Selected problems in special & general relativity
Cosmic-ray muon survival
The ratio $R(\mu/e)$ of muon neutrinos to electron neutrinos measured at ground level from cosmic radiation is $R(\mu/e) = 2$ at low energies. These neutrinos come from pion decays in the upper atmosphere: $\pi \to \mu + \nu_\mu,\; \mu \to e + \nu_\mu + \nu_e.$ At higher energies this ratio grows, because relativistic muons reach the ground before completing the second decay. In the muon's rest frame the lifetime is $\tau_0 = 2.2\,\mu$s. Find the smallest muon energy that allows the muons to reach the ground from $h = 10$ km before decaying substantially. Use $m_\mu c^2 = 106$ MeV.
Setup. "Reaches the ground before decaying substantially" means surviving one lab-frame lifetime over $h$: $h = v\tau = \beta c\gamma\tau_0$.
Using $\beta\gamma = \sqrt{\gamma^2-1}$: $\beta\gamma = h/(c\tau_0) = 10^4/660 \approx 15.15$, so $\gamma \approx 15.18$.
$$\boxed{\,E_{\min} = \gamma m_\mu c^2 \approx 1.6\,\text{GeV}.\,}$$
Without time dilation a muon would travel only $c\tau_0 = 660$ m before decaying — short of 10 km by a factor of ~15. The 1958 Frisch–Smith experiment on Mt. Washington confirmed this directly.
4-velocities and $\gamma$ as an inner product
(a) In an inertial frame $S$ an object has 3-velocity $\vec u$. A frame $S'$ moves in the negative $x$-direction relative to $S$ with relative speed $v'$. Write down the 4-velocities of the object and of $S'$ in $S$. (b) Show that $\gamma$ of an object in any frame equals the inner product of its 4-velocity with the 4-velocity of an observer at rest in that frame. (c) Express the $\gamma$ factor of the object in (a) in frame $S'$.
(a) $U^\mu = \gamma_u(c,\vec u)$ for the object; $V^\mu = \gamma_{v'}(c,-v',0,0)$ for an observer at rest in $S'$ as measured in $S$.
(b) In the observer's rest frame $\tilde S$, $V^\mu = (c,\vec 0)$ and $U^\mu = \gamma_{\tilde u}(c,\vec{\tilde u})$, so $U\!\cdot\!V = \gamma_{\tilde u}\,c^2$. Since $U\!\cdot\!V$ is a Lorentz scalar, $\gamma_{\tilde u} = (U\!\cdot\!V)/c^2$ in any frame.
(c) $U\!\cdot\!V = \gamma_u\gamma_{v'}(c^2 + u_x v')$, so
$$\boxed{\;\gamma_{u'} = \gamma_u\gamma_{v'}\!\left(1+\frac{u_x v'}{c^2}\right).\;}$$
Checks: if $\vec u=0$ then $\gamma_{u'}=\gamma_{v'}$; if $\vec u=-v'\hat x$ then $\gamma_{u'}=1$. One Lorentz-invariant contraction $U\!\cdot\!V$ encodes the velocity-addition algebra.
Light signal and rescue ship
You are traveling at relative speed $c/4$ toward a space station; when you send a light signal for help, the distance to the station in your frame is 1 light-day. The station dispatches a rescue ship at speed $3c/4$ relative to itself. How long, on your clock, does (a) the light signal take to reach the station, and (b) the rescue ship to reach you?
Work in your rest frame. Place yourself at $x=0$, station at $x=L_0=1$ ly approaching at $-c/4$.
Light-signal phase. $ct_1 = L_0 - (c/4)t_1 \Rightarrow t_1 = 4L_0/(5c) = 0.8$ day. Station at $x_1 = 4L_0/5$ at that moment.
Rescue ship velocity in your frame. Velocity-addition: $u_\text{res} = (-3c/4 - c/4)/(1+3/16) = -16c/19$.
Travel time and total. $\Delta t_2 = x_1/|u_\text{res}| = (4L_0/5)/(16c/19) = 19L_0/(20c) = 0.95$ day.
$$\boxed{\;t_\text{total} = t_1+\Delta t_2 = \frac{7L_0}{4c} = 1.75\,\text{day} = 42\,\text{h}.\;}$$
Relativistic pursuit at constant proper acceleration
A police officer at rest in $S$ pursues a spaceship passing at constant velocity $v>u$, accelerating with constant proper acceleration $a$. Find (a) the pursuit duration on the criminal's clock, (b) on the officer's clock, and (c) the relative velocity at catch-up.
With rapidity $\eta_p = a\tau_p/c$ for the officer, hyperbolic motion gives $t_S = (c/a)\sinh\eta_p$, $x_S=(c^2/a)(\cosh\eta_p-1)$. Setting $x_S = v\,t_S$ with $v = c\tanh\eta$ yields, via half-angle identities, $\tanh(\eta_p/2)=\tanh\eta$, hence $\boxed{\eta_p = 2\eta}$.
(a) $\tau_c = t_S/\gamma_v = (2c/a)\sinh\eta = 2v\gamma_v/a$.
(b) $\tau_p = 2c\eta/a = (c/a)\ln[(c+v)/(c-v)]$.
(c) Relative rapidity at catch-up: $\eta_p-\eta=\eta$, so $v_\text{rel} = v$. The officer catches the criminal moving relative to him at exactly the criminal's $S$-speed — a clean consequence of additive rapidities.
Pion decay and atmospheric muon flight
Super-Kamiokande measures atmospheric neutrinos via $\pi^+ \to \mu^+ + \nu_\mu,\; \mu^+ \to e^+ + \bar\nu_\mu + \nu_e$. (a) For decay at rest, find the antimuon kinetic energy and the neutrino 3-momentum (neglect $m_\nu$). (b) Mean flight distance of an antimuon in the pion rest frame before its decay (mean lifetime $\tau_\mu$).
(a) Conservation $m_\pi c^2 = \sqrt{(m_\mu c^2)^2+(pc)^2}+pc$ gives $$\boxed{\;p_{\nu_\mu} = \frac{(m_\pi^2-m_\mu^2)c}{2m_\pi},\quad T_{\mu^+} = \frac{(m_\pi-m_\mu)^2 c^2}{2m_\pi}.\;}$$ With $m_{\pi^+}c^2=139.57$ MeV, $m_\mu c^2 = 105.66$ MeV: $p_\nu c \approx 29.8$ MeV, $T_{\mu^+} \approx 4.12$ MeV. The antimuon is non-relativistic in the pion frame; almost all the released energy goes to the neutrino.
(b) $\beta_\mu\gamma_\mu = p_\mu/(m_\mu c) = (m_\pi^2-m_\mu^2)/(2 m_\pi m_\mu) \approx 0.282$. With $c\tau_\mu \approx 659$ m, $$\boxed{\;\langle d\rangle = \beta_\mu\gamma_\mu\,c\tau_\mu \approx 186\,\text{m}.\;}$$
Velocity as a function of displacement under constant electric force
A particle of mass $m$ and charge $q$ moves in a constant electric field $E$ starting from rest. Find $v(r)$.
Work-energy: $\mathcal{E} = mc^2 + qEr$. With $\mathcal{E}^2 = (pc)^2+(mc^2)^2$ and $v = pc^2/\mathcal{E}$: $$\boxed{\;v(r) = c\sqrt{1 - \left(\frac{mc^2}{mc^2+qEr}\right)^{\!2}}.\;}$$
Limits. Newtonian ($qEr\ll mc^2$): $v\approx\sqrt{2qEr/m}$. Ultra-relativistic ($qEr\gg mc^2$): $v\to c$. The worldline is a hyperbola in spacetime asymptoting to the light cone — constant proper acceleration $\alpha = qE/m$, the same hyperbolic motion that defines Rindler observers (Problem 2.28).
Field around a current-carrying wire in a frame moving along the wire
Current $I$ flows through a straight uncharged conductor. Determine the electromagnetic field in frame $K'$ moving parallel to the wire with velocity $v$, two ways: (a) transforming $F^{\mu\nu}$; (b) transforming the current 4-vector $J^\mu$.
(a) Wire along $\hat z$, $\vec B = (\mu_0 I/2\pi\rho)\hat y$ at perpendicular distance $\rho$. Boost gives $\vec E' = \gamma\vec v\times\vec B = -(\gamma v\mu_0 I/2\pi\rho)\hat x$ and $\vec B' = \gamma\vec B$.
(b) $J^\mu = (0,0,0,j_z)$ in $K$. Boost: $\rho' = -\gamma v j_z/c^2 = -\gamma v I/(c^2 \cdot 2\pi\rho)$ (line charge), $I' = \gamma I$. Coulomb + Ampère reproduce the same $\vec E'$, $\vec B'$.
Physical interpretation. Boosting along a neutral wire with drifting electrons makes the positive and negative carriers length-contract by different amounts; charge neutrality breaks. This is Purcell's classic demonstration that magnetism is relativistic electrostatics.
Plane waves in Lorenz gauge: transversality of E and B
In Lorenz gauge $\partial_\mu A^\mu = 0$, Maxwell's equations decouple to $\Box A^\nu = \mu_0 J^\nu$. For a current-free plane wave $A^\mu = \varepsilon^\mu e^{ik\cdot x}$, show $\vec E\!\cdot\!\vec k = \vec B\!\cdot\!\vec k = 0$.
$\Box A^\nu = 0$ gives $k^2 = 0$ (null dispersion). Lorenz: $k_\mu\varepsilon^\mu = 0$.
$F^{\mu\nu} = i(k^\mu\varepsilon^\nu - k^\nu\varepsilon^\mu)e^{ik\cdot x}$. The magnetic field $\vec B = -i\vec k\times\vec\varepsilon\,e^{ik\cdot x}$: $\vec B\!\cdot\!\vec k = -i\vec k\!\cdot\!(\vec k\times\vec\varepsilon) = 0$. Electric: $\vec E = i(\vec k\varepsilon^0 - \omega\vec\varepsilon)e^{ik\cdot x}$, $\vec E\!\cdot\!\vec k = i\varepsilon^0|\vec k|(c|\vec k| - \omega) = 0$ using the null condition. Both transversality relations follow from $k^2 = 0$ and $k\!\cdot\!\varepsilon = 0$.
Lorentz invariants of a free EM plane wave
Compute $F_{\mu\nu}F^{\mu\nu}$ and $\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$ for a free plane wave. Interpret.
With $F^{\mu\nu} \propto k^\mu\varepsilon^\nu - k^\nu\varepsilon^\mu$ and $k^2 = k\!\cdot\!\varepsilon = 0$, both invariants vanish identically. In terms of fields: $F_{\mu\nu}F^{\mu\nu} = 2(B^2 - E^2/c^2)$ and $\epsilon F F = -8\vec E\!\cdot\!\vec B/c$. So $$\boxed{\;|\vec E| = c|\vec B|\quad\text{and}\quad \vec E\perp\vec B.\;}$$
Null Lorentz class. No frame trivialises a free EM plane wave to pure $\vec E$ or pure $\vec B$; both invariants stay zero in every frame. Free EM waves enjoy the larger 15-parameter conformal symmetry, not just Lorentz — reflecting photon masslessness and absence of an intrinsic length scale.
$\vec E\!\cdot\!\vec B$ invariance, orthogonality, $\vec E\times\vec B \propto \vec k$
(a) Show $\vec E\!\cdot\!\vec B$ is Lorentz-invariant. (b) Frame-independence of $\vec E\perp\vec B$. (c) For free plane waves, $\vec E\perp\vec B$. (d) $\vec E\times\vec B = A\vec k$ with $A\neq 0$.
(a) $F_{\mu\nu}{}^*F^{\mu\nu} = -(4/c)\vec E\!\cdot\!\vec B$, manifestly a Lorentz scalar.
(b) Orthogonality $\Leftrightarrow \vec E\!\cdot\!\vec B = 0$, invariant by (a).
(c) $\vec E\!\cdot\!\vec B = 0$ by (a) + Problem 1.121, or directly via triple products $\vec k\!\cdot\!(\vec k\times\vec\varepsilon) = 0$.
(d) Expanding $\vec E\times\vec B = (\vec k\varepsilon^0 - \omega\vec\varepsilon)\times(\vec k\times\vec\varepsilon)$ with BAC-CAB, applying $k^2=0$ and $k\!\cdot\!\varepsilon = 0$, the $\vec\varepsilon$-coefficient kills (proportional to $k^2$); the $\vec k$-coefficient survives as $\omega(\varepsilon\!\cdot\!\varepsilon)$. So $\boxed{\vec E\times\vec B = \omega(\varepsilon\!\cdot\!\varepsilon)\vec k\,e^{2ik\cdot x} \equiv A\vec k}.$ Poynting flux along $\vec k$ — energy streams in the propagation direction, as required.