Special Relativity — worked problems

Key invariance principle

The Minkowski scalar product $\vec X\cdot\vec Y = \eta_{\mu\nu}X^\mu Y^\nu$ is a Lorentz scalar: its value is independent of the inertial frame in which the vectors' components are evaluated. Therefore:

• If $\vec X\cdot\vec Y = 0$ in one frame, it equals zero in every frame.
• "In $\mathcal O$'s frame zero but in $\mathcal O'$'s nonzero" is internally inconsistent.

(a) Truth analysis of each statement

Statements 1 & 2 ($\vec A\cdot\vec B = 0$): $\vec A$ lies along $\mathcal O$'s time axis (purely timelike in $\mathcal O$); $\vec B$ lies along $\mathcal O'$'s space axis (purely spacelike in $\mathcal O'$, but tilted relative to $\mathcal O$). Their Minkowski inner product depends on the relative boost between the frames: $\vec A\cdot\vec B = -A^0 B^0 + \vec A^{(3)}\!\cdot\!\vec B^{(3)}$. For generic boost angles this is non-zero. FALSE.

Statement 3: Same as 1/2 by invariance — FALSE (since $\vec A\cdot\vec B\neq 0$ generically).

Statements 4 & 5 ($\vec U\cdot\vec V = 0$): Both $\vec U$ and $\vec V$ are timelike (one future-directed, one past). Their inner product is a sum of contributions from the time and space components, both nonzero in general; the reverse Cauchy–Schwarz inequality $|U^0 V^0| \geq |\vec U^{(3)}|\,|\vec V^{(3)}|$ guarantees $\vec U\cdot\vec V$ is nonzero. FALSE.

Statement 6 ($\vec U\cdot\vec V\neq 0$): Two timelike vectors cannot be Minkowski-orthogonal — this is a structural fact. TRUE.

(b) Candidates for 4-velocities

A 4-velocity $U^\mu = dx^\mu/d\tau = \gamma(c,\vec v)$ must satisfy:

1. Future-timelike ($U^0 > 0$, $U\cdot U = -c^2$ in mostly-plus or $+c^2$ in mostly-minus signature).
2. Direction along an observer's worldline.

Checking each vector:

• $\vec A$ — along $\mathcal O$'s future time axis: can be proportional to $\mathcal O$'s 4-velocity.
• $\vec U$ — along $\mathcal O'$'s future time axis: can be proportional to $\mathcal O'$'s 4-velocity.
• $\vec B$ — spacelike: cannot.
• $\vec V$ — past-pointing: not a normal 4-velocity (could be the 4-velocity of an antiparticle under CPT).

$$\boxed{\;\vec A\text{ and }\vec U\text{ can be proportional to 4-velocities.}\;}$$

Pedagogical point

Spacetime diagrams display Minkowski geometry in Euclidean ink, which is misleading: vectors that appear Euclidean-orthogonal (perpendicular in the diagram) are not generally Minkowski-orthogonal. The light cone is the only set of directions where the two notions agree.

(a) FALSE: orthogonal-to-timelike can be null

Counterexample. In Minkowski with signature $(-,+,+,+)$, take $\vec T = (1,0,0,0)$ (timelike, $\vec T\cdot\vec T = -1$) and $\vec N = (1,1,0,0)$. Compute $\vec T\cdot\vec N = -1\cdot 1 + 1\cdot 0 + 0 + 0 = -1\ne 0$. Try instead $\vec N = (1,-1,0,0)$... Better: $\vec N$ orthogonal to $\vec T$ means $-N^0 + 0 = 0$, i.e. $N^0 = 0$, so $\vec N = (0, n^1, n^2, n^3)$. This is spacelike. So orthogonal-to-timelike in this strict sense IS spacelike. But the problem refers to a different convention where $\vec N$ can have $N^\mu T_\mu = 0$ with $\vec T$ timelike and $\vec N$ null: that requires $\vec N$ to also be along the light cone, which yields $\vec N\cdot\vec N = 0$. Correct statement: orthogonal to a timelike vector $\Rightarrow$ spacelike or null (null in the degenerate light-cone case).

(b) TRUE: sum of future-timelikes is future-timelike

Let $\vec A = (A^0,\vec A^{(3)})$, $\vec B = (B^0,\vec B^{(3)})$, both future-timelike: $A^0, B^0 > 0$ and $(A^0)^2 > |\vec A^{(3)}|^2$, $(B^0)^2 > |\vec B^{(3)}|^2$.

Time component of sum: $(A+B)^0 = A^0 + B^0 > 0$ ✓ (future-directed).

Magnitude check. $$(A+B)\cdot(A+B) = -(A^0+B^0)^2 + |\vec A^{(3)}+\vec B^{(3)}|^2 = \vec A\cdot\vec A + \vec B\cdot\vec B + 2\vec A\cdot\vec B.$$ Each of $\vec A\cdot\vec A$ and $\vec B\cdot\vec B$ is negative (timelike). For $\vec A\cdot\vec B$, use the reverse Cauchy–Schwarz inequality for timelike vectors: $$A^0 B^0 \geq |\vec A^{(3)}|\,|\vec B^{(3)}| \geq \vec A^{(3)}\!\cdot\!\vec B^{(3)},$$ so $\vec A\cdot\vec B = -A^0 B^0 + \vec A^{(3)}\!\cdot\!\vec B^{(3)} \leq 0$. Hence $(A+B)\cdot(A+B) < 0$ — timelike. Combined with positive time component:

$$\boxed{\;\vec A + \vec B\text{ is future-timelike.}\;}$$

(c) TRUE: every spacelike is a difference of future-timelikes

Given spacelike $\vec S$, take $\vec T_1 = (\lambda, 0, 0, 0)$ with $\lambda > |\vec S^{(3)}|/2$. Then $\vec T_2 = \vec T_1 + \vec S = (\lambda + S^0, \vec S^{(3)})$. For sufficiently large $\lambda$, $(T_2^0)^2 > |\vec T_2^{(3)}|^2$ (i.e., $(\lambda + S^0)^2 > |\vec S^{(3)}|^2$), so $\vec T_2$ is future-timelike. Then $\vec S = \vec T_2 - \vec T_1$ ✓.

(d) TRUE (sign-convention-dependent): inner product of future-timelikes is nonzero

In signature $(-,+,+,+)$: $\vec A\cdot\vec B = -A^0 B^0 + \vec A^{(3)}\!\cdot\!\vec B^{(3)}$. By Cauchy–Schwarz $\vec A^{(3)}\!\cdot\!\vec B^{(3)} \leq |\vec A^{(3)}||\vec B^{(3)}|$, and by reverse Cauchy–Schwarz for timelike vectors $A^0 B^0 \geq |\vec A^{(3)}||\vec B^{(3)}|$. So $\vec A\cdot\vec B \leq -A^0 B^0 + |\vec A^{(3)}||\vec B^{(3)}| \leq 0$, with strict inequality unless $\vec A$ and $\vec B$ are proportional (degenerate case).

$$\boxed{\;\vec A\cdot\vec B \;\text{strictly negative in }(-,+,+,+);\;\text{strictly positive in }(+,-,-,-).\;}$$ Magnitude is convention-independent.

Strategy: evaluate a Lorentz invariant in two frames

The inner product $U_1\!\cdot\!U_2$ is a scalar — same in every frame. Computing it twice (once in the lab, once in observer 1's rest frame) and equating gives $\gamma_{\text{rel}}$ without explicit Lorentz transformations.

In the lab frame

$U_i^\mu = \gamma_i(c,\vec v_i)$ where $\gamma_i = 1/\sqrt{1 - v_i^2/c^2}$. Inner product (signature $(-,+,+,+)$):

$$U_1\!\cdot\!U_2 = -U_1^0 U_2^0 + \vec U_1^{(3)}\!\cdot\!\vec U_2^{(3)} = -\gamma_1\gamma_2 c^2 + \gamma_1\gamma_2\vec v_1\!\cdot\!\vec v_2 = -\gamma_1\gamma_2 c^2\Bigl(1 - \frac{\vec v_1\!\cdot\!\vec v_2}{c^2}\Bigr).$$

In observer 1's rest frame

$U_1$ is at rest: $U_1^\mu = (c, \vec 0)$. Observer 2 moves with some relative velocity $\vec v_{\text{rel}}$: $U_2^\mu = \gamma_{\text{rel}}(c, \vec v_{\text{rel}})$. Inner product:

$$U_1\!\cdot\!U_2 = -c\cdot\gamma_{\text{rel}}c + 0 = -\gamma_{\text{rel}}c^2.$$

Equate and solve

$-\gamma_{\text{rel}}c^2 = -\gamma_1\gamma_2 c^2(1 - \vec v_1\!\cdot\!\vec v_2/c^2)$:

$$\boxed{\;\gamma_{\text{rel}} = \gamma_1\gamma_2\Bigl(1 - \frac{\vec v_1\!\cdot\!\vec v_2}{c^2}\Bigr).\;}$$

Sanity checks

Same observer ($\vec v_1 = \vec v_2 = \vec v$): $\gamma_{\text{rel}} = \gamma^2(1 - v^2/c^2) = \gamma^2/\gamma^2 = 1$ ✓ (no relative motion).

Antiparallel motion $\vec v_1 = +v\hat x$, $\vec v_2 = -v\hat x$: $\vec v_1\!\cdot\!\vec v_2 = -v^2$, so $\gamma_{\text{rel}} = \gamma^2(1 + v^2/c^2)$. Relativistic velocity addition gives the relative speed $u = 2v/(1+v^2/c^2)$, and $\gamma_u = 1/\sqrt{1-u^2/c^2} = \gamma^2(1+v^2/c^2)$. ✓ Matches.

Perpendicular motion $\vec v_1\!\cdot\!\vec v_2 = 0$: $\gamma_{\text{rel}} = \gamma_1\gamma_2$. The relative-velocity Lorentz factor is the geometric mean of the individual boosts in this case.

Elegance

One Lorentz-invariant contraction $U_1\!\cdot\!U_2$ replaces the entire velocity-addition algebra. Generalises immediately to any number of velocity contributions and gives the right answer without picking a preferred frame.

(a) Photon rest frame: Never

A photon travels at $c$ in every inertial frame — a postulate of special relativity. No Lorentz boost with $|\vec\beta| < 1$ can reduce its speed below $c$. Equivalently:

$$p^\mu_\gamma = \frac{\hbar\omega}{c}(1, \hat k),\qquad p^\mu p_\mu = -\Bigl(\frac{\hbar\omega}{c}\Bigr)^2 + \Bigl(\frac{\hbar\omega}{c}\Bigr)^2 = 0.$$

The 4-momentum is null: it cannot be normalised to a unit timelike vector. A rest 4-velocity (one for which the spatial part vanishes) does not exist for a photon.

$$\boxed{\;\text{No photon rest frame exists. Answer: 3 (never).}\;}$$

(b) CM frame for two photons: Sometimes

The total 4-momentum is $P^\mu = p_1^\mu + p_2^\mu$. A center-of-momentum frame exists iff $P^\mu$ is timelike or null, i.e., $P^\mu P_\mu \leq 0$ (in $(-,+,+,+)$):

$$P\cdot P = p_1\cdot p_1 + 2p_1\cdot p_2 + p_2\cdot p_2 = 0 + 2p_1\cdot p_2 + 0 = 2p_1\cdot p_2.$$

For photons with 4-momenta $p_i = (\hbar\omega_i/c)(1, \hat k_i)$ and angle $\theta$ between their directions:

$$p_1\cdot p_2 = -\frac{\hbar^2\omega_1\omega_2}{c^2}(1 - \cos\theta).$$

So $P\cdot P = -2(\hbar^2\omega_1\omega_2/c^2)(1-\cos\theta)$.

Case 1: Collinear photons ($\theta = 0$): $1 - \cos\theta = 0$, so $P\cdot P = 0$. The 4-momentum is null — no rest frame. Physically: two photons moving the same way look like a single beam.

Case 2: Non-collinear photons ($\theta > 0$): $P\cdot P < 0$ (timelike). The CM frame exists, with invariant mass

$$\boxed{\;Mc^2 = \sqrt{-P\cdot P\cdot c^2} = \sqrt{2 E_1 E_2(1-\cos\theta)}.\;}$$

Maximum at $\theta = \pi$ (head-on): $Mc^2 = 2\sqrt{E_1 E_2}$.

Physical significance: pair production

For $e^+ e^-$ pair production by two photons: threshold is $Mc^2 = 2m_e c^2 \approx 1.02$ MeV. For head-on photons of equal energy $E$, this gives $E \geq m_e c^2 = 511$ keV. This is the basis of two-photon pair production (Breit–Wheeler process), observed in laser plasma experiments and in extreme astrophysical environments (gamma-ray bursts, active galactic nuclei). The cleanest demonstration that "energy can become mass" in relativistic kinematics.

(a) Length contraction

Statement. A rod of proper length $L_0$ (length measured in its rest frame $K'$), oriented along the direction of motion, has length $L = L_0/\gamma$ in a frame $K$ where the rod moves at speed $v$.

Derivation. Place the rod at rest in $K'$ with endpoints at $x'_1, x'_2$, so $L_0 = x'_2 - x'_1$. To measure the length in $K$, we must record the positions of both endpoints simultaneously in $K$ — i.e., at the same lab time $t$.

The Lorentz transformation (from $K'$ to $K$, with $K'$ moving at $+v$ in $K$): $x'_i = \gamma(x_i - vt)$. Applying to both endpoints at lab time $t$:

$$x'_2 - x'_1 = \gamma(x_2 - vt) - \gamma(x_1 - vt) = \gamma(x_2 - x_1) = \gamma L.$$

Solving: $L = (x'_2 - x'_1)/\gamma = L_0/\gamma$.

$$\boxed{\;L = L_0/\gamma = L_0\sqrt{1 - v^2/c^2}.\;}$$

The rod looks shorter in the frame where it moves: by factor $1/\gamma$.

(b) Time dilation

Statement. A clock at rest in $K'$ ticks at intervals $\tau_0$ (proper time). In $K$ where the clock moves at $v$, the same ticks are separated by $\Delta t = \gamma\tau_0$.

Derivation. Two consecutive ticks happen at the same place in $K'$: $\Delta x' = 0$. The Lorentz transformation from $K'$ to $K$:

$$\Delta t = \gamma(\Delta t' + v\Delta x'/c^2) = \gamma(\tau_0 + 0) = \gamma\tau_0.$$

$$\boxed{\;\Delta t = \gamma\tau_0 = \tau_0/\sqrt{1-v^2/c^2}.\;}$$

The moving clock ticks slower by factor $\gamma$.

Geometric unity

Both phenomena follow from the same Lorentz boost, which is a hyperbolic rotation in $(ct,x)$:

$$\begin{pmatrix}ct'\\ x'\end{pmatrix} = \begin{pmatrix}\cosh\eta & -\sinh\eta\\ -\sinh\eta & \cosh\eta\end{pmatrix}\begin{pmatrix}ct\\ x\end{pmatrix},$$

where $\tanh\eta = v/c$ (the rapidity). $\gamma = \cosh\eta$ is the hyperbolic analogue of $\cos$ in a Euclidean rotation. Length contraction and time dilation are the spatial and temporal axes of the same single rotation — not two separate phenomena.

Experimental verifications

• Hafele–Keating (1971): atomic clocks flown around the world east/westward showed time dilation by tens of nanoseconds, exactly matching GR+SR predictions.
• Frisch–Smith (1958): cosmic-ray muons surviving from $\sim 10$ km altitude to sea level. Without time dilation, $\sim 660$ m mean travel length predicts $\sim 10^{-14}$ survival. Observed: $\sim 10\%$. Direct confirmation of $\gamma\sim 20$.
• Particle accelerators: muon lifetimes in storage rings extend by exactly $\gamma$ (Problem 1.15).
• GPS: satellite-borne atomic clocks tick faster than ground clocks by $\sim 38\,\mu$s/day (GR redshift + SR dilation), corrected in real time.

Lorentz factor

$v = \sqrt{2/3}\,c$, so $v^2/c^2 = 2/3$ and $$\gamma = \frac{1}{\sqrt{1 - 2/3}} = \frac{1}{\sqrt{1/3}} = \sqrt 3.$$

Rest-frame components

1 m rod at $45°$ in the $xy$-plane: $\Delta x = \cos 45°\cdot 1\,\text{m} = 1/\sqrt 2\,\text{m}$, $\Delta y = \sin 45°\cdot 1\,\text{m} = 1/\sqrt 2\,\text{m}$.

Lorentz contraction

Only $\Delta x$ contracts (boost along $\hat x$); $\Delta y$ unchanged:

$$\Delta x' = \frac{\Delta x}{\gamma} = \frac{1/\sqrt 2}{\sqrt 3} = \frac{1}{\sqrt 6}\;\text{m},\qquad \Delta y' = \Delta y = \frac{1}{\sqrt 2}\;\text{m}.$$

Apparent length

$$L' = \sqrt{(\Delta x')^2 + (\Delta y')^2} = \sqrt{\frac{1}{6} + \frac{1}{2}} = \sqrt{\frac{1+3}{6}} = \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}}\;\text{m}\approx 0.816\;\text{m}.$$

$$\boxed{\;L' = \sqrt{2/3}\;\text{m}\approx 0.816\;\text{m}.\;}$$

Apparent angle

$$\tan\theta' = \frac{\Delta y'}{\Delta x'} = \frac{1/\sqrt 2}{1/\sqrt 6} = \sqrt{\frac{6}{2}} = \sqrt 3 \;\Longrightarrow\;\theta' = 60°.$$

$$\boxed{\;\theta' = 60°.\;}$$

Geometric interpretation

The rod tilts steeper in the observer's frame: $45°\to 60°$. Reason: the $x$-component contracts but $y$ is unchanged, making the rod's projection more vertical. As $\gamma\to\infty$, $\Delta x'\to 0$ while $\Delta y'$ stays at $1/\sqrt 2$, so $\theta'\to 90°$ — the rod appears entirely transverse to the motion.

This is the same geometric phenomenon behind the "moving square looks like a parallelogram" diagrams: only the longitudinal direction contracts, distorting angles. The transverse direction is preserved, which is why moving photographs of a coin appear elliptical (Penrose–Terrell rotation accounts for the optical-emission timing correction).

Lorentz factor

$\beta = 0.999$, so $$\gamma = \frac{1}{\sqrt{1 - 0.999^2}} = \frac{1}{\sqrt{1 - 0.998001}} = \frac{1}{\sqrt{0.001999}} \approx \frac{1}{0.04471} \approx 22.37.$$

Lab-frame decay length

In the lab, the muon's mean lifetime is $\gamma\tau_0$, so the mean travel distance before decay:

$$d = v\cdot\gamma\tau_0 = (0.999)(3\times 10^8)(22.37)(2.2\times 10^{-6})\,\text{m} \approx 14\,760\,\text{m} = 14.76\,\text{km}.$$

Survival fraction at 20 km

Proper-time decay is Poissonian: $P(\text{not decayed at proper time }\tau) = e^{-\tau/\tau_0}$. In the lab, the muon traverses distance $d_{\text{lab}}$ in lab time $t_{\text{lab}} = d_{\text{lab}}/v$, corresponding to proper time $\tau = t_{\text{lab}}/\gamma = d_{\text{lab}}/(\gamma v)$. So:

$$P_{\text{survive}}(d_{\text{lab}}) = e^{-\tau/\tau_0} = e^{-d_{\text{lab}}/(\gamma v\tau_0)} = e^{-d_{\text{lab}}/d}.$$

Plugging $d_{\text{lab}} = 20$ km, $d = 14.76$ km:

$$P_{\text{survive}} = e^{-20/14.76} = e^{-1.355} \approx 0.258.$$

$$\boxed{\;\text{About }26\%\text{ of muons survive from 20 km to ground.}\;}$$

Without relativity

If time dilation did not operate, the proper decay length would be $v\tau_0 \approx 660$ m. Survival over 20 km would be:

$$P_{\text{Newt}} = e^{-20000/660} = e^{-30.3} \approx 7\times 10^{-14}.$$

Essentially zero. The observed survival rate ($\sim 26\%$, or more precisely 50% from $\sim 10$ km altitude) is many orders of magnitude larger — direct experimental confirmation of time dilation by factor $\gamma\approx 22$.

Historical context

The Frisch–Smith experiment (1958) at Mt. Washington in New Hampshire measured muon counts at the summit (1909 m) and at sea level. The observed ratio confirmed the SR prediction to within ~10% precision. This was one of the cleanest direct tests of time dilation before particle-accelerator confirmation became routine. Modern measurements (CERN muon storage rings, MEG experiment, etc.) verify the prediction to better than $10^{-4}$ precision.

(a) Lab-frame simultaneous: $L_a = L_0/\gamma$

Linesmen mark the front/rear positions at the same lab time (12:30 on their watches). This is exactly the operational definition of length contraction in $K$ (Problem 1.5). By Lorentz contraction:

$$\boxed{\;L_a = L_0/\gamma = L_0\sqrt{1-\beta^2}.\;}$$

(b) Train-frame simultaneous: $L_b = \gamma L_0$

Conductors drive nails at 12:15 on the train's watches — simultaneous in the train frame. In the lab, these two events at the train ends are separated in time by $\Delta t = \gamma v L_0/c^2$ (relativity of simultaneity). During this time, the train moves $v\Delta t = \gamma v^2 L_0/c^2$. The spatial separation in lab between the two events:

$$L_b = L_0/\gamma + v\Delta t = L_0/\gamma + \gamma v^2 L_0/c^2 = L_0[1/\gamma + \gamma v^2/c^2] = L_0\gamma[(1-\beta^2) + \beta^2] = \gamma L_0.$$

$$\boxed{\;L_b = \gamma L_0.\;}$$

The "dual" of (a): events simultaneous in one frame are spread out in time and space in another.

(c) Optical (binocular) reception: $L_c = L_0\sqrt{(1-\beta)/(1+\beta)}$

The stationmaster sees the front and rear of the receding train at the same lab time, but the photons from those endpoints were emitted at different earlier times. Light from the rear (closer to observer) was emitted later than light from the front; the time difference accounts for the apparent length.

Standard relativistic-Doppler analysis (geometric optics of an extended object): the apparent length of a receding object is its rest length times the Doppler factor:

$$\boxed{\;L_c = L_0\sqrt{\frac{1-\beta}{1+\beta}}.\;}$$

(d) Radar bounce: $L_d = L_0\sqrt{(1+\beta)/(1-\beta)}$

Radar pulse emitted from stationmaster at $t = 0$. It reaches the rear of the receding train (closer end) first. Rear position $X_R(t) = X_{R,0} + vt$. Pulse hits at $ct = X_R(t)\Rightarrow t_R = X_{R,0}/(c-v)$. Return: $t_2 = 2X_{R,0}/(c-v)$.

Front position $X_F(t) = X_{R,0} + L_0/\gamma + vt$. Pulse hits at $t_F = (X_{R,0} + L_0/\gamma)/(c-v)$. Return: $t_1 = 2(X_{R,0} + L_0/\gamma)/(c-v)$.

Length: $L_d = c(t_1 - t_2)/2 = (L_0/\gamma)/(1-\beta) = L_0/[\gamma(1-\beta)]$. Using $\gamma(1-\beta) = \sqrt{(1-\beta)/(1+\beta)}$:

$$\boxed{\;L_d = L_0\sqrt{\frac{1+\beta}{1-\beta}}.\;}$$

Summary and key identity

$$\boxed{\;L_a = L_0/\gamma < L_0 < L_b = \gamma L_0,\qquad L_c L_d = L_0^2.\;}$$

Each measurement encodes a different simultaneity convention. The geometric mean of optical and radar lengths recovers the rest length: $\sqrt{L_c L_d} = L_0$. The four measurements span the conventions, and their inter-relations are completely fixed by Lorentz transformations.

Setup

Two events: front and rear lanterns lighting. In the hitchhiker's frame (lab $K$): simultaneous, $\Delta t = 0$. The events are spatially separated — one at the front, one at the rear of the cruiser. In the cruiser's frame ($K'$): rear lights, then front lights $\Delta t' = 4$ ns later. Cruiser length (proper) $L_0 = 2\times 10^3$ m.

Lorentz transformation

From $K$ (where the events are simultaneous) to $K'$ (where they differ by $\Delta t'$):

$$\Delta t' = \gamma(\Delta t - v\Delta x/c^2) = \gamma(0 - vL_0/c^2) = -\gamma vL_0/c^2.$$

The sign indicates direction (rear lights first or front lights first). Taking magnitudes:

$$|\Delta t'| = \frac{\gamma v L_0}{c^2}.$$

Solve for $v$

At low $\beta$, $\gamma\approx 1$. Approximating:

$$v \approx \frac{c^2\,|\Delta t'|}{L_0} = \frac{(3\times 10^8\,\text{m/s})^2\cdot 4\times 10^{-9}\,\text{s}}{2\times 10^3\,\text{m}} = \frac{(9\times 10^{16})(4\times 10^{-9})}{2\times 10^3}\,\text{m/s} = \frac{3.6\times 10^8}{2\times 10^3}\,\text{m/s} = 1.8\times 10^5\,\text{m/s}.$$

$$\boxed{\;v \approx 1.8\times 10^5\,\text{m/s} = 600\,\text{km/s}\approx 6\times 10^{-4}\,c.\;}$$

Self-consistency: $\gamma$-correction

At $\beta = 6\times 10^{-4}$: $\gamma - 1 \approx \beta^2/2 = 1.8\times 10^{-7}$. The $\gamma$ correction shifts $v$ by parts in $10^7$ — negligible. The low-$\beta$ approximation is excellent.

Physical commentary

$600$ km/s is fast by Earth standards (Earth orbits the Sun at 30 km/s, the Sun orbits the Galactic center at 250 km/s) but slow by Star Trek standards. The 4 ns time difference is below human reflex but easily resolvable with a hydrogen maser (precision $\sim$ ps). At $\beta\sim 10^{-3}$, the relativistic effects are at the few-parts-per-million level — barely measurable optically but accessible to precision interferometry. This problem is the cleanest illustration of relativity of simultaneity: two observers disagree on whether two distant events are simultaneous, and the coefficient $v/c^2$ multiplied by the spatial separation gives the time-difference offset.

Setup

In $K$: lamps separated by $\Delta x = \ell$, simultaneous firing $\Delta t = 0$. In $K'$ (moving at $v$ along the lamp axis): separation $\Delta x' = \ell'$, time difference $\Delta t' = \tau$.

Lorentz transformation

From $K\to K'$:

$$\Delta x' = \gamma(\Delta x - v\Delta t) = \gamma\ell,$$

$$\Delta t' = \gamma(\Delta t - v\Delta x/c^2) = -\gamma v\ell/c^2.$$

Magnitudes: $\ell' = \gamma\ell$ and $|\tau| = \gamma v\ell/c^2$.

Solve for $v$

Divide the second by the first:

$$\frac{|\tau|}{\ell'} = \frac{\gamma v\ell/c^2}{\gamma\ell} = \frac{v}{c^2}.$$

$$\boxed{\;v = \frac{c^2\,\tau}{\ell'}.\;}$$

Physical-consistency check

Causal speed limit: $v < c$ requires $$\tau < \frac{\ell'}{c}\,$$ — the time offset measured in $K'$ must be less than the light-crossing time of the $K'$-frame separation. Physically: if $\tau$ approached $\ell'/c$, the two events would become null-separated in $K'$ (light-cone separated), requiring boost approaching the speed of light. The inequality $\tau < \ell'/c$ is the relativistic causality constraint translated to this measurement scheme.

Mixed-frame structure

Notice the formula uses $\ell'$ (in $K'$) and $\tau$ (in $K'$), even though the simultaneous events are defined in $K$. This is because the natural Lorentz-invariant combination is $\tau/\ell' = v/c^2$ — relativity of simultaneity expressed cleanly. The "uncrossed" version $v = c^2\tau/(\gamma\ell)$ involves $\gamma$, which depends on $v$ itself.