Peskin Schroeder - An introduction to quantum field theory
Chapter 2 - The Klein-Gordon field
Problem 2-1 : Classical electromagnetism (with no sources) follows from the action
\begin{equation} S = \int d^4 x (-\frac{1}{4} F_{\mu\nu} F^{\mu\nu}) \end{equation}where $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$.
- Derive Maxwell's equations as the Euler-Lagrange equations of this action, treating the components $A_\mu(x)$ as the dynamical variables. Write the equations in standard form by identifying $E^i = -F^{0i}$ and $\epsilon^{ijk} B^k = -F^{ij}$.
- Construct the energy-momentum tensor for this theory. Note that the usual procedure does not result in a symmetric tensor. To remedy that, we can add
Solution: For problem a : Let's first consider the derivatives of the Lagrangian with respect to the field and its derivatives. For the field itself, we have the relation
\begin{equation} \frac{\partial A^\mu}{\partial A^\nu} = \delta^\mu_\nu,\ \frac{\partial \partial_\alpha A^\mu}{\partial \partial_\beta A^\nu} = \delta^\beta_\alpha \delta^\mu_\nu \end{equation}with the cross derivatives of the field by its derivatives and vice-versa being $0$. As the metric tensor commutes with derivatives here, we can also derive
\begin{equation} \frac{\partial A_\mu}{\partial A^\nu} = \frac{\partial \eta_{\alpha\mu}A^\alpha}{\partial A^\nu} = \eta_{\alpha\mu} \delta^\alpha_\nu = \eta_{\nu\mu} \end{equation}and similarly for derivatives,
\begin{equation} \frac{\partial \partial_\alpha A_\mu}{\partial (\partial_\beta A^\nu)} = \eta_{\sigma\mu} \frac{\partial \partial_\alpha A^\sigma}{\partial (\partial_\beta A^\nu)} = \eta_{\sigma\mu} \delta^\beta_\alpha \delta^\sigma_\nu = \eta_{\nu\mu} \delta^\beta_\alpha \end{equation}Any number of other variants can be derived using this method.
The derivatives of the electromagnetic tensor $F$ are trivially $0$ with respect to the field, and with respect to its derivatives :
\begin{eqnarray} \frac{\partial F_{\mu\nu}}{\partial (\partial_\beta A^\sigma)} &=& \frac{\partial}{\partial (\partial_\beta A^\sigma)} \partial_\mu A_\nu - \frac{\partial}{\partial \partial_\beta A^\sigma} \partial_\nu A_\mu\\ &=& \eta_{\nu\sigma} \delta^\beta_\mu - \eta_{\mu\sigma} \delta^\beta_\nu \end{eqnarray}The derivative of the covariant version of this tensor will be
\begin{eqnarray} \frac{\partial F^{\mu\nu}}{\partial \partial_\beta A^\sigma} &=& \eta^{\gamma\mu} \eta^{\delta\nu} \frac{\partial F_{\gamma\delta}}{\partial \partial_\beta A^\sigma} \\ &=& \eta^{\gamma\mu} \eta^{\delta\nu} (\eta_{\delta\sigma} \delta^\beta_\gamma - \eta_{\gamma\sigma} \delta^\beta_\delta )\\ &=& \eta^{\beta\mu} \delta^\nu_\sigma - \eta^{\beta\nu} \delta^\mu_\sigma \end{eqnarray}