Cosmology - Friedman Equation Problem

[Tangent87]

I am doing question A1/16 part (i) at the top of page 13 here:

http://www.maths.cam.ac.uk/undergrad/pastpapers/2001/Part_2/list_II.pdf

I am stuck on the last part where we're given the equation of state $$P=\sigma\rho c^2$$ and we have to find a(t) and SHOW a(0)=0. I will work through to where I've got to and the problem I encounter:

So first of all if we substitute the equation of state into the fluid equation we find:

$$\frac{\dot{\rho}}{\rho}=-3(\sigma+1)\frac{\dot{a}}{a}$$

Solving this we get

$$\rho=Aa^{-3(\sigma+1)}$$ where A is a constant. Then we substitute this into the equation for k and use k=0 as they've told us to obtain:

$$\frac{\dot{a^2}}{a^2}=\frac{8\pi G}{3}\rho=\frac{8\pi G}{3}Aa^{-3(\sigma+1)}$$

Thus,

$$\frac{\dot{a}}{a}=\left(\frac{8\pi G}{3}A\right)^{1/2}a^{-\frac{3}{2}(\sigma+1)}$$

We can then use a(t_0)=1 (which means $$\dot{a}(t_0)=H_0$$) to find $$A^{1/2}=H_0\left(\frac{3}{8\pi G}\right)^{1/2}$$

We then integrate and finally we get:

$$\frac{1}{\frac{3}{2}\sigma+3/2}a^{\frac{3}{2}\sigma+3/2}=H_0t+const$$

But what do we do now? Because the only way I see to get rid of the const is to assume a(0)=0 which is want we want to SHOW! What am I missing?

 

[grey_earl]

Simply use a(t_0) = 1 again to determine the unknown constant. Even if I'm not sure where this comes from.

 

[Tangent87]

Simply use a(t_0) = 1 again to determine the unknown constant. Even if I'm not sure where this comes from.

But then we'll have a non-zero constant and so there's no way that a(0) will be 0? If we're to have any chance of showing a(0)=0 then we have to have than const being 0 a priori.

 

[grey_earl]

Have you done the calculus? What do you get for the constant?

 

[Tangent87]

Have you done the calculus? What do you get for the constant?

If we use a(t_0)=1 again we'll get $$const=\frac{1}{\frac{3}{2}\sigma+3/2}-H_0t_0$$. And so $$a^{\frac{3}{2}\sigma+3/2}=(\frac{3}{2}\sigma+3/2)H_0t+1-H_0t_0(\frac{3}{2}\sigma+3/2)$$

And so a(0) is definitely not 0, maybe there's a mistake in my derivation but I don't see anything?

 

[grey_earl]

So a(0) = 0 if and exactly if $$H_0 t_0 (3/2 \sigma +3/2) = 1$$, which determines t_0 if you have measured $$\sigma$$ and H_0 (by astronomical observations). And that's all.

BTW: I think in the next exercise they ask you to calculate the age of the universe, no? For a dust universe, σ=0, and so t_0 = 2/3 1/H_0, which is smaller than 1/H_0, so the universe is younger than Hubble makes us believe :)

 

[Tangent87]

So a(0) = 0 if and exactly if $$H_0 t_0 (3/2 \sigma +3/2) = 1$$, which determines t_0 if you have measured $$\sigma$$ and H_0 (by astronomical observations). And that's all.

Hmm yeah I noticed that too but I'm still not sure about it because you can't determine $$H_0 t_0 (3/2 \sigma +3/2) = 1$$ from any of the previous calculations, like you said you just have to define it that way but it seems like it's just kind of assuming what you want to show.

 

[grey_earl]

True, but neither you can $$\dot a(t_0) = H_0$$, or at least I don't see how. But assuming there is no error in your calculations (which I don't think, to me they look ok), without assuming this, you cannot show a(0) = 0. It's assuming one or the other.

 

[Tangent87]

True, but neither you can $$\dot a(t_0) = H_0$$, or at least I don't see how. But assuming there is no error in your calculations (which I don't think, to me they look ok), without assuming this, you cannot show a(0) = 0. It's assuming one or the other.

No I disagree, we not assuming $$\dot a(t_0) = H_0$$ since by definition $$H(t)=\frac{\dot a(t)}{a(t)}$$ so $$H_0=H(t_0)=\frac{\dot a(t_0)}{a(t_0)}=\frac{\dot a(t_0)}{1}=\dot{a(t_0)}$$

 

[grey_earl]

Ok, I see your point. How do you define t_0 ?

 

[Tangent87]

Ok, I see your point. How do you define t_0 ?

t_0 is just the time "today" (see the question)

 

[grey_earl]

Well, but that's no definition like you gave for H_0. From your solution, for every value of the free constant you will have a time t_1 where a(t_1) = 0, namely t_1 = -const/H_0. Then you just shift the time variable by a constant amount (namely, t_1) and have a(0) = 0, and "today" is t_0+t_1 = t_0^new. That's just the usual freedom to make a linear transformation of coordinates, i.e. choose units and zero point. You can specify this by giving two conditions on the time coordinate, p.ex. a(t_0) = 1 and $$\dot a_0(t_0) = H_0$$, but then a(0) is just what is given by your formula. If you want it to be zero, that's an extra restriction relating H_0, t_0 and σ, which doesn't follow from the maths, but the exact definition of those variables. So you define "today", i.e. t_0 by saying that a(0) = 0, which gives you the above restriction.

 

[fzero]

If we use a(t_0)=1 again we'll get $$const=\frac{1}{\frac{3}{2}\sigma+3/2}-H_0t_0$$. And so $$a^{\frac{3}{2}\sigma+3/2}=(\frac{3}{2}\sigma+3/2)H_0t+1-H_0t_0(\frac{3}{2}\sigma+3/2)$$

And so a(0) is definitely not 0, maybe there's a mistake in my derivation but I don't see anything?

You can use the above equation to show that there is some $$t_{BB}$$ for which $$a(t_{BB}) = 0$$. It is purely a convention to shift the time axis so that $$t_{BB}=0$$. Then $$t_0$$ is the age of the universe.

 

[Tangent87]

Ok thanks all, it makes sense now.