Conditions for spacetime to have flat spatial slices

[PeterDonis]

In another thread,

https://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]

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]

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]

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]

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". 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]

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.

 

[PAllen]

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]

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.

"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]

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]

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]

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]

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]

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]

(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]

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 spatial slice only if global diagonalization is possible. Isnt that a fairly specialized condition for a metric to have?

 

[Mentz114]

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 spatial 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 dx^u -> sqrt(g_uu)dx^u 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]

Roughly speaking, by "time" I mean "a coordinate that's timelike". 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]

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?)

----

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.

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

 

[Passionflower]

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]

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 spatial 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]

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]

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]

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]

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 spatial angle between a pair of null geodesics.

 

[Passionflower]

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 spatial 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]

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]

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]

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]

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]

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]

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]

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 "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]

What we automatically imagine, is the view "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]

What we automatically imagine, is the view "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.)

I wanted to cover this analogy a little further before moving on. We imagine three observers, a man standing on the top of a hill, overlooking a wide landscape, a plane flying at about 4 or 5 miles, and an observer on the moon.

In the man's point-of-view, he looks at the hills and valleys around him and sees curvature he is going to have to deal with, because he is walking. On the moon, the observer sees a round earth.

But in between, you have an observer who looks down and basically sees a flat surface until it disappears into the clouds in the distance.

So there is a scope where small-scale curvature is evident, and a scope where large scale curvature is evident, and in between, a scope where, though both can be detected, neither curvature is really obvious.

You can watch thehttp://video.google.com/videoplay?docid=8842256077873416888#" , and estimate that the small-scale curvature effects would probably disappear if you were looking at a patch of ground about 10^3.5 meters across (about 3 km). The large scale curvature begins to become obvious between 10^6.5 meters (about 3000 km). At scales between 10^3.5 and 10^6.5 meters, the Earth basically looks flat.

Is this analogy applicable to General Relativity? We have a curvature at the small scale (on the scale of stars and planets) that curve space in their region. But if we pull back, we can take a photo from the distance, at the scale of a solar system, or a galaxy, or perhaps even a galaxy cluster, and treat it as flat space.

Then we move yet further out, to where we can observe 1 million parsecs, (3.26 million light years) and this curvature amounts to something on the order of 70 km/second

http://upload.wikimedia.org/wikipedia/commons/2/2f/Local_Group.JPG
The image shows a region about 1 megaparsec in radius; 2 megaparsecs across​

First of all, to notice motion of 70 km/second on this scale would take thousands of years. (Using doppler effect it's not so difficult, but I'm talking about side-to-side motion.) At 70 km/second it takes ~4300 years to go just one light-year. But we're looking at a map on the scale of millions of light years.

Furthermore, 70 km/second per megaparsec is just 0.023% of the speed of light.

I don't know exactly how one calculates the relation between hubble's constants and curvature in General Relativity. However, if I can relate this velocity to Special Relativity, you could go out an awful lot of megaparsecs before this curvature becomes noticeable. Special Relativistic effects such as time-dilation become just barely noticeable at around 10% of the speed of light. That would be around 400 million megaparsecs or 1.4 billion light years.

So where the span over which the Earth appears flat is from about 3 kilometers* to 3000 kilometers, the span over which the universe looks flat is between say 1 Astronomical unit* to 1.4 billion light-years. So the difference between the local curvature and the global curvature based on Earth geological features is a factor around a thousand, while the difference between the local curvature and the global curvature based on general relativity features is a factor of around a trillion.

*There are some enormous features of the earth, such as the grand canyon and the large mountain ranges which will not look flat on the scale of 3 kilometers. There are also enormous features of the universe, such as black holes where you'd have to back away further than 1 AU to have them look flat.

 

[JDoolin]

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 let's 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.