Exercises — Time dilation — derivation, twin paradox

WHAT: $\beta =0.8$ ko formula mein plug karo. WHY: $\gamma$ woh single number hai jo batata hai ki time kitna stretch hoga; humein yeh pehle chahiye hota hai. $\gamma =\frac{1}{\sqrt{1-0.8^2}}=\frac{1}{\sqrt{1-0.64}}=\frac{1}{\sqrt{0.36}}=\frac{1}{0.6}\approx \mathrm{1.667.}$ Answer: $\gamma \approx 1.667$.

WHAT: Identify karo ki dono events (trip ka start, trip ka end) ek hi jagah par kahan hote hain. WHY: Proper time woh interval hai jo ek single clock measure karta hai jo dono events par present ho — yahan, wristwatch ship ke saath travel karti hai, isliye "start" aur "end" dono us watch ke paas hi hote hain. Wristwatch ki reading, $5$ saal, proper time $\Delta t_0$ hai. Earth ka $8$ saal dilated $\Delta t$ hai. Check: proper time chhota wala hota hai, aur $5<8$. ✓ Answer: $\Delta t_0=5$ saal.

WHAT: $\gamma$ nikalo, phir multiply karo. $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{0.64}}=\mathrm{1.25.}$ WHY multiply? Lab moving particle ki clock ko slow run hote dekhta hai, isliye uski lifetime longer lagti hai. Longer = $\gamma \ge 1$ se multiply karo. $\Delta t=\gamma \Delta t_0=1.25\times 2.6\times 10^{-8}\text{s}=3.25\times 10^{-8}\text{s}.$ Answer: $\Delta t\approx 3.25\times 10^{-8}\text{s}$.

WHAT: Events (Earth chhodi, pahunche) dono ship par hain, isliye ship proper time $\Delta t_0$ measure karti hai; Earth dilated $\Delta t=10$ yr measure karta hai. $\gamma =\frac{1}{\sqrt{1-0.36}}=\mathrm{1.25.}$ WHY divide? Hum dilated time jaante hain aur proper time chahiye, isliye $\Delta t=\gamma \Delta t_0$ ko invert karo: $\Delta t_0=\frac{\Delta t}{\gamma }=\frac{10}{1.25}=8\text{years}.$ Answer: Ship par $8$ saal.

WHAT: $\gamma =2$ ko $v$ ke liye solve karo. WHY is tarah invert karte hain? Yahan jis cheez ka jawab chahiye woh speed hai, lekin speed $\gamma$ ke denominator mein ek square root ke andar chhipi hai. Toh hum formula ko ek operation at a time peel karte hain — reciprocal, phir square, phir ${\beta }^2$ isolate — har step ko undo karte hue jo $\beta$ ke around wrap hai. Physically hum pooch rahe hain: "clock ko exactly half rate par tick karne ke liye kitni fast fly karni chahiye?" — speed unknown hai, $\gamma$ given hai. $2=\frac{1}{\sqrt{1-{\beta }^2}}\Rightarrow \sqrt{1-{\beta }^2}=\frac{1}{2}\Rightarrow 1-{\beta }^2=\frac{1}{4}\Rightarrow {\beta }^2=\frac{3}{4}.$ $\beta =\frac{\sqrt{3}}{2}\approx \mathrm{0.866.}$ Answer: $v\approx 0.866c$. (Note karo: tick-time sirf double karne ke liye bhi tumhe already light speed ka $87\%$ chahiye — dilation ek high-speed effect hai.)

WHAT (a): Naive distance $=v\Delta t_0$. WHY (a): Hum yeh deliberately galat compute karte hain pehle — yeh assume karte hue (jaise Newton karta) ki muon ki khud ki $2.2\mu s$ lifetime ground frame mein bhi available time hai. Yeh us galat assumption par rely karta hai ki time universal hai. Point yeh hai ki contradiction expose karo: yeh baseline woh hai jo relativity ko fix karni chahiye. $d_{\text{naive}}=0.98\times (3\times 10^8)\times (2.2\times 10^{-6})\approx 6.47\times 10^2\text{m}\approx 0.65\text{km}.$ Yeh $10$ km se kaafi kam hai — toh classically muon humtak kabhi nahi pahunchna chahiye. WHAT (b): $\gamma$ nikalo, lifetime dilate karo, distance recompute karo. WHY (b): Muon ki clock ground frame mein slow run karti hai, isliye ground se muon $\gamma$ times zyada jeeta lagta hai, jisse use distance cover karne ke liye kaafi zyada time milta hai. $\gamma =\frac{1}{\sqrt{1-0.98^2}}=\frac{1}{\sqrt{1-0.9604}}=\frac{1}{\sqrt{0.0396}}\approx \mathrm{5.025.}$ $\Delta t=\gamma \Delta t_0=5.025\times 2.2\mu s\approx 11.05\mu s.$ $d=v\Delta t=0.98\times (3\times 10^8)\times (11.05\times 10^{-6})\approx 3.25\times 10^3\text{m}\approx 3.25\text{km}.$ WHY it matters: Dilation ke saath muon kai km cover karta hai, isliye ek badi fraction actually ground tak pahunchti hai — yeh classic experimental proof hai (dekho Muon Decay Experiment). Answer: (a) $\approx 0.65$ km; (b) $\Delta t\approx 11.05\mu s$, $d\approx 3.25$ km.

WHAT: Muon ke frame mein woh rest mein hai aur $10$ km ka atmosphere $\gamma$ se contracted hokar rush karta hai (dekho Length Contraction). $L=\frac{L_0}{\gamma }=\frac{10\text{km}}{5.025}\approx 1.99\text{km}.$ WHY: Muon proper time $2.2\mu s$ experience karta hai; us time mein contracted distance $0.98c$ par past fly karti hai: $d_{\text{muon}}=v\Delta t_0=0.98\times (3\times 10^8)\times (2.2\times 10^{-6})\approx 0.647\text{km}.$ Contracted atmosphere thickness $1.99$ km se compare karte hue, muon ek fraction $0.647/1.99\approx 0.325$ cover karta hai — wahi fraction jo ground observer ne compute ki thi ($3.25$ km of $10$ km $=0.325$). Dono frames physical outcome par agree karte hain. ✓ Answer: contracted distance $\approx 1.99$ km; muon usme se $\approx 0.65$ km cross karta hai — ground calculation ke saath consistent.

WHAT (a): Round-trip distance $=12\text{ly}$ at $0.6c$: $\Delta t_A=\frac{12\text{ly}}{0.6\text{ly/yr}}=20\text{years}.$ WHY (a): Time = distance ÷ speed. Yeh Twin A ka apna frame hai, jahan A rest mein hai, isliye A bas ship ko fixed $12\text{ly}$ path $0.6\text{ly/yr}$ par cover karte dekhta hai — ek ordinary "distance over speed" bilkul kisi bhi everyday travel-time ki tarah, kyunki A kabhi frames nahi change karta. WHAT (b): $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{0.64}}=1.25$. Phir $\Delta t_B=\frac{\Delta t_A}{\gamma }=\frac{20}{1.25}=16\text{years}.$ WHY (b): B ki clock B ke saath ride karti hai, isliye dono trip-events (Earth chhodi, Earth par vapas aaye) us clock ke paas hi hote hain — yeh proper time measure karta hai, chhota wala. Hum dilated Earth time $\Delta t_A$ jaante hain aur proper time chahiye, isliye hum $\gamma$ se divide karte hain (jaise L2·Q2 mein). WHAT (c): $\Delta t_A-\Delta t_B=20-16=4\text{years}.$ WHY (c): Dono twins end mein ek hi jagah (Earth) par phir milte hain, isliye hum directly unke logged times subtract kar sakte hain real, permanent age gap nikalne ke liye. Sirf aisi reunion hi age comparison meaningful banati hai. WHY asymmetric: Sirf B star par turn around karta hai (inertial frame change karta hai, force feel karta hai). deepdives/dd-physics-2.3.29-d4-s02.png dekho — B ka world-path bent hai, A ka straight hai, aur spacetime mein straight path woh hai jo sabse zyada time clock karta hai. Answer: (a) $20$ yr; (b) $16$ yr; (c) B $4$ saal younger hai.

WHAT: B ke frame mein $6\text{ly}$ ka gap contracted hai: $L=\frac{6\text{ly}}{1.25}=4.8\text{ly}.$ WHY: B still baitha hai jabki star contracted $4.8\text{ly}$ ke across $0.6c$ par uski taraf rush karta hai: $t_{\text{out}}=\frac{4.8\text{ly}}{0.6\text{ly/yr}}=8\text{years}.$ Yeh exactly $\Delta t_B=16$ yr ka aadha hai. ✓ Length contraction aur time dilation ek hi Lorentz Transformations ke do faces hain. Answer: $8$ saal outbound, $16/2$ se match karta hua.

WHAT (a): Yahan hum do speeds combine karte hain: $v_1=0.8c$ (ship P past Earth) aur $v_2=0.5c$ (probe past ship P). Hum $u$ chahiye, probe ki speed Earth ke relative mein. Newtonian addition dega $v_1+v_2=0.8c+0.5c=1.3c>c$ — impossible. Relativistic rule use karo (dekho Relativistic Velocity Addition): $u=\frac{v_1+v_2}{1+\frac{v_1{v}_2}{c^2}}=\frac{0.8c+0.5c}{1+(0.8)(0.5)}=\frac{1.3c}{1.4}\approx 0.9286c.$ WHY: Denominator $1+v_1{v}_2/c^2$ result ko $c$ se neeche rakhta hai — koi bhi light se aage nahi nikal sakta. WHAT (b): Probe ki Earth-frame speed pata hone par, $\beta =0.9286$: $\gamma =\frac{1}{\sqrt{1-0.9286^2}}=\frac{1}{\sqrt{1-0.8623}}=\frac{1}{\sqrt{0.1377}}\approx \mathrm{2.695.}$ WHY (b): $\gamma$ hamesha us observer ke relative speed use karta hai jiske clock ke baare mein hum poochh rahe hain. Hum jaanna chahte hain ki probe ki clock Earth se dekhe jaane par kitni slow run karti hai, isliye hum probe ki Earth-frame speed $u$ feed karni hogi — $v_1$ ya $v_2$ nahi, jo doosre frames ke relative speeds hain. Sirf $u$ answer karta hai "Earth probe ka time kitna stretch hota hua dekhta hai?" Answer: (a) $\approx 0.929c$; (b) $\gamma \approx 2.70$.

WHAT (a): $v=0\Rightarrow \beta =0\Rightarrow \gamma =1/\sqrt{1-0}=1$, isliye $\Delta t=\Delta t_0$. WHY (a): Ek clock jo tumhare relative rest mein hai uske paas koi diagonal light-path nahi hai lengthen hone ke liye — photon bas seedha upar neeche jaata hai jaise clock ke apne frame mein. Extra path nahi toh extra time nahi, isliye dilated time proper time par collapse ho jaata hai. Yeh sanity check hai ki relativity ordinary Newtonian time ko apne slow-speed limit ke roop mein contain karta hai. WHAT (b): Jaise $v\to c$, $\beta \to 1$, isliye $1-{\beta }^2\to 0^{+}$ aur $\gamma \to \infty$. Phir $\Delta t=\gamma \Delta t_0\to \infty$. WHY (b): Physically, moving clock ki ek tick infinite observer-time legi — clock (aur uske upar har process) frozen appear hoga. Yahi reason hai ki koi massive object kabhi $c$ tak pahunch nahi sakta: aisa karne ke liye infinite time-stretch ki demand hogi, jo badle mein infinite energy require karta hai. Light khud koi proper time experience nahi karti. deepdives/dd-physics-2.3.29-d4-s01.png ka steep right edge phir se dekho. Answer: (a) $\gamma =1$, koi dilation nahi; (b) $\gamma \to \infty$, clock frozen appear hota hai.

WHAT: Chhote $\beta$ ke liye, expand karo $\gamma =(1-{\beta }^2{)}^{-1/2}\approx 1+\frac{1}{2}{\beta }^2$. WHY expansion? Everyday speeds par ${\beta }^2$ bahut chhota hota hai; ek linear (Taylor) approximation kaafi accurate hai aur catastrophic rounding se bachata hai. $\frac{1}{2}{\beta }^2=10^{-12}\Rightarrow {\beta }^2=2\times 10^{-12}\Rightarrow \beta =\sqrt{2\times 10^{-12}}\approx 1.414\times 10^{-6}.$ $v=\beta c\approx 1.414\times 10^{-6}\times 3\times 10^8\approx 424\text{m/s}.$ WHY it matters: Yeh ek jet-airliner ki speed hai — real flying-clock experiments exactly is level par dilation confirm karte hain. Answer: $v\approx 424\text{m/s}$.