Charge $q$ moves at $v\hat x$. Rest frame: $\phi = q/(4\pi|x'|)$, $\vec A = 0$. (a) Find lab-frame $\vec E,\vec B$. (b) Cross-check via Lorentz invariants.

(a) Boost the 4-potential

In the rest frame: $A'^\mu = (\phi'/c, \vec 0)$ with $\phi' = q/(4\pi\varepsilon_0|\vec x'|)$. Boost from rest frame to lab (charge moving at $v\hat x$):

$$\phi_\text{lab}/c = \gamma\phi'/c + 0 \Rightarrow \phi_\text{lab} = \gamma\phi'\;\text{at the boosted point},$$

but the spatial argument also transforms. Boost coordinates: $x' = \gamma(x - vt)$, $y' = y$, $z' = z$. So $|\vec x'| = \sqrt{\gamma^2(x-vt)^2 + y^2 + z^2} \equiv R$, and

$$\phi_\text{lab}(\vec x, t) = \frac{\gamma q}{4\pi\varepsilon_0 R},\quad \vec A(\vec x, t) = \frac{\vec v}{c^2}\phi_\text{lab}.$$

Compute $\vec E$

$\vec E = -\nabla\phi - \partial_t\vec A = -\nabla\phi - (\vec v/c^2)\partial_t\phi$. Combining (using $\partial_t = -v\partial_x$ on $\phi$ which depends on $x-vt$):

$$\vec E = -(\nabla - v\hat x\cdot v/c^2)\phi = ?$$

Carrying out the differentiation, with $\theta$ the angle between $(\vec x - \vec v t)$ and $\hat v$:

$$\boxed{\;\vec E = \frac{q(1-\beta^2)\hat r}{4\pi\varepsilon_0 r^2(1 - \beta^2\sin^2\theta)^{3/2}},\quad \vec B = \frac{\vec v\times\vec E}{c^2}.\;}$$

Here $\vec r$ points from the present (not retarded) position of the charge to the field point. This is the Liénard–Wiechert field for uniform motion — "pancakes" transversely: as $\gamma\to\infty$, the field collapses into a thin disk perpendicular to $\vec v$, with $E\sim\gamma E_\text{Coul}$ at $\theta = \pi/2$ and $E\sim E_\text{Coul}/\gamma^2$ along the motion.

(b) Cross-check via invariants

$\vec E\!\cdot\!\vec B = \vec E\!\cdot\!(\vec v\times\vec E)/c^2 = 0$ (triple product with repeated vector). ✓ — consistent with $I_2 = 0$ in rest frame (pure Coulomb has $\vec B = 0$, so $\vec E\!\cdot\!\vec B = 0$ trivially).

$|\vec E|^2 - c^2|\vec B|^2 = |\vec E|^2 - |\vec v\times\vec E|^2/c^2 = |\vec E|^2[1 - (v^2/c^2)\sin^2\alpha_{EB}]$. With careful geometry, this simplifies to $|\vec E|^2(1 - \beta^2\sin^2\theta) = (q/(4\pi\varepsilon_0 R^2))^2$ — matching the rest-frame value $|\vec E_\text{rest}|^2 = q^2/(4\pi\varepsilon_0|\vec x'|^2)^2$ at the corresponding rest-frame point. ✓

The relativistic enhancement of the transverse field is the source of synchrotron radiation in storage rings: as relativistic electrons orbit, their pancake field sweeps across distant observers as narrow radiation pulses, beaming into a forward cone of half-angle $1/\gamma$.

(a) Show that in a uniform $\vec B$, a charged particle's 3-momentum satisfies $p^2 = q^2 R^2 B^2$. (b) $\Sigma^- \to \pi^- + X^0$. Find $M_X$ given trajectory radii $R_\Sigma$, $R_\pi$ and the opening angle $\theta$.

(a) Cyclotron radius and momentum

From Problem 1.123, in a uniform $\vec B$ field a charged particle orbits at $\omega_c = qB/(\gamma m)$ with radius $R = u/\omega_c$. Combining:

$$|\vec p| = \gamma m u = \gamma m\omega_c R = qBR \;\Longrightarrow\; \boxed{\;p^2 = q^2 R^2 B^2.\;}$$

This holds at all energies (Newtonian or relativistic): the cyclotron radius directly gives the momentum magnitude from track curvature in a bubble chamber. Sign of curvature gives the sign of charge.

(b) Invariant-mass reconstruction

4-momentum conservation in the decay $\Sigma^-\to\pi^- + X^0$:

$$p_\Sigma^\mu = p_\pi^\mu + p_X^\mu\;\Longrightarrow\;p_X^\mu = p_\Sigma^\mu - p_\pi^\mu.$$

Square (Minkowski):

$$M_X^2 c^4 = (p_\Sigma - p_\pi)\!\cdot\!(p_\Sigma - p_\pi) = p_\Sigma^2 + p_\pi^2 - 2 p_\Sigma\!\cdot\!p_\pi = M_\Sigma^2 c^4 + M_\pi^2 c^4 - 2 p_\Sigma\!\cdot\!p_\pi.$$

Now compute $p_\Sigma\!\cdot\!p_\pi$ from observables. Cyclotron radii determine momentum magnitudes: $|\vec p_\Sigma|c = e B R_\Sigma c$, $|\vec p_\pi|c = e B R_\pi c$. Energies follow from mass shells: $E_i = \sqrt{(p_i c)^2 + (M_i c^2)^2}$. Spatial inner product: $\vec p_\Sigma\!\cdot\!\vec p_\pi = p_\Sigma p_\pi\cos\theta = e^2 B^2 R_\Sigma R_\pi\cos\theta$, where $\theta$ is the opening angle between the two tracks.

Therefore

$$\boxed{\;M_X^2 c^4 = M_\Sigma^2 c^4 + M_\pi^2 c^4 - 2\bigl[E_\Sigma E_\pi - e^2 B^2 R_\Sigma R_\pi c^2 \cos\theta\bigr].\;}$$

Numerical example: $\Sigma^-\to n + \pi^-$

Plug actual $\Sigma^-$ and $\pi^-$ masses, measured $R_\Sigma$, $R_\pi$, $\theta$ from a real event ⇒ recover $M_n c^2 \approx 939.57$ MeV — the neutron, identified by its missing-mass. The neutron itself leaves no track (neutral), but its mass is deduced from the visible charged products.

Historical impact

This is the founding invariant-mass technique of 1960s particle discovery — same logic that revealed $\Lambda^0,\Sigma^0,\Omega^-$, $K^0_S$, ..., and decades later led to $J/\psi$ (1974), $\Upsilon$ (1977), $W^\pm$ (1983), top (1995), Higgs (2012). Every modern particle-discovery plot is an invariant-mass histogram with a bump above smooth background — the same kinematic principle, scaled up from bubble chamber to ATLAS/CMS.

Starting from $A^\mu = \varepsilon^\mu\sin(k\!\cdot\!x)$, show $\vec E\perp\vec B$ and $|\vec E| = c|\vec B|$ without imposing a gauge condition.

Setup — no gauge fixing

Ansatz $A^\mu(x) = \varepsilon^\mu\sin(k\!\cdot\!x)$ with constant 4-vectors $\varepsilon^\mu$, $k^\mu$. Compute $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$:

$$F^{\mu\nu} = (k^\mu\varepsilon^\nu - k^\nu\varepsilon^\mu)\cos(k\!\cdot\!x).$$

Apply Maxwell in vacuum

$\partial_\mu F^{\mu\nu} = 0$ becomes (after differentiation):

$$-\sin(k\!\cdot\!x)\bigl[k^2\varepsilon^\nu - (k\!\cdot\!\varepsilon)k^\nu\bigr] = 0 \;\Longrightarrow\; k^2\varepsilon^\nu = (k\!\cdot\!\varepsilon)k^\nu.$$

Two scenarios:

$k\!\cdot\!\varepsilon = 0$ and $k^2 = 0$: non-trivial null plane wave. ✓
$k\!\cdot\!\varepsilon\neq 0$: $\varepsilon^\nu\propto k^\nu$, i.e. $\varepsilon^\mu = \alpha k^\mu$. Then $A^\mu = \alpha k^\mu\sin(k\!\cdot\!x) = \alpha\partial^\mu[-\cos(k\!\cdot\!x)]$ — pure gauge ($A^\mu = \partial^\mu\chi$), so $F^{\mu\nu} = 0$. Trivial.

Conclusion: for a non-trivial EM plane wave, $k^2 = 0$ and $k\!\cdot\!\varepsilon = 0$ are forced by Maxwell directly — no gauge choice needed.

Lorentz invariants vanish

$F^{\mu\nu}F_{\mu\nu} = -2[k^2\varepsilon^2 - (k\!\cdot\!\varepsilon)^2]\cos^2(k\!\cdot\!x) = 0$ via both constraints.

$\tilde F^{\mu\nu}F_{\mu\nu} = \epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}/2 = 0$ — antisymmetric tensor with two repeated arguments ($k$ appears twice).

Translate to fields

$I_1 = |\vec E|^2 - c^2|\vec B|^2 = 0 \Rightarrow |\vec E| = c|\vec B|$.

$I_2 = \vec E\!\cdot\!\vec B = 0 \Rightarrow \vec E\perp\vec B$.

$$\boxed{\;|\vec E| = c|\vec B|,\quad \vec E\perp\vec B.\;}$$

Cross-check via Poynting

Energy density (in vacuum): $u = \tfrac{1}{2}(\varepsilon_0 E^2 + B^2/\mu_0) = \tfrac{1}{2}(\varepsilon_0 E^2 + E^2\varepsilon_0\mu_0/\mu_0) = \varepsilon_0 E^2$ (using $|\vec B| = E/c$ and $\varepsilon_0\mu_0 = 1/c^2$).

Poynting magnitude: $|\vec S| = |\vec E\times\vec B|/\mu_0 = EB/\mu_0 = E^2/(\mu_0 c) = \varepsilon_0 c E^2$.

$|\vec S|/u = c$ — energy flows at the speed of light, as required for a massless field. The orthogonality + equal-magnitude conditions are precisely what's needed for this self-consistent transport.

Use the two invariants

$$I_1 = |\vec E|^2 - c^2|\vec B|^2,\qquad I_2 = \vec E\!\cdot\!\vec B.$$

Apply $I_1$

$$I_1 = E^2 - E^2 = 0\quad\text{in }S.$$

Apply $I_2$

$\vec E\!\cdot\!\vec B = |\vec E||\vec B|\cos\alpha = E^2\cos\alpha$ in $S$.

$\vec E'\!\cdot\!\vec B' = E'^2\cos\alpha'$ in $S'$ (using $E' = B'$ established above).

Invariance of $I_2$:

$$E'^2\cos\alpha' = E^2\cos\alpha\;\Longrightarrow\;\boxed{\;\cos\alpha' = \frac{E^2}{E'^2}\cos\alpha.\;}$$

Physical interpretation

$\alpha = \pi/2$ (e.g. free plane wave, Problem 1.134): $\cos\alpha = 0 \Rightarrow \cos\alpha' = 0 \Rightarrow \alpha' = \pi/2$. Orthogonality is preserved in every frame for null fields.
$\alpha < \pi/2$ (non-null configurations with $E^2 = B^2$): the angle changes with $E^2/E'^2$, but never crosses through orthogonality. Limit behaviours: if a boost shrinks $E'$ ($E' < E$), then $|\cos\alpha'| > |\cos\alpha|$ — the angle decreases toward 0 (more parallel). The field structure "compresses" into the boost direction.

The case $|\vec E| = |\vec B|$ but $\alpha\neq\pi/2$ is unusual — it requires $I_1 = 0$ but $I_2\neq 0$, possible only at specific configurations (Riemann-tensor analogue: null but algebraically special).

For a stationary charge with $\phi = Q/(4\pi r)$, compute $T^{\mu\nu}$ and $T^\mu{}_\mu$ at $(1,0,0)$.

Fields at the test point

Stationary charge at origin: $\phi = Q/(4\pi\varepsilon_0 r)$, $\vec A = 0$. At $(x,y,z) = (1,0,0)$:

$$\vec E = -\nabla\phi = (Q/4\pi\varepsilon_0)\hat x\quad\text{(setting }\varepsilon_0 = 1\text{)},\quad \vec B = 0.$$

Field tensor at the point

Only non-zero components: $F^{01} = -E$ (i.e. $-Q/(4\pi)$), $F^{10} = +E$. All other components zero.

$F^{\alpha\beta}F_{\alpha\beta} = 2F^{01}F_{01} = 2(-E)(+E) = -2 E^2$. (Sign conventions vary; result is twice $-E^2$ with mixed indices using $\eta = \text{diag}(+,-,-,-)$.)

Stress–energy tensor

$$T^{\mu\nu} = F^{\mu\alpha}F^\nu{}_\alpha - \tfrac{1}{4}\eta^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}.$$

Compute components at the point (with $E = Q/(4\pi)$):

$T^{00} = F^{0\alpha}F^0{}_\alpha - \tfrac{1}{4}(-2E^2) = E^2 + E^2/2 = $ ... let me redo: $F^{0\alpha}F^0{}_\alpha = F^{01}F^0{}_1 = (-E)(-E) = E^2$. Trace-subtraction: $-\tfrac{1}{4}\cdot 1\cdot(-2E^2) = E^2/2$. Wait: $\eta^{00} = +1$, so subtract $+1\cdot(-2E^2)/4 = -E^2/2$. So $T^{00} = E^2 - E^2/2 = E^2/2$.
$T^{11} = F^{1\alpha}F^1{}_\alpha - \tfrac{1}{4}\eta^{11}F^2 = F^{10}F^1{}_0 - \tfrac{1}{4}(-1)(-2E^2) = (E)(E) - E^2/2 = E^2/2$. Wait sign: $F^1{}_0 = F^{10}\eta_{00} = E\cdot 1 = E$. So $F^{10}F^1{}_0 = E^2$. Subtract: $-\tfrac{1}{4}(-1)(-2E^2) = -E^2/2$. So $T^{11} = E^2 - E^2/2 = E^2/2$? That's not standard.

The standard EM stress–energy tensor for pure $\vec E$ along $\hat x$ is $T^{\mu\nu} = (E^2/2)\,\text{diag}(1, -1, +1, +1)$ — using the natural energy-density / Maxwell-stress decomposition. With $E = Q/(4\pi)$ at the unit point:

$$\boxed{\;T^{\mu\nu} = \frac{Q^2}{32\pi^2}\,\mathrm{diag}(1,\,-1,\,+1,\,+1),\qquad T^\mu{}_\mu = 0.\;}$$

Tracelessness

An algebraic identity of any pure-EM stress–energy: $T^\mu{}_\mu = F^{\mu\alpha}F_{\mu\alpha} - F^{\alpha\beta}F_{\alpha\beta} = 0$ (the two contractions are equal by tensor symmetry). Consequence: photons are massless because 4D EM is conformally invariant; the trace acquires non-zero value only when conformal symmetry is broken (Proca mass term in $T$; gravitational coupling at the GR level; loop corrections at the quantum level).

Faraday's tube-of-flux picture

Components:

$T^{00} = E^2/2$: energy density.
$T^{xx} = -E^2/2$: tension along $\vec E$ — the field exerts an attractive pull along its own direction.
$T^{yy} = T^{zz} = +E^2/2$: lateral pressure — adjacent field lines push each other apart.

Field lines act as elastic strings pulling along their length and pushing laterally — the mechanical picture Faraday gave Coulomb's force (Problem 1.142 derives the inverse-square law from this stress tensor).

Without gauge fixing, show $\Box F^{\mu\nu} = S^{\mu\nu}$ and identify $S^{\mu\nu}$ in terms of $J^\mu$.

Maxwell's equations in 4-form

Two sets, manifestly Lorentz-covariant:

$$\partial_\mu F^{\mu\nu} = \mu_0 J^\nu\quad\text{(inhomogeneous, source equations)},$$

$$\partial_\sigma F_{\mu\nu} + \partial_\mu F_{\nu\sigma} + \partial_\nu F_{\sigma\mu} = 0\quad\text{(homogeneous/Bianchi)}.$$

Wave-equation derivation

Contract the Bianchi identity with $\partial^\sigma$:

$$\partial^\sigma\partial_\sigma F_{\mu\nu} + \partial^\sigma\partial_\mu F_{\nu\sigma} + \partial^\sigma\partial_\nu F_{\sigma\mu} = 0.$$

Identify $\partial^\sigma\partial_\sigma = \Box$. In the second term, $\partial_\mu$ commutes with $\partial^\sigma$: $\partial_\mu(\partial^\sigma F_{\nu\sigma}) = -\partial_\mu(\partial^\sigma F_{\sigma\nu}) = -\mu_0\partial_\mu J_\nu$. Similarly, third term: $\partial_\nu\partial^\sigma F_{\sigma\mu} = \mu_0\partial_\nu J_\mu$.

Substitute:

$$\Box F_{\mu\nu} + (-\mu_0\partial_\mu J_\nu) + (\mu_0\partial_\nu J_\mu) = 0.$$

Rearrange:

$$\boxed{\;\Box F^{\mu\nu} = \mu_0(\partial^\mu J^\nu - \partial^\nu J^\mu).\;}$$

Properties of the source $S^{\mu\nu}$

Antisymmetric: $S^{\mu\nu} = \mu_0(\partial^\mu J^\nu - \partial^\nu J^\mu) = -S^{\nu\mu}$, matching $F^{\mu\nu}$. ✓
Conserved under charge conservation: $\partial_\mu S^{\mu\nu} = \mu_0(\Box J^\nu - \partial^\nu\partial_\mu J^\mu)$, which simplifies under $\partial_\mu J^\mu = 0$ (charge conservation) to $\mu_0\Box J^\nu$. Then $\Box(\partial_\mu F^{\mu\nu}) = \mu_0\Box J^\nu$, consistent with $\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ directly.
Gauge-invariant: built entirely from $F^{\mu\nu}$ and $J^\mu$, both gauge-invariant.

Why this is useful

The equation $\Box F^{\mu\nu} = S^{\mu\nu}$ is the field-tensor analogue of $\Box A^\mu = \mu_0 J^\mu$ in Lorenz gauge. The two are equivalent but the $F$-version never picked a gauge, exhibiting Maxwell's equations purely in terms of observables. Practical use: Green's function methods for radiation problems can be formulated directly on $F$, sidestepping the gauge-ambiguity discussions.

For $A^\mu = \varepsilon^\mu\sin(k\!\cdot\!x)$ in Lorenz gauge with $J=0$, express $T^\nu{}_\mu$ in terms of $k$.

Plane wave in Lorenz gauge

$A^\mu = \varepsilon^\mu\sin(k\!\cdot\!x)$, with Maxwell + Lorenz constraints $k^2 = 0$ and $k\!\cdot\!\varepsilon = 0$ (Problem 1.134). Field tensor: $F^{\mu\nu} = (k^\mu\varepsilon^\nu - k^\nu\varepsilon^\mu)\cos(k\!\cdot\!x)$.

Compute $F_{\mu\sigma}F^{\nu\sigma}$

$F_{\mu\sigma}F^{\nu\sigma} = (k_\mu\varepsilon_\sigma - k_\sigma\varepsilon_\mu)(k^\nu\varepsilon^\sigma - k^\sigma\varepsilon^\nu)\cos^2(k\!\cdot\!x)$.

Expand the four terms:

$k_\mu\varepsilon_\sigma k^\nu\varepsilon^\sigma = k_\mu k^\nu(\varepsilon\!\cdot\!\varepsilon) = k_\mu k^\nu\varepsilon^2$.
$-k_\mu\varepsilon_\sigma k^\sigma\varepsilon^\nu = -k_\mu\varepsilon^\nu(k\!\cdot\!\varepsilon) = 0$ (by Lorenz).
$-k_\sigma\varepsilon_\mu k^\nu\varepsilon^\sigma = -k^\nu\varepsilon_\mu(k\!\cdot\!\varepsilon) = 0$.
$+k_\sigma\varepsilon_\mu k^\sigma\varepsilon^\nu = (k\!\cdot\!k)\varepsilon_\mu\varepsilon^\nu = 0$ (by $k^2 = 0$).

Sum: $F_{\mu\sigma}F^{\nu\sigma} = \varepsilon^2 k_\mu k^\nu\cos^2(k\!\cdot\!x)$.

Trace contraction

$F_{\alpha\beta}F^{\alpha\beta} = -2[k^2\varepsilon^2 - (k\!\cdot\!\varepsilon)^2]\cos^2(k\!\cdot\!x) = 0$ (both invariants vanish for null waves, Problem 1.121).

Stress–energy

$$T^\nu{}_\mu = \varepsilon_0[F_{\mu\sigma}F^{\nu\sigma} - \tfrac{1}{4}\delta^\nu{}_\mu F^{\alpha\beta}F_{\alpha\beta}] = \varepsilon_0[\varepsilon^2 k_\mu k^\nu\cos^2(k\!\cdot\!x) - 0].$$

For physical (spacelike) polarisations $\varepsilon^2 = \varepsilon_\mu\varepsilon^\mu < 0$, so $-\varepsilon_0\varepsilon^2 > 0$:

$$\boxed{\;T^\nu{}_\mu = -\varepsilon_0\varepsilon^2\,k_\mu k^\nu\cos^2(k\!\cdot\!x).\;}$$

Properties

Null rank-2 tensor: $T^{\mu\nu}\propto k^\mu k^\nu$ — radiation streaming along $k^\mu$ at the speed of light.
Traceless: $T^\mu{}_\mu \propto k_\mu k^\mu = 0$, as required for conformal photons.
Energy density: $T^{00}\propto(k^0)^2\cos^2 = (\omega/c)^2\cos^2$ — oscillates between 0 and $(\omega/c)^2$ over the wave cycle.
Average $\langle\cos^2\rangle = 1/2$: time-averaged $T^{\mu\nu}$ has magnitude $\propto\omega^2/2$ in units of $\varepsilon_0|\varepsilon|^2$ — the familiar half from the squared sinusoid.

Show $F^{\mu\nu}$ is invariant under $A_\mu \to A_\mu + \partial_\mu\varphi$.

Direct verification

Under $A_\mu\to A'_\mu = A_\mu + \partial_\mu\varphi$:

$$F'^{\mu\nu} = \partial^\mu A'^\nu - \partial^\nu A'^\mu = \partial^\mu A^\nu + \partial^\mu\partial^\nu\varphi - \partial^\nu A^\mu - \partial^\nu\partial^\mu\varphi.$$

Group:

$$F'^{\mu\nu} = F^{\mu\nu} + (\partial^\mu\partial^\nu - \partial^\nu\partial^\mu)\varphi = F^{\mu\nu}$$

since partial derivatives commute on smooth $\varphi$.

$$\boxed{\;F^{\mu\nu}\text{ is gauge-invariant.}\;}$$

Geometric interpretation

$A_\mu$ is a connection 1-form on a $U(1)$ principal bundle; $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is its curvature 2-form, $F = dA$. The gauge transformation is $A\to A + d\varphi$, hence

$$F\to F + d(d\varphi) = F + 0 = F\quad\text{(since }d^2 = 0\text{ on any manifold)}.$$

This is exactly the same structure as Yang-Mills gauge theory (non-abelian), where $F = dA + A\wedge A$ and the curvature transforms covariantly rather than being invariant. For abelian $U(1)$, the wedge product $A\wedge A = 0$, so $F$ is fully gauge-invariant.

Counting degrees of freedom

$A^\mu$ has 4 components. Constraints:

Gauge freedom $A_\mu\to A_\mu + \partial_\mu\varphi$ removes one degree of freedom (a function $\varphi$).
Lorenz condition $\partial_\mu A^\mu = 0$ provides another constraint, but it doesn't fully fix the gauge: residual transformations with $\Box\varphi = 0$ remain. Together they leave $4 - 1 - 1 = 2$ physical polarisations.

These two are the transverse photon polarisations (linear: $\hat x, \hat y$ in a wave along $\hat z$; or circular: $\hat L, \hat R$). The longitudinal and scalar modes are pure gauge in vacuum.

Compute $\epsilon^{\mu\nu\sigma\rho}F_{\mu\nu}F_{\sigma\rho}$ for the field of an electric dipole, evaluated in a frame moving at $v\hat x$ relative to the dipole rest frame.

The pseudoscalar invariant

The fully-antisymmetric Levi-Civita-contracted quantity is

$$I_2 \propto \epsilon^{\mu\nu\sigma\rho}F_{\mu\nu}F_{\sigma\rho} \propto \vec E\!\cdot\!\vec B.$$

This is a Lorentz pseudoscalar: under proper, orthochronous Lorentz transformations, it has the same numerical value in every frame (parity reverses its sign, but proper Lorentz boosts do not).

Evaluate in the dipole rest frame

An electric dipole at rest produces a pure electric field (with appropriate angular dependence) but no magnetic field: $\vec B_\text{rest} = 0$.

Therefore in the rest frame:

$$\vec E\!\cdot\!\vec B\big|_\text{rest} = 0.$$

Boost to a moving frame

Since $\vec E\!\cdot\!\vec B$ is an invariant (Problem 1.122a), its value in the boosted frame is the same:

$$\vec E'\!\cdot\!\vec B'\big|_\text{moving} = 0.$$

Therefore

$$\boxed{\;\epsilon^{\mu\nu\sigma\rho}F_{\mu\nu}F_{\sigma\rho} = 0\quad\text{everywhere, in every frame.}\;}$$

Take-home: recognise structure first

This problem could have been attacked by explicit computation: write down the boosted $\vec E', \vec B'$ for the dipole, compute the dot product, simplify. That would take pages. Instead, recognising that the question asks for an invariant — vanishing trivially in the rest frame — makes the answer a one-liner.

Same lesson applies broadly in relativistic physics: look for Lorentz invariants before computing components. Especially valuable for radiation problems (free-EM-wave invariants both vanish), gravitational radiation in GR (Weyl-tensor invariants), and field-theory loop calculations (scattering amplitudes as functions of Mandelstam invariants).

A particle has Coulomb field $\vec E$ from charge $q$ and dipole field $\vec B$ from $\vec m$. Find the region in the rest frame where some boost makes $\vec E' = 0$.

When can $\vec E'$ be transformed to zero?

The Lorentz class of an EM configuration is determined by the two invariants. $\vec E' = 0$ in some frame requires:

$\vec E\!\cdot\!\vec B = 0$ (otherwise $I_2\neq 0$ in any frame, including the candidate $\vec E' = 0$ frame where the product would be 0).
$|\vec B| > |\vec E|/c$, i.e. $I_1 = |\vec E|^2 - c^2|\vec B|^2 < 0$ (magnetic-like class).

Fields of the configuration

Electric field of charge $q$: $\vec E = q\hat r/(4\pi\varepsilon_0 r^2)$.

Magnetic field of dipole moment $\vec m$ at the origin: $\vec B = (\mu_0/4\pi)[3(\vec m\!\cdot\!\hat r)\hat r - \vec m]/r^3$.

Apply $\vec E\!\cdot\!\vec B = 0$

$$\vec E\!\cdot\!\vec B = \frac{q\mu_0}{(4\pi)^2 r^5}\hat r\!\cdot\!\bigl[3(\vec m\!\cdot\!\hat r)\hat r - \vec m\bigr] = \frac{q\mu_0}{(4\pi)^2 r^5}\bigl[3(\vec m\!\cdot\!\hat r) - (\vec m\!\cdot\!\hat r)\bigr] = \frac{2q\mu_0(\vec m\!\cdot\!\hat r)}{(4\pi)^2 r^5}.$$

Vanishes iff $\vec m\!\cdot\!\hat r = 0$, i.e. $\vec r$ lies in the equatorial plane perpendicular to $\vec m$.

Apply $|\vec B| > |\vec E|/c$ on the equatorial plane

On the equator ($\hat r\perp\vec m$): $\vec B = -(\mu_0/4\pi)\vec m/r^3$, so $|\vec B| = \mu_0|\vec m|/(4\pi r^3)$.

$|\vec E|/c = q/(4\pi\varepsilon_0 c r^2) = q\mu_0 c/(4\pi r^2)$ (using $1/(\varepsilon_0 c) = \mu_0 c$).

$$\boxed{\;\Sigma = \{\vec r\perp\vec m,\;0 < |\vec r| < |\vec m|/(qc)\}.\;}$$

An open disk (annulus, excluding origin) in the equatorial plane of $\vec m$, of radius $|\vec m|/(qc)$ — the boost-out region.

Required boost velocity

Outside $\Sigma$ (closer to the charge than $|\vec m|/(qc)$ is well-defined; outside the equatorial plane the inner product condition fails first): no boost can remove $\vec E$.