Simultaneity — relativity of simultaneity: Worked examples

2.3.27 · D3 · Physics › Modern Physics › Simultaneity — relativity of simultaneity

Examples se pehle, aao har symbol ko dobara samjhein taaki thanda padha hua reader bhi pehli line se follow kar sake.

Definition Chhe symbols, seedhe shabdon mein

$v$ — frame $S^{'}$ ki speed frame $S$ ke relative, ek seedhi line ke along measure ki gayi jise hum $x$ -axis kehte hain. Socho "train platform ke past kitni tezi se slide karti hai." Yahan hamesha $c$ ka fraction hoti hai.
$c$ — speed of light, $3 \times 1 0^{8}$ metres per second. Woh ek speed jis par har observer agree karta hai. Yahi wajah hai ki yeh poora effect exist karta hai (dekho Postulates of Special Relativity).
$Δ x = x_{2} - x_{1}$ — do events ke beech ka spatial gap, frame $S$ mein measure kiya gaya (woh frame jahan dono ek saath hote hain). Agar event 2, event 1 se $+ x$ direction mein aage hai, toh $Δ x$ positive hai.
$Δ t = t_{2} - t_{1}$ — do events ke beech ka time gap frame $S$ mein measure kiya gaya. Is poore page par (jahan hum aur nahi kehte wahan tak) hamare events $S$ mein simultaneous hain, matlab $Δ t = 0$ . Yeh symbol yaad rakho; yeh causality example mein wapas aata hai.
$γ$ (gamma) — stretch factor $\frac{1}{1 - v ^{2} / c ^{2}}$ . Yeh hamesha $\geq 1$ hota hai. $v = 0$ par yeh exactly $1$ hai; jaise $v \to c$ , yeh infinity tak blow up karta hai. Yeh batata hai "is speed par time aur space kitna warp hota hai?" (dekho Time Dilation, Length Contraction).
$Δ t^{'} = t_{2}^{'} - t_{1}^{'}$ — unhi do events ke beech ka time gap, lekin ab moving frame $S^{'}$ mein measure kiya gaya. Agar yeh negative nikle, toh event 2, $S^{'}$ mein event 1 se pehle hua.

Scenario matrix

Is topic ka har sawaal inhi cells mein se kisi ek mein rehta hai. Neeche ke examples us cell ke saath label kiye gaye hain jise woh fill karte hain.

#	Case class	Kya test ho raha hai	Example
A	$Δ x > 0$ , ordinary $v$	positive separation, $Δ t^{'}$ ka sign	Ex 1
B	$Δ x < 0$	negative separation sign flip karta hai	Ex 2
C	$Δ x = 0$ (degenerate)	same place → koi disagreement nahi	Ex 3
D	$v = 0$ (degenerate)	koi relative motion nahi → koi disagreement nahi	Ex 3
E	$v \to c$ (limiting)	$γ \to \infty$ , gap blow up karta hai	Ex 4
F	Real-world word problem	words ko symbols mein translate karo	Ex 5
G	"Kaun pehle / kitna" (train)	leading-clocks-lag apply karo	Ex 6
H	Exam twist: wrong-frame $Δ x$ trap	kis frame ka separation use karna hai	Ex 7
I	Ordering / causality check	spacelike vs timelike, kya order flip ho sakta hai?	Ex 8

Hum poore mein $c = 3 \times 1 0^{8} m/s$ use karte hain.

Picture padhna: "now" kaise tilt hoti hai

Numbers se pehle, dekho simultaneity of simultaneity ek spacetime diagram par kaisi dikhti hai — ek graph jisme position across aur time upar hota hai.

Figure — Simultaneity — relativity of simultaneity

Worked examples

Worked example Example 1 — Cell A: positive separation

Ek train $L = 240 m$ lambi hai. Yeh $+ x$ direction mein $v = 0.6 c$ par move karti hai. Platform frame $S$ mein, back (event 1, at $x_{1}$ ) aur front (event 2, at $x_{2} = x_{1} + 240$ ) par lightning simultaneously girti hai (toh $S$ mein $Δ t = 0$ ). Train frame $S^{'}$ mein $Δ t^{'}$ nikalo.

Forecast: Front motion ki direction mein hai, toh "leading clock lags" kehta hai ki front event train frame mein earlier register hona chahiye. Guess: $Δ t^{'}$ negative hai, ek microsecond ka fraction. Figure par dekho ki dono dots tilted magenta "now" line se kahan milte hain.

$Δ x = x_{2} - x_{1} = + 240 m$ , aur $Δ t = 0$ ( $S$ mein simultaneous). Yeh step kyun? Humein ek sign convention fix karni hai aur key assumption confirm karni hai. Event 2 = front, $+ x$ mein aage, toh $Δ x$ positive hai; simultaneous matlab boxed formula apply hota hai.
$γ = \frac{1}{1 - 0. 6 ^{2}} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25$ . Yeh step kyun? Formula ko $γ$ chahiye; yeh raw speed ko warp factor mein convert karta hai.
$Δ t^{'} = - \frac{γ v Δ x}{c ^{2}} = - \frac{( 1.25 ) ( 0.6 c ) ( 240 )}{c ^{2}} = - \frac{( 1.25 ) ( 0.6 ) ( 240 )}{c} = - \frac{180}{c}$ . Yeh step kyun? $c$ ka ek factor $v = 0.6 c$ aur $c^{2}$ ke beech cancel hota hai, metres over $c$ bachta hai.
$Δ t^{'} = - \frac{180}{3 \times 1 0 ^{8}} = - 6.0 \times 1 0^{- 7} s = - 0.60 μ s$ . Yeh step kyun? Metres ko $c$ se divide karo, seconds milte hain.

Verify: Sign negative hai → front (event 2) $S^{'}$ mein pehle hua. Yeh "leading clocks lag" aur figure mein magenta line ke tilt se match karta hai. Units: $m / (m/s) = s$ . ✅

Worked example Example 2 — Cell B: negative separation sign flip karta hai

Wahi train, wahi $v = 0.6 c$ , events ab bhi $S$ mein simultaneous ( $Δ t = 0$ ), lekin ab hum front ko event 1 aur back ko event 2 label karte hain. Toh event 2 peeche hai. $Δ t^{'}$ nikalo.

Forecast: Humne sirf relabel kiya — koi physics nahi badla. Magnitude $0.60 μ s$ rehni chahiye, lekin kyunki $Δ x$ ab doosri taraf point karta hai, sign positive ho jaana chahiye.

$Δ x = x_{2} - x_{1} = x_{back} - x_{front} = - 240 m$ . Yeh step kyun? Event 2 ab back hai, jo chhote $x$ par hai, toh difference negative hai.
$γ = 1.25$ (unchanged — speed nahi badla). Yeh step kyun? $γ$ sirf $v$ par depend karta hai, labels par nahi.
$Δ t^{'} = - \frac{( 1.25 ) ( 0.6 c ) ( - 240 )}{c ^{2}} = + \frac{180}{c} = + 6.0 \times 1 0^{- 7} s = + 0.60 μ s$ . Yeh step kyun? Formula ka minus times $Δ x$ ka minus plus deta hai.

Verify: $Δ t^{'} > 0$ matlab event 2 (back) baad mein hota hai, yaani front pehle hota hai — exactly wahi physical story jo Example 1 mein thi, swapped labels ke saath. Magnitude identical. ✅ Lesson: sign bilkul tumhare is choice mein hai ki "2" event kaun sa hai.

Worked example Example 3 — Cells C & D: do degenerate cases

(C) Do firecrackers $S$ mein same point par blast hote hain: $Δ x = 0$ , $Δ t = 0$ , $v = 0.6 c$ . (D) Do firecrackers $S$ mein $240 m$ door simultaneously ( $Δ t = 0$ ) blast hote hain, lekin $S^{'}$ ki zero speed hai: $Δ x = 240$ , $v = 0$ . Dono mein $Δ t^{'}$ nikalo.

Forecast: Dono mein exactly zero disagreement hona chahiye — ek kyunki disagree karne ke liye koi separation hi nahi hai, doosre mein kyunki disagreement cause karne ke liye koi motion hi nahi hai.

(C) $Δ t^{'} = - \frac{γ ( 0.6 c ) ( 0 )}{c ^{2}} = 0$ . Yeh step kyun? Poora effect $Δ x$ ke proportional hai. $Δ x$ khatam, effect khatam. Same-place events sab ke liye simultaneous hote hain.
(D) $γ = 1/ 1 - 0 = 1$ , aur $Δ t^{'} = - \frac{( 1 ) ( 0 ) ( 240 )}{c ^{2}} = 0$ . Yeh step kyun? $v = 0$ ke saath frames ek hi frame hain, toh simultaneity preserved hoti hai.

Verify: Dono zero. Yeh do cells tumhare sanity anchors hain — jo bhi formula tum likhte ho woh in limits mein no-disagreement par collapse hona chahiye. ✅

Worked example Example 4 — Cell E: ultra-relativistic limit

$v \to c$ Ek fixed separation $Δ x = 240 m$ lo jisme events $S$ mein simultaneous hain ( $Δ t = 0$ ), aur dekho $Δ t^{'}$ kaise badhta hai jab $v$ climb karta hai: $0.6 c$ , $0.9 c$ , $0.99 c$ , aur limiting trend jaise $v \to c$ .

Forecast: $γ$ blow up karta hai jaise $v \to c$ , toh disagreement bina kisi ceiling ke unboundedly badhni chahiye — infinity approach karte hue, kisi finite ceiling tak nahi pahunchi.

$v = 0.6 c$ par: $γ = 1.25$ , $∣Δ t^{'} ∣ = \frac{( 1.25 ) ( 0.6 ) ( 240 )}{c} = \frac{180}{c} = 6.0 \times 1 0^{- 7} s$ . Yeh step kyun? Example 1 se baseline.
$v = 0.9 c$ par: $γ = 1/ 1 - 0.81 = 1/ 0.19 \approx 2.294$ . $∣Δ t^{'} ∣ = \frac{( 2.294 ) ( 0.9 ) ( 240 )}{c} = \frac{495.6}{c} \approx 1.652 \times 1 0^{- 6} s$ . Yeh step kyun? Wahi formula, zyada speed — number climb karte dekho.
$v = 0.99 c$ par: $γ = 1/ 1 - 0.9801 \approx 7.089$ . $∣Δ t^{'} ∣ = \frac{( 7.089 ) ( 0.99 ) ( 240 )}{c} = \frac{1684.3}{c} \approx 5.614 \times 1 0^{- 6} s$ . Yeh step kyun? Jump notice karo — $c$ ke qareeb, product $γ v$ tezi se upar jaata hai.
Jaise $v \to c$ : $γ \to \infty$ , toh $∣Δ t^{'} ∣ \to \infty$ . Yeh step kyun? $1 - v^{2} / c^{2} \to 0$ denominator mein; kuch bhi gap ko bound nahi karta.

Verify: Orange curve dekho — yeh tezi se upar jaati hai aur dashed $v = c$ wall ke qareeb sky ki taraf shoot karti hai. Teen coloured dots exactly steps 1–3 ke values hain. Disagreement unbounded hai, step 4 confirm karta hai. ✅

Worked example Example 5 — Cell F: real-world word problem

Ek $12, 000 km$ lamba fibre-optic cable Earth par $+ x$ direction mein rakha hai. Do maintenance pulses bheji jaati hain taaki, Earth ke frame mein, woh dono ends mein same instant par enter hon ( $Δ t = 0$ Earth's frame mein). Ek spacecraft cable ke upar se front-to-back, $v = 0.5 c$ par fly karta hai. Spacecraft ke frame mein, kaun se end ka pulse pehle enter karta hai, aur kitne se?

Forecast: Spacecraft $+ x$ mein move karta hai (far end, event 2 ki taraf). Leading clock lags → far end pulse ship frame mein pehle enter hona chahiye. Magnitude: separation bahut badi hai ( $1.2 \times 1 0^{7} m$ ), toh bada gap expect karo — tens of milliseconds.

$Δ x = 1.2 \times 1 0^{7} m$ , $Δ t = 0$ Earth's frame mein (near end = event 1 at $x_{1}$ , far end = event 2 at $x_{2} > x_{1}$ ). Yeh step kyun? km ko m mein convert karo, sign set karo (far end bade $x$ par), aur note karo ki simultaneity hold karti hai toh boxed formula apply hota hai.
$γ = 1/ 1 - 0.25 = 1/ 0.75 \approx 1.1547$ . Yeh step kyun? Speed $0.5 c$ hai; warp factor nikalo.
$Δ t^{'} = - \frac{( 1.1547 ) ( 0.5 c ) ( 1.2 \times 1 0 ^{7} )}{c ^{2}} = - \frac{( 1.1547 ) ( 0.5 ) ( 1.2 \times 1 0 ^{7} )}{3 \times 1 0 ^{8}}$ . Yeh step kyun? Plug in karo; ek $c$ cancel karo.
$Δ t^{'} = - \frac{6.928 \times 1 0 ^{6}}{3 \times 1 0 ^{8}} \approx - 2.31 \times 1 0^{- 2} s \approx - 23.1 ms$ . Yeh step kyun? $c$ se divide karo; seconds mein land karo, phir milliseconds mein convert karo.

Verify: Negative sign → far end (event 2) pehle, forecast se match karta hai. Magnitude $\approx 23$ ms — badi kyunki cable enormous hai, exactly jaise predict kiya gaya. Units: $m / (m/s) = s$ . ✅

Worked example Example 6 — Cell G: "kaun pehle, kitna" (train, forecast-then-verify)

Ek $500 m$ lambi train ek station se $v = 0.8 c$ par guzarti hai. Station kehta hai front aur back ke dono doors simultaneously khulte hain ( $Δ t = 0$ station frame mein). Train frame mein, kaun sa door pehle khulta hai, aur kitne se?

Forecast: Front leading end hai; "leading clock lags" matlab front event moving frame mein earlier time assign hota hai. Toh front door pehle khulta hai. Guess magnitude: couple of microseconds. Figure par, front dot tilted magenta "now" line ke neeche baitha hai.

$Δ x = + 500 m$ , $Δ t = 0$ (front = event 2, $+ x$ mein aage; station frame mein simultaneous). Yeh step kyun? Sign set karo taaki front event 2 ho aur assumption confirm karo.
$γ = 1/ 1 - 0.64 = 1/ 0.36 = 1/0.6 \approx 1.6667$ . Yeh step kyun? $0.8 c$ par standard warp factor.
$Δ t^{'} = - \frac{( 1.6667 ) ( 0.8 c ) ( 500 )}{c ^{2}} = - \frac{( 1.6667 ) ( 0.8 ) ( 500 )}{c} = - \frac{666.7}{c}$ . Yeh step kyun? Ek $c$ cancel karo; minus track karo.
$Δ t^{'} = - \frac{666.7}{3 \times 1 0 ^{8}} \approx - 2.22 \times 1 0^{- 6} s = - 2.22 μ s$ . Yeh step kyun? $c$ se divide karo.

Verify: Negative → front (event 2) earlier → front door pehle, corrected forecast aur figure mein tilt se match karta hai. Yeh parent ke Einstein-train story ka number mein resolution hai. ✅

Worked example Example 7 — Cell H: wrong-frame trap

Ek rod apne rest frame $S^{'}$ mein $200 m$ lambi hai. Yeh tumhare past $v = 0.6 c$ par fly karta hai. Dono ends par tumhare frame $S$ mein simultaneously ( $Δ t = 0$ $S$ mein) strike hoti hai. Ek classmate compute karta hai $Δ t^{'} = - γ v (200) / c^{2}$ . Unhone kya galat kiya, aur sahi answer kya hai?

Forecast: Formula $Δ x$ us frame mein maangta hai jahan events simultaneous hain — woh tumhara frame $S$ hai, rod ka rest frame nahi. $200 m$ rod ki rest length hai, toh $S$ mein rod length-contracted aur chhoти hai. Guess: classmate ka number exactly $γ$ se bada hai.

$S$ mein sahi $Δ x$ : rod contracted hai, $Δ x = \frac{200}{γ}$ . Yeh step kyun? Length Contraction se, ek rest length $L_{0}$ us frame mein $L_{0} / γ$ dikhti hai jahan woh move karta hai.
$γ = 1.25$ (Ex 1 ki tarah), toh $Δ x = 200/1.25 = 160 m$ . Yeh step kyun? Numerically length contract karo.
Sahi $Δ t^{'} = - \frac{γ v Δ x}{c ^{2}} = - \frac{( 1.25 ) ( 0.6 c ) ( 160 )}{c ^{2}} = - \frac{120}{c} = - 4.0 \times 1 0^{- 7} s = - 0.40 μ s$ . Yeh step kyun? Ab har quantity sahi frame mein hai.
Classmate ka galat value: $- \frac{( 1.25 ) ( 0.6 ) ( 200 )}{c} = - \frac{150}{c} = - 5.0 \times 1 0^{- 7} s$ — factor $200/160 = 1.25 = γ$ se bada. Yeh step kyun? Exactly dikhao ki mistake answer ko kaise inflate karti hai.

Verify: $0.50 μ s = γ \times 0.40 μ s = 1.25 \times 0.40 μ s$ . ✅ Trap real hai aur precisely $γ$ se off hai. Hamesha us frame ka separation use karo jahan events simultaneous hain.

Worked example Example 8 — Cell I: kya order actually reverse ho sakta hai? (causality check)

Frame $S$ mein do events simultaneous hain ( $Δ t = t_{2} - t_{1} = 0$ ) aur $Δ x = 240 m$ se separated hain. (a) Kya yeh spacelike ya timelike separated hain? (b) Kya koi observer event 1 ko event 2 cause karta dekh sakta hai?

Forecast: Simultaneous lekin separated matlab light zero time mein gap cross nahi kar sakti — toh yeh spacelike separated hone chahiye, aur koi causal link possible nahi hai. Unka time-order frame-dependent hai (flip ho sakta hai), lekin yeh harmless hai.

Spacetime interval squared: $s^{2} = (c Δ t)^{2} - (Δ x)^{2} = (c \cdot 0)^{2} - (240)^{2} = - (240)^{2} < 0$ . Yeh step kyun? Yahan $Δ t = 0$ woh time separation hai $S$ mein (woh symbol jo humne upar define kiya). Negative $s^{2}$ (is sign convention ke saath) ek spacelike separation define karta hai — dekho Causality and the Light Cone.
Unhe link karne ke liye, light ko $Δ t = 0$ seconds mein $240 m$ travel karna hoga — impossible, kyunki iske liye speed $\infty > c$ chahiye. Yeh step kyun? $c$ ya neeche ka koi signal unhe connect nahi kar sakta, toh koi bhi doosre ko cause nahi kar sakta.
Kyunki yeh spacelike hain, unka order frames ke beech flip kar sakta hai (Examples 1 & 2 ne already dikhaya ki $Δ t^{'}$ ka sign labelling ke saath change hota hai), aur yeh theek hai — causal kuch bhi kabhi stake par tha hi nahi. Yeh step kyun? Parent note ka promise confirm karta hai: sirf spacelike pairs flip hote hain, aur woh ek doosre ko influence nahi kar sakte, toh koi paradox nahi.

Verify: $s^{2} = - 57600 m^{2} < 0$ → spacelike, confirmed. Order-flip allowed, causality safe. ✅

Recall Quick self-test

Ek $300 m$ train ( $v = 0.6 c$ ) ke front aur back par platform frame mein simultaneously strike hoti hai. Train frame mein gap? ::: $∣Δ t^{'} ∣ = γ v Δ x / c^{2} = (1.25) (0.6 c) (300) / c^{2} = 225/ c = 7.5 \times 1 0^{- 7} s = 0.75 μ s$ ; front pehle strike karta hai. Kaun si assumption $Δ t^{'} = - γ v Δ x / c^{2}$ ko valid banati hai? ::: Events $S$ mein simultaneous hain, yaani $Δ t = 0$ ; warna poora $γ (Δ t - v Δ x / c^{2})$ use karo. Formula mein kis frame ka $Δ x$ jaata hai? ::: Woh frame jahan do events simultaneous hain. Jaise $v \to c$ aur $Δ x$ fixed hai, $Δ t^{'}$ ka kya hota hai? ::: Yeh without bound badhta hai kyunki $γ \to \infty$ . Agar $Δ x = 0$ hai, toh har frame mein $Δ t^{'}$ kya hai? ::: Zero — same-place events absolutely simultaneous hote hain.

Parent par wapas jao: Simultaneity — relativity of simultaneity.