# Derivatives with upper and lower indices

I'm studying classical and quantum field theory, but evaluating derivatives of fields (scalar and/or vector) described with upper and lower indices is somewhat new to me. I'm trying to evaluate

$$\frac{\partial}{\partial A_\mu} (F_{\mu \nu}) \text{,} \tag{1}$$

where $$F_{\mu \nu}= \partial_{\mu} A_{\nu}- \partial_{\nu} A_{\mu}.\tag{2}$$

Just for confirmation, is the result of the above partial derivative just simply $- \partial_{\nu}$?

I'm trying to use the Euler-Lagrange equation of classical field theory to derive the equations of motion for a Lagrangian density that describes an abelian Higgs model with given potential $- \lambda(\phi^{\dagger} \phi - a^2)^2$ for $\lambda, a$ arbitrary constants and $\phi$ a complex-valued scalar field, but the evaluation of derivatives with upper and lower indices makes it a lot messier. Is there a better or quicker method?

• Comment to the question (v2): The first expression (1) with the partial derivative vanishes identically. Is that what you want to know? – Qmechanic Sep 19 '15 at 17:48
• Yes, that's right. – Libertron Sep 19 '15 at 17:49
• You can't get easily around the upper and lower indices, as they encode transformation information that is relevant for relativistic invariance. – Sebastian Riese Sep 19 '15 at 18:04

First, you should be careful with your choice of indices. What you have written in Eq. 1 implies a summation over $\mu$ that I don't think you actually want. It is true that $$\frac{\partial F_{\mu\nu}}{\partial A_{\sigma}}=0,$$ but that is just because $F_{\mu\nu}$ depends on the derivatives of $A_{\mu}$ and not $A_{\mu}$ explicitly. Let's work out a more full example to see what's going on with the upper and lower indices. Suppose we have the Lagrangian $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}^2.$$ Now, we want to calculate the Euler-Lagrange equations with respect to the gauge field $A_{\mu}$: $$\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A_{\nu})} =\frac{\partial\mathcal{L}}{\partial A_{\nu}}.$$ Recall that indices are raised and lowered using the metric (in this example Minkowski metric $\eta$) $$A^{\mu} = \eta^{\mu\nu}A_{\nu}.$$ Let's start by rewriting the Lagrangian $$\begin{split} \mathcal{L}&=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=-\frac{1}{4}\eta^{\mu\sigma}\eta^{\nu\rho}F_{\mu\nu}F_{\sigma\rho} \\ &= -\frac{1}{4}\eta^{\mu\sigma}\eta^{\nu\rho}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}) (\partial_{\sigma}A_{\rho}-\partial_{\rho}A_{\sigma}) \end{split}$$ Now, we are already using $\mu,\nu,\rho$ and $\sigma$ as dummy summation indices, so it would be unwise of us to reuse any of these when computing the Euler-Lagrange equations. We compute $$\begin{split} \frac{\partial\mathcal{L}}{\partial(\partial_{\kappa}A_{\lambda})} &=-\frac{1}{4}\eta^{\mu\sigma}\eta^{\nu\sigma}\left[(\delta^{\kappa}_{\mu}\delta^{\lambda}_{\nu} - \delta^{\lambda}_{\mu}\delta^{\kappa}_{\nu})(\partial_{\sigma}A_{\rho}-\partial_{\rho}A_{\sigma}) \right. \\ &\left.+ (\delta^{\kappa}_{\sigma}\delta^{\lambda}_{\rho} - \delta^{\lambda}_{\sigma}\delta^{\kappa}_{\rho})(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}) \right] \\ &= -\frac{1}{4}\left[(\eta^{\kappa\sigma}\eta^{\lambda\rho}-\eta^{\kappa\rho}\eta^{\lambda\sigma})(\partial_{\sigma}A_{\rho}-\partial_{\rho}A_{\sigma}) \right. \\ & + \left. (\eta^{\mu\kappa}\eta^{\nu\lambda}-\eta^{\mu\lambda}\eta^{\nu\kappa}) (\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}) \right] \\ &= -(\partial^{\kappa}A^{\lambda} - \partial^{\lambda}A^{\kappa}) \\ &= -F^{\kappa\lambda} \end{split}$$ And again $$\frac{\partial\mathcal{L}}{\partial A_{\lambda}} = 0,$$ since $F$ has no explicit dependence on $A$. Thus $$\partial_{\kappa}F^{\kappa\lambda}=0.$$