# Partial derivative of Lagrangian density for vector field

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 $-F^{\alpha\beta}$.

• Comment to the question(v2): Apparently $A_{\mu}\leftrightarrow C_{\mu}$. Oct 24, 2012 at 23:30

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

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

• yeah, that's exactly what i did. it always comes down to these minor mistakes. thanks a bunch! Oct 25, 2012 at 6:35

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.

• The question is probably about deriving Noether currents, where the sign of the derivative matters for getting positive energy. Obviously the sign doesn't matter for equations of motion. Oct 24, 2012 at 20:51