(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) $\omega = qB/(\gamma m)$, $|\vec p| = \gamma m\omega R = qBR$. The relativistic cyclotron radius gives the magnitude of $\vec p$ from observables.

(b) 4-momentum conservation: $p_X^\mu = p_\Sigma^\mu - p_\pi^\mu$. $$\boxed{\;M_X^2 c^4 = M_\Sigma^2 c^4 + M_\pi^2 c^4 - 2\bigl[E_\Sigma E_\pi - e^2 B^2 R_\Sigma R_\pi c^2 \cos\theta\bigr].\;}$$ For $\Sigma^- \to n + \pi^-$, plugging numbers recovers $M_n c^2 \approx 939.6$ MeV. This is the founding invariant-mass technique of 1960s particle discovery — same logic that led to $\Lambda^0,\Sigma^0,\Omega^-$ and, decades later, to the Higgs.

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.

Maxwell $\partial_\mu F^{\mu\nu} = 0$ gives $k^2 \varepsilon^\nu = (k\!\cdot\!\varepsilon) k^\nu$. If $k\!\cdot\!\varepsilon\neq 0$, $A^\mu$ is pure gauge; for a non-trivial wave, $k^2 = 0$ and $k\!\cdot\!\varepsilon = 0$ are forced by Maxwell directly (no gauge choice needed). Then both Lorentz invariants vanish: $F^2 \propto k^2\varepsilon^2 - (k\!\cdot\!\varepsilon)^2 = 0$, $F\tilde F = 0$ identically. Translating: $|\vec E| = c|\vec B|$ and $\vec E\!\cdot\!\vec B = 0$.

Cross-check via Poynting: $u = \tfrac{1}{2}(\epsilon_0 E^2 + B^2/\mu_0) = \epsilon_0 E^2$; $|\vec S| = EB/\mu_0 = \epsilon_0 c E^2$; so $|\vec S|/u = c$ — energy at the speed of light.

$I_1 \propto B^2 - E^2 = 0$ — invariant, so $E' = B'$ in $S'$ too. $I_2 \propto \vec E\!\cdot\!\vec B = E^2\cos\alpha$ — invariant, so $E'^2\cos\alpha' = E^2\cos\alpha$. $$\boxed{\;\cos\alpha' = \frac{E^2}{E'^2}\cos\alpha.\;}$$ At $\alpha = \pi/2$ (plane-wave configuration of Problem 1.134), orthogonality is preserved in every frame.

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

At $(1,0,0)$: $\vec E = (Q/4\pi)\hat x$, $\vec B = 0$. With $F^{01} = -E$ and $F^{\alpha\beta}F_{\alpha\beta} = -2E^2$: $$\boxed{\;T^{\mu\nu} = \frac{Q^2}{32\pi^2}\,\mathrm{diag}(1,-1,1,1),\qquad T^\mu{}_\mu = 0.\;}$$

Tracelessness is an algebraic identity of any pure-EM tensor: $T^\mu{}_\mu = -F^{\mu\alpha}F_{\mu\alpha} + F^{\alpha\beta}F_{\alpha\beta} = 0$. Photons are massless because 4D EM is conformally invariant; the trace acquires non-zero value only when conformal symmetry is broken (Proca mass term, gravitational coupling).

Faraday's tube-of-flux interpretation. $T^{xx} = -E^2/2$ (tension along $\vec E$), $T^{yy} = T^{zz} = +E^2/2$ (lateral pressure). Field lines act as elastic strings pulling along their length and pushing each other apart laterally — the mechanical picture of Coulomb's law from Problem 1.142.

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

Bianchi: $\partial_\sigma F_{\mu\nu} + \partial_\mu F_{\nu\sigma} + \partial_\nu F_{\sigma\mu} = 0$. Contract with $\partial^\sigma$ and use $\partial^\sigma F_{\sigma\nu} = \mu_0 J_\nu$: $$\boxed{\;\Box F^{\mu\nu} = \mu_0(\partial^\mu J^\nu - \partial^\nu J^\mu).\;}$$ Antisymmetric (matching $F$), conserved when $\partial_\mu J^\mu = 0$. Gauge-invariant by construction — never picked a gauge.

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

Maxwell + Lorenz: $k^2 = 0$, $k\!\cdot\!\varepsilon = 0$. Then $F_{\mu\sigma}F^{\nu\sigma} = \varepsilon^2 k_\mu k^\nu \cos^2(k\!\cdot\!x)$ and $F^{\alpha\beta}F_{\alpha\beta} = 0$, so the trace-subtraction drops out: $$\boxed{\;T^\nu{}_\mu = -\epsilon_0\varepsilon^2 k_\mu k^\nu \cos^2(k\!\cdot\!x).\;}$$ Null rank-2 tensor $\propto k_\mu k^\nu$ — radiation streaming along $k^\mu$. Trace $\propto k^\mu k_\mu = 0$, as required for conformal photons.

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

$F^{\mu\nu} \to F^{\mu\nu} + (\partial^\mu\partial^\nu - \partial^\nu\partial^\mu)\varphi = F^{\mu\nu}$ — partials commute on smooth $\varphi$.

Geometrically: $A_\mu$ is a $U(1)$ connection, $F_{\mu\nu}$ its curvature. The gauge transformation is $F = dA \to F$ under $A \to A + d\varphi$ since $d^2 = 0$. Of $A_\mu$'s four components, the gauge orbit accounts for one, the Lorenz condition another, leaving the two physical photon polarisations.

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 quantity is a Lorentz pseudoscalar, equal to the same number in every frame: it's proportional to $\vec E\!\cdot\!\vec B$. In the dipole rest frame $\vec B = 0$, so $\vec E\!\cdot\!\vec B = 0$. Therefore $$\boxed{\;\epsilon^{\mu\nu\sigma\rho}F_{\mu\nu}F_{\sigma\rho} = 0\quad\text{everywhere, in every frame.}\;}$$ Recognise structure before computing: the answer is a one-liner from invariance alone.

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 $\vec E' = 0$.

Condition: $\vec E\!\cdot\!\vec B = 0$ and $|\vec B| > |\vec E|/c$ (magnetic-like Lorentz class).

$$\boxed{\;\Sigma = \{\vec r\perp\vec m,\;0 < |\vec r| < c|\vec m|/q\}\;}$$ — an open disk in the equatorial plane. The required boost velocity is $|\vec u|/c = E/(cB) < 1$ exactly inside this disk.

Two charges $+q$ at separation $d$. Integrate $T^{\mu\nu}$ over the median plane to find the force.

On the median plane $\vec E_x = 0$, $E^2 = q^2(y^2+z^2)/(4\pi^2\varepsilon_0^2 r^6)$ with $r^2 = (d/2)^2 + y^2 + z^2$. Stress: $T^{xx} = -\tfrac{1}{2}\varepsilon_0 E^2$.

Integral: $\int (\rho^2/r^6)\,dA = 2\pi/d^2$. Hence $$\boxed{\;F^x = -\frac{q^2}{4\pi\varepsilon_0 d^2}\;}$$ — exactly Coulomb repulsion, recovered from the field's transverse pressure on the median plane without ever touching the charges. Faraday's tube-of-flux mechanism made rigorous.

Special Relativity — worked problems

Bubble chamber: cyclotron radii and invariant-mass reconstruction

Plane-wave: $|\vec E| = c|\vec B|$ and $\vec E\perp\vec B$ without gauge fixing

Angle between $\vec E$ and $\vec B$ when $|\vec E| = |\vec B|$

EM stress–energy tensor of a point charge (rest frame)

Sourced wave equation for $F^{\mu\nu}$

Stress–energy of an EM plane wave

Gauge invariance of the field tensor

Lorentz pseudoscalar $F\tilde F$ for an electric dipole

Region where $\vec E$ can be transformed to zero (monopole + dipole)

Maxwell stress between two like charges