Canonical momentum of a 4-vector field

Question

In a four-vector field theory, we have a given Lagrangian: $$\mathscr{L} = C_{1} (\partial_{\nu} A_{\mu}) (\partial^{\nu} A^{\mu}) + C_2 (\partial_{\nu} A_{\mu}) (\partial^{\mu} A^{\nu}) + C_3 A_{\mu} A^\mu.$$ From that we should compute the canonical momentum: $$\pi_\nu = \frac{\partial \mathscr{L}}{\partial \partial_0A^\nu}.$$

My question here is:

Can someone please explain me how to take that derivative? Because I do not understand how to derivate these terms and I do not find an example where this is done for a 4-vector field explicitly.

Answer 1

Since this is very homework-like, a full solution should not be presented.

Nevertheless, OP seems to generally have confusion regarding all the 4-vector indices and seems to want a more detailed explanation.

For those reasons, I will show how to compute the derivative in great detail, for just part of the Lagrangian:

$$\mathscr{L}_1 = C_{1} (\partial_{\nu} A_{\mu}) (\partial^{\nu} A^{\mu})$$

Also, in order to avoid confusion between free and summed indices, I will compute $$ \frac{\partial \mathscr{L}_1}{\partial (\partial_0 A^\rho)}\;, $$ where I am using $\rho$ instead of $\nu$, since $\nu$ is already being used as a summed index in the Lagrangian.

First, rewrite $\mathscr{L}_1$ as: $$ C_1(\partial_{\nu} A^{\alpha}) g_{\mu\alpha}g^{\nu\beta} (\partial_{\beta} A^{\mu}) $$ so that all the derivatives have the lower index (just like in the derivative of the Lagrangian we are interested in) and all the fields have upper index.

Then use: $$ \frac{\partial \partial_\gamma A^{\delta}}{\partial \partial_0 A^{\rho}} = \delta^{0}_\gamma\delta^{\delta}_\rho $$

And use the product rule to see that: $$ \frac{\partial \mathscr{L}_1}{\partial (\partial_0 A^\rho)} =C_1\frac{\partial}{\partial (\partial_0 A^\rho)}\left((\partial_{\nu} A^{\alpha}) g_{\mu\alpha}g^{\nu\beta} (\partial_{\beta} A^{\mu})\right) $$ $$ =C_1\left( \delta^{0}_\nu\delta^{\alpha}_\rho g_{\mu\alpha}g^{\nu\beta}(\partial_{\beta} A^{\mu}) +(\partial_{\nu} A^{\alpha}) g_{\mu\alpha}g^{\nu\beta}\delta^{0}_\beta\delta^{\mu}_\rho \right) $$ $$ =C_1\left(\partial^0 A_\rho + \partial^0A_\rho\right) = 2C_1\partial^0 A_\rho $$

So, the final result: $$ \frac{\partial \mathscr{L}_1}{\partial (\partial_0 A^\rho)} = 2C_1\partial^0 A_\rho $$

Answer 2

There are two contributions to $\pi_\nu$, one each from the $C_i$ terms. The first term gives $2\partial^0 A_\nu$, since we can rewrite a term as $C_1\partial_\mu A^\nu \partial^\mu A_\nu$. (This uses a combination of index swapping and index-height changing.) And with a similar treatment of the last term, we arrive at $\pi_\nu =2(C_1\partial^0 A_\nu + C_2\partial_\nu A^0)$.