# Derivation of non decaying mode in cosmology

1. Jul 22, 2015

### smallgirl

1. The problem statement, all variables and given/known data
Derive the following result : $\phi_k=-C_1(k)\frac{\dot{H}(t)}{H(t)}$.

2. Relevant equations

$\phi_k=C_1(k)\Bigg(1-\frac{H}{a}\int\limits^{t}a(t)dt\Bigg)$

$a(t)= a(t_i)\exp\Bigg(H_i(t-t_i)+\dot{H}_i\frac{(t^2-t_i^2)}{2}\Bigg)$

3. The attempt at a solution
So I stuck a(t) into $\phi_k$ and then expanded the exponential and integrated. Just ended up with a mess really.

2. Jul 23, 2015

### fzero

I assume that $\phi$ is the metric perturbation and that you mean
$$\phi_k = - C_1(k) \frac{\dot{H}}{H^{2}}.$$
Note the $H^2$ in the denominator.

If so, there is kind of a bizarre trick to use here. First, the fact that $H = \dot{a}/a$, implies that
$$a dt =H^{-1} da .~~~(*)$$
The idea then is to integrate by parts to show that
$$\int a dt = \int H^{-1} da = H^{-1} a- \int a~ d \left( H^{-1} \right).$$
It seems weird, but this is formally ok to do. Now, we want to write
$$d \left( H^{-1} \right) = \frac{ d \left( H^{-1} \right)}{dt} dt$$
and use (*) again to write the $dt$ integral as one over $da$. Then it's one more integration by parts to get the term you want, plus another integral that you can see is small for $\dot{H} \ll H^2$.

I had to give a lot of the problem away to explain the trick, but it's probably confusing enough that you still have a lot of work to do for it. I'll try to help with followup questions as needed.

3. Jul 23, 2015

### smallgirl

Hey,

Thank you so much for the help but to solve the problem I have to put a(t) into $\phi_k$ or have I missed something? I know that at some point I need to do an expansion in $\frac{\dot{H_i}}{H^2_i}$

So following what you have said I have $d(H^{-1})=\frac{d(H^{-1})}{dt}$

and $dt=\frac{H^{-1}}{a}da\longrightarrow d(H^{-1})=\frac{d(H^{-1})}{dt}H^{-1}\frac{da}{a}$ which looks hideous. Do I then stick this into the second term on the RHS of the integral? and then integrate by parts?

Last edited: Jul 23, 2015
4. Jul 23, 2015

### fzero

With this method of doing the problem, we don't need to know anything special about $a(t)$ other than $H = \dot{a}{a}$. So you don't need to use the expression for $a$ that you quoted.

Yes, integrate by parts
$$\int u dv = uv - \int v du$$
with $u=(H^{-1}/a)( d(H^{-1})/dt)$ and $v=a$. The integrated part gives the expression that you were asked to show. The remaining integral is a mess of time derivatives that you should be able to argue is much smaller than the leading term.

5. Jul 23, 2015

### smallgirl

Ahh thank you! Will have a work through it and see what happens.

6. Jul 23, 2015

### smallgirl

Ok, so I've looked at it, and something doesn't seem quite right, I am integrating over da, and so if I set v=a, then dv=da but then what happens to my da and what happens to the second a in the integral?

7. Jul 23, 2015

### fzero

You've got a stray $1/a$, so let me back up a step to where we had
$$\int a~d\left(H^{-1}\right) = \int a~\frac{d\left(H^{-1}\right)}{dt} dt = \int \frac{d\left(H^{-1}\right)}{dt} H^{-1} da.$$
We have to integrate this by parts, which gives
$$\int \frac{d\left(H^{-1}\right)}{dt} H^{-1} da = \frac{d\left(H^{-1}\right)}{dt} H^{-1} a- \int a ~d\left( \frac{d\left(H^{-1}\right)}{dt} H^{-1} \right) .$$
The integrand on the RHS should be thought of as $d(\cdots) = [ d(\cdots)/dt] dt$.

8. Jul 23, 2015

### smallgirl

So I have now

$\int \frac{d\left(H^{-1}\right)}{dt} H^{-1} da = \frac{d\left(H^{-1}\right)}{dt} H^{-1} a- \int a ~\frac{d}{dt}\left( \frac{d\left(H^{-1}\right)}{dt} H^{-1} \right)dt$

And I want to integrate the integral on the RHS by parts. To do this, I (assume)
$dv=a$ and

$u=\frac{d}{dt}\Bigg(\frac{d(H^{-1})}{dt}H^{-1}\Bigg)$

9. Jul 23, 2015

### fzero

Whenever you have $a dt$ you want to rewrite it as $H^{-1} da$. If you keep integrating by parts you get a series with terms like
$$\left( H^{-1} \frac{d}{dt} \right)^n H^{-1}.$$
This series is what you want to truncate to the first term.

10. Jul 25, 2015

### smallgirl

Thank you so much, after doing the integrals about 4 times to see the series appear, I was able to see what I wanted. Thank you so much for your help and patience.