Magnetic Fields in Relativistic MHD
In relativistic MHD, the magnetic field 4-vector $b_\mu$ and electric field $e_\mu$ in the fluid frame is defined through the 4-vector $u^{\mu}$ and EM 2-form $F_{\mu\nu}$.
$$\begin{aligned} b^\mu &=& u_{\nu}*F^{\mu\nu}\\ e^\mu &=& u_{\nu}F^{\mu\nu} \end{aligned}$$where $*$ means the Hodge-dual of the 2-form. It is obvious that $u_{\mu}e^{\mu}=0$ and $u_{\mu}e^{\mu}=0$. MHD condition requires that $e^{\mu}=0=u_{\nu}F^{\mu\nu}$. Using the antisymmetric tensor, we have
$$\begin{equation} b^{\mu} = \frac{1}{2}\epsilon^{\mu\nu\kappa\lambda}u_{\nu}F_{\lambda\kappa}. \end{equation}$$Contract with $\epsilon_{\mu\alpha\beta\gamma}$ on both sides, and using the equality from (3.50h) Misner, Thorn, Wheeler
$$\begin{equation} \delta^{\nu\kappa\lambda}_{\alpha\beta\gamma} = -\epsilon^{\mu\nu\kappa\lambda}\epsilon_{\mu\alpha\beta\gamma}. \end{equation}$$$\delta^{\nu\kappa\lambda}_{\alpha\beta\gamma} $ takes the value $1$ if $\nu\kappa\lambda$ is an even permutation of ${\alpha\beta\gamma}$, and $-1$ for odd permutation, $0$ otherwise. Hence,
$$\begin{equation} \epsilon_{\mu\alpha\beta\gamma}b^{\mu} = -\frac{1}{2}\delta^{\nu\kappa\lambda}_{\alpha\beta\gamma}u_{\nu}F_{\lambda\kappa}. \end{equation}$$Now contract both sides with $u^{\alpha}$, and using the fact that $u^{\mu}u_{\mu}=-1$,$u_{\nu}F^{\mu\nu}=0$, only nonzero term on the right hand side come from $\alpha=\nu$, therefore
$$\begin{equation} F_{\beta\gamma}=\epsilon_{\beta\gamma\alpha\mu}u^{\alpha}b^{\mu}. \end{equation}$$$$\begin{aligned} *F^{\mu\nu}&=&\frac{1}{2}\epsilon^{\beta\gamma\mu\nu}\epsilon_{\beta\gamma\alpha\rho}u^{\alpha}b^{\rho}\\ &=&-\delta^{\mu\nu}_{\alpha\rho}u^{\alpha}b^{\rho}\\ &=&-(\delta^{\mu}_{\alpha}\delta^{\nu}_{\rho}-\delta^{\nu}_{\alpha}\delta^{\mu}_{\rho})u^{\alpha}b^{\rho}\\ &=&b^{\mu}u^{\nu}-b^{\nu}u^{\mu} \end{aligned}$$There might be a difference in sign in the last expression which does not matter. The Maxwell $*F^{\mu\nu}_{;\nu}=0$ is not affected by a minus sign on both sides. The fluid equation $T^{\mu\nu}_{;\nu}=0$.
$$\begin{equation} T^{\mu\nu}_{EM} = F^{\mu\alpha}F_{\nu\alpha}-\frac{1}{4}g^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta} \end{equation}$$is quadratic in $F^{\mu\nu}$.
