2
$\begingroup$

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

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Very beautiful answer. Thankyou $\endgroup$
    – sawan kt
    Feb 10, 2021 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.