Covariant Form of Ohm's Law | Yucheng (Luca) Jin

For a general 4-velocity, written as $u^\mu = \gamma(c,\mathbf{v})$, show that \[F_{\mu\nu}u^\nu = \gamma\begin{pmatrix}-\mathbf{E}\cdot\mathbf{v}/c\\[4pt] \mathbf{E}+\mathbf{v}\times\mathbf{B}\end{pmatrix}.\tag{1.1}\]

In the rest-frame of a conducting medium, Ohm's law states that $\mathbf{J}=\sigma\mathbf{E}$, where $\sigma$ is the conductivity and $\mathbf{J}$ is the 3-current. Assuming that $\sigma$ is a Lorentz scalar, show that Ohm's law can be written covariantly as \[J^\mu + \frac{1}{c^2}(J^\nu u_\nu)u^\mu = \sigma F^{\mu\nu}u_\nu,\tag{1.2}\] where $J^\mu$ is the 4-current and $u^\mu$ is the (uniform) 4-velocity of the medium. If the medium moves with 3-velocity $\mathbf{v}$ in some inertial frame, show that the current in that frame is \[\mathbf{J} = \rho\mathbf{v} + \sigma\gamma\!\left(\mathbf{E} + \mathbf{v}\times\mathbf{B} - \frac{1}{c^2}(\mathbf{v}\cdot\mathbf{E})\mathbf{v}\right),\tag{1.3}\] where $\rho$ is the charge density. Simplify this formula, given that the charge density vanishes in the rest-frame of the medium.

It is easy to verify that \[F_{\mu\nu}u^\nu = \gamma\begin{pmatrix} 0 & E_x/c & E_y/c & E_z/c \\ -E_x/c & 0 & -B_z & B_y \\ -E_y/c & B_z & 0 & -B_x \\ -E_z/c & -B_y & B_x & 0 \end{pmatrix} \begin{pmatrix} c\\v_x\\v_y\\v_z \end{pmatrix} = \gamma\begin{pmatrix}-\mathbf{E}\cdot\mathbf{v}/c\\ \mathbf{E}+\mathbf{v}\times\mathbf{B}\end{pmatrix}.\tag{1.4}\]

In the rest-frame, $u^\mu = (c,0,0,0)$. Then, consider \[J^\mu + \frac{1}{c^2}(J^\nu u_\nu)u^\mu = \begin{pmatrix} \rho c\\\mathbf{J} \end{pmatrix}+\frac{1}{c^2}\begin{pmatrix} -\rho c &\mathbf{J} \end{pmatrix}\cdot\begin{pmatrix} c\\0 \end{pmatrix}\begin{pmatrix} c\\0 \end{pmatrix} = \begin{pmatrix} 0\\\mathbf{J} \end{pmatrix},\tag{1.5}\] and \[\sigma F_{\mu\nu}u^\nu = \begin{pmatrix} 0\\\sigma\mathbf{E}. \end{pmatrix}\tag{1.6}\] Thus, we have verified that \[J^\mu + \frac{1}{c^2}(J^\nu u_\nu)u^\mu = \sigma F^{\mu\nu}u_\nu\tag{1.7}\] is indeed the covariant representation of Ohm's law.

For a moving medium, the equation gives \[\begin{pmatrix} \rho c+\gamma^2(-\rho c^2+\mathbf{J}\cdot\mathbf{v})\\ \mathbf{J} +\gamma^2(-\rho c^2+\mathbf{J}\cdot\mathbf{v})\mathbf{v}/c \end{pmatrix} =\gamma \sigma\begin{pmatrix} \mathbf{E}\cdot\mathbf{v}/c\\\mathbf{E}+\mathbf{v}\times\mathbf{B} \end{pmatrix}.\tag{1.8}\] Combining the two equations, we find \[\mathbf{J} = \rho\mathbf{v} + \sigma\gamma\!\left(\mathbf{E} + \mathbf{v}\times\mathbf{B} - \frac{1}{c^2}(\mathbf{v}\cdot\mathbf{E})\mathbf{v}\right).\tag{1.9}\]

Finally, given $\rho_0 = 0$, then $J^\mu u_\mu = -\rho_0 c^2$. This implies \[-\rho c^2 + \mathbf{J}\cdot\mathbf{v} = 0,\tag{1.10}\] Taking the dot product of $\mathbf{J}$ with $\mathbf{v}$ gives \[\rho v^2 + \sigma\gamma\left(\mathbf{v}\cdot\mathbf{E} - \frac{v^2}{c^2}\mathbf{v}\cdot\mathbf{E}\right) = \rho c^2.\tag{1.11}\] Rearranging, we find \[\rho = \frac{\sigma \gamma}{c^2}\,\mathbf{v}\cdot\mathbf{E}.\tag{1.12}\] Substituting back, the current becomes \[\mathbf{J} = \sigma\gamma\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right).\tag{1.13}\]