Time dilation — derivation, twin paradox: Visual walkthrough

2.3.29 · D2 · Physics › Modern Physics › Time Dilation — Derivation, Twin Paradox

Hume sirf do facts chahiye, dono Special Relativity Postulates se:

Step 1 — Tool banao: ek light clock

KYA. Hum light se ek clock banate hain. Do mirrors ek doosre ke saamne hain, unke beech ek vertical gap hai jise hum $L$ kehte hain (sirf unke beech ki height — ek length, metres mein naapi gayi). Light ka ek flash upar uthta hai, top mirror se takraata hai, aur wapas neeche aata hai. Hum ek poori round-trip = ek "tick" kehte hain.

YEH TOOL KYUN. Ek normal clock (gears, springs) chhupa leta hai ki kyun yeh slow hota hai. Ek light clock apni ticking ko seedha $c$ se jodata hai — wahi cheez jo Rule 2 fix karta hai. Agar light ko behave karna hai, toh clock ko bhi karna hoga.

TASVEER. Flash bottom mirror se nikalta hai, top ko chhoota hai, wapas aata hai. Emission aur final catch same jagah hoti hai — yaad rakho, yeh baad mein kaam aayega.

Figure — Time dilation — derivation, twin paradox

Step 2 — Rider ka view: light seedha upar-neeche jaata hai

KYA. Clock par baitho aur uske saath ride karo. Tumhare liye, clock move nahi kar raha. Flash seedha upar jaata hai $L$ distance, phir seedha neeche $L$ distance. Light ka total path: $2 L$ .

KYUN. Speed = distance divided by time. Hum distance jaante hain ( $2 L$ ) aur speed bhi ( $c$ ). Toh ek tick ka time hum par impose ho jaata hai:

$rider ka tick-time Δ t_{0} = \frac{2 L light ka path}{light ki speed c}$

Baad ke liye $L$ handy rakhne ke liye rearrange karte hain:

$L = \frac{c Δ t _{0}}{2}$

TASVEER. Ek seedhi vertical line upar, ek seedhi vertical line neeche — kuch bhi slanted nahi, kyunki is frame mein mirrors still hain.

Step 3 — Tumhara view: clock urta jaata hai, light diagonal jaati hai

KYA. Ab zameen par khado aur dekho wahi clock tumhare paas se sideways $v$ speed se guzar raha hai. Ek tick ke dauran clock sideways drift karta hai jabki flash bounce karta hai. Toh flash tumhare liye seedha upar nahi ja sakta — jab tak woh top mirror tak pahunche, top mirror aage badh chuka hota hai. Tum light ko ek slanted diagonal upar, phir ek slanted diagonal neeche travel karte dekhhte ho: ek tent " $Λ$ " shape.

KYUN. Rider aur tum usi physical flash ko dekhte ho, lekin tum straight up ke baare mein alag raay rakhte ho kyunki clock uske neeche se khisak gaya. Yeh ek hi honest tasveer hai jo clock ke move karne ke saath consistent hai.

TASVEER. Do diagonal chalk strokes top mirror par milte hue — flash ka path jaise tum dekhte ho.

Step 4 — Tick ko aadha karo aur ek triangle banao

KYA. Sirf tick ki pehli aadhi dekho — flash jaata hai upar. Us half-tick mein, teen lengths aati hain, aur woh ek right triangle banati hain:

Vertical leg — flash ko gap cross karna hi tha, toh yeh $L$ utha.
Horizontal leg — clock sideways $v \cdot \frac{Δ t}{2}$ drift kiya (speed $\times$ half the tick-time).
Hypotenuse — woh slanted path jo light ne actually liya.

HYPOTENUSE KI LENGTH $c \cdot \frac{Δ t}{2}$ KYUN HAI AUR ISSE ZYADA NAHI. Yahan Rule 2 apna kaam karta hai. Tum expect kar sakte ho ki light longer diagonal par "faster jaaye" — lekin woh nahi ja sakti. Tumhe bhi iski speed exactly $c$ maapni padegi. Toh diagonal length = speed $\times$ time = $c \cdot \frac{Δ t}{2}$ .

TASVEER. Ek right triangle: chalk-blue vertical $L$ , chalk-pink horizontal $v Δ t /2$ , pale-yellow slanted hypotenuse $c Δ t /2$ , neeche ke corner par right angle marked.

Step 5 — $L$ daalo aur apna time $Δ t$ solve karo

KYA. Step 2 mein humne $L = \frac{c Δ t _{0}}{2}$ nikala tha. Ise substitute karo taaki equation sirf times ( $Δ t$ , $Δ t_{0}$ ) aur speeds ( $c$ , $v$ ) ki baaat kare:

$\frac{c ^{2} Δ t ^{2}}{4} = L^{2} rewritten \frac{c ^{2} Δ t _{0}^{2}}{4} + \frac{v ^{2} Δ t ^{2}}{4}$

KYUN. Hum rider ke tick $Δ t_{0}$ aur tumhare tick $Δ t$ ke beech relationship chahte hain. $L$ ko hatane se exactly wahi milta hai.

Quarters clear karne ke liye har term ko $4$ se multiply karo: $c^{2} Δ t^{2} = c^{2} Δ t_{0}^{2} + v^{2} Δ t^{2}$

$Δ t$ terms ko ek taraf ikkattha karo (yahi woh unknown hai jo hum akela chahte hain): $c^{2} Δ t^{2} - v^{2} Δ t^{2} = c^{2} Δ t_{0}^{2} ⟹ Δ t^{2} (c^{2} - v^{2}) = c^{2} Δ t_{0}^{2}$

$(c^{2} - v^{2})$ se divide karo, phir upar aur neeche $c^{2}$ se divide karo: $Δ t^{2} = \frac{c ^{2} Δ t _{0}^{2}}{c ^{2} - v ^{2}} = \frac{Δ t _{0}^{2}}{1 - v ^{2} / c ^{2}}$

Square root lo:

TASVEER. Triangle re-drawn jisme substitution dikhayi gayi hai, arrows har geometric side se final boxed formula mein uski role tak point karte hue.

Step 6 — $γ$ kya karta hai: sabhi speeds par stretch factor dekho

KYA. $γ$ sirf ek number hai jo is par depend karta hai ki clock kitna fast move kar raha hai. Chalo isse har possible speed par dekhte hain, standstill se lekar almost light-speed tak.

KYUN SAB CASES COVER KARTE HAIN. Ek formula jise tum extremes par picture nahi kar sakte, woh aisa formula hai jis par tum trust nahi karte. Toh hum slow end, everyday end, aur ferocious end check karte hain.

$v = 0$ (clock at rest): $γ = 1/ 1 - 0 = 1$ , toh $Δ t = Δ t_{0}$ . Dono agree — koi dilation nahi. ✅ Sanity wapas aayi.
$v = 0.01 c$ (ek fast jet, phir bhi "slow"): $γ \approx 1.00005$ — ek lakh mein paanch parts ka stretch. Invisible. Isliye Newton "kaam karta tha."
$v = 0.6 c$ : $γ = 1/ 1 - 0.36 = 1/0.8 = 1.25$ . Moving clock $25%$ slow tick karta hai.
$v \to c$ (light-speed ke kareeb): $1 - v^{2} / c^{2} \to 0$ , toh $γ \to \infty$ . Moving clock tumhare view mein almost freeze ho jaata hai.
$v > c$ ? Tab $1 - v^{2} / c^{2} < 0$ aur square root imaginary hoga — nonsense. Yeh maths ka polite tarika hai tumhe batane ka: kuch bhi light se faster nahi ja sakta.

TASVEER. $γ$ -versus- $v$ curve: chhote $v$ ke liye flat aur $1$ ke kareeb, phir $v$ ke $c$ ki taraf badhne par infinity tak rocket karta hai ( $v = c$ par ek vertical wall).

Step 7 — Aasman mein ek real clock: muon (worked case)

KYA. Ek muon apne frame mein sirf $Δ t_{0} = 2.2 μ s$ zinda rehta hai decay karne se pehle. Yeh atmosphere mein $v = 0.99 c$ se neeche race karta hai. Lab ise kitne der tak zinda dekhta hai?

KYUN. Yeh derivation ek aisi experiment mein cash out ho raha hai jo tum actually run kar sakte ho.

$γ = \frac{1}{1 - 0.9 9 ^{2}} = \frac{1}{1 - 0.9801} = \frac{1}{0.0199} \approx 7.09$ $Δ t = γ Δ t_{0} = 7.09 \times 2.2 μ s \approx 15.6 μ s$

TASVEER. Do number-lines side by side: muon ki apni chhoti lifetime bar vs. lab ki stretched-out bar — zameen tak pahunchne ke liye kaafi lambi.

Ek-tasveer summary

Upar sab kuch ek single triangle mein collapse ho jaata hai. Rider ka tick vertical side hai; tumhara tick hypotenuse hai; clock ki motion base hai. Usi light-speed $c$ par ek lambi hypotenuse matlab hai zyada time per tick — yahi time dilation hai, drawn.

Recall Feynman retelling — plain words mein poora walkthrough

Ek aisa clock banao jo do mirrors ke beech light bounce karke tick kare. Agar tum ise carry karo, light bas seedha upar-neeche jaati hai — chhota safar, jaldi tick. Agar main tumhe fly karte dekhuun, wahi light ek slanted zig-zag travel karna padta hai, kyunki mirrors sideways khisak gaye jabki light flight mein thi. Zig-zag lamba hai. Lekin light ko speed up karne ki permission nahi — mere liye bhi yeh $c$ par stuck hai. Same speed par lambi road matlab trip mein zyada time lagta hai. Toh tumhare clock ka har tick mere liye zyada time leta hai: tumhara clock, aur tumhare baare mein sab kuch, slow motion mein chalta hai. Up-trip ko ek right triangle ke roop mein draw karo — straight-up side, sideways-drift side, slanted-light side — aur Pythagoras tumhe exact stretch factor $γ$ de deta hai. Walking speed par $γ$ basically $1$ hai (kisi ko notice nahi hota); light-speed ke kareeb yeh infinity ki taraf blow up hota hai (clock almost freeze); light-speed se aage maths imaginary ho jaata hai, ise forbid karta hai. Woh single triangle hi poori theory hai.

Visual walkthrough — Time dilation — derivation, twin paradox