Homogeneous Maxwell equations from the Bianchi identity

Question

It is easily proven that:

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

Lots of sources say this equation implies the Homogeneous Maxwell equations: $\nabla \times \mathbf E + \dfrac{\partial \mathbf B}{\partial t} = 0$ and $\nabla \cdot \mathbf B = 0$.

But I cannot find a proof of this derivation.

This is clear from their definitons: $\mathbf B = \nabla \times \mathbf A$ and $\mathbf E = -\nabla \phi - \dot{\mathbf A}$, since $\nabla \cdot (\nabla \times \mathbf v) = 0$ and $\nabla \times (\nabla f) = 0$.

But how can I find these results from the Bianchi identity?

Answer 1

The Bianchi identity, or just the fact that $F_{\mu\nu}$ is antisymmetry, leads to:

$$\epsilon^{\alpha\beta\gamma\delta} \partial_\beta F_{\gamma\delta} = 0$$

Noting that $E^i = -F^{0i}$ and that $B^i = -\frac{1}{2}\epsilon_{ijk}F^{jk}$ where $i,j,k = 1,2,3$, then we can choose $\alpha = 0$, or $\alpha = i$.

For $\alpha = 0$, we have $\epsilon^{0ijk}\partial_i F_{jk} = 0$ so $\partial_i (\epsilon_{ijk}F^{jk}) = 0$ and $\nabla \cdot \mathbf B = 0$.

For $\alpha = i$, we have:

\begin{align*} \epsilon^{i0jk}\partial_0 F_{jk } + \epsilon^{ij0k}\partial_j F_{0k} + \epsilon^{ijk0}\partial_j F_{k0} &= 0 \\ \partial_0(\epsilon_{ijk}F^{jk}) + 2 \epsilon_{ijk} \partial_j F^{0k} &= 0 \\ 2\dfrac{\partial \mathbf B}{\partial t} + 2 \nabla \times \mathbf E &= 0 \\ \nabla \times \mathbf E + \dfrac{\partial \mathbf B}{\partial t} &= 0 \end{align*}