Partial derivative of Lagrangian density for vector field

Question

The lagrangian density of a massless vector field is

$ \mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$, where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$

Expanding out gives

$\mathcal{L} = -\frac{1}{2} \left( \partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu} - \partial_{\mu}A_{\nu}\partial^{\nu}A^{\mu} \right)$

In order to solve the equations of motion, at some point you need to take the derivative wrt $\partial_{\alpha}A_{\beta}$. As far as I can see, this should be, by recognizing

$\partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu} = g^{\mu\rho}g^{\nu\sigma}\partial_{\mu}A_{\nu}\partial_{\rho}A_{\sigma}$

and then fixing $(\mu,\nu)$ or $(\rho,\sigma)$ to $(\alpha,\beta)$, or similarly for the second term,

$\frac{\partial \mathcal{L}}{\partial(\partial_{\alpha}A_{\beta})} = -\partial^{\alpha}A^{\beta} + \partial^{\beta}A^{\alpha} = -F^{\alpha\beta}$

However, the particular example problem I'm working through will only lead to the final answer if the solution is $+F^{\alpha\beta}$. This positive solution is also what is found in the QED wikipedia page.

Can someone please explain why I have the wrong sign?

EDIT: In response to Dox below.

The exercise is actually for a massive vector field $C_{\mu}$, with

$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}m^{2}C_{\mu}C^{\mu}$

So the E-L equation I get is

$-\partial_{\alpha}F^{\alpha\beta} + m^{2}C^{\beta} = 0$

But both terms should actually be of the same sign. The exercise goes further to show some explicit relationship between $C_{0}$ and $C_{i}$, which I would be able to derive correctly if I had the proper signage in the E-L equation. Ultimately, my problem is still how to get $+F^{\alpha\beta}$ and not $-F^{\alpha\beta}$.

Answer 1

From the Lagrangian density

$$\mathcal{L} ~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}m^{2}C_{\mu}C^{\mu},$$

and with the identification $A_{\mu}\leftrightarrow C_{\mu}$, I get the proper sign in the equation of motion

$$+\partial_{\alpha}F^{\alpha\beta} + m^{2}C^{\beta} ~=~ 0.$$

You didn't forget about the minus sign in the Euler-Lagrange equation, did you?

Answer 2

If you are considering pure gauge theory, the equation of motion is

$\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\right)=0,$

which yields,

$-\partial_\mu F^{\mu\nu}=0.$

Of course, $\partial_\mu F^{\mu\nu}=0$ is the same equation of motion.