# Linear Gravity Field - Lagrangian

1. Aug 15, 2013

### ProfDawgstein

Another Exercise... Ohanian Exercise 6

The Euler-Lagrange Equation given:

$\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})} - \frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0$

$\frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0$, because there is no explicit $h^{\alpha\beta}$ dependence

Now, since we have $h$ here, which is the trace of $h^{\alpha\beta}$

$h = \eta^{\alpha\beta} h_{\alpha\beta} = \eta_{\alpha\beta} h^{\alpha\beta}$

Expanding all $h$ in the Lagrangian using

$\partial^\nu h = \eta_{\alpha\beta} \partial^\nu h^{\alpha\beta}$

$\partial_\nu h = \eta_{\alpha\beta} \partial_\nu h^{\alpha\beta}$

results in the 'new' Lagrangian

$\mathcal{L}_{(0)} = \frac{1}{4} \left( (\partial_\alpha h_{\mu\nu}) (\partial^\alpha h^{\mu\nu}) - 2 (\partial_\alpha h_{\mu\nu}) (\partial^\nu h^{\alpha\mu}) + 2 (\partial^\mu h_{\mu\nu}) (\partial^\nu \eta_{\alpha\beta} h^{\alpha\beta}) - (\partial_\nu \eta_{\alpha\beta} h^{\alpha\beta}) (\partial^\nu \eta_{\rho\sigma} h^{\rho\sigma}) \right)$

Now after doing lots of product rules and triple $\delta$'s I can't get 2nd order derivatives...

[Doing: $\frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})}$]

Replacing the $\partial_a h^{bc}$ with $\delta \delta \delta$

I always have something like $\eta \partial_{\alpha} h^{ab}$,

but never $\eta \partial \partial h^{ab}$.

And somehow I still end up making all terms look like $\partial_{\mu} h^{\alpha\beta}$ for the Euler-Lagrange derivatives.

If we use $\eta_{\mu\nu}$ twice on $h^{\alpha\beta}$ we shouldn't change the tensor

$h^{\alpha\beta} = h_{\alpha\beta}$ | (-1)(-1) = +1 and

for the partials I just use $\partial_\alpha = \eta_{\alpha\mu} \partial^\mu$ ?

I am getting like 80 indices in the Lagrangian, doing all the details... (a,b,sigma,tau,alpha,beta,mu,nu,rho,lambda,epsilon) [lol]

Also never had any 'symmetry problems', as he says in the HINT.

Hints to make things even simpler are always appreciated :)

I really don't want to do things like 'contravariant $\leftrightarrow$ covariant'.
(Staying consistent with the Langrangian derivative)

I really hope I didn't forget any product or chain rules...

Last edited: Aug 15, 2013
2. Aug 15, 2013

### fzero

The 2nd derivatives come from the partial derivative in the E-L equation

$$\underbrace{\frac{\partial}{\partial x^\mu}} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})} - \frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0.$$

I'm not sure if that is your main problem. I haven't gone through the algebra of this particular computation recently.

3. Aug 15, 2013

### ProfDawgstein

Wow, the obvious again...
I have seen way too many indices today, thanks alot!
Kind of stressed to finish this, not my own book...

I will do the whole thing again tomorrow, and will post my results here.

Any tips/hints to make the calculation easier?
I think my starting point should be correct.

$\mathcal{L}_{(0)} = \frac{1}{4} \left( (\partial_\alpha h_{\mu\nu}) (\partial^\alpha h^{\mu\nu}) - 2 (\partial_\alpha h_{\mu\nu}) (\partial^\nu h^{\alpha\mu}) + 2 (\partial^\mu h_{\mu\nu}) (\partial^\nu \eta_{\alpha\beta} h^{\alpha\beta}) - (\partial_\nu \eta_{\alpha\beta} h^{\alpha\beta}) (\partial^\nu \eta_{\rho\sigma} h^{\rho\sigma}) \right)$

Could have pulled the $\eta_{\alpha\beta}$ out, though...

I would love to know a few more tricks for this type of calculations :P

Also, does it help / is it a good idea
-to lower / raise the indices to the same level? (but I don't want to insert too many $\eta$'s)
-rewrite the euler equation to have 2 derivatives (1 with covariant and 1 with contravariant $\partial$'s) to reduce the $\eta$'s
-what about the $h_{,\nu}$ terms? any way to make that easier / better looking?

I want everything to be consistent, but not too complicated.

[I want to improve my 'style']

Last edited: Aug 15, 2013
4. Aug 15, 2013

### fzero

You could certainly take the derivatives term-by-term. Perhaps one convenient way of doing the calculation, that would simplify taking derivatives of the traces, would be to write the Lagrangian in the form

$$\mathcal{L}_{(0)} = C^{\alpha \mu \nu \beta \rho \sigma} \partial_\alpha h_{\mu\nu} \partial_\beta h_{\rho\sigma},$$

where $C^{\alpha \mu \nu \beta \rho \sigma}$ can be written as a sum of terms involving $\eta$s. Because of the symmetry, you really only take one derivative here and then contract $C^{\alpha \mu \nu \beta \rho \sigma}$ and your $\delta$s into a single factor of $\partial_\mu \partial_\beta h_{\rho\sigma}$.

5. Aug 16, 2013

### ProfDawgstein

Okay, after 1 hour of 'index porn' I got this result...

$\partial_\mu \partial^\mu h - \partial_\mu \partial_\nu h^{\mu\nu} = 0$

1) I treated $h^{\mu\nu}$ as symmetric
2) I contracted out the $\eta$'s
3) Divided by the common factors (0.5 in mine, 4 in 3.38)

If I do the same for (3.38) I get exactly the same as mine.

Does anybody know what this equation means?
The first part looks like a wave equation...

It took me about 3 pages with 20 lines each.

What I did:
1) make all terms look like $\partial_\mu h^{\alpha\beta}$
2) re-arrange -> $\eta$'s in front
3) do the lagrange derivative
4) do the $\delta$ contractions
5) do the $\eta$ contractions
6) contract out the remaining $\eta$'s
7) group terms
8) treat $h^{\alpha\beta}$ as symmetric
9) done

I don't know if it was allowed to treat $h^{\alpha\beta}$ as symmetric, if not the whole thing is a bit longer...

Without contracting out the $\eta$'s and not treating $h^{\alpha\beta}$ as symmetric I get this (divided by 1/2)

$\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \eta_{\alpha\beta} \partial_\mu \partial_\nu h^{\nu \mu} + \partial_\alpha \partial_\beta h - \eta_{\alpha\beta} \partial_\mu \partial^\mu h = 0$

-----------------------------------------------------------------------------------

EDIT:

It seems like I just derived $R = 0$.

Last edited: Aug 16, 2013
6. Aug 16, 2013

### fzero

This agrees with 3.38, so is the result that you wanted to obtain. The first equation seems to be a result of some incorrect manipulations. In particular, you seem to be setting

$$\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \partial_\alpha \partial_\beta h = 0$$

by symmetry, which is incorrect. Perhaps you can give a few more details on what makes you think these terms cancel amongst themselves.

That might be correct, but the Einstein field equation is

$$R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = k T_{\mu\nu},$$

so you end up with the extra terms in 3.38.

For some sense of how this becomes a wave equation, you might try to read through the discussion of the de Donder gauge here. It might also be discussed in another section of O'Hanian.

7. Aug 16, 2013

### ProfDawgstein

In this exercise $T_{\mu\nu} = 0$.

$\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \eta_{\alpha\beta} \partial_\mu \partial_\nu h^{\nu \mu} + \partial_\alpha \partial_\beta h - \eta_{\alpha\beta} \partial_\mu \partial^\mu h = 0$

I contracted this one with $\eta^{\alpha\beta}$ and then after renaming some indices I get

$- \partial_\mu \partial^\mu h - \frac{1}{2} \partial_\mu \partial_\nu h^{\mu\nu} + \frac{3}{2} \partial_\mu \partial_\nu h^{\nu\mu} = 0$

The order of the $\partial$'s shouldn't matter, so ($h_{\mu\nu} = h_{\nu\mu}$)

$\partial_\mu \partial^\mu h - \partial_\mu \partial_\nu h^{\mu\nu} = 0$

8. Aug 16, 2013

### fzero

OK, that's fine, but you have to remember that when you take the trace of an equation, the result does not contain as much information as the original equation. 3.38 is a set of 10 independent equations (accounting for symmetry of indices, but not diffeomorphism invariance), while the trace is a single linear combination of these. It turns out that one can use diffeomorphism invariance to relate the system of equations to a single wave equation for 2 polarizations of the gravity wave, but the details are a bit more subtle (as you can see in that link) than just taking a trace.

9. Aug 16, 2013

### ProfDawgstein

Oh yes, I know.
I just wanted to see what I get.
Taking the trace / contracting with $\eta$ made it easier to compare both equations.

If you have $h_{ab}$ or $h^{ab}$ shouldn't matter, right?
It just has to be on the same levels as in the other equation.

At the moment I am just happy that I finished this one :D

--------------------------------------------------------------------

Could I also have done something like this?

$\frac{\partial (\partial_a h^{bc})}{\partial (\partial_\mu h^{\alpha\beta})} = \delta^{\mu}_{a} \delta^{b}_{\alpha} \delta^{c}_{\beta}$

$\frac{\partial (\partial^a h_{bc})}{\partial (\partial^\mu h_{\alpha\beta})} = \delta^{a}_{\mu} \delta^{\alpha}_{b} \delta^{\beta}_{c}$

I didn't because it's not consistent with the Euler-Lagrange derivative, so I made all terms to look like

$\partial_\mu h^{\alpha\beta}$

Is there a nice way to do the whole thing notationally correct but simpler?

-> less $\eta$ contractions

Having this

$\mathcal{L}_{(0)} = \frac{1}{4} \left( (\partial_\alpha h_{\mu\nu}) (\partial^\alpha h^{\mu\nu}) - 2 (\partial_\alpha h_{\mu\nu}) (\partial^\nu h^{\alpha\mu}) + 2 (\partial^\mu h_{\mu\nu}) (\partial^\nu \eta_{\alpha\beta} h^{\alpha\beta}) - (\partial_\nu \eta_{\alpha\beta} h^{\alpha\beta}) (\partial^\nu \eta_{\rho\sigma} h^{\rho\sigma}) \right)$

and doing this

$\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})} - \frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0$.

I need to insert like 10 $\eta$'s to make that consistent with the EL derivative...

Could somebody show how to do this thing nicely, but keeping the whole thing together?

Thanks