Linear Gravity Field - Lagrangian

[ProfDawgstein]

Another Exercise... Ohanian Exercise 6

Exercise 6 Show that $\mathcal{L}_{(0)}$ leads to the field equation (3.38) with $T^{\mu\nu} = 0$ (Hint: When working out the partial derivatives of $\mathcal{L}_{(0)}$ with respect to $h^{\alpha\beta}_{\ \ \ ,\mu}$ you can treat $h_{\alpha\beta}$ as independent of $h_{\beta\alpha}$; that is, you need not worry about the symmetry of the tensor field. The terms in the Lagrangian (A.34) are already arranged in such a way that they directly lead to a symmetric field equation.)

(A.34) $\mathcal{L}_{(0)} = \frac{1}{4} \left( h_{\mu\nu ,\alpha} h^{\mu\nu ,\alpha} - 2 h_{\mu\nu ,\alpha} h^{\alpha\mu ,\nu} + 2 h_{\mu\nu}^{\ \ \ \ ,\mu} h^{,\nu} - h_{,\nu} h^{,\nu} \right)$

(3.38) $\partial_\lambda \partial^\lambda h^{\mu\nu} + \partial^\mu \partial^\nu h - (\partial_\lambda \partial^\nu h^{\mu\lambda} + \partial_\lambda \partial^\mu h^{\nu\lambda}) - \eta^{\mu\nu} \partial_\lambda \partial^\lambda h + \eta^{\mu\nu} \partial_\lambda \partial_\sigma h^{\lambda\sigma} = -k T^{\mu\nu} = 0$

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...

Thank you in advance.

 

[fzero]

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 $\eta \partial \partial h^{ab}$.

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.

 

[ProfDawgstein]

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.

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']

 

[fzero]

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']

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 $\partial_\mu \partial_\beta h_{\rho\sigma}$.

 

[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)

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

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EDIT:

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

 

[fzero]

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

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

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.

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

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.

 

[ProfDawgstein]

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.

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 ($h_{\mu\nu} = h_{\nu\mu}$)

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

 

[fzero]

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 ($h_{\mu\nu} = h_{\nu\mu}$)

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

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.

 

[ProfDawgstein]

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.

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

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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