Postulates of SR: Exercises — Hinglish deep dive

2.3.26 · D4 · Physics › Modern Physics › Postulates of SR

Definition Is poore page par use hone wale symbols (ek baar build karo, har jagah use karo)

$c = 3 \times 1 0^{8} m/s$ — vacuum mein speed of light. Postulate 2 ke mutabiq yeh har inertial (non-accelerating) observer ke liye same number hai.
$v$ — do inertial frames ke beech ki relative speed (jaise ek ship Earth se kitni tez guzarti hai). Hamesha $v < c$ .
$Δ t_{0}$ — proper time: do events ke beech ka woh time jo us frame mein measure kiya gaya ho jahan woh events ek hi jagah hote hain (jaise moving clock par hi). Woh events ke liye yeh sabse chhota possible time hota hai.
$Δ t$ — dilated time: unhi do events ke beech ka time jo kisi doosre inertial observer ne measure kiya ho. Hamesha $Δ t \geq Δ t_{0}$ .
$γ$ — Lorentz factor, woh "stretch factor" jo dono ko jodta hai: $γ = \frac{1}{1 - v ^{2} / c ^{2}} \geq 1, Δ t = γ Δ t_{0} .$ Kyunki $v < c$ hai, $v^{2} / c^{2} < 1$ hai, isliye square root 1 se chhota hai aur uska reciprocal $γ$ kam se kam 1 hai. Roz ki speeds par $γ \to 1$ aur Newton wapas aa jaata hai.
$u$ — kisi object ki speed ek moving frame ke andar measure ki gayi (jaise rocket ke andar koi ball phenki gayi, us rocket mein baithе insaan ne measure ki).
$u^{'}$ — usi object ki speed bahar wale observer ne measure ki (jaise Earth ne). Relativistic Velocity Addition rule inhe jodta hai: $u^{'} = \frac{u + v}{1 + uv / c ^{2}}$ .

Level 1 — Recognition

Recall Solution

(b) aur (d).

(a) Postulate 2 ka ulta hai — light ki speed source par depend nahi karti. Galat.
(b) Postulate 1 hai (Principle of Relativity). ✅
(c) Exactly woh Newtonian idea hai jo Einstein ne phek diya tha. Galat.
(d) Postulate 2 hai ( $c$ ki constancy). ✅

Recall Solution

Ground lab (approximately) inertial hai: akela chhoda gaya koi object wahi reh jaata hai ya straight line mein constant speed se chalta hai — Newton's first law hold karta hai, koi pseudo-forces nahi. Braking car inertial nahi hai: woh accelerate kar rahi hai, isliye ek free object (seat par rakhi ball) apne aap aage lurche jaata hai — ek pseudo-force appear hoti hai. SR ka Postulate 1 accelerating frames par laagu nahi hota.

Level 2 — Application

Recall Solution

Step 1 — $γ$ calculate karo. $v^{2} / c^{2} = 0.64$ ke saath: $γ = \frac{1}{1 - 0.64} = \frac{1}{0.36} = \frac{1}{0.6} = 1.6 \overline{6} .$ Step 2 — $Δ t = γ Δ t_{0}$ lagao (yahan $Δ t_{0} = 5 s$ ship ka proper time hai, $Δ t$ Earth ka measurement hai): $Δ t = 1.6 \overline{6} \times 5 = 8.33 s .$ Earth maapti hai ki moving clock slow chal rahi hai (zyada time beeta). Dekho Time Dilation.

Recall Solution

Bilkul $c$ ( $3 \times 1 0^{8} m/s$ ), na ki $1.95 c$ . Postulate 2: light ki speed frame-independent hai. Relativistic Velocity Addition formula se check karo. Pehle symbols define karo:

$u$ = object ki speed rocket ke frame ke andar measure ki gayi (yahan laser light, toh $u = c$ );
$v$ = rocket ke frame ki speed planet observer ke relative ( $v = 0.95 c$ );
$u^{'}$ = object ki speed planet observer ne measure ki (yahi chahiye hain hame). $u^{'} = \frac{u + v}{1 + uv / c ^{2}} = \frac{c + 0.95 c}{1 + ( c ) ( 0.95 c ) / c ^{2}} = \frac{1.95 c}{1 + 0.95} = \frac{1.95 c}{1.95} = c .$ Formula light ko force karke $c$ par wapas le aata hai.

Recall Solution

$v / c = 300/ (3 \times 1 0^{8}) = 1 0^{- 6}$ , toh $v^{2} / c^{2} = 1 0^{- 12}$ . Ab small-quantity expansion use karo. Maano $x \equiv v^{2} / c^{2}$ — yahan $x = 1 0^{- 12}$ , ek bahut chhota number. Kisi bhi chhote $x$ ke liye, $\frac{1}{1 - x} \approx 1 + \frac{1}{2} x$ , toh $γ - 1 \approx \frac{1}{2} x = \frac{1}{2} (1 0^{- 12}) = 5 \times 1 0^{- 13} .$ Bilkul negligible — isliye roz ki zindagi mein Newtonian physics kaam karti hai.

Level 3 — Analysis

Worked example L3.1 — Light-clock diagonal phir se banao

Ek light-clock (mirror separation $L$ ) ko Earth ke frame se dekha ja raha hai, jisme clock daayein $v$ speed se move kar rahi hai. Is Earth frame mein photon har half-tick mein ek diagonal travel karta hai kyunki top mirror sideways slide karta hai jab light udaan mein hoti hai. (Clock ke apne frame mein light seedha upar-neeche jaati — woh chhota, purely-vertical trip proper time $Δ t_{0}$ hai; woh diagonal trip jo hum yahan analyse kar rahe hain woh dilated time $Δ t$ hai.) Neeche diye figure ka use karke, EK diagonal ki length $L$ , $v$ , aur pura Earth-frame tick $Δ t$ ke terms mein likho — phir explain karo ki speed $c$ kyun use hoti hai, $c + v$ kyun nahi.

Figure — Earth-frame view of a moving light-clock. Do gray horizontal bars bottom aur top mirrors hain, teen moments par draw kiye gaye jab clock daayein slide karti hai. Ek blue double-arrow fixed vertical separation $L$ mark karta hai. Ek orange double-arrow bottom mein sideways shift $v Δ t /2$ mark karta hai jo clock half tick mein karti hai. Red line photon ki actual slanted path hai (upar phir neeche) — woh "diagonal" $d$ jiske liye hum solve kar rahe hain.

Recall Solution

Ek pura tick $Δ t$ mein mirror daayein $v Δ t$ move karta hai; aadhe tick mein woh $v Δ t /2$ move karta hai (har triangle ka orange base). Vertical side $L$ hai (blue). Pythagoras se, ek diagonal (red) hai $d = L^{2} + (\frac{v Δ t}{2})^{2} .$ Speed $c$ kyun? Postulate 2: photon ki speed har inertial frame mein $c$ hai, including Earth's frame jisme clock move karti hai. Path lamba ho gaya, lekin speed compensate karne ke liye nahi badhi — toh tick ko zyada waqt lagta hai, yaani $c Δ t = 2 d$ . Woh extra time exactly time dilation hai. (Poora algebra → Time Dilation.)

Recall Solution

Dono sides square karo. Note karo $(2 L^{2} + (v Δ t /2)^{2})^{2} = 4 (L^{2} + \frac{v ^{2} Δ t ^{2}}{4}) = 4 L^{2} + v^{2} Δ t^{2}$ : $c^{2} Δ t^{2} = 4 L^{2} + v^{2} Δ t^{2} .$ $Δ t^{2}$ terms group karo: $Δ t^{2} (c^{2} - v^{2}) = 4 L^{2} .$ Square root lo. Equation do roots deti hai, $Δ t = \pm 2 L / c^{2} - v^{2}$ ; hum sirf positive root rakhte hain kyunki $Δ t$ ek elapsed time hai aur negative nahi ho sakta (negative root yahan unphysical hai). Toh $Δ t = \frac{2 L}{c ^{2} - v ^{2}} .$ Root se $c$ factor karo: $c^{2} - v^{2} = c 1 - v^{2} / c^{2}$ , toh $Δ t = \frac{2 L}{c 1 - v ^{2} / c ^{2}} = \frac{2 L / c}{1 - v ^{2} / c ^{2}} .$ Ab $Δ t_{0} = 2 L / c$ substitute karo: $Δ t = \frac{Δ t _{0}}{1 - v ^{2} / c ^{2}} = γ Δ t_{0} . ■$

Recall Solution

(a) $v^{2} / c^{2} = 0.9604$ , toh $1 - v^{2} / c^{2} = 0.0396$ aur $γ = 1/ 0.0396 = 5.025$ . $Δ t = γ Δ t_{0} = 5.025 \times 2.2 μ s = 11.06 μ s .$ (b) Earth ke frame mein distance $= v Δ t = 0.98 \times (3 \times 1 0^{8}) \times 11.06 \times 1 0^{- 6}$ $= 2.94 \times 1 0^{8} \times 1.106 \times 1 0^{- 5} = 3251 m \approx 3.25 km .$ Dilation ke bina muon sirf $0.98 c \times 2.2 μ s \approx 647 m$ cover karta — woh kabhi zameen par nahi pahunchta. Phir bhi muons detectors tak pahunchte hain: time dilation ka experimental proof. (Muon ke apne nazar se, atmosphere shrink hoti hai — Length Contraction — do frames, ek hi physics.)

Level 4 — Synthesis

Recall Solution

Global box ke symbols use karte hue: $u = 0.6 c$ (ball ki speed frame $B$ ke andar measure ki gayi), $v = 0.6 c$ ( $B$ ki speed Earth ke relative), aur $u^{'}$ = Earth ka measurement. Galilean (yahan galat): $u + v = 0.6 c + 0.6 c = 1.2 c$ — impossible, light se tez. Relativistic Relativistic Velocity Addition: $u^{'} = \frac{u + v}{1 + uv / c ^{2}} = \frac{1.2 c}{1 + 0.36} = \frac{1.2 c}{1.36} = 0.882 c .$ Phir bhi $c$ se neeche — relativity sab kuch sub-luminal rakhta hai. Denominator mein correction term $uv / c^{2} = 0.36$ exactly wahi hai jo excess ko "kha" jaata hai.

Recall Solution

Dono ke liye $γ = 1/ 1 - 0.36 = 1/0.8 = 1.25$ — situation symmetric hai. Har ek doosre ki moving clock ko $\times 1.25$ se slow ticking maanta hai (yaani doosre ki clock ke har 1 s ke liye, apni clock par $1.25$ s beette hain). Koi contradiction nahi kyunki "moving clock slow hai" ek statement hai ek frame mein measure ki gayi do events aur doosre mein proper time ke baare mein. Alice aur Bob alag alag pairs of synchronised clocks aur simultaneity ki alag alag notions use karte hain (woh classical idea jo Einstein ne sacrifice kiya tha). "Kaun actually zyada jawaan hai" tay karne ke liye unhe saath wapas aana hoga — jisme acceleration chahiye (ek turnaround), jo symmetry tod deta hai. Dekho Lorentz Transformations.

Recall Solution

Maxwell's Equations ek fixed light speed $c$ predict karte hain. Inme electricity aur magnetism ke constants hain, aur ek single wave speed $c$ nikalta hai — bina is baat ke ki kaun dekh raha hai.
Postulate 1 kehta hai ki woh equations har inertial frame mein same dikhni chahiye. Kyun? Postulate 1 kehta hai physics ke laws sabhi inertial frames mein identical hain; Maxwell's equations physics ke laws hain, toh koi bhi inertial observer unhe alag nahi likh sakta. Isliye har inertial observer, same equations lagate hue, same $c$ paata hai.
Agar light ko koi medium ("ether") chahiye hota, toh Earth ki us medium se guzarhat ek direction-dependent "ether wind" produce karti — $c$ direction ke saath badlta.
Michelson–Morley (1887) ne woh wind dhoondhi aur kuch nahi paya: $c$ har direction mein, har season mein same tha.
Nateeja: koi preferred medium nahi hai aur koi special frame nahi hai. $c$ sabhi inertial observers ke liye same hai — yahi Postulate 2 hai.

Level 5 — Mastery

Recall Solution

Ship ka proper time $Δ t_{0} = Δ t / γ$ hai, toh deficit hai $Δ t - Δ t_{0} = Δ t (1 - \frac{1}{γ}) = Δ t (1 - 1 - v^{2} / c^{2}) .$ Chhote $v / c$ ke liye: $x \equiv v^{2} / c^{2}$ maano, $1 - x \approx 1 - \frac{1}{2} x$ , toh $1 - 1 - x \approx \frac{1}{2} x = \frac{1}{2} v^{2} / c^{2}$ . Deficit ko $1 s$ set karo: $86400 \times \frac{1}{2} \frac{v ^{2}}{c ^{2}} = 1 \Rightarrow \frac{v ^{2}}{c ^{2}} = \frac{2}{86400} = 2.315 \times 1 0^{- 5} .$ $\frac{v}{c} = 2.315 \times 1 0^{- 5} = 4.81 \times 1 0^{- 3} .$ Toh $v \approx 0.00481 c \approx 1.44 \times 1 0^{6} m/s$ . Kyunki $v / c ≪ 1$ hai, approximation safe hai (exact $γ$ sirf 6th decimal mein alag hai).

Recall Solution

(a) Har leg $6$ light-years $0.6 c$ par cover karti hai, toh $6/0.6 = 10$ Earth-years lagte hain. Jaana aur aana $Δ t = 2 \times 10 = 20$ Earth-years deta hai. (b) $γ = 1/ 1 - 0.36 = 1/0.8 = 1.25$ . Traveller ki clock journey ke liye proper time record karti hai (apna frame), jo chhota hota hai: $Δ t_{0} = \frac{Δ t}{γ} = \frac{20}{1.25} = 16 years .$ (c) Traveller $20 - 16 = 4$ saal se jawaan hai. Apparent symmetry (har koi claim kar sakta tha ki doosra move kar raha hai) is liye toot jaati hai kyunki traveller turnaround par accelerate karta hai — woh inertial frames change karta hai, jabki Earth twin poore waqt ek inertial frame mein rehti hai. Woh frame-switch physically real hai; Lorentz Transformations confirm karte hain ki traveller ka total elapsed time chhota hai. (Yeh famous twin paradox hai — dekho Time Dilation.)

Recall Solution

Total energy: $E = γ m c^{2} = 5.025 \times 105.7 MeV = 531.1 MeV .$ Kinetic energy: $K = (γ - 1) m c^{2} = 4.025 \times 105.7 MeV = 425.4 MeV .$ Wahi $γ$ jo muon ki lifetime stretch karta hai, uski energy bhi multiply karta hai — ek factor, do consequences. Dekho Mass-Energy Equivalence.

Connections

Postulates of SR — woh parent note jise yeh page drill karta hai.
Time Dilation, Length Contraction, Lorentz Transformations, Relativistic Velocity Addition, Mass-Energy Equivalence — upar use ki gayi machinery.
Michelson-Morley Experiment, Maxwell's Equations, Galilean Relativity — historical/logical roots.

Recall Self-test checklist

Kya tum bina dekhe dono postulates bata sakte ho? ::: Same laws (sabhi inertial frames) + same light speed (sabhi inertial observers). Multiply ya divide by $γ$ karein doosre frame ka time paane ke liye? ::: Multiply karo: $Δ t = γ Δ t_{0}$ , hamesha lamba. Light-clock ka diagonal kyun nahi matlab ki light $c$ se zyada tez jaati hai? ::: Lamba path zyada time se pay hota hai, kyunki speed $c$ par locked hai. Twin symmetry kya todta hai? ::: Turnaround acceleration — sirf ek twin inertial frames change karta hai.