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 $F_{\mu\nu}$. \(\begin{eqnarray} b^\mu &=& u_{\nu}*F^{\mu\nu}\\ e^\mu &=& u_{\nu}F^{\mu\nu} \end{eqnarray}\) 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 \(\begin{equation} F_{\beta\gamma}=\epsilon_{\beta\gamma\alpha\mu}u^{\alpha}b^{\mu}. \end{equation}\)

\[\begin{eqnarray} *F^{\mu\nu}&=&\frac{1}{2}\epsilon^{\beta\gamma\mu\nu}\epsilon_{\beta\gamma\alpha\rho}u^{\alpha}b^{\rho}\\ &=&-\delta^{\mu\nu}_{\alpha\rho}u^{\alpha}b^{\rho}\\ &=&-(\delta^{\mu}_{\alpha}\delta^{\nu}_{\rho}-\delta^{\nu}_{\alpha}\delta^{\mu}_{\rho})u^{\alpha}b^{\rho}\\ &=&b^{\mu}u^{\nu}-b^{\nu}u^{\mu} \end{eqnarray}\]

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 \(F^{\mu\nu}\).

Xinyu Li
Xinyu Li
Postdoctoral Fellow

My research areas are theoretical high energy astrophysics and cosmology. The topics I am working on include magnetars, neutron star mergers, dark matter and large-scale structure of the universe.