Canonical momentum of a 4-vector field

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.

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

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 $\pi_\nu =2(C_1\partial^0 A_\nu + C_2\partial_\nu A^0)$.

• Okay thank you, can you please say me how you come to the solution of the first term? Because I really do not know how to compute these 4 derivatives. Maybe you can show me the full computation way in steps as a picture or something? That would be really helpful for my understanding :) Commented Apr 19, 2018 at 13:10
• @Armani42 To differentiate $\partial_\mu A^\nu\partial^\mu A_\nu$ with respect to $\partial_0 A^\nu$, just use the project rule. The first factor gives $\delta_\mu^0$, so the contribution is $\partial^0 A_\nu$. The second factor gives $g^{\mu 0}$. giving another copy of the same result. It looks complicated, but it's just a variation on $\partial_x y^2 = 2x\partial_x y$.
– J.G.
Commented Apr 19, 2018 at 13:39
• so the first term with C1 I've been written out into a vector and differentiated it. But now I have a problem with the C2 term because there we have $(\partial_{nu} A_{mu})(\partial^{mu} A^{nu})$. So the indices of A and the derivatives are not the same anymore so how to deal with that? In the link I post, you can find a picture of how I calculated the C1 term: ibb.co/eMLoRn So if you look at the last vector, you see what I mean, I now have the $\partial_{mu}$ and then the $\partial^{nu}$ so my question: What has to be done that I can go on computing and come to your result(C2)? Commented Apr 20, 2018 at 8:11
• @Armani42 Rewrite the term as $C_2\partial_\rho A_\mu\partial^\mu A^\rho$, so by the product rule the derivative is $C_2(\delta_\rho^0 \eta_{\mu\nu}\partial^\mu A^\rho+\partial_\rho A_\mu \eta^{\mu 0}\delta^\rho_\nu)$. Simplify that.
– J.G.
Commented Apr 20, 2018 at 8:14
• okay, thank you but how can I write that into my 4-vector on the picture? I just want to make it in detail that I really understand what is going on. Commented Apr 20, 2018 at 8:18