Exercise 1.11

Verify that (1.98) is indeed equivalent to (1.97), and that they are both equivalent to the last two equations in (1.93).

Exploiting the antisymmetry of $F_{\mu \nu}$, we find that $$ \partial_{[\mu}F_{\nu\lambda]}=\frac{1}{6}\left(\underbrace{ \partial_{\mu}F_{\nu\lambda}-\partial_{\mu}F_{\lambda\nu} }_{2\times\partial_{\mu}F_{\nu\lambda} }+\overbrace{ \partial_{\nu}F_{\lambda \mu}-\partial_{\nu}F_{\mu\lambda} }^{2\times\partial_{\nu}F_{\lambda \mu} }+\underbrace{ \partial_{\lambda}F_{\mu \nu}-\partial_{\lambda}F_{\nu\mu} }_{2\times\partial_{\lambda}F_{\mu \nu} } \right)=0 $$ is indeed equivalent to $$ \partial_{\mu}F_{\nu\lambda}+\partial_{\nu}F_{\lambda \mu}+\partial_{\lambda}F_{\mu \nu} =0 $$

We recall that $$ F_{\mu \nu}= \begin{pmatrix} 0 & -E_{1} & -E_{2} & -E_{3} \\ E_{1} & 0 & B_{3} & -B_{2} \\ E_{2} & -B_{3} & 0 & B_{1} \\ E_{3} & B_{2} & -B_{1} & 0 \end{pmatrix} \qquad F^{\mu \nu}= \begin{pmatrix} 0 & E_{1} & E_{2} & E_{3} \\ -E_{1} & 0 & B_{3} & -B_{2} \\ -E_{2} & -B_{3} & 0 & B_{1} \\ -E_{3} & B_{2} & -B_{1} & 0 \end{pmatrix} $$ 

• For $\mu=\nu=\lambda=0$, all terms in the LHS are equal to 0, the equation is trivial. $$ \partial_{0}\cancelto{ 0 }{ F_{00} }+\partial_{0}\cancelto{ 0 }{ F_{0 0} }+\partial_{0}\cancelto{ 0 }{ F_{00} }=0 $$ • For $\mu=\nu=0$ and $\lambda=i$, the LHS vanishes as well. $$ \begin{align} \partial_{0}F_{0i}+\partial_{0}F_{i 0}+\partial_{i}\cancelto{ 0 }{ F_{0 0} }&=0 \\ \partial_{0}E_{i}+\partial_{0}(-E_{i})&=0 \\ \partial_{0}(\cancel{ E_{i}-E_{i} })&=0 \end{align} $$

• For $\mu=i$, $\nu=j$ and $\lambda=0$ $$ \begin{align} \partial_{i}F_{j0}+\partial_{j}F_{0 i}+\partial_{0}F_{i j}&=0 \\ -\partial_{i}F^{j0}-\partial_{j}F^{0 i}+\partial_{0}F^{i j}&=0 \\ %\swarrow \quad\; &(F^{ij}=F_{ij}) \\ \underbrace{ \partial_{i}E_{j}-\partial_{j}E_{i} }_{\;\;\,\searrow}+\tilde{\epsilon}^{ijk}\partial_{0}B_{k}&=0 \\ \overbrace{ \tilde{\epsilon}^{ijk}\partial_{j}E_{k} }^{ } +\tilde{\epsilon}^{ijk}\partial_{0}B_{k}&=0 \end{align} $$ The $\tilde{\epsilon}^{ijk}$ multiplies the two terms in the LHS either by 0, 1 or -1. In the first case, the equation is trivial. The two other cases are equivalent to each other. We have to keep the first $\tilde{\epsilon}$ to sum over the two $E$ terms. However, we can remove it from the $B$ term: the common $i$ index enforces the signs in the entire LHS. $$ \tilde{\epsilon}^{ijk}\partial_{j}E_{k}+\partial_{0}B^{i}=0 $$ We find the third part of (1.93).

• For $\mu=i$, $\nu=j$ and $\lambda=k$ $$ \begin{align} \partial_{i}F_{jk}+\partial_{j}F_{k i}+\partial_{k}F_{i j}&=0 \\ \tilde{\epsilon}_{ljk}\partial_{i}B^{l}+\tilde{\epsilon}_{ilk}\partial_{j}B^{l}+\tilde{\epsilon}_{ijl}\partial_{k}B^{l}&=0 \\ \tilde{\epsilon}_{ijk}(\partial_{i}B^{i}+\partial_{j}B^{j}+\partial_{k}B^{k})&=0 \\ 3\times\partial_{i}B^{i}&=0 \\ \partial_{i}B^{i}&=0 \\ \end{align} $$ we find the fourth part of (1.93).