Pre-Quantum Electrodynamics

Maxwell's Equations in Relativistic Notation red.rem.Me

Let us consider a charge density of moving sources. For an infinitesimal cloud of volume \(V\) containing charge \(Q\) moving at velocity \({\boldsymbol u}\), we have \[ \rho = \frac{Q}{V}, \hspace{10mm} {\boldsymbol J} = \rho {\boldsymbol u}. \] In the rest frame of the charges, we have the proper density \[ \rho_0 = \frac{Q}{V_0} \] in terms of the rest volume \(V_0\), which is related to the perceived volume \(V\) in the original frame by \[ V = \sqrt{1 - u^2/c^2} V_0. \] We thus obtain \[ \rho = \rho_0 \frac{1}{\sqrt{1 - u^2/c^2}}, \hspace{10mm} {\boldsymbol J} = \rho_0 \frac{{\boldsymbol u}}{\sqrt{1 - u^2/c^2}}. \] Recognizing the proper velocity, we thus get that charge density and current density can together form the

{\bf Current density 4-vector} \[ J^\mu = \rho_0 \eta^\mu, \hspace{10mm} J^\mu = \left( c\rho, J_x, J_y, J_z \right) \]

The continuity equation (\ref{eq:continuity}) takes the simple form

{\bf Continuity equation} \[ \frac{\partial J^\mu}{\partial x^\mu} = 0 \]

while similarly the notation simplifies for

{\bf Maxwell's equations} \[ \frac{\partial F^{\mu \nu}}{\partial x^\nu} = \mu_0 J^\mu, \hspace{10mm} \frac{\partial G^{\mu \nu}}{\partial x^\nu} = 0 \]

In terms of \(F^{\mu \nu}\) and the proper velocity \(\eta^\mu\), we also have the

{\bf Minkowski force on a charge \(q\)} \[ K^\mu = q F^{\mu \nu} \eta_\nu \]

whose vector components are \[ {\boldsymbol K} = \frac{q}{\sqrt{1 - u^2/c^2}} \left( {\boldsymbol E} + {\boldsymbol u} \times {\boldsymbol B} \right) \] which becomes the Lorentz force law when remembering (\ref{eq:MinkowskiForce}).

We had \[ {\boldsymbol E} = -{\boldsymbol \nabla} V - \frac{\partial {\boldsymbol A}}{\partial t}, \hspace{10mm} {\boldsymbol B} = {\boldsymbol \nabla} \times {\boldsymbol A}. \] We can group the potentials together into a 4-vector: \[ A^\mu = \left( V/c, A_x, A_y, A_z \right). \] The field tensor is then expressed as \[ F^{\mu \nu} = \frac{\partial A^\nu}{\partial x_\mu} - \frac{\partial A^\mu}{\partial x_\nu}. \] The inhomogeneous Maxwell equation becomes \[ \frac{\partial F^{\mu \nu}}{\partial x^\nu} = \mu_0 J^\mu ~~\longrightarrow~~ \frac{\partial}{\partial x_\mu} \left(\frac{\partial A^\nu}{\partial x^\nu} \right) - \frac{\partial}{\partial x_\nu} \left( \frac{\partial A^\mu}{\partial x^\nu} \right) = \mu_0 J^\mu \] We can now exploit gauge invariance \[ A^\mu ~~\longrightarrow~~ {A^\mu}^\prime = A^\mu + \frac{\partial \lambda}{\partial x_\mu} \] which leaves \(F^{\mu \nu}\) invariant. In particular, we can choose the Lorenz gauge (\ref{eq:InhomogeneousMaxwellLorenzGauge}) here expressed as \[ \frac{\partial A^\mu}{\partial x^\mu} = 0. \] Defining the

{\bf d'Alembertian operator} \[ \square^2 \equiv \frac{\partial}{\partial x_\nu} \frac{\partial}{\partial x^\nu} = {\boldsymbol \nabla}^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \]

we obtain the final form of

{\bf Maxwell's equations (Lorenz gauge, 4-vector notation)} \[ \square^2 A^\mu = -\mu_0 J^\mu \]

which is the most aesthetically pleasing and practical form of these equations.




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Author: Jean-Sébastien Caux

Created: 2022-02-21 Mon 20:41