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Interpreting the Supernova Data 
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#19
Sep1010, 07:44 AM

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#20
Sep1010, 07:47 AM

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#21
Sep1010, 09:11 AM

PF Gold
P: 706

[tex]\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*}[/tex] 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. 


#22
Sep1010, 10:45 AM

PF Gold
P: 706

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 [tex]x\geq c t[/tex] But the point is, you would still have the blackbody 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, 


#23
Sep1010, 01:42 PM

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#24
Sep1010, 02:31 PM

PF Gold
P: 706

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 eventintervals. But the second, if I'm not mistaken, "the Einstein Tensor," is a tensor designed to operate on a momentum fourvector. 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.) 


#25
Sep1010, 02:42 PM

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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.



#26
Sep1010, 06:46 PM

PF Gold
P: 706

[tex]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}[/tex] 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. 


#27
Sep1010, 09:35 PM

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http://en.wikipedia.org/wiki/Milne_model Generally what you do first is compute the Christoffel Symbols([itex]\Gamma_{ab}^c[/itex]), which are derivatives of the metric, then from the Christoffel symbols you compute the Riemann curvature tensor ([itex]R^a_{bcd}[/itex]), 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. 


#28
Sep1110, 07:44 AM

PF Gold
P: 706

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. 


#29
Sep1110, 11:01 AM

PF Gold
P: 706

It appears someone has gone out and deleted the wikipedia article about the twosphere universe since I posted the link this morning. I still had the screen up in another window, so I'll repost what was on the page when I saw it.
[tex]ds^2 = dt^2t^2(dr^2+\sinh^2{r} d\Omega^2)\ [/tex] The twosphere 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: [tex] ds^2 = (cdt)^2  dr^2 [/tex] ...which is the same metric that results from having no matter present. 


#30
Sep1110, 12:39 PM

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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. 


#31
Sep1110, 02:06 PM

P: 3,010

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? 


#32
Sep1110, 02:49 PM

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P: 4,782

[tex]r' = t \sinh (r)[/tex] [tex]t' = t \cosh (r)[/tex] ...you can show that the two metrics are identical. This is done by identifying [itex]r'[/itex] and [itex]t'[/itex] as the Minkowski coordinates, and [itex]r[/itex] and [itex]t[/itex] as the Milne coordinates. The angular coordinates are the same in either case. 


#33
Sep1110, 11:10 PM

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#34
Sep1110, 11:15 PM

P: 6,863

Can you do that? 


#35
Sep1110, 11:18 PM

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.



#36
Sep1210, 01:41 AM

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