Exercises — Length contraction — derivation

(a). Proper length $L_0$ us frame mein measure hoti hai jahan object rest mein ho. Astronaut stick pakde hua hai, toh stick unke frame mein rest mein hai — woh $L_0$ measure karte hain. Ground observer ise move karte dekhta hai, isliye woh contracted $L<L_0$ measure karta hai.

Jhooth. Sirf woh dimension jo motion ki direction mein hai contract karti hai. $v$ ke perpendicular dono dimensions unchanged rehti hain. Ek cube jo apne $x$-axis ke saath move kar raha hai ek rectangular box ban jaata hai, sirf $x$ mein squash hota hai.

Factor 1 (koi contraction nahi). $v=0$ par, $\gamma =1/\sqrt{1-0}=1$, toh $L=L_0/1=L_0$. Tumhare frame mein rest mein rakha object apni poori proper length par measure hota hai.

Step 1 — WHY pehle $\gamma$ nikalo: contraction hai $L=L_0/\gamma$, toh shrink karne se pehle $\gamma$ chahiye. $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{1-0.36}}=\frac{1}{\sqrt{0.64}}=\frac{1}{0.8}=\mathrm{1.25.}$ Step 2 — divide karo (Earth moving ship dekhta hai, isliye woh chhoti wali hai): $L=\frac{150}{1.25}=120\text{ m}.$ Answer: 120 m. Sanity check: 150 se chhota ✔.

Atmosphere kis frame mein rest mein hai? Earth's — toh 60 km proper length hai. Muon ke frame mein atmosphere rush karke past hoti hai, isliye contracted hai. $\gamma =\frac{1}{\sqrt{1-0.995^2}}=\frac{1}{\sqrt{1-0.990025}}=\frac{1}{\sqrt{0.009975}}\approx \mathrm{10.01.}$ $L=\frac{60}{10.01}\approx 5.99\text{ km}.$ Answer: ≈ 6.0 km. Muon sirf lagbhag ek-daswa hissa thickness "dekhta" hai — isi tarah woh decay hone se pehle ground tak pahunch jaata hai.

$L=L_0/\gamma$ aur $L=L_0/2$ ka matlab hai $L_0/\gamma =L_0/2$, toh $\gamma =2$. Answer: $\gamma =2$. (Reference ke liye yeh $v=\sqrt{3}/2c\approx 0.866c$ par hota hai.)

Yahan $L$ diya hai aur $L_0$ unknown hai — hum formula invert karte hain: $L_0=\gamma L$. $\gamma =1/\sqrt{1-0.36}=1/0.8=1.25,L_0=1.25\times 0.72=0.90\text{ m}.$ Answer: 0.90 m. Note karo yeh measured 0.72 m se zyada hai — proper length maximum hoti hai ✔.

Step 1 — ratio form use karo, kyunki hum dono lengths jaante hain: $\frac{L}{L_0}=\sqrt{1-v^2/c^2}⟹0.28=\sqrt{1-v^2/c^2}.$ Step 2 — dono sides square karo ($v^2/c^2$ ko root ke neeche se nikaalne ke liye): $0.0784=1-v^2/c^2⟹v^2/c^2=\mathrm{0.9216.}$ 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 $=1-0.99499=0.00501$, yaani lagbhag $0.50\%$. Answer: ≈ $0.50\%$. Light speed ke daswe hisse par bhi effect bahut chota hai — isliye hamari daily life mein hum ise kabhi notice nahi karte.

(a) $\gamma =1/\sqrt{1-0.64}=1/0.6=1.6\overline{6}$. Toh $L=200/1.6\overline{6}=120\text{ m}.$ (b) "Ek point se completely pass karna" matlab hai poori measured length $L$ (contracted 120 m — yahi woh length hai jo station actually dekhta hai) speed $v$ par sweep karti hai: $t=\frac{L}{v}=\frac{120}{0.8\times (3\times 10^8)}=\frac{120}{2.4\times 10^8}=5.0\times 10^{-7}\text{ s}.$ Answer: (a) 120 m, (b) $5.0\times 10^{-7}$ s ($0.50\mu$s). Key subtlety: hum contracted length use karte hain, 200 m nahi, kyunki station moving ship measure karta hai.

$\gamma =1/\sqrt{1-0.98^2}=1/\sqrt{1-0.9604}=1/\sqrt{0.0396}\approx 5.025$. (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 contract hokar $L=10/5.025\approx 1.99$ km ho jaati hai: $t_{\text{muon}}=\frac{1990}{0.98\times 3\times 10^8}\approx 6.77\times 10^{-6}\text{ s}.$ (c) Ratio: $t_{\text{Earth}}/t_{\text{muon}}=3.40\times 10^{-5}/6.77\times 10^{-6}\approx 5.02=\gamma$ ✔. Dono frames muon ki survival ko same $\gamma$ se explain karte hain — Earth muon ki clock ka time dilation use karta hai, muon atmosphere ka length contraction use karta hai. Do views, ek physics.

Sirf $v$ ke saath wali edge contract karti hai; baaki do $2.0$ m rehti hain (perpendicular dimensions unchanged hoti hain). $\gamma =1.25,a_{\parallel }=2.0/1.25=1.6\text{ m},a_{\perp }=2.0\text{ m (each)}.$ $V=1.6\times 2.0\times 2.0=6.4{\text{ m}}^3.$ Answer: $6.4$ m³ (rest volume $8.0$ m³ tha, sirf ek factor $1/\gamma$ se kam hua).

$\gamma =1/\sqrt{1-0.866^2}=1/\sqrt{1-0.75}=1/\sqrt{0.25}=2.0$. (a) Barn frame: pole contract hokar $L=10/2=5.0$ m ho jaati hai, aur $5.0<6$, toh haan, momentarily poori pole dono doors ke beech fit ho jaati hai. (b) Pole frame: ab barn move kar rahi hai, contract hokar $6/2=3.0$ m ho jaati hai, jabki pole poori $10$ m hai. Clearly $10>3$, toh is frame mein pole fit nahi hoti. Resolution: "Fit" ka matlab hai dono ends ek hi waqt andar hain. Dono frames is baat par disagree karte hain ki woh instant kaunsi hai. Barn frame mein dono door-events (front far door tak pahuncha, back near door se guzri) simultaneous hain; pole frame mein yeh nahi hain — near door close hoti hai aur reopen karti hai usse pehle ki far end pahunche. Koi physical contradiction nahi: har frame ka "fit" events ke alag simultaneous pairs refer karta hai. Yeh Relativity of simultaneity hi saara kaam kar rahi hai — contraction ka wahi root cause.

(a) Jaise $v\to c$, $v^2/c^2\to 1$, toh $\sqrt{1-v^2/c^2}\to 0$ aur $\gamma \to \infty$. Isliye $L=L_0/\gamma \to 0$: object zero length ki taraf contract hota hai (ek flat "pancake"). Mass wali koi cheez kabhi $c$ tak nahi pahunchti, toh $L$ kabhi actually 0 nahi hota. (b) Jaise $v\to 0$, $\gamma \to 1$ aur $L\to L_0$: poori proper length, koi contraction nahi — everyday world.

Har frame same proper rod measure karta hai, toh dono independently calculate karo. 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) chhoti rod measure karta hai, jaise expected hai — zyada speed, zyada contraction. Difference $=9.6-7.2=2.4$ m. Answer: B chhota hai; dono mein 2.4 m ka farq hai.