Expressing Maxwell's equations in tensor form using Electromagnetic field strength tensor [closed]

Question

I have yet another derivation question from Carroll's General Relativity textbook. Given the electromagnetic field strength tensor is of the form: $$ F_{\mu\upsilon} = \left( \begin{matrix} 0 & -E_1 & -E_2 & -E_3\\ E_1 & 0 & B_3 & -B_2\\ E_2 & -B_3 & 0 & B_1\\ E_3 & B_2 & -B_1 & 0\\ \end{matrix} \right) = -F_{\upsilon\mu}$$

The Maxwell's equations are expressed in component notation: $$ \bar{\epsilon}^{ijk}\partial_jB_k - \partial_0E^i = J^i\\ \partial_iE^i = J^0\\ \bar{\epsilon}^{ijk}\partial_jE_k + \partial_0B^i = 0\\ \partial_iB^i = 0.$$

Given that the field strength tensor can be written in the two tensor equations $F^{0i} = E^i$ and $F^{ij} = \bar{\epsilon}^{ijk}B_k$, how do I reduce the last two equations to the form, $$ \partial_{\lambda} F_{\mu \nu} + \partial_{\mu} F_{\nu \lambda}+ \partial_{\nu} F_{\lambda \mu} = 0,\qquad \mu,\nu,\lambda=0,1,2,3 $$

Answer 1

Since $F_{ij} = \epsilon_{ijk}B^k$ one has $\epsilon^{lij} F_{ij} = \epsilon^{lij}\epsilon_{ijk}B^k = 2\delta_{lk} B^k$, hence $B^l = \frac{1}{2} \epsilon^{lij} F_{ij}$.

The third Maxwell's equation in OP's question can be expressed with the field strength tensor $F_{\mu\nu}$ according to

\begin{align} 0 &= \epsilon^{ijk} \partial_j E_k + \partial_0 B^i \\ &= -\epsilon^{ijk} \partial_j F_{k0} + \frac{1}{2} \epsilon^{ikl} \partial_0 F_{kl}\\ &= -2\epsilon^{ijk} \partial_j F_{k0} + \epsilon^{ijk} \partial_0 F_{jk}\\ &=-\epsilon^{ijk} \partial_0 F_{jk} + \epsilon^{ijk} \partial_j F_{k0} - \epsilon^{ijk} \partial_j F_{0k}\ . \end{align}

The Bianchi identity, i.e., the relation $\partial_\mu F_{\alpha\beta}+ \partial_\alpha F_{\beta\mu}+ \partial_\beta F_{\mu\alpha} =0$, can be compactly arrange with a four-dimensional totally antisymmetric Levi-Civita symbol $$\epsilon^{\alpha\beta\gamma\delta} \partial_\beta F_{\gamma\delta}=0$$ with the property $\epsilon^{0\beta\gamma\delta}=\epsilon^{ijk}$.

Replacing the three-dimensional Levi-Civita symbol we obtain \begin{align} 0&=-\epsilon^{ijk} \partial_0 F_{jk} + \epsilon^{ijk} \partial_j F_{k0} - \epsilon^{ijk} \partial_j F_{0k}\\ &=\epsilon^{i0jk} \partial_0 F_{jk} + \epsilon^{ij0k} \partial_j F_{k0} + \epsilon^{ijk0} \partial_j F_{0k}\\ &=\epsilon^{i\beta\gamma\delta} \partial_\beta F_{\gamma\delta}\ . \end{align}

The still missing case $\alpha=0$ follows immediately from the last Maxwell's equation: \begin{align} 0&=\partial_i B^i\\ &=\epsilon^{ijk} \partial_i F_{jk}\\ &=\epsilon^{0\beta\gamma\delta}\partial_\beta F_{\gamma\delta}\ , \end{align} which eventually reduces to the Bianchi identity.

PS: note the sign for covariant and contravariant indices, e.g. $F_{0i}=\eta_{00}\eta_{ij}F^{0j}=-F^{0i}$.

Answer 2

Ah, I've been here before(I got help from a professor offsite)! The matrix multiplication representation of electromagnetism in general relativity is $$\begin{matrix}\vec{f}=\widetilde{F}\overleftarrow{J}&\widetilde{F}=\nabla\vec{A}-\left(\nabla\vec{A}\right)^T\\\overleftarrow{J}=\frac{1}{\mu_0}\nabla\cdot\underline{F}=\nabla\cdot\underline{\mathcal{D}}&\underline{\mathcal{D}}=\frac{1}{\mu_0}\underline{F}\\\end{matrix}$$ where $$\begin{matrix}\underline{M}={\widetilde{g}}^{-1}\widetilde{M}{\widetilde{g}}^{-1}&\left(\nabla\widetilde{M}\right)_{ijk}=\left(\frac{\partial M_{jk}}{\partial x_i}\right)&\left(\widetilde{M}\right)_i=\left[\begin{matrix}M_{i1}\\M_{i2}\\M_{i3}\\M_{i4}\\\end{matrix}\right]\\\widetilde{M}=\widetilde{g}\underline{M}\widetilde{g}&\nabla\cdot\widetilde{M}=\sum_{i}\left(\frac{\partial\left(\widetilde{M}\right)_i}{\partial x_i}\right)&\nabla\cdot\left(f\widetilde{M}\right)=f\left(\nabla\cdot\widetilde{M}\right)+{\widetilde{M}}^T\nabla f\\\end{matrix}$$ A covariant matrix is represented by $\widetilde{M}$, a contravariant matrix is represented by $\underline{M}$, a covariant vector is represented by $\vec{A}$, and a contravariant vector is represented by $\overleftarrow{J}$.

Due to how the Faraday tensor is defined with the magnetic vector potential, we can prove $$\left(\frac{\partial F_{\mu\nu}}{\partial x_\lambda}\right)+\left(\frac{\partial F_{\nu\lambda}}{\partial x_\mu}\right)+\left(\frac{\partial F_{\lambda\mu}}{\partial x_\nu}\right)=\left(\frac{\partial}{\partial x_\lambda}\left(\left(\frac{\partial A_\nu}{\partial x_\mu}\right)-\left(\frac{\partial A_\mu}{\partial x_\nu}\right)\right)\right)+\left(\frac{\partial}{\partial x_\mu}\left(\left(\frac{\partial A_\lambda}{\partial x_\nu}\right)-\left(\frac{\partial A_\nu}{\partial x_\lambda}\right)\right)\right)+\left(\frac{\partial}{\partial x_\nu}\left(\left(\frac{\partial A_\mu}{\partial x_\lambda}\right)-\left(\frac{\partial A_\lambda}{\partial x_\mu}\right)\right)\right)=\left(\left(\frac{\partial^2A_\nu}{\partial x_\mu\partial x_\lambda}\right)-\left(\frac{\partial^2A_\nu}{\partial x_\mu\partial x_\lambda}\right)\right)+\left(\left(\frac{\partial^2A_\mu}{\partial x_\nu\partial x_\lambda}\right)-\left(\frac{\partial^2A_\mu}{\partial x_\nu\partial x_\lambda}\right)\right)+\left(\left(\frac{\partial^2A_\lambda}{\partial x_\mu\partial x_\nu}\right)-\left(\frac{\partial^2A_\lambda}{\partial x_\mu\partial x_\nu}\right)\right)=0$$

\begin{align} &\frac{\partial F_{\mu\nu}}{\partial x_\lambda}+\frac{\partial F_{\nu\lambda}}{\partial x_\mu}+\frac{\partial F_{\lambda\mu}}{\partial x_\nu}= \nonumber\\ &\frac{\partial}{\partial x_\lambda}\left(\frac{\partial A_\nu}{\partial x_\mu}-\frac{\partial A_\mu}{\partial x_\nu}\right)+\frac{\partial}{\partial x_\mu}\left(\frac{\partial A_\lambda}{\partial x_\nu}-\frac{\partial A_\nu}{\partial x_\lambda}\right)+\frac{\partial}{\partial x_\nu}\left(\frac{\partial A_\mu}{\partial x_\lambda}-\frac{\partial A_\lambda}{\partial x_\mu}\right)= \tag{01}\label{01}\\ &\left(\frac{\partial^2A_\nu}{\partial x_\mu\partial x_\lambda}-\frac{\partial^2A_\nu}{\partial x_\mu\partial x_\lambda}\right)+\left(\frac{\partial^2A_\mu}{\partial x_\nu\partial x_\lambda}-\frac{\partial^2A_\mu}{\partial x_\nu\partial x_\lambda}\right)+\left(\frac{\partial^2A_\lambda}{\partial x_\mu\partial x_\nu}-\frac{\partial^2A_\lambda}{\partial x_\mu\partial x_\nu}\right)=0 \nonumber \end{align}