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The Schrödinger lagrangian for complex fields is

$$L=\frac{1}{2m}(D_i \psi)^* Di \psi - \frac{i}{2} \left[\psi ^* D_0 \psi - (D_o \psi)^* \right] - \frac{1}{4}F^{\mu \nu}F_{\mu \nu}$$

Where $D_\mu = \partial _\mu + ieA_\mu$ is the covariant derivative and $F^{\mu \nu}$ is the field strength tensor. The equations of motion for the radiation fields take the form

$$\partial _\mu \left[\frac{\partial L}{\partial(\partial_\mu A_\nu)} -\right] - \frac{\partial L}{\partial A_\nu} = 0$$

We know that $A_\mu=(A_o,A_i)$. How do we express the derivative $\frac{\partial L}{\partial A_\nu}$ in terms of the $A_o$ and $A_i$ components in order to compute the derivatives $\frac{\partial L}{\partial A_0}$ and $\frac{\partial L}{\partial A_i}$?

Edit: I already use the chain rule and compute the derivatives

$$\frac{\partial L}{\partial A_\nu}=eJ^0 \delta^\nu _0 +eJ^i \delta^\nu _i=eJ^\nu$$

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closed as unclear what you're asking by ACuriousMind, Kyle Kanos, Ryan Unger, Gert, Sebastian Riese Apr 17 '16 at 20:59

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    $\begingroup$ I'm not clear on exactly what you're trying to do. Are you trying to "break up" the derivative into its temporal and spatial components? If so why? Could you give an example of where this would be useful? $\endgroup$ – Jold Apr 7 '16 at 2:44
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    $\begingroup$ Possible duplicate of physics.stackexchange.com/q/20647 $\endgroup$ – David Z Apr 7 '16 at 12:00
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Making use of Einstein's convention so that we sum implicitly over the repeated index in a single term, we have

$\frac{\partial{L}}{\partial{A_{\nu}}}=\frac{\partial{L}}{\partial{A_{\mu}}}\frac{\partial{A_{\mu}}}{\partial{A_{\nu}}}$

directly by the chain rule, where the greek index goes from $0$ to $n$ ($n$ being the total number of variables). Now, the RHS is a summation and we can, trivially, rearrange the equation separating the "temporal" term:

$\frac{\partial{L}}{\partial{A_{\nu}}}=\frac{\partial{L}}{\partial{A_{\mu}}}\frac{\partial{A_{\mu}}}{\partial{A_{\nu}}}=\frac{\partial{L}}{\partial{A_{0}}}\frac{\partial{A_{0}}}{\partial{A_{\nu}}}+\frac{\partial{L}}{\partial{A_{i}}}\frac{\partial{A_{i}}}{\partial{A_{\nu}}}$

where $i$ goes from $1$ to $n$. Of course that, since you defined $A_{\mu}$ as a n-vector, its components are most likely independent, therefore

$\frac{\partial{A_{\mu}}}{\partial{A_{\nu}}}=\delta^{\mu}_{\nu}$

where $\delta^{\mu}_{\nu}$ is equal to $1$ when both index are the same and $0$ otherwise. So, aplying the chain rule, as I did, is kind of useless.

Sorry if this wasn't your question, was the only thing I could think of.

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