Exercises — Length contraction — derivation

(a). Proper length $L_0$ is measured in the frame where the object is at rest. The astronaut holds the stick, so it is at rest in their frame — they measure $L_0$. The ground observer sees it move, so they measure the contracted $L<L_0$.

False. Only the dimension along the direction of motion contracts. The two dimensions perpendicular to $v$ are unchanged. A cube moving along its $x$-axis becomes a rectangular box, squashed only in $x$.

Factor 1 (no contraction). At $v=0$, $\gamma =1/\sqrt{1-0}=1$, so $L=L_0/1=L_0$. An object at rest in your frame is measured at its full proper length.

Step 1 — WHY find $\gamma$ first: contraction is $L=L_0/\gamma$, so we need $\gamma$ before shrinking. $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{1-0.36}}=\frac{1}{\sqrt{0.64}}=\frac{1}{0.8}=\mathrm{1.25.}$ Step 2 — divide (Earth sees the moving ship, so it is the shorter one): $L=\frac{150}{1.25}=120\text{ m}.$ Answer: 120 m. Sanity check: shorter than 150 ✔.

Whose frame is the atmosphere at rest in? Earth's — so 60 km is the proper length. In the muon's frame the atmosphere rushes past, hence contracted. $\gamma =\frac{1}{\sqrt{1-0.995^2}}=\frac{1}{\sqrt{1-0.990025}}=\frac{1}{\sqrt{0.009975}}\approx \mathrm{10.01.}$ $L=\frac{60}{10.01}\approx 5.99\text{ km}.$ Answer: ≈ 6.0 km. The muon "sees" only about a tenth of the thickness — that is how it reaches the ground before decaying.

$L=L_0/\gamma$ and $L=L_0/2$ means $L_0/\gamma =L_0/2$, so $\gamma =2$. Answer: $\gamma =2$. (For reference this happens at $v=\sqrt{3}/2c\approx 0.866c$.)

Here $L$ is given and $L_0$ is the unknown — we invert the formula: $L_0=\gamma L$. $\gamma =1/\sqrt{1-0.36}=1/0.8=1.25,L_0=1.25\times 0.72=0.90\text{ m}.$ Answer: 0.90 m. Note it is longer than the measured 0.72 m — proper length is the maximum ✔.

Step 1 — use the ratio form, since we know both lengths: $\frac{L}{L_0}=\sqrt{1-v^2/c^2}⟹0.28=\sqrt{1-v^2/c^2}.$ Step 2 — square both sides (to free $v^2/c^2$ from under the root): $0.0784=1-v^2/c^2⟹v^2/c^2=\mathrm{0.9216.}$ Step 3 — root: $v/c=\sqrt{0.9216}=0.96$. Answer: $v=0.96c$.

$\frac{L}{L_0}=\sqrt{1-0.01}=\sqrt{0.99}\approx \mathrm{0.99499.}$ Contraction $=1-0.99499=0.00501$, i.e. about $0.50\%$. Answer: ≈ $0.50\%$. Even at a tenth of light speed the effect is tiny — this is why we never notice it in daily life.

(a) $\gamma =1/\sqrt{1-0.64}=1/0.6=1.6\overline{6}$. So $L=200/1.6\overline{6}=120\text{ m}.$ (b) "Completely pass a point" means the whole measured length $L$ (the contracted 120 m — this is the length the station actually sees) sweeps past at speed $v$: $t=\frac{L}{v}=\frac{120}{0.8\times (3\times 10^8)}=\frac{120}{2.4\times 10^8}=5.0\times 10^{-7}\text{ s}.$ Answer: (a) 120 m, (b) $5.0\times 10^{-7}$ s ($0.50\mu$s). Key subtlety: we use the contracted length, not 200 m, because the station measures the moving ship.

$\gamma =1/\sqrt{1-0.98^2}=1/\sqrt{1-0.9604}=1/\sqrt{0.0396}\approx 5.025$. (a) Earth frame: distance 10 km, speed $0.98c$: $t_{\text{Earth}}=\frac{10000}{0.98\times 3\times 10^8}\approx 3.40\times 10^{-5}\text{ s}.$ (b) Muon frame: atmosphere contracts to $L=10/5.025\approx 1.99$ km: $t_{\text{muon}}=\frac{1990}{0.98\times 3\times 10^8}\approx 6.77\times 10^{-6}\text{ s}.$ (c) Ratio: $t_{\text{Earth}}/t_{\text{muon}}=3.40\times 10^{-5}/6.77\times 10^{-6}\approx 5.02=\gamma$ ✔. The two frames explain the muon's survival with the same $\gamma$ — Earth uses time dilation of the muon's clock, the muon uses length contraction of the atmosphere. Two views, one physics.

Only the edge along $v$ contracts; the other two stay $2.0$ m (perpendicular dimensions are untouched). $\gamma =1.25,a_{\parallel }=2.0/1.25=1.6\text{ m},a_{\perp }=2.0\text{ m (each)}.$ $V=1.6\times 2.0\times 2.0=6.4{\text{ m}}^3.$ Answer: $6.4$ m³ (rest volume was $8.0$ m³, reduced by the single factor $1/\gamma$).

$\gamma =1/\sqrt{1-0.866^2}=1/\sqrt{1-0.75}=1/\sqrt{0.25}=2.0$. (a) Barn frame: the pole is contracted to $L=10/2=5.0$ m, and $5.0<6$, so yes, momentarily the whole pole fits between the two doors. (b) Pole frame: now the barn moves, contracting to $6/2=3.0$ m, while the pole is a full $10$ m. Clearly $10>3$, so in this frame the pole does not fit. Resolution: "Fits" means both ends are inside at the same instant. The two frames disagree on which instant that is. In the barn frame the two door-events (front reaches far door, back passes near door) are simultaneous; in the pole frame they are not — the near door closes and reopens before the far end arrives. No physical contradiction: each frame's "fit" refers to a different pair of simultaneous events. This is Relativity of simultaneity doing all the work — the same root cause as contraction itself.

(a) As $v\to c$, $v^2/c^2\to 1$, so $\sqrt{1-v^2/c^2}\to 0$ and $\gamma \to \infty$. Therefore $L=L_0/\gamma \to 0$: the object contracts toward zero length (a flat "pancake"). Nothing with mass ever reaches $c$, so $L$ never actually hits 0. (b) As $v\to 0$, $\gamma \to 1$ and $L\to L_0$: full proper length, no contraction — the everyday world.

Each frame measures the same proper rod, so compute both independently. Frame A: ${\gamma }_A=1/0.8=1.25\Rightarrow L_A=12/1.25=9.6$ m. Frame B: ${\gamma }_B=1/0.6=1.6\overline{6}\Rightarrow L_B=12/1.6\overline{6}=7.2$ m. Frame $B$ (faster) measures the shorter rod, as expected — more speed, more contraction. Difference $=9.6-7.2=2.4$ m. Answer: B is shorter; they differ by 2.4 m.