# Interpreting the Supernova Data

by JDoolin
Tags: big bang, inflation, metric, milne, supernova
P: 4,782
 Quote by TrickyDicky 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.
P: 4,782
 Quote by JDoolin 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, (Direct Link) 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.
PF Gold
P: 706
 Quote by Chalnoth 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*}

 Quote by Chalnoth 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 here 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.
PF Gold
P: 706
 Quote by Chalnoth 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 here:

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

Standard Model

Milne Model

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 here,

 Quote by me 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.
P: 4,782
 Quote by JDoolin There are some cases where tensors can be used as matrices;
While they can be represented as matrices, they are different mathematical objects.

 Quote by JDoolin 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.
PF Gold
P: 706
 Quote by Chalnoth 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.)
 Sci Advisor P: 4,782 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.
PF Gold
P: 706
 Quote by 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.
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.
P: 4,782
 Quote by JDoolin 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.

 Quote by JDoolin 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.
PF Gold
P: 706
 Quote by Chalnoth 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)

 Quote by wikipedia article 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 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.
PF Gold
P: 706
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:

$$ds^2 = (cdt)^2 - dr^2$$

...which is the same metric that results from having no matter present.
P: 4,782
 Quote by JDoolin 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/Friedma...3Walker_metric

...with a(t) = t, and k = -1.
P: 3,010
 Quote by 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. 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?
P: 4,782
 Quote by TrickyDicky 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.
P: 6,863
 Quote by JDoolin 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.
P: 6,863
 Quote by JDoolin 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?
 P: 6,863 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.