Exercise 1.13

Consider adding to the Lagrangian for electromagnetism an additional term of the form $\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu \nu}F^{\rho\sigma} $.
(a) Express $\mathcal{L}'$ in terms of $\mathbf{E}$ and $\mathbf{B}$.
(b) Show that including $\mathcal{L}'$ does not affect Maxwell's equations. Can you think of a deep reason for this?

$$ F^{\mu \nu}= \begin{pmatrix} 0 & E_{1} & E_{2} & E_{3} \\ -E_{1} & 0 & B_{3} & -B_{2} \\ -E_{2} & -B_{3} & 0 & B_{1} \\ -E_{3} & B_{2} & -B_{1} & 0 \end{pmatrix} $$ (a) Given the antisymmetry of $F$ $(F^{\mu \nu}=-F^{\nu \mu})$, we notice that $$ \begin{align} \tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma} &=\tilde{\epsilon}_{\nu\mu\rho\sigma}F^{\nu\mu}F^{\rho\sigma} \\ &=\tilde{\epsilon}_{\mu\nu\sigma\rho}F^{\mu\nu}F^{\sigma\rho} \\ &=\tilde{\epsilon}_{\nu\mu\sigma\rho}F^{\nu\mu}F^{\sigma\rho} \\ &=\tilde{\epsilon}_{\rho\sigma\mu\nu}F^{\rho\sigma}F^{\mu\nu} \\ &=\tilde{\epsilon}_{\sigma\rho\mu\nu}F^{\sigma\rho}F^{\mu\nu} \\ &=\tilde{\epsilon}_{\rho\sigma\nu\mu}F^{\rho\sigma}F^{\nu\mu} \\ &=\tilde{\epsilon}_{\sigma\rho\nu\mu}F^{\sigma\rho}F^{\nu\mu} \end{align} $$ so we have only 3 terms to consider. $$ \begin{align} \mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu \nu}F^{\rho\sigma} &=8(F^{01}F^{23}-F^{02}F^{13}+F^{03}F^{12}) \\ &=8\Big(E_{1}B_{1}-E_{2}(-B_{2})+E_{3}B_{3}\Big) \\ &=8\Big(E_{1}B_{1}+E_{2}B_{2}+E_{3}B_{3}\Big) \\ &=8\;\mathbf{E}\cdot\mathbf{B} \end{align} $$ (b) We have $$ \mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu \nu}F^{\rho\sigma} =\tilde{\epsilon}_{\mu\nu\rho\sigma}(\partial^{\mu}A^{\nu}-\partial ^{\nu}A^{\mu})(\partial ^{\rho}A^{\sigma}-\partial ^{\sigma}A^{\rho}) $$ Straightforwardly $$ \frac{ \partial \mathcal{L}' }{ \partial A_{\nu} } =0 $$ Less straightforwardly $$ \begin{align} \frac{ \partial \mathcal{L}' }{ \partial (\partial_{\mu}A_{\nu}) } &=\frac{ \partial (\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu \nu}F^{\rho\sigma} ) }{ \partial (\partial_{\mu}A_{\nu}) } \\ &=\eta^{\alpha \mu}\eta^{\beta \nu}\eta^{\gamma \rho}\eta^{\lambda \sigma}\tilde{\epsilon}_{\mu\nu\rho\sigma}\left[ \left( \frac{ \partial F_{\alpha\beta} }{ \partial (\partial_{\mu}A_{\nu}) } \right)F_{\gamma\lambda} + F_{\alpha\beta}\left( \frac{ \partial F_{\gamma\lambda} }{ \partial (\partial_{\mu}A_{\nu}) } \right)\right] \\ &=\eta^{\alpha \mu}\eta^{\beta \nu}\eta^{\gamma \rho}\eta^{\lambda \sigma}\tilde{\epsilon}_{\mu\nu\rho\sigma}\left[ \left( \delta ^{\mu}_{\alpha}\delta ^{\nu}_{\beta}-\delta ^{\mu}_{\beta}\delta ^{\nu}_{\alpha} \right)F_{\gamma\lambda} +\left( \delta ^{\mu}_{\gamma}\delta ^{\nu}_{\lambda}-\delta ^{\mu}_{\lambda}\delta ^{\nu}_{\gamma} \right)F_{\alpha\beta}\right] \\ &=\left( \tilde{\epsilon}_{\alpha\beta\rho\sigma}-\tilde{\epsilon}_{\beta\alpha\rho\sigma}\right)\eta^{\alpha \mu}\eta^{\beta \nu}\eta^{\gamma \rho}\eta^{\lambda \sigma}F_{\gamma\lambda} + \left( \tilde{\epsilon}_{\gamma\lambda\rho\sigma}-\tilde{\epsilon}_{\lambda\gamma\rho\sigma}\right)\eta^{\alpha \mu}\eta^{\beta \nu}\eta^{\gamma \rho}\eta^{\lambda \sigma}F_{\alpha\beta} \\ &=\left( \tilde{\epsilon}_{\alpha\beta\rho\sigma}-\tilde{\epsilon}_{\beta\alpha\rho\sigma}\right)\eta^{\alpha \mu}\eta^{\beta \nu}F^{\rho\sigma} + \left( \tilde{\epsilon}_{\gamma\lambda\rho\sigma}-\tilde{\epsilon}_{\lambda\gamma\rho\sigma}\right)\eta^{\gamma \rho}\eta^{\lambda \sigma}F^{\mu \nu} \\ &=2\,(\tilde{\epsilon}_{\alpha\beta\rho\sigma}\eta^{\alpha \mu}\eta^{\beta \nu}F^{\rho\sigma} + \tilde{\epsilon}_{\gamma\lambda\rho\sigma}\eta^{\gamma \rho}\eta^{\lambda \sigma}F^{\mu \nu}) \\ &=2\,(\underbrace{ \tilde{\epsilon}^{\mu \nu}{}_{\rho\sigma}F^{\rho\sigma} }_{ =\,0 } + \cancelto{ 0 }{ \tilde{\epsilon}^{\rho\sigma}{}_{\rho\sigma} }F^{\mu \nu}) =0 \end{align} $$ Therefore $$ \underbrace{ \frac{ \partial (\mathcal{L}+\mathcal{L}') }{ \partial A_{\nu} } }_{ J^{\nu} } -\partial_{\mu}\underbrace{ \left( \frac{ \partial (\mathcal{L}+\mathcal{L}') }{ \partial (\partial_{\mu}A_{\nu}) } \right) }_{ -F^{\mu \nu} }=0 \qquad \implies \qquad \partial_{\mu}F^{\nu \mu}=J^{\nu} $$ This symmetry is probably related to a conserved quantity, I don't yet see which.