% vim: tw=50 % 09/03/2023 09AM \subsection{The $4$-potential} We previously wrote \[ \bf{E} = -\nabla \phi - \pfrac{\bf{A}}{t} ,\] in terms of scalar and vector potentials. What does this mean for $F_{\mu\nu}$? \begin{align*} F_{i0} &= \frac{E_i}{c} \\ &= v-\frac{1}{c} \pfrac{\phi}{x_i} - \frac{1}{c} \pfrac{A_i}{t} \\ &= \partial_i \left( -\frac{\phi}{c} \right) - \partial_0 A_i \\ F_{ij} &= \eps_{ijk} B_k \\ &= \eps_{ijk} \eps_{klm} \partial_l A_m \\ &= \partial_i A_j - \partial_j A_i \end{align*} We can interpret $F_{\mu\nu}$ as the antisymmetric derivative \[ \boxed{F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu} \] of the covector \[ A_\mu = \left( -\frac{\phi}{c}, \bf{A} \right) \] The related $4$-vector is the \emph{4-potential} \[ A^\mu = \left( \frac{\phi}{c}, \bf{A} \right) .\] A gauge transformation \[ \tilde{\bf{A}} = \bf{A} + \nabla \chi, \qquad \tilde{\phi} = \phi - \pfrac{\chi}{t} \] corresponds to replacing $A_\mu$ with \[ \tilde{A}_\mu = A_\mu + \partial_\mu \chi \] i.e. adding $\partial_\mu \chi = \left( \frac{1}{c} \pfrac{\chi}{t}, \nabla \chi \right)$ to $A_\mu = \left( -\frac{\phi}{c}, \bf{A} \right)$. This leaves $F_{\mu\nu}$ unaffected, because \[ \partial_\mu(\partial_\nu \chi) - \partial_\nu(\partial_\mu \chi) = 0 .\] The spacetime equivalent of the Coulomb gauge is the \emph{Lorentz gauge} in which \[ \partial_\mu A^\mu = 0 .\] If $\partial_\mu A^\mu \neq 0$ then we can make a gauge transformation such that $\partial_\mu \tilde{A}^\mu = 0$ by choosing $\chi$ to be the solution of the forced wave equation \[ -\Box \chi = \partial_\mu A^\mu \] (Note that $\partial_\mu \partial^\mu \chi = \eta^{\mu\nu} \partial_\mu \partial_\nu \chi = \Box \chi$). \subsection{Lorentz transformation of $\mathbf{E}$ and $\mathbf{B}$} Since $F_{\mu\nu}$ is a $(0, 2)$ tensor, it transforms as \[ F'_{\mu\nu} = (\lambda^{-1})\indices{^\rho_\mu} (\lambda^{-1})\indices{^\sigma_\nu} F_{\rho\sigma} \] In matrix notation, \[ F' = (\lambda^{-1})^\top F(\Lambda^{-1}) \] For a Lorentz boots in standard configuration, \[ \Lambda^{-1} = \begin{pmatrix} \gamma & \gamma \beta & 0 & 0 \\ \gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \] Hence (exercise): \begin{align*} F'_{01} &= \gamma^2 (F_{01} + \beta^2 F_{10} + \beta(F_{00} + F_{11}) \\ &= \gamma^2 (1 - \beta^2) F_{01} \\ &= F_{01} \\ F'_{02} &= \gamma(F_{02} + \beta F_{12}) \\ F'_{03} &= \gamma(F_{03} + \beta F_{13}) \\ F'_{12} &= \gamma(F_{12} + \beta F_{02}) \\ F'_{13} &= \gamma(F_{13} + \beta F_{03}) \\ F'_{23} &= F_{23} \end{align*} From which we deduce \begin{equation*} \boxed{ \begin{aligned} E_x' &= E_x &B_x' &= B_x \\ E_y' &= \gamma(E_y - vB_z) &B_y' &= \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\ E_z' &= \gamma(E_z + vB_y) \qquad &B_z' &= \gamma \left( B_z - \frac{v}{c^2} E_y \right) \end{aligned} } \end{equation*} More generally if $\bf{v}$ is the velocity of $S'$ with respect to $S$, \begin{equation*} \boxed{ \begin{aligned} \bf{E}'_{\parallel} &= \bf{E}_{\parallel} &\bf{B}'_{\parallel} &= \bf{B}_{\parallel} \\ \bf{E}'_{\perp} &= \gamma(\bf{E}_{\perp} + \bf{v} \times \bf{B}_{\perp}) \qquad &\bf{B}'_{\perp} &= \gamma \left( \bf{B}_{\perp} - \frac{1}{c^2} \bf{v} \times \bf{E}_{\perp} \right) \end{aligned} } \end{equation*} where we decompose $\bf{E}$ and $\bf{B}$ into components parallel and perpendicular to $\bf{v}$: \[ \bf{E} = \bf{E}_\parallel + \bf{E}_\perp \qquad \bf{B} = \bf{B}_\parallel + \bf{B}_\perp \] $F_{\mu\nu}$ can be double raised to make \[ F^{\mu\nu} = \begin{pmatrix} 0 & E_x / c & E_y / c & E_z / c \\ -E_x / c & 0 B_z & -B_y \\ -E_y / c & -B_z & 0 & B_x \\ -E_z / c & B_y & -B_x & 0 \end{pmatrix} = -F^{\mu\nu} \] Thus \begin{align*} F^{00} &= 0 \\ F^{0i} &= \frac{E_i}{c} \\ F^{i0} &= -\frac{E_i}{c} \\ F^{ij} &= \eps_{ijk} B_k \end{align*} We can find two Lorentz-invariant combinations of $\bf{E}$ and $\bf{B}$ by forming scalars from $F_{\mu\nu}$: \begin{align*} F_{\mu\nu} F^{\mu\nu} &= F_{00} F^{00} + F_{0i} F^{0i} + F_{i0} F^{i0} + F_{ij} + F_{ij} \\ &= 0 + \left( -\frac{E_i}{c} \right) \frac{E_i}{c} + \frac{E_i}{c} \left( -\frac{E_i}{c} \right) + \eps_{ijk} B_k \eps_{ijl} B_l \\ &= -2 \frac{|\bf{E}|^2}{c^2} + 2\delta_{kl} B_k B_l \\ &= -\frac{2}{c^2} \left(|\bf{E}|^2 - c^2 |\bf{B}|^2\right) \\ \eps_{\mu\nu\rho\sigma} F^{\mu\nu} F^{\rho\sigma} &= \eps_{0ijk} F^{0i} F^{jk} + \eps_{i0jk} F^{i0} F^{jk} + \eps_{ij0k} F^{ij} F^{0k} + \eps_{ijk0} F^{ij} F^{k0} \\ &= \eps_{ijk} \frac{E_i}{c} \eps_{jkl} B_l - \eps_{ijk} \left( -\frac{E_i}{c} \right) \eps_{jkl} B_l + \eps_{ijk}\eps_{ijl} B_l \frac{E_k}{c} - \eps_{ijk} \eps_{ijl} B_l \left( -\frac{E_k}{c} \right) \\ &= 4\eps_{ijk} \frac{E_i}{c} \eps_{jkl} B_l \\ &= \frac{8}{c} \delta_{il} E_i B_l \\ &= \frac{8}{c} \bf{E} \cdot \bf{B} \end{align*} where we used $\eps_{0ijk} = \eps_{ijk}$ and permutations. Thus $|\bf{E}|^2 - c^2|\bf{B}|^2$ is a scalar and $\bf{E} \cdot \bf{B}$ is a pseudoscalar. For example, if $\bf{E} \perp \bf{B}$ in one inertial frame then they are perpendicular in any frame. \begin{example*}[Boosted line charge] From \hyperlink{subsubsection.2.1.2}{Section 2.1.2}, the electric field of a line charge $\lambda$ per unit length along the $x$-axis is \[ \bf{E} = \frac{\lambda}{2\pi\eps_0 (y^2 + z^2)} (0, y, z) \] while $\bf{B} = \bf{0}$ for static charge. Under a standard Lorentz boost, \begin{align*} \bf{E}' &= \frac{\gamma\lambda}{2\pi\eps_0(y^2 + z^2)}(0, y, z) \\ &= \frac{\gamma\lambda}{2\pi\eps_0(y'^2 + z'^2}(0, y', z') \\ \bf{B}' &= \frac{\gamma v}{c^2} \frac{\lambda}{2\pi\eps_0(y^2 + z^2)} (0, z, -y) \\ &= \frac{\mu_0 \gamma \lambda v}{2\pi (y'^2 +z'^2)} (0, z', -y') \end{align*} $\bf{E}'$ can be understood as the electric field of a line charge $\lambda' = \gamma\lambda$ per unit length along the $x'$-axis. (Length contraction enhances the charge density). $\bf{B}'$ can be understood as the magnetic field due to a current $I' = -\lambda' v$ along the $x'$-axis. (The line charge moves with velocity $-v$ in $S'$). \begin{center} \includegraphics[width=0.6\linewidth] {images/ff37b234be6111ed.png} \end{center} \end{example*}