# Newtonian Friedmann Equation, Referance frame, Homogeneity

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1. Jul 13, 2015

### RyanH42

Hi all
I want to ask a question about NFE(Newtonian Friedmann Equation).I know that NFE is not usefull to describe universe.But we can have a general idea about universe to use that formula.

I know that the only spacial coordinate system is CMB referance frame and NFE is derived from that point(taking RF(Referance Frame) CMB).

My question,
Is Universe homogeneius and isotrophic somewhere else ?
I mean Is there any other point which we can say "Universe homogenic and isotrophic" ?

If theres any other point, and we want to calculate NFE(Referance frame will be our new point) it will be also same as NFE (CMB referance frame taken) isnt it ?

2. Jul 13, 2015

### Staff: Mentor

Do you have a reference that describes what this is?

3. Jul 13, 2015

### RyanH42

4. Jul 13, 2015

### Staff: Mentor

These references talk about the Friedmann equation, yes. But you said "Newtonian Friedmann Equation". Where does that term come from? If you are basing it on the fact that the hyperphysics page says "a simplified, non-relativistic version based on Newton's laws", note that the page is incorrect in that statement, as far as I can see: the equations it gives are the correct relativistic ones for cold matter, i.e., matter whose pressure is negligible, and they are derived from Einstein's Field Equation, not from Newton's laws.

5. Jul 13, 2015

### RyanH42

I said like that cause as you said I thought it derived from Newtons law.But I see it wrong what would be the right term to describe that equation NRFE(non-relativistic friedmann equation)?

6. Jul 13, 2015

### Staff: Mentor

The right term is "Friedmann equations". There's no such thing as a "non-relativistic" vs. a "relativistic" version. All that changes is what assumptions you make about the energy density $\rho$, the pressure $p$, the curvature constant $k$, and the cosmological constant $\Lambda$ that appear in the equations. If you assume that $p = 0$ and $\Lambda = 0$ (which is what the hyperphysics page you linked to does), that doesn't mean you're using a "non-relativistic" version; it just means you've assumed that those quantities are zero (or at least negligible) for the particular scenario you're analyzing.

7. Jul 13, 2015

### RyanH42

How can $p=0$ ? $p=wcρ$ isnt it so you mean $p_m+p_r=0$ ?
The other things are clear thanks for that.
So what about my questions ?

8. Jul 13, 2015

### Staff: Mentor

If ordinary matter has a sufficiently low temperature, its pressure is negligible in comparison with its energy density, so it can be assumed to be zero as a very good approximation. On cosmological scales, the ordinary matter in our universe meets this criterion.

Yes, and if $w = 0$, then $p = 0$. For ordinary "cold" matter, $w = 0$; that's what I was describing above. For radiation, $w = 1/3$, and for dark energy, $w = -1$ (at least in the simplest form of a cosmological constant).

They appeared to me to depend on your belief that the "Newtonian Friedmann equation" was something different from the ordinary one. Since it isn't, I would recommend rethinking your questions and, if necessary, asking them again.

9. Jul 13, 2015

### RyanH42

Ok I understand why $p=0$ thanks again.
The derivation of Friedmann Equation is different.Here the derivation
$1/2mV^2-mMG/r=U$
$V^2-2MG/r=2U/m$
$H^2R^2-8πGR^2/3=-k$
$H^2-8πG/3=-k/R^2$
This derivation is wrong I guess.But this is the way that I learn.The problem is This works only when universe is homogeneius and isotrophic.But this situation exist only CMB referance frame.Is there any other referance frame which universe is still homogeneius and isotrophic.So I can use this equation again.To desribe universe.

I dont know FR works all referance frames or only CMB frame(I think it works only CMB frame)

If my derivation is wrong whats the right one ? The answer will be GR but My derivation is also true isnt it ? It cant be coincidence

10. Jul 13, 2015

### Staff: Mentor

Where is this from?

11. Jul 13, 2015

### Staff: Mentor

Homogeneity and isotropy are properties of spacelike slices of the universe, not the universe as a whole. There is only one family of spacelike slices that have these properties: they are the spacelike slices of constant time in the usual FRW coordinates. But the slices having those properties is not a function of the coordinates; they would have those properties in any coordinates you choose. They just wouldn't be slices of constant time in other coordinates. (The main reason FRW coordinates are chosen is to make those slices be slices of constant coordinate time, to make the math simpler.)

The Einstein Field Equations, which is what the Friedmann equations are derived from, work in any frame. But they don't necessarily look equally simple in every frame. You could do cosmology in some other frame besides the usual FRW coordinates, but the math would be more complicated.

12. Jul 13, 2015

### RyanH42

These pics from a book Andrew Little An Introduction to Modern Cosmology and Leonard Susskind Cosmology Lectures

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13. Jul 13, 2015

### RyanH42

I understand that we dont need any homogeneity or isotrophic universe to derive Friedmann equation.But the math will be complicated

14. Jul 13, 2015

### Staff: Mentor

Note the statement early on: "The Newtonian derivation is, however, some way from being completely rigorous." That's a fancy way of saying it's not really valid. As the book notes: "general relativity is required to fully patch it up", i.e., to do a valid derivation. I would guess that the reason the book chose to do things this way is that the author(s) thought Newtonian gravity would be more familiar to the reader than GR, so they gave a derivation that, while not valid, at least would make the Friedmann equation seem plausible (as long as you're willing to accept an invalid derivation as a heuristic argument).

15. Jul 13, 2015

### RyanH42

Ok.I got the idea thanks.I need to learn GR immidiatly.

16. Jul 13, 2015

### RyanH42

We need homogenity and isotrophy but the time slices will be different.So that makes the rquation complicated.Am I right ?

Last edited: Jul 13, 2015
17. Jul 13, 2015

### Chalnoth

Correct, it will be more complicated if you don't have homogeneity or isotropy. In fact, I don't think anybody has figured out how to produce an exact solution to Einstein's equations without using isotropy as an assumption*, and homogeneity adds another symmetry that can be used to reduce the number of necessary parameters further.

Once you have the solution, it's possible to translate any results you want into any coordinate system you choose.

* Edit: Well, I suppose rotating systems such as the Kerr metric may be an exception, depending upon how you define isotropy (the Kerr metric is isotropic far from the origin).

Last edited: Jul 13, 2015
18. Jul 13, 2015

### RyanH42

Thank you Chalnoth and PeterDonis.I get my answers for all questions.

19. Jul 15, 2015

### RyanH42

I am confused about referance frame.Whats the "job" of referance frame in Friedmann Equation ? Why we need it ?

20. Jul 15, 2015

### Staff: Mentor

"Reference frame" here means the coordinates you choose. We choose a particular set of coordinates (the "comoving" coordinates in which the FRW metric is usually written) in which to write down the Friedmann equation because it makes the equation look as simple as possible.