Ok, I tried sometimes already, however I can't see my mistake. What I need to do is to find the field equation of

$$ L = -(\partial_\mu A^\nu)(\partial_\nu A^\mu) + \frac{m^2}{2} A_\mu A^\mu + \frac{\lambda}{2} (\partial_\mu A^\mu)^2 $$

trough the Euler-Lagrange equation

$$ \partial_\nu \left( \frac{\partial L}{\partial (\partial_\nu \phi_\rho)} \right) - \frac{\partial L}{\partial \phi_\rho}=0 $$

Here are the steps that I'm doing.

First I'm lowering every index in the Lagrangian density

$$ L = -g^{\alpha \nu}g^{\beta \mu} (\partial_\mu A_\alpha)(\partial_\nu A_\beta) + \frac{m^2}{2} g^{\gamma \mu} A_\mu A_\gamma +\frac{\lambda}{2} (\partial_\mu A_\delta) (\partial_\nu A_\epsilon) $$

Now, solving the first part $$ \begin{align} \frac{\partial L}{\partial (\partial_\eta A_\phi)} & = \\ & = -g^{\alpha \nu}g^{\beta \mu} \left( \frac{\partial (\partial_\mu A_\alpha)}{\partial (\partial_\eta A_\phi} \partial_\nu A_\beta + \frac{\partial (\partial_\nu A_\beta)}{\partial (\partial_\eta A_\phi} \partial_\mu A_\alpha \right) + \frac{\lambda}{2} \left( \frac{\partial (\partial_\mu A_\delta)}{\partial (\partial_\eta A_\phi} \partial_\nu A_\epsilon + \\ \frac{\partial (\partial_\nu A_\epsilon)}{\partial (\partial_\eta A_\phi} \partial_\mu A_\delta\right) \\ & = -g^{\alpha \nu}g^{\beta \mu} (\delta^\eta_\mu \delta^\phi_\alpha \partial_\nu A_\beta + \delta^\eta_\nu \delta^\phi_\beta \partial_\mu A_\alpha) + \frac{\lambda}{2} g^{\delta \mu}g^{\epsilon \nu} (\delta^\eta_\mu \delta^\phi_\delta \partial_\nu A_\epsilon + \delta^\eta_\nu \delta^\phi_\epsilon \partial_\mu A_\delta) \\ & = -g^{\phi \nu} g^{\beta \eta} \partial_\nu A_\beta - g^{\phi \mu} g^{\alpha \eta} \partial_\mu A_\alpha + \frac{\lambda}{2} g^{\eta \phi} g^{\epsilon \nu} \partial_\nu A_\epsilon + \frac{\lambda}{2} g^{\delta \mu} g^{\eta \phi} \partial_\mu A_\delta \\ & = -\partial^\phi A^\eta - \partial^\phi A^\eta + \frac{\lambda}{2} g^{\eta \phi} \partial^\epsilon A_\epsilon + \frac{\lambda}{2} g^{\eta \phi} \partial^\delta A_\delta \\ & =-2 \partial^\phi A^\eta + \lambda g^{\eta \phi} \partial^\delta A_\delta \end{align} $$ Then

$$ \begin{align} \partial _\eta\left(\frac{\partial L}{\partial (\partial_\eta A_\phi)} \right) & = - 2 \partial_\eta \partial^\phi A^\eta + \lambda g^{\eta \phi} \partial_\eta \partial^\delta A_\delta \\ & = -2 \Box A^\eta + \lambda g^{\eta \phi} \Box A_\delta \end{align} $$

The second part of E-L equation:

$$ \begin{align} \frac{\partial L}{\partial A_\rho} & = \frac{m^2}{2} g^{\gamma \mu} \left( \frac{\partial A_\mu}{\partial A_\rho} A_\gamma + \frac{\partial A_\gamma}{\partial A_\rho} A_\mu \right) \\ & = \frac{m^2}{2} g^{\gamma \mu} (\delta^\rho_\mu A_\gamma + \delta^\rho_\gamma A_\mu) \\ & = \frac{m^2}{2} g^{\gamma \rho} A_\gamma + \frac{m^2}{2} g^{\rho \mu} A_\mu \\ & = \frac{m^2}{2}(A^\rho + A^\rho) \\ & = m^2 A^\rho \end{align} $$

Finally, my final solution would be:

$$ \boxed{ -2 \Box A^\eta + \lambda g^{\eta \phi} \Box A_\delta -m^2 A^\rho = 0 } $$

Which I can easily see that's wrong because of the excess of free indexes, but I can't find where I'm making the mistakes...

Thanks in advance!


Correcting the D'Alembertian term:

$$ \boxed{ - 2 \partial_\eta \partial^\phi A^\eta + \lambda g^{\eta \phi} \partial_\eta \partial^\delta A_\delta -m^2 A^\rho = 0 } $$

Still with two free indexes...


closed as off-topic by Kyle Kanos, John Rennie, Martin, ACuriousMind, user10851 Jul 9 '15 at 21:42

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  • $\begingroup$ As a general rule, if a final expression contains mismatched indices, you must have made a mistake somewhere in the middle. You just have to go through line by line looking for a place where indices don't match. $\endgroup$ – Javier Jul 8 '15 at 19:21
  • 4
    $\begingroup$ Isn't this a "check my work" question? $\endgroup$ – DanielSank Jul 8 '15 at 20:59
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/192849/2451 $\endgroup$ – Qmechanic Jul 12 '15 at 11:49

I similarly find $$\frac {\partial \mathcal L}{\partial(\partial_\eta A_\phi)} = \lambda ~ g^{\eta\phi} (\partial_\mu A^\mu) - 2 ~\partial^\phi A^\eta .$$But now when you bring $\partial_\eta$ over this expression you have to not skip steps when you write out$$\partial_\eta~ \frac {\partial \mathcal L}{\partial(\partial_\eta A_\phi)} = \partial_\eta~(\lambda ~ g^{\eta\phi} (\partial_\mu A^\mu) - 2 ~\partial^\phi A^\eta) .$$There is of course a special connection $\partial$ by which $\partial_\alpha g^{\beta\gamma} = 0$ so you don't have to worry so much about the product rule, but you will nevertheless get$$\lambda ~ \partial^\phi \partial_\mu A^\mu - 2 ~\partial^\phi \partial_\eta A^\eta = (\lambda - 2) ~\partial^\phi ~\partial_\eta ~A^\eta.$$

The remaining inconsistency appears to be due to your replication of $\phi$ and $\rho$, which need to be the same, hence the full equation is$$(\lambda - 2) ~\partial^\phi ~\partial_\eta ~A^\eta - m^2 A^\phi = 0.$$This has no "free indexes": everything which should be free (which is to say, $\phi$) is left free and everything else (which is to say, $\eta$) is involved in a contraction.


First of all, you have some indices that randomly get lowered or raised. I'd advise you check for consistency.

But mostly, it seems that you have the wrong definition for $\Box$.
In your work, it seems that you think this is correct: $$\partial^\phi\partial_\eta A^\eta=\Box A^\eta$$ And that cannot be true, since your indices don't match. On your left side you have only $\phi$ as a free index, and $\eta$ as a dummy index, and on your right you have $\eta$ as a free index.
The correct definition of the D'Alembertian is: $$\Box=\partial^\mu\partial_\mu$$ You can see this mistake, for example, in your second $\partial _\eta\left(\frac{\partial L}{\partial (\partial_\eta A_\phi)} \right)$ term, where you have three free indices after your $\Box$ definition, and one before it.

  • $\begingroup$ I already had a hunch that I was using the D'alembertian wrongly, but I wrote it in here to check. However I did not understand what do you mean by "indices that randomly get lowered or raised", first I lowered EVERY index that I had, when I've finished the calculation I re-raised it to leave in the simplest form possible. $\endgroup$ – Edison Cesar Jul 8 '15 at 19:06
  • $\begingroup$ I mean, for example, that you have a partial derivative with a lower index phi, which then gets turned into an upper index randomly in the next line. (In calculating the first term) $\endgroup$ – Omry Jul 8 '15 at 19:11
  • $\begingroup$ Well, that was a legitimate typo only. Corrected already! Thanks for spotting it! $\endgroup$ – Edison Cesar Jul 8 '15 at 19:18

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