Special Relativity — worked problems

(a) The two 4-velocities in $S$

The 4-velocity of a particle is $U^\mu = dx^\mu/d\tau = \gamma(c, \vec u)$ where $\gamma = (1-u^2/c^2)^{-1/2}$. Applying this to the object,

$$U^\mu = \gamma_u(c,\,\vec u),\qquad \gamma_u = (1 - u^2/c^2)^{-1/2}.$$

The frame $S'$ moves at $-v'\hat x$ relative to $S$. An observer at rest in $S'$ has 3-velocity $-v'\hat x$ in $S$, so

$$V^\mu = \gamma_{v'}(c,\,-v',0,0),\qquad \gamma_{v'} = (1 - v'^2/c^2)^{-1/2}.$$

(b) $\gamma$ from the inner product

Evaluate $U\!\cdot\!V = U^\mu V_\mu = U^0 V^0 - \vec U\!\cdot\!\vec V$ in the rest frame $\tilde S$ of the observer, where $\tilde V^\mu = (c,\vec 0)$ and the object has 4-velocity $\tilde U^\mu = \gamma_{\tilde u}(c,\vec{\tilde u})$:

$$U\!\cdot\!V = \gamma_{\tilde u}\,c^2 - 0 = \gamma_{\tilde u}\,c^2.$$

$U\!\cdot\!V$ is a Lorentz scalar — the same number in every frame. So in any frame,

$$\boxed{\;\gamma_{\tilde u} = \frac{U\!\cdot\!V}{c^2}.\;}$$

(c) Apply to frame $S'$

Compute $U\!\cdot\!V$ in $S$ using the components above:

$$U\!\cdot\!V = \gamma_u\gamma_{v'}\bigl[c\cdot c - \vec u\!\cdot\!(-v'\hat x)\bigr] = \gamma_u\gamma_{v'}(c^2 + u_x v').$$

Dividing by $c^2$,

$$\boxed{\;\gamma_{u'} = \gamma_u\gamma_{v'}\!\left(1+\frac{u_x v'}{c^2}\right).\;}$$

Sanity checks

• $\vec u = 0$: object at rest in $S$ ⇒ in $S'$ it moves at $-v'\hat x$, giving $\gamma_{u'} = \gamma_{v'}$. ✓
• $\vec u = -v'\hat x$: object is at rest in $S'$, so $\gamma_{u'} = 1$; formula gives $\gamma_u\gamma_{v'}(1 - v'^2/c^2) = \gamma_{v'}/\gamma_{v'} \cdot 1 = 1$. ✓

A single Lorentz-invariant contraction $U\!\cdot\!V$ encodes the full relativistic velocity-addition algebra, cross-referenced with Problem 1.3.

Setup

Work entirely in your rest frame to avoid juggling proper-time corrections (you measure both intervals on your single clock). Place yourself at $x = 0$. The station approaches you at $-c/4$; in your frame its current distance is $L_0 = 1$ light-day.

Phase 1: light signal reaches the station

Light leaves at $t = 0$ from $x = 0$ moving at $+c$. The station's position is $x_\text{stn}(t) = L_0 - (c/4)t$. Intersection:

$$ct_1 = L_0 - \tfrac{c}{4}t_1 \;\Longrightarrow\; t_1 = \frac{4 L_0}{5c} = 0.8\text{ day.}$$

Station location at reception: $x_1 = c t_1 = 4L_0/5 = 0.8$ ly.

Phase 2: rescue-ship velocity in your frame

The ship leaves the station at speed $3c/4$ "relative to itself" — i.e. relative to the station, in the station's frame. The station has lab velocity $-c/4$ and the ship moves at $-3c/4$ in the station frame (toward you, i.e. in the $-\hat x$ direction). Relativistic velocity addition:

$$u_\text{res} = \frac{-3c/4 - c/4}{1 + (3c/4)(c/4)/c^2} = \frac{-c}{1 + 3/16} = -\frac{16}{19}c.$$

Phase 3: rescue-ship travel time

Ship starts from $x_1 = 4L_0/5$, moves at $|u_\text{res}| = 16c/19$ toward $x = 0$:

$$\Delta t_2 = \frac{x_1}{|u_\text{res}|} = \frac{4L_0/5}{16c/19} = \frac{19 L_0}{20 c} = 0.95\text{ day.}$$

Total time on your clock

$$\boxed{\;t_\text{total} = t_1 + \Delta t_2 = \frac{4L_0}{5c} + \frac{19L_0}{20c} = \frac{7 L_0}{4c} = 1.75\text{ day} = 42\text{ h.}\;}$$

Newtonian shortcut would have given $u_\text{res}^\text{Newt} = -c$, conflicting with light-speed limit; the relativistic answer differs by ~16% — significant for any space-rescue scheduler.

Setup

The officer (subscript $p$) starts at rest in $S$ at $t=0$ and undergoes hyperbolic motion with constant proper acceleration $a$. The criminal (subscript $c$) cruises at constant velocity $v > 0$ along $+\hat x$. We parametrise by the officer's rapidity $\eta_p = a\tau_p/c$ (where $\tau_p$ is the officer's proper time).

Officer's worldline

Hyperbolic motion is the integral of $\dot x = c\tanh\eta_p,\;\gamma_p = \cosh\eta_p$, giving

$$t_S = \frac{c}{a}\sinh\eta_p,\qquad x_S = \frac{c^2}{a}(\cosh\eta_p - 1).$$

Catch-up condition

Criminal at $x = v t$. Set $x_S = v\,t_S$:

$$\frac{c^2}{a}(\cosh\eta_p - 1) = v\cdot\frac{c}{a}\sinh\eta_p \;\Longrightarrow\; \frac{\cosh\eta_p - 1}{\sinh\eta_p} = \frac{v}{c}.$$

Half-angle identities: $\cosh\eta_p - 1 = 2\sinh^2(\eta_p/2)$ and $\sinh\eta_p = 2\sinh(\eta_p/2)\cosh(\eta_p/2)$, so

$$\tanh(\eta_p/2) = v/c = \tanh\eta,\quad\text{where}\quad \eta \equiv\operatorname{arctanh}(v/c).$$

$$\boxed{\;\eta_p = 2\eta.\;}$$

(a) Criminal's proper time

$\tau_c = t_S/\gamma_v$ where $\gamma_v = \cosh\eta$ and $t_S = (c/a)\sinh(2\eta) = (2c/a)\sinh\eta\cosh\eta$:

$$\boxed{\;\tau_c = \frac{2c\sinh\eta}{a} = \frac{2v\gamma_v}{a}.\;}$$

(b) Officer's proper time

$$\tau_p = c\eta_p/a = 2c\eta/a = \boxed{\;\frac{c}{a}\ln\!\frac{c+v}{c-v}.\;}$$

(c) Relative velocity at catch-up

Rapidities subtract along boost-aligned worldlines: $\eta_\text{rel} = \eta_p - \eta = 2\eta - \eta = \eta$, so

$$\boxed{\;v_\text{rel} = c\tanh\eta = v.\;}$$

The officer catches the criminal moving relative to him at exactly the criminal's $S$-speed — a clean consequence of additive rapidities. In Newtonian mechanics with constant acceleration, $v_\text{rel}$ at catch-up would equal $v$ as well (officer reaches twice the criminal's speed), but only the rapidity formulation makes this transparent relativistically.

(a) The Rindler metric

Differentiate the coordinate transformation:

$$dt = \sinh(aT)\,dX + aX\cosh(aT)\,dT,\quad dx = \cosh(aT)\,dX + aX\sinh(aT)\,dT.$$

Substitute into $ds^2 = dt^2 - dx^2 - dy^2 - dz^2$:

$$dt^2 - dx^2 = a^2 X^2(\cosh^2 - \sinh^2)\,dT^2 - (\cosh^2 - \sinh^2)\,dX^2 + (\text{cross terms})\cdot 2\sinh\cosh\,(dT\,dX).$$

The cross terms have coefficient $aX\sinh\cosh\cdot\cosh - aX\cosh\cdot\sinh = 0$. Using $\cosh^2 - \sinh^2 = 1$:

$$\boxed{\;ds^2 = a^2 X^2\,dT^2 - dX^2 - dY^2 - dZ^2.\;}$$

The time-time component $g_{TT} = a^2 X^2$ depends on $X$ — gravitational redshift between Rindler observers, the equivalence-principle prototype for Unruh and Hawking radiation.

(b) Photon 4-wavevector in Rindler coordinates

Apply the Jacobian $\partial(T,X)/\partial(t,x)$ at the event $(t_0,x_0)$, equivalent to $(T_0,X_0)$ with $t_0 = X_0\sinh(aT_0)$, $x_0 = X_0\cosh(aT_0)$:

$$\frac{\partial T}{\partial t} = \frac{\cosh(aT_0)}{aX_0},\quad \frac{\partial T}{\partial x} = -\frac{\sinh(aT_0)}{aX_0},\quad \frac{\partial X}{\partial t} = -\sinh(aT_0),\quad \frac{\partial X}{\partial x} = \cosh(aT_0).$$

Acting on $k^\mu = (\omega,\omega\cos\theta,0,\omega\sin\theta)$ in Minkowski coordinates:

$$\boxed{\;k'^T = \frac{\omega[\cosh(aT_0) - \cos\theta\sinh(aT_0)]}{aX_0},\quad k'^X = \omega[\cos\theta\cosh(aT_0) - \sinh(aT_0)],\quad k'^Y = 0,\quad k'^Z = \omega\sin\theta.\;}$$

The $1/(aX_0)$ in $k'^T$ is the Rindler redshift in disguise: observers at large $X$ see lower frequency.

(c) Proper time on the worldline

With $X = X_0$, $Y = vT$, $Z = 0$: $dX = 0$, $dY = v\,dT$, so

$$d\tau^2 = (a^2 X_0^2 - v^2)\,dT^2.$$

Integrating from $T=0$ to $T_0$:

$$\boxed{\;\tau = T_0\sqrt{a^2 X_0^2 - v^2}.\;}$$

Real proper time requires $v < aX_0$ — the local Rindler speed limit. The Rindler horizon at $X = 0$ would force $\tau \to 0$, signaling the boundary of the accelerated wedge.

Locate the event $t'=0$ on the worldline

The boost $t' = \gamma(t - vx)$ gives $t' = 0 \iff t = vx$. Substituting into $t^2 - x^2 = -1/\alpha^2$:

$$v^2 x^2 - x^2 = -1/\alpha^2 \;\Longrightarrow\; x^2(1-v^2) = 1/\alpha^2 \;\Longrightarrow\; x = \gamma/\alpha,\quad t = v\gamma/\alpha.$$

Lab velocity and acceleration at that event

From $t^2 - x^2 = -1/\alpha^2$, differentiate: $t = x\dot x$ so $\dot x = t/x = v$ — the particle is momentarily at rest in $S'$. Differentiate again: $1 = \dot x^2 + x\ddot x = v^2 + (\gamma/\alpha)\ddot x$, giving

$$\ddot x = \frac{1-v^2}{\gamma/\alpha} = \frac{\alpha}{\gamma^3}.$$

Transform to $S'$

Coordinate acceleration transforms as $a' = \ddot x/[\gamma^3(1 - v\dot x)^3]$ (derived from boosting $dv/dt$). With $\dot x = v$: $(1 - v\dot x)^3 = (1-v^2)^3 = 1/\gamma^6$, so

$$a' = \gamma^3\cdot\frac{\alpha}{\gamma^3}\cdot\gamma^6\cdot\frac{1}{\gamma^6} = \alpha.$$

Wait — cleaner: $a' = \ddot x/[\gamma^3(1-v\dot x)^3] = (\alpha/\gamma^3)/[\gamma^3 \cdot 1/\gamma^6] = (\alpha/\gamma^3)\cdot(\gamma^3) = \alpha$.

$$\boxed{\;a' = \alpha.\;}$$

This is the definition of proper acceleration $\alpha$: the acceleration measured in the instantaneous rest frame. Equivalence-principle backbone of GR — Rindler observers feel constant $\alpha$, indistinguishable locally from a uniform gravitational field.

Setup

Hyperbolic motion with proper acceleration $\alpha$ starting from rest at the origin has worldline

$$x_B(t) = \sqrt{1/\alpha^2 + t^2} - 1/\alpha\,,\quad t\ge 0$$

(so that $x_B(0) = 0$, $\dot x_B(0) = 0$, $\alpha_\text{proper} = \alpha$). Light signal sent from $A$ at lab time $t_0$ travels along $x = t - t_0$.

Intersection

Set $x_B(t) = t - t_0$:

$$\sqrt{1/\alpha^2 + t^2} = t - t_0 + 1/\alpha.$$

Square both sides ($t > t_0$ for the signal to reach $B$):

$$1/\alpha^2 + t^2 = (t - t_0 + 1/\alpha)^2.$$

Expanding: $t^2 + 1/\alpha^2 = t^2 - 2t(t_0 - 1/\alpha) + (t_0 - 1/\alpha)^2$, which gives

$$t_\text{rec} = \frac{(t_0 - 1/\alpha)^2 - 1/\alpha^2}{2(t_0 - 1/\alpha)}\cdot(-1) \cdot\frac{1}{1} = \frac{t_0(2/\alpha - t_0)}{2(1/\alpha - t_0)} = \frac{t_0(2 - \alpha t_0)}{2\alpha(1-\alpha t_0)/\alpha^2}\cdot\alpha = \frac{t_0(2-\alpha t_0)}{2(1-\alpha t_0)}\cdot\frac{1}{1}.$$

Cleanly: $t_\text{rec} = t_0(2-\alpha t_0)/[2(1-\alpha t_0)]$, real and positive only when $\alpha t_0 < 1$.

Convert to $B$'s proper time

For hyperbolic motion, $\tau_B = \alpha^{-1}\sinh^{-1}(\alpha t)$. Therefore

$$\boxed{\;\tau_B = \frac{1}{\alpha}\sinh^{-1}\!\left[\frac{\alpha t_0(2 - \alpha t_0)}{2(1 - \alpha t_0)}\right],\quad t_0 < 1/\alpha.\;}$$

The Rindler horizon

For $t_0 \ge 1/\alpha$: $t_0$ lies on or beyond the asymptotic null ray $x = t - 1/\alpha$ that the worldline approaches but never crosses. The light signal lies in the part of Minkowski space outside the Rindler wedge — it never reaches $B$.

This is the equivalence-principle analogue of a black-hole event horizon. With quantum vacuum fluctuations, the same redshift structure generates thermal Unruh radiation at $T = \alpha\hbar/(2\pi k_B c)$, exactly mirroring Hawking radiation at the Schwarzschild horizon.

Worldline and 4-velocity

Choose lab coordinates so the orbit lies in the $xy$-plane: $\vec x(t) = R(\cos\omega t, \sin\omega t, 0)$ with angular frequency $\omega = v/R$. Lab velocity $\vec u = R\omega(-\sin\omega t,\cos\omega t,0)$ has constant magnitude $|\vec u| = v$, so $\gamma$ is constant. The 4-velocity:

$$U^\mu = \gamma\bigl(c,\; v(-\sin\omega t, \cos\omega t, 0)\bigr).$$

4-acceleration

$A^\mu = dU^\mu/d\tau = \gamma\,dU^\mu/dt$. Time-derivative of the spatial part: $-\gamma v\omega(\cos\omega t,\sin\omega t,0) = -\gamma v\omega\,\hat r$. So

$$A^\mu = \gamma\bigl(0,\;-\gamma v\omega\hat r\bigr) = (0,\;-\gamma^2 v\omega\,\hat r).$$

Magnitude (spacelike, since $A\!\cdot\!U = 0$):

$$|\vec A| = \gamma^2 v\omega = \gamma^2 v^2/R.$$

Proper acceleration and orbital proper period

The magnitude of the 4-acceleration is the proper acceleration:

$$\boxed{\;\alpha = \gamma^2 v^2/R.\;}$$

Lab period $T_\text{lab} = 2\pi R/v$. Proper period contracts:

$$\boxed{\;T_\text{proper} = T_\text{lab}/\gamma = \frac{2\pi R}{\gamma v}.\;}$$

Physical interpretation

The Newtonian centripetal acceleration is $v^2/R$; the proper value is enhanced by $\gamma^2$ — the relativistic-mass enhancement of effective inertia. At LEP ($e^-$ at 100 GeV, $\gamma\sim 2\times 10^5$, $R\approx 4.3$ km), $\alpha \sim 4\times 10^{14}$ m/s² in the electron's own frame — producing copious synchrotron radiation. Same physics as Problem 1.15 (muon storage rings): $1/\gamma$ proper-period gives many lab orbits per proper lifetime.

The 4-force decomposition

For a particle whose internal energy $M$ can change, the relativistic Newton equation reads

$$F^\mu = \frac{d(M U^\mu)}{d\tau} = \frac{dM}{d\tau}U^\mu + M\,\frac{dU^\mu}{d\tau} = \frac{dM}{d\tau}U^\mu + MA^\mu,$$

where $A^\mu = dU^\mu/d\tau$ is the 4-acceleration. Since $U\!\cdot\!U = c^2$ is constant, $A\!\cdot\!U = 0$ — the 4-acceleration is always orthogonal to the 4-velocity.

Projection onto $U$

Take the inner product with $U_\mu$:

$$F\!\cdot\!U = \frac{dM}{d\tau}\,U\!\cdot\!U + M\,A\!\cdot\!U = \frac{dM}{d\tau}\cdot c^2 + 0.$$

(Here I set $c=1$ to match the problem statement, so $U\!\cdot\!U = 1$.) Given $F = fU$: $F\!\cdot\!U = f\,U\!\cdot\!U = f$. Therefore

$$\boxed{\;\frac{dM}{d\tau} = f.\;}$$

Projection orthogonal to $U$

From the decomposition: $MA^\mu = F^\mu - (dM/d\tau)U^\mu = fU^\mu - fU^\mu = 0$, so

$$\boxed{\;A^\mu = 0,\quad\alpha = 0.\;}$$

Physical interpretation

A 4-force parallel to $U$ deposits all its energy as rest mass (internal heating, excitation) — no trajectory bending. The particle continues on a geodesic of Minkowski space while its rest energy grows linearly with proper time. Examples: thermal absorption in a thin opaque target moving with the particle; gain of binding energy without ejecta.

Duality. Compare with Problem 1.53 (4-force purely spatial, $F = (0,\vec b)$): all energy goes to acceleration, none to mass change — the orthogonal projection.

Compute $F\!\cdot\!U$

With $U^\mu = \gamma(c, \vec u)$ and $F^\mu = (0,\vec b)$:

$$F\!\cdot\!U = F^\mu U_\mu = 0\cdot\gamma c - \vec b\!\cdot\!\gamma\vec u = -\gamma\vec b\!\cdot\!\vec u.$$

(With signature $(+,-,-,-)$; for $(-,+,+,+)$ the sign flips, but the physics is the same.) Using the Problem 1.52 derivation $F\!\cdot\!U = c^2\,dm/d\tau$ (with $M\to m$):

$$\boxed{\;\frac{dm}{d\tau} = \frac{\vec b\!\cdot\!\vec u\,\gamma}{c^2}.\;}$$

Proper acceleration

Decompose $F^\mu = (dm/d\tau)U^\mu + mA^\mu$. In the instantaneous rest frame, $\vec u = 0$, so $dm/d\tau = 0$ and $F^\mu_\text{rest} = mA^\mu_\text{rest}$. The spatial part of $F^\mu$ in the rest frame is $\vec b_\text{rest}$, with magnitude $|\vec b_\text{rest}| = |\vec b|$ (since the spatial part of a 4-vector with $F^0 = 0$ in some frame is not Lorentz-invariant in magnitude in general — but for this specific configuration $F\!\cdot\!F = -|\vec b|^2$ is invariant). So

$$\boxed{\;\alpha = |\vec b|/m.\;}$$

Pure-force condition

"Pure" means $dm/d\tau = 0$ — all the 4-force goes into acceleration, none into mass change. This requires

$$\vec b\!\cdot\!\vec u = 0 \;\Longleftrightarrow\; \vec b\perp\vec u.$$

Magnetic force example

The magnetic Lorentz 3-force is $\vec b = q\vec u\times\vec B$, which automatically satisfies $\vec b\!\cdot\!\vec u = q(\vec u\times\vec B)\!\cdot\!\vec u = 0$. So magnetic forces are exactly pure forces: $dm/d\tau = 0$ — magnetic forces never do work. The covariant version: the Lorentz 4-force $F^\mu = qF^{\mu\nu}U_\nu$ satisfies $F\!\cdot\!U = qF^{\mu\nu}U_\mu U_\nu = 0$ by antisymmetry of $F^{\mu\nu}$. Dual to Problem 1.52 (parallel force = pure mass change).

Acceleration transformation formulae

The general transformation of 3-acceleration from rest frame to lab is

$$a'_\parallel = \frac{a_\parallel}{\gamma^3(1 - vu_\parallel/c^2)^3},\qquad a'_\perp = \frac{a_\perp + (v/c^2)(\vec u\times\vec a)_\perp\,(\dots)}{\gamma^2(1 - vu_\parallel/c^2)^3}.$$

In the instantaneous rest frame $\vec u = 0$, so the $(1 - vu_\parallel/c^2)^3 = 1$ factor disappears. The rest-frame components of proper acceleration $a_0$ at angle $\theta$ to the boost direction are $a_\parallel = a_0\cos\theta$, $a_\perp = a_0\sin\theta$, so

$$a'_\parallel = \frac{a_0\cos\theta}{\gamma^3},\quad a'_\perp = \frac{a_0\sin\theta}{\gamma^2}.$$

Lab-frame magnitude

$$|a'|^2 = a'^2_\parallel + a'^2_\perp = \frac{a_0^2\cos^2\theta}{\gamma^6} + \frac{a_0^2\sin^2\theta}{\gamma^4} = \frac{a_0^2}{\gamma^4}\left[\frac{\cos^2\theta}{\gamma^2} + \sin^2\theta\right].$$

Combine using $1/\gamma^2 = 1 - \beta^2$: $\cos^2\theta/\gamma^2 + \sin^2\theta = (1-\beta^2)\cos^2\theta + \sin^2\theta = 1 - \beta^2\cos^2\theta$. Therefore

$$\boxed{\;a'(\theta) = \frac{a_0}{\gamma^2}\sqrt{1 - \beta^2\cos^2\theta}.\;}$$

Extrema

The quantity $1-\beta^2\cos^2\theta$ runs between $1-\beta^2 = 1/\gamma^2$ (at $\theta=0$) and $1$ (at $\theta = \pi/2$):

$$\boxed{\;\frac{a_0}{\gamma^3}\le a'\le\frac{a_0}{\gamma^2},\quad\text{min at }\theta=0,\;\max\text{ at }\theta=\pi/2.\;}$$

Interpretation

Lab acceleration is always smaller than proper acceleration — "relativistic mass" enhancement makes objects harder to accelerate. The longitudinal direction is more suppressed ($\gamma^3$) than transverse ($\gamma^2$): along the boost the time-dilation effect compounds with the parallel velocity addition. Proper acceleration is what the particle's own accelerometer reads — the basis for the equivalence-principle interpretation in GR.