# CMB Reference Frame

1. Jun 7, 2015

### Mysteryciel

I know that in physics we can change referance frame.The Newtonian Friedmann equation is
$H^2-{8πGp /3}={-k/a^2}$
I know that this equation derived from CMB referance frame.Now can we change this referance frame ?
(And If you can give me an article about this issue I will be very happy,I need exact proof)
Thanks

Last edited: Jun 7, 2015
2. Jun 7, 2015

### Staff: Mentor

In a different reference frame, the universe is not homogeneous and isotropic any more. The parameters H and a do not make sense there (unless you transform back to the CMB frame but then you gained nothing).

3. Jun 8, 2015

### RyanH42

How can you proof that idea ? Is there any article about that issue ?
I didnt understand something If universe is homogeneous and isotropic how can it be homogeneous and isotropic anymore when we change referance frame.
Homogeneous and isotropic things are relative ?
If its true Is there any picture which I can visulaize it ?
Thanks

4. Jun 8, 2015

### Staff: Mentor

Which one? The first statement is trivial.

If you are at rest relative to the atmosphere, it can be the same in every direction. If you move relative to the atmosphere, you feel wind coming from one direction. This is clearly non-isotropic.
The same applies to the CMB. If you move in one direction, you'll see photons coming from there blueshifted, while photons coming from the opposite side are redshifted.

5. Jun 8, 2015

### ChrisVer

That's too far stretched. The anisotropies are pretty small (of the order of the dipole CMB anisotropy we observe).

6. Jun 8, 2015

### RyanH42

Who is right ?

7. Jun 9, 2015

### ChrisVer

@mfb is right. But what I say is that the break of isotropy is very small and gave the order of it (dipole anisotropy of the CMB). The difference in temperatures we observe (the red/blue shift @mfb mentioned) due to our relative velocity to the CMB and so the corresponding anisotropy is of $\frac{\delta T}{T} \sim 10^{-3}$. That means that the "wind" you will feel is pretty weak... from the one side you measure a temperature 2.724K and from the other you measure 2.726K.

At the same time the relative velocity can affect the measuring of distances and so how we define the Hubble constant H: $v=cz=Hd$. The deviation >=5% from this equation due to some extra reasonable velocity addition (v~350km/s) is affecting close distance galaxies (5Mpc or less). For larger distances the deviation is even less than 5% and that's why I said that it's far stretched.

Last edited: Jun 9, 2015
8. Jun 9, 2015

### RyanH42

Is that small affect will change Newtonian Friedmann equation ($H^2-{8πGp /3}={-k/a^2}$) ?

Is there will be an extra term v in Newtonian Friedmann equation you said ?
Is there any article about "change in referance frame in Newtonian Friedmann Equation"

9. Jun 9, 2015

### ChrisVer

No it won't be so simple...the equation will just deviate a little from this form because the universe is not exactly isotropic and homogeneous, so the metric of the spacetime you will have to use must contain this information... This is not a Newtonian Friedmann Equation, although you might have come across its derivation from Newtonian dynamics arguments. The equation is a solution of the Einstein Field Equations (EFE) for the Friedmann Robertson Walker (FRW) metric. Since the metric will change a little (you can do that perturbatively I guess), this solution will also get some perturbative terms.

10. Jun 9, 2015

### RyanH42

Are you sure about that ?? In my equation $ρ=ρ_m+ρ_r+p_Λ$ so that equation must be true and I assumed $c=1$

The derivation is $1/2mv^2-MmG/r=U$ and $k=-U/2m$

The metric changes cause homogenity and isotrophy changes.And this affects FRW metric.Ok,but I want to keep things simple.So thats the reason I used NFE(Newtonian Friedmann Equation) NFE is much more simpler.This equation begins assuming homogenic and isotrophic universe.But we changed it.So our equation cannot be true anymore I understand.
Is there any book which I can learn these kind of things ?

11. Jun 9, 2015

### ChrisVer

The fact that the Newtonian arguments give you the right answer is just a coincidence. So it's not simple, it's oversimplification to the borders of false. Since this type of derivation is wrong, I don't think you can find any textbook that deals with this problem through Newtonian mechanics.

In newtonian mechanics, if you would like for your problem not to be isotropic you would have to change the things that you initially took isotropic. For example the masses should have some directional arguments and the gravitational potential would be changed (the masses also get some radial dependence) because of inhomogeneity.

12. Jun 9, 2015

### RyanH42

Why its wrong ? I think its good to proof the NFE.
I have textbook which do the same thing as me "derivation part".Maybe its not good to solve this problem using NFE equation but I think its enough to other solutions.

13. Jun 9, 2015

### ChrisVer

It's wrong because it gives the Friedmann equation by coincidence . The right way is to derive it from general relativity. The textbook you have probably is introductory or does not want to get into the details of derivations but talk about the results.
What you solved was a mechanical problem, and as such if you want to change the input you should expect different output.

14. Jun 9, 2015

### RyanH42

Ok,I understand.thanks.In general relativity referance frame will affect universe type or $Ω_k$ isnt it ?

And I want to thank you so much.I think you are best teacher (I dont now you are a teacher or not ).Thank you and thank you again .

15. Jun 9, 2015

### ChrisVer

I don't know... I guess the Friedman equations will need extra terms...

I'm not a teacher (thank god )

16. Jun 9, 2015

### RyanH42

I want to learn cosmology with GR Is there any book or video or pdf which I can study.? I know simple calculus. And what do you think about Leonard Susskind GR an cosmology lectures. I want to do make these kind of things
Thanks

17. Jun 9, 2015

### Staff: Mentor

There is no point in setting up a FLRW-like metric for a reference frame that moves relative to the CMB. It is possible, sure, but things are so much easier if you use the CMB rest frame and transform everything later to predict your actual observations.

18. Jun 9, 2015

### RyanH42

I have just curious what would be happen then..I learned NFE so I dont know GR -FRLW metric.I want to learn them so thats why I asked a textbook or something like that.

19. Jun 9, 2015

### Staff: Mentor

Which textbook?

20. Jun 10, 2015

### RyanH42

An Introduction to Modern Cosmology Andrew Liddle
If you search in google you will find pdf.You can see there.Chapter 3 book page 17

21. Jun 10, 2015

### ChrisVer

The GR problem is found in every GR textbook that contains cosmology.
Now it's just an exercise to get the RW metric from $ds^2 = dt^2 - a^2(t) \Big(\frac{dr^2}{1-kr^2} + r^2 d \theta^2 + r^2 \sin^2 \theta d \phi^2 \Big)$ and derive everything up to the Ricci scalar. It's better to move to comoving coordinates where the metric becomes $ds^2 = dt^2 - a^2(t) (dx^2 + dy^2 + dz^2)$ or $g_{\mu \nu} = diag(1, -a^2(t),-a^2(t), -a^2(t) )$.
The Einstein equations are then:
$R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} \mathcal{R} = 8 \pi GT_{\mu \nu} \Big(- \Lambda g_{\mu \nu} \Big)$
Where the $R_{\mu \nu}$ is the Ricci tensor, $\mathcal{R}$ is the Ricci scalar, $T_{\mu \nu}$ is the energy momentum tensor that takes the information of the matter content. In given coordinates it can be diagonal with elements the energy density $\rho$ and the momenta $p$:
$T_{\mu \nu} = diag( \rho, p , p ,p)$.

You can have a look in S. Dodelson's textbook "Modern Cosmology" (whole chapter 2.1) since he demonstrates some derivations.

22. Jun 10, 2015

### RyanH42

You made a great job.But I dont know metric or even basics GR look this https://www.physicsforums.com/threads/text-book-for-cosmology.818370/
Thank you.