Relativistic Electrodynamics

Theory, worked examples and 18 previous-year questions with full solutions — IIT JAM, GATE, CSIR NET-JRF and JEST, 2020 onwards.

Theory

Physical motivation

Maxwell's equations are already fully compatible with special relativity — Einstein was led to relativity precisely by studying them. This chapter reveals the deep unity of \(\vec{E}\) and \(\vec{B}\): they are not separate fields but two faces of a single relativistic object, the electromagnetic field tensor \(F^{\mu\nu}\). A pure electric field in one frame may appear as a mix of \(\vec{E}\) and \(\vec{B}\) in another. This is the ultimate unification of electrostatics and magnetostatics.

Special Relativity — Essential Toolkit

Lorentz factor and relativistic kinematics

\begin{equation} \gamma = \frac{1}{\sqrt{1-\beta^2}},\qquad \beta=\frac{v}{c},\qquad \gamma\geq1 \end{equation} Time dilation: \(\Delta t = \gamma\Delta t_0\)

Length contraction: \(L = L_0/\gamma\)

Relativistic momentum: \(\vec{p} = \gamma m\vec{v}\)

Energy: \(E = \gamma mc^2\); \(E^2 = (pc)^2+(mc^2)^2\)

Lorentz transformation (frame \(S'\) moves at \(v\) along \(\xhat\))

\begin{align} x' &= \gamma(x-vt) & t' &= \gamma(t-vx/c^2)\\ y' &= y & z' &= z \end{align} Inverse: replace \(v\to-v\).

Four-vectors

A four-vector \(A^\mu=(A^0,A^1,A^2,A^3)\) transforms like \((ct,x,y,z)\). Invariant: \(A^\mu A_\mu = (A^0)^2-(A^1)^2-(A^2)^2-(A^3)^2\) (using metric \(g_{\mu\nu}=\text{diag}(+,-,-,-)\)).

Four-velocity: \(U^\mu = \gamma(c,\vec{v})\); \(U^\mu U_\mu=c^2\)

Four-momentum: \(p^\mu = m U^\mu = (E/c,\vec{p})\); \(p^\mu p_\mu=(mc)^2\)

Four-current: \(J^\mu = (\rho c,\vec{J})\); \(\partial_\mu J^\mu=0\) (continuity)

Four-potential: \(A^\mu = (V/c,\vec{A})\)

Lorentz Transformation of Fields

The key result: \(\vec{E}\) and \(\vec{B}\) are components of a single tensor and mix under a Lorentz boost.

Transformation of \(\vec{E}\) and \(\vec{B}\)

Frame \(S'\) moves with velocity \(\vec{v}=v\xhat\) relative to \(S\): \begin{align} E_x' &= E_x & B_x' &= B_x \\[4pt] E_y' &= \gamma(E_y - vB_z) & B_y' &= \gamma\!\left(B_y + \frac{v}{c^2}E_z\right) \\[4pt] E_z' &= \gamma(E_z + vB_y) & B_z' &= \gamma\!\left(B_z - \frac{v}{c^2}E_y\right) \end{align} Compact form (components parallel/perpendicular to \(\vec{v}\)): \begin{align} \vec{E}'_\parallel &= \vec{E}_\parallel & \vec{B}'_\parallel &= \vec{B}_\parallel \\[4pt] \vec{E}'_\perp &= \gamma(\vec{E}+\vec{v}\times\vec{B})_\perp & \vec{B}'_\perp &= \gamma\!\left(\vec{B}-\frac{\vec{v}\times\vec{E}}{c^2}\right)_\perp \end{align}

Electric and magnetic fields unify

A pure electric field in \(S\) (\(\vec{B}=\mathbf{0}\)) appears as both \(\vec{E}'\) and \(\vec{B}'\) in \(S'\). Magnetism is fundamentally a relativistic effect of electricity — there is only one field, seen from different frames.

Example 8.1 B field from a moving charge

A charge \(q\) moves along \(+\xhat\) with velocity \(v\). In its rest frame \(S'\), the field is purely electric: \(\vec{E}'=kq/r'^2\,\hat{r}'\), \(\vec{B}'=\mathbf{0}\).

Transform to lab frame \(S\): \[ \vec{B}_\perp = \gamma\frac{\vec{v}}{c^2}\times\vec{E}'_\perp \approx \frac{\vec{v}\times\vec{E}}{c^2} = \frac{\vec{v}\times\vec{E}}{c^2} = \frac{1}{c^2}\vec{v}\times\vec{E}. \] But \(\vec{E}=kq/r^2\,\hat{r}\) in the non-relativistic limit, so: \[ \vec{B} = \frac{\muz}{4\pi}\frac{q\vec{v}\times\hat{r}}{r^2} = \frac{\muz}{4\pi}\frac{q\vec{v}\times\vec{r}}{r^3}. \] \[\boxed{\text{The Biot-Savart law is a relativistic effect of Coulomb's law!}}\]

Example 8.2 Transformation of a uniform field

In frame \(S\): \(\vec{E}=E_0\yhat\), \(\vec{B}=\mathbf{0}\). Frame \(S'\) moves at \(v=0.6c\) along \(\xhat\). Find \(\vec{E}'\) and \(\vec{B}'\).

\(\gamma = 1/\sqrt{1-0.36} = 1/0.8 = 1.25\).

\(E_x'=E_x=0\); \(E_y'=\gamma(E_y-vB_z)=1.25\cdot E_0\); \(E_z'=\gamma(E_z+vB_y)=0\).

\(B_x'=0\); \(B_y'=\gamma(B_y+vE_z/c^2)=0\); \(B_z'=\gamma(B_z-vE_y/c^2)=-\gamma vE_0/c^2=-1.25\times0.6E_0/c=-0.75E_0/c\). \[ \boxed{\vec{E}'=1.25E_0\,\yhat,\quad \vec{B}'=-\frac{0.75E_0}{c}\,\zhat} \] A purely electric field in \(S\) appears as both \(\vec{E}'\) and \(\vec{B}'\) in \(S'\).

Example 8.3 Pure magnetic field in one frame

In \(S\): \(\vec{E}=\mathbf{0}\), \(\vec{B}=B_0\zhat\). Frame \(S'\) moves at \(v\) along \(\xhat\).

\(E_y'=\gamma(0-vB_0)=-\gamma vB_0\); \(E_z'=0\); \(E_x'=0\).

\(B_z'=\gamma(B_0-0)=\gamma B_0\); \(B_y'=0\); \(B_x'=0\). \[ \boxed{\vec{E}'=-\gamma vB_0\,\yhat,\quad \vec{B}'=\gamma B_0\,\zhat} \] A pure magnetic field generates an electric field in a moving frame.

Electromagnetic Field Tensor

The six components of \(\vec{E}\) and \(\vec{B}\) combine into an antisymmetric rank-2 tensor:

The Electromagnetic Field Tensor \(F^{\mu\nu}\)

\begin{equation} 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} \end{equation} Antisymmetric: \(F^{\mu\nu}=-F^{\nu\mu}\), so \(F^{\mu\mu}=0\). Six independent components: three for \(\vec{E}\), three for \(\vec{B}\).

Dual tensor \(\tilde{F}^{\mu\nu}\)

\begin{equation} \tilde{F}^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma} = \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z/c & -E_y/c \\ B_y & -E_z/c & 0 & E_x/c \\ B_z & E_y/c & -E_x/c & 0 \end{pmatrix} \end{equation} Obtained from \(F^{\mu\nu}\) by \(\vec{E}/c\to\vec{B}\), \(\vec{B}\to-\vec{E}/c\) (duality transformation).

Field tensor properties

[noitemsep]
\(F^{\mu\nu}\) is antisymmetric: independent components = \(4\times3/2=6\)
Two Lorentz invariants: \(F^{\mu\nu}F_{\mu\nu} = 2(B^2-E^2/c^2)\) and \(F^{\mu\nu}\tilde{F}_{\mu\nu} = -4\vec{E}\cdot\vec{B}/c\)
If \(\vec{E}\perp\vec{B}\) in one frame, they are perpendicular in all frames
If \(E=cB\) in one frame, \(E'=cB'\) in all frames

Maxwell's Equations in Covariant Form

Covariant Maxwell's Equations

\begin{align} \partial_\nu F^{\mu\nu} &= \muz J^\mu \tag{M1+M4 covariant} \\[6pt] \partial_\mu\tilde{F}^{\mu\nu} &= 0 \tag{M2+M3 covariant} \end{align} where \(\partial_\mu = (\partial/\partial(ct),\partial/\partial x, \partial/\partial y,\partial/\partial z)\) and \(J^\mu=(\rho c,\vec{J})\).

Equation (the figure) in components gives M1 and M4; Equation (the figure) gives M2 and M3.

Elegance of covariant form

Eight Maxwell equations (4 differential equations, each with 3 components) compress into just 2 tensor equations. The symmetry between them reflects the underlying duality between \(\vec{E}\) and \(\vec{B}\).

Covariant Lorentz force law

\begin{equation} m\frac{dU^\mu}{d\tau} = qF^{\mu\nu}U_\nu \end{equation} where \(U^\mu=\gamma(c,\vec{v})\) is the four-velocity and \(\tau\) is proper time. This is the relativistic generalization of \(\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})\).

Four-potential and field tensor

\begin{equation} F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu,\qquad A^\mu = (V/c,\vec{A}) \end{equation} Gauge transformation: \(A^\mu\to A^\mu+\partial^\mu\lambda\) (leaves \(F^{\mu\nu}\) invariant).

Lorenz gauge: \(\partial_\mu A^\mu=0\), giving \(\Box^2 A^\mu=-\muz J^\mu\).

Lorentz Invariants

Two Lorentz scalar invariants of the EM field

\begin{align} \mathcal{I}_1 &= F^{\mu\nu}F_{\mu\nu} = 2\left(B^2-\frac{E^2}{c^2}\right) \\[4pt] \mathcal{I}_2 &= F^{\mu\nu}\tilde{F}_{\mu\nu} = -\frac{4}{c}\vec{E}\cdot\vec{B} \end{align} These have the same value in all inertial frames.

Physical consequences of invariants

If	Then in every frame
\(E>cB\) in some frame	\(E>cB\) always (cannot boost to \(E=0\))
\(E<cB\) in some frame	\(E<cB\) always (cannot boost to \(B=0\))
\(E=cB\) in some frame	\(E=cB\) always (e.g., plane wave)
\(\vec{E}\perp\vec{B}\) in some frame	\(\vec{E}\perp\vec{B}\) always
\(\vec{E}\parallel\vec{B}\) in some frame	\(\vec{E}\parallel\vec{B}\) always

Example 8.4 Checking invariants

In frame \(S\): \(\vec{E}=E_0\yhat\), \(\vec{B}=\mathbf{0}\). In frame \(S'\) (from an earlier example): \(\vec{E}'=1.25E_0\yhat\), \(\vec{B}'=-0.75E_0/c\,\zhat\).

Check \(\mathcal{I}_1\): In \(S\): \(2(0-E_0^2/c^2)=-2E_0^2/c^2\).

In \(S'\): \(2((0.75E_0/c)^2-(1.25E_0)^2/c^2) = 2(0.5625-1.5625)E_0^2/c^2=-2E_0^2/c^2\). ✓

Check \(\mathcal{I}_2\): In \(S\): \(-4\vec{E}\cdot\vec{B}/c=0\). In \(S'\): \(-4\vec{E}'\cdot\vec{B}'/c=-4(1.25E_0)(0)\cdot(-0.75E_0/c)/c=0\). ✓

Example 8.5 Can we boost to \(\vec{B}=0\)?

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Chapter 8 Relativistic Electrodynamics