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In Heaviside-Lorentz units the Maxwell's equations are:

$$\nabla \cdot \vec{E} = \rho $$ $$ \nabla \times \vec{B} - \frac{\partial \vec{E}}{\partial t} = \vec{J}$$ $$ \nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 $$ $$ \nabla \cdot \vec{B} = 0$$

From EM Lagrangian: $$\mathcal{L} = \frac{-1}{4} F^{\mu \nu}F_{\mu\nu} + J^\mu A_\mu$$

I can derive the first two equations from the variation of the action integral: $S[A] = \int \mathcal{L} \, d^4x$. Is it possible to derive the last two equations from it?

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Maybe I'm being clueless as usual, but the up voted answers so far haven't answered the question, which asked for a derivation using variation of the electromagnetic action. –  Larry Harson Jul 19 '13 at 20:51
@LarryHarson, to be perfectly pedantic, the question does not, in fact, ask for a derivation but, rather, asks is it possible? –  Alfred Centauri Jul 19 '13 at 21:02
@LarryHarson Since you seem pretty annoyed by the absence of clarity of the present answers, I would like to make a sum-up of them. So the question is is it possible to derive the Maxwell's equations without source from the action $S(A)$. The answer is YOU DON'T NEED, because they are *contained* in the writing $S(A)$: they are definition of the fields in terms of the potentials. That's what everybody below have tried to say: since $F=dA$, then $dF=0$, called the Bianchi equality. –  FraSchelle Jul 20 '13 at 7:24
@LarryHarson Otherwise how could you start from an action $S(A)$ depending on the potential, and end with equations for fields, if you don't choose $B=\nabla\times A$ and $E=\partial A + \nabla\phi$ ? These two field <-> potential relations\definitions impose $\nabla\cdot B =0$ and $\partial B + \nabla\times E =0$. –  FraSchelle Jul 20 '13 at 7:26
@Oaoa yes, you and the others are right - thanks. –  Larry Harson Jul 22 '13 at 13:21
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3 Answers

up vote 12 down vote accepted

Assume for simplicity that the speed of light $c=1$. The existence of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$ alone implies that the source-free Maxwell equations $$\vec{\nabla} \cdot \vec{B} ~=~ 0 \qquad ``\text{no magnetic monopole"}$$

$$ \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\qquad ``\text{Faraday's law"}$$

are already identically satisfied. To prove them, just use the definition of the electric field

$$\vec{E}~:=~-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t},$$

and the magnetic field


in terms of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$.

The above is more naturally discussed in a manifestly Lorentz-covariant notation. OP might also find this Phys.SE post interesting.

Thus, to repeat, even before starting varying the Maxwell action $S[A]$, the fact that the action $S[A]$ is formulated in terms the gauge $4$-potential $A^{\mu}$ means that the source-free Maxwell equations are identically satisfied. Phrased differently, since the source-free Maxwell equations are manifestly implemented from the very beginning in this approach, varying the Maxwell action $S[A]$ will not affect the status of the source-free Maxwell equations whatsoever.

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Can you please elaborate? I have not understood it. –  Ome Jul 19 '13 at 19:30
-1 The question asked if it's possible to derive the last two equations using variation of the electromagnetic action. –  Larry Harson Jul 19 '13 at 21:13
@LarryHarson you have not seriously downvoted one of Qmechanic's well explained and nicely written answers, or have you? This is hilarious, instead of downvoting him you should better try to learn from what he says. Anyway, I cancelled it :-P –  Dilaton Jul 19 '13 at 21:22
@Qmechanic: In your answer, the last paragraph is awesome. I have just loved it. :D Sometimes, I just envy you. These things are so clear to you! :P –  Ome Jul 19 '13 at 22:54
@Larry Harson: The OP has already told you what he wanted. Now you should reverse your downvote on this answer. EDIT: You actually downvoted after the clarification? This DOES answer the question. –  Dimensio1n0 Jul 22 '13 at 13:53
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Note that since $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, we have the equation $$ \partial_\mu F_{\nu\alpha} + \partial_\alpha F_{\mu\nu} + \partial_\nu F_{\alpha\mu} = 0 $$ This equation is called the Bianchi Identity. This equation is separate from the equations of motion one obtains from varying the action. It can be shown that the Bianchi equation is equivalent to the last two equations you have mentioned in your question.

NOTE: To elaborate on @Qmechanic's answer, the very fact that $F_{\mu\nu}$ can be written as $\partial_\mu A_\nu - \partial_\nu A_\mu$ is itself a consequence of the last two equations.

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-1 The question asked if it's possible to derive the last two equations using variation of the electromagnetic action. –  Larry Harson Jul 19 '13 at 21:12
@LarryHarson you know that it is completely legitimate to write partial answers to a question? Downvoting all answers that do not completely answer the question is almost trolling ... –  Dilaton Jul 19 '13 at 21:27
@LarryHarson: Since the OP has clarified what he wanted, your downvote was unjustified Keep checking this answer daily to see if it has been edited so that you can reverse your downvote. . EDIT: I see that you downvoted after the clarificaiton... How does this not answer the question ? . –  Dimensio1n0 Jul 22 '13 at 13:55
@Dimension10 and prahar: yeah, I'll upvote this when it gets edited. –  Larry Harson Jul 22 '13 at 13:59
@LarryHarson: I've edited the <s>question</s> ANSWER now. Please reverse your downvote . –  Dimensio1n0 Aug 14 '13 at 12:15
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From a geometric perspective, the last two equations are a consequence of:

(1)F = dA (Faraday tensor is the exterior derivative of the four-potential)

(2) dd = 0 (the exterior derivative of the exterior derivative vanishes)


(3) dF = 0

which gives the last two equations in your question.

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@LarryHarson: How does this not address the question? . –  Dimensio1n0 Jul 22 '13 at 13:59
I've removed my downvote. –  Larry Harson Jul 22 '13 at 14:04
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