Special Relativity — worked problems

Decompose the rod

Rest frame components: $\Delta x = \ell\cos\theta$, $\Delta z = \ell\sin\theta$.

Lorentz contraction

Boost along $\hat x$: only the $x$-component contracts; the perpendicular $z$-component is unchanged.

$$\Delta x' = \frac{\Delta x}{\gamma} = \frac{\ell\cos\theta}{\gamma},\qquad \Delta z' = \Delta z = \ell\sin\theta.$$

Apparent length

$$\ell' = \sqrt{(\Delta x')^2 + (\Delta z')^2} = \ell\sqrt{\frac{\cos^2\theta}{\gamma^2} + \sin^2\theta}.$$

Simplify using $1/\gamma^2 = 1 - \beta^2$:

$$\frac{\cos^2\theta}{\gamma^2} + \sin^2\theta = (1-\beta^2)\cos^2\theta + \sin^2\theta = \cos^2\theta + \sin^2\theta - \beta^2\cos^2\theta = 1 - \beta^2\cos^2\theta.$$

$$\boxed{\;\ell' = \ell\sqrt{1 - \beta^2\cos^2\theta}.\;}$$

Limiting cases

$\theta = 0$ (rod along boost): $\ell' = \ell\sqrt{1-\beta^2} = \ell/\gamma$ — full contraction. ✓

$\theta = \pi/2$ (rod transverse): $\ell' = \ell$ — no contraction. ✓

$\theta = \pi/4$ (diagonal): $\ell' = \ell\sqrt{1 - \beta^2/2}$ — intermediate.

As $\beta\to 1$ (ultra-relativistic): $\ell' \to \ell|\sin\theta|$. Only the transverse component survives; the rod looks asymptotically perpendicular to the motion. The longitudinal projection vanishes faster than $1/\gamma$ because of the angle factor.

Apparent angle

$$\tan\theta' = \frac{\Delta z'}{\Delta x'} = \frac{\ell\sin\theta}{\ell\cos\theta/\gamma} = \gamma\tan\theta.$$

The rod always tilts more transverse in the moving frame: $\theta' > \theta$ for $0 < \theta < \pi/2$. This is the general $\theta$-version of Problem 1.6 (where $\theta = 45°\to\theta' = 60°$).

Setup

In $K$ (lab frame where the events are simultaneous): $\Delta t = 0$, $\Delta x = L$. $K'$ moves at $-u$ along $\hat x$ relative to $K$, equivalently $K$ moves at $+u$ in $K'$.

Lorentz transformation

Transform from $K$ to $K'$ (boost at velocity $+u$ along $\hat x$):

$$\Delta x' = \gamma(\Delta x - u\Delta t) = \gamma L,$$

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

The minus sign indicates which event is earlier (depends on direction). Taking magnitudes: $|\Delta t'| = \gamma uL/c^2$.

Solve for $L$

$$L = \frac{c^2|\Delta t'|}{\gamma u}.$$

Substituting $\gamma = 1/\sqrt{1-u^2/c^2}$:

$$\boxed{\;L = \frac{c^2\Delta t'}{u}\sqrt{1 - \frac{u^2}{c^2}} = \frac{c^2\Delta t'}{\gamma u}.\;}$$

Distance in $K'$

$L' = \gamma L = c^2\Delta t'/u$. Notice the clean form: in $K'$, the formula $v = c^2\Delta t'/\Delta x'$ holds without the $\gamma$ correction — same as Problem 1.10. The structure $\Delta t'/\Delta x' = v/c^2$ encodes relativity of simultaneity at the same level of generality.

Consistency check

$L' > L$ by factor $\gamma$ — the moving observer sees the events spatially farther apart than the lab observer. This is the "rest-frame length expansion" seen also in Problem 1.8(b).

Setup

In $S$: $\Delta t = 0$ (simultaneous), $\Delta x = 10$ ly. In $S'$ (moving along $\hat x$ relative to $S$): $\Delta t' = 1$ year, $\Delta x' = ?$.

Find velocity

From Lorentz transformation: $|\Delta t'| = \gamma v\,\Delta x/c^2 = \gamma\beta\cdot(10\,\text{ly}/c) = \gamma\beta\cdot 10\,\text{yr}$. Setting equal to 1 yr:

$$\gamma\beta\cdot 10 = 1\;\Longrightarrow\;\gamma\beta = 0.1.$$

Solve for $\beta$: $\gamma\beta = \beta/\sqrt{1-\beta^2}$, so $$\frac{\beta^2}{1-\beta^2} = 0.01\;\Longrightarrow\;\beta^2 = \frac{0.01}{1.01}\;\Longrightarrow\;\beta = \frac{1}{\sqrt{101}}\approx 0.0995.$$

$$\boxed{\;v \approx 0.0995\,c\approx 2.98\times 10^7\,\text{m/s}.\;}$$

Distance in $S'$

$\gamma = \sqrt{101/100} = \sqrt{1.01}\approx 1.005$. From the Lorentz transformation:

$$\Delta x' = \gamma\Delta x - \gamma v\Delta t = \gamma\cdot 10\,\text{ly} = \sqrt{101}\,\text{ly}\approx 10.05\,\text{ly}.$$

$$\boxed{\;\Delta x' = \sqrt{101}\,\text{ly}\approx 10.05\,\text{ly}.\;}$$

Interpretation

$\gamma\approx 1.005$ — the relativistic correction is only 0.5%. The 1-year temporal offset in $S'$ comes almost entirely from the simultaneity offset $v\Delta x/c^2 \approx \beta\cdot 10$ yr, not from significant length dilation. This is the kinematic regime where Lorentz physics first becomes visible: low $\beta$, but distances measured in years (or light-years) make the simultaneity coefficient $v/c^2$ accumulate to detectable effects.

Setup

"Reaches the ground before decaying substantially" means traveling distance $h = 10$ km in one lab-frame mean lifetime $\gamma\tau_0$:

$$h = v\cdot\tau_{\text{lab}} = (\beta c)(\gamma\tau_0).$$

Solve for $\gamma$

$\beta\gamma = h/(c\tau_0)$:

$$\beta\gamma = \frac{10^4\,\text{m}}{(3\times 10^8\,\text{m/s})(2.2\times 10^{-6}\,\text{s})} = \frac{10^4}{660} \approx 15.15.$$

Use the identity $\beta\gamma = \sqrt{\gamma^2 - 1}$ to extract $\gamma$:

$$\gamma^2 - 1 = (15.15)^2 = 229.5\;\Longrightarrow\;\gamma = \sqrt{230.5}\approx 15.18.$$

(Very close to $\beta\gamma$ — indicating $\beta\approx 1$, ultra-relativistic regime.)

Minimum energy

$$E_{\min} = \gamma m_\mu c^2 = 15.18\times 106\,\text{MeV} \approx 1.6\,\text{GeV}.$$

$$\boxed{\;E_{\min}\approx 1.6\,\text{GeV.}\;}$$

Without time dilation

A muon at $v\approx c$ would travel only $c\tau_0 = 660$ m before decaying — short of 10 km by a factor of $\sim 15$. The required factor of $\sim 15$ in penetration depth is provided entirely by the Lorentz factor $\gamma$, which dilates the lab-frame lifetime. This was the central insight of the 1958 Frisch–Smith experiment: counting muons at Mt. Washington (1909 m) and at sea level gave a ratio consistent with time dilation, not with the Newtonian prediction.

Connection to the $\nu_\mu/\nu_e$ ratio

The problem's context is the atmospheric neutrino ratio $R(\mu/e) = 2$ at low energies. Above $\sim 1.6$ GeV, the parent muons reach the ground before completing the second decay $\mu\to e\nu_\mu\nu_e$, so only the first decay ($\pi\to\mu\nu_\mu$) contributes — eliminating the electron neutrinos and driving $R(\mu/e)$ upward. This high-energy enhancement was the empirical signature that led to the atmospheric-neutrino anomaly (Super-Kamiokande), eventually establishing neutrino oscillations (Nobel 2015).

Lorentz factor and rapidity

At total energy $E = 1$ GeV, $$\gamma = \frac{E}{m_\mu c^2} = \frac{1000\,\text{MeV}}{106\,\text{MeV}}\approx 9.43.$$ $$\beta\gamma = \sqrt{\gamma^2 - 1} = \sqrt{88.9 - 1} = \sqrt{87.9}\approx 9.38.$$

(Already ultra-relativistic: $\beta\approx 0.994$.)

Lab-frame decay length

$$d = \beta c\cdot\gamma\tau_0 = \beta\gamma\cdot c\tau_0 = 9.38\times(3\times 10^8)(2.2\times 10^{-6})\,\text{m} = 9.38\times 660\,\text{m}\approx 6190\,\text{m} = 6.19\,\text{km}.$$

Circumference and turns

$C = 2\pi R = 2\pi\cdot 50\,\text{m} \approx 314.2\,\text{m}$.

$$N = \frac{d}{C} = \frac{6190}{314.2}\approx 19.7.$$

$$\boxed{\;N\approx 19.7\;\text{turns.}\;}$$

Physical context: muon $g-2$

This is the principle of muon storage rings: relativistic time dilation extends muon lifetimes by factor $\gamma$, allowing many orbits before decay. The actual Fermilab $g-2$ experiment runs at $\gamma\approx 29.3$ ($E = 3.094$ GeV), giving $N\approx 60$ turns and ample precession periods for extracting the muon's anomalous magnetic moment. The 2021 announcement of $a_\mu^{\text{exp}} = 0.00116592061(41)$ deviates from Standard Model predictions by $4.2\sigma$ — a hint of new physics.

Engineering implications

At $\gamma = 9.43$, the muon orbit decays via a lab-frame Larmor precession. The muon's proper time per orbit is $T_{\text{proper}} = 2\pi R/(\gamma v)\approx 1.12\times 10^{-7}$ s — well within $\tau_0 = 2.2\mu$s, allowing ~20 orbits per proper lifetime. At higher energies, the orbit count grows linearly with $\gamma$; at the LHC's TeV scales, charged particles can orbit millions of times before decay.

Rest-frame setup

3–4–5 right triangle: legs $a = 3\ell$, $b = 4\ell$, hypotenuse $c = 5\ell$ (Pythagoras: $9 + 16 = 25$ ✓). Right angle between $a$ and $b$. Rest-frame area: $A = \tfrac12 ab = \tfrac12\cdot 3\ell\cdot 4\ell = 6\ell^2$.

(a) Boost parallel to side $a$

Place $a$ along $\hat x$, $b$ along $\hat y$. Only $\hat x$-components contract:

$a' = a/\gamma = 3\ell/\gamma$; $b' = b = 4\ell$ (perpendicular, unchanged); $c' = \sqrt{(a/\gamma)^2 + b^2} = \ell\sqrt{9/\gamma^2 + 16}$.

The right angle is preserved (perpendicular axes stay perpendicular under boost along one of them):

$$A' = \tfrac12 a'b' = \tfrac12\cdot\frac{3\ell}{\gamma}\cdot 4\ell = \frac{6\ell^2}{\gamma}.$$

(b) Boost parallel to hypotenuse $c$

Decompose each side along $\hat c$ (boost direction) and $\hat c_\perp$. The unit vectors are $\hat c = (3\hat x + 4\hat y)/5$, $\hat c_\perp = (-4\hat x + 3\hat y)/5$.

Each side decomposes; the $\hat c$-component contracts by $1/\gamma$, the $\hat c_\perp$-component is unchanged.

Going through this carefully (or just using the general result below), the legs $a$ and $b$ rotate by some angle and shrink anisotropically, but the area still shrinks by $1/\gamma$:

$$A' = \frac{6\ell^2}{\gamma}\;\text{(same as case a)}.$$

Area scaling is direction-independent

$$\boxed{\;A' = \frac{6\ell^2}{\gamma}\quad\text{(both directions of boost).}\;}$$

Why? Length contraction is a 1D contraction by $1/\gamma$ along the boost direction. Area is a 2-form: $dA = dx\wedge dy$. Under a boost, one direction shrinks by $1/\gamma$, the other is unchanged. The product (area) always shrinks by the same factor $1/\gamma$, regardless of which 1D direction we choose to call "boost direction" inside the 2D plane. This generalises to volumes ($1/\gamma$) and 4-volumes (invariant) — the structure of Lorentz transformations as area-preserving maps in the perpendicular planes.

The specific shape change differs between (a) (rectangle-like reduction) and (b) (skew reduction along the hypotenuse), but the integrated area drops by exactly the same factor. Useful general principle for any 2D figure boosted in any direction.

Geometric setup

The observer sits at the spatial origin. The pole, at height $h$ above the line of motion, moves along $-\hat x$. The observer receives light at the same moment from three points on the pole, perceiving them at angles $\theta_F = \pi/3$ (front), $\theta_M = \pi/4$ (mark), $\theta_E = \pi/6$ (end) measured from $\hat x$.

For each photon, the apparent position at the moment of emission has $x = h\cot\theta$ (basic geometry: $\tan\theta = h/x$).

Apparent positions

$$x_F^{\text{app}} = h\cot(\pi/3) = h\cdot\frac{1}{\sqrt 3} = \frac{h}{\sqrt 3},$$ $$x_M^{\text{app}} = h\cot(\pi/4) = h,$$ $$x_E^{\text{app}} = h\cot(\pi/6) = h\sqrt 3.$$

Apparent distances

Front-to-mark:

$$|FM|^{\text{app}} = x_M^{\text{app}} - x_F^{\text{app}} = h - \frac{h}{\sqrt 3} = h\cdot\frac{\sqrt 3 - 1}{\sqrt 3}.$$

Front-to-end (full pole):

$$|FE|^{\text{app}} = x_E^{\text{app}} - x_F^{\text{app}} = h\sqrt 3 - \frac{h}{\sqrt 3} = h\cdot\frac{3 - 1}{\sqrt 3} = \frac{2h}{\sqrt 3}.$$

Ratio

$$r = \frac{|FM|^{\text{app}}}{|FE|^{\text{app}}} = \frac{(\sqrt 3 - 1)/\sqrt 3}{2/\sqrt 3} = \frac{\sqrt 3 - 1}{2}.$$

$$\boxed{\;r = \frac{\sqrt 3 - 1}{2}\approx 0.366.\;}$$

The mark sits at ~37% of the length from the front, not at the midpoint.

Why the result is not 1/2: optical snapshot

Photons from different parts of the pole were emitted at different proper times: light from the receding part of the pole was emitted earlier than light from the approaching part. The observer reconstructs an "apparent" geometry that mixes the pole's motion with finite light-travel-time corrections.

This is the Penrose–Terrell rotation applied to a 1-dimensional rod. The general phenomenon: fast-moving objects don't look length-contracted; they look rotated. A sphere flying past at relativistic speed still looks like a sphere, just rotated by some angle that depends on the viewing geometry. This was an important conceptual correction to the naive picture of length contraction as visual squashing.

(a) Does the string break? YES.

This is the celebrated Bell spaceship paradox. Both ships have identical worldlines in the original frame (just shifted by 40 km), so their lab-frame separation stays 40 km throughout the maneuver. But:

In the ships' eventual common rest frame (after both stop), the lab separation $L_{\text{lab}} = 40$ km is a length-contracted measurement, so the ships' proper separation in that frame is

$$L_{\text{rest}} = \gamma\cdot L_{\text{lab}} = \gamma\cdot 40\,\text{km}.$$

As $\gamma$ grows during acceleration, the rest-frame separation grows past the initial 40 km. A real physical string of rest length 40 km cannot stretch — it breaks.

The conceptual point

Bell's CERN colleagues initially disagreed: surely identical acceleration in the lab means identical separation everywhere? The resolution is the relativity of simultaneity:

In the leading ship's instantaneous co-moving frame, the trailing ship's clock runs differently. What the leading ship calls "now-when-I-have-velocity-$v$" doesn't correspond to "now-when-the-trailing-ship-has-velocity-$v$." The trailing ship, in the leading ship's reckoning, lags behind in its acceleration history. So in the leading ship's frame, the trailing ship is further behind than the lab-frame 40 km.

The two ships have identical trajectories in the lab but different simultaneity slices in the co-moving frame — the source of the apparent contradiction and its resolution.

(b) Numerical separation in leading ship's frame

After $t_0 = 30$ s with $a = c/50$ s$^{-1}$:

$$v = at_0 = \frac{c}{50}\cdot 30\,\text{s} = 0.6c.$$

Lorentz factor:

$$\gamma = \frac{1}{\sqrt{1 - 0.36}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25.$$

After the engines stop, both ships move at the same constant velocity in lab, separated by 40 km in lab. In the leading ship's rest frame, the separation is dilated by $\gamma$:

$$\Delta x' = \gamma\cdot L_{\text{lab}} = 1.25\cdot 40\,\text{km} = 50\,\text{km}.$$

$$\boxed{\;\Delta x' = 50\,\text{km}.\;}$$

Physical reality of the rupture

The string, with rest length 40 km, would need to span 50 km in the leading ship's frame — physically impossible without stretching. The 10 km excess represents the rest-frame elongation that the string cannot accommodate, so it ruptures. The rupture is a real, frame-independent event — the string ends are at distinct spacetime points, not at the same point seen differently from different frames.

Setup in Wolf's frame

Train of rest length $L$ moves at $+v$ in Wolf's frame. Lab-frame length: $L_{\text{lab}} = L/\gamma$. At $t = 0$, lightning strikes Wolf at $x = 0$. Einstein is at the middle of the train, also at $x = 0$ at $t = 0$. Lab positions of train ends at $t = 0$: front at $L/(2\gamma)$, rear at $-L/(2\gamma)$.

When do light pulses reach the reflectors?

The front reflector is moving away (receding); the rear is moving toward Wolf (approaching). Light propagates outward at $\pm c$ from the strike point.

Front: $ct_F = L/(2\gamma) + vt_F$ (light catches up to receding reflector). Solving: $t_F = L/[2\gamma c(1-\beta)]$.

Rear: $-ct_B = -L/(2\gamma) + vt_B$ (light moving in $-\hat x$ catches approaching reflector). Solving: $t_B = L/[2\gamma c(1+\beta)]$.

Since $1-\beta < 1+\beta$, we have $t_F > t_B$ — the rear reflector is hit first in Wolf's frame.

Both reflections meet Einstein at one event

Track the reflected photons:

Rear-reflection: emitted at $(t_B, -L/[2\gamma(1+\beta)] + vt_B)$, propagates in $+\hat x$ at speed $c$.

Front-reflection: emitted at $(t_F, L/[2\gamma(1-\beta)])$, propagates in $-\hat x$ at speed $c$.

Einstein's worldline: $x_E(t) = vt$.

Setting up the catch-up conditions and solving (algebra), both reflections meet Einstein at $$(t, x) = \Bigl(\frac{L\gamma}{c}, \frac{vL\gamma}{c}\Bigr).$$

This is one and the same spacetime event.

Answers

(a) YES — the reception event at Einstein's position is a single spacetime point. Single events are frame-invariant: every observer (including Wolf) agrees the two reflections reach Einstein simultaneously.

$$\boxed{\;\text{Wolf sees both reflections reach Einstein simultaneously.}\;}$$

(b) NO — the two reflection events (at the front and rear of the train) are different spacetime points, spatially separated. Their time ordering depends on the frame. In the train's rest frame they are simultaneous (Einstein sees them at the same time at the same place); in Wolf's frame, the rear event happens first.

$$\boxed{\;\text{Wolf disagrees: reflections happened at different times in his frame.}\;}$$

(c) NO — Wolf sits at $x = 0$. The rear-reflection (closer to Wolf, and happening earlier) reaches him first; the front-reflection arrives later. Wolf sees the two reflections at different times.

$$\boxed{\;\text{Reflections do NOT reach Wolf simultaneously.}\;}$$

Pedagogical principle

The seeming paradox dissolves once one separates:

• "One event seen by two observers" — frame-invariant. A single spacetime point is a single spacetime point.
• "Two events compared across frames" — frame-dependent. Time ordering, simultaneity, and spatial separation all change with the boost.

Einstein's original 1905 train thought experiment was constructed exactly to expose this distinction. It is the seed of all subsequent SR confusions and their resolutions.

Setup

Two rockets at relative velocity $u$. Slow rocket (rest length $2L$) is the long one; fast rocket (rest length $L$) is the short one and overtakes. Two events:

• $A$: front of fast rocket reaches rear of slow.
• $B$: front of fast rocket reaches front of slow.

The "witness" of both events is the front of the fast rocket — the same physical object.

In the slow rocket's frame

Slow rocket at rest, length $2L$. The fast rocket's front sweeps from $x = 0$ (rear of slow) to $x = 2L$ (front of slow) at speed $u$:

$$\Delta t_{\text{slow}} = \frac{2L}{u}.$$

In the fast rocket's frame

The fast rocket's front is fixed at $x' = 0$ (say). The slow rocket moves past at $-u$, with its lab length contracted to $2L/\gamma_u$. The contracted slow rocket sweeps past $x' = 0$ at speed $u$:

$$\Delta t_{\text{fast}} = \frac{2L/\gamma_u}{u} = \frac{2L}{\gamma_u\,u}.$$

Compare

$$\frac{\Delta t_{\text{fast}}}{\Delta t_{\text{slow}}} = \frac{1}{\gamma_u} < 1.$$

$$\boxed{\;\Delta t_{\text{slow}} > \Delta t_{\text{fast}}\;\;\text{(by factor }\gamma_u\text{).}\;}$$

The observer in the long (slow) rocket measures the longer time.

Why? Proper time vs coordinate time

In the fast rocket's frame, events $A$ and $B$ both occur at the same spatial point (front of fast rocket). The interval $\Delta t_{\text{fast}}$ is therefore a proper time — measured by a single clock at one position.

In the slow rocket's frame, events $A$ and $B$ occur at different spatial points (rear and front, separated by $2L$). The interval $\Delta t_{\text{slow}}$ is a coordinate-time interval between events at different places — necessarily longer than the proper time by $\gamma_u$.

General rule: proper time is the shortest time interval between two events, measured by whoever sees both events at the same place.

Spacetime interval $\Delta s^2 = c^2\Delta t^2 - \Delta x^2$ is Lorentz-invariant. For the fast frame ($\Delta x' = 0$): $\Delta s^2 = c^2\Delta t_{\text{fast}}^2$. For the slow frame ($\Delta x = 2L$): $\Delta s^2 = c^2\Delta t_{\text{slow}}^2 - 4L^2$. Setting equal:

$$\Delta t_{\text{fast}}^2 = \Delta t_{\text{slow}}^2 - 4L^2/c^2 \;\Longrightarrow\;\Delta t_{\text{fast}}^2 = \Delta t_{\text{slow}}^2(1 - u^2/c^2),$$

using $u = 2L/\Delta t_{\text{slow}}$. So $\Delta t_{\text{fast}} = \Delta t_{\text{slow}}/\gamma_u$ ✓ — confirming the result invariantly.