking, this is false, although it's sometimes used colloquially to describe what I was describing above, that events very close to the big bang happened throughout the universe (in the sense of causal disconnection I gave above).

In the MMD, the light from the big bang is gone... long gone. It might hit something OUTSIDE our universe, but it is not coming back in. What we are seeing is not light from the big bang, but light from matter billions of years after the big bang, (time dilated so that it seems instants after the big bang.) (Technically, we see light from the Hydrogen Recombination era, thousands of years after the big bang).

As I noted above, the FRW models do not claim that we can see light (or any other signal) "from the big bang" itself. The earliest *photons* we can see in the FRW models are, as you say, those from the time of recombination. However, the FRW model would predict that we could see other radiation from earlier times (e.g., neutrinos from the electroweak phase transition, or gravitational waves from even earlier times). We don't currently have any way of testing such predictions because of our poor ability to detect any kind of radiation other than electromagnetic.

You're right, for the most part. Milne does bring this up if you have the e-book. He found it rather troubling, as do I, that there was nothing preventing other objects from being outside. Unless they are other Big Bang's though, there isn't much to worry about. The model really is infinite in mass and energy; it's not "really really big; so big we might as well call it infinite." in other words, before something from outside got to you, it would have to pass an infinite number of particles. And if you think of a particle right at the edge, then it has to pass an infinite number of particles before it gets to that particle at the edge, etc.

I reached this point in the e-book today, as it happens. I still find the "infinite density" part of the model physically unreasonable, but I agree that *if* you stipulate that the density goes to infinity at the "photon shell", there would be no possibility of anything coming in from outside the shell.

That being said, I don't know what would happen if we had two Big Bangs in the same Minkowski Spacetime. It would be the unstoppable object meeting the unstoppable object. Just as in GR, you can't model the infinite curvature at t=0, I'm not sure how to model two planes of infinite density colliding at the speed of light.

This is one reason (but hardly the only reason) that I find the infinite density physically unreasonable.

(By the way, outside this sphere, there may well be infinite curvature, because you have the infinite density coming AT you from ONE direction, instead of the infinite density going AWAY from you in ALL directions. But inside, due to symmetry, I still argue that there is NO curvature.)

Another reason I find the infinite density physically unreasonable is that it should result in infinite curvature at the "photon shell", which, aside from any other objections, would contradict the initial assumption of a flat background Minkowski spacetime. Even if spacetime *inside* the shell were flat (which it could be since that's a general result for inside a symmetrical spherical shell even in Newtonian gravity), the *complete* spacetime in which everything is embedded could not be.

Let me tell you what I think the point of his argument is, and then you can tell me I'm missing the point again. We can have a ball of particles that are moving apart. The point of his argument is that this ball he's talking about are the test particles. It doesn't matter whether these test particles are all comoving or if they are all moving apart. They are all going to follow the laws of physics. In regions where the Einstein Field Equations apply, they're going to follow the Einstein Field Equations. So, this should certainly cover situations like the Schwartszchild metric, or the Kerr Metric, or anyplace where we know the general matter distribution in the space around us; we can use that general matter distribution to figure out how to solve for the motions of the matter.

However, if we disagree on the general distribution of the matter around us, for instance, if I think the matter around us approaches an infinite density within a finite distance in all directions, while you think that the density of the universe is the same in all directions, but the majority of it is not yet causally connected to us, then we probably aren't going to agree on what the Einstein Field Equations say, or necessarily on whether they should even be applied.

This is much closer, I think, but I still have a couple of comments:

(1) According to General Relativity, there are no situations where the EFE does not apply. It always does. There are certainly a wide variety of particular *solutions* to the EFE, among which are the various spacetimes we've been discussing (Schwarzschild, Kerr, FRW, etc.), and which specific solution applies in a particular case will depend on the distribution of matter. But the EFE, as the equation to be solved, applies in every case. So if we disagree on the distribution of matter, we may well disagree on which specific solution to the EFE to apply, but if we accept GR, we *have* to agree that the EFE applies. If you don't accept that, you don't accept GR.

(The only caveat to the above is what I've said before about spacetime singularities: there the EFE itself tells us it can't apply. But we can get as close to the singularities as we like and still apply the EFE.)

(2) Specifying a "matter distribution" in order to solve the EFE can be done in a variety of ways; it can, as you say, be "general", but you may not be appreciating just how general it can be. For example, to obtain the FRW solutions, we specify: "The matter distribution is a perfect fluid, and we'll write the solution in coordinates in which that fluid is isotropic." That's all. Similarly, to obtain the Schwarzschild solution, we specify: "There is no matter--the stress-energy tensor is identically zero--and the solution must be spherically symmetric." (As you can see, often our "specification" takes the form of symmetry properties that the solution must satisfy.)

George Jones

Nov23-10, 12:32 PM

Heh, good phrase. Can you give a reference? I've read a fair amount of Penrose's writing (at least his writing for the lay reader) and I haven't come across this one.

Look at pages 189-190 in the hardcover edition of Penrose's Road to Reality.

The diagram below is Schwarschild spacetime in Painleve(-Gullstrand) coodinates \left(T,R\right), which are related to the usual Schwarzschild coordinates \left(t,r\right) by

\begin{equation*}
\begin{split}
T &= t+4M\left( \sqrt{\frac{r}{2M}}+\frac{1}{2}\ln \left| \frac{\sqrt{\frac{r}{2M}}-1}{\sqrt{\frac{r}{2M}}+1}\right| \right)\\
R &= r.
\end{split}
\end{equation*}

On the diagram, the R-axis is horizontal and the T-axis is vertical, the black line is the worldline of an observer who freely falls radially from rest at infinity, and the event horizon is the vertical line R = 1. At three events on the observer's worldline (outside the horizon, on the horizon, and inside the horizon) I have plotted forward light cones in red. In green, I have plotted lines of constant Schwarzschild t that go through the worldline events inside and outside the horizon. These green lines would be horizontal lines on a Schwarzschild \left(t,r\right) grid which has t as the vertical axis and r as the horizontal axis.The green lines both asymptotically approach the event horizon as T \rightarrow -\infty.
Of course, coordinate time in these coordinates is not actual proper time for any observer except at r=infinity. You used an interesting phrasing earlier:

"suppose we have a coordinate system where coordinate time directly represents the proper time of some family of observers"

Do you mean something other than equals for "directly represents"? Or are you thinking of some simple transform of the standard Schwarzschild coordinates that normalized t to equal tau ?

Suppose the observer's watch is set such that it reads zero when the observer "hits" the singularity R = 0. Then, the events on the observer's (black) worldline all have their T coordinates equal to the observer's watch readings.
Just to check that I've got this right, after some digestion, here's my grasp of these four items, slightly out of order:

2) The transformation between t and T doesn't "change the direction" of the integral curves; it just reparametrizes them.

Yes. Sorry, I going to be a bit pedantic. More pecisely, for every T-integral curve parametrized by T (T-integral curves don't have to be parametrized by T), there is a t-integral curve parametrized by t (t-integral curves don't have to be parametrized by t) such that the two integral curves differ by a constant shift of curve parameter. Even without shiftting parameter, however, every T-integral curve is already a t-integral curve (and vice versa).
This means that, in both coordinate systems, we can use the integral curves of "time" to uniquely label spatial points (each curve has constant values of r = R, \theta, and \varphi).

Yes.
For the rest of this, I'll leave out the angular coordinates (assume them held constant) and only talk about "time" and "radius".

1) The lines of constant T, which are integral curves of \partial_{R}, "cut at a different angle" from the lines of constant t, which are integral curves of \partial_{r}. So even though the integral curves of "time" stay the same, the "spatial slices" cut through them can be different if they're cut at different angles.

Yes.
3) It looks to me like the 4-velocity you gave, which gives us 4), also gives us 3), since it makes it obvious that the 4-velocity is *not* just \partial_{T}, so the integral curves of the 4-velocity can't be the same as the integral curves of \partial_{T}. The integral curves of the 4-velocity are "tilted inward", while the integral curves of \partial_{T} (and thus, of course, \partial_{t}) are "vertical".

Yes.
4) Since the 4-velocity "tilts inward", the integral curves of \partial_{R}, to be orthogonal to them, must "tilt downward" relative to the integral curves of \partial_{r}, which are "horizontal".
Oops, I think this should be "tilt upward", since this is spacetime, not space, so "orthogonal" works differently.

On a \left(t,r\right) diagram, the nature of an integral curve of \partial_{R} depends on which side of the horizon the curve lies. Here, I think you mean outside the horizon. On an \left(t,r\right) diagram, integral curves of \partial_{R} look like the natives of the green curves below.

5) The integral curves of \partial_{R} are always spacelike (horizontal lines outside all lightcones below). Contrast this with the integral curves of \partial_{r}, which are spacelike outside the horizon (right green line outside lightcone) and timelike inside (left green line inside lightcone).

6) \left(T,R\right) coodinates are defined on the horizon (unlike \left(t,r\right) coodintes), with T lightlike on the horizon (part of red lightcone is vertical).

7) T is timelike outside the horizon, but all (four) Painleve coordinates are spacelike inside the horizon!

http://img832.imageshack.us/img832/4121/painlevegullstrand.jpg

PAllen

Nov23-10, 01:17 PM

Suppose the observer's watch is set such that it reads zero when the observer "hits" the singularity R = 0. Then, the events on the observer's (black) worldline all have their T coordinates equal to the observer's watch readings.

You responded this way to my observation:

"Of course, coordinate time in these coordinates is not actual proper time for any observer except at r=infinity. You used an interesting phrasing earlier:"

I should have said "any constant r observer" rather than "any observer". I thought the context was clear (implied by r=infinity, for example), but perhaps not. Anyway, thanks, it is useful to know this.

PeterDonis

Nov23-10, 03:42 PM

Look at pages 189-190 in the hardcover edition of Penrose's Road to Reality.

Hmm, I have that book but I must have missed the quote. I'll see if I can dig it out of the pile of books awaiting shelving... :redface:

Here, I think you mean outside the horizon.

Yes, I did. Thanks for the detailed response and diagram, it makes things very clear.

7) T is timelike outside the horizon, but all (four) Painleve coordinates are spacelike inside the horizon!

I hadn't thought about this before, but it looks to me like this follows from the Painleve metric, since the coefficient g_{TT} is the same as the coefficient g_{tt} in the Schwarzschild metric, and changes sign for r < 2M.

This does bring up another question, which is whether such a coordinate system really "qualifies" as a coordinate system, since there is no timelike coordinate inside the horizon. Is there some way to tell "by inspection" that the system is still "OK" (where "OK" means something like "covers everything we want it to cover"--in this case that would be the exterior and the future interior)? Or does one just need to turn the crank and look at the integral curves of the coordinates in some other system, e.g., Kruskal, that is known to cover the entire manifold?

JDoolin

Nov24-10, 07:45 AM

Ok, good, that makes it clear what the lightcone in your model means--except that this...

...is *not* correct if what you said above is true. There is no "horizontal plane through (0, 0)" in the FWD. See my comments further below.

This is correct, because, as I just noted, there is *no* "horizontal line" in the FWD (I said this in a previous post as well). The FWD includes lines that get as close to "horizontal" as you like, but none that are exactly horizontal. Yes, that means that this...

...is true; that's usually how "conformal" diagrams work. (A Mercator projection of the Earth's surface, for example, is "conformal" in this sense--it maps the North and South poles to horizontal lines, not points.) But that does *not* mean that this...

is true. Consider the Mercator projection again: the lines of longitude (great circles through the poles) are mapped to vertical lines, which *appear* to meet the poles at "different places". But that's an artifact of the "infinite distortion" that the projection makes at the poles. In the same way, the *apparent* "multiple light rays" from the initial singularity in the CPD are an artifact of the "infinite distortion" that this diagram makes at the initial singularity.

How do we deal with this? The only really consistent way is to accept that these "conformal" diagrams *cannot* actually represent the singularities (just as we don't actually use the Mercator projection at the poles). What the "multiple light rays" in the conformal diagram are actually indicating is that, at very short times after the initial singularity, worldlines which emerged from that singularity "in different directions" will be causally disconnected; the more "different" the initial directions are, the longer it will take for the worldlines to become causally connected again. The light that is reaching is now from "close to the big bang" is coming from worldlines that emerged from the big bang in a direction that was "more different" from ours than light that reached us from close to the big bang some time ago.

I realize the above is a somewhat vague description; when I have more time I can try to make it more precise if needed.

Strictly speaking, this is false, although it's sometimes used colloquially to describe what I was describing above, that events very close to the big bang happened throughout the universe (in the sense of causal disconnection I gave above).

As I noted above, the FRW models do not claim that we can see light (or any other signal) "from the big bang" itself. The earliest *photons* we can see in the FRW models are, as you say, those from the time of recombination. However, the FRW model would predict that we could see other radiation from earlier times (e.g., neutrinos from the electroweak phase transition, or gravitational waves from even earlier times). We don't currently have any way of testing such predictions because of our poor ability to detect any kind of radiation other than electromagnetic.

I reached this point in the e-book today, as it happens. I still find the "infinite density" part of the model physically unreasonable, but I agree that *if* you stipulate that the density goes to infinity at the "photon shell", there would be no possibility of anything coming in from outside the shell.

This is one reason (but hardly the only reason) that I find the infinite density physically unreasonable.

Another reason I find the infinite density physically unreasonable is that it should result in infinite curvature at the "photon shell", which, aside from any other objections, would contradict the initial assumption of a flat background Minkowski spacetime. Even if spacetime *inside* the shell were flat (which it could be since that's a general result for inside a symmetrical spherical shell even in Newtonian gravity), the *complete* spacetime in which everything is embedded could not be.

This is much closer, I think, but I still have a couple of comments:

(1) According to General Relativity, there are no situations where the EFE does not apply. It always does. There are certainly a wide variety of particular *solutions* to the EFE, among which are the various spacetimes we've been discussing (Schwarzschild, Kerr, FRW, etc.), and which specific solution applies in a particular case will depend on the distribution of matter. But the EFE, as the equation to be solved, applies in every case. So if we disagree on the distribution of matter, we may well disagree on which specific solution to the EFE to apply, but if we accept GR, we *have* to agree that the EFE applies. If you don't accept that, you don't accept GR.

(The only caveat to the above is what I've said before about spacetime singularities: there the EFE itself tells us it can't apply. But we can get as close to the singularities as we like and still apply the EFE.)

(2) Specifying a "matter distribution" in order to solve the EFE can be done in a variety of ways; it can, as you say, be "general", but you may not be appreciating just how general it can be. For example, to obtain the FRW solutions, we specify: "The matter distribution is a perfect fluid, and we'll write the solution in coordinates in which that fluid is isotropic." That's all. Similarly, to obtain the Schwarzschild solution, we specify: "There is no matter--the stress-energy tensor is identically zero--and the solution must be spherically symmetric." (As you can see, often our "specification" takes the form of symmetry properties that the solution must satisfy.)

This last post comes off as though you are correcting me, rather than acknowledging that we really are talking about two distinct theories. Do you understand that I am describing TWO DIFFERENT models yet?

The Milne Model (Milne Minkowski Diagram MMD), where the Big Bang is ONE EVENT, and the horizontal plane in the FWD is mapped to the light-cone.

The Standard Model (Comoving Particle Diagram CPD), where the Big Bang is MANY EVENTS, and the horizontal plane in the FWD does not exist.

I appreciate your objection to Milne Model, that it has "infinite density" and that is a problem for you. t least I can see that you are actually familiar with the model. But infinite density is a natural extension of lorentz contraction and time dilation as rapidities go from -infinity to infinity. It's not something that is simply assumed; it is something that follows logically from Special Relativity.

I also appreciate that the EFE's start with a symmetry. The Milne Model has the following symmetry: It is the ONLY distribution of matter which is invariant under Lorentz Transformation.

Finally, I wanted to ask whether this transformation (where a single event gets mapped to multiple locations) happens in other metrics; schwarszchild, Kerr, etc.? You seem comfortable with it, as though obviously, it happens at the big bang, but is there anything other than the imperfect analogy with the Mercator Projection which makes it follow naturally or inevitably?

PeterDonis

Nov24-10, 10:04 AM

This last post comes off as though you are correcting me, rather than acknowledging that we really are talking about two distinct theories. Do you understand that I am describing TWO DIFFERENT models yet?

I haven't been sure, because some of the things you've been saying seem to imply that there is at least a correspondence between the models; for example, this:

The Milne Model (Milne Minkowski Diagram MMD), where the Big Bang is ONE EVENT, and the horizontal plane in the FWD is mapped to the light-cone.

The Standard Model (Comoving Particle Diagram CPD), where the Big Bang is MANY EVENTS, and the horizontal plane in the FWD does not exist.

If the Milne model is really supposed to be a "different model", then where does the "horizontal plane in the FWD is mapped to the light-cone" come from? That correspondence between the FWD and a "light cone" diagram, for example the "SR" diagram on Ned Wright's page, *only* applies if we are talking about the FRW model with k = -1 and zero density (as Wright says on his web page). In fact, that correspondence is the basis for the claim that the Milne model is a special case of the FRW models (the case with k = -1 and zero density), which you reject. If you are not talking about the FRW model but about some other model, then I don't see how you're coming up with a correspondence between your "light cone" diagram and the FWD.

But infinite density is a natural extension of lorentz contraction and time dilation as rapidities go from -infinity to infinity. It's not something that is simply assumed; it is something that follows logically from Special Relativity.

Only if you also assume that spacetime can be flat with a non-zero stress-energy tensor. See next comment.

I also appreciate that the EFE's start with a symmetry. The Milne Model has the following symmetry: It is the ONLY distribution of matter which is invariant under Lorentz Transformation.

First of all, I assume you mean "globally invariant under Lorentz Transformation," since any distribution which is proportional to the metric (e.g., a cosmological constant or a scalar field) will be *locally* invariant under Lorentz Transformation.

Second, and more important, your model assumes that you can have a non-zero distribution of matter without affecting the spacetime geometry. Or, if you insist on avoiding any "geometric" interpretation, your model assumes that you can have a non-zero distribution of matter without any tidal gravity effects--that any pair of freely moving particles in your model will have the same constant relative velocity (each one as seen by the other) for all time. This is known to be false--for example, because of the curvature in the Hubble diagram that we discussed before.

To be clear, I am not talking here about "local" effects such as bending of light by the Sun, but about "global" effects, about the relative velocity of "freely moving" particles on a cosmological scale (the ones whose worldlines are straight lines radiating out from the Big Bang event in your diagram). In GR, such particles, even though they are freely falling (they feel no acceleration), can have relative velocities that vary with time. This shows up in our observations as a variation in the "Hubble constant"--the slope of the curve in the Hubble diagram--with time. According to the Milne model, this is impossible--this should be obvious from the fact that, as you say, the Milne model is based on logical deductions from SR, since in SR there can be no such variation with time in the relative velocity of freely falling objects (i.e., objects moving on inertial worldlines). This is why GR was necessary--because in the presence of gravity (i.e., when the effect of mass-energy on the behavior of inertial worldlines is significant), freely falling objects can change their relative velocity with time (in other words, tidal gravity is present), and SR cannot account for that.

You can see this "curvature of freely falling worldlines" in Ned Wright's diagram of the "critical density" case (the "FPD" version). Notice that in that diagram, the worldlines radiating out from the Big Bang curve inward towards each other--unlike the "zero density" diagram, where they are straight. This is the effect of non-zero mass-energy (i.e., gravity) on freely falling worldlines (or "spacetime geometry" in the usual terminology). The usual pop-science way of describing this is that "the gravitational attraction of the mass-energy in the universe causes the expansion of the universe to slow down." (This terminology was invented before we discovered that, for the last few billion years or so, the expansion has actually been "speeding up", which is why dark energy has been added to the "standard" cosmological model--Ned Wright's diagrams don't cover that case, although I believe he discusses it elsewhere on his cosmology site.)

Finally, I wanted to ask whether this transformation (where a single event gets mapped to multiple locations) happens in other metrics; schwarszchild, Kerr, etc.? You seem comfortable with it, as though obviously, it happens at the big bang, but is there anything other than the imperfect analogy with the Mercator Projection which makes it follow naturally or inevitably?

Mathematically, it's fairly easy to construct transformations that do weird things like this. For example, the transformations used to construct Penrose diagrams map various points or lines at "infinity" to finite coordinate values (see the Wikipedia page here: http://en.wikipedia.org/wiki/Penrose_diagram). There's nothing inconsistent about them; you just have to get used to how they work.

As far as other metrics are concerned, yes, there are transformations often used in GR that have similar effects. For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point, as you can see by transforming to Kruskal coordinates, where that entire line becomes the single point at the center of the diagram. (Here I've been ignoring the angular coordinates; when we put them back in, the "point" is actually a 2-surface.) This transformation also maps the "point" at t = infinity in Schwarzschild coordinates to an entire null line (the 45-degree line between regions I and II in the diagram with a yellow background on the Wikipedia page here: http://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates); this null line, the "future horizon", is where all the interesting physics at the horizon actually happens, and it is "invisible" in Schwarzschild coordinates, which often leads to confusion if those coordinates are taken too literally.

The fact that stuff like this can happen is a big reason why physicists are hesitant to attribute too much meaning to coordinates; you always have to check the physical invariants to see what's really going on. For example, I asserted just now that the apparent "line" at the horizon in Schwarzschild coordinates is actually just a point--or, if we include the angular coordinates, what appears to be a 3-surface is actually just a 2-surface. How do I know this is right? (Put another way, how do I know that the description in Kruskal coordinates is the "right" one physically?) Because I can compute the physical 3-volume of the apparent 3-surface, using the metric, and find that it is zero (because the metric coefficient g_{tt} is zero at r = 2M in Schwarzschild coordinates). A similar computation in FRW coordinates shows me that the initial singularity is, physically, a point (because a(t) is zero there, so the spatial metric vanishes), even though it looks like a line (actually a 3-surface, if we include the angular coordinates) in the "conformal" diagram. (Here I do really mean a literal point--zero dimensions--unlike the horizon of a black hole, which is physically a 2-surface--we can compute its area and find that it's non-zero, because the spatial part of the metric doesn't vanish completely. In the FRW case, the entire spatial metric vanishes at the initial singularity.)

Passionflower

Nov24-10, 11:59 AM

For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point,...

A point, please explain? Are you saying that all travelers from different times on a given radial angle meet at the same time at this point?

For example, I asserted just now that the apparent "line" at the horizon in Schwarzschild coordinates is actually just a point--or, if we include the angular coordinates, what appears to be a 3-surface is actually just a 2-surface. How do I know this is right? (Put another way, how do I know that the description in Kruskal coordinates is the "right" one physically?) Because I can compute the physical 3-volume of the apparent 3-surface, using the metric, and find that it is zero (because the metric coefficient g_{tt} is zero at r = 2M in Schwarzschild coordinates).
What volume are you computing?

Could you please explain a bit more what you mean here, I am not sure I can agree here, but I probably misunderstand.

PeterDonis

Nov24-10, 01:05 PM

A point, please explain? Are you saying that all travelers from different times on a given radial angle meet at the same time at this point?

No, as I noted elsewhere in the post, it's actually a 2-surface when the angular coordinates are taken into account. Also, I was *not* saying that any traveler crossing the horizon passes through this point; as I noted further on, all the actual physics at the horizon is on the "future horizon" null line that runs at 45 degrees up and to the right from the center point in the Kruskal diagram. That's where worldlines crossing the horizon go, and they can cross at any one of an infinite number of different events.

What volume are you computing?

The "3-volume" spanned by r = 2M, t = minus infinity to plus infinity, theta = 0 to pi, phi = 0 to 2 pi. Since the metric coefficient g_{tt} is zero at r = 2M, the integral corresponding to this 3-volume vanishes, indicating that what looks like a 3-volume in Schwarzschild coordinates is actually, at most, a 2-surface. (We can verify that it is, in fact, a 2-surface and not something with even fewer dimensions by, for example, integrating over the full range of angular coordinates at the "point" at the center of the Kruskal diagram, which gives the nonzero area of the horizon.)

Passionflower

Nov24-10, 02:07 PM

That's where worldlines crossing the horizon go, and they can cross at any one of an infinite number of different events.

If the worldlines don't cross there then there must be a dimension to separate them right? How do you explain that?

The "3-volume" spanned by r = 2M, t = minus infinity to plus infinity, theta = 0 to pi, phi = 0 to 2 pi. Since the metric coefficient g_{tt} is zero at r = 2M, the integral corresponding to this 3-volume vanishes, indicating that what looks like a 3-volume in Schwarzschild coordinates is actually, at most, a 2-surface. (We can verify that it is, in fact, a 2-surface and not something with even fewer dimensions by, for example, integrating over the full range of angular coordinates at the "point" at the center of the Kruskal diagram, which gives the nonzero area of the horizon.)
Perhaps you could show me the formulas you use, lets say we take the "volume" (I am still not sure what physical volume you are talking about) of r=4M and r=2M so we can all see the difference between a surface at r=4M and r=2M.

Just to remind everybody, the r-coordinate is not a radius or a measure of distance, the r-coordinate is instead a function of an area!

So r is defined as:

r = {1 \over 2}\,\sqrt {{\frac {A}{\pi }}}

Let's keep that in mind.

JDoolin

Nov24-10, 04:19 PM

I haven't been sure, because some of the things you've been saying seem to imply that there is at least a correspondence between the models; for example, this:

If the Milne model is really supposed to be a "different model", then where does the "horizontal plane in the FWD is mapped to the light-cone" come from? That correspondence between the FWD and a "light cone" diagram, for example the "SR" diagram on Ned Wright's page, *only* applies if we are talking about the FRW model with k = -1 and zero density (as Wright says on his web page). In fact, that correspondence is the basis for the claim that the Milne model is a special case of the FRW models (the case with k = -1 and zero density), which you reject. If you are not talking about the FRW model but about some other model, then I don't see how you're coming up with a correspondence between your "light cone" diagram and the FWD.

If I recall correctly, yes, the Milne model has k=0, a=1. It is Minkowski Spacetime with a density that approaches infinity as you go out from the center toward r=c*t, and it has a density that approaches infinity as to go toward t=0. That is not at all compatible with a zero density.

If anyone thinks they have correctly modeled the Milne universe with a zero density, they are just fooling themselves, and using some kind of circular reasoning or a straw-man argument.

There are two variables in the Friedmann Walker Diagram, the horizontal variable is a "space-like" variable, and the vertical is a "time-like" variable. To map to the Comoving Particle Diagram, I'm not sure exactly how it is done, but I think the vertical "time component" is just mapped straight over, while the horizontal "space component" is some form of velocity * distance. I may be wrong, but I *think* it is the integral of the changing scale factor with respect to the "cosmological time."

The horizontal variable is SPACE, and the vertical variable is TIME; some kind of "Absolute" or "cosmological" time, which really doesn't exist in the realm of Special Relativity.

What the Milne model does is treats the horizontal variable in the FWD as sort of a "rapidity-space" The mapping from the FWD to the MMD assumes that the meaning of the FWD "space-like" variable is distance = rapidity * proper time. For a set of particles all coming from an event (0,0), giving the rapidity and proper time for a particle uniquely defines its position in space and time. To map from the FWD to the MMD, you are simply mapping: (d'=rapidity*proper time,t'=proper time) to (d=space, t=time).

To map from the FWD to the CPD, you are mapping (d'="Stretchy" Velocity * Cosmological Time, t'=Cosmological Time) to (d=Space,t=Cosmological Time).

I'll see if I can express this as mathematically and unambiguously as I can, so that if I'm wrong it can be corrected.

\begin{matrix} FWD \mapsto CPD \text{ as }(d\int a(\tau)d\tau,\tau)\mapsto(d,\tau) \\ d=Proper Distance = Cosmological Distance \\ \tau=Proper Time=CosmologicalTime \\ a(\tau)=ScaleFactor \end{matrix}
On the other hand, the Milne mapping looks like this:

\begin{matrix} FWD \mapsto MMD \text{ as }(\varphi \cdot\tau,\tau)\mapsto(v \cdot t,t) \\ \varphi=rapidity \\ \tau=proper time \\ v = velocity \\ t = time \end{matrix}

As you can see, the Milne mapping is linear; there's no changing scale factor. The relation between rapidity and velocity and distance, time, and proper time is the same as is usually given in Special Relativity.

Rapidities between -infinity and +infinity map to velocities between -c and +c. So the horizontal plane (representing infinite rapidity) in the Friedmann Walker Diagram maps to the light-cone in the Milne Minkowski Diagram.

So after any corrections to the FWD to CPD mapping, I wonder if you are yet convinced that we are talking about two different mappings? Can you see that the rapidity=infinity line is included in the Milne Model? Can you see that the distance vs. time relation is fundamentally different? Can you see how an infinite density naturally results from this mapping? Can you see how the Milne Model has only one event at x=0,t=0, while the CPD has an infinite number of events at x=0,t=0?

Only if you also assume that spacetime can be flat with a non-zero stress-energy tensor. See next comment.

First of all, I assume you mean "globally invariant under Lorentz Transformation," since any distribution which is proportional to the metric (e.g., a cosmological constant or a scalar field) will be *locally* invariant under Lorentz Transformation.

Yes, I mean "globally invariant"

Second, and more important, your model assumes that you can have a non-zero distribution of matter without affecting the spacetime geometry. Or, if you insist on avoiding any "geometric" interpretation, your model assumes that you can have a non-zero distribution of matter without any tidal gravity effects--that any pair of freely moving particles in your model will have the same constant relative velocity (each one as seen by the other) for all time. This is known to be false--for example, because of the curvature in the Hubble diagram that we discussed before.

I may have implied that "any pair" of freely moving particles would have the same constant relative velocity, but I must back off on that. But it is not "any pair." It is a specific set of particles which are completely undisturbed after the moment of the Big Bang which will maintain a constant relative velocity.

You have the "Relativity, Gravitation, and World Structure" e-book; check section 112, the list of properties, item 10, Milne says:

"Every particle of the system is in uniform radial motion outward from any arbitrary particle O of the system, and the acceleration of every particle in the system is zero. But the acceleration of a freely projected particle, other than the given particles, is not zero."

I think what is happening is that every particle that is at the center of mass remains at the center of mass, but if you move away from that center of mass, you'll be attracted toward it.

The unusual thing is that accelerating toward the center of mass actually causes the center to move away from you rather than toward you. That results in inflation.

To be clear, I am not talking here about "local" effects such as bending of light by the Sun, but about "global" effects, about the relative velocity of "freely moving" particles on a cosmological scale (the ones whose worldlines are straight lines radiating out from the Big Bang event in your diagram). In GR, such particles, even though they are freely falling (they feel no acceleration), can have relative velocities that vary with time. This shows up in our observations as a variation in the "Hubble constant"--the slope of the curve in the Hubble diagram--with time. According to the Milne model, this is impossible--this should be obvious from the fact that, as you say, the Milne model is based on logical deductions from SR, since in SR there can be no such variation with time in the relative velocity of freely falling objects (i.e., objects moving on inertial worldlines).

This is the elephant in the room that I wanted to talk about earlier. Acceleration.
The Big Bang is one explosion, but particle decay should lead to secondary explosions. If you have secondary (and tertiary, etc.) explosions of matter, this naturally leads to variations in the "Hubble Constant" based on logical deductions from SR. Our local Hubble Constant is based not on the age of the universe, but on the time since the most recent explosion. Only the most distant Hubble Constant tells you the age of the universe.

This is why GR was necessary--because in the presence of gravity (i.e., when the effect of mass-energy on the behavior of inertial worldlines is significant), freely falling objects can change their relative velocity with time (in other words, tidal gravity is present), and SR cannot account for that.

You can see this "curvature of freely falling worldlines" in Ned Wright's diagram of the "critical density" case (the "FPD" version). Notice that in that diagram, the worldlines radiating out from the Big Bang curve inward towards each other--unlike the "zero density" diagram, where they are straight. This is the effect of non-zero mass-energy (i.e., gravity) on freely falling worldlines (or "spacetime geometry" in the usual terminology). The usual pop-science way of describing this is that "the gravitational attraction of the mass-energy in the universe causes the expansion of the universe to slow down." (This terminology was invented before we discovered that, for the last few billion years or so, the expansion has actually been "speeding up", which is why dark energy has been added to the "standard" cosmological model--Ned Wright's diagrams don't cover that case, although I believe he discusses it elsewhere on his cosmology site.)

The pop-science way of describing these things is not nearly sufficient to give any hint to me about what they are actually measuring. I'm interested in whatever data they've gathered to determine that the "expansion of the universe is speeding up" but I think it more likely that whatever the effect, the cause is much more likely to be our galaxy's acceleration toward the receding center of mass, rather than some universal cosmological scale factor increasing at a faster rate.

Mathematically, it's fairly easy to construct transformations that do weird things like this. For example, the transformations used to construct Penrose diagrams map various points or lines at "infinity" to finite coordinate values (see the Wikipedia page here: http://en.wikipedia.org/wiki/Penrose_diagram). There's nothing inconsistent about them; you just have to get used to how they work.

As far as other metrics are concerned, yes, there are transformations often used in GR that have similar effects. For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point, as you can see by transforming to Kruskal coordinates, where that entire line becomes the single point at the center of the diagram. (Here I've been ignoring the angular coordinates; when we put them back in, the "point" is actually a 2-surface.) This transformation also maps the "point" at t = infinity in Schwarzschild coordinates to an entire null line (the 45-degree line between regions I and II in the diagram with a yellow background on the Wikipedia page here: http://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates); this null line, the "future horizon", is where all the interesting physics at the horizon actually happens, and it is "invisible" in Schwarzschild coordinates, which often leads to confusion if those coordinates are taken too literally.

The fact that stuff like this can happen is a big reason why physicists are hesitant to attribute too much meaning to coordinates; you always have to check the physical invariants to see what's really going on. For example, I asserted just now that the apparent "line" at the horizon in Schwarzschild coordinates is actually just a point--or, if we include the angular coordinates, what appears to be a 3-surface is actually just a 2-surface. How do I know this is right? (Put another way, how do I know that the description in Kruskal coordinates is the "right" one physically?) Because I can compute the physical 3-volume of the apparent 3-surface, using the metric, and find that it is zero (because the metric coefficient g_{tt} is zero at r = 2M in Schwarzschild coordinates). A similar computation in FRW coordinates shows me that the initial singularity is, physically, a point (because a(t) is zero there, so the spatial metric vanishes), even though it looks like a line (actually a 3-surface, if we include the angular coordinates) in the "conformal" diagram. (Here I do really mean a literal point--zero dimensions--unlike the horizon of a black hole, which is physically a 2-surface--we can compute its area and find that it's non-zero, because the spatial part of the metric doesn't vanish completely. In the FRW case, the entire spatial metric vanishes at the initial singularity.)

I'm afraid I'm at a loss for following most of this. I'm not entirely convinced you have anywhere a single event which is mapped to multiple places (except for the FRW case where a(t) shrinks to zero).

If we could focus on the nature of the Schwarzschild coordinates; I'm afraid you'll have to go completely remedial to explain it to me, because I don't know the equations for the Schwarzschild coordinates or the reasoning. All I know is that time slows down as you get near the surface, and this causes light to turn toward the planet and get a higher frequency. When you're talking about the "line" at the horizon which is actually just a point, or the three-surface which is really just a two-surface, I'm not sure precisely what is meant.

PeterDonis

Nov24-10, 04:51 PM

If the worldlines don't cross there then there must be a dimension to separate them right? How do you explain that?

There is--it's the Kruskal V coordinate (in the null U, V formulation of Kruskal coordinates, which uses V to label events along the future horizon).

Perhaps you could show me the formulas you use, lets say we take the "volume" (I am still not sure what physical volume you are talking about) of r=4M and r=2M so we can all see the difference between a surface at r=4M and r=2M.

The 3-volume is a 3-volume in 4-dimensional spacetime; the only difference between it and a "normal" spatial 3-volume is that one of the dimensions is the time coordinate. All you do is integrate the appropriate volume measure over the appropriate range of coordinates. The volume measure is just the product of the distance measures along each coordinate; each distance measure is the square root of the appropriate metric coefficient times the coordinate differential. (Some of this may be simpler because the Schwarzschild metric is diagonal; I'd have to go back and do some review to remind myself of what, if any, complications arise when dealing with a non-diagonal metric such as the Kerr metric.) So for a constant value of r, the "volume integral" looks like:

V = \int_{t_{1}}^{t_{2}} \sqrt{- g_{tt}} dt \int_{0}^{\pi} \sqrt{g_{\theta \theta}} d\theta \int_{0}^{2 \pi} \sqrt{g_{\varphi \varphi}} d\varphi = 4 \pi r^{2} \sqrt{1 - \frac{2M}{r}} \left( t_{2} - t_{1} \right)

where I've included the minus sign in front of {g_{tt}} because of the opposite sign of that metric coefficient. At r = 4M, this integral gives a positive value (how large a value depends on the range of t we choose--we can make it infinite by letting t cover its full range of - \infty to \infty), but at r = 2M, the integral vanishes identically.

Just to remind everybody, the r-coordinate is not a radius or a measure of distance, the r-coordinate is instead a function of an area!

Yes, it is, which makes the integral over the angular coordinates very easy to do at a constant value of r (that's why there's just 4 \pi r^{2} in the integral above, instead of some more complicated function of r). But the r coordinate is still being used to *label* events along the radial dimension, which is different from the two angular dimensions. It's not a direct measure of radial distance because the metric coefficient g_{rr} varies, but it's still a coordinate in the radial direction.

Passionflower

Nov24-10, 05:14 PM

There is--it's the Kruskal V coordinate (in the null U, V formulation of Kruskal coordinates, which uses V to label events along the future horizon).

So then if you agree that different observers at different times can pass the event horizon at the same physical location then clearly there must be a line and not a point for a given theta and phi?

I have to get back on your volume calculation, I am not very encouraged by what I see (but clearly that must be my shortcoming).

PeterDonis

Nov24-10, 05:54 PM

If I recall correctly, yes, the Milne model has k=0, a=1.

Not if it's a different model from the FRW models; the k and a parameters apply to those models. The FRW model that is claimed to be an alternate formulation of the Milne model has k = -1, a(t) = t (linear variation, starting with a = 0 at t = 0, the Big Bang) in the FRW parameter system; but you don't agree that that model actually is an alternate formulation of the Milne model.

It is Minkowski Spacetime with a density that approaches infinity as you go out from the center toward r=c*t, and it has a density that approaches infinity as to go toward t=0. That is not at all compatible with a zero density.

If anyone thinks they have correctly modeled the Milne universe with a zero density, they are just fooling themselves, and using some kind of circular reasoning or a straw-man argument.

No, they are pointing out the same physical issue that I pointed out in my last post, that a nonzero density will curve freely falling worldlines, so you can't have Minkowski spacetime with a non-zero density. See below.

So after any corrections to the FWD to CPD mapping, I wonder if you are yet convinced that we are talking about two different mappings? Can you see that the rapidity=infinity line is included in the Milne Model? Can you see that the distance vs. time relation is fundamentally different? Can you see how an infinite density naturally results from this mapping? Can you see how the Milne Model has only one event at x=0,t=0, while the CPD has an infinite number of events at x=0,t=0?

I understand everything you're saying about your diagram of the Milne model. You're not correct that the CPD has an infinite number of events at x = 0, t = 0; it's only one event., and it has coordinates t = 0, but x, y, z undefined (in the same way the longitude of the South Pole is undefined), and it looks like a line in the diagram only because we need to have a reference for the "limit point" of all the vertical worldlines. As you'll see from my comments near the end of this post, I'm afraid I can't really do justice to the mathematical side of this, but I'll try to briefly explain what I think it's supposed to mean physically.

The "conformal" diagram of the FRW model (your "CPD") is intended to make it easy to see the causal structure of the spacetime, and that's all--it highly distorts everything else in order to ensure that light rays always travel on 45 degree lines, so it's easy to see the boundaries of light cones. Pick two worldlines that are widely separated in the conformal diagram, and follow them back towards the "initial singularity" line at the bottom. You'll see that the past light cones of those two worldlines stop overlapping while they're still a fair distance away from the initial singularity. If you pick two worldlines that are closer, their past light cones stop overlapping closer to the singularity. As worldlines get closer and closer, their past light cones get closer and closer to the singularity before they stop overlapping. But for any time after the initial singularity, there will be some "conformal separation" (separation horizontally in the conformal diagram) of worldlines for which the past light cones have stopped overlapping. The conformal diagram was constructed to make all this visually obvious, but the price you pay is making the single event at t = 0 (all space coordinates undefined) look like a line. Actually, the boundary line itself is *not* part of the spacetime; it's just a convenient way of showing, visually, that the vertical worldlines "end" at the initial singularity (which, technically, is not part of the spacetime either, as I've said before--the density and curvature are infinite there, so physically we expect new physics, perhaps quantum gravity, to take over before that point is reached).

I may have implied that "any pair" of freely moving particles would have the same constant relative velocity, but I must back off on that. But it is not "any pair." It is a specific set of particles which are completely undisturbed after the moment of the Big Bang which will maintain a constant relative velocity.

You have the "Relativity, Gravitation, and World Structure" e-book; check section 112, the list of properties, item 10, Milne says:

"Every particle of the system is in uniform radial motion outward from any arbitrary particle O of the system, and the acceleration of every particle in the system is zero. But the acceleration of a freely projected particle, other than the given particles, is not zero."

Yes, thanks for including that quote as it clarifies the point I was trying to make. The "specific set of particles" you mention is the set of "given particles" in Milne's terminology. In your diagram, they appear as straight lines radiating out from the big bang event, all within the "envelope" of the light cone. The point I am making is that in the presence of gravity (i.e., with a non-zero density), GR predicts that those lines cannot be straight lines! As I noted before, in the diagram of the "critical density" FRW model on Ned Wright's page, you see the corresponding lines curve inward towards each other, whereas in the "zero density" model, they are straight.

So here we have a clear case where standard GR and the Milne model make different physical predictions: Milne's model says that we can have a non-zero density in the universe but still have a family of "given particles" that maintain constant relative velocity for all time; GR says that's not possible and predicts that with non-zero density we will see the relative velocities even of "comoving" particles (the ones that correspond to Milne's "given" particles) vary with time.

The Hubble diagram provides a way to test this; but since you commented on that with regard to "inflation", I'll comment further on it below.

The unusual thing is that accelerating toward the center of mass actually causes the center to move away from you rather than toward you. That results in inflation.

This is the elephant in the room that I wanted to talk about earlier. Acceleration. The Big Bang is one explosion, but particle decay should lead to secondary explosions. If you have secondary (and tertiary, etc.) explosions of matter, this naturally leads to variations in the "Hubble Constant" based on logical deductions from SR. Our local Hubble Constant is based not on the age of the universe, but on the time since the most recent explosion. Only the most distant Hubble Constant tells you the age of the universe.

Saying that "our local Hubble constant is based not on the age of the universe, but on the time since the most recent explosion" is not based on direct observation but on a conclusion from your model; you need to give me some way of directly testing this assertion vs. the contrary assertion of standard GR, which is that the "Hubble Constant" at any distance indicates the general "expansion of the universe" at the epoch corresponding to that distance. Since we observe variation in the observed Hubble constant with distance (the curvature in the Hubble diagram), the standard GR model (the FRW model) infers that the relative velocity of "comoving" observers can change with time.

It seems as though you're saying that the Milne model explains the curvature in the Hubble diagram by "secondary explosions". The main problem I see with this is that it should not produce curvature in the diagram; it should produce anisotropy--different slopes of the Hubble plot in different directions (corresponding to different epochs for each "secondary explosion"), but each plot should be a straight line, no curvature (because once the explosion happens, the debris, so to speak, travels in straight lines, with each piece having a constant relative velocity with respect to all other pieces--each "piece" being one of the set of "given particles" created by that particular explosion). We don't see anything like this; we see curvature in the diagram, but it's isotropic.

Even if you push all the secondary explosions back into the "inflation" era, they should still produce anisotropies in, for example, the CMBR, that would have to be larger than those we've observed--at least it seems that way to me, but since your model isn't quantified, it's hard to tell what it predicts for this. But in any case, if all the secondary explosions occurred during the inflation era, they wouldn't affect the Hubble diagram at all, since that only goes back as far as we can see galaxies, quasars, and other objects that have measurable redshifts. So the Hubble diagram, in this case, would indicate the general expansion of the universe, and should *not* have any curvature according to the Milne model.

I'm afraid I'm at a loss for following most of this. I'm not entirely convinced you have anywhere a single event which is mapped to multiple places (except for the FRW case where a(t) shrinks to zero).

If we could focus on the nature of the Schwarzschild coordinates; I'm afraid you'll have to go completely remedial to explain it to me, because I don't know the equations for the Schwarzschild coordinates or the reasoning. All I know is that time slows down as you get near the surface, and this causes light to turn toward the planet and get a higher frequency.

That would take us pretty far afield, and there's a lot of material out there already about Schwarzschild spacetime. (Sorry to punt, more or less, but I just don't think I can do justice to the topic or give you a proper treatment of it in a discussion thread; it really does need to be studied carefully offline, at your own pace and following your own thread of inquiry.) Instead, let me just post a couple of links:

http://casa.colorado.edu/~ajsh/schwp.html

http://www.phy.syr.edu/courses/modules/LIGHTCONE/schwarzschild.html

Unfortunately, the PhysicsForums library doesn't appear to have anything on Schwarzschild spacetime (at least, not anything I could find with a quick search).

Also, all the textbooks on GR spend a fair bit of time talking about Schwarzschild spacetime because it's used so much.

(I'm also pretty much punting at this point, I'm afraid, on the general question of transformations that do weird things like map points to lines, lines to points, etc. As I said in the previous post you quoted, I think it's a mistake to get too wrapped up in what a geometry "looks like" in a particular coordinate system. I'd rather focus on the physics and what is actually observed, which is what I was trying to do with my comments about tidal gravity, curvature in the Hubble diagram, etc. There is a vast body of mathematics behind these various coordinate transformations, but again, I just don't think I can do justice to it here; it's something that really has to be studied at your own pace. Unfortunately I can't point you to any particular texts in this area--the GR textbooks go into this somewhat, but they don't really do it with the sort of rigor that mathematicians do.)

When you're talking about the "line" at the horizon which is actually just a point, or the three-surface which is really just a two-surface, I'm not sure precisely what is meant.

See my previous post in response to Passionflower for a little more info about this point.

PeterDonis

Nov24-10, 05:57 PM

So then if you agree that different observers at different times can pass the event horizon at the same physical location then clearly there must be a line and not a point for a given theta and phi?

Yes--it's the 45 degree line that goes up and to the right from the center point of the Kruskal diagram. (This line has U = constant, V > 0 in the null Kruskal coordinates.) All worldlines that cross the horizon into the black hole cross somewhere on that line, at some positive V coordinate.

Passionflower

Nov24-10, 06:36 PM

Yes--it's the 45 degree line that goes up and to the right from the center point of the Kruskal diagram. (This line has U = constant, V > 0 in the null Kruskal coordinates.) All worldlines that cross the horizon into the black hole cross somewhere on that line, at some positive V coordinate.
Ok so you are saying that for a given theta and phi that an observer A can cross the EH before observer an B. That implies that r=rs, phi=0, theta=0 must be a line not a point. How else would you explain it?

PeterDonis

Nov24-10, 06:55 PM

Ok so you are saying that for a given theta and phi that an observer A can cross the EH before observer an B. That implies that r=rs, phi=0, theta=0 must be a line not a point. How else would you explain it?

It is a line, but the line does not appear in the Schwarzschild chart (either exterior or interior). The line is the one I described; U = 0, V > 0 in the null Kruskal coordinates. (The full "horizon" is actually the pair of crossing lines U = 0 and V = 0, but the portion I've described is where all the worldlines going into the black hole from "region I", the "normal" exterior, cross.) This line also has the r coordinate 2M, as can be seen from the implicit equation for r in terms of U and V, as given for example on the Wikipedia page:

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates.

I'm using U, V here to refer to what the Wiki page calls the "light cone variant" of the Kruskal coordinates (this is its term for null coordinates, apparently--they put tildes over U, V for these coordinates and use U, V without tildes to denote the spacelike and timelike Kruskal coordinates, which I'm more used to seeing as X, T, or sometimes R, T, as the Wiki page briefly comments). In these coordinates, r is given implicitly by

UV = \left( 1 - \frac{r}{2M} \right) e^{\frac{r}{2M}}

So if either U = 0 or V = 0, r = 2M. But this pair of crossing lines (U = 0 and V = 0) does *not* have a well-defined Schwarzschild t coordinate. The equation for t is either

tanh \left( \frac{t}{4M} \right) = \frac{V + U}{V - U}

outside the horizon, or

tanh \left( \frac{t}{4M} \right) = \frac{V - U}{V + U}

inside the horizon; but if either U = 0 or V = 0 (i.e., on the horizon), both of these equations have no solution (because the tanh function would have to have a value of +/- 1, and it never does, it only asymptotes to those values).

Passionflower

Nov24-10, 07:03 PM

It is a line
Well good I agree with that. So now we extend this line in the direction of phi and theta what do we get?

Now earlier you wrote:

For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point,
So do you agree it is a line or do I misunderstand here?

PeterDonis

Nov24-10, 07:05 PM

if either U = 0 or V = 0 (i.e., on the horizon), both of these equations have no solution (because the tanh function would have to have a value of +/- 1, and it never does, it only asymptotes to those values).

I should add, to clarify, that if both U = 0 *and* V = 0 (i.e, at the "center point" of the diagram"), both expressions give 0 / 0 on the RHS, and we would have to take a limit somehow to see if the tanh function could satisfy either equation. The usual assumption seems to be that this can somehow be done in such a way as to be able to assign *any* Schwarzschild t value we like to the center point, from - infinity to infinity (but still at r = 2M).

I haven't seen anything specific in the books I've read on how this could be done, but here's one possible way I can think of: if we assume that either V is some function of U that's linear in U, or U is some function of V that's linear in V, then we can apply L'Hopital's rule and differentiate top and bottom to obtain some constant value on the RHS; and by suitable definition of the functions we can make that constant value basically be anything from -1 to 1, so in effect the center point can map to *any* Schwarzschild t value from - infinity to infinity (at r = 2M).

PeterDonis

Nov24-10, 07:07 PM

Well good I agree with that. So now we extend this line in the direction of phi and theta what do we get?

A spacetime 3-volume. But again, this 3-volume does not appear in the Schwarzschild chart; only a single 2-surface picked out of it does (the one corresponding to the "center point" of the Kruskal diagram). To see the full 3-volume you have to use a chart that covers the "future horizon" line, such as the Kruskal chart. (The ingoing Eddington-Finkelstein chart or the Painleve chart would also work.)

Passionflower

Nov24-10, 07:11 PM

A spacetime 3-volume. But again, this 3-volume does not appear in the Schwarzschild chart; only a single 2-surface picked out of it does (the one corresponding to the "center point" of the Kruskal diagram).
Yes, but did you claim just the opposite?

PeterDonis

Nov24-10, 07:22 PM

Yes, but did you claim just the opposite?

No. Here's everything I said in my post #127 that first responded to this question from you:

No, as I noted elsewhere in the post, it's actually a 2-surface when the angular coordinates are taken into account. Also, I was *not* saying that any traveler crossing the horizon passes through this point; as I noted further on, all the actual physics at the horizon is on the "future horizon" null line that runs at 45 degrees up and to the right from the center point in the Kruskal diagram. That's where worldlines crossing the horizon go, and they can cross at any one of an infinite number of different events.

The "3-volume" spanned by r = 2M, t = minus infinity to plus infinity, theta = 0 to pi, phi = 0 to 2 pi. Since the metric coefficient g_{tt} is zero at r = 2M, the integral corresponding to this 3-volume vanishes, indicating that what looks like a 3-volume in Schwarzschild coordinates is actually, at most, a 2-surface. (We can verify that it is, in fact, a 2-surface and not something with even fewer dimensions by, for example, integrating over the full range of angular coordinates at the "point" at the center of the Kruskal diagram, which gives the nonzero area of the horizon.)

And I've been repeating ever since that yes, there is a line, but the line is *not* covered by the Schwarzschild chart--what looks like a line in the Schwarzschild chart is actually just a single point, which is at the center of the Kruskal diagram. The line, the "real" one that worldlines going into the hole actually cross, is the line going up and to the right at 45 degrees from the center point. And when you add in the angular coordinates, "point" becomes "2-surface" and "line" becomes "3-volume".

Passionflower

Nov24-10, 07:26 PM

Here is your original posting (I boldfaced the relevant parts):

No, as I noted elsewhere in the post, it's actually a 2-surface when the angular coordinates are taken into account. Also, I was *not* saying that any traveler crossing the horizon passes through this point; as I noted further on, all the actual physics at the horizon is on the "future horizon" null line that runs at 45 degrees up and to the right from the center point in the Kruskal diagram. That's where worldlines crossing the horizon go, and they can cross at any one of an infinite number of different events.

The "3-volume" spanned by r = 2M, t = minus infinity to plus infinity, theta = 0 to pi, phi = 0 to 2 pi. Since the metric coefficient g_{tt} is zero at r = 2M, the integral corresponding to this 3-volume vanishes, indicating that what looks like a 3-volume in Schwarzschild coordinates is actually, at most, a 2-surface. (We can verify that it is, in fact, a 2-surface and not something with even fewer dimensions by, for example, integrating over the full range of angular coordinates at the "point" at the center of the Kruskal diagram, which gives the nonzero area of the horizon.)
In both parts you say it is actually a 2-surface and not a 3-surface seems indicated by Schw. coordinates. But one posting ago you agreed it is actually a 3-surface? I must be seriously mistaken.

PeterDonis

Nov24-10, 07:38 PM

In both parts you say it is actually a 2-surface and not a 3-surface seems indicated by Schw. coordinates. But one posting ago you agreed it is actually a 3-surface? I must be seriously mistaken.

You are. Here's the full relevant part of my post #125, which originally prompted your question (but you only quoted part of it in your question):

For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point, as you can see by transforming to Kruskal coordinates, where that entire line becomes the single point at the center of the diagram. (Here I've been ignoring the angular coordinates; when we put them back in, the "point" is actually a 2-surface.) This transformation also maps the "point" at t = infinity in Schwarzschild coordinates to an entire null line (the 45-degree line between regions I and II in the diagram with a yellow background on the Wikipedia page here: http://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates); this null line, the "future horizon", is where all the interesting physics at the horizon actually happens, and it is "invisible" in Schwarzschild coordinates, which often leads to confusion if those coordinates are taken too literally.

I've stated from the very beginning that "point" and "2-surface" refer only to the "apparent line" from t = - infinity to infinity in *Schwarzschild* coordinates, which is physically just a point (or 2-surface if angular coordinates are included) and appears that way at the center of the Kruskal diagram, and that there *is* a "real" line (or 3-volume if angular coordinates are included) at the horizon that is covered by Kruskal coordinates but which does *not* appear in Schwarzschild coordinates.

Passionflower

Nov24-10, 07:43 PM

which is physically just a point
Didn't you just agree that, for instance when two observers passing it after each other, it is a line?

PeterDonis

Nov24-10, 07:52 PM

Didn't you just agree that, for instance when two observers passing it after each other, it is a line?

No. Go back and read my posts again, carefully, and see exactly what I've referred to as a "point" and what I've referred to as a "line", and look at a Kruskal diagram and see how the two relate to each other. Then fill in a couple of timelike worldlines on the diagram, crossing the future horizon line at different events. Where are those events placed, in relation to the point at the center? Do they pass through that point?

Passionflower

Nov24-10, 07:54 PM

No.
Ok, so you are saying it is a point?

So then I ask you again, observer A passes this point before observer B, clearly their worldlines do not cross there, how do you explain that if there is no line but a point?

PeterDonis

Nov24-10, 07:59 PM

Ok, so you are saying it is a point?

So then I ask you again, observer A passes this point before observer B, clearly their worldlines do not cross there, how do you explain that if there is no line but a point?

Did you do what I asked in my last post, looking at the Kruskal diagram? You will find that you can draw two timelike worldlines that cross the future horizon line at two different events (points), but neither of those events are the center point of the diagram (the "point" which appears as a line in Schwarzschild coordinates). Please do that before asking this question again.

Passionflower

Nov24-10, 08:06 PM

Did you do what I asked in my last post, looking at the Kruskal diagram? You will find that you can draw two timelike worldlines that cross the future horizon line at two different events (points), but neither of those events are the center point of the diagram (the "point" which appears as a line in Schwarzschild coordinates). Please do that before asking this question again.
I am not talking about coordinate charts but about physics. An observer A passes the EH at the same spatial location before an observer B, what does that imply?

Too bad Peter, I think you are trying to avoid my questions.

PeterDonis

Nov24-10, 09:45 PM

I am not talking about coordinate charts but about physics. An observer A passes the EH at the same spatial location before an observer B, what does that imply?

Um, that spacetime has a time dimension? I'm not trying to avoid anything, I just don't understand what point you're trying to make. I've already agreed that observer A and observer B, in your example just above, would cross the horizon (the line U = 0, V > 0 that goes up and to the right at a 45 degree angle from the center of the Kruskal diagram) at two different events. Neither of those events is the point at the center of the diagram.

If it's the bit about "the same spatial location" that's bothering you, I've also already said that the "future horizon" line I described has r = 2M all along it. I gave the function relating r to the Kruskal U and V that makes that clear. It's true that on the Kruskal diagram, the two events where A and B cross the horizon have different *Kruskal* coordinates (different V, or different Kruskal spatial coordinate X or R, if you use those coordinates instead of U and V to label events in the diagram). That's because the Kruskal chart is a different chart that labels events differently. What's the problem?

Passionflower

Nov24-10, 10:04 PM

What's the problem?
Do you think r=rs, phi=0, theta=0 is a line or a point in spacetime?

PeterDonis

Nov24-10, 10:10 PM

Do you think r=rs, phi=0, theta=0 is a line or a point in spacetime?

You said you weren't talking about coordinate charts, but about physics. Physically, I've agreed several times that there is a line at the horizon (for a given angular direction in space), so that different infalling observers (or ingoing light rays) can cross the horizon at different events. Whether or not that line is covered by a given coordinate chart depends on the chart, so if you want an answer to your question exactly as you posed it, you'll need to specify which chart the coordinates you gave relate to.

Passionflower

Nov24-10, 10:35 PM

You said you weren't talking about coordinate charts, but about physics. Physically, I've agreed several times that there is a line at the horizon (for a given angular direction in space), so that different infalling observers (or ingoing light rays) can cross the horizon at different events. Whether or not that line is covered by a given coordinate chart depends on the chart, so if you want an answer to your question exactly as you posed it, you'll need to specify which chart the coordinates you gave relate to.
I am talking about this statement:

For example, I asserted just now that the apparent "line" at the horizon in Schwarzschild coordinates is actually just a point--or, if we include the angular coordinates, what appears to be a 3-surface is actually just a 2-surface. How do I know this is right?
So with actually you mean 'coordinates' and not 'physically'?

But I am confused by this statement:

For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point
Doesn't this say the opposite to what we now agree on?

PeterDonis

Nov24-10, 10:45 PM

So with actually you mean 'coordinates' and not 'physically'?

No, the "actually" there meant "physically", because, as I've noted several times, the actual, physical line at the horizon is not covered by the Schwarzschild chart. Let me go ahead and answer the question you posed in your last post but one, exactly as you posed it, in terms of the Painleve and Schwarzschild charts, since both use the r coordinate directly.

In terms of the Painleve chart, the answer to your question exactly as you posed it is "a line". What appears as a line in this chart, the line theta = phi = constant, r = 2M, T = minus infinity to plus infinity (I'll use capital T for Painleve time to avoid confusion with Schwarzschild time t), is actually, physically, a line--it's the "future horizon" line we've been talking about, which different infalling observers can cross at different events.

In terms of the Schwarzschild chart, however, the answer to your question exactly as you posed it is "a point". What *appears* as a line in this chart, the line theta = phi = constant, r = 2M, t = minus infinity to plus infinity (small t this time), is actually, physically, just a single point, *not* a line.

George Jones and I had an exchange about the relationship between the Painleve chart and the Schwarzschild chart earlier in this thread: his posts #80 and #121 have good information (and a helpful diagram in the latter post).

http://www.physicsforums.com/showpost.php?p=2988151&postcount=80

http://www.physicsforums.com/showpost.php?p=3000266&postcount=121

Passionflower

Nov24-10, 11:07 PM

In terms of the Schwarzschild chart, however, the answer to your question exactly as you posed it is "a point". What *appears* as a line in this chart, the line theta = phi = constant, r = 2M, t = minus infinity to plus infinity (small t this time), is actually, physically, just a single point, *not* a line.

Well there is the rub, you speak about Schw. coordinates and seems to assign physical attributes to it. The line physically exists in spacetime but Schw. coordinates do not cover this line, that does not mean it is not physically there is just means that the chart has limitations.

PeterDonis

Nov24-10, 11:58 PM

The line physically exists in spacetime but Schw. coordinates do not cover this line, that does not mean it is not physically there is just means that the chart has limitations.

Which is exactly what I've been saying all along. That was my whole point in bringing up the example in the first place.

PeterDonis

Nov25-10, 12:12 AM

But I am confused by this statement:

For example, in Schwarzschild coordinates, there appears to be an entire infinite line at the horizon, r = 2M, t = minus infinity to plus infinity, that actually, physically, is just a point.

Doesn't this say the opposite to what we now agree on?

I saw this edit to your post #151 just now--it must have appeared while I was editing one of mine. No, it doesn't say the opposite. I've stated what the physical point and line are several times, but maybe a quick summary will help:

The Physical Line: The "future horizon" line. This appears as:

-- A vertical line in the Painleve chart (r = 2M, T = minus infinity to infinity)

-- A 45 degree line up and to the right in the Kruskal chart (U = 0, V > 0).

-- Does *not* appear in the Schwarzschild chart (it's shoved up to "plus infinity", off the chart).

The Physical Point: This appears as:

-- A vertical line in the Schwarzschild chart (r = 2M, t = minus infinity to infinity).

-- The "center point" in the Kruskal chart (U = 0, V = 0).

-- Does *not* appear in the Painleve chart (it's shoved down to "minus infinity", off the chart).

Passionflower

Nov25-10, 12:15 AM

I saw this edit to your post #151 just now--it must have appeared while I was editing one of mine. No, it doesn't say the opposite. I've stated what the physical point and line are several times, but maybe a quick summary will help:

The Physical Line: The "future horizon" line. This appears as:

-- A vertical line in the Painleve chart (r = 2M, T = minus infinity to infinity)

-- A 45 degree line up and to the right in the Kruskal chart (U = 0, V > 0).

-- Does *not* appear in the Schwarzschild chart (it's shoved up to "plus infinity", off the chart).

The Physical Point: This appears as:

-- A vertical line in the Schwarzschild chart (r = 2M, t = minus infinity to infinity).

-- The "center point" in the Kruskal chart (U = 0, V = 0).

-- Does *not* appear in the Painleve chart (it's shoved down to "minus infinity", off the chart).
Ok, I see, and now understand what the confusion was. Looks like we agree. :)

JDoolin

Nov25-10, 09:41 AM

There are two variables in the Friedmann Walker Diagram, the horizontal variable is a "space-like" variable, and the vertical is a "time-like" variable. To map to the Comoving Particle Diagram, I'm not sure exactly how it is done, but I think the vertical "time component" is just mapped straight over, while the horizontal "space component" is some form of velocity * distance. I may be wrong, but I *think* it is the integral of the changing scale factor with respect to the "cosmological time."

The horizontal variable is SPACE, and the vertical variable is TIME; some kind of "Absolute" or "cosmological" time, which really doesn't exist in the realm of Special Relativity.

What the Milne model does is treats the horizontal variable in the FWD as sort of a "rapidity-space" The mapping from the FWD to the MMD assumes that the meaning of the FWD "space-like" variable is distance = rapidity * proper time. For a set of particles all coming from an event (0,0), giving the rapidity and proper time for a particle uniquely defines its position in space and time. To map from the FWD to the MMD, you are simply mapping: (d'=rapidity*proper time,t'=proper time) to (d=space, t=time).

To map from the FWD to the CPD, you are mapping (d'="Stretchy" Velocity * Cosmological Time, t'=Cosmological Time) to (d=Space,t=Cosmological Time).

I'll see if I can express this as mathematically and unambiguously as I can, so that if I'm wrong it can be corrected.

\begin{matrix} FWD \mapsto CPD \text{ as }(d\int a(\tau)d\tau,\tau)\mapsto(d,\tau) \\ d=Proper Distance = Cosmological Distance \\ \tau=Proper Time=CosmologicalTime \\ a(\tau)=ScaleFactor \end{matrix}
On the other hand, the Milne mapping looks like this:

\begin{matrix} FWD \mapsto MMD \text{ as }(\varphi \cdot\tau,\tau)\mapsto(v \cdot t,t) \\ \varphi=rapidity \\ \tau=proper time \\ v = velocity \\ t = time \end{matrix}

As you can see, the Milne mapping is linear; there's no changing scale factor. The relation between rapidity and velocity and distance, time, and proper time is the same as is usually given in Special Relativity.

Rapidities between -infinity and +infinity map to velocities between -c and +c. So the horizontal plane (representing infinite rapidity) in the Friedmann Walker Diagram maps to the light-cone in the Milne Minkowski Diagram.

There is a very clear difference between the two mappings, but I am still uncertain of the mapping from the FWD to the MMD. I used an integration where I think it may have been unnecessary, but I'm pretty sure the scale factor is invoked in the mapping from FWD to MMD.

\begin{matrix} FWD \mapsto CPD \text{ as }(d a(\tau)\tau,\tau)\mapsto(d,\tau) \\ d=Proper Distance = Cosmological Distance \\ \tau=Proper Time=CosmologicalTime \\ a(\tau)=ScaleFactor \end{matrix}

Is there anyone who can verify or correct this mapping?

JDoolin

Nov25-10, 10:24 AM

Ok, I see, and now understand what the confusion was. Looks like we agree. :)

Alright, that was a nice discussion, but over my head, I'm afraid. But regardless of the complexity of the mathematics, there are some things that are fundamental.

The issue is whether there is a difference between a single event, and an infinite number of events occurring in an environment which has been scaled to zero.

If you represent anything in polar coordinates, then you can represent the origin by any number of possibilities: (0,0 radians) (0,1 radian) (0,2 radians) etc. It is the only point on the circle that is non-uniquely defined in polar coordinates. Yet, obviously, there is only one point. Yet it would not take too much work to distort that coordinate system so that that point turned into a horizontal line, and the concentric circles became horizontal lines, and the radial lines became vertical.

Now, you could map this coordinate system using a scale factor a(r)=r, which would tell you the ACTUAL arc-length of each segment.

You can have an infinite number of lines coming out of that point, but all of those lines meet at the event, this is ONE EVENT. But if you try to say those lines don't meet at that point; it is just the scale factor reduces to zero, making them look like they meet, then you're talking about an infinite number of events.

I feel like Peter is trying to have it both ways, claiming that with the singularity, that the Big Bang is, at the same time, one event, and an infinite number of events. You're violating a most basic law of logic, because you can't have something be both true, and untrue at the same time.

The two models resolve the mystery in two different ways. The Milne Model says there is one event with many particles coming out. The Standard Model says there are many events, reduced in scale to zero, so that it only LOOKS like one event.

Peter is saying there are examples of this happening all the time in General Relativity where single events become multiple events or vice versa, but you have to do more than handwaving to convince me that you have successfully violated the law of the excluded middle, and proven it mathematically.

PeterDonis

Nov25-10, 10:41 AM

You can have an infinite number of lines coming out of that point, but all of those lines meet at the event, this is ONE EVENT. But if you try to say those lines don't meet at that point; it is just the scale factor reduces to zero, making them look like they meet, then you're talking about an infinite number of events.

I feel like Peter is trying to have it both ways, claiming that with the singularity, that the Big Bang is, at the same time, one event, and an infinite number of events. You're violating a most basic law of logic, because you can't have something be both true, and untrue at the same time.

The two models resolve the mystery in two different ways. The Milne Model says there is one event with many particles coming out. The Standard Model says there are many events, reduced in scale to zero, so that it only LOOKS like one event.

No; you're misunderstanding the standard FRW model. I've been saying all along that, physically, the initial singularity in the FRW models is *one event*. I've also been saying all along that in the "conformal" diagram, that one event *appears* to be a line, but physically, it's still just a point (one event). I've never claimed anything else. Please read carefully what I've posted; I've tried to be careful about making these distinctions, between the actual, physical invariant objects (points, lines, etc.) and the *appearances* in various coordinate systems, which may not reflect the actual, physical reality.

You seem to believe that somehow, because the FRW model achieves the initial singularity by reducing a scale factor to zero, the intial singularity is really an infinite number of events. That's not correct. I showed earlier how you can tell that it's just one event: look at the spatial "volume element" at a constant time t, which is just the square root of the product of the spatial metric coefficients. (As I cautioned before, this looks this simple only because the FRW metric is diagonal; in a non-diagonal metric things are more complicated.) Since *all three* of the spatial metric coefficients are multiplied by the scale factor a(t), if a(t) goes to zero, the volume element vanishes identically, and this happens whether we try to compute a 3-dimensional volume, a 2-dimensional surface area, or even a 1-dimensional length. That means that, physically, the "volume" t = 0 (the initial singularity) is actually just a single point, with zero dimensions, regardless of how it *appears* in some coordinate system.

Peter is saying there are examples of this happening all the time in General Relativity where single events become multiple events or vice versa, but you have to do more than handwaving to convince me that you have successfully violated the law of the excluded middle, and proven it mathematically.

Once again, I've *never said* that single events actually, physically, become multiple events. Obviously that's absurd. I've *always said* that some transformations can make single events *appear* to be lines instead of points, or conversely, it can make actual, physical lines (or surfaces or volumes) effectively "invisible" because of some strangeness in a particular coordinate system. But obviously you can't change the actual, physical nature of an event (a point), or a line, or a surface, or a volume, by changing coordinates. I've said that before too, in almost exactly those words. Once again, please read carefully what I've posted.