Time dilation — derivation, twin paradox: Worked examples

2.3.29 · D3 · Physics › Modern Physics › Time dilation — derivation, twin paradox

Shuru karne se pehle, symbols yaad kar lo taaki kuch bhi bina samjhe use na ho:

Definition Do times aur speed ratio (re-anchored)

$c$ = speed of light in vacuum, lagbhag $3.0 \times 1 0^{8}$ metres per second. Yeh har observer ke liye same number hai (postulate 2) — wahi constancy poori wajah hai time dilation ki.
$Δ t_{0}$ = proper time — do events ke beech ka woh time jo ek aisi clock se measure hota hai jo dono events par present ho (woh usi clock ke frame mein same place par hote hain). Socho: jo cheez move kar rahi hai uska personal time.
$Δ t$ = woh time jo doosra observer un same do events ke liye measure karta hai, clock ko fly past karte dekhte hue.
$β$ (Greek "beta") $= \frac{v}{c}$ speed as a fraction of light-speed hai. Toh $β = 0.6$ matlab "light-speed ke $60%$ par move kar raha hai". Yeh sirf $v / c$ ko ek tidy symbol mein pack karta hai jo neeche har jagah dikhta hai.
$γ = \frac{1}{1 - v ^{2} / c ^{2}} = \frac{1}{1 - β ^{2}} \geq 1$ "stretch factor" hai. Kyunki $γ \geq 1$ , hamesha $Δ t \geq Δ t_{0}$ hoga: moving clock slowly ticks karta hai watcher ki clock se.

Scenario matrix

Har time-dilation question aslaan inhi cells mein se ek hota hai. Neeche ke examples mein cell label kiya gaya hai.

Cell	Kya vary hota hai / trap kya hai	Kaun sa example
A. $Δ t_{0}$ se $Δ t$ nikalna	moving clock ka apna time pata hai, watcher ka chahiye	Ex 1
B. $Δ t$ se $Δ t_{0}$ nikalna	watcher ka time pata hai, moving clock ka chahiye	Ex 2
C. Do times se $v$ (ya $γ$ ) nikalna	inverse problem — speed ke liye solve karo	Ex 3
D. Degenerate: $v = 0$	clock at rest → no dilation	Ex 4
E. Limiting: $v \to c$	$γ \to \infty$ , time freeze ho jaata hai	Ex 4
F. Low-speed check: $v ≪ c$	$γ \approx 1$ , Newton wapas aata hai	Ex 4
G. Real-world word problem	identify karo kaunsa frame $Δ t_{0}$ hold karta hai	Ex 5 (muon), Ex 6 (GPS-style)
H. Round-trip / twin twist	asymmetry, kaun turn around karta hai	Ex 7
I. Exam trap: wrong variable	watcher-time ko $Δ t_{0}$ mein plug karna	Ex 8

"Same-place" test sabse aasaan worldline diagram par dikhta hai — ek picture jisme position across (space) hai aur time upar, jahan ek clock line trace karti hai jaise woh move karti hai. Examples se pehle yeh study karo.

Figure — Time dilation — derivation, twin paradox

Cell A — moving clock ka time pata hai, watcher ka nikalo

Worked example Example 1 — spaceship pilot ki stopwatch (Cell A)

Ek pilot ki onboard stopwatch $Δ t_{0} = 4.0$ years measure karti hai $v = 0.6 c$ ( $β = 0.6$ ) par journey ke liye. Earth par usi journey ke liye kitne saal guzarte hain?

Forecast: Earth par zyada saal honge ya kam? (Guess: pilot moving clock hai, toh Earth zyada dekhegi — moving clock slow chali, toh usne kam log kiya.)

$γ$ compute karo: $γ = \frac{1}{1 - β ^{2}} = \frac{1}{1 - 0.36} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25$ . Yeh step kyun? $γ$ dono times ke beech bridge hai; hum ise hamesha pehle $β$ se build karte hain.
$Δ t_{0}$ identify karo: stopwatch ship par hai, departure aur arrival dono events par present — ship ke frame mein same place. Toh $Δ t_{0} = 4.0$ yr. Yeh step kyun? Yeh "same place" test hai — yeh decide karta hai kaun sa number $Δ t_{0}$ hai.
Multiply karo: $Δ t = γ Δ t_{0} = 1.25 \times 4.0 = 5.0$ years on Earth. Yeh step kyun? Hum watcher ka (Earth ka) time chahte hain, aur $Δ t = γ Δ t_{0}$ proper time ko enlarge karta hai.

Verify: $Δ t = 5.0 > Δ t_{0} = 4.0$ ✓ (moving clock slow chali, toh Earth ki clock ne zyada log kiya). Units: years in, years out ✓.

Cell B — watcher ka time pata hai, moving clock ka nikalo

Worked example Example 2 — Earth clock diya hua hai, ship clock chahiye (Cell B)

Earth $Δ t = 10$ years measure karta hai ek probe ke liye jo $v = 0.8 c$ ( $β = 0.8$ ) par fly kar raha hai. Probe ki apni clock par kitna time guzarta hai?

Forecast: (Guess: probe moving clock hai, toh use 10 saal se kam log karna chahiye.)

$γ = \frac{1}{1 - β ^{2}} = \frac{1}{1 - 0.64} = \frac{1}{0.36} = \frac{1}{0.6} = 1.6 \overline{6}$ . Yeh step kyun? Hamesha ki tarah same bridge factor.
Do events (launch, arrival) probe par hote hain → probe ki clock dono par present hai → probe $Δ t_{0}$ hold karta hai. Yeh step kyun? Phir se same-place test: yeh batata hai ki hum $Δ t_{0}$ ke liye solve karna chahiye, multiply nahi karna.
$Δ t = γ Δ t_{0}$ ko rearrange karke $Δ t_{0} = \frac{Δ t}{γ} = \frac{10}{1.6 6} = 6.0$ years milo. Yeh step kyun? Yahan watcher's time known quantity hai, toh proper time par wapas shrink karne ke liye $γ$ se divide karte hain.

Verify: $6.0 < 10$ ✓ (probe ne kam log kiya). Consistency check: $γ \times 6 = 1.6 \overline{6} \times 6 = 10$ ✓.

Cell C — inverse problem, speed ke liye solve karo

Worked example Example 3 — woh kitni tez ja rahe the? (Cell C)

Ek crew $Δ t_{0} = 3.0$ years log karti hai jabki mission control $Δ t = 5.0$ years log karta hai. Speed $v$ nikalo.

Forecast: (Guess: times kaafi alag hain, toh $v$ $c$ ka bada fraction hoga.)

Ratio banao: $γ = \frac{Δ t}{Δ t _{0}} = \frac{5.0}{3.0} = 1.6 \overline{6}$ . Yeh step kyun? Dono times known hain, aur unka ratio seedha $γ$ hai — abhi $v$ ki zaroorat nahi.
$γ$ ki definition undo karo. $γ = \frac{1}{1 - β ^{2}}$ se start karo, dono sides square karo: $γ^{2} = \frac{1}{1 - β ^{2}}$ . Yeh step kyun? Unknown $β$ ek square root ke andar trap hai; squaring ise free karta hai.
$β^{2}$ ke liye solve karo: $1 - β^{2} = \frac{1}{γ ^{2}} \Rightarrow β^{2} = 1 - \frac{1}{γ ^{2}}$ . Toh $v = c 1 - 1/ γ^{2}$ . Yeh step kyun? Algebraic isolation — hum $v$ akela chahte hain.
Plug in karo: $\frac{1}{γ ^{2}} = \frac{1}{( 5/3 ) ^{2}} = \frac{9}{25} = 0.36$ , toh $β^{2} = 1 - 0.36 = 0.64$ , giving $β = 0.8$ , yaani $v = 0.8 c$ . Yeh step kyun? Nice numbers confirm karte hain ki humne ek "movie-friendly" speed choose ki.

Verify: $v = 0.8 c$ par, $γ = 1/ 1 - 0.64 = 1/0.6 = 1.6 \overline{6}$ , aur $1.6 \overline{6} \times 3 = 5$ ✓.

Cells D, E, F — teen limits (ek figure, teen readings)

$γ$ ka poora behaviour ek curve mein hai. Ise teen tarike se padho.

Worked example Example 4 — teen limiting cases (Cells D, E, F)

Teen speeds par $γ$ (aur hence moving clock ke saath kya hota hai) evaluate karo: $v = 0$ , $v \to c$ , aur $v = 0.01 c$ .

Forecast: (Guess: rest par kuch nahi hoga; light-speed ke paas time almost ruk jaayega; slow speed par almost kuch nahi hoga.)

Cell D — $v = 0$ (clock at rest):

$γ = \frac{1}{1 - 0} = 1$ . Toh $Δ t = 1 \cdot Δ t_{0} = Δ t_{0}$ . Yeh step kyun? Tumhare relative rest par ek clock tumhari apni clock hai — koi stretching nahi. Curve ke left end par white dot dekho: $γ$ flat $1$ se shuru hota hai.

Cell E — $v \to c$ (light speed ke paas pahunchna): 2. Jab $v \to c$ , $β^{2} \to 1$ , toh $1 - β^{2} \to 0^{+}$ , aur $γ = 1/ tiny \to \infty$ . Phir $Δ t = γ Δ t_{0} \to \infty$ : moving clock ki ek tick tumhare liye forever leti hai — time frozen dikhta hai. Yeh step kyun? Yeh curve ki amber wall hai — $v = c$ par vertical asymptote. Yahi wajah hai ki mass wali koi cheez $c$ nahi pahunchti: infinite time-stretch chahiye hogi.

Cell F — $v = 0.01 c$ (everyday speed): 3. $γ = 1/ 1 - 1 0^{- 4} \approx 1 + \frac{1}{2} (1 0^{- 4}) = 1.00005$ . Yeh step kyun? Approximation $\frac{1}{1 - x} \approx 1 + \frac{x}{2}$ use karo tiny $x$ ke liye. Yeh curve ki near-flat lower shelf hai — $5$ parts per hundred thousand ki dilation, bilkul invisible.

Verify: $γ (0) = 1$ exactly ✓; $γ$ $v \to c$ par unbounded badhta hai ✓; $γ (0.01 c) \approx 1.00005$ exact $1/ 0.9999 = 1.0000500 \dots$ se match karta hai ✓.

Recall Tiny

$x$ ke liye $\frac{1}{1 - x} \approx 1 + \frac{x}{2}$ kyun? (Ex 4 aur Ex 6 mein use hua) Answer ::: Yeh binomial approximation $(1 + u)^{n} \approx 1 + n u$ hai $∣ u ∣ ≪ 1$ ke liye. Yahan $\frac{1}{1 - x} = (1 - x)^{- 1/2}$ , toh $u = - x$ aur $n = - \frac{1}{2}$ , giving $1 + (- \frac{1}{2}) (- x) = 1 + \frac{x}{2}$ . Higher-order terms $x^{2}$ size ke hain, negligible jab $x$ tiny ho.

Cell G — real-world word problems (sahi frame dhundho)

Worked example Example 5 — woh muon jo ground tak nahi pahunchna chahiye (Cell G)

Muons atmosphere mein oopar form hote hain — ek typical production height $H \approx 15$ km lo (stratosphere mein lagbhag $10$ km plus slant path, cosmic-ray muon studies ke liye standard textbook value). Unka proper lifetime $Δ t_{0} = 2.2 μ s$ hai, $v = 0.99 c$ ( $β = 0.99$ ) par move karte hain. Newtonian tarike se ek muon sirf $c Δ t_{0} \approx (3 \times 1 0^{8}) (2.2 \times 1 0^{- 6}) \approx 660$ m travel karta hai decay se pehle. Lab (ground) frame mein ek typically kitna travel karta hai — aur itne saare $H$ puri $15$ km cross karke detectors tak kyun pahunchte hain?

Forecast: (Guess: lab frame mein muon kaafi zyada dair jeeta hai, toh uski typical range $γ$ ke factor se badhti hai.)

$γ = 1/ 1 - 0.9 9^{2} = 1/ 1 - 0.9801 = 1/ 0.0199 \approx 7.089$ . Yeh step kyun? Diye gaye speed se bridge factor build karo.
Kaun sa frame $Δ t_{0}$ hold karta hai? Muon ki apni clock "born" aur "decay" dono events par present hai → $2.2 μ s$ proper time hai. Lab (ground) watcher hai. Yeh step kyun? Same-place test: birth aur death muon par hote hain.
Lab lifetime: $Δ t = γ Δ t_{0} = 7.089 \times 2.2 μ s \approx 15.6 μ s$ . Yeh step kyun? Hum jaanna chahte hain muon ground ke nazariye se kitna jeeta hai — Cell A multiply.
Typical lab-frame range: $d = v Δ t = 0.99 (3 \times 1 0^{8}) (15.6 \times 1 0^{- 6}) \approx 4.63 \times 1 0^{3}$ m $\approx 4.6$ km — sirf $0.66$ km se upar. Yeh step kyun? Dilated time ko physical range mein convert karta hai.

Ek subtlety — woh phir bhi puri $H = 15$ km kaise cross karte hain? $4.6$ km ek proper lifetime ke liye mean range hai; yeh koi wall nahi hai. Radioactive decay exponential hai, toh muons ka ek fraction kai lifetimes tak survive karta hai. Dilation ke bina production height $H = 15$ km par survival fraction $e^{- H /660 m} = e^{- 15000/660} \approx e^{- 22.7} \approx 1 0^{- 10}$ hoti — essentially zero. Dilation ke saath relevant exponent dilated range use karta hai, $e^{- 15000/4630} \approx e^{- 3.24} \approx 0.039$ — lagbhag $4%$ survive karte hain, ek hundred-million-fold boost jo "none detected" ko "plenty detected" mein badal deta hai.

Verify: $4.6$ km $≫ 0.66$ km ✓; survival ratio $e^{- 3.24} / e^{- 22.7} = e^{19.5} \approx 3 \times 1 0^{8}$ ✓ — exponent mein $γ$ ka factor hi muons ko $H = 15$ km ke through le jaata hai. Yeh observed muon flux at ground level se match karta hai. (Muon ke apne frame mein distance shrink hoti hai — Length Contraction dekho — same physics, doosri seat se dekha gaya.)

Worked example Example 6 — ek fast-orbit clock drift karta hai (Cell G, GPS-style)

Ek clock $v = 4.0 \times 1 0^{3}$ m/s par (roughly orbital speed) ground time ke ek din, $Δ t = 86400$ s ke liye move karta hai. Sirf velocity time dilation ki wajah se moving clock kitna peeche reh jaata hai?

Forecast: (Guess: kuch km/s par, $β$ tiny hai, toh offset sirf microseconds mein hoga.)

$β = v / c = \frac{4.0 \times 1 0 ^{3}}{3.0 \times 1 0 ^{8}} = 1.333 \times 1 0^{- 5}$ , toh $β^{2} = 1.778 \times 1 0^{- 10}$ . Yeh step kyun? Choti speeds par hum precision rakkhne ke liye seedha $β^{2}$ ke saath kaam karte hain.
Elapsed ground time per moving clock ka time lost: $Δ t - Δ t_{0} = Δ t (1 - \frac{1}{γ}) \approx Δ t \cdot \frac{1}{2} β^{2}$ . Yeh step kyun? Binomial approximation se (Ex 5 se pehle collapsible dekho): $γ = (1 - β^{2})^{- 1/2} \approx 1 + \frac{1}{2} β^{2}$ , toh $\frac{1}{γ} \approx 1 - \frac{1}{2} β^{2}$ , fractional lag $\frac{1}{2} β^{2}$ deta hai. Yahi kaam karne ka ek route hai — do almost-equal seconds ka raw difference rounding mein dub jaata.
Plug in karo: $Δ t \cdot \frac{1}{2} β^{2} = 86400 \times \frac{1}{2} \times 1.778 \times 1 0^{- 10} \approx 7.68 \times 1 0^{- 6}$ s $\approx 7.7 μ s$ per day. Yeh step kyun? Fractional lag ko actual clock offset mein convert karta hai.

Verify: Order of magnitude — microseconds per day — satellite clock corrections mein real velocity contribution hai ✓. Units: $s \times (dimensionless) = s$ ✓.

Cell H — round-trip twin twist

Worked example Example 7 — stopover wale twins (Cell H)

Twin B $6$ light-years door ek star tak (Earth frame) $v = 0.6 c$ ( $β = 0.6$ ) par jaata hai, phir same speed par wapas aata hai. Twin A ghar rehta hai. Har twin ki aging nikalo, aur asymmetry confirm karo.

Forecast: (Guess: B kam age hoga; gap kuch saal ka hoga.)

Earth (frame S, poora ek inertial frame) round trip time karta hai: out-and-back distance $= 12$ ly at $0.6 c$ , toh $Δ t_{A} = \frac{12}{0.6} = 20$ years. Yeh step kyun? Twin A kabhi frames nahi change karta, toh distance/speed A ki aging cleanly deta hai.
$γ = 1/ 1 - 0.36 = 1/0.8 = 1.25$ . Yeh step kyun? B ki clock ke liye bridge.
B ki clock departure, turnaround, aur return par present hai — har straight leg par B proper time hold karta hai. Toh $Δ t_{B} = \frac{Δ t _{A}}{γ} = \frac{20}{1.25} = 16$ years. Yeh step kyun? Cell B (divide) leg-by-leg apply hota hai; turnaround yahan instantaneous hai.
Asymmetry check: sirf B ne turn around kiya (inertial frames switch kiye, rocket ka push feel kiya). A ek frame mein raha. Toh situation symmetric nahi hai aur yeh legitimate hai ki B kam age hota hai. Yeh step kyun? Yahi paradox ka resolution hai — turning twin woh younger hota hai.

Verify: $Δ t_{B} = 16 < Δ t_{A} = 20$ ✓; B $4$ saal younger return karta hai. Consistency: $16 \times 1.25 = 20$ ✓. (Ek deeper accounting invariant Spacetime Interval use karta hai, jis par har frame agree karta hai.)

Asymmetry ek shape fact hai: B ki worldline bent hai, A ki straight. Ise drawn dekho:

Cell I — exam trap

Worked example Example 8 — deliberately mislabelled question (Cell I)

"Ek spaceship Earth se $v = 0.6 c$ par guzarti hai. Earth observers do Earth landmarks ke beech trip ko $8.0$ s time karte hain. Ek student likhta hai $Δ t = γ \times 8.0 = 1.25 \times 8 = 10$ s ship ke liye. Kya student sahi hai?"

Forecast: (Guess: nahi — unhone galat label kiya hai kaun sa time proper hai.)

Do events locate karo: ship landmark 1 se guzarti hai, phir landmark 2 se. Yeh Earth par do alag jagahon par hote hain, lekin ship par same place par (ship ki nose par). Yeh step kyun? Same-place test sab decide karta hai, aur yahan yeh naive assumption flip kar deta hai.
Isliye ship proper time hold karta hai, aur Earth reading $8.0$ s dilated $Δ t$ hai. Yeh step kyun? Student ne assume kiya tha ki Earth $Δ t_{0}$ hold karta hai — classic Cell I error (parent note ka Mistake 2 dekho).
Sahi computation: $Δ t_{0}^{(ship)} = \frac{Δ t}{γ} = \frac{8.0}{1.25} = 6.4$ s. Yeh step kyun? Hum divide karein, multiply nahi — ship single clock hai jo dono passing-events par present hai, toh proper time hold karta hai aur do synchronized Earth clocks se kam seconds read karta hai.

Verify: $6.4 < 8.0$ ✓ (single moving ship-clock do synchronized Earth clocks se kam read karta hai). Student ka $10$ s galat hai precisely isliye ki unhone wrong number multiply kiya: unhone Earth ke $8.0$ s ko proper time treat kiya jabki woh actually dilated time hai.

Recall Har example mein

$Δ t_{0}$ kaun sa frame hold karta hai? Ex 1 ship ::: ship (events ship par) → Earth ke liye multiply Ex 2 probe ::: probe → Earth ke 10 yr se divide Ex 5 muon ::: muon (uspar born & decay hota hai) → lab ke liye multiply Ex 8 trap ::: ship (dono landmark-passings uski nose par) → divide, multiply nahi

Worked examples — Time dilation — derivation, twin paradox