Euler-Lagrange Tensor Equations I need to vary w.r.t $a_{\alpha \beta} $
$\frac{\partial L}{\partial_{\mu}(\partial_{\mu}{a_{\alpha\beta}})}-\frac{\partial L}{\partial {a_{\alpha \beta}}}$ (1)
I am looking at varying the term in the Lagrangian of $\frac{1}{3}A^{\mu} \partial_{\mu}\Phi $
where $A^{\beta}=\partial_k {a^{k\beta}} $
$\frac{\partial L}{\partial_{\mu}(\partial_{\mu}{a_{\alpha\beta}})}-\frac{\partial L}{\partial {a_{\alpha \beta}}}$
My working so far:
(Expect it to yield a corresponding term in the EoMs as: $\frac{1}{3}\partial^m \partial^{n} \Phi-\frac{1}{4}\eta^{\alpha \beta} \partial^f \partial_f \Phi$ [2] )
I don't think I've ever done this properly, so my apologies, but my guess is want to keep things as general as possible with the indicies in (1) and those used on the $a_{\alpha \beta} $ tensor so (also, with the first term in (1)- is the index $\mu$ are they both supposed to be the same or is it better to keep it more general, something like:$\frac{\partial L}{\partial_{\nu}(\partial_{\nu}{a_{bc}})}$?) :
$ \frac{1}{3}A^{\mu} \partial_{\mu}\Phi= \frac{1}{3}\partial_c a^{c \mu} \partial_{\mu} \Phi, $
So if I say I am varying it w.r.t $\partial_{k}a^{mn}$ for the first term of (1), (obviously second term of (1) not relevant):
first I lower the indices on : $\frac{1}{3}\partial_c \eta^{wc} \eta^{zu}a_{wz}\partial_{\mu} \Phi $
then, (this is the bit I'm more unsure of- get deltas from  requiring  the indices on the derivative and tensor to match...)when varying wrt $\partial_{k}a^{mn}$ :
$\frac{1}{3}\delta_{c,k}\delta_{m,w}\delta_{n,z}\eta^{wc}\eta^{zu}\partial_{\mu}\Phi $
$=\frac{1}{3}\eta^{mk}\eta^{n\mu}\partial_{\mu}\Phi $
so for $\frac{\partial L}{\partial_k(\partial_{k}{a_{mn}})}$ get:
$\partial_k\frac{1}{3}\eta^{mk}\eta^{n\mu}\partial_{\mu}\Phi $
$= \frac{1}{3}\partial^m\partial^n\Phi $
And so I have got the first term of [2] but not the second term. :(
Thanks.
 A: You have to be much more carefully with the notation and indices. First of all, the correct Euler-Lagrange equations for fields look like
$$\frac{\partial\mathcal{L}}{\partial a_{\alpha\beta}}=\partial_{\mu}\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}a_{\alpha\beta})}$$
where there is a sum over $\mu$ on the right-hand side. In your expression, you have something like $\frac{\partial\mathcal{L}}{\partial_{\mu} (\partial_{\mu}a_{\alpha\beta})}$, but this is not correct. (Note that already the Einstein sum convention does not work in your expression, since you always have to sum over a covariant and a contravariant index). If you like to write it in short-hand notation, then you would have to write $\frac{\partial\mathcal{L}}{\partial^{\mu} (\partial_{\mu}a_{\alpha\beta})}$, because $\partial_{\mu}=\frac{\partial}{\partial x^{\mu}}$.
So, your Lagrangian is $\mathcal{L}=\frac{1}{3}A^{\mu}\partial_{\mu}\Phi=\frac{1}{3}(\partial_{\nu}a^{\nu\mu})(\partial_{\mu}\Phi)$
First of all, it is easy to see that the left-hand side of the Euler-Lagrange equation is zero, since the Lagrangian does not depend on $a_{\alpha\beta}$, but only on its derivative. For the second, term, we first of all compute $\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}a_{\alpha\beta})}$. Here you have to be very careful to use different indices:
$$\frac{\partial}{\partial(\partial_{\mu}a_{\alpha\beta})}\frac{1}{3}(\partial_{\nu}a^{\nu\sigma})(\partial_{\sigma}\Phi)=\frac{1}{3}(\partial_{\sigma}\Phi)\frac{\partial}{\partial(\partial_{\mu}a_{\alpha\beta})}\eta^{\lambda\nu}\eta^{\kappa\sigma}(\partial_{\nu}a_{\lambda\kappa})=\frac{1}{3}(\partial_{\sigma}\Phi)\eta^{\lambda\nu}\eta^{\kappa\sigma}\frac{\partial (\partial_{\nu}a_{\lambda\kappa})}{\partial(\partial_{\mu}a_{\alpha\beta})}=\frac{1}{3}(\partial_{\sigma}\Phi)\eta^{\lambda\nu}\eta^{\kappa\sigma}\delta^{\mu}_{\nu}\delta^{\alpha}_{\lambda}\delta^{\beta}_{\kappa}=\frac{1}{3}(\partial^{\beta}\Phi)\eta^{\alpha\mu}$$
As a next step, we have to take the derivativ of this expression and sum over $\mu$, which yields
$$\partial_{\mu}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}a_{\alpha\beta})}=\partial_{\mu}\frac{1}{3}(\partial^{\beta}\Phi)\eta^{\alpha\mu}=\frac{1}{3}(\partial^{\alpha}\partial^{\beta}\Phi).$$
Hence, to sum up, the Euler-Lagrange equations corresponding to your Lagrangian are
$$\partial^{\alpha}\partial^{\beta}\Phi=0.$$
This is different to your claimed result. But in general, your claimed result does not really make sense, because there have to be something wrong with the indices. The first term has as free indices $m$ and $n$, whereas the second term has $\alpha$ and $\beta$. So, I think you should check the resource where you found this solution.
