# Conformal time

1. Jan 24, 2007

### Logarythmic

1. The problem statement, all variables and given/known data
A closed Friedmann universe contains a single perfect fluid with an equation of state of the form $$p=w\rho c^2$$. Transforming variables to conformal time $$\tau$$ using $$dt=a(t)d\tau$$, show that the variable $$y=a^{(1+3w)/2}$$ is described by a simple harmonic equation as a function of $$\tau$$. Hence argue that all closed Friedmann models with a given equation of state have the same conformal lifetime.

2. The attempt at a solution
Please give me a starter. I haven't got a clue...

2. Jan 24, 2007

### Dick

You'll want to start by writing down the Friedmann equations with curvature term +1, changing time derivatives to conformal time derivatives and substitute your energy/pressure relation. Do all this busy work and then think about the problem again.

3. Jan 24, 2007

### Logarythmic

So first the Friedmann eq. is

$$\left(\frac{\dot{a}}{a_0}\right)^2=H^2_0 \Omega_{0w} \left(\frac{a_0}{a}\right)^{1+3w}$$

and changing the time derivative like

$$\frac{da}{dt}=\frac{1}{a}\frac{da}{d\tau}$$

giving

$$\left(\frac{da}{d\tau}\right)^2=a^2a_0^2H_0^2\Omega_{0w}\left(\frac{a_0}{a}\right)^{1+3w}=H_0^2\Omega_{0w}a^{1-3w}a_0^{3+3w}=H_0^2\Omega_{0w}a^{1-3p/\rho c^2}a_0^{3+3p/\rho c^2}$$.

I still don't get it.

4. Jan 24, 2007

### Dick

I don't get it either. That doesn't look like the first Friedmann equation to me. Where's rho? Where's the curvature term? You've already done a bunch of substitutions.

5. Jan 24, 2007

### Logarythmic

I have used the Friedmann equation

$$\left(\frac{\dot{a}}{a_0}\right)^2 - \frac{8\pi}{3}G\rho\left(\frac{a}{a_0}\right)^2 = H_0^2\left(1-\frac{\rho_0}{\rho_{0c}}\right) = H_0^2(1-\Omega_0) = -\frac{Kc^2}{a_0^2}$$

and

$$\rho a^{3(1+w)} = const. = \rho_{0w}a_0^{3(1+w)}$$

to get

$$\left(\frac{\dot{a}}{a_0}\right)^2=H^2_0 \left[\Omega_{0w} \left(\frac{a_0}{a}\right)^{1+3w} + (1-\Omega_{0w}) \right]$$

Then, for a curved model, the last term is negligible so

$$\left(\frac{\dot{a}}{a_0}\right)^2=H^2_0 \Omega_{0w} \left(\frac{a_0}{a}\right)^{1+3w}$$

This is all from Coles Cosmology.

6. Jan 24, 2007

### Dick

You are assuming the evolution is a power law in a(t) and it's not. The curvature term is not negligible. You are going to want to find an equation of the form b'=Kb where ' denotes the derivative wrt conformal time.

7. Jan 24, 2007

### Logarythmic

Well, then I'm back to zero and still stuck.

8. Jan 24, 2007

### Dick

Ok, try this. Go back to the REAL Friedmann equation for H^2. We know this is a closed universe so it will reach a point of maximum expansion and then collapse. So pick t_0 to be the time of maximum expansion, so H_0=0, may as well take a_0=1 to make a choice of scale. Now put in your dependence of rho on and fix the constants at t_0. Now what does the equation look like?

9. Jan 24, 2007

### Dick

Well. I actually worked through this in detail for practice. It does work. Substitute p=w*rho. Solve the H^2 eqn for rho and substitute it into the a'' one. Express the t derivatives in terms of conformal derivatives. Finally substitute y^(2/(3*w+1)) for a. You will get a y'' term, a (y')^2 term and a y^2 term. If you have actually managed to do all of the math correctly (it took me a few times), you will find the (y')^2 terms magically cancel (hence the choice of exponent). The remaining equation looks like a harmonic oscillator. Yahooo!

10. Jan 25, 2007

### Logarythmic

Uhm, what equations are H^2 and a''?

11. Jan 25, 2007

### Dick

The H^2 equation is the one you have been working with, just involving rho. The a'' equation is the other Friedmann equation also involving the pressure and second derivative of the scale factor. If you substitute for the pressure in terms of rho you can combine them and eliminate the rho. Then work with that equation. Do NOT discard the curvature term.

12. Jan 25, 2007

### Logarythmic

And where did you get the a'' eq.?

13. Jan 25, 2007