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.

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 $\pi_\nu =2(C_1\partial^0 A_\nu + C_2\partial_\nu A^0)$.
• @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. Apr 19 '18 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)? – Armani42 Apr 20 '18 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. Apr 20 '18 at 8:14