Equivalent form of Bianchi identity in electromagnetism

Question

In electromagnetism, we can write the Bianchi identity in terms of the field strength tensor $F_{\mu \nu}$ as,

$$ \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{\nu \lambda}+ \partial_{\nu} F_{\lambda \mu} = 0,\qquad \mu,\nu,\lambda=0,1,2,3. \tag{1}$$

Now, in a textbook I am reading (Classical Covariant Fields - Burgess), the Bianchi identity is given as,

$$ \sum_{j,k=1}^3\epsilon_{ijk} \partial_j E_k + \partial_t B_i = 0,\qquad i=1,2,3.\tag{2a}$$

and $$\sum_{i=1}^3\partial_i B_i = 0.\tag{2b} $$

However, I am struggling to see how these two forms are equivalent, i.e. starting from one equation (1), how can we arrive at the two others (2)?

Answer 1

Let's start by contracting the first equation with the 4-dimensional totally antisymmetric tensor $\epsilon^{\alpha\lambda\mu\nu}$. Thanks to the properties of $\epsilon^{\alpha\lambda\mu\nu}$ we then have

$$ \epsilon^{\alpha\lambda\mu\nu} \partial_{\lambda} F_{\mu\nu} = 0 . $$

Next we separate the Faraday tensor into its temporal (0) and spatial (1,2,3) components. This gives us the electric and magnetic fields: $F_{00}=0$; $F_{k0}=-F_{0k}=E_k$; $F_{ij}=\epsilon_{ijk} B^k$. If we set $\alpha=0$ we get $$ \epsilon^{0ijk} \partial_{i} F_{jk} = \epsilon^{ijk} \partial_{i} (\epsilon_{jkp} B^p) = 2 \partial_{i} B^i = 2 \nabla\cdot\mathbf{B} = 0 , $$ because $\epsilon^{ijk} \epsilon_{jkp}=2\delta_p^i$. For $\alpha=i$ we have: $$ \epsilon^{ij0k} \partial_{j} F_{0k} + \epsilon^{ijk0} \partial_{j} F_{k0} + \epsilon^{i0jk} \partial_{0} F_{jk} = 0 . $$

$$ - 2 \epsilon^{ijk} \partial_{j} E_{k} - \epsilon^{ijk} \partial_{0} (\epsilon_{jkp} B^p) = 0 . $$

$$ \epsilon^{ijk} \partial_{j} E_{k} + \partial_{0} B^i = 0 . $$

Answer 2

In reply to the additional query (posed in a comment): without loss of generality, assume i = 1. Then: $$ϵ^{ijk} ϵ_{jkp} = ϵ^{123}ϵ_{23p} + ϵ^{132}ϵ_{32p} = ϵ_{23p} - ϵ_{32p} = 2 ϵ_{23p} = 2 δ^1_p = 2 δ^i_p.$$