1. Mar 22, 2005

### Rev Prez

Looking at the Friedman solution for a fluid with a given pressure and energy density, we have:

$$H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho + \frac{\Lambda}{3} - \frac{k}{a^2}$$

Would someone mind walking me through how we arrive at a?

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2. Mar 22, 2005

### Crosson

From the assumption that the universe is isotropic we are lead immediately to the Robertson-Walker metric:

$$ds^2 = c^2dt^2 + a(t) [ dr^2 + S_k(r) d\Omega^2 ]$$

K is a constant either -1, 0, +1.

If k = +1 , the function S(r) = sin(r).

If k = -1, S(r) = sinh(r).

k = 0 implies S(r) = r.

By the way, omega is shorthand for some angular metric components, look it up under "hypersphere" at www.mathworld.com

Anyway, from this metric, along with the stress-energy tensor for a perfect fluid:

$$T_{\alpha\beta} = P g_{\alpha\beta} - (P - \rho)U_\alpha U_\beta$$

(P is pressure, u is four velocity, rho is density, and g is the metric tensor)

We insert both of these guys in to the einstein field equations (remember the einstein tensor G is just a bunch of derivatives of the RW metric):

$$G_{\alpha\beta} = \frac{8 \pi G}{c^2} T_{\alpha\beta}$$

Then solve for a(t) to get the freidmann equation. It is all very do-able, but it takes at least half an hour.

3. Mar 22, 2005

### Rev Prez

Okay.

1. $$d\Omega = d\theta^2 + \sin^2(\theta)d\phi$$.

2. Shouldn't $$S_k(r)$$ take on the same units as dr (as $$d\Omega$$ is dimensionless)? In which case, as is indicated here, we should multiply by the mag of the curvature radius. I'm not being facetious, but its easy to get lost when you're not quite clear on the unit constant convention used.

3. The codomain of k is restricted to -1, 0, 1, or is this just a convenient choice since we're more or less concerned with whether the large scale structure is closed, flat or hyperbolic?

Now this is where I should've started asking questions. What process do you go through starting with "homogenous, isotropic universe" to get a tensor description of it? I understand intuitively what it means, but I would love to know how you reason out a perfect fluid's formal representation.

I can guess that:
1. Because FLRW is for an isotropic, homogenous universe, P and $$\rho$$ are assumed constant...

...and...

2. for an observer at rest with the fluid, the equation reduces to $$T_{\alpha\beta} = P g_{\alpha\beta}$$

But where do you get $$(P - \rho)$$?

Thanks for your help. I'll digest this and get back to the thread if I have more questions.

Rev Prez

Last edited: Mar 22, 2005
4. Mar 22, 2005

### Crosson

It is much easier to look at the stress energy tensor for a perfect fluid in matric form than as:

$$T_{\alpha\beta} = P g_{\alpha\beta} - (P - \rho)U_\alpha U_\beta$$

But I don't know how to Latex format a matrix. I will describe the components in the rest frame of the fluid:

$$T_{00} = \rho c^2$$

$$T_{11} = P_x$$

$$T_{22} = P_y$$

$$T_{33} = P_z$$

And the rest (off diagonal) are zero. Remember that U is the four velocity, and in the rest frame of the fluid this is equal to c in the time direction. Also, remember pressure is stress.

Yes, an isotropic (simply connected) shape can only have one of three curvatures.

5. Mar 22, 2005

### Rev Prez

Okay, I got it. So check me here. In the fluid's rest frame, which a choice of units c = 1, the equation gives us just the energy density, correct?

Rev Prez

Last edited: Mar 22, 2005
6. Mar 23, 2005

### Crosson

What are you using for g ? In the rest frame of the particle the metric tensor (g) is simply the flat metric of minkowski spacetime (16 components, all of them zero except g00 = -1 and g11=g22=g33=1).

U is a vector which is all zero accept for the zeroth component (time) which is c = 1.

Using these and the abstract equation, you should be able to construct the components of the stress-energy tensor of a perfect fluid (in its rest frame). (the diagonal terms I described.)

7. Mar 23, 2005

### Rev Prez

Just g00 = -1, everything else set to zero.

Since

$$g_{\alpha\beta} = -U_{\alpha}U_{\beta$$

in the rest frame,

$$T_{\alpha\beta} = Pg_{\alpha\beta} - (P - \rho)U_{\alpha}U_{\beta}$$

reduces to

$$T_{\alpha\beta} = \rho(U_{\alpha}U_{\beta})$$

where all that remains is the energy density component in the time direction.

Right?

Rev Prez

8. Mar 23, 2005

### Rev Prez

I think I got it.

You will end up with $$diag(\rho, p, p, p)$$ for constant pressure and energy density and with c = 1. The $$U_{\alpha}U_{\beta}$$ term will give us the energy density in the time dimension and the metric term will give us the pressure in space.

Rev Prez