06 Jun 2025

Relativistic Electrodynamics

Four-current, four-potential, field tensor, Lorentz transformations of fields, covariant Maxwell equations, and the Lorentz four-force.

mj-16 relativity electromagnetic-field-tensor maxwell-equations lorentz-force

Use coordinates $x^\mu=(ct,x,y,z)$ and metric $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$. Let frame $S’$ move with speed $v$ along $+x$ relative to $S$. With $\beta=v/c$ and $\gamma=(1-\beta^2)^{-1/2}$,

\[\begin{pmatrix}ct'\\x'\\y'\\z'\end{pmatrix} =\underbrace{\begin{pmatrix} \gamma&-\beta\gamma&0&0\\ -\beta\gamma&\gamma&0&0\\ 0&0&1&0\\0&0&0&1 \end{pmatrix}}_{\Lambda} \begin{pmatrix}ct\\x\\y\\z\end{pmatrix}.\]

Every contravariant four-vector transforms by the same matrix.

Charge density and current density

Local charge conservation is

\[\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf J=0.\]

Define the four-current

\[J^\mu=(c\rho,J_x,J_y,J_z), \qquad \partial_\mu=\left(\frac1c\partial_t,\partial_x,\partial_y,\partial_z\right).\]

Then $\partial_\mu J^\mu=0$. Applying $J’^\mu=\Lambda^\mu{}_{\nu}J^\nu$ gives

\[\boxed{ \rho'=\gamma\left(\rho-\frac{vJ_x}{c^2}\right), \qquad J'_x=\gamma(J_x-v\rho), \qquad J'_y=J_y,\quad J'_z=J_z. }\]

The mixture of $\rho$ and $J_x$ is required because charge density depends on the simultaneity slice used to measure volume, although the total charge of an isolated system is invariant.

Scalar and vector potentials

The scalar and vector potentials form

\[A^\mu=\left(\frac\phi c,A_x,A_y,A_z\right).\]

Therefore

\[\boxed{ \phi'=\gamma(\phi-vA_x), \qquad A'_x=\gamma\left(A_x-\frac{v\phi}{c^2}\right), \qquad A'_y=A_y,\quad A'_z=A_z. }\]

The Lorenz gauge becomes the Lorentz scalar condition $\partial_\mu A^\mu=0$. In that gauge the potential equations combine as

\[\partial_\mu\partial^\mu A^\nu=\mu_0J^\nu, \qquad \partial_\mu\partial^\mu =\frac1{c^2}\partial_t^2-\nabla^2.\]

Electromagnetic field tensor

Define

\[F^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu.\]

With the stated metric and potential convention,

\[F^{\mu\nu}= \begin{pmatrix} 0&-E_x/c&-E_y/c&-E_z/c\\ E_x/c&0&-B_z&B_y\\ E_y/c&B_z&0&-B_x\\ E_z/c&-B_y&B_x&0 \end{pmatrix}, \qquad F^{\mu\nu}=-F^{\nu\mu}.\]

The tensor transformation law is

\[F'^{\mu\nu}=\Lambda^\mu{}_{\alpha} \Lambda^\nu{}_{\beta}F^{\alpha\beta}.\]

The six independent contractions are

\[\begin{aligned} F'^{01} &=\gamma^2F^{01}+\beta^2\gamma^2F^{10} =\gamma^2(1-\beta^2)F^{01} =-\frac{E_x}{c},\\ F'^{02} &=\Lambda^0{}_{\alpha}\Lambda^2{}_{\beta}F^{\alpha\beta} =\gamma F^{02}-\beta\gamma F^{12}\\ &=-\frac\gamma c(E_y-vB_z) =-\frac{E'_y}{c},\\ F'^{03} &=\gamma F^{03}-\beta\gamma F^{13} =-\frac\gamma c(E_z+vB_y),\\ F'^{23}&=F^{23}=-B_x,\\ F'^{13} &=-\beta\gamma F^{03}+\gamma F^{13} =\gamma\left(B_y+\frac{vE_z}{c^2}\right),\\ F'^{12} &=-\beta\gamma F^{02}+\gamma F^{12} =-\gamma\left(B_z-\frac{vE_y}{c^2}\right). \end{aligned}\]

Using $F’^{0i}=-E’_i/c$, $F’^{23}=-B’_x$, $F’^{13}=B’_y$, and $F’^{12}=-B’_z$ gives

\[\boxed{ \begin{aligned} E'_x&=E_x, &E'_y&=\gamma(E_y-vB_z), &E'_z&=\gamma(E_z+vB_y),\\ B'_x&=B_x, &B'_y&=\gamma\left(B_y+\frac{vE_z}{c^2}\right), &B'_z&=\gamma\left(B_z-\frac{vE_y}{c^2}\right). \end{aligned}}\]

The signs follow from the declared boost direction. Reversing the frame transformation means replacing $v$ by $-v$.

For \(\mathbf E=E_0\hat{\mathbf y}\), \(\mathbf B=0\), the equations give \(E'_y/E_0=\gamma\) and \(cB'_z/E_0=-\gamma\beta\).

Maxwell equations in covariant form

The inhomogeneous Maxwell equations are the four components of

\[\boxed{\partial_\mu F^{\mu\nu}=\mu_0J^\nu.}\]

For $\nu=0$,

\[\partial_iF^{i0}=\frac1c\nabla\cdot\mathbf E =\mu_0c\rho \quad\Longrightarrow\quad \nabla\cdot\mathbf E=\frac\rho{\epsilon_0}.\]

For a spatial index $\nu=i$,

\[\partial_0F^{0i}+\partial_jF^{ji}=\mu_0J^i\]

is precisely $\nabla\times\mathbf B-\partial_t\mathbf E/c^2=\mu_0\mathbf J$. The homogeneous equations are

\[\boxed{ \partial_\lambda F_{\mu\nu} +\partial_\mu F_{\nu\lambda} +\partial_\nu F_{\lambda\mu}=0, }\]

which contain $\nabla\cdot\mathbf B=0$ and $\nabla\times\mathbf E+\partial_t\mathbf B=0$.

Under a Lorentz transformation, $\partial’\mu=(\Lambda^{-1})^\alpha{}{\mu}\partial_\alpha$, $F’^{\mu\nu}=\Lambda^\mu{}{\rho}\Lambda^\nu{}{\sigma}F^{\rho\sigma}$, and $J’^\nu=\Lambda^\nu{}_{\sigma}J^\sigma$. Contracting the first two relations cancels one pair of inverse matrices, leaving

\[\partial'_\mu F'^{\mu\nu}=\mu_0J'^\nu.\]

This establishes the covariance of Maxwell’s equations.

Lorentz force and its transformation

For four-velocity $U^\mu=\gamma_u(c,\mathbf u)$ and four-momentum $p^\mu=(E/c,\mathbf p)$, define the four-force

\[\boxed{K^\mu=\frac{dp^\mu}{d\tau}=qF^{\mu\nu}U_\nu.}\]

Since $U_\nu=\gamma_u(c,-\mathbf u)$, its time component is

\[K^0=\gamma_u\frac q c\mathbf E\cdot\mathbf u =\gamma_u\frac1c\frac{dE}{dt},\]

and the spatial components are

\[\mathbf K=\gamma_uq(\mathbf E+\mathbf u\times\mathbf B) =\gamma_u\frac{d\mathbf p}{dt}.\]

Hence the ordinary force law is

\[\boxed{\frac{d\mathbf p}{dt}=q(\mathbf E+\mathbf u\times\mathbf B).}\]

Because $F^{\mu\nu}$ and $U_\nu$ are tensors of complementary rank, $K’^\mu=\Lambda^\mu{}_{\nu}K^\nu$. The Lorentz force therefore transforms covariantly with the same field transformations derived above.

The boost-matrix multiplication, all six field transformations, and the four-force components are verified in the Maxima worksheet; every printed residual is zero.