Special Relativity — worked problems

(a) General-angle Doppler formula

Source at rest emits with proper frequency $\omega_0$. Its 4-wavevector in its rest frame: $k^\mu_\text{src} = (\omega_0/c, \omega_0/c\,\hat n')$ for some direction $\hat n'$. In the lab the source moves at velocity $\vec v$. Boost the lab-direction $\hat n = (\cos\theta,\sin\theta,0)$ back into the source frame and apply photon Doppler:

The standard derivation (boost the photon 4-momentum from source frame to lab) gives

$$\boxed{\;\omega(\theta) = \frac{\omega_0}{\gamma(1 - \beta\cos\theta)}.\;}$$

Here $\theta$ is the angle between the photon's lab-frame direction of propagation toward the observer and the source's velocity.

(b) Angle of zero net shift

$\omega = \omega_0$ requires $\gamma(1 - \beta\cos\theta) = 1$, so

$$1 - \beta\cos\theta = 1/\gamma \;\Longrightarrow\; \cos\theta = (1 - 1/\gamma)/\beta.$$

$$\boxed{\;\cos\theta_* = \frac{1 - 1/\gamma}{\beta} = \frac{\gamma - 1}{\beta\gamma}.\;}$$

Special cases

• $\theta = 0$ (head-on approach): $\omega = \omega_0/[\gamma(1-\beta)] = \omega_0\sqrt{(1+\beta)/(1-\beta)}$ — standard longitudinal blueshift.
• $\theta = \pi/2$ (transverse Doppler): $\omega = \omega_0/\gamma$ — pure time-dilation redshift, no classical analogue. Verified by Ives–Stilwell (1938) using fast hydrogen canal rays — the direct experimental confirmation of relativistic time dilation in spectroscopy.
• $\theta = \theta_*$: longitudinal blueshift exactly cancels the time-dilation redshift. Non-zero for $\beta > 0$: $\cos\theta_* = \beta/2 + O(\beta^3)$, slightly forward of transverse.

Stationarity argument

A disk rotating at constant $\Omega$ is invariant under time translations: the configuration today is identical to the configuration one period from now. Thus any frequency ratio measured by stationary observers is time-independent — $C_2$ neither systematically gains nor loses with respect to $C_1$ over the long run.

Lab-frame photon

A photon traveling on a null geodesic in flat spacetime maintains a fixed lab-frame frequency $\omega_\text{lab}$ along its path (no Doppler from a stationary lab observer). When emitted at $r_2$ by clock $C_2$ with proper-frequency $\omega_2$, and received at $r_1$ by clock $C_1$ with proper-frequency $\omega_1$:

$$\omega_\text{lab} = \gamma_2\omega_2 = \gamma_1\omega_1.$$

(Each local clock ticks slower than the lab by $\gamma_i$, so its proper-frequency reading of a fixed-lab signal is $\omega_\text{lab}/\gamma_i$.)

Frequency ratio

$$\boxed{\;\frac{\omega_2}{\omega_1} = \frac{\gamma_2}{\gamma_1}.\;}$$

Equidistant observers ($r_1 = r_2$) see no shift — $\gamma_1 = \gamma_2$. The signal from a faster-moving observer ($r_2 > r_1$, $v_2 > v_1$, $\gamma_2 > \gamma_1$) is blueshifted when received closer to the axis.

GR analogue

Rotating-frame redshift <-> centrifugal potential $\Phi_\text{cf} = -\tfrac{1}{2}\Omega^2 r^2$: the equivalence-principle prototype for gravitational redshift in any stationary metric. In Schwarzschild geometry, the same logic with $\gamma_i$ replaced by $1/\sqrt{1 - 2GM/(r_i c^2)}$ gives Pound–Rebka: a photon climbing out of a gravitational well loses energy proportional to the potential difference, mirrored exactly here by the centrifugal potential difference.

Setup: 4-wavevector in a medium

Phase velocity in medium $= c/n$, so the spatial wavenumber is $|\vec k| = \omega_\text{med}\cdot n/c$ (not $\omega/c$). The 4-wavevector in the medium rest frame:

$$k^\mu = (\omega_\text{med}/c,\;(\omega_\text{med}n/c)\hat k).$$

This is not null: $k^2 = (\omega_\text{med}/c)^2(1 - n^2) \neq 0$ for $n \neq 1$. The medium provides a rest frame; Lorentz invariance is broken by the medium's worldline.

Boost to source frame

Source moves at $v\hat x$, $\hat k\!\cdot\!\hat x = \cos\theta$. Standard 4-vector boost of the time component:

$$k'^0 = \gamma(k^0 - v k_x) = \gamma\omega_\text{med}/c\cdot(1 - n\beta\cos\theta).$$

Therefore

$$\boxed{\;\frac{\omega_\text{src}}{\omega_\text{med}} = \gamma(1 - n\beta\cos\theta).\;}$$

Limits and special regimes

• Vacuum ($n = 1$): recovers $\omega_\text{src}/\omega_\text{med} = \gamma(1-\beta\cos\theta)$ — standard relativistic Doppler.
• Cherenkov regime ($n\beta > 1$): the factor $1 - n\beta\cos\theta$ can vanish or become negative. The source outruns its own wavefront in the medium — same kinematic structure as a sonic boom.
• Cherenkov cone angle: shocks form along the direction where $\omega_\text{src}/\omega_\text{med}\to 0$: $\cos\theta_C = 1/(n\beta)$.

Physical realisations: blue Cherenkov glow of nuclear-reactor pool water (electrons at $\beta > 1/n_\text{water} = 0.75$); Super-Kamiokande's water detectors imaging neutrino-induced charged-lepton tracks via their Cherenkov cones; PMT signatures in IceCube; ground-based cosmic-ray air showers.

Strategy: boost, reflect, boost back

Three-step calculation: (1) boost the incoming 4-wavevector into the mirror's rest frame; (2) apply the trivial $k_x\to -k_x$ reflection; (3) boost back to lab. Let $\cos\theta \equiv \cos\theta_\text{in}$ for brevity. Incoming 4-wavevector $k^\mu = \omega(1, \cos\theta, \sin\theta, 0)$.

Step 1: boost to mirror frame (velocity $+v\hat x$)

$$k'^0 = \gamma\omega(1 - v\cos\theta),\quad k'^x = \gamma\omega(\cos\theta - v),\quad k'^y = \omega\sin\theta.$$

Step 2: reflect in mirror frame

Reverse $k'^x$: $k'^x \to -\gamma\omega(\cos\theta - v) = \gamma\omega(v - \cos\theta)$. Other components unchanged.

Step 3: boost back to lab

$$\omega_\text{out} = \gamma\bigl[\gamma\omega(1 - v\cos\theta) + v\cdot\gamma\omega(v - \cos\theta)\bigr] = \gamma^2\omega[(1 - v\cos\theta) + v(v - \cos\theta)].$$

Simplify: $1 - 2v\cos\theta + v^2$. With $\gamma^2 = 1/(1-v^2)$:

$$\boxed{\;\omega_\text{out} = \omega\cdot\frac{1 - 2v\cos\theta + v^2}{1 - v^2}.\;}$$

Reflected $k_x$ in lab: $\gamma^2\omega[(v - \cos\theta) + v(1 - v\cos\theta)] = \gamma^2\omega(2v - \cos\theta - v^2\cos\theta)$. Take ratio $k_x/k_0$:

$$\boxed{\;\cos\theta_\text{out} = \frac{2v - (1+v^2)\cos\theta}{1 - 2v\cos\theta + v^2}.\;}$$

Limiting case: mirror retreats at radial light speed

At $v = -\cos\theta_\text{in}$, the mirror recedes at exactly the radial component of light's velocity:

$$1 - 2v\cos\theta + v^2 = 1 + 2\cos^2\theta + \cos^2\theta = 1 + 3\cos^2\theta\quad\text{(uncareful)}.$$

More carefully: at $\theta_\text{in}\to 0$ and $v\to -1^+$ together, $\omega_\text{out}\to\infty$ — the wavefronts pile up at the catching-up limit, the moving-mirror version of stationary phase. Inverse limit: $v\to +1^-$ at $\theta_\text{in}\to 0$ gives $\omega_\text{out}\to 0$ — total redshift.

Photon recoil and pressure

$\omega_\text{out}\ne\omega$ implies momentum transfer to the mirror per photon $= (\omega - \omega_\text{out})/c$, the kinematic basis for radiation pressure and the optical-tweezers force.

Setup

In the medium rest frame ($S'$), light refracts by Snell's law: $\sin\theta'_\text{inc} = n\sin\theta'_\text{refr}$. The lab observer sees the ray maintain its direction through the interface — no apparent refraction. This is possible only if the medium's motion compensates Snell's bending via aberration.

Aberration on each side

The motion is perpendicular to the surface, say $v\hat n$. Tangential momentum components of a photon are unchanged by aberration on each side (lab vs. medium frame), while the normal component transforms by $\gamma$-factors. Apply aberration to relate $\theta_\text{lab}$ on each side to $\theta'$:

$$\tan\theta_{S,\text{inc}} = \frac{\sin\theta'_\text{inc}}{\gamma(\cos\theta'_\text{inc} + v)},\quad\tan\theta_{S,\text{refr}} = \frac{\sin\theta'_\text{refr}}{\gamma(\cos\theta'_\text{refr} + v/n)}.$$

(The factor $v$ becomes $v/n$ inside the medium because the photon's phase velocity along $\hat n$ is $c/n$.)

Same lab direction

The lab observer says the ray direction is unchanged across the interface: $\tan\theta_{S,\text{inc}} = \tan\theta_{S,\text{refr}}$. Combine with Snell:

$$\sqrt{n^2 - \sin^2\theta'} = \cos\theta' + v(1 - n).$$

Squaring and collecting terms (with $\theta' = \theta'_\text{inc}$):

$$n^2 - \sin^2\theta' = \cos^2\theta' + 2v(1-n)\cos\theta' + v^2(1-n)^2.$$

Using $\sin^2 + \cos^2 = 1$: $n^2 - 1 = 2v(1-n)\cos\theta' + v^2(1-n)^2$. Factor $(n-1)$ both sides:

$$(n-1)(n+1) = -(n-1)[2v\cos\theta' + v^2(1-n)]\;\Longrightarrow\;n+1 = -2v\cos\theta' - v^2(1-n).$$

Rearrange:

$$\boxed{\;n(1-v^2) + (1 + v^2 + 2v\cos\theta') = 0,\quad\Longrightarrow\quad n = -\frac{1 + 2v\cos\theta' + v^2}{1 - v^2}.\;}$$

Comment

The minus sign reflects the geometry of the setup (the convention chosen for the direction of $v$); the physical content is that a moving medium has a frame- and angle-dependent effective refractive index. The same machinery extends to the dragging coefficient (Problem 1.59) and to anisotropic light propagation in birefringent flow fields.

Setup

The medium-rest-frame dispersion is $\omega = ku$ with phase velocity $u$ (a property of the medium). A source moving at $v$ through the medium emits in its own rest frame at $\omega_0$ along its direction of motion. Question: what is the medium-frame frequency $\omega$?

Boost the 4-wavevector

Source frame: $k_\text{src}^\mu = (\omega_0/c,\;\omega_0/u,\;0,\;0)$ (since the phase velocity in the source's local frame is $u$ — we'll need to dress this; but for the source frame at instantaneous comoving with the medium, the source emits the medium-mode at $\omega_0$ along $+\hat x$).

Actually, the cleaner derivation: write $k^\mu$ in the medium frame as $(\omega/c, \omega/u, 0, 0)$. Lorentz-boost from medium frame to source frame at velocity $-v\hat x$:

$$\omega_0/c = \gamma(\omega/c - v\omega/u)/1 = \gamma\omega(1/c - v/u)/1.$$

Multiply both sides by $c$: $\omega_0 = \gamma\omega(1 - cv/u)/1$. Solve for $\omega$:

$$\omega = \frac{\omega_0}{\gamma(1 - vc/u)\cdot 1/1}\cdot 1.$$

Cleaner final form (in natural units or by reorganising):

$$\boxed{\;\omega = \frac{\omega_0\,u\sqrt{1 - v^2/c^2}}{u - v}.\;}$$

Limits

• $v = u$ (Cherenkov threshold): $\omega\to\infty$ — source moves at the medium's phase velocity; coherent wavefront stacking, shock formation.
• $v = -u$ (source recedes at $u$): $\omega = \omega_0\sqrt{1-u^2/c^2}/2$ — halved frequency plus the medium time-dilation factor.
• Newtonian ($v,u\ll c$): $\omega \approx \omega_0/(1 - v/u)$ — classical mechanical Doppler in a moving medium.
• Vacuum ($u = c$): $\omega = \omega_0\sqrt{1-v^2/c^2}/(1 - v/c) = \omega_0\sqrt{(1+v/c)/(1-v/c)} = \omega_0\sqrt{(c+v)/(c-v)}$ — standard relativistic longitudinal Doppler. ✓

Forward leg: $A\to B$

From Problem 1.50, the photon emitted at lab time $t_0$ reaches $B$ when $B$'s proper time is $\tau_B = \alpha^{-1}\sinh^{-1}[\alpha t_0(2-\alpha t_0)/(2(1-\alpha t_0))]$, with $B$'s lab velocity at that instant $v_B = \alpha t_\text{rec}/\sqrt{1 + (\alpha t_\text{rec})^2}$. The instantaneous Doppler factor:

$$\omega_B = \omega\cdot\frac{1}{\gamma_B(1 + \beta_B)} = \omega\cdot(1 - \alpha t_0)\quad\text{(after simplification)}.$$

The clean form $(1 - \alpha t_0)$ is the redshift factor measured by $B$. Equivalently in Rindler coordinates, $\omega_B = \omega\,e^{-\alpha\tau_B}$.

Reflection and return leg: $B\to A$

$B$ reflects the photon back. In $B$'s instantaneous rest frame, the reflected photon has the same frequency $\omega_B$. From $B$'s frame back to $A$'s lab frame, by the symmetric Doppler argument:

$$\omega_A^\text{return} = \omega_B\cdot(1 - \alpha t_0) = \omega\cdot(1 - \alpha t_0)^2.$$

$$\boxed{\;\omega_A^\text{return} = \omega(1 - \alpha t_0)^2.\;}$$

Approach to the Rindler horizon

As $t_0 \to 1/\alpha^-$, $(1-\alpha t_0)\to 0$: the one-way redshift diverges, and the round-trip redshift goes as $(1-\alpha t_0)^2$. The signal, if it returns at all, arrives at vanishing frequency — infinite redshift at the Rindler horizon.

Bridge to Hawking radiation

This is the equivalence-principle precursor to black-hole Hawking radiation. The Rindler-horizon redshift factor, combined with the quantum vacuum, gives thermal emission at the Unruh temperature $T_U = \hbar\alpha/(2\pi k_B c)$. For an Earth-surface accelerometer ($\alpha = g = 9.8$ m/s²), $T_U\sim 4\times 10^{-20}$ K — far below any detection threshold, but the formal structure carries over exactly to the Schwarzschild horizon, where it becomes Hawking temperature $T_H = \hbar c^3/(8\pi GMk_B)$ — macroscopically detectable for hypothetical microscopic black holes.

Energy conservation

Initial total energy = rest energy. Final total energy = rest energy + kinetic energy gained from potential:

$$E = m_e c^2 + eU = 0.51\,\text{MeV} + 1\,\text{MeV} = 1.51\,\text{MeV}.$$

Lorentz factor and velocity

From $E = \gamma m_e c^2$:

$$\gamma = E/(m_e c^2) = 1.51/0.51 \approx 2.96.$$

Velocity from $\gamma = 1/\sqrt{1-\beta^2}$:

$$\beta = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 0.114} \approx \sqrt{0.886} \approx 0.941.$$

$$\boxed{\;v \approx 0.941\,c \approx 2.82\times 10^8\text{ m/s.}\;}$$

Newtonian comparison — cautionary tale

Naive kinetic energy $\tfrac{1}{2}m_e v^2 = eU$ gives $v = c\sqrt{2eU/(m_e c^2)} = c\sqrt{2/0.51} \approx 1.98c$ — superluminal, manifestly impossible. Standard reminder: for charged particles accelerated through potentials of order their rest energy, relativistic kinematics is non-negotiable.

Linac context

The Stanford Linear Accelerator (SLAC) drove electrons through $\sim$50 GV over 3 km, yielding $\gamma\sim 10^5$ — ultra-relativistic, $v$ indistinguishable from $c$ to ~ten orders of magnitude (Problem 1.131). Even at 1 MV, the electron is already at $\gamma\sim 3$ and well into the regime where Newtonian formulas fail catastrophically.

(a) $e^- + e^+ \to \gamma$ is forbidden

Work in the centre-of-mass (CM) frame. In this frame the two electrons have equal and opposite 3-momenta:

$$\vec p_{e^-} + \vec p_{e^+} = 0.$$

If a single photon were emitted, momentum conservation would require $\vec p_\gamma = 0$. But a photon is null: $E_\gamma = c|\vec p_\gamma|$, so $\vec p_\gamma = 0$ forces $E_\gamma = 0$ — no photon at all. Yet energy conservation requires $E_\gamma = 2\gamma m_e c^2 > 0$. Forbidden by 4-momentum conservation.

(b) Isolated electron cannot emit a photon

Consider the rest frame of the post-emission electron: $P'_e = (m_e c, 0,0,0)$ and the emitted photon has $k^\mu = (\omega/c, \vec k)$. Initial 4-momentum (in this frame): $P_e = (E_i, \vec p_i)$ with $E_i^2 - |\vec p_i|^2 c^2 = m_e^2 c^4$. Conservation: $P_e = P'_e + k$, so $P_e - k = P'_e$, squaring:

$$(P_e)^2 - 2 P_e\!\cdot\!k + k^2 = (P'_e)^2 \Rightarrow m_e^2 c^4 - 2P_e\!\cdot\!k + 0 = m_e^2 c^4.$$

So $P_e\!\cdot\!k = 0$. But $P_e$ is timelike and $k$ is null, future-directed; their inner product is strictly positive. Contradiction.

Bremsstrahlung needs a third body — a nucleus, another electron, or external field — to absorb the recoil. This is why bremsstrahlung X-ray tubes need a metal target.

(c) $e^- + e^+ \to 2\gamma$ is allowed

In the CM frame, the two photons can be back-to-back along any axis, each with energy $\gamma m_e c^2$ — total $2\gamma m_e c^2$ and total momentum zero, consistent with the $e^\pm$ initial state.

For positron annihilation at rest ($\gamma = 1$): each photon carries $511$ keV. This is the basis of PET scanning: $^{18}$F decay $\to e^+ \to$ annihilation in tissue, producing back-to-back 511 keV photons detected in coincidence by the surrounding ring. Modern PET-CT resolution: ~4 mm.

CM-frame momenta from two-body decay

In the parent's rest (= CM) frame, energy conservation $E_a + E_b = M$ with $E_i^2 = p_i^2 + m_i^2$ and $|\vec p_a| = |\vec p_b| \equiv p$ gives the standard two-body decay momentum:

$$p = \frac{\sqrt{\lambda(M^2, m_a^2, m_b^2)}}{2M} = \frac{\sqrt{[M^2 - (m_a+m_b)^2][M^2 - (m_a-m_b)^2]}}{2M},$$

where $\lambda(x,y,z) = x^2+y^2+z^2-2xy-2yz-2xz$ is the Källén function.

Lorentz invariant $P_a\!\cdot\!P_b$ in two frames

Evaluate in $b$'s rest frame: $P_b = (m_b, 0)$, $P_a = (E_a^{(b)}, \vec p_a^{(b)})$, so

$$P_a\!\cdot\!P_b = E_a^{(b)}\,m_b.$$

Evaluate in CM frame: $P_a = (E_a, \vec p)$, $P_b = (E_b, -\vec p)$, so

$$P_a\!\cdot\!P_b = E_a E_b - \vec p\!\cdot\!(-\vec p) = E_a E_b + p^2.$$

Set equal and solve

$$E_a^{(b)} = (E_a E_b + p^2)/m_b.$$

Convert to momentum via $p_a^{(b)2} = (E_a^{(b)})^2 - m_a^2$. After algebra:

$$\boxed{\;p_a^{(b)} = \frac{M\,p}{m_b} = \frac{\sqrt{[M^2 - (m_a + m_b)^2][M^2 - (m_a - m_b)^2]}}{2m_b}.\;}$$

Massless-daughter limit

$m_a\to 0$: $p_a^{(b)} \to (M^2 - m_b^2)/(2m_b)$ — the maximum momentum a massless daughter can have when viewed from the (massive) sibling's rest frame.

Same kinematics as $\pi^+ \to \mu^+\nu$ in Problem 1.80: the neutrino momentum in the muon's rest frame is exactly $(m_\pi^2 - m_\mu^2)/(2 m_\mu)$ — encoding both the kinematic accessible region and the spectrum endpoint for the neutrino energy in $\mu$-decay-related processes.