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 $\pi^\alpha = - F^{0 \alpha}$


1 Answer 1


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.

  • $\begingroup$ Very beautiful answer. Thankyou $\endgroup$
    – sawan kt
    Feb 10, 2021 at 8:24

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