# Interpreting the Supernova Data

1. Sep 8, 2010

### JDoolin

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

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" [Broken]. 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.

Last edited by a moderator: May 4, 2017
2. Sep 9, 2010

### Chalnoth

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.

3. Sep 9, 2010

### phyzguy

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

4. Sep 9, 2010

### JDoolin

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

5. Sep 9, 2010

### JDoolin

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?

Last edited by a moderator: Apr 25, 2017
6. Sep 9, 2010

### Chalnoth

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.

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.

Last edited by a moderator: Apr 25, 2017
7. Sep 9, 2010

### George Jones

Staff Emeritus
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

8. Sep 9, 2010

### Chalnoth

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

9. Sep 9, 2010

### George Jones

Staff Emeritus
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.

10. Sep 9, 2010

### Chalnoth

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.

11. Sep 9, 2010

### JDoolin

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?

Last edited: Sep 9, 2010
12. Sep 9, 2010

### JDoolin

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.

Last edited: Sep 9, 2010
13. Sep 9, 2010

### JDoolin

That is simply not true.

14. Sep 9, 2010

### Chalnoth

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.

Last edited: Sep 9, 2010
15. Sep 10, 2010

### 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.

16. Sep 10, 2010

### Chalnoth

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.

17. Sep 10, 2010

### 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.

18. Sep 10, 2010

### 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, http://upload.wikimedia.org/wikipedia/en/f/f8/Milne_Model.jpg" [Broken] 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.

Last edited by a moderator: May 4, 2017
19. Sep 10, 2010

### Chalnoth

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.

20. Sep 10, 2010

### 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.

Last edited by a moderator: May 4, 2017
21. Sep 10, 2010

### JDoolin

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*}

I can see http://en.wikipedia.org/wiki/Stress-energy_tensor" [Broken] 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.

Last edited by a moderator: May 4, 2017
22. Sep 10, 2010

### JDoolin

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" [Broken]:

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

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" [Broken],

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.

Last edited by a moderator: May 4, 2017
23. Sep 10, 2010

### 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.

24. Sep 10, 2010

### JDoolin

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.)

Last edited: Sep 10, 2010
25. Sep 10, 2010

### 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.