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Special Relativity and Electromagnetism

Maxwell’s equations in covariant form using the field tensor FμνF^{\mu\nu}:

μFμν=μ0Jν(inhomogeneous)\partial_\mu F^{\mu\nu} = \mu_0 J^\nu \quad \text{(inhomogeneous)}

λFμν+μFνλ+νFλμ=0(homogeneous / Bianchi identity)\partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0 \quad \text{(homogeneous / Bianchi identity)}

The electromagnetic field tensor:

Fμν=(0Ex/cEy/cEz/cEx/c0BzByEy/cBz0BxEz/cByBx0)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}

The dual tensor: F~μν=12ϵμνρσFρσ\tilde{F}^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}.

The Lorentz force: fμ=qFμνuνf^\mu = qF^{\mu\nu}u_\nu where uν=γ(c,v)u^\nu = \gamma(c, \mathbf{v}) is the four-velocity.

Under a boost with velocity vv along the xx-axis:

Ex=Ex,Bx=BxE'_x = E_x, \quad B'_x = B_x

Ey=γ(EyvBz),By=γ ⁣(By+vc2Ez)E'_y = \gamma(E_y - vB_z), \quad B'_y = \gamma\!\left(B_y + \frac{v}{c^2}E_z\right)

Ez=γ(Ez+vBy),Bz=γ ⁣(Bzvc2Ey)E'_z = \gamma(E_z + vB_y), \quad B'_z = \gamma\!\left(B_z - \frac{v}{c^2}E_y\right)

Key insight: E\mathbf{E} and B\mathbf{B} mix under Lorentz transformations. What appears as a pure electric field in one frame becomes a mixture of electric and magnetic fields in another. There is no frame-independent distinction between E\mathbf{E} and B\mathbf{B}.

Invariants: E2c2B2E^2 - c^2B^2 and EB\mathbf{E}\cdot\mathbf{B} are Lorentz invariants. A pure radiation field (E=cBE = cB, EB\mathbf{E}\perp\mathbf{B}) satisfies both invariants being zero.

12.3 Electromagnetic Field Momentum and Angular Momentum

Section titled “12.3 Electromagnetic Field Momentum and Angular Momentum”

Field momentum density:

g=Sc2=ε0E×B\mathbf{g} = \frac{\mathbf{S}}{c^2} = \varepsilon_0\mathbf{E} \times \mathbf{B}

Field angular momentum: L=r×gd3r\mathbf{L} = \int \mathbf{r} \times \mathbf{g}\, d^3r.

Conservation: \frac{d}{dt}\left(\mathbf{p}_{\text{mech} + \mathbf{p}_{\text{field}\right) = 0}}.

For a charge and a magnetic monopole (if they exist), the field angular momentum L=qgr^/(4π)\mathbf{L} = -qg\hat{\mathbf{r}}/(4\pi) is quantised in units of /2\hbar/2Leading to the Dirac charge quantisation condition eg=n/2eg = n\hbar/2.

Quantity3-vector form4-vector / tensor form
Potentialϕ\phi, A\mathbf{A}Aμ=(ϕ/c,A)A^\mu = (\phi/c, \mathbf{A})
FieldsE\mathbf{E}, B\mathbf{B}Fμν=μAννAμF^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu
Charge-currentρ\rho, J\mathbf{J}Jμ=(cρ,J)J^\mu = (c\rho, \mathbf{J})
Force densityρE+J×B\rho\mathbf{E} + \mathbf{J}\times\mathbf{B}fμ=FμνJνf^\mu = F^{\mu\nu}J_\nu
Energy-momentumu=12(ε0E2+B2/μ0)u = \frac{1}{2}(\varepsilon_0 E^2 + B^2/\mu_0), S\mathbf{S}TμνT^{\mu\nu}
  • Assuming E\mathbf{E} and B\mathbf{B} transform independently. They do not; the field tensor transforms as a whole under Lorentz boosts. A pure electric field in one frame becomes a mixture in another.
  • Confusing the dual tensor F~μν\tilde{F}^{\mu\nu} with FμνF^{\mu\nu}. The dual swaps electric and magnetic fields (EcB\mathbf{E} \to c\mathbf{B}, BE/c\mathbf{B} \to -\mathbf{E}/c) and is used in the homogeneous Maxwell equation μF~μν=0\partial_\mu \tilde{F}^{\mu\nu} = 0.
  • Forgetting that FμνF^{\mu\nu} is antisymmetric. This antisymmetry encodes the fact that there are six independent field components (three for E\mathbf{E}, three for B\mathbf{B}).
  • Misapplying the Lorentz force formula. The relativistic Lorentz force fμ=qFμνuνf^\mu = qF^{\mu\nu}u_\nu gives the four-force, not the three-force. The spatial components reduce to dp/dt=q(E+v×B)d\mathbf{p}/dt = q(\mathbf{E} + \mathbf{v}\times\mathbf{B}) in the non-relativistic limit.

Problem 1. Show that EB\mathbf{E}\cdot\mathbf{B} is a Lorentz invariant.

Solution. EB\mathbf{E}\cdot\mathbf{B} is proportional to 14F~μνFμν\frac{1}{4}\tilde{F}^{\mu\nu}F_{\mu\nu}. Since this is a full contraction of two tensors, it is a scalar and thus invariant. Explicitly: F~μνFμν=4EB/c\tilde{F}^{\mu\nu}F_{\mu\nu} = -4\mathbf{E}\cdot\mathbf{B}/c. Under any Lorentz transformation, both FμνF^{\mu\nu} and F~μν\tilde{F}^{\mu\nu} transform as tensors, so their contraction is invariant. \blacksquare

Problem 2. Derive the transformation of the Poynting vector under a Lorentz boost.

Solution. S=E×B/μ0\mathbf{S} = \mathbf{E} \times \mathbf{B} / \mu_0 transforms as part of the energy-momentum tensor TμνT^{\mu\nu}. The components T0i=Si/cT^{0i} = S_i/c transform under a boost: Sx=SxS'_x = S_x, Sy=γ(Syvu)S'_y = \gamma(S_y - v u), Sz=γ(Sz+vu)S'_z = \gamma(S_z + v u) where uu is the energy density. This shows that energy flux in one frame contributes to energy density in another. \blacksquare

  • Particle physics: The covariant formulation is essential for quantum electrodynamics (QED), where FμνF^{\mu\nu} couples to the Dirac field via minimal coupling μDμ\partial^\mu \to D^\mu.
  • Plasma physics: Relativistic plasmas require the covariant formulation for correct treatment of high-energy particle motion in strong electromagnetic fields.
  • Astrophysics: Pulsar electrodynamics and magnetar fields involve enormous Lorentz factors where the field transformation laws govern radiation emission mechanisms.
  • Accelerator physics: The design of particle accelerators requires precise knowledge of how electromagnetic fields appear in the rest frame of relativistic particle bunches.
Worked Example 12.1: Fields of a Moving Point Charge

A point charge qq at rest at the origin has E=qr^/(4πε0r2)\mathbf{E} = q\hat{\mathbf{r}}/(4\pi\varepsilon_0 r^2), B=0\mathbf{B} = 0.

In a frame moving with velocity vv along the xx-axis, the fields at the boosted position are:

Ey=γqy4πε0(r2+γ2v2t2)3/2,Bz=vc2EyE'_y = \gamma\frac{qy'}{4\pi\varepsilon_0(r'^2 + \gamma^2 v^2 t'^2)^{3/2}}, \quad B'_z = -\frac{v}{c^2}E'_y

At t=0t' = 0: E\mathbf{E}' is still radial (from the instantaneous position) but with an enhanced transverse component by factor γ\gamma. The magnetic field is B=v×E/c2\mathbf{B}' = -\mathbf{v} \times \mathbf{E}'/c^2Circulating around the direction of motion.

The Poynting vector S=E×B/μ0\mathbf{S}' = \mathbf{E}' \times \mathbf{B}'/\mu_0 is nonzero even for a uniformly moving charge (it points outward and forward, indicating energy flow in the direction of motion).

For ultrarelativistic motion (γ1\gamma \gg 1): the fields are concentrated in a thin disk of angular width 1/γ\sim 1/\gamma around the plane perpendicular to the motion. This is the basis of synchrotron radiation patterns.