Exercises — Simultaneity — relativity of simultaneity

Hum kya test kar rahe hain: kya spatial separation hi disagreement drive karti hai. Yahaan $\Delta x=0$ hai. Simultaneity gap ko kaam karne ke liye nonzero $\Delta x$ chahiye, lekin zyada directly yeh events timelike hain (same jagah, real time gap → light easily connect kar sakti hai). Timelike order sab frames mein preserve hoti hai. Toh haan, rocket ke liye same order — pehla event pehla hi rehta hai. Rocket lamba gap dekhta hai (woh Time Dilation hai), lekin kabhi flip nahi hoga.

Hum kya test kar rahe hain: numbers plug kiye bina sign padhna. Figure s01 par map karo: A back B hai (at $x=0$), B front F hai (at $x=+500$). Amber clock dekho — front clock lag karta hai, toh $S^{\prime }$ mein front event B pehle hota hai. Signed formula se check karo (event 2 $=$ B, event 1 $=$ A): $\Delta x=x_B-x_A=+500>0$, toh $\Delta t^{\prime }=t_B^{\prime }-t_A^{\prime }=-\gamma v\Delta x/c^2<0$. Negative signed gap matlab $t_B^{\prime }<t_A^{\prime }$: B pehle hota hai. ✅

Step 1 — $\gamma$ nikalo (kyun: formula ko chahiye). $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{1-0.36}}=\frac{1}{\sqrt{0.64}}=\frac{1}{0.8}=1.25$ Step 2 — signed gap formula mein plug karo. Event 2 $=$ front (bada $x$), event 1 $=$ back. $\Delta x=+400\text{m}$ woh separation hai platform frame $S$ mein jahan events simultaneous hain — exactly wahi jo formula chahta hai. $\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}=-\frac{(1.25)(0.6c)(400)}{c^2}=-\frac{300}{c}$ Step 3 — numbers. $\Delta t^{\prime }=-\frac{300}{3\times 10^8}=-1.0\times 10^{-6}\text{s},|\Delta t^{\prime }|=1.0\mu \text{s}$ Negative sign (Figure s01 mein amber clock lag karta hai) kehta hai train frame mein front bolt pehle record hoti hai; magnitude $1.0\mu \text{s}$ hai.

Pehle reasoning: dono $\gamma$ aur $v$ increase hote hain, toh gap barhega. $\gamma =1/\sqrt{1-0.64}=1/0.6=5/3\approx 1.6667$. $\Delta t^{\prime }=-\frac{(5/3)(0.8c)(400)}{c^2}=-\frac{(5/3)(0.8)(400)}{3\times 10^8}=-\frac{533.3}{3\times 10^8}\approx -1.78\times 10^{-6}\text{s}$ Toh $|\Delta t^{\prime }|\approx 1.78\mu \text{s}$ — $0.6c$ par ke $1.0\mu \text{s}$ se bada, jaisa predict kiya tha. Sign abhi bhi negative: front bolt pehle.

Kyun alag alag compute karein: yeh dikhata hai ki gap koi trick nahi hai — yeh full Lorentz Transformation $t^{\prime }=\gamma (t-vx/c^2)$ se directly aata hai. $\gamma =1/\sqrt{1-0.25}=1/\sqrt{0.75}=1.1547$. $t_A^{\prime }=\gamma \left(0-\frac{(0.5c)(0)}{c^2}\right)=0$ $t_B^{\prime }=\gamma \left(0-\frac{(0.5c)(600)}{c^2}\right)=-\gamma \frac{0.5\cdot 600}{c}=-1.1547\cdot \frac{300}{3\times 10^8}$ $t_B^{\prime }=-1.1547\times 1.0\times 10^{-6}=-1.15\times 10^{-6}\text{s}$ Difference: $\Delta t^{\prime }=t_B^{\prime }-t_A^{\prime }=-1.15\times 10^{-6}\text{s}$ (yani $|\Delta t^{\prime }|=1.15\mu \text{s}$). Negative → B (leading event, Figure s01 mein amber clock) $S^{\prime }$ mein A se pehle hota hai. ✅ Rule se match karta hai.

Subtlety: flashes rod frame $S$ mein simultaneous hain, toh formula mein $\Delta x$ rest length (proper length) $L_0$ honi chahiye, na ki contracted ground-frame 300 m. Step 1 — length contraction undo karo $L_0$ paane ke liye. Length contraction: $L=L_0/\gamma$, toh $L_0=\gamma L$. $\gamma =1/\sqrt{1-0.64}=1/0.6=5/3$. $L_0=\frac{5}{3}\times 300=500\text{m}$ Step 2 — signed gap formula apply karo. Event 2 $=$ front end par flash (bada $x$, motion ki direction), event 1 $=$ rear flash, toh $\Delta x=+L_0=+500\text{m}$. $\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}=-\frac{(5/3)(0.8c)(500)}{c^2}=-\frac{(5/3)(0.8)(500)}{3\times 10^8}=-\frac{666.7}{3\times 10^8}$ $\Delta t^{\prime }\approx -2.22\times 10^{-6}\text{s},|\Delta t^{\prime }|=2.22\mu \text{s}$ Kaun si flash pehle? Sign negative hai, toh $t_{\text{front}}^{\prime }<t_{\text{rear}}^{\prime }$: ground frame $S^{\prime }$ mein front (leading) flash pehle hoti hai, $2.22\mu \text{s}$ se. Yeh fir Figure s01 ka amber clock hai.

Frame bookkeeping (page ke upar ke convention ke saath). Convention hai: $S$ woh frame hai jahan do events simultaneous hain. Yahaan nose- aur tail-"same reading" events ship mein simultaneous hain, toh is problem ke liye ship $S$ ka role play karta hai aur ground $S^{\prime }$ ka, ship ke relative $-x$ direction mein $v=0.6c$ par move karte hue (doosre words mein, ship ground se aage $+x$ mein jaata hai). Hum sirf "$S$" simultaneous frame ko bolte hain — yeh exactly page ke upar ka rule hai, consistently apply kiya gaya. $\gamma =1.25$, $\Delta x=L_0=120\text{m}$ (separation $S$ mein, yaani ship mein). $|\Delta t^{\prime }|=\frac{\gamma v\Delta x}{c^2}=\frac{(1.25)(0.6c)(120)}{c^2}=\frac{90}{c}=\frac{90}{3\times 10^8}=3.0\times 10^{-7}\text{s}=0.30\mu \text{s}$ Kaun aage hai? "Leading clocks lag" se (Figure s01 mein amber clock), nose (front, ship motion ki direction mein) clock peeche padh raha hai, toh tail clock aage hai ground se dekhe jaane par $0.30\mu \text{s}$ se. ✅

(a) Classify karo (kyun: sirf spacelike events ka hi order flip ho sakta hai — Figure s02 mein tilted line aur Spacetime Diagrams dekho). $\Delta x=900\text{m}$, $\Delta t=2\times 10^{-6}\text{s}$, toh $c\Delta t=3\times 10^8\times 2\times 10^{-6}=600\text{m}$. Kyunki $|\Delta x|=900>c\Delta t=600$, events spacelike hain — light ek se doosre tak nahi pahunch sakti. Order-flip allowed hai. (b) Full transform (dono events mein nonzero $t$ hai, toh general signed formula use karo, simultaneous shortcut nahi): $\gamma =1.25$. $\Delta t^{\prime }=\gamma \left(\Delta t-\frac{v\Delta x}{c^2}\right)=1.25\left(2\times 10^{-6}-\frac{(0.6c)(900)}{c^2}\right)$ $\frac{(0.6)(900)}{c}=\frac{540}{3\times 10^8}=1.8\times 10^{-6}\text{s}$ $\Delta t^{\prime }=1.25(2\times 10^{-6}-1.8\times 10^{-6})=1.25\times 0.2\times 10^{-6}=2.5\times 10^{-7}\text{s}$ Positive aur chhota: B abhi bhi A ke baad hai, koi flip nahi. Figure s02 mein $S^{\prime }$ ki simultaneity line abhi B ke past tilt nahi hui hai. $v$ barhao toh flip ho sakti hai: flip ke liye chahiye $v\Delta x/c^2>\Delta t$, yani $v>c^2\Delta t/\Delta x$. Numerically $c^2\Delta t/\Delta x=(9\times 10^{16})(2\times 10^{-6})/900=2.0\times 10^8\text{m/s}=\frac{2}{3}c$.

Condition: $\Delta t^{\prime }=0\Rightarrow \gamma (\Delta t-v\Delta x/c^2)=0$. Kyunki $\gamma \ne 0$, hamen chahiye $\Delta t=\frac{v\Delta x}{c^2}\Rightarrow v=\frac{c^2\Delta t}{\Delta x}=\frac{(3\times 10^8{)}^2(2\times 10^{-6})}{900}$ $v=\frac{9\times 10^{16}\times 2\times 10^{-6}}{900}=\frac{1.8\times 10^{11}}{900}=2.0\times 10^8\text{m/s}$ $c$ ke fraction ke roop mein: $v/c=2.0\times 10^8/3\times 10^8=0.667$, yani $v=\frac{2}{3}c<c$. Figure s02 mein yeh exactly woh tilt hai jo $S^{\prime }$ ki simultaneity line ko A aur B dono se guzaarta hai. Ek valid sub-light frame exist karta hai exactly isliye kyunki pair spacelike hai ($|\Delta x|>c\Delta t$). Agar woh timelike hote, toh yeh formula $v>c$ demand karta — impossible — aur exactly isliye causal order kabhi flip nahi ho sakti.

Step 1 — $\gamma$ (kyun: ultra-relativistic hai, bada $\gamma$ expect karo). $\gamma =\frac{1}{\sqrt{1-0.995^2}}=\frac{1}{\sqrt{1-0.990025}}=\frac{1}{\sqrt{0.009975}}=\frac{1}{0.099875}\approx 10.01$ Step 2 — signed gap formula event 2 $=$ woh detector jis ki taraf muon ud raha hai (bada $x$), $\Delta x=+600\text{m}$ ground frame $S$ mein jahan woh simultaneous hain: $\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}=-\frac{(10.01)(0.995c)(600)}{c^2}=-\frac{(10.01)(0.995)(600)}{3\times 10^8}$ $\Delta t^{\prime }\approx -1.99\times 10^{-5}\text{s},|\Delta t^{\prime }|\approx 19.9\mu \text{s}$ (b) Sign: negative → leading detector (bada $x$, jis ki taraf muon ud raha hai) muon frame mein pehle fire hota hai — phir wahi Figure s01 ka amber clock. Space mein ek modest 600 m gap enormous $\gamma$ aur near-$c$ speed ki wajah se $\sim 20\mu \text{s}$ ke time gap mein blow up ho jaata hai. Wahi physics jaise Time Dilation muon-lifetime result mein hai, simultaneity ki side se dekha gaya.

Yeh kyun important hai: $\Delta x$ ka sign sirf ek labelling choice hai; physics (kaun sa detector pehle fire karta hai) nahi badalni chahiye. Yeh woh missing edge case hai. $\Delta t^{\prime }=-\frac{\gamma v\Delta x}{c^2}=-\frac{(10.01)(0.995c)(-600)}{c^2}=+\frac{(10.01)(0.995)(600)}{3\times 10^8}\approx +1.99\times 10^{-5}\text{s}$ Ab $\Delta t^{\prime }=t_2^{\prime }-t_1^{\prime }>0$, matlab $t_2^{\prime }>t_1^{\prime }$: event 2 (trailing detector) baad mein fire hota hai, yani front/leading detector abhi bhi pehle fire karta hai. L5.2 se identical physics — sirf bookkeeping label flip hua. Rule: positive $\Delta x$ (event 2 lead karta hai) negative $\Delta t^{\prime }$ deta hai (event 2 pehle); negative $\Delta x$ (event 2 trail karta hai) positive $\Delta t^{\prime }$ deta hai (event 1 pehle). Kisi bhi taraf, leading event pehle hota hai — exactly Figure s01 ka amber clock.

Disagreement ka trigger ::: nonzero spatial separation $\Delta x$ (simultaneous frame mein). Formula ki hidden assumption ::: events $S$ mein simultaneous hain, yani $\Delta t=0$. $\Delta t^{\prime }$ ka sign vs order ::: $\Delta t^{\prime }=t_2^{\prime }-t_1^{\prime }<0$ matlab event 2 (bada $x$) pehle hota hai. Agar tum labels flip karo taki $\Delta x<0$ ho ::: $\Delta t^{\prime }$ sign flip karta hai lekin leading event abhi bhi pehle fire karta hai. Order kab flip ho sakti hai ::: sirf spacelike pairs ke liye ($|\Delta x|>c|\Delta t|$). Kaun sa clock peeche padh raha hai ::: leading wala (front, motion ki direction mein) — amber clock, Figure s01.