What Are the Key Questions About Supernova Data and the Modified Milne Model?

[TrickyDicky]

From the Minkowski metric with this coordinate transform

<br /> r \to t \sinh r<br />

<br /> t \to t \cosh r<br />
Here's what I get since: \cosh^2{r}-\sinh^2{r}=1

ds^2 = dt^2\cosh^2{r}-dt^2\sinh^2{r}-t^2\sinh^2{r} d\Omega^2 =dt^2(\cosh^2{r}-\sinh^2{r})-t^2\sinh^2{r} d\Omega^2=dt^2-t^2(\sinh^2{r} d\Omega^2)​

 

[Chalnoth]

From the Minkowski metric with this coordinate transform

<br /> r \to t \sinh r<br />

<br /> t \to t \cosh r<br />
Here's what I get since: \cosh^2{r}-\sinh^2{r}=1

ds^2 = dt^2\cosh^2{r}-dt^2\sinh^2{r}-t^2\sinh^2{r} d\Omega^2 =dt^2(\cosh^2{r}-\sinh^2{r})-t^2\sinh^2{r} d\Omega^2=dt^2-t^2(\sinh^2{r} d\Omega^2)​

Oh, okay, looks like you're not carrying the product rule through.

r \to t \sinh r

Leads to:
dr \to \sinh r dt + t \cosh r dr

I'm sure you can figure out the rest.

 

[TrickyDicky]

Oh, okay, looks like you're not carrying the product rule through.

r \to t \sinh r

Leads to:
dr \to \sinh r dt + t \cosh r dr

I'm sure you can figure out the rest.

Oops, I must have forgotten the little calculus I know.

Thanks

 

[Chalnoth]

Oops, I must have forgotten the little calculus I know.

Thanks

No worries :) I made a few mistakes in doing this myself before I finally got it right. When you posted, I had to go back to make sure I didn't make a mistake myself...

 

[JDoolin]

Makes the math simpler.



General relativity says that you can chose whatever clocks and rulers you want and you'll get the same answer. If you want a ruler that shrinks as you move it. That's fine. If you want a clock that speeds up or slows down, that's also fine.

The easily analogy is that you can draw a diagram with square graph paper, or you can draw the diagram with polar coordinates. It's all the same.


The metric does not change, but it's perfectly fine to do a coordinate transform. The whole point of relativity is that you give me a metric. I can do certain coordinate transforms that I want, and the metric stays the same.

It is okay to change from Cartesian coordinates to Spherical coordinates, because you are talking about fundamentally different variables. (x,y,z) \to (r,\theta,\phi) is a legitimate transformation, because the first and second sets of coordinates are describing the same information in the different ways:

\begin {matrix}<br /> z=r cos \theta \\<br /> x=r sin \theta cos \phi\\<br /> y=r sin \theta sin \phi<br /> \end {matrix}<br />​

This is something true about the mathematical relationships between the radius, the polar angle, and the azimuthal angle.

In Special Relativity, also, you can perform the coordinate transformation:

<br /> \begin{bmatrix}<br /> c t&#039; \\ x&#039; \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> \cosh\phi &amp;-\sinh\phi \\<br /> -\sinh\phi &amp; \cosh\phi \\<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> c t \\ x \end{bmatrix}<br />​

This is also okay, because on the left hand side, we have t' and x'. These are the coordinates of events in another inertial reference frame. t' and x' are fundamentally different from t and x, so it is okay that they have different forms. (You cannot, by the way, say "you want a ruler that shrinks as you move it." or "you want a clock that speeds up or slows down." There are explicit relationships that are already determined.)

Mapping from (x,y,z) \to (r,\theta,\phi) or (x,t) \to (x&#039;,t&#039;) is mathematically and physically valid. However, when you say:

<br /> \begin {matrix}<br /> t\to t \cosh r\\<br /> r \to t \sinh r<br /> \end {matrix}<br />​

You are replacing distance with distance, and time with time. Certainly, you preserve all of the information by doing so, but you do not preserve the shape.

For instance: all of the following represent coordinate transformations of the earth.
a) http://en.wikipedia.org/wiki/Peters_map
b) http://en.wikipedia.org/wiki/Albers_projection
c) http://en.wikipedia.org/wiki/Mercator_projection
d) http://en.wikipedia.org/wiki/Globe

Three of these transformations significantly affect the shape of the earth, while the fourth only affects the size and position. The globe represents the true shape, and the others represent convenient distortions of the shape depending on the purpose.. However, they still do not actually map (x,y,z) \to (x,y,z).

If you would like to claim:

<br /> \begin {matrix}<br /> t&#039; \to t \cosh r\\<br /> r&#039; \to t \sinh r<br /> \end {matrix}<br />​

...then I can ask you why you think this is a convenient distortion of the shape, and what was your purpose in making that distortion. But to claim

<br /> \begin {matrix}<br /> t\to t \cosh r\\<br /> r \to t \sinh r<br /> \end {matrix}<br />​

... is to claim that t is mathematically different than t, and r is mathematically different from r. This is not legitimate.

Milne, by the way, was also quite disturbed at Eddington's "scale factors" and spent quite some effort in pointing out how ridiculous it was. If Milne knew that a model named after him had been saddled with such a thing, I think he would roll over in his grave.

 

[Chalnoth]

Your resistance to this coordinate change is truly amusing. It is possible in General Relativity to make any coordinate transformation at all, provided the coordinate transformation is one-to-one in some region. That is, the only limitation opposed on coordinate transformations is that they don't throw away information.

As long as your coordinate transformation satisfies this, any calculation you might ever do regarding a physical observable will give the same answer. The use of different systems of coordinates is merely a mathematical convenience of no physical meaning whatsoever. The real world, after all, doesn't have numbers written on it.

 

[JDoolin]

Your resistance to this coordinate change is truly amusing. It is possible in General Relativity to make any coordinate transformation at all, provided the coordinate transformation is one-to-one in some region. That is, the only limitation opposed on coordinate transformations is that they don't throw away information.

As long as your coordinate transformation satisfies this, any calculation you might ever do regarding a physical observable will give the same answer. The use of different systems of coordinates is merely a mathematical convenience of no physical meaning whatsoever. The real world, after all, doesn't have numbers written on it.

I pointed out a genuine flaw in your reasoning, and you dismissed it by pointing and laughing.

I have no resistance to saying

<br /> <br /> \begin {matrix}<br /> t&#039; \to t \cosh r\\<br /> r&#039; \to t \sinh r<br /> \end {matrix}<br /> <br />​

...if you can explain what you mean by t, t', r, r'.

But, what you are claiming, is:

<br /> <br /> \begin {matrix}<br /> t \to t \cosh r\\<br /> r \to t \sinh r<br /> \end {matrix}<br /> <br />​

This is nonsense, and it certainly wasn't what Milne ever meant.

 

[Chalnoth]

I pointed out a genuine flaw in your reasoning, and you dismissed it by pointing and laughing.

I have no resistance to saying

<br /> <br /> \begin {matrix}<br /> t&#039; \to t \cosh r\\<br /> r&#039; \to t \sinh r<br /> \end {matrix}<br /> <br />​

...if you can explain what you mean by t, t', r, r'.

But, what you are claiming, is:

<br /> <br /> \begin {matrix}<br /> t \to t \cosh r\\<br /> r \to t \sinh r<br /> \end {matrix}<br /> <br />​

This is nonsense, and it certainly wasn't what Milne ever meant.

You're really making a big deal out of a small lack of rigor? You may note that I actually did use the more correct notation in my first post about this: https://www.physicsforums.com/showpost.php?p=2876300&postcount=32

 

[JDoolin]

You're really making a big deal out of a small lack of rigor? You may note that I actually did use the more correct notation in my first post about this: https://www.physicsforums.com/showpost.php?p=2876300&postcount=32

Let us see if we can find the physical definitions for t', r', t, and r, and then we can have a sensible discussion.

 

[Chalnoth]

Simply give me the physical definitions for t', r', t, and r, and then we can have a sensible discussion.

There is no physical definition for any particular coordinates in General Relativity. These are not physical entities, just labels we place on the system.

However, in this case the primed coordinates are the Minkowski coordinates, with the unprimed coordinates being the Milne coordinates. The Milne coordinates can be thought of as the inside of a light cone in Minkowski space-time. There is a plot here that shows the shape:
http://world.std.com/~mmcirvin/milne.html#time

In any case, coordinates are not physical things, and are merely chosen for their convenience for a particular problem. The Milne coordinates are exactly equivalent to the Minkowski coordinates.

 

[JDoolin]

There is no physical definition for any particular coordinates in General Relativity. These are not physical entities, just labels we place on the system.

However, in this case the primed coordinates are the Minkowski coordinates, with the unprimed coordinates being the Milne coordinates. The Milne coordinates can be thought of as the inside of a light cone in Minkowski space-time. There is a plot here that shows the shape:
http://world.std.com/~mmcirvin/milne.html#time

In any case, coordinates are not physical things, and are merely chosen for their convenience for a particular problem. The Milne coordinates are exactly equivalent to the Minkowski coordinates.

If you have Mathematica, you can paste this into it. Otherwise, treat it as pseudocode.

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

What "the metric" is doing is converting a homogeneous group of comoving particles into a set of particles which are separated by an equipartition of rapidity. (i.e. they start together at a point, and are flying away from each other.)

These two things are in no way the same. Milne's model is flying apart. The standard model is standing still. There's no way to claim they're both the same. In Milne's model, the particles were all at the same point at t=0. In the Standard Model, all the particles were at different points at t=0.

This is analogous to a mercator projection of the earth. On the Mercator Projection, the north and south poles occupy the same space as the equator. In real life, the north and south poles are points. And there is no confusion about which form is real.

 

[TrickyDicky]

These two things are in no way the same. Milne's model is flying apart. The standard model is standing still. There's no way to claim they're both the same. In Milne's model, the particles were all at the same point at t=0. In the Standard Model, all the particles were at different points at t=0.

In post n. 40 you derived yourself the metric of the Milne model and it turned out it was Minkowski metric, so what s your point?

Is this some kind of joke?

 

[JDoolin]

A good way to re-write the equations in a sensible manner would be as follows:

\begin {matrix}<br /> t = t&#039; \cosh \phi\\<br /> r = t&#039; \sinh \phi<br /> \end {matrix}<br />​

where t and r are the position and time (in the reference frame of a stationary observer.) where a particle which has traveled at constant rapidity \phi, from the Big Bang event (t=0, r=0) reaches the proper age of t',

The use of a fictional reference frame where the particles are comoving is entirely unnecessary.

 

[JDoolin]

While Milne was attempting to show how ridiculous Eddington's ideas were, he gave an equation which would map comoving world-lines (t \in \lbrace 0, \infty \rbrace,r=constant) to world-lines that were moving away from a single event at a constant velocity. (t',r')

\begin {matrix}<br /> t&#039; \to t \cosh r\\<br /> r&#039; \to t \sinh r<br /> \end {matrix}<br />​

The equation was nonsense, and Milne's point was that it was nonsense. (To make it legitimate, the r term should be rapidity--not a distance. The t term refers to the proper time since the (0,0) event of the particle.)

However, because his point was also that Eddington's ideas were ridiculous, the Eddington followers latched onto the very equation that Milne was describing as nonsense, and began calling it The Milne Model.

The Minkowski metric and the real Milne metric are equivalent.

ds^2=dt^2-dx^2-dy^2-dz^2​

However when you map in the nonsense equation,

<br /> \begin {matrix}<br /> t \to t \cosh r\\<br /> r \to t \sinh r<br /> \end {matrix}<br />​

you "derive" the metric given on Wikipedia for the Milne Model:

ds^2 = dt^2-t^2(dr^2+\sinh^2{r} d\Omega^2)​

where

d\Omega^2 = d\theta^2+\sin^2\theta d\phi^2​

This metric is no longer equivalent to the Minkowski Metric.

Why is it important not to change metrics?

Your distance and time are fundamentally different things than rapidity and proper time of a distant particle who has maintained constant velocity since (t=0,r=0). If I have to try to use the Lorentz transformations but am only allowed to use the initial rapidity of the particle, and its proper age assuming that it remained at that initial rapidity, I won't be able to do any good physics at all.

If you have a set of comoving galaxies, and treat it in Minkowski spacetime, then when an observer changes velocity, you'll have length contraction of the entire universe. In other words, there is one unique velocity at which the universe appears to be "at rest."

If you have a set of particles, all equipartitioned by rapidity, all coming from a single event, and treat the system in Minkowski Spacetime, the result is a Lorentz Invariant expanding sphere. Meaning, if an observer accelerates, no matter how large the \Delta v, he will continue to be inside an expanding spherical shape. This means there is no "special" velocity where the universe appears spherical. There is also no "special" particle within this system who can say "only I am at the center." No matter how fast the observer is going, the universe will look like a sphere. And no matter which particle you pick, it looks like it is in the center.

Jonathan

 

[Chalnoth]

So you can't use Lorentz transformations. So what? Any physical quantity you could ever calculate will give you the exact same result in either coordinate system.

An example of a physical quantity, by the way, would be the time required for a light beam to bounce off some far-away mirror and return.

 

[JDoolin]

So you can't use Lorentz transformations. So what? Any physical quantity you could ever calculate will give you the exact same result in either coordinate system.

An example of a physical quantity, by the way, would be the time required for a light beam to bounce off some far-away mirror and return.

Yeah. Unfortunately, I ended with what I realized was my weakest point. I deleted the line, but then I realized you had already responded. Sorry about that.

However, since you brought it up, though you make the case that you "can't use Lorentz Transformations" that is a cop-out. Changing the metric does not release you from the Lorentz Transformation--it only changes the form of the Lorentz Transformation.

 

[Chalnoth]

Yeah. Unfortunately, I ended with what I realized was my weakest point. I deleted the line, but then I realized you had already responded. Sorry about that.

Fair enough, but I still don't see how a change in coordinates is something to argue against. If they're useful, they're useful. If not, not. A coordinate change doesn't change anything measurable.

However, since you brought it up, though you make the case that you "can't use Lorentz Transformations" that is a cop-out. Changing the metric does not release you from the Lorentz Transformation--it only changes the form of the Lorentz Transformation.

Yes.

 

[George Jones]

If you have a set of particles, all equipartitioned by rapidity, all coming from a single event, and treat the system in Minkowski Spacetime,

How is this a solution to Einstein's equation for general relativity?

 

[TrickyDicky]

How is this a solution to Einstein's equation for general relativity?

Obviously, it's not. Milne never accepted GR.

 

[JDoolin]

Hmmm. I'd better start distinguishing between the Minkowski-Milne model and the Friedman-Milne model. The Minkowski-Milne model describes an infinite number of particles flying apart from a single event into pre-existing "Minkowski" space.

The Friedman-Milne model is a mapping from one spacetime where all of the particles are comoving to another spacetime where all of the particles are flying apart. Which of these two spacetimes is the one where the Minkowski Metric applies? And which one of them is what you think of as the "true" metric?

How is this a solution to Einstein's equation for general relativity?

I don't entirely understand the Einstein Field Equations or what they are for. They are the "equations you solve to do General Relativity" and have something to do with gravity.

I still am stuck, conceptually, on how taking a derivative of the scale factor, has any meaningful relationship to gravity. Part of the problem is that I stubbornly insist that the scale factor is constant. From my perspective, of course, it appears you are stubbornly insisting the scale factor is NOT constant, though I cannot fathom your reason to suppose it is changing.

Certainly I have seen pictures of earth-colored balls causing dents in a sheet, and then other balls will roll down to them. At one time, I actually thought "aha!" but over time I realized this had no explanatory power whatsoever. All that model does is turn the source of the gravity perpendicular to the plane of motion. This would require a fourth spatial dimension if it were a valid description.

If you want to see where I'm at in understanding the Einstein Field Equations, go back into this thread and read posts 11, 14, 21, 23-28.

From my own explorations, I am rather swayed that only time is affected by gravity. For instance, from my (not entirely complete) analysis of the Rindler coordinate problem, it seems to me that the deeper a clock is in a gravitational well, the slower it will tick. Though I'm still working on it, I currently suspect this slowing in time also slows the speed of light. As long as the speed of light goes slower, and not faster, then all of the event-intervals associated with that disturbed light ray become timelike (which means they won't make causality problems.)

But the rocket in the Rindler problem is actually exactly the same length to a person on board the rocket as it is to an inertial observer with whom the rocket is instantaneously at rest.

If the rocket appears to be the same length to both parties, this means that "acceleration" does not cause a warping of space--hence I would expect that gravity does not either.

As such, I would propose a theory of gravity which merely slows the clocks (and possibly the speed of light) in gravitational wells, but does not affect the scale of space.

I don't know whether such a thing is compatible with the Einstein Field Equations. There are apparently 10 Einstein field equations, so if it is compatible, perhaps this would reduce their number, and simplify them greatly.

Jonathan

 

[Chalnoth]

I still am stuck, conceptually, on how taking a derivative of the scale factor, has any meaningful relationship to gravity. Part of the problem is that I stubbornly insist that the scale factor is constant. From my perspective, of course, it appears you are stubbornly insisting the scale factor is NOT constant, though I cannot fathom your reason to suppose it is changing.

One way to look at it is this. Let's imagine that we want to answer the question, "What is the most general type of metric we can write down that is both homogeneous and isotropic?"

First of all, if it is to be isotropic, the metric must not have any off-diagonal components. That is, there are no dxdy or drd\theta components.

Now, if we multiply the entire metric by any function, it doesn't change the physics, so we can arbitrarily choose the dt^2 component to have no pre-factors. Now, to make things simple, we'll work in Euclidean space for the three spatial components, and ask what sorts of metric factors they can pick up. Well, since we demand isotropy, we know that whatever function we choose, we must place the same function in front of every spatial component of the metric. Otherwise we would be picking out a specific direction in space.

Now this function we place in front of the other components of the metric can obviously be a function of time and retain homogeneity and isotropy. Naively we wouldn't think, however, that it could be a function of space. But it does turn out that there is a specific choice of function that does depend upon space which still obeys homogeneity and isotropy: constant spatial curvature.

So our general homogeneous, isotropic metric becomes:

ds^2 = dt^2 - {a^2(t) \over 1 - k(x^2 + y^2 + z^2)}(dx^2 + dy^2 + dz^2)

So we automatically get a scale factor that depends upon time just by asking what the most general homogeneous, isotropic metric can be. It then becomes an exercise in math to determine what this metric does in General Relativity, and we are led inexorably to the Friedmann equations.

Certainly I have seen pictures of earth-colored balls causing dents in a sheet, and then other balls will roll down to them. At one time, I actually thought "aha!" but over time I realized this had no explanatory power whatsoever. All that model does is turn the source of the gravity perpendicular to the plane of motion. This would require a fourth spatial dimension if it were a valid description.

This is just a visualization of the curvature. General Relativity requires no extra dimensions to describe the curvature of space-time, but we can't very well visualize the curvature without artificially adding an extra dimension.

What happens in General Relativity, though, is that so-called "test particles" always follow paths that mark the shortest space-time distance between two points in space-time. These hypothetical test particles are objects which respond to the space-time curvature but don't affect it. They are a good approximation to reality whenever you're tracking the path of an object that is much less massive/energetic than the sources of the gravitational field it's traveling in.

Now, in flat space-time, the shortest path between any two events is always a straight line. This means that in flat space-time, objects always move with constant speed in a constant direction.

So when we see an object like the Moon orbiting the Earth, that means there is a massive departure from flat space-time surrounding the Earth: instead of going in a straight line, the Moon goes in a circle! This can be visualized as space-time being sort of a rubber sheet and the Earth providing an indentation on that sheet, an indentation which the Moon follows, but this is just a visualization because we simply can't visualize four-dimensional space-time curvature directly.

One thing that we know from General Relativity, however, is that the only way you can have flat space-time, which is the case for Minksowki/Milne space-time, is if the universe is empty. If you take the above homogeneous, isotropic metric, for example, the Milne metric pops out as the metric you get when you set the energy density of the universe to zero.

 

[twofish-quant]

This metric is no longer equivalent to the Minkowski Metric.

Yes it is. Let's define some light beams. When you have a beam of light then ds = 0, and you'll find that the curves for which ds=0 are the same. Once you have a grid of light beams, then you can start describing the path of an object in reference to different light beams, and if you change from one coordinate to another, you'll find that the paths are the same.

The particles in your universe don't know anything about r or t. They can only do experiments by sending light beams over to each other or describing their location with respect to light beams, and you'll find that those are the same.

 

[twofish-quant]

IYou are replacing distance with distance, and time with time. Certainly, you preserve all of the information by doing so, but you do not preserve the shape.

The information about the shape is in the ds equation. When you change your coordinates, then the distance equation changes so that the shapes are the same.

Three of these transformations significantly affect the shape of the earth, while the fourth only affects the size and position.

They only change the shape if you throw away the metric equation.

 

[twofish-quant]

What "the metric" is doing is converting a homogeneous group of comoving particles into a set of particles which are separated by an equipartition of rapidity. (i.e. they start together at a point, and are flying away from each other.)

No you aren't. You are just replacing one piece of graph paper with one that has different lines. Now if you have particles that follow the lines of one piece of graph paper, and then you change the graph paper radically, then it's no longer going to follow the lines on the other piece.

But that doesn't matter.

These two things are in no way the same. Milne's model is flying apart. The standard model is standing still. There's no way to claim they're both the same.

Different pieces of graph paper. Beams of light will travel along lines in which ds=0.

 

[twofish-quant]

Hmmm. I'd better start distinguishing between the Minkowski-Milne model and the Friedman-Milne model. The Minkowski-Milne model describes an infinite number of particles flying apart from a single event into pre-existing "Minkowski" space.

You are using the work "metric" in a way that I don't understand.

In SR, you can use any set of coordinates you want to describe a physical situation. The important number is the "space-time distance" between two events, and two observers will always agree on that. If you have a beam of light, the coordinates through which the beam of light goes through is always going to be ds=0.

Everything else is just graph paper.

Now if you are proposing something different, that's fine, but you aren't talking about metrics.

But it doesn't matter...

Also to relate this to observational cosmology. It's really all rather unimportant when you compare to observations. The only thing that you care about is how quickly the universe expands. Whether it expands according to GR, SR, or something else isn't important. Once you get an equation for how quickly the universe expands, then you see how sound waves go through the expanding universe, and you get a lumpiness factor.

Now it turns out that you can punch in numbers to your computer programs in which the universe expands in exactly the same way that the Milne model says it should, and you find that the universe expands too quickly. The faster the universe expands, the quicker it cools and the more deuterium you end up with. Also the faster the universe the further sound waves can to before they stall...

http://cmb.as.arizona.edu/~eisenste/acousticpeak/acoustic_physics.html

The important thing to point out is that *these* calculations only involve gas physics, gravity only enters as far as it tells you how the quickly universe expands.

 

[JDoolin]

One way to look at it is this. Let's imagine that we want to answer the question, "What is the most general type of metric we can write down that is both homogeneous and isotropic?"

First of all, if it is to be isotropic, the metric must not have any off-diagonal components. That is, there are no dxdy or drd\theta components.

Now, if we multiply the entire metric by any function, it doesn't change the physics, so we can arbitrarily choose the dt^2 component to have no pre-factors. Now, to make things simple, we'll work in Euclidean space for the three spatial components, and ask what sorts of metric factors they can pick up. Well, since we demand isotropy, we know that whatever function we choose, we must place the same function in front of every spatial component of the metric. Otherwise we would be picking out a specific direction in space.

To me, claiming that the space is stretching represents a HUGE change in the physics. To me, claiming that Lorentz Transformations are not valid in cosmology represents a HUGE change in the physics. If it did not represent a change in the physics then we would not be arguing with each other. We would be saying to one another: "ah, yes, that's another perfectly valid way to look at it."

For the Milne-Minkowski model, I would suggest that we should consider the view of this planet from a distant galaxy traveling away at 90, or 99% of the speed of light. If the alien is asked to "compute the speed of the clock in on earth," For a good approximation, he may freely neglect the rotational velocity of the arms of the Milky Way Galaxy. And the effect of the Earth's gravity on the speed of the clock will be even more negligible than that. The small effects of general relativity will be tiny compared to the effects of Special Relativity.

But I frequently hear proponents of the "standard model" say that the effects of Special Relativity are only a local effect. (since all the galaxies are comoving, I gather, there is no time-dilation or desynchronization between the galaxies.) This is simply not true in the Milne-Minkowski model--where you must consider the relativity of simultaneity. This represents another HUGE change in the physics based on the metric.

Now this function we place in front of the other components of the metric can obviously be a function of time and retain homogeneity and isotropy. Naively we wouldn't think, however, that it could be a function of space. But it does turn out that there is a specific choice of function that does depend upon space which still obeys homogeneity and isotropy: constant spatial curvature.

Why is your goal to find a metric where homogeneity and isotropy are retained? Why don't you, instead, make the goal to find a distribution of matter in which homogeneity and isotropy are retained?

This is what Milne already has found--a distribution of matter in Minkowski Space that is both homogeneous and isotropic. Isn't the only reason that Friedmann etc. continued to look for a "metric" because they erroneously denied that Milne's model was homogeneous and isotropic?

So our general homogeneous, isotropic metric becomes:

ds^2 = dt^2 - {a^2(t) \over 1 - k(x^2 + y^2 + z^2)}(dx^2 + dy^2 + dz^2)

So we automatically get a scale factor that depends upon time just by asking what the most general homogeneous, isotropic metric can be. It then becomes an exercise in math to determine what this metric does in General Relativity, and we are led inexorably to the Friedmann equations.

We should check the possibility that the variety of "metrics" you are creating may well be ways to map a stationary or comoving distribution of matter into a variety of homogeneous isotropic moving distributions of matter.

If so, there may be some compatibility between what we are each talking about, and I strongly suspect there is.

This is just a visualization of the curvature. General Relativity requires no extra dimensions to describe the curvature of space-time, but we can't very well visualize the curvature without artificially adding an extra dimension.

What happens in General Relativity, though, is that so-called "test particles" always follow paths that mark the shortest space-time distance between two points in space-time. These hypothetical test particles are objects which respond to the space-time curvature but don't affect it. They are a good approximation to reality whenever you're tracking the path of an object that is much less massive/energetic than the sources of the gravitational field it's traveling in.

In this area, I will not argue with you. When you're talking about local gravitational effects, I can entertain the idea of a non-constant metric. But it has to be a mapping from one view to another view--for instance the free-falling view, vs. the view from the ground, vs. the view from orbit, vs. the view from the center of the planet.

The variables must represent different physical quantities before and after the "metric" is applied.

I think the case has been made for the local effects of gravity, but from afar, all these local effects will simply manifest themselves as a slowing of the speed of light. All of the events can still be mapped to a Minkowskian global metric. The large scale global metric does not need to adjust for these modified light-like intervals, for we already have many examples of materials (glass, water, etc) slowing the speed of light.

Now, in flat space-time, the shortest path between any two events is always a straight line. This means that in flat space-time, objects always move with constant speed in a constant direction.

So when we see an object like the Moon orbiting the Earth, that means there is a massive departure from flat space-time surrounding the Earth: instead of going in a straight line, the Moon goes in a circle! This can be visualized as space-time being sort of a rubber sheet and the Earth providing an indentation on that sheet, an indentation which the Moon follows, but this is just a visualization because we simply can't visualize four-dimensional space-time curvature directly.

One thing that we know from General Relativity, however, is that the only way you can have flat space-time, which is the case for Minksowki/Milne space-time, is if the universe is empty. If you take the above homogeneous, isotropic metric, for example, the Milne metric pops out as the metric you get when you set the energy density of the universe to zero.

I'm pretty sure you are still applying the Friedman/Milne logic. In the Friedman/Milne model, you pretend that you don't need to worry about the relativity of simultaneity, because all the galaxies are comoving.

But remember, in the Minkowski/Milne model, we have already found a homogeneous, isotropic distribution of matter, without any change in "metric" at all. Since the distribution is isotropic, no matter how much matter or energy there is, it should all balance out--there's no net force in any direction, no matter how much "matter density" or "energy density" you have.

You have said the Milne model introduces an "explosion" which you find unaesthetic. But I think this is more aesthetically pleasing than what the standard model offers: In the standard model, everything in the universe appeared all at once, at t=0, uniformly distributed through space, all perfectly stationary with each other, but in a universe with a scale factor of zero.

So, instead of a single event creating all the matter in the universe, the standard model offers an infinite number of events, all occurring at the same time, at different places, but in the same place because the scale factor was zero.

Perhaps you find the point "explosion" idea unaesthetic, but do you really think it is more bizarre than the standard model's tiny infinite universe?

 

[JDoolin]

Now it turns out that you can punch in numbers to your computer programs in which the universe expands in exactly the same way that the Milne model says it should, and you find that the universe expands too quickly.

I need more detail here. Exactly how did they make this analysis that Milne's model universe would expand too quickly? Was this after or before they decided Milne's model had no matter in it?

The outer radius of the Minkowski/Milne's universe would expand at a speed of precisely the speed of light, though, as I've mentioned elsewhere, to an accelerating observer, the twin paradox manifests itself as universal inflation.

As for the local expansion, that would be determined, approximately, by an equipartition of rapidity, and the scale of the partition would be determined somehow by Planck's constant, and the mass of the primordial particles. If the size of those particles were extremely large, this velocity would be extremely low. I don't think you can say exactly how fast the Milne model would expand, unless you know the nature of the first particles, and how fast they moved away from each other.

In the context of the Minkowski/Milne model, what I would recommend is determine how fast the universe appears to expand, locally, and then, from that they could determine the size of this primordial particle.

In my original post, I said...

The reason I wish to modify the Milne model is to add two or three major events. These events are sudden accelerations of our galaxy or explosions of the matter around our galaxy, while the universe was still very dense, well before our galaxy actually spread out into stars.

The possibility had occurred to me that some of these events might be the quantum decay processes of gargantuan primordial particles.

 

[Chalnoth]

To me, claiming that the space is stretching represents a HUGE change in the physics. To me, claiming that Lorentz Transformations are not valid in cosmology represents a HUGE change in the physics.

Here's the thing: if you work with purely Newtonian gravity and work out how an expanding universe would behave, you get the same answer. So arguing against the expanding universe requires arguing that the behavior of gravity changes drastically on large distance scales. And we don't have any evidence of that.

What's more, we do have ample evidence against the Milne cosmology. Nobody is disagreeing that the Milne cosmology is different. It's just that the Milne cosmology is ruled out by observation.

But I frequently hear proponents of the "standard model" say that the effects of Special Relativity are only a local effect. (since all the galaxies are comoving, I gather, there is no time-dilation or desynchronization between the galaxies.) This is simply not true in the Milne-Minkowski model--where you must consider the relativity of simultaneity. This represents another HUGE change in the physics based on the metric.

Yes, because in General Relativity, you can use Minkowski space-time to describe the local region about any point. But if you try to apply special relativity globally, you start getting the wrong answers pretty quickly. Now, many of the same effects you see in Special Relativity still exist in General Relativity, it's just that the details differ. You may think of General Relativity only talking about effects due to the local galaxy, but in cosmology it also adds effects due to the intervening curvature between us and a far-away galaxy. Of course, you have to go very far because the cosmological curvature is very small, but when you get out to a few billion light years, the differences start to become significant.

Why is your goal to find a metric where homogeneity and isotropy are retained? Why don't you, instead, make the goal to find a distribution of matter in which homogeneity and isotropy are retained?

Well, if the distribution of matter obeys homogeneity and isotropy, then the particular solution to the Einstein equations must also obey the same symmetries. Thus we write down a metric that obeys homogeneity and isotropy in order to reduce the number of degrees of freedom, to make the system easier to solve. In this case, it reduces to a function of time (the scale factor) and constant parameter (the spatial curvature). The relationship between these and a homogeneous, isotropic matter distribution leads us, through the Einstein field equations, to the Friedmann equations.

This is what Milne already has found--a distribution of matter in Minkowski Space that is both homogeneous and isotropic. Isn't the only reason that Friedmann etc. continued to look for a "metric" because they erroneously denied that Milne's model was homogeneous and isotropic?

So? It's observationally wrong.

You have said the Milne model introduces an "explosion" which you find unaesthetic.

It's not "unaesthetic". It's observationally wrong.

 

[twofish-quant]

I need more detail here. Exactly how did they make this analysis that Milne's model universe would expand too quickly?

OK. Let's forget about a theory of gravity. You just give some equations telling me how you think the universe is behaving and then I run them through a simulation that just simulates the behavior of gas under the conditions that you gave me.

The three things that I can get out of that simulations are:

1) the composition of the universe from nuclear reaction rates
2) the lumpiness factors of the cosmic microwave background
3) the lumpiness factors of the galaxies

So let's have things expand at a constant rate, and let's not be concerned about how that happens. What you find is that the universe cools very quickly and so you end up without burning deuterium. The second thing that you find is that the sound waves travel further before they run into each other and so you end up with a universe that is much less lumpy.

The important thing about these things is that you are very limited as to the amount of weird physics that you can put in. Gas is gas. Nuclear reactions are nuclear reactions. What happens is that you put in all of the known physics, it doesn't work either. At that point you ask yourself what you have to do to get things to work, and you find that things work out if you put in just the right about of dark matter and dark energy.

In the context of the Minkowski/Milne model, what I would recommend is determine how fast the universe appears to expand, locally, and then, from that they could determine the size of this primordial particle.

If I'm understanding the Milne model, things are expanding at a constant rate, so you just take the current Hubble expansion and then assume that there is no slowdown.

 

[twofish-quant]

To me, claiming that the space is stretching represents a HUGE change in the physics.

Curiously the fact that space "bends" is something that you can test experimentally with spacecraft .

Anyone if you find GR weird as a theory of gravity and want to propose a new one, that's find. There is an entire industry of physicists proposing alternative theories of gravity. However, if you want to apply any new theory to the universe, you have to deal with the observational constraints that I've mentioned. You tell me how the universe expands, you push the numbers into your favorite nucleosynthesis and lumpiness factor code, and I tell you if that will work or not.

The two things that the standard models get right are the deuterium abundances and the existence of the first acoustic peak.

Why is your goal to find a metric where homogeneity and isotropy are retained? Why don't you, instead, make the goal to find a distribution of matter in which homogeneity and isotropy are retained?

Because you don't get the right deuterium abundances and the first acoustic peak.

You have said the Milne model introduces an "explosion" which you find unaesthetic. But I think this is more aesthetically pleasing than what the standard model offers: In the standard model, everything in the universe appeared all at once, at t=0, uniformly distributed through space, all perfectly stationary with each other, but in a universe with a scale factor of zero.

No it doesn't. The standard model of cosmology says *NOTHING* about what happened pre-inflation. I have to put this in bold because this is something people get wrong. With current observations you can get to the inflationary period, but what happened before is *NOT* part of the standard model.

So, instead of a single event creating all the matter in the universe, the standard model offers an infinite number of events, all occurring at the same time, at different places, but in the same place because the scale factor was zero.

No it doesn't. The standard model says *NOTHING* about how things behaved at t=0.