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Interpreting the Supernova Data

 Quote by Chalnoth Don't ask me. These are just the coordinates for the Milne cosmology, which we know is wrong, so I don't see any reason to look into it further.
Except for the fact that they are not the scale for the Milne cosmology. Someone lied. Milne used the same metric as Minkowski.

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 Quote by JDoolin I don't want to lose the main question of my original post. So far, no-one has given me an answer to the questions of my original post. I hoped I could find someone who knew how to obtain the following data about the supernovae . a) right ascension b) declination c) luminosity distance d) redshift
Right ascension and declination are coordinates on the sky. Declination is degrees from the equator as seen from Earth, with negative values used for stars in the southern hemisphere. Right ascension is a value given in hours, typically using the J2000 standard epoch. That is, the hour gives the hour of the day on Jan. 1, 2000 where the star would have been directly overhead.

These values are typically listed for each supernova when a supernova experiment's results are reported. For instance, here is the SNLS first-year release:
http://xxx.lanl.gov/abs/astro-ph/0510447

The table including the supernovae is at the end. The redshift of each supernova is measured by careful measurement of its spectrum. The magnitudes of each supernova are also listed, and the relationship to the luminosity distance is described in the paper.

 Quote by JDoolin I disagree. The standard model cosmology does not have that good of a track record. One need only skim through an article about dark energy to see that it introduces more questions than it answers.
Introducing more questions than it answers is the norm in science, though. This is what learning new things is all about! The fact remains that the parameters in Lambda-CDM are measured to be the same to within experimental errors no matter which set of cosmological observations we use.

 Quote by JDoolin For instance, in the Milne model, inflation arises very naturally from acceleration. You take a sphere expanding at the speed of light, and perform a Lorentz Transformation around any event after the Big Bang. All you need is collisions or explosions to get the necessary Delta V.
No, this won't work, because you won't get the right expansion rate, and you definitely won't come remotely close to a nearly-homogeneous universe.

Look, you can't just say, "Hey, look, in this model the words I use make it sound sort of like it works like this other model! Therefore they're the same!" This isn't the way science works. You have to dig into the details of the model and see what it actually says, not just look at superficial behavior.

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 Quote by Chalnoth The parameters in Lambda-CDM are: 1. Normal matter density. 2. Dark matter density (assumed to be zero-temperature). 3. Cosmological constant value. 4. Scalar spectral index (an inflation parameter). 5. Optical distance to the surface of last scattering. 6. Hubble constant.
I'm apologize. I have never seen that analysis. I do recall seeing an article which essentially did a curve fitting of the data to some Bessel function, and they were surprised at how big the 17th coefficient or something other was.

In the case that you have a model with these six parameters, then I have to admit that is more interesting.

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 Quote by JDoolin I'm apologize. I have never seen that analysis. I do recall seeing an article which essentially did a curve fitting of the data to some Bessel function, and they were surprised at how big the 17th coefficient or something other was.
Yes, this has been done. This sort of analysis doesn't interest most scientists.

 Quote by JDoolin In the case that you have a model with these six parameters, then I have to admit that is more interesting.
I'm glad. You can see how these parameters affect the power spectrum here:
http://space.mit.edu/home/tegmark/cmb/movies.html

Note that these animations were made before WMAP, and the data on the plot is a combination of pre-WMAP data. He also adds the effects of a couple of extra parameters that are not in the simplest-case analysis, so you can see what they do.

 Quote by JDoolin However, why is the frequency of the light redshifted? Why is it that this light can still be seen? In the standard model, the the events happened long ago, (the plane of last scattering is no longer happening anywhere in the universe) but the light is just now reaching us. But in the Milne model the redshift is entirely due to the recession velocity of the plane of last-scattering. And the fact that this plane can still be seen is due to time-dilation. The events are still happening; the plane of last-scattering is still there, moving away from us at nearly the speed of light.
Six of one. Half dozen of the other. In GR those explanations are quite equivalent. You can set your coordinate system so that time moves at a constant rate and space is expanding, or you can set your system so that space is fixed and time is slowing down. The really are the same explanation.

 On your other detail, you are asking "Can the Milne model predict the lumpiness." Of course not! We can see it, and perhaps hope to find an explanation for it, but the idea that you could or should predict the lumpiness of the CMBR from a metric is mere conceit.
Lamba-CDM predicts the lumpiness of the universe. Once you have a figure for how quickly the universe expands and you make some assumptions about what the universe is made of, then you can rather easily calculate how pressure waves go through the universe, and get lumpiness coefficients.

You can do the same calculation with the Milne model. People have done that calculation and you get nowhere near the observed universe.

 If you have a map, you can develop a real theory: With WMAP and COBE, for instance, they could see that the light was a thermal spectrum, and then they were able to come up with a phenomenon that caused it.
Other way around. The CMB was discovered in 1965. The calculations for how to calculate the lumpiness factor of the universe were done in the 1980's. At that point, you have a group of scientists go to Congress and then lobby for money to send up a spacecraft designed to measure the lumpiness of the universe.

 Quote by JDoolin And why would you make that replacement?
Makes the math simpler.

 Time is not flowing slower as you go out from the center. Distance is not getting smaller as you go out from the center.
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 when the particles are moving away from one another.
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.

 Quote by Chalnoth Introducing more questions than it answers is the norm in science, though. This is what learning new things is all about! The fact remains that the parameters in Lambda-CDM are measured to be the same to within experimental errors no matter which set of cosmological observations we use.
Coming up with useful questions is more important than coming up with answers. Once you come up with the questions, you then build instruments to answer those questions.

 Look, you can't just say, "Hey, look, in this model the words I use make it sound sort of like it works like this other model! Therefore they're the same!" This isn't the way science works. You have to dig into the details of the model and see what it actually says, not just look at superficial behavior.
In particular, cosmic inflation is defined as some dark energy field dumping energy that increases the expansion rate of the universe. It's really inconsistent with the Milne cosmology by definition. In particular, in most inflationary models, the expansion of the universe was exponential and approaches the de Sitter model.
 When I apply this transformation to the minkowski metric: $$r \to t \sinh r$$ $$t \to t \cosh r$$ I don't get this $$ds^2 = dt^2-t^2(dr^2+\sinh^2{r} d\Omega^2)\$$ But this $$ds^2 = dt^2-t^2(\sinh^2{r} d\Omega^2)\$$ I lose the dr^2 term, what am I doing wrong?

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 Quote by TrickyDicky When I apply this transformation to the minkowski metric: $$r \to t \sinh r$$ $$t \to t \cosh r$$ I don't get this $$ds^2 = dt^2-t^2(dr^2+\sinh^2{r} d\Omega^2)\$$ But this $$ds^2 = dt^2-t^2(\sinh^2{r} d\Omega^2)\$$ I lose the dr^2 term, what am I doing wrong?
Perhaps a sign error? The derivative of the hyperbolic sine and cosine are both positive. Maybe you accidentally made the derivative of cosh negative? Alternatively, maybe it comes from not carrying the product rule through for both $dr$ and $dt$?
 From the Minkowski metric with this coordinate transform $$r \to t \sinh r$$ $$t \to t \cosh r$$ Here's what I get since: $$\cosh^2{r}-\sinh^2{r}=1$$ $$ds^2 = dt^2\cosh^2{r}-dt^2\sinh^2{r}-t^2\sinh^2{r} d\Omega^2 =dt^2(\cosh^2{r}-\sinh^2{r})-t^2\sinh^2{r} d\Omega^2=dt^2-t^2(\sinh^2{r} d\Omega^2)$$

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 Quote by TrickyDicky From the Minkowski metric with this coordinate transform $$r \to t \sinh r$$ $$t \to t \cosh r$$ 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.

$$r \to t \sinh r$$

$$dr \to \sinh r dt + t \cosh r dr$$

I'm sure you can figure out the rest.

 Quote by Chalnoth 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

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 Quote by TrickyDicky 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...

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 Quote by twofish-quant 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:

$$\begin {matrix} z=r cos \theta \\ x=r sin \theta cos \phi\\ y=r sin \theta sin \phi \end {matrix}$$

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:

$$\begin{bmatrix} c t' \\ x' \end{bmatrix} = \begin{bmatrix} \cosh\phi &-\sinh\phi \\ -\sinh\phi & \cosh\phi \\ \end{bmatrix} \begin{bmatrix} c t \\ x \end{bmatrix}$$

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 $(x,t) \to (x',t')$ is mathematically and physically valid. However, when you say:

$$\begin {matrix} t\to t \cosh r\\ r \to t \sinh r \end {matrix}$$

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:

$$\begin {matrix} t' \to t \cosh r\\ r' \to t \sinh r \end {matrix}$$

...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

$$\begin {matrix} t\to t \cosh r\\ r \to t \sinh r \end {matrix}$$

... 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.
 Recognitions: Science Advisor 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.

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

I have no resistance to saying

$$\begin {matrix} t' \to t \cosh r\\ r' \to t \sinh r \end {matrix}$$

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

But, what you are claiming, is:
$$\begin {matrix} t \to t \cosh r\\ r \to t \sinh r \end {matrix}$$

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

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 Quote by JDoolin I pointed out a genuine flaw in your reasoning, and you dismissed it by pointing and laughing. I have no resistance to saying $$\begin {matrix} t' \to t \cosh r\\ r' \to t \sinh r \end {matrix}$$ ...if you can explain what you mean by t, t', r, r'. But, what you are claiming, is: $$\begin {matrix} t \to t \cosh r\\ r \to t \sinh r \end {matrix}$$ This is nonsense, and it certainly wasn't what Milne ever meant.