Length contraction — derivation: Question bank — Hinglish deep dive

2.3.30 · D5 · Physics › Modern Physics › Length contraction — derivation

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Kisi bhi trap se pehle, har letter ka matlab samjho jo neeche dikhega. Har ek letter ek picture ko naam deta hai, sirf ek idea ko nahi.

Definition Measuring cast (measuring karne wale characters)

$c$ — speed of light, lagbhag $300, 000$ km per second. Yeh universe ki fixed top speed hai; koi bhi information-carrying cheez isse beat nahi kar sakti. Isko ek aisa ruler-and-clock socho jis par poora cosmos agree karta hai.
$v$ — speed of the object measure karne wale observer ke relative mein, hamesha $c$ ka koi fraction. Ek rod ko apni taraf se daayin (right) taraf slide karta hua socho: $v$ kitni tez slide karta hai woh hai.
$L_{0}$ — proper length: rod ki length us frame mein measure ki gayi jahan rod still khadi hai. Uske apne ends kabhi move nahi karte, toh tum unhe aaram se padhh sakte ho. Yeh woh sabse bada number hai jo koi bhi paa sakta hai.
$L$ — contracted length: rod ki length us frame mein measure ki gayi jahan woh speed $v$ se move karti hai. Tumhe dono ends ko ek hi instant mein pakadna hoga. Yeh chhota nikalta hai.
$γ$ (Greek "gamma") — Lorentz factor, ek pure number $\geq 1$ jo batata hai ki speed $v$ par relativity measurements ko kitni strongly distort karti hai.

Figure — Length contraction — derivation

Curve ko left se right padhho: $γ$ ordinary speeds par $1$ ke paas rehta hai (isliye highway par humein contraction kabhi nahi dikhti) aur sirf $v \approx 0.8 c$ ke baad upar rocket karta hai. Woh steep tail wahi hai jahan saari "weird" relativity rehti hai.

Moving rod chhota kyun nikalta hai — $L = L_{0} / γ$ ke peeche ki picture

Poora subject ek hi idea hai: ek moving rod measure karne ke liye tumhe dono ends ko ek saath mark karna hota hai, aur "ek saath" alag-alag observers ke liye alag-alag hota hai. Chalte hain dekhte hain yeh hota kaise hai.

Hum do frames use karte hain (ek frame bas ek observer ki rulers aur clocks ki grid hai):

$S$ = ground, jahan rod speed $v$ se right slide karti hai.
$S^{'}$ = rod ka apna frame, rod ke saath ride karta hua, jahan rod still khadi hai length $L_{0}$ ke saath.

Lorentz transformation ek ground position-and-time $(x, t)$ ko rod-frame position $x^{'}$ mein convert karta hai: $x^{'} = γ (x - v t)$

Intuition vt-cancellation, ek spacetime diagram par dekhi gayi

Neeche ki figure time upar, position daayan plot karti hai. Rod ke dono ends do slanted world-lines trace karte hain (woh paths jo woh spacetime mein sweep karte hain). $S$ mein $L$ measure karne ke liye, ground observer dono ends ko ek horizontal dotted line par mark karta hai — ek single instant $t_{1} = t_{2} = t$ . Har end ko transformation mein feed karo: $x_{2}^{'} = γ (x_{2} - v t), x_{1}^{'} = γ (x_{1} - v t)$ Subtract karo, aur kyunki same $t$ dono mein hai, $v t$ pieces identical hain aur cancel ho jaate hain: $L_{0} = x_{2}^{'} - x_{1}^{'} = γ (x_{2} - x_{1}) = γ L \Rightarrow L = \frac{L _{0}}{γ}$ Yeh cancellation algebra ki kismat nahi hai — yeh simultaneity hai jo kaam kar rahi hai. Measuring line ko slanted kar do (alag times) aur $v t$ terms aur match nahi karenge.

Kyunki $γ \geq 1$ hai, usse divide karne par $L_{0}$ sirf chhota ho sakta hai. Isliye moving rod hamesha chhoti measured hoti hai, aur $L_{0}$ (kabhi divide nahi hota) maximum rehta hai.

Perpendicular sizes kyun nahi badalti — ring-through-hole ki picture

Ek slanted rod rotate bhi kyun karta hai — angled-rod ki picture

True or false — justify karo

Recall

Ek moving rod ko ek force physically squeeze karti hai, isliye woh chhoti measure hoti hai. ::: False — koi force act nahi karta. Rod apne frame mein unstressed hai; shortening frame-dependent simultaneity ka ek geometric consequence hai, compression nahi. Proper length $L_{0}$ woh sabse badi length hai jo koi bhi observer kabhi bhi us object ke liye measure karega. ::: True — $L_{0}$ rest frame mein measure ki jaati hai, aur har doosra frame ise $γ \geq 1$ se divide karta hai, toh saari measured lengths $L \leq L_{0}$ satisfy karti hain. Agar observer A, B ki rod ko chhoti measure kare, toh B ko A ki rod badi measure karni chahiye cheezein consistent rakhne ke liye. ::: False — B bhi A ki rod ko chhota measure karta hai. Koi contradiction nahi hai kyunki A aur B ki "ek saath" ki definition alag hai, toh woh alag pairs of events measure karte hain. Apni length ke perpendicular move karne wali ek rod bhi us length along contract karti hai. ::: False — sirf woh dimension jo motion ke parallel hai wahi contract karta hai. Ek rod jo tumhare past sideways le jaai jaaye woh apni poori length rakhti hai. $v = 0$ par length contraction formula phir bhi apply hota hai aur $L = L_{0}$ deta hai. ::: True — $v = 0$ ke saath, $γ = 1$ , toh $L = L_{0} /1 = L_{0}$ . Formula smoothly "no motion, no contraction" mein reduce ho jaata hai. Jaise-jaise $v \to c$ , measured length zero ke paas pahunch jaati hai. ::: True — $γ \to \infty$ , toh $L = L_{0} / γ \to 0$ . Light speed par ek rod ki length uski motion ke along zero measure hogi (real objects ke liye unreachable, lekin yahi limiting trend hai). Length contraction aur time dilation dono proper quantity ko $γ$ se multiply karte hain. ::: False — length $γ$ se divide hoti hai (chhoti ho jaati hai) jabki time $γ$ se multiply hota hai (bada ho jaata hai); proper length maximum hai, proper time minimum hai. Ek fast rod ki photograph literally ise exactly $1/ γ$ factor se squished dikhayegi. ::: False — measurement $1/ γ$ deta hai, lekin photograph alag times par different parts se nikla light record karti hai, rotation/distortion effects add karte hue. Contraction simultaneous marking ke baare mein hai, raw camera images ke nahi.

Error dhundho

Recall

"Rod $S$ mein move karti hai, toh main abhi front end record karta hun aur thodi der baad back end, phir positions subtract karta hun." ::: Error — endpoints ko $S$ mein simultaneously record karna hoga. Agar tum wait karte ho, rod readings ke beech move karti hai aur tumhe length nahi, nonsense milta hai. "Contract karne ke liye, main multiply karta hun: $L = γ L_{0}$ , kyunki $γ$ relativity factor hai." ::: Error — isse $L$ , rest length se bada ho jaata hai. Proper length maximum hai, toh tumhe divide karna hoga: $L = L_{0} / γ$ . "Lorentz step mein $v t$ terms cancel ho jaate hain kyunki $v$ dono ends ke liye same hai." ::: Error — woh cancel hote hain kyunki $t_{1} = t_{2}$ (same time), jo do $v t$ terms ko identical banata hai. Sirf equal $v$ kaafi nahi; simultaneity hi kaam karti hai. "Muon apna khud ka decay lifetime shrink hota dekha, isliye woh ground tak pahunch paata hai." ::: Error — muon ke frame mein atmosphere contract hoti hai (cross karne ke liye chhoti distance), muon ka khud ka lifetime nahi. Lifetime shrinking Earth-frame view hai time dilation ke through — do frames, ek physics. " $L_{0} = γ L$ ka matlab hai ki rest length contracted wali hai." ::: Error — $L_{0} = γ L$ sahi se kehta hai $L = L_{0} / γ$ , toh $L$ (moving-frame measurement) woh chhoti wali hai. $L_{0}$ abhi bhi sabse badi hai. "Dono observers ek doosre ki rod ko chhoti measure karte hain, yeh relativity ke principle ka violation hai." ::: Error — yeh principle ko uphold karta hai: physics dono frames mein same dikhti hai, toh har ek symmetrically doosre ko contracted measure karta hai. Symmetry hi point hai, paradox nahi.

Why questions

Recall

Moving rod ki length measure karne ke liye simultaneity kyun zaroori hai? ::: Kyunki length dono ends ke beech ka gap hai ek hi instant par; agar ends alag times par record ki jaayein toh moving rod readings ke beech shift ho jaati hai, gap corrupt ho jaata hai. Proper length maximum kyun hai, minimum kyun nahi? ::: Rest frame mein ends kabhi move nahi karte, toh koi simultaneity subtlety reading ko shrink nahi karti. Har moving frame ki simultaneity mismatch ek piece subtract karti hai, chhota value deti hai. Length aur time $γ$ ke under opposite directions mein scale kyun karte hain? ::: Kyunki proper length sabse bada possible measurement hai jabki proper time sabse chhota, toh moving observers length reduced dekhte hain ( $\div γ$ ) lekin elapsed time increased ( $\times γ$ ). Perpendicular dimension contract kyun nahi karta? ::: Ek consistency argument: ek ring jo ek matching hole se guzar rahi hai woh dono frames mein fit honi chahiye. Agar perpendicular sizes badalte, toh ek frame ise jam hote dekhta aur doosra nikal jaate — ek single physical event ke liye impossible. $γ$ hamesha $\geq 1$ kyun hota hai? ::: Kyunki real objects ke liye $v^{2} / c^{2} < 1$ hai, toh $1 - v^{2} / c^{2}$ 0 aur 1 ke beech hai, aur uske square root ka reciprocal kam se kam 1 hai. Yeh guarantee karta hai ki contraction kabhi rod ko lengtha nahi karta. Length contraction ko "real" kyun kaha jaata hai agar koi force nahi hai? ::: "Real" ka matlab hai ki us frame ka har observer consistently ise measure karta hai aur iske physical consequences hain (jaise muons ka ground tak pahunchna). Yeh measurement ki ek genuine property hai, bas mechanical squeeze nahi. Hum simply yeh kyun nahi keh sakte ki moving rod "looks" chhoti hai, yaani yeh ek optical illusion hai? ::: Kyunki yeh ek measured coordinate difference hai jo ideal simultaneous survey dwara record ki gayi hai, light-travel ya viewing angle se independent — yeh us careful correction ke baad survive karti hai jo kisi bhi optical illusion ko remove kar deta.

Edge cases

Recall

$L$ ka kya hota hai jab $v$ ek tiny everyday speed hai, jaise ek car? ::: $v^{2} / c^{2}$ unimaginably small hai, toh $γ \approx 1$ aur $L \approx L_{0}$ — contraction real hai lekin detect karne ke liye bahut zyaada chhoti, jo everyday experience se match karta hai. Apni motion ke exactly perpendicular point karne wali rod ki contracted length kya hogi? ::: Uski poori $L_{0}$ — $v$ ke perpendicular koi contraction nahi hoti, sirf along hoti hai, toh yeh rod apni rest length par measured hogi. Ek rod angle par move karti hai (partly along, partly across motion). Kya hota hai? ::: Sirf uska component jo motion ke along hai wahi contract karta hai; perpendicular component fixed rehta hai, toh rod apni rest orientation ke relative dono shortened aur rotated dikhti hai (upar angled-rod figure dekho). Agar do observers $v$ par pass karte hain aur har ek doosre ki identical rod measure karta hai, toh kya unhe same number milta hai? ::: Haan — symmetry se har ek doosre ki rod $L_{0} / γ$ measure karta hai, same numeric factor, koi contradiction nahi kyunki woh alag simultaneous events mark kar rahe hain. Kya fast speed par move karta ek spherical ball sphere measure hoga? ::: Nahi — yeh ek ellipsoid measure hoga, motion along $1/ γ$ se flattened jabki dono transverse diameters full size rehte hain. Formally, photon-speed ( $v = c$ ) rod ki kya length hogi? ::: Formula $L \to 0$ deta hai, lekin massive rod ke liye $v = c$ unreachable hai, toh yeh trend ke baare mein ek limiting statement hai, koi physical object nahi.

Connections

Length contraction — derivation — parent derivation jinhe yeh traps stress-test karte hain.
Relativity of simultaneity — yahan almost har "paradox" ki root cause.
Lorentz transformation — jahan $v t$ -cancellation subtlety rehti hai.
Lorentz factor gamma — kyun $γ \geq 1$ shortening force karta hai.
Time dilation — kai contrasts mein use hua opposite-direction twin.
Muon decay experiment — frame-swap trap ke peeche ka real-world case.
Spacetime interval — woh invariant jo dono effects preserve karte hain.