Exercises — Lorentz transformation — derivation

Kya: $\gamma =1/\sqrt{1-v^2/c^2}$ mein plug in karo. Kyun: $\gamma$ woh single number hai jo control karta hai ki space aur time kitna stretch hote hain; har relativity problem yahan se start hoti hai. $\gamma =\frac{1}{\sqrt{1-0.6^2}}=\frac{1}{\sqrt{1-0.36}}=\frac{1}{\sqrt{0.64}}=\frac{1}{0.8}=\mathrm{1.25.}$ Answer: $\gamma =1.25$.

Extra piece $-\frac{vx}{c^2}$ term hai. Yeh kya karta hai: yeh time-shift ko where event hota hai ($x$) par depend karata hai. Do events alag jagah par lekin same lab time $t$ par alag primed times lete hain — yeh Relativity of simultaneity ka seed hai. Overall $\gamma$ doosra departure hai (jo sab kuch multiply karta hai).

$\frac{v^2}{c^2}=\frac{(1.5{)}^2}{(3\times 10^8{)}^2}=\frac{2.25}{9\times 10^{16}}=2.5\times 10^{-17}.$ Yeh unimaginably tiny hai, isliye $\gamma =1/\sqrt{1-2.5\times 10^{-17}}\approx 1$. Matlab: Lorentz transform Galileo par wapas collapse ho jaata hai, $x^{\prime }\approx x-vt$, $t^{\prime }\approx t$. Nayi theory purani ko contain karti hai — jaisa kisi bhi sahi generalization ko karna chahiye.

Kya: dono master equations mein substitute karo. $x^{\prime }=\gamma (x-vt)=1.25(2-0.6\cdot 5)=1.25(2-3)=1.25\times (-1)=-1.25\text{ls}.$ $t^{\prime }=\gamma \left(t-\frac{vx}{c^2}\right)=1.25(5-0.6\cdot 2)=1.25(5-1.2)=1.25\times 3.8=4.75\text{s}.$ Numbers kyun: $vx/c^2=0.6\times 2=1.2$ s kyunki $c=1$. Answer: $(x^{\prime },t^{\prime })=(-1.25\text{ls},4.75\text{s})$. Event moving origin ke peeche hai ($x^{\prime }<0$) aur $S^{\prime }$ mein baad mein hota hai.

Kya: clock moving origin par hai, isliye $x^{\prime }=0$ ke saath inverse time equation use karo: $t=\gamma \left(t^{\prime }+\frac{v{x}^{\prime }}{c^2}\right)\stackrel{x^{\prime }=0}{\to }\Delta t=\gamma \Delta t^{\prime }=1.25\times 1=1.25\text{s}.$ $x^{\prime }=0$ kyun matter karta hai: Time dilation usi clock ke ticks compare karta hai; woh clock $S^{\prime }$ mein stationary hai, isliye uski position wahan kabhi nahi badlati. Answer: $\Delta t=1.25$ s. Lab moving clock ko slowly tick karte dekhta hai (har tick par zyada lab time).

Kya: lab mein length measure karne ka matlab hai $S$ mein dono ends simultaneously note karna, isliye $\Delta t=0$. $x^{\prime }=\gamma (x-vt)$ ko difference ki tarah use karo: $\Delta x^{\prime }=\gamma (\Delta x-v\Delta t)\stackrel{\Delta t=0}{\to }{L}_0=\gamma L\Rightarrow L=\frac{L_0}{\gamma }=\frac{4}{1.25}=3.2\text{ls}.$ $\Delta t=0$ kyun: agar tum ends alag times par note karo toh moving rod shift ho gayi hoti, measurement corrupt ho jaata. Answer: $L=3.2$ ls (contracted). Dekho Length contraction.

Kya: har event ka time transform karo. $t_1^{\prime }=\gamma \left(t_1-\frac{v{x}_1}{c^2}\right)=1.25(0-0)=0.$ $t_2^{\prime }=\gamma \left(t_2-\frac{v{x}_2}{c^2}\right)=1.25(0-0.6\cdot 3)=1.25\times (-1.8)=-2.25\text{s}.$ Isko padhna: $t_2^{\prime }=-2.25$ s, $t_1^{\prime }=0$ se pehle hai, isliye $S^{\prime }$ mein door wala explosion (event 2) pehle hota hai, $2.25$ s se. Kyun: $-vx/c^2$ term bade $x$ ko bada negative time-shift deta hai — simultaneity absolute nahi hai. Dekho Relativity of simultaneity. Neeche figure mein tilted line of simultaneity dekho.

$S$ mein: $\Delta t=0$, $\Delta x=3$ ls. $c=1$ ke saath: $s^2=(1{)}^2(0{)}^2-(3{)}^2=-9{\text{ls}}^2.$ $S^{\prime }$ mein: L3.1 se, $\Delta t^{\prime }=t_2^{\prime }-t_1^{\prime }=-2.25$ s. Aur $\Delta x^{\prime }=\gamma (\Delta x-v\Delta t)=1.25(3-0)=3.75$ ls. $s^{\prime 2}=(1{)}^2(-2.25{)}^2-(3.75{)}^2=5.0625-14.0625=-9{\text{ls}}^2.$ Answer: dono $s^2=-9{\text{ls}}^2$ dete hain. Negative kyun hai: negative $s^2$ ka matlab hai events space-like separated hain — koi signal unhe connect nahi kar sakta, yahi reason hai ki unka time-order frames ke beech flip ho sakta hai. Dekho Spacetime interval.

Kya: relativistic addition rule use karo (inverse form, $v$ add karo): $u=\frac{u^{\prime }+v}{1+u^{\prime }v/c^2}=\frac{0.5c+0.6c}{1+(0.5)(0.6)}=\frac{1.1c}{1.3}\approx 0.846c.$ $1.1c$ kyun nahi: naive addition $c$ exceed kar deta! Denominator $1+u^{\prime }v/c^2=1.3$ ise tame karta hai. Answer: $u\approx 0.846c$, abhi bhi $c$ se kam.

Kya: muon ki lifetime proper time hai (apne clock par measured, $\Delta x^{\prime }=0$). Lab ise dilated dekhta hai: $\gamma =\frac{1}{\sqrt{1-0.98^2}}=\frac{1}{\sqrt{1-0.9604}}=\frac{1}{\sqrt{0.0396}}\approx \mathrm{5.025.}$ $\Delta t_{\text{lab}}=\gamma \tau =5.025\times 2.2\mu \text{s}\approx 11.06\mu \text{s}.$ Lab distance: $d=v\Delta t_{\text{lab}}=0.98\times (3\times 10^8)\times 11.06\times 10^{-6}\text{m}\approx 3251\text{m}\approx 3.25\text{km}.$ Naive distance (dilation ignore karke): $v\tau =0.98\times 3\times 10^8\times 2.2\times 10^{-6}\approx 647$ m. Farq kyun: time dilation ke bina muon "ground tak nahi pahuncha chahiye"; $\gamma \approx 5$ factor use $\sim 5\times$ door travel karne deta hai. Yeh ek real observed effect hai. Answer: $\approx 3.25$ km (vs $\approx 0.65$ km naive).

Step 1 (${\gamma }_1=1.25$): $x^{\prime }=1.25(4-0.6\cdot 5)=1.25(4-3)=1.25\text{ls},t^{\prime }=1.25(5-0.6\cdot 4)=1.25(5-2.4)=3.25\text{s}.$ Step 2 (same $\gamma =1.25$, primed coords par apply hota hai): $x^{\prime \prime }=1.25(1.25-0.6\cdot 3.25)=1.25(1.25-1.95)=1.25(-0.7)=-0.875\text{ls},$ $t^{\prime \prime }=1.25(3.25-0.6\cdot 1.25)=1.25(3.25-0.75)=1.25(2.5)=3.125\text{s}.$ (b) Combined velocity: $v_{\text{tot}}=\frac{0.6c+0.6c}{1+0.36}=\frac{1.2c}{1.36}\approx 0.88235c,{\gamma }_{\text{tot}}=\frac{1}{\sqrt{1-0.88235^2}}\approx \mathrm{2.125.}$ Lab se directly single boost: $x^{\prime \prime }=2.125(4-0.88235\cdot 5)=2.125(4-4.41176)=2.125(-0.41176)\approx -0.875\text{ls},$ $t^{\prime \prime }=2.125(5-0.88235\cdot 4)=2.125(5-3.52941)=2.125(1.47059)\approx 3.125\text{s}.$ Answer: dono routes $(x^{\prime \prime },t^{\prime \prime })\approx (-0.875\text{ls},3.125\text{s})$ dete hain. Kyun: ek axis ke along Lorentz boosts ek group form karte hain — do in a row ek bade boost ke barabar hain, jiska speed velocity addition se set hota hai ($1.2c$ nahi!).

Kya: velocity differentials ka ratio hai, isliye dono transforms ke differentials lo. $d{x}^{\prime }=\gamma (dx-vdt),d{t}^{\prime }=\gamma \left(dt-\frac{vdx}{c^2}\right).$ Differentials kyun: $u^{\prime }=d{x}^{\prime }/d{t}^{\prime }$ by definition hai, aur $\gamma$'s ratio mein cancel ho jaayenge — yahi elegance hai. $u^{\prime }=\frac{d{x}^{\prime }}{d{t}^{\prime }}=\frac{\gamma (dx-vdt)}{\gamma \left(dt-\frac{vdx}{c^2}\right)}=\frac{dx-vdt}{dt-\frac{vdx}{c^2}}.$ Ab top aur bottom ko $dt$ se divide karo (taaki $dx/dt=u$ appear ho): $u^{\prime }=\frac{\frac{dx}{dt}-v}{1-\frac{v}{c^2}\frac{dx}{dt}}=\boxed{\frac{u-v}{1-uv/c^2}}.$ Check: $u=c$ set karo: $u^{\prime }=\frac{c-v}{1-v/c}=\frac{c-v}{(c-v)/c}=c$. Light $c$ par rehti hai. $✓$

Kya: transforms substitute karo aur expand karo. $c^2{t}^{\prime 2}=c^2{\gamma }^2{\left(t-\frac{vx}{c^2}\right)}^2={\gamma }^2\left(c^2{t}^2-2vxt+\frac{v^2{x}^2}{c^2}\right).$ $x^{\prime 2}={\gamma }^2(x-vt{)}^2={\gamma }^2\left(x^2-2vxt+v^2{t}^2\right).$ Subtract karo — cross terms $-2vxt$ cancel ho jaate hain (yahi reason hai ki interval survive karta hai): $c^2{t}^{\prime 2}-x^{\prime 2}={\gamma }^2\left[c^2{t}^2+\frac{v^2{x}^2}{c^2}-x^2-v^2{t}^2\right].$ $t^2$ aur $x^2$ ke hisaab se group karo: $={\gamma }^2\left[c^2{t}^2\left(1-\frac{v^2}{c^2}\right)-x^2\left(1-\frac{v^2}{c^2}\right)\right]={\gamma }^2\left(1-\frac{v^2}{c^2}\right)(c^2{t}^2-x^2).$ Final blow: ${\gamma }^2(1-v^2/c^2)=1$ by definition of $\gamma$, isse milta hai: $c^2{t}^{\prime 2}-x^{\prime 2}=c^2{t}^2-x^2.✓$ Yeh invariance Minkowski diagram aur Spacetime interval ka geometric heart hai.

(a) $\gamma =1/\sqrt{1-0.64}=1/\sqrt{0.36}=1/0.6=1.667$. $L=\frac{L_0}{\gamma }=\frac{100}{1.667}=60\text{m}.$ (b) Train frame mein dono doors same time par khulte hain $\Delta t^{\prime }=0$, $\Delta x^{\prime }=L_0=100$ m se separated. Platform par inverse time equation se transform karo: $\Delta t=\gamma \left(\Delta t^{\prime }+\frac{v\Delta x^{\prime }}{c^2}\right)=\gamma \frac{v\Delta x^{\prime }}{c^2}.$ Numbers plug karo ($v=0.8c$, $c=3\times 10^8$): $\Delta t=1.667\times \frac{0.8\times 3\times 10^8\times 100}{(3\times 10^8{)}^2}=1.667\times \frac{0.8\times 100}{3\times 10^8}=1.667\times 2.667\times 10^{-7}\text{s}.$ $\Delta t\approx 4.44\times 10^{-7}\text{s}=0.444\mu \text{s}.$ Answer: platform length $60$ m measure karta hai aur doors ko $\approx 0.444\mu \text{s}$ apart khulte dekhta hai — front aur back events, train par simultaneous, platform par simultaneous nahi hain. Yahi har "pole-in-barn" paradox ke peeche ka engine hai.

$v=0.6c$ ke liye $\gamma$? ::: $1.25$ Galilean simultaneity kaunsa term todta hai? ::: $-vx/c^2$ Time dilation ke liye condition? ::: same clock, $\Delta x^{\prime }=0\Rightarrow \Delta t=\gamma \Delta t^{\prime }$ Length contraction ke liye condition? ::: lab mein simultaneous, $\Delta t=0\Rightarrow L=L_0/\gamma$ $v=0.6c$ frame ke andar $u^{\prime }=0.5c$ ki velocity (lab mein)? ::: $\approx 0.846c$ Interval $s^2=c^2\Delta t^2-\Delta x^2$ hai... ::: saare inertial frames mein invariant