Length contraction — derivation: Visual walkthrough

2.3.30 · D2 · Physics › Modern Physics › Length contraction — derivation

Step 1 — Do observers, do rulers

KYA. Hum do points of view set up karte hain (jinhe frames kehte hain). Ek frame bas "koi apna ruler aur apna set of clocks lekar, sab aapas mein agree karke" hota hai. Ek frame ground/lab ka hai — hum ise $S$ kehte hain. Doosra rod ka khud ka hai — hum ise $S^{'}$ kehte hain (bolo "S-prime").

KYUN. Poora effect do observers ke beech ek disagreement hai. Ek hi observer se disagreement nahi dekh sakte, isliye hamesha do frames chahiye: ek jahan rod still hai, aur ek jahan woh tezi se guzarti hai.

PICTURE. Figure mein, rod (blue) apne frame $S^{'}$ mein still khadi hai. Ground observer (frame $S$ mein) ise speed $v$ se daayein jaate hue dekhta hai. Speed $v$ metres per second mein measure hoti hai; $c$ light ki speed hai, cosmic speed limit.

Figure — Length contraction — derivation

Step 2 — Ek chalti cheez ki "length" ka matlab kya hai

KYA. Length = (front end ki position) minus (back end ki position). Hum motion ki direction ke saath positions ko $x$ likhte hain. Toh length do $x$ values ka farq hai: $x_{front} - x_{back}$ .

KYUN. Rest mein rod ke liye yeh boring hai — ends nahi chalte, kabhi bhi padh lo. Lekin chalti rod ke liye dono ends ko ek hi instant par pakadna hoga. Agar front ki abhi photo lo aur back ki ek second baad, toh rod beech mein aage khiski aur tumhara jawab kachra hai.

PICTURE. Do panels. Left: dono ends ek hi time par snap kiye — ek honest length (green bracket). Right: front jaldi snap kiya, back baad mein — rod move kar gayi, ek fake length di (red bracket). Shabd " $t$ " ek time reading hai; ek hi $t$ par do readings ko simultaneous kehte hain.

Step 3 — Proper length, $S^{'}$ mein padhi gayi

KYA. Rod ke apne frame $S^{'}$ mein dono ends fixed positions par baithe hain, jinhe hum $x_{1}^{'}$ (back) aur $x_{2}^{'}$ (front) kehte hain. Prime $^{'}$ bas matlab hai " $S^{'}$ mein measure kiya gaya." Unka farq proper length hai: $L_{0} = front, in S^{'} x_{2}^{'} - back, in S^{'} x_{1}^{'} .$

KYUN. $S^{'}$ mein rod still khadi hai, isliye uske ends kabhi nahi chalte — yahan simultaneity ki zaroorat bhi nahi. Yeh "sachi, at-rest" length hai, define karne mein sabse aasaan.

PICTURE. Blue rod $S^{'}$ mein parked, back end $x_{1}^{'}$ label ki, front end $x_{2}^{'}$ , aur unke beech $L_{0}$ mark karta yellow bracket.

Step 4 — Frames ke beech ka bridge: Lorentz transformation

KYA. $S$ mein coordinates $S^{'}$ mein kaise badte hain? Seedha subtraction se nahi (woh purani Newtonian physics hogi). Sahi rule hai Lorentz transformation: $x^{'} = γ (position in S x - shift for motion v t), γ = \frac{1}{1 - v ^{2} / c ^{2}} .$

Term by term padho:

$x$ — event kahan hai, ground ruler $S$ par.
$v t$ — time $t$ tak $S^{'}$ ka origin kitna khiска hai; ise subtract karna moving frame par re-centre karta hai.
$γ$ (gamma) — stretch factor jo units theek karta hai taaki light dono frames mein speed $c$ se chale.

KYUN simple $x^{'} = x - v t$ nahi? Kyunki naive Galilean rule secretly assume karta hai dono frames ek universal clock share karte hain. Experiments kehte hain aisa nahi hai — light sabke liye $c$ par travel karti hai. Iske saath consistent rule sirf Lorentz wala hai, jo $γ$ se multiply karta hai. Hum ise precisely isliye use karte hain kyunki yeh woh transformation hai jo light ki speed fixed rakhti hai. Dekho Lorentz transformation aur Lorentz factor gamma.

PICTURE. $γ$ ek curve ke roop mein: $v = 0$ par $1$ hai (koi relativity nahi) aur $v \to c$ jaate hi infinity ki taraf uchhalti hai. Woh utha hua curve hi wajah hai ki fast cheezein ajeeb behave karti hain.

Step 5 — $S$ mein measurement: dono ends ek instant par

KYA. Ground observer dono ends ko ground positions $x_{1}$ (back) aur $x_{2}$ (front) par mark karta hai, ek hi ground time $t$ par: $t_{1} = t_{2} = t .$ Har end ko Lorentz rule mein daalo: $x_{1}^{'} = γ (x_{1} - v t), x_{2}^{'} = γ (x_{2} - v t) .$

KYUN. Yeh ek line, $t_{1} = t_{2} = t$ , length contraction ki poori physics hai. Yeh Step-2 rule (simultaneous readings) ko transformation ke andar enforce karti hai. Dekho aage kya hota hai.

PICTURE. Ek spacetime sketch: rod ke dono ends ki worldlines (space aur time mein paths), aur ek horizontal "same-time-in- $S$ " slice dono ko ek instant $t$ par kaatti hai. Do crossing points $x_{1}$ aur $x_{2}$ hain.

Step 6 — Subtract karo: shift cancel hoti dekho

KYA. Step 5 ki dono end-equations subtract karo: $x_{2}^{'} - x_{1}^{'} = γ [(x_{2} - v t) - (x_{1} - v t)] = γ (x_{2} - x_{1}) .$

KYUN. $v t$ term dono equations mein identical hai — sirf isliye ki $t_{1} = t_{2}$ . Toh jab subtract karte hain, $- v t$ aur $- v t$ bilkul cancel ho jaate hain. Agar do readings alag times par hoti, $- v t_{1}$ aur $- v t_{2}$ nahi cancel hote, aur koi clean length nahi milti. Yahi cancellation hai jahan simultaneity apna kaam karta hai.

PICTURE. Dono equations ek ke upar ek rakhi hain, $- v t$ terms highlight ki hain aur ek arrow dikhata hai ki woh annihilate ho jaate hain, $γ (x_{2} - x_{1})$ bachta hai.

Ab differences ko rename karo using woh physically kya hain:

$x_{2}^{'} - x_{1}^{'} = L_{0}$ — proper length (Step 3).
$x_{2} - x_{1} = L$ — woh length jo ground ne actually measure ki.

Toh: $L_{0} = γ L .$

Step 7 — Solve karo, aur shrinking padho

KYA. Measured length $L$ ke liye $L_{0} = γ L$ rearrange karo: $L = \frac{L _{0}}{γ} = L_{0} 1 - \frac{v ^{2}}{c ^{2}}$

Term by term:

$L_{0}$ — sabse badi (rest) length, upar baithi hai kyunki baaki sab chhoti hain.
$γ \geq 1$ — kisi $\geq 1$ se divide karna $L_{0}$ ko sirf chhhota kar sakta hai.
$1 - v^{2} / c^{2}$ — $0$ aur $1$ ke beech ek number; $L_{0}$ ko isse multiply karna rod ko chhhota karta hai.

KYUN multiply nahi, divide? Kyunki $L_{0}$ maximum define hai. Koi bhi moving observer kam measure karega, toh $L_{0}$ fraction ke upar hona chahiye. (Compare karo Time dilation se, jahan proper time minimum hai, toh woh upar multiply hoti hai.)

PICTURE. Measured length $L$ speed ke against $L_{0}$ ke fraction ke roop mein: low speed par $L_{0}$ ke paas flat, $v \to c$ jaate hi $0$ par gir jaata hai.

Step 8 — Har case: limits check karo

KYA & KYUN. Jo formula tumhe trust ho woh apne edges par sensibly behave karna chahiye. Hum teen regimes test karte hain.

$v = 0$ (rod at rest): $γ = 1$ , toh $L = L_{0}$ . Koi motion nahi, koi contraction nahi. ✔
Chhota $v$ (rozana ki speeds): $v / c \approx 0$ , toh $1 - v^{2} / c^{2} \approx 1$ ; $L \approx L_{0}$ . Isliye hum daily life mein contraction kabhi notice nahi karte — gaadiyaan bahut slow hain.
$v \to c$ (light speed): $γ \to \infty$ , toh $L \to 0$ . Rod zero thickness par crush hoti dikhegi. Koi massive object $c$ tak nahi pahunch sakta, toh yeh ek limit hai, koi event nahi.
Perpendicular direction (degenerate case): derivation ne sirf $x$ , motion ki direction ko kabhi touch kiya. Heights aur widths ( $y, z$ directions) kabhi enter nahi hui, toh woh contract nahi hoti. Ek fast chalti ball apni motion ke saath ek pancake mein squash ho jaati hai lekin apni poori height rakhti hai.

PICTURE. Teen rods $v = 0$ , $v = 0.6 c$ , $v = 0.98 c$ par motion ke saath shrink ho rahi hain jabki unki heights ek jaisi rehti hain — dono shrink aur untouched perpendicular size dikhaate hue.

Ek-picture summary

Upar sab kuch ek hi spacetime diagram mein: rod ki dono ends ki worldlines, ground ka "same-time" slice (chhota $L$ deta hai), rod-frame ka "same-time" slice (bada $L_{0}$ deta hai), aur dono slices ke beech ka tilt — woh tilt hi relativity of simultaneity hai, shrinking ka asli source.

Recall Poore walkthrough ki Feynman retelling

Socho ek train tumhare paas se tez guzar rahi hai. Theek se kehne ke liye "train itni lambi hai," tum aur main dono agree karte hain ki front aur back ko bilkul ek hi instant par pakadna hoga — ek photo, do nahi. Theek hai. Ab yahan Einstein ka twist hai: train ke passengers aur tum is baat par agree nahi karte ki "bilkul ek hi instant" ka matlab kya hai. Jab tum train ki history ko apne "now" ke idea mein slice karte ho, tum ise thoda tilted angle par kaatte ho compare karke ki passengers apna "now" kaise slice karte hain. Woh tilt poori kahaani hai. Kyunki tumhara slice aur unka slice train ke dono ends ko alag pairs of moments par cross karte hain, tum train ka ek chhota hissa bracket karte ho compared to passengers jo apni train ka bracket karte hain. Kisi ne ise dabaya nahi, koi force uss par nahi padi — tum dono bas "now" ke baare mein disagree karte ho, aur ek chalti cheez ke liye "now" ke baare mein disagree karna automatically matlab hai uski length ke baare mein disagree karna. Gamma se divide karo, aur har baar chhota number bahar aata hai.

Active recall

Connections

Length contraction — derivation — parent note seedhi algebra ke saath.
Lorentz transformation — Steps 4–6 mein use kiya bridge equation.
Lorentz factor gamma — Step 4 ka $γ$ curve.
Relativity of simultaneity — summary figure mein tilt, asli wajah.
Time dilation — mirror-image effect ( $\times γ$ ).
Muon decay experiment — jahan yeh contraction real mein observe hoti hai.
Spacetime interval — invariant jo dono slices preserve karte hain.