Postulates of SR: Worked examples — Hinglish deep dive

2.3.26 · D3 · Physics › Modern Physics › Postulates of SR

Pehle, ek symbol jo hum baar baar use karenge. Neeche har jagah, $β$ (Greek "beta") sirf ek nickname hai:

The scenario matrix

Is chapter ka har relativity problem in cells mein se kisi ek mein aata hai. Neeche ke examples cell ke hisaab se labelled hain.

Cell	Kya special hai	Example
A — everyday slow speed	$β ≪ 1$ , $γ \approx 1$ ; Newton wapas aata hai	Ex 1
B — moderate speed	$β \sim 0.6$ , clean $γ$	Ex 2
C — length, time nahi	stretch ki jagah shrink (Length Contraction)	Ex 3
D — velocities forward	light + rocket, kya yeh $c$ exceed kar leta hai?	Ex 4
E — velocities backward	negative $v$ , sign handling	Ex 5
F — the $v \to c$ limit	degenerate: $γ \to \infty$	Ex 6
G — real-world word problem	muon ground tak pahunchta hai	Ex 7
H — exam twist / two effects	time aur length dono ka ek saath consistency	Ex 8

Cell A — everyday slow speed

Worked example Ex 1 (Cell A) — Ek jet ki clock

Ek jet $v = 300 m/s$ par fly karti hai (roughly Mach 0.9). Uske paas ek clock hai. Ground $1$ ghante $= 3600 s$ measure karne ke baad, jet ki clock kitni peeche hai?

Forecast: guess karo — nanoseconds? seconds? minutes?

$β$ compute karo. $β = v / c = 300/ (3 \times 1 0^{8}) = 1 0^{- 6}$ . Yeh step kyun? $β$ hi woh akela knob hai jis par $γ$ depend karta hai; pehle yahi nikalo.
$β^{2}$ compute karo. $β^{2} = 1 0^{- 12}$ . Kyun? $γ$ mein $β^{2}$ use hota hai, aur yeh itna chhota hai ki hum aage approximate kar sakte hain.
$γ$ approximate karo. Chhote $x$ ke liye, $\frac{1}{1 - x} \approx 1 + \frac{1}{2} x$ . Toh $γ \approx 1 + \frac{1}{2} (1 0^{- 12}) = 1 + 5 \times 1 0^{- 13}$ . Yeh step kyun? $1 0^{- 12}$ ko calculator mein daalne se $γ = 1.000 \dots$ milta hai aur tiny difference kho jaata hai. Linear approximation meaningful part ko preserve karti hai. Yahi exactly wajah hai ki Newton 200 saal tak kaam karta raha — correction 13th decimal mein chhupa hota hai.
Time gap. Ground ka time $γ$ factor se longer hota hai, toh lag hai $(γ - 1) \times Δ t_{0}$ . Yahan sabse clean tarika hai ki bolein jet ki apni clock $Δ t_{0} = Δ t / γ \approx 3600 (1 - 5 \times 1 0^{- 13})$ read karti hai, toh yeh $3600 \times 5 \times 1 0^{- 13} = 1.8 \times 1 0^{- 9} s$ peeche pad jaati hai. Kyun? $(γ - 1)$ fractional slowdown hai; elapsed time se multiply karo.

Answer: approximately $1.8 \times 1 0^{- 9} s = 1.8 nanoseconds$ peeche.

Verify: units — $[dimensionless] \times [s] = s$ ✓. Sanity — poore ek ghante mein nanoseconds, isliye tumhe kabhi notice nahi hota; relativity hamesha on rehti hai lekin $β = 1 0^{- 6}$ par invisible hoti hai.

Cell B — moderate speed, clean numbers

Cell C — length stretch ki jagah shrink hoti hai

Yeh woh geometry hai jo logon ko trip kara deti hai. Time stretch hota hai ( $\times γ$ ) lekin length shrink hoti hai ( $\div γ$ ). Figure dikhata hai kyun wahi $γ$ dono taraf kheenchta hai.

Worked example Ex 3 (Cell C) — Ek rod fly karta hai

Ek rod apne frame mein $L_{0} = 100 m$ lamba hai (Length Contraction, proper length). Yeh $v = 0.6 c$ par tumhare paas se guzarta hai. Tum ise kitna measure karoge?

Forecast: $100 m$ se lamba ya chhota?

$β = 0.6$ , $β^{2} = 0.36$ , $1 - β^{2} = 0.64$ , $= 0.8$ , $γ = 1.25$ . Kyun? Pehle wali hi ladder; $γ$ ka recipe kabhi nahi badalta.
Multiply nahi, divide karo. $L = L_{0} / γ = 100/1.25 = 80 m$ . Yahan divide kyun? Motion ki direction mein length contract hoti hai. Proper length $L_{0}$ sabse bada measure hota hai jo koi bhi karta hai; moving observer use squeezed dekhta hai. Figure mein blue bar dekho — wahi $γ$ , ulti direction.

Answer: $80 m$ .

Verify: $L < L_{0}$ ✓ (contraction, dilation nahi). Cross-check karo ki dono effects $γ$ share karte hain: time $\times 1.25$ jaata, length $\times (1/1.25) = 0.8$ — reciprocals, jaisa figure dikhata hai.

Cell D — velocities forward (kya light $c$ beat kar leti hai?)

Everyday velocities add hoti hain: $3$ m/s train par $2$ m/s chalte ho toh ground $5$ dekhta hai. Relativity uski jagah Relativistic Velocity Addition use karta hai:

Cell E — velocities backward (sign handling)

Formula kisi bhi sign ke liye kaam karta hai — bas minus saath le chalte ho. Backward motion matlab $v < 0$ ya $u < 0$ .

Worked example Ex 5 (Cell E) — Do rockets, opposite directions

Rocket A tumhare relative right mein $v = 0.8 c$ par move kar raha hai. Woh A ke frame mein peeche $u = - 0.5 c$ par ek probe fire karta hai. Tum probe ki kitni speed dekhte ho?

Forecast: $0.3 c$ (naive $0.8 - 0.5$ )? Ya kuch thoda alag?

Sign ke saath substitute karo. $u^{'} = \frac{- 0.5 c + 0.8 c}{1 + \frac{( - 0.5 c ) ( 0.8 c )}{c ^{2}}} = \frac{0.3 c}{1 - 0.40}$ . Denominator mein minus kyun rakha? $uv$ ka product ab negative hai, toh relativistic tax subtract karta hai — sign faithfully track karna zaroori hai.
Simplify karo. $= \frac{0.3 c}{0.60} = 0.5 c$ . Kyun? Arithmetic; $0.3/0.6 = 0.5$ .

Answer: $+ 0.5 c$ (abhi bhi right mein move kar raha hai, bas A se slower).

Verify: result magnitude $0.5 c < c$ ✓ (light-speed kabhi exceed nahi hoti). Galilean guess $0.3 c$ tha; relativistic answer $0.5 c$ alag hai — proof ki minus sign ke saath bhi formula real kaam karta hai.

Cell F — degenerate limit $v \to c$

Light-speed par kya hota hai? Figure mein $γ$ wall ki taraf chadhta hua plot kiya gaya hai.

Worked example Ex 6 (Cell F) — Light-speed ki taraf push karna

$β = 0.9$ par $γ$ compute karo, phir $0.99$ , phir $0.999$ . $β \to 1$ hone par pattern kya hai?

Forecast: kya $γ$ level off hoga, ya blow up karega?

$β = 0.9$ : $1 - 0.81 = 0.19$ , $= 0.436$ , $γ = 2.29$ . Kyun? Wahi ladder; root ko shrink hote dekho.
$β = 0.99$ : $1 - 0.9801 = 0.0199$ , $= 0.141$ , $γ = 7.09$ . Kyun? Root abhi chhota ho gaya, toh uska reciprocal jump karta hai.
$β = 0.999$ : $1 - 0.998 = 0.001999$ , $= 0.0447$ , $γ = 22.4$ . Kyun? Jab $β^{2} \to 1$ , root ke neeche wali number $\to 0$ , toh $γ \to \infty$ .

Answer: $γ \to \infty$ jab $β \to 1$ . Time dilation infinite ho jaata hai; $c$ tak pahunchne ke liye infinite energy chahiye (Mass-Energy Equivalence). Isliye massive objects kabhi light-speed tak nahi pahunch sakte — $β = 1$ par wall unreachable hai. Figure mein red asymptote dekho.

Verify: exactly $β = 1$ par, $1 - β^{2} = 0$ aur $γ = 1/0$ undefined hai — degenerate case genuinely forbidden hai, postulates mein " $<$ " se consistent.

Cell G — real-world word problem

Worked example Ex 7 (Cell G) — Woh muon jo survive nahi karna chahiye tha

Muons atmosphere mein $\sim 10 km$ upar bante hain, $v = 0.98 c$ par neeche aate hain, aur apne frame mein sirf $Δ t_{0} = 2.2 μ s$ jeete hain. Bina relativity ke, woh kitni door jaate hain? Time dilation ke saath, kya woh ground tak pahunchte hain?

Forecast: kya woh poore $10$ km tak pahunchenge?

Naive distance (no relativity). $d = v Δ t_{0} = 0.98 \times (3 \times 1 0^{8}) \times (2.2 \times 1 0^{- 6}) \approx 647 m$ . Kyun? Classical distance = speed × lifetime. Sirf $0.65$ km — woh upar hi mar jaate.
$γ$ compute karo. $β = 0.98$ , $β^{2} = 0.9604$ , $1 - β^{2} = 0.0396$ , $= 0.199$ , $γ \approx 5.03$ . Kyun? Lab muon ki clock ko slow chalte dekhti hai, uski life extend hoti hai.
Hamara frame mein dilated lifetime. $Δ t = γ Δ t_{0} = 5.03 \times 2.2 = 11.06 μ s$ . $\times$ kyun? $2.2 μ s$ proper time hai; ground frame zyada measure karta hai.
Dilation ke saath distance. $d = v Δ t = 0.98 \times (3 \times 1 0^{8}) \times (11.06 \times 1 0^{- 6}) \approx 3250 m \approx 3.25 km$ . Kyun? Wahi distance formula, lekin dilated lifetime ke saath.

Answer: classically $\approx 0.65$ km (fail); relativistically $\approx 3.25$ km — kaafi saare detectors tak pahunchte hain, aur yeh measured hai. Time dilation ka real experimental proof.

Verify: $3250/647 = γ$ ? $3250/647 \approx 5.02 \approx γ$ ✓ — extra reach exactly $γ$ factor hai.

Cell H — exam twist: dono effects agree karne chahiye

Worked example Ex 8 (Cell H) — Wahi journey, do viewpoints

Ek ship $L_{0} = 6 \times 1 0^{9} m$ ka ek lane cross karti hai (ground dwara measured) $v = 0.6 c$ par. (a) Ground clock: crossing kitni der mein hoti hai? (b) Ship crew ki clock: unke liye kitna time? Dikhao ki dono views consistent hain.

Forecast: kaun si clock kam read karegi — ground ki ya ship ki?

Ground time. $t = L_{0} / v = \frac{6 \times 1 0 ^{9}}{0.6 \times 3 \times 1 0 ^{8}} = \frac{6 \times 1 0 ^{9}}{1.8 \times 1 0 ^{8}} = 33.3 s$ . Kyun? Ground poori lane length aur apna time measure karta hai — plain distance/speed.
$γ$ . $β = 0.6 \Rightarrow γ = 1.25$ (Ex 3 se). Kyun? Frames ke beech convert karne ke liye chahiye.
Dilation se ship time. Crew ek hi jagah ek clock hai → unka crossing time proper time hai: $t_{ship} = t / γ = 33.3/1.25 = 26.7 s$ . Divide kyun? Moving crew ki clock ground ke hisaab se slow chalti hai; equivalently unka apna elapsed time chhota hota hai.
Length contraction se cross-check. Crew lane ko contracted dekhti hai: $L = L_{0} / γ = 6 \times 1 0^{9} /1.25 = 4.8 \times 1 0^{9} m$ . Unka time = $L / v = 4.8 \times 1 0^{9} / (1.8 \times 1 0^{8}) = 26.7 s$ . Yeh kyun kiya? Do alag effects (dilation vs contraction) wahi ship time dene chahiye. Dete hain — yahi relativity ki internal consistency hai.

Answer: ground $33.3 s$ , ship $26.7 s$ ; dono methods agree karte hain.

Verify: $33.3/26.7 = 1.25 = γ$ ✓, aur length- aur time-based ship computations dono $26.7 s$ par match karte hain ✓.

Recall Quick self-test

Ek rod $β = 0.6$ par $50$ m proper length ka hai; tum kya length measure karoge? ::: $50/1.25 = 40$ m Ek clock $β = 0.8$ par $4 μ s$ proper read karti hai; lab kya read karega? ::: $γ = 5/3$ , toh $6.67 μ s$ $0.99 c$ ship se fire ki gayi light — tumhare liye uski speed? ::: exactly $c$ Jab $β \to 1$ , $γ \to$ ? ::: $\infty$

Connections

Parent: Postulates of SR
Time Dilation · Length Contraction · Lorentz Transformations · Relativistic Velocity Addition · Michelson-Morley Experiment · Mass-Energy Equivalence · Galilean Relativity