Canonical momentum ##\pi^\rho## of the electromagnetic field

[Riotto]

In David Tong's QFT notes (see http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf , page 131, Eq. 6.38) the expression for canonical momentum $\pi^0$ is given by $\pi^0=-\partial_\rho A^\rho$ while my calculation gives $\pi^\rho=-\partial_0 A^\rho$ so that $\pi^0=-\partial_0 A^0$. Is it wrong in Tong's note?

 

[Demystifier]

It depends on the choice of $\alpha$ in (6.37). Tong writes "We will use $\alpha=1$." You might have been using $\alpha=\infty$, but even then your result is wrong because $F_{\mu\nu}$ is antisymmetric so $F_{00}=\partial_0A_0-\partial_0A_0=0$.

 

[Riotto]

Can you show a few lines of computation because I cannot figure out how are you getting that result. No, I am using $\alpha=1$ .

 

[Riotto]

It depends on the choice of $\alpha$ in (6.37). Tong writes "We will use $\alpha=1$." You might have been using $\alpha=\infty$, but even then your result is wrong because $F_{\mu\nu}$ is antisymmetric so $F_{00}=\partial_0A_0-\partial_0A_0=0$.

Can you show a few lines of computation because I cannot figure out how are you getting that result. No, I am using $\alpha=1.$

 

[Antarres]

So you have Lagrangian of the form($\alpha = 1$):
$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$
Calculation of canonical momenta is as follows:
$$\pi^\mu \equiv \frac{\partial\mathcal{L}}{\partial \dot{A}_\mu} = -\frac{1}{2}\frac{\partial F_{\nu\rho}}{\partial\dot{A}_{\mu}}F^{\nu\rho} - \partial_\nu A^\nu\eta^{\mu 0}$$
The first term gives:
$$\frac{\partial F_{\nu\rho}}{\partial\dot{A}_{\mu}} = \frac{\partial}{\partial (\partial_0 A_{\mu})}(\partial_\nu A_{\rho} - \partial_\rho A_{\nu}) = \delta^0_\nu \delta^\mu_\rho - \delta^0_\rho \delta^\mu_\nu$$

Finally we obtain:
$$\pi^\mu = F^{\mu 0} - \eta^{\mu 0}\partial_\nu A^\nu$$

This agrees with Tong, so there you go, that's the calculation. It's just basic differentiation though, so I don't know what was the problem there.