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## Main Question or Discussion Point

Another Exercise... Ohanian Exercise 6

##\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.

The Euler-Lagrange Equation given:Book said:Exercise 6Show 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##

##\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.

Last edited: