Whether you can get the first couple of Maxwell equations from a variational principle? In the second volume of the Landau theoretical physics said that it is impossible.

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    $\begingroup$ In which section do they say that? $\endgroup$ – Ján Lalinský Jun 17 '14 at 6:23
  • $\begingroup$ classical electrodynamics in the 4 section $\endgroup$ – user50617 Jun 17 '14 at 6:26
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    $\begingroup$ Please add details, such as which Maxwells equations you are talking about and the section of Landau in the question. $\endgroup$ – user7757 Jun 17 '14 at 6:53
  • $\begingroup$ Ok, how can you get the first couple of Maxwell equations from a variational principle? I talking about the equation $dF=0$, where d is internal differential. it is in geometric form. How see that the $dF=0$ is the consequence of the variational principle? $\endgroup$ – user50617 Jun 17 '14 at 6:54
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/71611/2451 $\endgroup$ – Qmechanic Jun 17 '14 at 7:13

The Maxwell Lagrangian is given by,


where $F_{\mu\nu}$ is the field-strength of the gauge field, or alternatively may be interpreted as the curvature of a $U(1)$ Lie algebra valued connection, $A_{\mu}$. By applying the variational principle we obtain,

$$\partial_\mu F^{\mu\nu}=0$$

in vacuum. In terms of the electric and magnetic fields,

$$\nabla \cdot \vec{E}=0 \quad \quad \partial_t \vec{E}=\nabla \times \vec{B}$$

we recover two of Maxwell's equations. Notice, in differential form language, $F=dA$, i.e. the curvature is an exact form, and all exact forms are also closed under the operation of exterior differentiation, i.e.

$$dF=d^2 A=0$$

Converting the above expression to a tensor equation, using the standard definition,

$$d\omega^{(n)}_{a_1 \dots a_n}=\frac{1}{n!} \left( \partial_{[a_1} \omega_{\dots a_n]}\right)$$

recovers the tensor form of the Bianchi identity,

$$\partial_\lambda F_{\mu\nu}+\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}=0$$

from which the two remaining Maxwell equations follow:

$$\nabla \cdot \vec{B}=0\quad \quad \partial_t \vec{B}=-\nabla\times \vec{E}$$

Recall, given the spin connection $\omega$, by Cartan's second structure equation, the curvature form is,

$$\mathcal{R}=d\omega + \omega \wedge \omega$$

However, the Lie group $U(1)$ is Abelian, and the structure constants vanish, hence the above simplifies,


which is completely analogous to the definition of the electromagnetic field strength. Other gauge groups may not possess the same field-strength. For example, in quantum chromodynamics, $SU(3)$ is non-Abelian, and the extra term does not vanish; in tensor form:

$$G_{\mu\nu}^a=\partial_\mu A^a_\nu - \partial_\nu A^a_\mu + gf^{a}_{bc}A^{b}_\mu A^{c}_\nu$$

  • $\begingroup$ In what sense is $F_{\mu \nu}$ as the curvature of $A_{\mu}$? $\endgroup$ – user7757 Jun 17 '14 at 8:43
  • $\begingroup$ @ramanujan_dirac: See the updated answer. $\endgroup$ – JamalS Jun 17 '14 at 9:24

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