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)?
Best Answer
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: $$ \epsilon^{ij0k} \partial_{j} F_{0k} + \epsilon^{ijk0} \partial_{j} F_{k0} + \epsilon^{i0jk} \partial_{0} F_{jk} = 0 . $$
$$ - 2 \epsilon^{ijk} \partial_{j} E_{k} - \epsilon^{ijk} \partial_{0} (\epsilon_{jkp} B^p) = 0 . $$
$$ \epsilon^{ijk} \partial_{j} E_{k} + \partial_{0} B^i = 0 . $$