Lorentz transformation — derivation: Visual walkthrough

2.3.28 · D2 · Physics › Modern Physics › Lorentz transformation — derivation

Hum dekh rahe hain ki do observers ek hi event (ek flash, ek click — space mein ek point, ek hi instant pe) ko kaise label karte hain. Ek observer frame $== S ==$ mein baitha hai (ise lab kaho). Doosra frame $== S^{'} ==$ mein sawaar hai, jo shared $x$ -axis ke saath ek steady speed $== v ==$ se khisakta hai. Hum sideways directions $y, z$ ko alag rakhenge — motion $x$ ke saath hai, isliye $y^{'} = y$ , $z^{'} = z$ aur yeh kabhi nahi badalte.

Step 1 — Ek event do grids mein rehta hai

KYA: Hum ek event ko $S$ grid par rakhte hain aur uske $S^{'}$ readings poochte hain. KYUN: Poori derivation ek hi sawal hai — $(t, x)$ diya hua hai, $(t^{'}, x^{'})$ kya hain? Pehle yeh dekhna zaroori hai ki dono frames ek hi reality describe kar rahe hain, sirf do alag tarike se label kiya hua. PICTURE: Neeche, kala dot ek single flash hai. Navy grid $S$ hai; chalte observer alag (tilted) grid draw karte, lekin abhi ke liye bas note karo: ek hi dot, do label-sets. Neeche-baayein corner ka origin dot shared handshake $t = t^{'} = 0$ hai.

Step 2 — Naive guess aur yeh kyun toot-ta hai

KYA: Hum $S^{'}$ ka origin (point $x^{'} = 0$ ) draw karte hain jo $S$ se travel kar raha hai. KYUN: Chalte observer apne origin par baithte hain; $S$ mein woh origin $x = v t$ par hai. Toh hum jo shift subtract karte hain woh exactly $v t$ hai.

Guess ko term by term padho: $x^{'} = lab position x - how far S^{'} has slid by time t v t, t^{'} = t .$

PICTURE: Violet line chalte origin ka path $x = v t$ hai. Ek light pulse (orange) origin se $t = 0$ par nikalta hai aur fixed speed $c$ par travel karta hai. Galileo ki flat clocks ke saath, ground pulse ko $c$ speed se dekhta lekin train wala $c - v$ dekhta — dono orange slopes alag hain. Yeh Michelson–Morley experiment se contradict karta hai, jisne paya ki light ki speed kabhi nahi badalti. Toh $t^{'} = t$ hi galti hai.

Step 3 — Straight lines ko straight rehne do (linearity)

KYA: Hum sabse general straight-line-preserving relation likhte hain. KYUN: Map ko moda toh ek gliding particle ek observer ko accelerated lagta — physics frames ke beech alag ho jaati, jo allowed nahi hai. (Setup mein handshake ki wajah se, linear map ko koi added constant chahiye nahi — point $0, 0$ pehle se $0, 0$ pe map hota hai.)

$x^{'} = A x + B t, t^{'} = D x + E t .$

Term by term: $A$ position stretch karta hai, $B$ time mix karta hai, $D$ position ko naye time mein mix karta hai, $E$ time stretch karta hai. Chaar unknowns — inhe track karo: Step 4, $B = - A v$ fix karta hai; Step 6, $A = γ$ fix karta hai; aur Step 7 poora system solve karta hai taaki $D$ aur $E$ baahir aayein $D = - γ v / c^{2}$ (position-into-time mixing jo simultaneity tod-ta hai) aur $E = γ$ (time stretch). Hum har ek ko phir se name karenge jis moment wo pin ho jaayega.

PICTURE: Left panel — ek curved map ek straight worldline ko moda karta hai (physics toot-ti hai). Right panel — ek linear map use straight rakhta hai. Isliye hum zarooran linear form use karte hain.

Step 4 — $B$ pin karo: chalti origin $v$ par sawaar hai

KYA: $x^{'} = 0$ set karo ( $S^{'}$ ka origin) aur dekho yeh $S$ mein kahan baithta hai. KYUN: Definition ke hisaab se $S^{'}$ ka origin $v$ par $S$ se travel karta hai, yaani $x = v t$ par baitha hai (yeh shared handshake se measure hota hai, toh koi extra constant nahi). Yeh $B$ ko $A$ se baandh-ta hai.

$x^{'} = A x + B t = 0$ se milta hai $x = - \frac{B}{A} t$ . $x = v t$ se match karke: $- \frac{B}{A} = v \Rightarrow B = - A v \Rightarrow x^{'} = A (x - v t) .$

Term by term: poora bracket $(x - v t)$ hai "chalti origin se measured position"; $A$ ek not-yet-known overall stretch hai (Galileo ne secretly $A = 1$ assume kiya tha). Unknown $B$ ab named hai: $B = - A v$ .

PICTURE: Violet worldline $x = v t$ exactly woh dots ka set hai jo chalti observer $x^{'} = 0$ kehti hai. Uske daayein sab kuch $x^{'} > 0$ hai.

Step 5 — Symmetry se inverse free mein milta hai

KYA: Inverse map likho. KYUN: Agar dono directions alag factors use karte, ek frame physically special hota. Symmetry yeh forbid karta hai. (Shared handshake ki wajah se inverse mein bhi koi constant term nahi hai.)

$x = A (x^{'} + v t^{'}) .$

Term by term: same $A$ , aur $+ v$ replace karta hai $- v$ ko kyunki $S$ , $S^{'}$ ki nazar se doosri taraf move karta hai. (Yeh " $v \to - v$ " swap exactly woh sign convention hai jo setup mein note kiya tha — inverse koi naya formula nahi, sirf wahi formula hai $v$ ka sign flip karke.)

PICTURE: Do mirrored observers ek doosre ki taraf point karte hue; har ek doosre ko speed $v$ par door jaate dekhta hai, aur har ek identical factor $A$ use karta hai. Setup bilkul symmetric hai.

Step 6 — Light postulate solve karta hai $A = γ$ ke liye

KYA: Ek light pulse shared origin se $t = t^{'} = 0$ (handshake) par bhejo. Yeh $S$ mein $x = c t$ obey karta hai aur $S^{'}$ mein $x^{'} = c t^{'}$ — dono mein same speed $c$ . KYUN: Yeh ek experimental demand hai (Michelson–Morley experiment) jo Galileo ko theek karta hai. Yahi derivation ko relativistic banata hai.

$x = c t$ ko $x^{'} = A (x - v t)$ mein daalo: $x^{'} = A t (c - v)$ . $x^{'} = c t^{'}$ ko $x = A (x^{'} + v t^{'})$ mein daalo: $c t = A t^{'} (c + v)$ . Dono equations multiply karo ( $x^{'} = c t^{'}$ use karke left sides $c^{2} t t^{'}$ dete hain), phir $t t^{'}$ cancel karo:

$c^{2} = A^{2} (c - v) (c + v) = A^{2} (c^{2} - v^{2}) .$

Solve karo (positive root lete hain, kyunki $A > 0$ : grids flip nahi hain): $A^{2} = \frac{c ^{2}}{c ^{2} - v ^{2}} = \frac{1}{1 - v ^{2} / c ^{2}} \Rightarrow A = \frac{1}{1 - v ^{2} / c ^{2}} \equiv γ .$

$γ$ ke andar term by term: $v^{2} / c^{2}$ hai "light-speed ke kitna close, squared"; ise 1 se subtract karo, root lo, flip karo. Unknown $A$ ab named hai: $A = γ$ . Dekho Time dilation.

PICTURE: Dono orange light lines har grid mein same slope (speed $c$ ) rakhte hain — yeh ek requirement hai jo $A$ ko $γ$ hone par majboor karti hai, 1 nahi. $γ$ versus $v / c$ ka curve dikhata hai ki slow speeds ke liye yeh 1 ke paas rehta hai aur $v = c$ ke paas rocket ki tarah upar jaata hai.

Step 7 — Time equation khud nikal aata hai (yahan $D$ aur $E$ name hote hain)

KYA: $x^{'} = γ (x - v t)$ ko inverse $x = γ (x^{'} + v t^{'})$ mein daalo aur $t^{'}$ ke liye solve karo. KYUN: Space rule pehle se hamare paas hai; wahi do relations, milke, force karte hain time rule ko. Koi nayi assumption nahi chahiye — aur yahan Step 3 ke last two unknowns $D$ aur $E$ finally apni values paate hain.

$x = γ (γ (x - v t) + v t^{'}) \Rightarrow γ v t^{'} = x (1 - γ^{2}) + γ^{2} v t .$

$v \neq = 0$ ke liye, $1 - γ^{2} = - γ^{2} v^{2} / c^{2}$ use karo aur $γ v$ se divide karo: $t^{'} = γ (t - \frac{v x}{c ^{2}}) = D (- \frac{γ v}{c ^{2}}) x + E γ t .$ (Aur $v = 0$ par yahi formula $t^{'} = t$ deta hai, toh yeh dono cases cover karta hai.)

Term by term: $t$ lab time hai; $\frac{v x}{c ^{2}}$ ek position-dependent time offset hai — $1/ c^{2}$ ek distance $x$ ko time mein convert karta hai. Step 3 ke coefficients read karke: $D = - γ v / c^{2}$ (position jo naye time mein mix hoti hai — woh term jo simultaneity tod-ti hai) aur $E = γ$ (time stretch). Charon unknowns $A, B, D, E$ ab determined hain.

PICTURE: Ek fixed lab time $t$ ke liye, $S^{'}$ -time depend karta hai kahan event hai (uske $x$ par): bahut door-right events (magenta) ko left events (violet) se bada negative offset milta hai. Same $t$ , different $t^{'}$ — yahi hai Relativity of simultaneity.

Step 8 — Andar chupi hui invariant: $c^{2} t^{2} - x^{2}$

KYA: Hamare do formulas lo aur $c^{2} t^{'2} - x^{'2}$ calculate karo. KYUN: Yeh linearity/light arguments ko baandh-ta hai: transform exactly woh shear hai jo is combination ko fixed rakhta hai — relativity ka geometric heart, aur woh cheez jo summary picture draw karegi.

$c^{2} t^{'2} - x^{'2} = c^{2} γ^{2} (t - \frac{v x}{c ^{2}})^{2} - γ^{2} (x - v t)^{2} = c^{2} t^{2} - x^{2} .$

Term by term: $v x t$ form ke cross terms cancel ho jaate hain; bacha hua $γ^{2} (1 - v^{2} / c^{2}) = 1$ baaki sab ko $c^{2} t^{2} - x^{2}$ par wapas collapse karta hai. Poori cancellation ke liye Verify block dekho.

PICTURE: Dotted navy curve un sabhi events ka set hai jinka same interval hai — ek hyperbola. Tilted $S^{'}$ grid events ko is curve ke saath saath slide karta hai lekin usse hatata nahi: interval frame-independent hai.

Step 9 — Har case: slow, fast, aur $v \to c$ ki wall

$v = 0$ (frames saath rest mein): $γ = 1$ , offset $v x / c^{2} = 0$ → $x^{'} = x$ , $t^{'} = t$ . Identity — jaisa hona chahiye (yeh Step 7 ka $v = 0$ branch hai).
$v ≪ c$ (roz ke cars): $v^{2} / c^{2} \approx 0$ , $γ \approx 1$ → $x^{'} \approx x - v t$ , $t^{'} \approx t$ . Hum Galilean transformation recover karte hain. Nayi theory purani ko contain karti hai.
$v < 0$ ( $S^{'}$ baayein move karta hai): identical formulas, $γ$ unchanged (depends on $v^{2}$ ); sirf $v t$ aur $v x / c^{2}$ terms ke signs flip hote hain. Koi nayi physics nahi.
$v$ close to $c$ (e.g. $0.9 c$ ): $γ$ blow up karta hai; space aur time heavily mix hote hain.
$v \to c$ : $γ \to \infty$ — grid shear hoti hai jab tak time aur space axes light line par fold nahi ho jaate. Mass wali koi cheez is wall tak nahi pahunchti.
$v \geq c$ : $1 - v^{2} / c^{2} \leq 0$ , root imaginary ya zero hai → koi real transform nahi. Isliye $c$ ek speed limit hai.

KYA / KYUN: Limits check karna prove karta hai ki formula har allowed jagah well-behaved hai aur jahan nahi pahunchna chahiye wahan forbid karta hai. PICTURE: $S^{'}$ axes (violet = constant $x^{'}$ ki line, magenta = constant $t^{'}$ ki line) orange light line ki taraf tilt hote hain jab $v$ $0.3 c$ se $0.9 c$ tak badhta hai. Yeh light line se sirf impossible limit $v = c$ mein milte hain.

Ek-picture summary

Upar sab kuch ek diagram mein jeeta hai — Minkowski diagram. Navy grid $S$ hai. Tilted violet/magenta grid $S^{'}$ hai. Orange 45° line light hai: yeh dono grids mein same slope rakhti hai — woh ek demand jo $A = γ$ set karti hai. Kisi bhi event ka $(x, t)$ seedhi navy grid se padho aur $(x^{'}, t^{'})$ tilted grid se. Bracket $(x - v t)$ space ka tilt hai; offset $v x / c^{2}$ time ka tilt hai; $γ$ dono ko scale karta hai. Quantity $c^{2} t^{2} - x^{2}$ (Spacetime interval Step 8 se) dono grids mein same number hai — woh hyperbola jo koi bhi shear nahi badal sakti.

Recall Feynman retelling — poora walkthrough seedhe shabdon mein

Do log ek hi light flash ko label karte hain, aur woh ek starting handshake par agree karte hain: jis instant unke origins guzarte hain, dono ghariyan zero padhti hain. Newton ne kaha: bas apna ruler sideways khisao aur same clock rakho. Lekin light speed badalne se mana karti hai, toh woh kahani toot-ti hai. Use theek karne ke liye hum insist karte hain ki map "straight-lines-stay-straight" (linear) ho, phir hum uske knobs ek ek karke pin karte hain: ek knob kehta hai chalti origin $v$ par drift karti hai; symmetry kehti hai return trip same knob use karti hai; aur light rule — light sabke liye $c$ hai — us knob ko ek precise number par squeeze karta hai jise hum $γ$ name karte hain. Ek baar $γ$ fixed ho jaata hai, time rule koi nayi idea nahi hai; woh automatically nikal aata hai (thodi si care ke saath ki agar dono frames bilkul move nahi kar rahe, toh hum zero se divide nahi karte — bas "same time" milta hai), aur yeh ek sneaky $v x / c^{2}$ term carry karta hai matlab ki door-door events ek hi time par nahi rehte jab tum move karo. Aur bhi gehri baat, ek combination $c^{2} t^{2} - x^{2}$ kabhi nahi badalta — woh asli aadhar hai. Speed kam karo aur sab kuch Newton mein pighal jaata hai; speed ko light ki taraf badhao aur grid shear hoti hai jab tak shear ho sake nahi — woh wall speed limit $c$ hai.

Recall Active recall

Woh kaun si ek demand hai jo stretch factor $A = γ$ fix karti hai? ::: Light ki speed dono frames mein $c$ equal hoti hai.
Map linear kyun hona chahiye? ::: Taaki constant-velocity (straight) worldlines dono frames mein straight rahein.
Simultaneity kon sa term tod-ta hai, aur kyun? ::: $t^{'}$ mein $- v x / c^{2}$ — yeh naye time ko position $x$ par depend banata hai.
Step 7 mein $v$ se divide karne se pehle kya check karna chahiye? ::: Ki $v \neq = 0$ ; $v = 0$ case seedha identity $x^{'} = x, t^{'} = t$ deta hai.
Kaunsi quantity har frame mein same hai? ::: Spacetime interval $c^{2} t^{2} - x^{2}$ .
$v \to c$ hone par $γ$ ka kya hota hai? ::: Yeh $\infty$ tak diverge karta hai; $v \geq c$ par koi real transform exist nahi karta.
Chaar linear coefficients $A, B, D, E$ pin hone ke baad kya hain? ::: $A = γ$ , $B = - γ v$ , $D = - γ v / c^{2}$ , $E = γ$ .
Hum $S^{'}$ ko negative $x$ -direction mein move karte kaise handle karte hain? ::: $v < 0$ use karo; wahi formulas valid hain aur $γ$ unchanged rehta hai kyunki yeh $v^{2}$ par depend karta hai.