# About Homogeneous Maxwell equation from EM Lagrangian

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?

• The fact that there exists a four-potential implies those equations, and this fact is built into the Lagrangian. – Javier Mar 16 '18 at 15:38
• Possible duplicate: physics.stackexchange.com/q/71611/2451 – Qmechanic Mar 16 '18 at 15:48

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.
• @Gourav: As mentioned in the comments above, if you're viewing $A_\mu$ as your "field variable", then the existence of $A_\mu$ implies the Bianchi identity automatically. This will be true even if the action is not minimized (i.e., $\partial^\mu F_{\mu \nu} \neq J_\nu$); once you accept the existence of $A_\mu$, then the Bianchi identity is in some sense more fundamental than the inhomogeneous equations. – Michael Seifert Mar 18 '18 at 15:41