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

[JDoolin]

I have been arguing the case that the universe may be a modified Milne model. Let me ask my questions first.

First, I am asking about graphs and data presented here:

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

Questions
1) The "binned data" appear to be points with constant redshift (y-axis), but with error-bars in the luminosity distance. (x-axis). How many supernovae make up each of the "bins." Better yet, is there a similar graph that simply shows one dot per data-point?
2) The remaining graphs on the page all refer to $\DELTA D M$. What is this quantity?
3) I believe this data appear http://arxiv.org/abs/0804.4142" . Is there a copy of this data anywhere in spreadsheet format?
4) Also, what are z, m(max B), s, c, mu? Are these sufficient to find Right Ascension, Declination, Luminosity Distance, and Redshift? Now, if you're curious about the modified Milne model:

In the Milne model, all of the universe explodes from a single event. A fixed "point" in space is stationary in only one reference frame. On the other hand, a fixed event can be the center of an expanding sphere in any and every reference frame.

The particles in the Milne universe are least dense in the center, and much more dense on the outside. Any observer in the Milne universe will be co-moving (but not necessarily co-located) with the center, in his own reference frame.

There is some argument that the Milne model can only represent an empty universe. I acknowledge that I have never understood this argument. Milne's own analysis was that there had to be an infinite amount of matter in the causally connected portion of the universe. The density of particles increases towards infinity at the outside edges.

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.

I found that an http://groups.google.com/group/sci.astro/msg/2751e0dc068c725c?hl=en" in the supernova data seemed consistent with this modified Milne Model.

 

[Chalnoth]

There is some argument that the Milne model can only represent an empty universe. I acknowledge that I have never understood this argument.

It's not that difficult. You take General Relativity with some reasonable choice of coordinates (e.g. spherical with constant curvature), set the stress-energy tensor to zero, and you get the Milne model.

 

[phyzguy]

No comment on your other questions, but DM refers to Distance Modulus, which is the difference between the apparent and absolute magnitudes.

 

[JDoolin]

No comment on your other questions, but DM refers to Distance Modulus, which is the difference between the apparent and absolute magnitudes.

Thank Goodness. With the references to the lambda-DM model, I was afraid DM stood for dark matter.

 

[JDoolin]

It's not that difficult. You take General Relativity with some reasonable choice of coordinates (e.g. spherical with constant curvature), set the stress-energy tensor to zero, and you get the Milne model.

Unfortunately, that's still word salad to me. I'm trying to get a handle on general relativity https://www.physicsforums.com/showthread.php?t=424686"

But, the analysis I'm doing so far is for an object undergoing constant acceleration, (or sitting on the ground). To understand this, I gather, you'd have to do a related analysis for an object in free-fall. Is your idea of "curvature" a description of the gravitational potential through space?

 

[Chalnoth]

Unfortunately, that's still word salad to me. I'm trying to get a handle on general relativity https://www.physicsforums.com/showthread.php?t=424686"

Well, to explain a bit more, the meat of General Relativity comes down to this equation (ignoring constants for clarity):
$$G_{\mu\nu} = T_{\mu\nu}$$

The right hand side of this equation is all about matter. Specifically, it's a tensor which includes contributions from energy density, momentum density, pressure, and shear of matter fields (shear is a twisting force). The left hand side of this equation is all about the curvature of space-time.

When you write down something like the Milne cosmology, what you're doing is specifying the format of the left hand side of this equation. But particular cosmology only works if you set the right hand side to zero, which means no matter (no energy, no momentum, no pressure).

Does that help any?

By the way, I'd also like to point out that you can exclude the Milne cosmology very easily by combining supernova observations with other cosmological observations.

But, the analysis I'm doing so far is for an object undergoing constant acceleration, (or sitting on the ground). To understand this, I gather, you'd have to do a related analysis for an object in free-fall. Is your idea of "curvature" a description of the gravitational potential through space?

In a way, kind of, yes, curvature is related to the gravitational potential. It's not quite that simple, but that's the general picture. Applying General Relativity to a situation as specific as that, though, would require a lot of work, and if you are a beginner in the field, I wouldn't even try just yet.

 

[George Jones]

There is some argument that the Milne model can only represent an empty universe. I acknowledge that I have never understood this argument.

It's not that difficult. You take General Relativity with some reasonable choice of coordinates (e.g. spherical with constant curvature), set the stress-energy tensor to zero, and you get the Milne model.

Unfortunately, that's still word salad to me.

Milne's universe is just an interesting coordinate system on a proper subset of Minkowski spacetime for special relativity. When Chalnoth says "constant curvature", he refers to the curvature of space. Since the Milne universe has no energy/matter content (and no dark energy), the Milne universe is a part of the flat spacetime of special relativity (zero spacetime curvature). For a demonstration of this, see

https://www.physicsforums.com/showthread.php?p=1757634#post1757634.

 

[Chalnoth]

Milne's universe is just an interesting coordinate system on a proper subset of Minkowski spacetime for special relativity. When Chalnoth says "constant curvature", he refers to the curvature of space. Since the Milne universe has no energy/matter content (and no dark energy), the Milne universe is a part of the flat spacetime of special relativity (zero spacetime curvature). For a demonstration of this, see

https://www.physicsforums.com/showthread.php?p=1757634#post1757634.

Actually, the Milne universe and special relativity are distinct, as the Milne universe has a constant spatial curvature.

 

[George Jones]

Actually, the Milne universe and special relativity are distinct, as the Milne universe has a constant spatial curvature.

The Milne universe is just a special coordinate patch on part of Minkowski spacetime. Spatial curvature of the Milne universe is just the intrinsic curvature of a particular 3-dimensional hypersurface in Minkowski spacetime, the hypersurface that results when one (Milne) coordinate is held constant.

For another similar example, use standard spherical coordinates for Minkowski space. Holding t and r constant results in a constant intrinsic curvature 2-dimensional surface in Minkowski spacetime, a sphere.

 

[Chalnoth]

The Milne universe is just a special coordinate patch on part of Minkowski spacetime. Spatial curvature of the Milne universe is just the intrinsic curvature of a particular 3-dimensional hypersurface in Minkowski spacetime, the hypersurface that results when one (Milne) coordinate is held constant.

For another similar example, use standard spherical coordinates for Minkowski space. Holding t and r constant results in a constant intrinsic curvature 2-dimensional surface in Minkowski spacetime, a sphere.

I guess that makes sense. It would have to work out that the total space-time curvature of a Milne universe is identically zero, because obviously the Minkowski metric is also a solution to an empty universe.

 

[JDoolin]

Well, to explain a bit more, the meat of General Relativity comes down to this equation (ignoring constants for clarity):
$$G_{\mu\nu} = T_{\mu\nu}$$

The right hand side of this equation is all about matter. Specifically, it's a tensor which includes contributions from energy density, momentum density, pressure, and shear of matter fields (shear is a twisting force). The left hand side of this equation is all about the curvature of space-time.

When you write down something like the Milne cosmology, what you're doing is specifying the format of the left hand side of this equation. But particular cosmology only works if you set the right hand side to zero, which means no matter (no energy, no momentum, no pressure).

Does that help any?

Not really. What do the matrices $$G_{\mu\nu} = T_{\mu\nu}$$ operate on? Do they input and output events, or do they input and output momentum vectors? What is the transform between? Transforming between what and what? The view from afar vs. the view nearby? The view in free-fall vs. the view from the ground? Ostensibly the matrices are four-by-four with either numbers or functions in each of the 16 positions, and should operate on explicit 1 by 4 vectors, which are also explicitly determined quantities.

I really can't fathom what the tensors are for, at all. The big question is, do they operate on momentum and energy of individual particles, or do they operate on the coordinate positions of events in space?

 

[JDoolin]

The Milne universe is just a special coordinate patch on part of Minkowski spacetime. Spatial curvature of the Milne universe is just the intrinsic curvature of a particular 3-dimensional hypersurface in Minkowski spacetime, the hypersurface that results when one (Milne) coordinate is held constant.

For another similar example, use standard spherical coordinates for Minkowski space. Holding t and r constant results in a constant intrinsic curvature 2-dimensional surface in Minkowski spacetime, a sphere.

Okay. Which way is the particle at (r,t) pulled by gravity? Toward the center, or away from the center, and why? (and by how much?)

As a hint, Milne claimed, a particle at rest (v=0) in this reference frame would be pulled toward the center. I do not recall how he reasoned this out, though. It was not entirely clear. I would have expected there to be no pull in either direction. Because a particle in the same position, but with v=r/t would be in the center, in its own reference frame, so would feel no such pull.

 

[JDoolin]

By the way, I'd also like to point out that you can exclude the Milne cosmology very easily by combining supernova observations with other cosmological observations.

That is simply not true.

 

[Chalnoth]

Not really. What do the matrices $$G_{\mu\nu} = T_{\mu\nu}$$ operate on? Do they input and output events, or do they input and output momentum vectors? What is the transform between? Transforming between what and what? The view from afar vs. the view nearby? The view in free-fall vs. the view from the ground? Ostensibly the matrices are four-by-four with either numbers or functions in each of the 16 positions, and should operate on explicit 1 by 4 vectors, which are also explicitly determined quantities.

These aren't matrices, and can't really be thought of in such terms. It's essentially a convenient method for expressing the connection between the curvature of space-time and matter. With the Milne cosmology, for instance, you are explicitly setting the curvature of space-time. So you can take the Milne cosmology as an input, and directly compute $G_{\mu\nu}$. It's not entirely trivial to do this, so I won't go into it here, but suffice it to say that when you do, you get zero.

Thus, by the Einstein field equations,

$$G_{\mu\nu} = T_{\mu\nu}$$
$$0 = T_{\mu\nu}$$

This means that every component of the stress-energy tensor is necessarily zero. Since the stress-energy tensor contains components related to energy density, momentum density, pressure, and stress, this means that energy density is zero, pressure is zero, momentum density is zero, and stress is zero. In other words, it's an empty universe.

If you want to get a slightly better idea of how you compute particle paths in a space time, what you need to understand is that the Einstein tensor $G_{\mu\nu}$ is a function of what is known as the metric, $g_{\mu\nu}$. The metric is a way of encoding the space-time distance along a path, and the motion of any particle is always the shortest space-time distance between its starting point and time and its ending point and time. For instance, if I consider a three dimensional metric with just one's along its diagonal, then this represents flat space (no time for now). This is equivalent to the equation:

$$ds^2 = dx^2 + dy^2 + dz^2$$

I can then use this metric to find the distance between any two points in flat space, and lo and behold, when I compute the shortest distance between any two points, I get a straight line.

 

[twofish-quant]

I'm not sure that the data will be useful to you. If you are looking for supernova *counts* rather than using them as standard candles, then I think the data will be useless since you can't easily remove the effects of stellar evolution or selection biases.

For example, you could have a lot of supernova at a certain distance because that happens to be the distance at which the first generation of stars blow up. Or not.

 

[Chalnoth]

That is simply not true.

I forgot to respond to this, but it is most definitely true that the Milne cosmology is completely ruled out by any combination of supernovae and any of the other major cosmological probes. WMAP data is a good example here.

 

[TrickyDicky]

$$G_{\mu\nu} = T_{\mu\nu}$$
$$0 = T_{\mu\nu}$$

Actyally these equations are used to get a vacuum solution, not the empty universe, so with them you can get the Schwarzschild metric for instance.

To get the Minkowski spacetime you actually have to make the Riemannian tensor vanish, not just the Einstein tensor.

 

[JDoolin]

I forgot to respond to this, but it is most definitely true that the Milne cosmology is completely ruled out by any combination of supernovae and any of the other major cosmological probes. WMAP data is a good example here.

As for the combination of supernovae and other cosmological probes, that's what I'm trying to determine. But as far as the WMAP data, Milne's model is certainly not ruled out.

http://en.wikipedia.org/wiki/Edward_Arthur_Milne

Click on the full-sized image, http://upload.wikimedia.org/wikipedia/en/f/f8/Milne_Model.jpg" and you'll see, in the very last line "The particles near the boundary tend toward invisibility as seen by the central observer, and fade into a continuous background of finite intensity."

Milne actually predicted a continuous background of finite intensity. He did not know precisely what wavelength it would be at. He did not know if we would ever have instruments sensitive enough to detect it. I imagine that to him, this prediction was actually an inconvenience, since it predicted something that had never been detected.

You're saying that the WMAP data rules out the Milne model, but to the contrary, the WMAP data resoundingly supports the Milne model. It is a long awaited vindication of the Milne model.

 

[Chalnoth]

Actyally these equations are used to get a vacuum solution, not the empty universe, so with them you can get the Schwarzschild metric for instance.

To get the Minkowski spacetime you actually have to make the Riemannian tensor vanish, not just the Einstein tensor.

While this is true, bear in mind that we're specifying the metric, not solving for it. The thing we're trying to solve for is the stress-energy tensor, not the metric or the Riemann tensor.

 

[Chalnoth]

As for the combination of supernovae and other cosmological probes, that's what I'm trying to determine. But as far as the WMAP data, Milne's model is certainly not ruled out.

http://en.wikipedia.org/wiki/Edward_Arthur_Milne

Click on the full-sized image, http://upload.wikimedia.org/wikipedia/en/f/f8/Milne_Model.jpg" and you'll see, in the very last line "The particles near the boundary tend toward invisibility as seen by the central observer, and fade into a continuous background of finite intensity."

Milne actually predicted a continuous background of finite intensity. He did not know precisely what wavelength it would be at. He did not know if we would ever have instruments sensitive enough to detect it. I imagine that to him, this prediction was actually an inconvenience, since it predicted something that had never been detected.

You're saying that the WMAP data rules out the Milne model, but to the contrary, the WMAP data resoundingly supports the Milne model. It is a long awaited vindication of the Milne model.

Er, no, it doesn't. The devil is in the details. From the Milne cosmology, you might be able to infer that there could potentially be a nearly-uniform background, but you cannot predict it would have a thermal spectrum, and you certainly can't predict that it would have anisotropies with the statistical properties that we observe.

 

[JDoolin]

These aren't matrices, and can't really be thought of in such terms.

There are some cases where tensors can be used as matrices; for instance

$$\begin{align*} ds^2&= (cdt, dx, dy, dz)\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix}cdt\\dx\\dy\\dz \end{pmatrix} \\ &= c^2 dt^2 - dx^2 - dy^2 - dz^2 \end{align*}$$

This means that every component of the stress-energy tensor is necessarily zero. Since the stress-energy tensor contains components related to energy density, momentum density, pressure, and stress, this means that energy density is zero, pressure is zero, momentum density is zero, and stress is zero. In other words, it's an empty universe.

I can see http://en.wikipedia.org/wiki/Stress-energy_tensor" that the tensor somehow relates to energy density, energy flux, shear stress, momentum density, and momentum flux.

These numbers along the diagonal (pressure and energy density) are not zero when you consider a gravity free region, but 1 or -1. I think, perhaps you are thinking of the elements in the Tensor as magnitudes, when you should be thinking of them as ratios.

 

[JDoolin]

Er, no, it doesn't. The devil is in the details. From the Milne cosmology, you might be able to infer that there could potentially be a nearly-uniform background, but you cannot predict it would have a thermal spectrum, and you certainly can't predict that it would have anisotropies with the statistical properties that we observe.

Don't look to the devil for details. :) He lies.

It is fairly well supported that this thermal spectrum is predicted from the phenomenon of "hydrogen recombination." The "surface of last scattering" is completely consistent with the Milne Model. The only difference is, in the Milne model, that surface is receding (as in all of the dx/dt is due to velocity, rather than any stretching of space effects.) If I understand correctly, the surface, as described in the standard model is simply "popping into the observable universe" due to the light from the surface just now overtaking the stretching of space, as described http://www.astro.ucla.edu/~wright/photons_outrun.html" :

Maybe I can put the two animations right next to each other:

Standard Model
[URL]http://www.astro.ucla.edu/~wright/cphotons.gif[/URL]

Milne Model
[PLAIN]http://upload.wikimedia.org/wikipedia/en/2/24/2d_milne_dist.jpg

Unfortunately I don't have this one animated, but the idea is that the outside particles can move away only at the speed of light; and within the Milne model, we don't consider the region beyond $$x\geq c t$$

But the point is, you would still have the black-body spectrum either way, because it would be coming from the same phenomenon.

As for the dipole anisotropy, I did a rough calculation of that http://www.spacetimeandtheuniverse.com/space-time-universe/2446-expanding-spacetime-relative-speed.html" ,

The actual temperature of this "surface of last scattering" is estimated to be about 3000 Kelvin, so the time dilation factor (gamma) ranges from 1098.3 (=3000/2.7315) to 1101.1 (=3000/2.7245)

As for the "other anisotropies" we can simply assume that the universe is a little bit lumpy; The surface of last scattering has varying velocities as you look around the sphere.

 

[Chalnoth]

There are some cases where tensors can be used as matrices;

While they can be represented as matrices, they are different mathematical objects.

These numbers along the diagonal (pressure and energy density) are not zero when you consider a gravity free region, but 1 or -1. I think, perhaps you are thinking of the elements in the Tensor as magnitudes, when you should be thinking of them as ratios.

No, they're zero. The Einstein tensor ($G_{\mu\nu}$) is computed from derivatives of the metric ($g_{\mu\nu}$). So when you use the metric for Minkowski space-time (or Milne space-time), the Einstein tensor is exactly zero.

 

[JDoolin]

While they can be represented as matrices, they are different mathematical objects.No, they're zero. The Einstein tensor ($G_{\mu\nu}$) is computed from derivatives of the metric ($g_{\mu\nu}$). So when you use the metric for Minkowski space-time (or Milne space-time), the Einstein tensor is exactly zero.

The devil does seem to be in the details. Let's see if we can extract the little monster.

We must be talking about two different things. One of them is a tensor which simplifies to a diagonal matrix {-1, 1, 1, 1} in the absence of gravity, while the other simplifies to a tensor of all zeros in the absence of gravity.

The first one is the one that actually affects the metric. The first tensor operates on differential event-intervals.

But the second, if I'm not mistaken, "the Einstein Tensor," is a tensor designed to operate on a momentum four-vector.

Edit: (I see now, what you're saying. One is the derivative of the other. But like velocity is a property of a particle, but distance is a property of space, I think the same argument should be made here. The Einstein Tensor does not affect the scale of space.)

 

[Chalnoth]

The Einstein tensor doesn't operate on anything. It is a particular mathematical representation of curvature, specifically the component of curvature that couples to matter (that is, from the Einstein tensor, you can constrain the properties of the matter distribution, and vice versa). In any location where the Einstein tensor is zero, there is no matter. And the Milne cosmology has an Einstein tensor equal to zero everywhere.

 

[JDoolin]

The Einstein tensor doesn't operate on anything. It is a particular mathematical representation of curvature, specifically the component of curvature that couples to matter (that is, from the Einstein tensor, you can constrain the properties of the matter distribution, and vice versa). In any location where the Einstein tensor is zero, there is no matter. And the Milne cosmology has an Einstein tensor equal to zero everywhere.

Hmmm. Trust but verify. What is needed to calculate the Einstein Tensor? The density function? The density function for the Milne model is:

$$n dx dy dz = \frac{B t dx dy dz}{c^3 \left(t^2-\frac{x^2+y^2+z^2}{c^2}\right)^2}$$​

Where n is the density at (x,y,z,t) and B is the density at (0,0,0,t)

Case 1) That density function creates an Einstein tensor equal to zero everywhere. In which case you actually can have matter along with a zero Einstein Tensor.

Case 2) That density function creates some other Einstein tensor... in which case the Milne cosmology has an Einstein tensor which is NOT equal to zero everywhere.

Either way, something has to give.

 

[Chalnoth]

Hmmm. Trust but verify. What is needed to calculate the Einstein Tensor? The density function?

Just the metric. You can see the Milne metric here:
http://en.wikipedia.org/wiki/Milne_model

Generally what you do first is compute the Christoffel Symbols($\Gamma_{ab}^c$), which are derivatives of the metric, then from the Christoffel symbols you compute the Riemann curvature tensor ($R^a_{bcd}$), then the Einstein tensor is computed from that.

The main issue here is that the Christoffel Symbols are rank 3 objects, and thus it can be functionally difficult to keep track of them (it's very, very easy to make mistakes here), and the Riemann curvature tensor is a rank 4 tensor.

Case 1) That density function creates an Einstein tensor equal to zero everywhere. In which case you actually can have matter along with a zero Einstein Tensor.

Case 2) That density function creates some other Einstein tensor... in which case the Milne cosmology has an Einstein tensor which is NOT equal to zero everywhere.

Either way, something has to give.

Milne's density function only works if the density is zero everywhere. Otherwise, the only possible way that this density function can be accurate is if General Relativity is wrong, so you'd have to propose an entirely different theory of gravity.

 

[JDoolin]

Just the metric. You can see the Milne metric here:
http://en.wikipedia.org/wiki/Milne_model

I fear someone has played a little joke on Wikipedia. (Misner, Thorne and Wheeler, perhaps)

Setting the spatial curvature and speed of light to unity the metric for a Milne universe can be expressed with hyperspherical coordinates as:

$$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\$$​

is the metric for a two-sphere.

The http://en.wikipedia.org/wiki/Two_Sphere_Universe." is not the Milne model. One should not start an analysis of any model with the assumption that the entire universe is rotating.

In any case, how can you start with the metric to figure out the metric? Surely you have to start with the distribution of matter, and derive the metric.

 

[JDoolin]

It appears someone has gone out and deleted the wikipedia article about the two-sphere universe since I posted the link this morning. I still had the screen up in another window, so I'll re-post what was on the page when I saw it.

This article is an orphan, as few or no other articles link to it. Please introduce links to this page from related articles; suggestions are available. (February 2009)
It has been suggested that this article or section be merged into Celestial spheres. (Discuss)
This article does not cite any references or sources.
Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. (September 2007)

The "Two Sphere Universe" is a term coined by Thomas Samuel Kuhn, in his book The Copernican Revolution. It was a conception of the astronomical heavens by ancient Greece. For most Greek astronomers and philosophers, from the 4th century BC on, the Earth was suspended at the geometric center of a much larger rotating sphere which carried the stars. The sun moved in the vast space between the Earth and the sphere of the stars. Outside of the outer sphere there was nothing at all. These two spheres (earthly and celestial) comprise the two-spheres, in the "two-sphere universe". This was not the only theory of the universe, but it was the one that gained the most adherents. It was this developed version of this theory that the medieval world inherited from the ancients.

Retrieved from "http://en.wikipedia.org/wiki/Two_Sphere_Universe"

As far as the actual two-sphere discussed by Misner, Wheeler, Thorne, (and erroneously attributing it to the Milne model) it makes the angular scale of space a function of a radial distance from the center, just as though whatever object you're looking at is length contracted as though it were traveling at a velocity caused by spinning around the center at radius r.

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

The two-sphere rotation metric described by the equation given above, has nothing to do with the Milne model. Milne model, if given in spherical coordinates, should have the metric:

$$ds^2 = (cdt)^2 - dr^2$$​

...which is the same metric that results from having no matter present.

 

[Chalnoth]

The two-sphere rotation metric described by the equation given above, has nothing to do with the Milne model. Milne model, if given in spherical coordinates, should have the metric:

$$ds^2 = (ct)^2 - r^2$$​

...which is the same metric that results from having no matter present.

There is more than one metric that one can write down that is consistent with having no matter present, but this is not one of them. What you have written isn't even a metric.

The metric given in the Wikipedia article, however, is accurate. This is just a special case of the FLRW metric in hyperspherical coordinates:
http://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

...with a(t) = t, and k = -1.

 

[TrickyDicky]

It appears someone has gone out and deleted the wikipedia article about the two-sphere universe since I posted the link this morning. I still had the screen up in another window, so I'll re-post what was on the page when I saw it.



As far as the actual two-sphere discussed by Misner, Wheeler, Thorne, (and erroneously attributing it to the Milne model) it makes the angular scale of space a function of a radial distance from the center, just as though whatever object you're looking at is length contracted as though it were traveling at a velocity caused by spinning around the center at radius r.

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

The two-sphere rotation metric described by the equation given above, has nothing to do with the Milne model. Milne model, if given in spherical coordinates, should have the metric:

$$ds^2 = (ct)^2 - r^2$$​

...which is the same metric that results from having no matter present.

You got it all wrong here, Milne's universe metric is the one given in wikipedia.

BTW, I wonder why is generally said that this metric is simply the Minkowski metric with a change of coordinates (from flat to hyperbolic) when it has another difference: the Milne metric has a scale factor=t, while minkowski spacetime is static. Or am I missing something?

 

[Chalnoth]

BTW, I wonder why is generally said that this metric is simply the Minkowski metric with a change of coordinates (from flat to hyperbolic) when it has another difference: the Milne metric has a scale factor=t, while minkowski spacetime is static. Or am I missing something?

I thought so too, but then I realized that the fact that the Milne cosmology is perfectly empty indicates that curvature must vanish means it has to just be Minkowski space-time in different coordinates. So I worked through it, and it turns out that if you use the following coordinate substitution:

$$r' = t \sinh (r)$$
$$t' = t \cosh (r)$$

...you can show that the two metrics are identical. This is done by identifying $r'$ and $t'$ as the Minkowski coordinates, and $r$ and $t$ as the Milne coordinates. The angular coordinates are the same in either case.

 

[twofish-quant]

Milne actually predicted a continuous background of finite intensity. He did not know precisely what wavelength it would be at. He did not know if we would ever have instruments sensitive enough to detect it. I imagine that to him, this prediction was actually an inconvenience, since it predicted something that had never been detected.

CMB was discovered in 1965. The big discovery of COBE was that the cosmic background radiation was not constant. What's WMAP has done is to give us high precision measurements of how the CMB varies.

 

[twofish-quant]

Don't look to the devil for details. :) He lies.

You are in the wrong game then. I can make any model work if I don't care about the details. It's getting the details right that's a pain.

The only difference is, in the Milne model, that surface is receding (as in all of the dx/dt is due to velocity, rather than any stretching of space effects.) If I understand correctly, the surface, as described in the standard model is simply "popping into the observable universe" due to the light from the surface just now overtaking the stretching of space

You understanding is wrong. The way that the standard model handles electron recombination is exactly the same as how the Milne model does.

But the point is, you would still have the black-body spectrum either way, because it would be coming from the same phenomenon.

Right.

As for the "other anisotropies" we can simply assume that the universe is a little bit lumpy; The surface of last scattering has varying velocities as you look around the sphere.

So can you with the Milne model calculate the exact spectrum of the lumpiness. With lamba-CDM, I can put in some parameters and get a fit to get the lumpiness.

Can you do that?

 

[twofish-quant]

Also, I think you can get the Milne model as a subset of the standard model. If you set all of the densities to zero in the standard cosmology, what you get is the Milne model.