# Linearised Gravitational Waves Derivation

1. May 3, 2012

### tomelwood

Hi
A topic came up in a lecture the other day about how if certain simplifications are made, then the Einstein equation reduces to a form of the wave equation.
When I look at derivations of how this happens, I get a little confused as to how this happens.
I think I'm posting it in the right place putting it here, as it's not strictly a homework question, since it is only to help my understanding of the course so far...

Looking at the website http://iopscience.iop.org/1367-2630/7/1/204/fulltext/#nj192710s2
I don't understand why you are allowed to take the step made in Equation (2.9) - i.e. why does putting a prime on the LHS simply mean that you can go ahead and "prime" all of the h in the RHS? If this is allowed, then I can see how the rest of that equality works, no problem.

Secondly, taking a step or two back, where it says substituting "h-bar" into (2.6) "and expanding", how does this simplify down to (2.7)? Because I can see how the terms where both indices are "downstairs" are changed, but what about the h's in the form of a (1,1) tensor? That is one upstairs and one downstairs? I tried applying the Minkowski metric to the (1,1) h, to try and get into "downstairs format" for me to work with, but I was left with some delta's.. Specifically (apologies in advance for appalling Latex):

$h^{c}_{a,bc}=\eta^{da}\bar{h}_{ab,bc}-\frac{1}{2}\eta_{ab}\eta^{ad}\bar{h}_{,bc}=\bar{h}^{d}_{b,bc}-\frac{1}{2}\delta\stackrel{d}{b}\bar{h}_{,bc}$

and I don't understand how to move on from here?

Any observations would be greatly appreciated.

Last edited: May 3, 2012
2. May 4, 2012

### clamtrox

Why wouldn't it work? I mean, you just change the coordinates $x'^{a} = x^{a} + \xi^{a}$. $h'_{ab}$ is the metric in the new gauge, so you just define your trace-reversed metric in the new gauge as $\bar{h}'_{a b}= h'_{ab} - 1/2 \eta_{a b} h'$

You are doing something very naughty there...

$$\partial_b \partial_c h^c_a = \partial_b \partial_c \bar{h}^c_a + \frac{1}{2} \partial_b \partial_c \delta^c_a h = \partial_b \partial_c \bar{h}^c_a + \frac{1}{2} \partial_b \partial_a h,$$ right?

3. May 8, 2012

### tomelwood

Oh of course. Thanks. OK. So now all that section makes sense, the final step is to show that the trace terms cancel out.
I've been trying to find out the intermediate steps going on, and have managed to solve the equation if I can show that:

$h_{,ab}=\frac{1}{2}\eta_{ca}h_{,b}^{c}+\frac{1}{2}\eta_{cb}h_{,a}^{c}$
where the superscript c's indicate partial differentiation with respect to c as well (I couldn't make the latex work for that bit)

The only problem is I can't do this, as I know you can't raise/lower indices of partials...
Thanks!

4. May 9, 2012

### clamtrox

Remember two things:
1) what's the covariant derivative of a scalar?
2) in perturbation theory, you can drop everything 2nd order. How big is $\Gamma^\mu_{\alpha \beta}$?