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I have studied some of the relevant Q&A here. Everything is quite satisfactory. But is there any way to prove homogeneous part of 4 Maxwell equation from Lagrangian formalism, i.e constructing the Lagrangian and applying Langranges formulation. I know about the path integral formalism also. But I am interested in the equations, $\vec{\nabla} \cdot \vec{B} ~=~ 0 \qquad ``\text{no magnetic monopole"} \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\qquad ``\text{Faraday's law"}$ Can I get them without using the Bianchi identity?

Qmechanic
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Gourav
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1 Answers1

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You can get these equations if you allow the introduction of a potential. If $\vec{B} = \vec{\nabla} \times \vec{A}$ and $\vec{E} = - \vec{\nabla} \phi - \partial \vec{A} /\partial t$, you can then assemble $\phi$ and $\vec{A}$ into a four-vector potential: $$ A^\mu = (\phi, \vec{A}). $$ It then follows that $$ F_{\mu \nu} = \frac{\partial A_\nu}{\partial x^\mu} - \frac{\partial A_\nu}{\partial x^\mu} $$ and the Bianchi identity then follows from the equality of mixed partials:
$$ \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{\nu \lambda} + \partial_{\nu} F_{\lambda \mu} = 0. $$

However, this is a bit of a cheat, since the statement that a (four-)vector potential exists is actually a stronger statement than the statement that the Bianchi identity holds. These two statements are equivalent if you're working in a space that is topologically trivial (like Minkowski space), but if you have "holes" in your spacetime you can end up with situation where the Bianchi identity holds but there does not exist a vector potential such that $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. This observation is the basis of a beautiful branch of mathematics called de Rham cohomology, which I never miss an opportunity to mention because I like it so much.

Michael Seifert
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