Lorentz transformation — derivation: Worked examples

2.3.28 · D3 · Physics › Modern Physics › Lorentz transformation — derivation

Shuru karne se pehle, units ke baare mein ek reminder taaki koi example aapko trip na kare.

Scenario matrix

Har Lorentz-transform problem is table ke ek cell mein aata hai. Right column us example ka naam batata hai jo us cell pe land karta hai.

Cell class	Kya special hai	Covered by
Zero velocity ( $v = 0$ )	$γ = 1$ : kuch bhi change nahi hona chahiye	Ex. 1
Low velocity ( $v ≪ c$ )	Galileo $x^{'} = x - v t$ pe reduce hona chahiye	Ex. 2
Event ahead ( $x > 0$ ), forward frame ( $v > 0$ )	standard positive case	Ex. 3
Event behind ( $x < 0$ ) ya backward frame ( $v < 0$ )	$- v x / c^{2}$ term ka sign flip ho jaata hai	Ex. 4
Simultaneous pair ( $Δ t = 0$ )	relativity of simultaneity, dono signs	Ex. 5
Limiting speed ( $v \to c$ )	$γ \to \infty$ , degenerate squeeze	Ex. 6
Real-world word problem	muon / spaceship, khud $x, t, v$ nikalo	Ex. 7
Exam twist (inverse + invariant)	$S^{'}$ se wapas $S$ mein jao, $s^{2}$ check karo	Ex. 8

Hum inhe order mein chalte hain. Minkowski diagram, Time dilation, Length contraction aur Relativity of simultaneity pages ko dhyan mein rakho — har example quietly unhe touch karta hai.

Worked examples

Worked example 1 · Zero-velocity sanity check — cell:

$v = 0$ Statement. Frame $S^{'}$ move nahi karta: $v = 0$ . Ek event $x = 4 ls$ , $t = 5 s$ pe hota hai. $(x^{'}, t^{'})$ nikalo.

Forecast: Compute karne se pehle guess karo — agar frames ek doosre ke relative move hi nahi karte, toh koi bhi coordinate change hona chahiye kya?

$γ$ compute karo. $γ = 1/ 1 - 0^{2} / c^{2} = 1/ 1 = 1$ . Yeh step kyun? Har cheez $γ$ se scale hoti hai; agar $γ = 1$ toh koi stretch hi nahi hai.
$x^{'} = γ (x - v t)$ apply karo. $v = 0$ ke saath: $x^{'} = 1 \cdot (4 - 0) = 4 ls$ . Yeh step kyun? $- v t$ correction zero ho jaata hai kyunki $v = 0$ hai.
$t^{'} = γ (t - v x / c^{2})$ apply karo. $v = 0$ ke saath: $t^{'} = 1 \cdot (5 - 0) = 5 s$ . Yeh step kyun? $- v x / c^{2}$ correction bhi zero ho jaata hai.

Verify: $(x^{'}, t^{'}) = (4, 5) = (x, t)$ . Identical frames identical coordinates dete hain — transform identity mein reduce ho jaata hai, exactly waisa hi jaisa physics demand karti hai. ✓

Worked example 2 · Low speed → Galileo — cell:

$v ≪ c$ Statement. Ek car frame $v = 30 m/s$ pe move kar raha hai. Ek event $x = 300 m$ , $t = 2 s$ pe hai. Dikhao ki $x^{'} \approx x - v t$ aur $t^{'} \approx t$ . ( $c = 3 \times 1 0^{8} m/s$ .)

Forecast: Kya aap expect karte ho ki car speeds pe koi relativistic difference dikhega?

$v^{2} / c^{2}$ estimate karo. $v^{2} / c^{2} = (30)^{2} / (3 \times 1 0^{8})^{2} = 900/ (9 \times 1 0^{16}) = 1 0^{- 14}$ . Yeh step kyun? Yahi chhota number square root ke andar baithta hai; agar yeh negligible hai, toh $γ \approx 1$ .
Toh $γ \approx 1$ 14 decimal places tak. Yeh step kyun? $1 - 1 0^{- 14} \approx 1$ , toh $γ$ practically 1 se alag nahi hai.
Space: $x^{'} \approx x - v t = 300 - 30 \cdot 2 = 240 m$ — plain Galilean answer. Yeh step kyun? $γ = 1$ ke saath Lorentz ka $x^{'}$ ही Galilean $x^{'}$ ban jaata hai.
Time: correction $v x / c^{2} = 30 \cdot 300/ (9 \times 1 0^{16}) = 1 0^{- 13} s$ — bilkul negligible hai, toh $t^{'} \approx t = 2 s$ . Yeh step kyun? $v x / c^{2}$ woh term hai jo universal time ko "break" karta hai; yahan yeh $1 0^{- 13} s$ hai, unmeasurable.

Verify: Naya theory purane ko limit ke roop mein contain karta hai (Galilean transformation recover hua). Units: $v x / c^{2}$ ki units $(m/s) (m) / (m/s)^{2} = s$ hain. ✓

Worked example 3 · Event ahead, forward-moving frame — cell:

$x > 0, v > 0$ Statement. $S^{'}$ , $v = 0.6 c$ pe move karta hai. Ek event $x = 10 ls$ , $t = 5 s$ pe hai. $(x^{'}, t^{'})$ nikalo.

Forecast: Frame $+ x$ ki taraf chase kar raha hai. Kya event ki primed distance 10 se zyada hogi ya kam? Kya uski primed time 5 se pehle hogi ya baad mein?

$γ$ compute karo. $γ = 1/ 1 - 0. 6^{2} = 1/ 1 - 0.36 = 1/ 0.64 = 1/0.8 = 1.25$ . Yeh step kyun? $γ$ dono outputs ko multiply karta hai.
Space. $c = 1$ use karte hain toh distances ls mein hain, times s mein: $x^{'} = γ (x - v t) = 1.25 (10 - 0.6 \cdot 5) = 1.25 (10 - 3) = 1.25 \cdot 7 = 8.75 ls$ . Yeh step kyun? $v t = 0.6 \cdot 5 = 3$ ls woh distance hai jo $S^{'}$ aage khiski hai; event $γ$ stretch se pehle sirf 7 ls aage hai.
Time. $t^{'} = γ (t - v x / c^{2}) = 1.25 (5 - 0.6 \cdot 10) = 1.25 (5 - 6) = 1.25 \cdot (- 1) = - 1.25 s$ . Yeh step kyun? $c = 1$ ke saath, $v x / c^{2} = 0.6 \cdot 10 = 6$ ; kyunki $x$ bada aur positive hai, yeh correction itna bada hai ki $t^{'}$ ko negative kar deta hai — $S^{'}$ mein event pehle hi ho chuka hai.

Verify: Neeche Minkowski diagram ke saath sign check: ek door-ahead event, ek forward observer ke liye, earlier primed time pe drag ho jaata hai. Positive $x$ , positive $v$ ⇒ subtract ⇒ time negative ja sakta hai. ✓

Worked example 4 · Event behind / backward frame — cell:

$x < 0$ ya $v < 0$ Statement. Same speed magnitude lekin ab $S^{'}$ doosri taraf move karta hai, $v = - 0.6 c$ , aur event origin ke peeche hai $x = - 10 ls$ , $t = 5 s$ . $(x^{'}, t^{'})$ nikalo.

Forecast: Do sign flips hain ( $v$ aur $x$ ). Kya woh time equation mein cancel ho jaate hain ya double up ho jaate hain?

$γ$ unchanged hai. $γ = 1/ 1 - (- 0.6)^{2} = 1/ 0.64 = 1.25$ . Yeh step kyun? $γ$ sirf $v^{2}$ pe depend karta hai, toh $v$ ka sign kabhi usse affect nahi karta — ek moving clock dono directions mein same slow hoti hai.
Space. $x^{'} = γ (x - v t) = 1.25 (- 10 - (- 0.6) (5)) = 1.25 (- 10 + 3) = 1.25 (- 7) = - 8.75 ls$ . Yeh step kyun? $- v t = + 3$ ab ho gaya (double negative); event peeche hi rehta hai, $- 8.75$ ls pe.
Time. $t^{'} = γ (t - v x / c^{2}) = 1.25 (5 - (- 0.6) (- 10)) = 1.25 (5 - 6) = - 1.25 s$ . Yeh step kyun? $(- 0.6) (- 10) = + 6$ : do minus signs multiply hokar plus dete hain, toh correction Ex. 3 jaisi hi hai. Yahi key lesson hai — yeh product $v x$ hai, na koi ek sign, jo time shift set karta hai.

Verify: Pure mirror symmetry se answer hai $(- 8.75 ls, - 1.25 s)$ : Ex. 3 ki poori picture left-to-right flip ho gayi. Sign discipline confirmed. ✓

Worked example 5 ·

$S$ mein simultaneous, $S^{'}$ mein nahi — cell: $Δ t = 0$ , $x$ ke dono signs Statement. $S$ mein, do flashes same time $t = 0$ pe hoti hain: flash A $x_{A} = + 8 ls$ pe, flash B $x_{B} = - 8 ls$ pe. Frame $S^{'}$ , $v = 0.6 c$ pe move karta hai. $S^{'}$ pehle kaunsi flash dekhta hai?

Forecast: Woh $S$ mein simultaneous hain. Guess karo: kya $S^{'}$ unhe bhi simultaneous kehta hai? Agar nahi, toh pehle kaun aata hai — aage wali ya peeche wali?

$γ = 1.25$ pehle jaisi. Yeh step kyun? Same speed, same factor.
Flash A ( $x = + 8$ ): $t_{A}^{'} = γ (0 - v x_{A} / c^{2}) = 1.25 (0 - 0.6 \cdot 8) = 1.25 (- 4.8) = - 6 s$ . Yeh step kyun? Sirf $- v x / c^{2}$ term bachta hai ( $t = 0$ ); positive $x$ se negative primed time milta hai.
Flash B ( $x = - 8$ ): $t_{B}^{'} = γ (0 - v x_{B} / c^{2}) = 1.25 (0 - 0.6 \cdot (- 8)) = 1.25 (+ 4.8) = + 6 s$ . Yeh step kyun? Negative $x$ sign flip kar deta hai, positive primed time deta hai.
Compare karo. $t_{A}^{'} = - 6 < t_{B}^{'} = + 6$ , toh $S^{'}$ mein flash A (aage wali) pehle hoti hai, B se poore 12 s pehle. Yeh step kyun? $+ x$ ki taraf move karne wala observer us side ke event se "pehle" milta hai apne time coordinate mein.

Verify: $Δ t = 0$ $S$ mein, lekin $Δ t^{'} = t_{B}^{'} - t_{A}^{'} = 12 s \neq = 0$ $S^{'}$ mein. Yahi exactly Relativity of simultaneity hai — simultaneity absolute nahi hai, aur ordering $x$ ke sign se set hoti hai. ✓

Worked example 6 · Limit

$v \to c$ — cell: degenerate / limiting speed Statement. $γ$ aur transform ko track karo jab $v$ badhta hai: $v = 0.8 c, 0.99 c, 0.9999 c$ . $γ$ ko kya hota hai, aur $x = 1 ls, t = 0$ pe event ko?

Forecast: Jab $v$ , $c$ ke paas pahunchta hai, kya $γ$ kisi finite ceiling ke paas jaata hai ya blow up ho jaata hai?

$v = 0.8 c$ : $γ = 1/ 1 - 0.64 = 1/ 0.36 = 1/0.6 = 1.6 \overline{6}$ . Yeh step kyun? Upar jaane se pehle baseline establish karo.
$v = 0.99 c$ : $γ = 1/ 1 - 0.9801 = 1/ 0.0199 \approx 7.089$ . Yeh step kyun? Root ke andar ka number 0 ki taraf shrink ho raha hai, toh $γ$ upar shoot karta hai.
$v = 0.9999 c$ : $γ = 1/ 1 - 0.99980001 = 1/ 0.00019999 \approx 70.71$ . Yeh step kyun? $c$ ke har step ke paas $1 - v^{2} / c^{2}$ ten-fold chhota ho jaata hai, $γ$ ten-fold... nahi, $\approx 10$ -fold bada ho jaata hai. Yeh diverge karta hai: $γ \to \infty$ jab $v \to c$ .
Event ka time $t^{'} = γ (0 - v x / c^{2}) = - γ v x / c^{2}$ magnitude mein unbounded badhta hai. Yeh step kyun? Mass wali koi cheez $v = c$ reach nahi kar sakti — transform singular ho jaata hai ( $γ$ ke andar division-by-zero), jo physics ka yeh kehna hai ki " $c$ ek wall hai."

Verify: Sequence $γ \in {1.667, 7.089, 70.71}$ monotone increasing aur unbounded hai. $v = c$ case genuinely undefined hai (denominator mein $0$ ), sahi degenerate behaviour. ✓

Worked example 7 · Real-world word problem — muon — cell: word problem

Statement. Ek muon atmosphere mein upar create hota hai aur seedha neeche $v = 0.98 c$ pe move karta hai. Apne frame mein yeh decay hone se pehle $τ = 2.2 μ s$ jeeta hai. Transform use karke, decay hone se pehle muon ground frame mein kitna door travel karta hai? ( $c = 3 \times 1 0^{8} m/s$ .)

Forecast: $0.98 c$ pe $2.2 μ s$ mein ek Newtonian ko $\approx 647 m$ milega. Guess karo: kya ground use zyada ya kam door jaate dekhega?

Frames identify karo. $S^{'}$ = muon (wahan woh rest mein hai), $S$ = ground. Do events: birth $(x^{'} = 0, t^{'} = 0)$ aur death $(x^{'} = 0, t^{'} = τ = 2.2 \times 1 0^{- 6} s)$ — $S^{'}$ mein same jagah kyunki muon apne frame mein still hai. Yeh step kyun? Us frame ko choose karna jahan muon rest mein hai, uski lifetime ko clean proper time $τ$ banata hai.
$γ$ . $γ = 1/ 1 - 0.9 8^{2} = 1/ 1 - 0.9604 = 1/ 0.0396 \approx 5.025$ . Yeh step kyun? Yahi factor muon ki short life ko ground time mein stretch karta hai.
Death ka ground time inverse $t = γ (t^{'} + v x^{'} / c^{2})$ se, $x^{'} = 0$ ke saath: $t = γ τ = 5.025 \times 2.2 \times 1 0^{- 6} \approx 1.106 \times 1 0^{- 5} s = 11.06 μ s$ . Yeh step kyun? Hum primed → unprimed ja rahe hain, toh inverse transform use karte hain; moving muon ki clock slow chali, toh ground clock zyada padhti hai (Time dilation).
Ground distance. $S$ mein muon travel karta hai $d = v t = 0.98 \times (3 \times 1 0^{8}) \times 1.106 \times 1 0^{- 5} \approx 3251 m \approx 3.25 km$ . Yeh step kyun? Distance = speed × ground time; dilated time use karne se real experimental range milti hai.

Verify: Naive answer $647 m$ vs relativistic $\approx 3251 m$ — factor hai $γ \approx 5.03$ , aur indeed $647 \times 5.025 \approx 3251$ . Yahi extra reach hai isliye muons ground tak pahunchte hain aur detect hote hain. Units: $(m/s) (s) = m$ . ✓

Worked example 8 · Exam twist — inverse transform + invariant interval — cell: inverse &

$s^{2}$ Statement. Tumhe ek event ke primed coordinates diye gaye hain: $x^{'} = 3 ls$ , $t^{'} = 5 s$ , $v = 0.6 c$ ke saath. (a) Inverse transform use karke $S$ mein $(x, t)$ nikalo. (b) Confirm karo ki Spacetime interval $s^{2} = c^{2} t^{2} - x^{2}$ dono frames mein match karta hai.

Forecast: Do independent checks. Kya primed values se compute kiya gaya interval unprimed values wale se match karega?

$γ = 1.25$ ( $v = 0.6 c$ se). Yeh step kyun? Pehle wale examples jaisi hi speed; reuse karo.
Inverse space $x = γ (x^{'} + v t^{'}) = 1.25 (3 + 0.6 \cdot 5) = 1.25 (3 + 3) = 1.25 \cdot 6 = 7.5 ls$ . Yeh step kyun? $S^{'}$ se $S$ wapas jaane ke liye $v$ ka sign flip ho jaata hai, minus plus ban jaata hai — inverse transform.
Inverse time $t = γ (t^{'} + v x^{'} / c^{2}) = 1.25 (5 + 0.6 \cdot 3) = 1.25 (5 + 1.8) = 1.25 \cdot 6.8 = 8.5 s$ . Yeh step kyun? Time equation ke liye bhi same sign-flip logic.
$S$ mein interval ( $c = 1$ ): $s^{2} = t^{2} - x^{2} = 8. 5^{2} - 7. 5^{2} = 72.25 - 56.25 = 16$ . Yeh step kyun? Unprimed values se candidate invariant compute karo.
$S^{'}$ mein interval: $s^{'2} = t^{'2} - x^{'2} = 5^{2} - 3^{2} = 25 - 9 = 16$ . Yeh step kyun? Primed values se same quantity compute karo.

Verify: $s^{2} = s^{'2} = 16$ — spacetime interval frame-independent hai, wahi geometric heart jo parent note ne promise kiya tha. Inverse coordinates aur invariance dono check out karte hain. ✓