# Canonical conjugate momenta of EM Field Lagrangian density

I have the EM Field Lagrangian density given as

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

where $$F^{\mu \nu}$$ is the Field strength tensor defined as $$F^{\mu \nu} = \partial^\mu A^\nu- \partial^\nu A^\mu$$, where $$A^\nu = (\phi/c, A^x,A^y A^z)$$

I need to find the canonical conjugate momenta ($$\pi^\alpha$$) w.r.t. this lagrangian density to find the primary constraint.

$$\pi^\alpha = \frac{\partial L}{\partial (\partial_0 A_\alpha)}$$

I am unable to solve it. please help me. ( I tried by expanding the field strength tensor and then solving it, but that didn't help)

The solution is $$\pi^\alpha = - F^{0 \alpha}$$

I'll do it step by step. First lower all the indices:

$$\begin{equation} -\frac{1}{4}F_{\mu \nu}F^{\mu\nu}=-\frac{1}{4}F_{\mu \nu}g^{\mu\rho}g^{\nu\sigma}F_{\rho\sigma} \end{equation}$$

then expand the products,

$$\begin{equation} -\frac{1}{4}(\partial_{\mu}A_{\nu}\partial_{\rho}A_{\sigma}-\partial_{\mu}A_{\nu}\partial_{\sigma}A_{\rho}-\partial_{\nu}A_{\mu}\partial_{\rho}A_{\sigma}+\partial_{\nu}A_{\mu}\partial_{\sigma}A_{\rho})g^{\mu\rho}g^{\nu\sigma} \end{equation}$$

Using the symmetry of the metric, I can rename the indices and then switch them, to get

$$\begin{equation} -\frac{1}{4}(2\partial_{\mu}A_{\nu}\partial_{\rho}A_{\sigma}-2\partial_{\mu}A_{\nu}\partial_{\sigma}A_{\rho})g^{\mu\rho}g^{\nu\sigma} \end{equation}$$

Now let us take a functional derivative with respect to $$\partial_{\lambda}A_{\alpha}$$

$$\begin{equation} \frac{\partial}{\partial(\partial_{\lambda}A_{\alpha})}(\partial_{\mu}A_{\nu}\partial_{\rho}A_{\sigma})g^{\mu\rho}g^{\nu\sigma}=(\delta^{\lambda}_{\mu}\delta^{\alpha}_{\nu}\partial{\rho}A_{\sigma}+\delta^{\lambda}_{\rho}\delta^{\alpha}_{\sigma}\partial_{\mu}A_{\nu})g^{\mu\rho}g^{\nu\sigma}=2\partial^{\lambda}A^{\alpha} \end{equation}$$

Similarly for the other term. Hence, $$\begin{equation} \frac{\partial}{\partial(\partial_{\lambda}A_{\alpha})}\left(-\frac{1}{4}F_{\mu \nu}g^{\mu\rho}g^{\nu\sigma}F_{\rho\sigma}\right)=-\left(\partial^{\lambda}A^{\alpha}-\partial^{\alpha}A^{\lambda}\right)=-F^{\lambda\alpha} \end{equation}$$

Now setting $$\lambda=0$$ gives the desired result.

• Very beautiful answer. Thankyou Feb 10, 2021 at 8:24