Relativistic electromagnetism

Relativistic electromagnetism combines electric and magnetic fields into Lorentz tensors.

For metric signature $(-,+,+,+)$, a particle of mass $m$ and charge $q$ coupled to a four-potential has the reparametrization-invariant action

Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, the interaction changes byFixed-endpoint equations of motion are therefore gauge invariant.

Varying the relativistic charged-particle action with fixed endpoints givesor the equivalent raised-index equation.

The four-potential combines the electric potential and magnetic vector potential as $A^{\mu }=(\varphi /c,\mathbf{A})$. The electromagnetic four-potential $A_{\mu }$ transforms underwhile the antisymmetric field tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is unchanged.

An electromagnetic multipole expansion expresses the potentials or fields of a localized source as successively smaller monopole, dipole, quadrupole, and higher moments in the ratio of source size to observation distance.

A retarded potential evaluates each source at the event on the source's past light cone. This accounts for propagation at the speed of light and enforces causality.

For a source point $y(s)$ and field event $(x,t)$, the retarded time solves

For $R(s)=|x-y(s)|$ and $|v|<c$, the function $F(s)=s+R(s)/c$ hasIt is strictly increasing, so the retarded-time equation has at most one solution.

The Lorenz gauge condition isor ${\partial }_{\mu }{A}^{\mu }=0$ in covariant notation.

An Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$ changes $\partial \cdot A$ by $\square \lambda$. Consequently gauge fixing by a quadratic function of $\partial \cdot A$ retains transformations satisfying $\square \lambda =0$, subject to the same boundary conditions. Eliminating those residual transformations is part of specifying an inverse with a particular Green function prescription.

The electromagnetic field tensor isIt packages the electric and magnetic fields into an antisymmetric rank-two Lorentz tensor and transforms as

The dual electromagnetic field tensor is obtained by applying the Hodge star operator to the electromagnetic field tensor. For metric signature $(+---)$ and ${ϵ}^{0123}=+1$, define ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. With $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, it has components ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. The homogeneous Maxwell equations become ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$.

The source-free electromagnetic Lagrangian density is quadratic in the electromagnetic field tensor. Its Euler-Lagrange field equation is ${\partial }_{\mu }{F}^{\mu \nu }=0$. Antisymmetry of $F$ makes the density invariant under gauge transformations, and contraction of its indices makes it a Lorentz scalar.

The source-free Maxwell Lagrangian gives $S=-\frac{1}{4}\int F_{\mu \nu }{F}^{\mu \nu }{d}^{4}x$ in mostly-minus Lorentzian signature. Its Euclidean action is the displayed positive quadratic functional of the electromagnetic field tensor. Variation, with boundary terms vanishing, yields ${\partial }_{\mu }{F}_{\mu \nu }=0$.

Because the electromagnetic field tensor is the exterior derivative $F=dA$ of a potential, $dF=d^{2}A=0$. For a torsion-free connection this iswhich is the covariant form of the two homogeneous Maxwell equations.

For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, the covariant Lorentz-force law isIts spatial and temporal components are

For smooth stationary fields on ${\mathbb{R}}^3$, set ${\Gamma }^i{}_{tt}=-E^i$ and ${\Gamma }^i{}_{tj}={\Gamma }^i{}_{jt}={ϵ}^i{}_{jk}{B}^k$, with all other Christoffel symbols zero. This torsion-free connection makes the equation $\ddot{\mathbf{x}}=\mathbf{E}+2\mathbf{B}\times \dot{\mathbf{x}}$ a geodesic equation with time as affine parameter. Arbitrary smooth fields suffice; a metric realization imposes additional potential conditions.

In uniform perpendicular electric and magnetic fields, the guiding centre of a charged particle drifts with ${\mathbf{v}}_D=\mathbf{E}\times \mathbf{B}/|\mathbf{B}{|}^2$. This velocity is independent of the charge and mass; cyclotron motion is superposed on it.

The covariant force on a charged particle is its charge times the electromagnetic field-strength tensor contracted with its four-velocity. Antisymmetry keeps the four-force orthogonal to that velocity and preserves the rest mass.

A wire carrying current $I$ in a magnetic field experiencesThis follows by summing the magnetic part of the Lorentz force over the moving charge carriers.

In a constant magnetic field $B\hat{z}$ with no electric field, a charged relativistic particle moves in a circle in the perpendicular plane. Its angular frequency with respect to proper time iswhereas its angular frequency in an inertial coordinate time is ${\omega }_t=qB/(\gamma m)$. The orbit radius at speed $v$ is

For $qB>0$, choose $A_y=Bx$ and parametrize one clockwise orbit by $x=R\cos ({\omega }_{\tau }\tau )$ and $y=-R\sin ({\omega }_{\tau }\tau )$. Over $\Delta \tau =2\pi /{\omega }_{\tau }$, the relativistic charged-particle action isThe same formula with $|{\omega }_{\tau }|$ applies for either orbit orientation.

The Lorentz force per unit volume is ${\mathbf{f}}_{\mathrm{L}}={\rho }_e\mathbf{E}+\mathbf{j}\times \mathbf{B}/c$ in Gaussian units.

In uniform perpendicular electric and magnetic fields, a charged particle's cyclotron orbit has guiding-centre velocityindependent of its charge and mass.

For $\mathcal{E}=\gamma m{c}^2$, the temporal component of the covariant Lorentz-force equation givesThe magnetic field does no work because $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})=0$.

Combining $d(\gamma m\mathbf{v})/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ with $m{c}^{2}d\gamma /dt=q\mathbf{E}\cdot \mathbf{v}$ gives

For signature $(-,+,+,+)$ and a consistent orientation,Overall signs depend on tensor and orientation conventions, while their vanishing does not.

A null field has both electromagnetic invariants zero:For a nonzero constant null field, the mixed tensor is nilpotent of index three,

For metric signature $(-,+,+,+)$,In particular, $T^{00}=\frac{1}{2}({ϵ}_0{E}^2+B^2/{\mu }_0)$.

For every null vector $k$, put $q^{\rho }=k_{\mu }{F}^{\mu \rho }$. Antisymmetry gives $q\cdot k=0$, so $q$ has nonnegative norm and

Lorentz boosts mix transverse electric and magnetic field components.

Two equal opposite line charge densities moving in opposite directions have zero net line charge but a nonzero total current $I$. Under a longitudinal Lorentz boost, the charge-current four-vector gives ${\lambda }^{\prime }=\gamma (\lambda -vI/c^2)=-\gamma vI/c^2$. Equivalently, for lab densities $n$ and speeds $\pm u$, their separate boosted number densities are $\gamma n(1\mp vu/c^2)$. Their imbalance creates the boosted radial electric field, consistently with Lorentz transformation of electromagnetic fields and Gauss's law.

If a plane is stationary in one frame and an observer moves normally to it with velocity $\mathbf{v}$, thenThe second relation accounts for the magnetic-field discontinuity moving through the observer's frame.