Special Relativity — worked problems

4-momentum boost

For a particle with $P^\mu = (E,p,0,0)$ in $S$ viewed from a frame $S'$ moving at $v\hat x$:

$$E' = \gamma(E - v p),\qquad p' = \gamma\left(p - \frac{v E}{c^2}\right).$$

Velocity in $S'$

By definition $u' = p' c^2/E'$. Using $u = p c^2/E$ in $S$:

$$u' = \frac{p' c^2}{E'} = \frac{\gamma(p - vE/c^2)c^2}{\gamma(E - vp)} = \frac{pc^2/E - v}{1 - vp/E\cdot c^2/c^2\cdot 1} = \frac{u - v}{1 - uv/c^2}.$$

$$\boxed{\;u' = \frac{u - v}{1 - uv/c^2}\;}$$

— the velocity-addition formula, derived here from 4-momentum (consistent with the 4-velocity derivation in Problem 1.2).

Energy in $S'$

Substitute $p = uE/c^2$ into the energy transformation:

$$E' = \gamma(E - v\cdot uE/c^2) = \gamma E(1 - uv/c^2).$$

$$\boxed{\;E' = \gamma E(1 - uv/c^2).\;}$$

Photon limit ($u = c$)

$E' = \gamma E(1 - v/c) = \gamma E\frac{1-\beta}{1}\cdot\frac{1}{1} = E\sqrt{(1-\beta)/(1+\beta)}$ — using $\gamma(1-\beta) = \sqrt{(1-\beta)/(1+\beta)}$. This is the relativistic longitudinal Doppler shift in 4-momentum guise. The same formula applies to the frequency (since $E = \hbar\omega$) and the wavelength of light, derived here purely from 4-vector kinematics with no wave-mechanical input.

Setup

$F^\mu = f(1,1,0,0)$ is null ($F\!\cdot\!F = 0$). Newton's second law in 4-form: $dP^\mu/d\tau = F^\mu$, with $P^\mu = (E,p,0,0)$. Initial conditions: at $\tau = 0$, $P^\mu(0) = (m_0,0,0,0)$ (rest).

Energy and momentum along $\tau$

$$\frac{dE}{d\tau} = f \Rightarrow E(\tau) = m_0 + f\tau,\qquad \frac{dp}{d\tau} = f \Rightarrow p(\tau) = f\tau.$$

Mass evolution

The instantaneous mass is from the mass-shell condition $m^2 = E^2 - p^2$:

$$m^2(\tau) = (m_0 + f\tau)^2 - (f\tau)^2 = m_0^2 + 2 m_0 f\tau.$$

$$\boxed{\;m(\tau) = \sqrt{m_0^2 + 2 m_0 f\tau}.\;}$$

Unlike the constant-rest-mass hyperbolic motion (Problem 1.131), here mass grows — the null 4-force does work that goes partly into kinetic energy and partly into internal energy.

Lab time

The relation $dt/d\tau = E/m = \gamma$ gives

$$t(\tau) = \int_0^\tau\frac{m_0 + f\tau'}{\sqrt{m_0^2 + 2m_0 f\tau'}}\,d\tau'.$$

Substitute $u = m_0^2 + 2m_0 f\tau'$, $du = 2m_0 f\,d\tau'$, $m_0 + f\tau' = (m_0 + u/(2m_0)\cdot 1/1)$... cleaner: differentiate the proposed antiderivative $F(\tau) = (2m_0 + f\tau)\sqrt{m_0^2 + 2m_0 f\tau}/(3m_0 f)$ and verify $F'(\tau) = (m_0 + f\tau)/\sqrt{m_0^2 + 2m_0 f\tau}$. Adjusting for $F(0)$:

$$\boxed{\;t(\tau) = \frac{(2m_0 + f\tau)\sqrt{m_0^2 + 2m_0 f\tau} - 2m_0^2}{3m_0 f}.\;}$$

Limits

• Non-relativistic ($f\tau\ll m_0$): expand $\sqrt{1 + 2f\tau/m_0}\approx 1 + f\tau/m_0$. Then $m\approx m_0$, $t\approx\tau$, $p\approx f\tau$ — Newtonian constant-force motion.
• Ultra-relativistic ($f\tau\gg m_0$): $m\sim\sqrt{2m_0 f\tau}\to\infty$ — mass-gaining trajectory. The 4-force keeps depositing internal energy faster than it kicks momentum.

Dual to constant-rest-mass hyperbolic motion (Problem 1.131): a timelike 4-force conserves rest mass; a null 4-force grows it.

Setup

The particle moves with velocity $\vec u$ at angle $\theta$ measured from the observer's $-\hat x$ direction in the observer's frame. The observer moves at $v\hat x$ in the lab frame, so the lab measures both the particle (3-velocity components $u\cos\theta - v$ along $\hat x$, $u\sin\theta$ along $\hat y$ — subject to relativistic transformation) and the observer.

Velocity-addition for general directions

The vector velocity-addition formula from boost along $\hat x$:

$$u'_\parallel = \frac{u_\parallel + v}{1 + u_\parallel v/c^2},\qquad u'_\perp = \frac{u_\perp}{\gamma(1 + u_\parallel v/c^2)}.$$

With $u_\parallel = u\cos\theta$, $u_\perp = u\sin\theta$ in the observer's frame:

$$u'_x = \frac{u\cos\theta + v}{1 + uv\cos\theta/c^2},\qquad u'_y = \frac{u\sin\theta}{\gamma(1 + uv\cos\theta/c^2)}.$$

Magnitude and angle in lab

$$|u'|^2 = \frac{(u\cos\theta + v)^2 + u^2\sin^2\theta/\gamma^2}{(1 + uv\cos\theta/c^2)^2}.$$

Use $1/\gamma^2 = 1 - v^2/c^2$. Expand the numerator: $u^2\cos^2\theta + 2uv\cos\theta + v^2 + u^2\sin^2\theta(1 - v^2/c^2) = u^2 + 2uv\cos\theta + v^2 - u^2v^2\sin^2\theta/c^2$. Therefore

$$\boxed{\;u'(\theta) = \frac{\sqrt{u^2 + 2uv\cos\theta + v^2 - u^2v^2\sin^2\theta/c^2}}{1 + uv\cos\theta/c^2},\quad \tan\theta' = \frac{u\sin\theta/\gamma}{u\cos\theta + v}.\;}$$

Limits

• Ultra-relativistic ($u\to c$): numerator $\to\sqrt{c^2 + 2cv\cos\theta + v^2 - v^2\sin^2\theta} = \sqrt{c^2 + 2cv\cos\theta + v^2\cos^2\theta} = c + v\cos\theta$. Denominator: $1 + v\cos\theta/c$. Ratio: $u' = c$. ✓ Light-speed preserved in every frame.
• Newtonian ($u,v\ll c$): drop the $v^2/c^2$ and $uv/c^2$ corrections: $u' \to\sqrt{u^2 + 2uv\cos\theta + v^2} = |\vec u + v\hat x|$ — Galilean vector sum.
• Aberration of light ($u = c$): $\tan\theta' = c\sin\theta/[\gamma(c\cos\theta + v)]$, the formula behind stellar aberration (Bradley 1727).

Lab kinematics

$x = at^2/2 \Rightarrow v(t) = at,\; \ddot x = a$. The lab observer sees ordinary uniform acceleration. Reach $v_0$ at $t_0 = v_0/a$.

Proper-time integral

$$d\tau = \sqrt{1 - v^2/c^2}\,dt = \sqrt{1 - (at/c)^2}\,dt.$$

Substitute $\sin\phi = at/c$, $d\phi\cdot c/a = \cos\phi\,d\phi\cdot c/a = dt$ scaled. After integration:

$$\tau(t_0) = \int_0^{t_0}\sqrt{1 - (at/c)^2}\,dt = \frac{c}{2a}\bigl[(at_0/c)\sqrt{1 - (at_0/c)^2} + \arcsin(at_0/c)\bigr].$$

With $at_0 = v_0$:

$$\boxed{\;\tau(t_0) = \frac{1}{2a}\bigl[v_0\sqrt{1 - v_0^2/c^2} + c\arcsin(v_0/c)\bigr].\;}$$

Proper acceleration

The transformation of coordinate to proper acceleration (Problem 1.54 with $\theta = 0$): $\alpha = \gamma^3 a_\parallel$. Here $a_\parallel = a$ and $\gamma(t) = 1/\sqrt{1 - (at/c)^2}$, so

$$\boxed{\;\alpha(t) = \frac{a}{(1 - (at/c)^2)^{3/2}}.\;}$$

Divergence

As $at\to c$, $\alpha\to\infty$: an external agent would need to deliver infinite proper acceleration to maintain constant lab acceleration near light speed. Physically impossible for a real engine.

Duality with hyperbolic motion

Hyperbolic motion (Problem 1.131, 1.49) has constant proper acceleration $\alpha$ and $dv/dt\to 0$ as $v\to c$ — the natural relativistic analogue of "constant force". Constant coordinate acceleration (this problem) requires unboundedly growing proper acceleration. Newtonian intuition fails near light speed: one of these two must give, and the physical realization is hyperbolic motion.

Velocity addition

In the water's rest frame, light moves at $u_0 = c/n$ along $+\hat x$. The water moves at $v$ along $+\hat x$ in the lab frame. By relativistic velocity addition:

$$u = \frac{u_0 + v}{1 + u_0 v/c^2} = \frac{c/n + v}{1 + v/(nc)}.$$

First-order expansion in $v/c$

Use $(1 + x)^{-1}\approx 1 - x$ for $|x| = v/(nc) \ll 1$:

$$u \approx (c/n + v)\left(1 - \frac{v}{nc}\right) = \frac{c}{n} + v - \frac{c}{n}\cdot\frac{v}{nc} - \frac{v^2}{nc}.$$

Drop $O(v^2/c)$:

$$u\approx \frac{c}{n} + v\left(1 - \frac{1}{n^2}\right).$$

$$\boxed{\;u \approx u_0 + k\,v,\quad k = 1 - 1/n^2.\;}$$

Historical context

Fresnel (1818) postulated this "ether drag" coefficient $k$ ad hoc to fit Arago's null result for stellar aberration through moving water. Fizeau (1851) measured it interferometrically by sending light counter-propagating through a moving water column: for $n = 1.33$, $k = 0.437$, exactly as observed.

SR derives what Fresnel had to assume — Einstein cited Fizeau (along with stellar aberration and Michelson–Morley) as one of the three pre-1905 classical results that already pointed toward relativity. The dragging coefficient is purely kinematic; no ether is needed.

Setup

Light travels in $+\hat x$ with speed $u_0 = c/n$ in the water's rest frame. The water moves at $v\hat y$ in the lab. Boost the photon 3-velocity from the water frame to lab using velocity-addition along $\hat y$:

$$u_{x,\text{lab}} = \frac{u_0}{\gamma_v(1 + 0)} = u_0/\gamma_v,\qquad u_{y,\text{lab}} = \frac{0 + v}{1 + 0} = v.$$

Lab-frame speed magnitude

$$|u|^2 = u_{x,\text{lab}}^2 + u_{y,\text{lab}}^2 = u_0^2/\gamma_v^2 + v^2 = u_0^2(1 - v^2/c^2) + v^2 = u_0^2 + v^2(1 - u_0^2/c^2).$$

Take square root:

$$|u| = u_0\sqrt{1 + (v^2/u_0^2)(1 - u_0^2/c^2)}\approx u_0 + \frac{v^2(1 - u_0^2/c^2)}{2u_0} = u_0 + O(v^2).$$

$$\boxed{\;|u| \approx u_0 + \mathcal O(v^2/c^2)\quad\text{—no linear drag.}\;}$$

Why no linear term?

The Fresnel formula gives $u = u_0 + (1-1/n^2)v$ along the propagation direction. When the medium flows perpendicular to the light, the longitudinal contribution to drag vanishes; only the transverse aberration remains, which is second order.

Connection to Michelson–Morley

Same structure as the transverse Doppler effect (Problem 1.65): null at first order in $v/c$. This is the kinematic reason Michelson–Morley failed to detect a first-order cross-axis ether wind — they were measuring a quantity already known to be second-order even classically (Fresnel-Fizeau picture). The full SR account is consistent: transverse perturbations of light contribute at $O(v^2/c^2)$ only.

Compute the redshift

$$z = \frac{\lambda_\text{obs}}{\lambda_\text{rest}} - 1 = \frac{3600}{1215} - 1 \approx 1.963.$$

Relativistic longitudinal Doppler

For purely radial recession, $\lambda_\text{obs}/\lambda_\text{rest} = \sqrt{(1+\beta)/(1-\beta)}$, so

$$(1+z)^2 = \frac{1+\beta}{1-\beta}.$$

With $(1+z)^2 = 2.963^2 \approx 8.78$:

$$\frac{1+\beta}{1-\beta} = 8.78 \;\Longrightarrow\; 1 + \beta = 8.78(1-\beta) \;\Longrightarrow\; 9.78\beta = 7.78 \;\Longrightarrow\; \beta \approx 0.795.$$

$$\boxed{\;v \ge 0.795\,c \approx 2.39\times 10^8\text{ m/s.}\;}$$

Why a lower bound?

Any transverse component of velocity contributes a time-dilation redshift $\gamma$-factor in addition to the longitudinal $\sqrt{(1+\beta)/(1-\beta)}$. If the recession is partly transverse, the same observed $z$ requires a larger total $v$. Pure radial motion is the minimum-$v$ solution.

Modern context

Schmidt's 1965 discovery: 3C 9 is a quasar, an active galactic nucleus at $z\sim 2$ — comoving distance ~10 Gly, look-back time ~10 Gyr. The deeper modern interpretation is that quasar redshifts are cosmological, not kinematic Doppler: the universe's expansion stretches wavelengths $\lambda\to a(t_\text{now})\lambda_\text{em}/a(t_\text{em})$. The "velocity" $v = 0.795c$ is the formal Doppler equivalent; the actual recession (Hubble flow) at $z=2$ corresponds to $v > c$ in proper distance terms — not a violation, since cosmological expansion is not a local boost.

Identify the 4-wavevector

The phase of a plane wave is a Lorentz scalar: $\Phi = -k_\mu x^\mu$. Read from $E(x) = E_0\sin[2\pi(x^1/\lambda - vt)] = E_0\sin(2\pi x^1/\lambda - 2\pi v t)$:

$$k_\mu x^\mu = \omega t - k_x x^1\;\Longrightarrow\; \omega = 2\pi v,\;k_x = 2\pi/\lambda.$$

With phase velocity $c$ (vacuum EM wave), $v\lambda = c$, so $\omega = c k_x$. Therefore

$$k^\mu = (\omega/c,\;\omega/c,\;0,\;0).$$

Lightlike condition

$$k^2 = k^\mu k_\mu = (\omega/c)^2 - (\omega/c)^2 - 0 - 0 = 0.\;\checkmark$$

Doppler shift in rapidity form

Boost along $+\hat x$ by rapidity $\theta$ (so $\cosh\theta = \gamma$, $\sinh\theta = \gamma\beta$, $\tanh\theta = \beta$):

$$k'^0 = \cosh\theta\,k^0 - \sinh\theta\,k^1 = (\omega/c)(\cosh\theta - \sinh\theta) = (\omega/c)e^{-\theta}.$$

Therefore

$$\boxed{\;\omega' = \omega\,e^{-\theta} = \omega\sqrt{\frac{1-\beta}{1+\beta}}.\;}$$

Why rapidity is "natural"

Rapidities are additive under successive boosts along the same axis: $\theta_\text{tot} = \theta_1 + \theta_2$. So under a chain of Lorentz boosts, frequencies multiply: $\omega' = \omega e^{-\sum\theta_i}$. The Doppler factor is the exponential of the rapidity, just as the time-dilation factor $\gamma = \cosh\theta$ exhibits the additive-rapidity structure trigonometrically.

Newtonian limit $\beta\ll 1$: $\theta\approx\beta$, $\omega'\approx\omega(1-\beta)$ — Doppler reduced to first order.

Burst duration and redshift

A pulse emitted with rest-frame duration $\tau_0$ has its arrival-time spread by the same factor that stretches wavelengths:

$$\frac{\tau_\text{obs}}{\tau_0} = \frac{\lambda_\text{obs}}{\lambda_0} = 1 + z.$$

This follows because both observed events (start, end) are equidistant in rest-frame proper time, and the photon-travel time scales identically with the metric stretching factor.

Apply

Given $\tau_\text{obs}/\tau_0 = 2$:

$$\boxed{\;z = 1,\quad \lambda_\text{obs} = 2\lambda_0.\;}$$

Equivalent Doppler velocity

From $(1+z)^2 = (1+\beta)/(1-\beta)$: $4 = (1+\beta)/(1-\beta) \Rightarrow \beta = 3/5 = 0.6$.

Cosmological interpretation

At $z = 1$, the universe was approximately half its present age — the source is at cosmic age $\sim 6$ Gyr, and the light has been traveling about 8 Gyr (look-back time depends on cosmological parameters; for $\Lambda$CDM with $H_0 = 67$ km/s/Mpc).

The fact that both arrival-time spreading and spectral redshift carry the same $(1+z)$ factor is a canonical signature of cosmological distance — verified in GRBs, type Ia supernova light curves, quasar variability. If the redshift were purely Doppler (special-relativistic), both effects would still scale identically; what distinguishes the cosmological interpretation is the Hubble flow's correlation with distance, not the individual time-dilation factor.

(a) Relative velocity by velocity addition

In the person's frame, object 1 moves at $+c/2$ and object 2 at $-c/2$. The velocity of object 1 in object 2's rest frame is

$$u'_1 = \frac{c/2 - (-c/2)}{1 - (c/2)(-c/2)/c^2} = \frac{c}{1 + 1/4} = \frac{4c}{5}.$$

Equivalently from object 1's perspective: $u'_2 = -4c/5$. Magnitude:

$$\boxed{\;|u'| = 4c/5 = 0.8c.\;}$$

Newtonian addition would give $|u'| = c$, but that would violate the universal speed limit. The relativistic correction reduces it to $4c/5$.

(b) Doppler-shifted ruby laser

Object 1 emits at rest-frame wavelength $\lambda_0 = 694.3$ nm. Object 2 sees it approaching at $\beta = 4/5$, so by the relativistic blueshift formula:

$$\lambda' = \lambda_0\sqrt{\frac{1-\beta}{1+\beta}} = 694.3\,\text{nm}\sqrt{\frac{1/5}{9/5}} = 694.3\,\text{nm}\sqrt{1/9} = \frac{694.3}{3}\,\text{nm}.$$

$$\boxed{\;\lambda' \approx 231\,\text{nm}\quad(\text{deep UV}).\;}$$

Frequency factor 3 — ruby red shifts past visible into the UV. The ruby laser's signature 694.3 nm red light, blueshifted by $\beta = 0.8$, would be invisible to the naked eye on object 2 (UV) and could cause ionising biological damage at the same incident intensity. The same principle (and reversed) explains the extreme reddening of light from receding cosmological sources.