Length contraction — derivation: Worked examples

2.3.30 · D3 · Physics › Modern Physics › Length contraction — derivation

Shuru karne se pehle, do lengths ko plain words mein yaad karo:

==Proper length $L_{0}$ == — us frame mein length jahan object still baitha hai. Koi bhi jo value measure kare, yahi sabse badi hoti hai.
==Measured length $L$ == — us frame mein length jahan object speed $v$ se saasamne se guzarta hai. Hamesha $L \leq L_{0}$ .
== $γ$ (gamma)== — "stretch/shrink dial," $γ = 1/ 1 - v^{2} / c^{2}$ . Yeh hamesha $\geq 1$ hota hai. (Dekho Lorentz factor gamma.)

Neeche sab kuch sirf in teeno quantities aur ek relation $L = L_{0} / γ$ hai jo alag-alag sawaalon ka saamna karta hai.

The scenario matrix

Length contraction ke koi "quadrants" nahi hote jaise angle ke hote hain — lekin iske apne alag-alag cases ka ek family hai. Yeh har woh cell hai jise hume cover karna hai:

#	Cell (case class)	Kya vary karta hai	Covered by
A	Forward direction — $L_{0}$ aur $v$ diye hain, $L$ nikalo	ek known rod ko shrink karo	Ex 1
B	Inverse direction — $L$ aur $L_{0}$ diye hain, $v$ nikalo	speed ke liye solve karo	Ex 2
C	Zero-speed limit ( $v = 0$ )	degenerate: $γ = 1$	Ex 3
D	Light-speed limit ( $v \to c$ )	limiting behaviour: $L \to 0$	Ex 3
E	Slow / everyday speed ( $v ≪ c$ )	yeh kyun nahi dikta hame	Ex 4
F	Symmetry / "who sees whom" — dono frames	reciprocity, koi paradox nahi	Ex 5
G	Perpendicular dimension	woh direction jo contract nahi hoti	Ex 6
H	Real-world word problem	muon / astronomy	Ex 7
I	Exam twist — time ke saath mixed, ya "contracted distance travelled"	effects combine karo	Ex 8
J	Diagonal object — motion rod ki length ke along nahi hai	sirf $x$ -part shrink hota hai	Ex 9

Neeche ke figures high-contrast style mein draw kiye hain: white pe black lines, aur ek red object wahi hai jo hum us figure mein actually measure kar rahe hain.

Cell A — Forward direction

Worked example Example 1 — Ek known rod ko shrink karo (Cell A)

Ek metre-scale rod ki proper length $L_{0} = 10$ m hai aur yeh lab ke saamne se $v = 0.8 c$ speed se guzarti hai. Lab isko kitni lambi measure karti hai?

Forecast: $0.8 c$ bahut fast hai, toh noticeable shrink expect karo — aage padhne se pehle 5 aur 7 m ke beech kuch guess karo.

Figure mein, upar wala black bar rest-frame rod hai (poora $10$ m, iske end-caps $0$ aur $10$ pe); neeche wala red bar jisme chota $v$ -arrow hai, wahi rod hai jaisi lab measure karti hai — deliberately $6$ pe rok ke draw ki gayi hai taaki tum woh missing $4$ m dekh sako. Jab hum isko compute karte hain toh us red bar pe nazar rakho.

Step 1 — $γ$ build karo. $γ = \frac{1}{1 - 0. 8 ^{2}} = \frac{1}{1 - 0.64} = \frac{1}{0.36} = \frac{1}{0.6} = \frac{5}{3} \approx 1.667.$ Yeh step kyun? $γ$ woh dial hai jo batata hai kitna chota hoga. Jab tak yeh nahi pata, hum kuch bhi shrink nahi kar sakte.

Step 2 — Divide karo (kabhi multiply mat karo). $L = \frac{L _{0}}{γ} = \frac{10}{1.667} = 6.0 m .$ Divide kyun? Lab rod ko move karte dekha hai, toh lab contracted (choti) length measure karti hai. Proper length $L_{0}$ maximum hai, isliye yeh fraction ke upar honi chahiye. Woh $6$ m exactly wahin hai jahan figure mein red bar khatam hoti hai.

Verify: $6.0 < 10$ ✔ — chota, jaise chahiye tha. Units: metres ÷ (dimensionless $γ$ ) = metres ✔.

Cell B — Inverse direction

Worked example Example 2 — Speed nikalo (Cell B)

Ek rod rest mein $1.00$ m lambi hai, lekin lab isko $0.60$ m measure karti hai. Yeh kitni fast move kar rahi hai?

Forecast: Yeh 60% tak shrink hua — yeh bada shrink hai, isliye $v$ badi fraction of $c$ honi chahiye. $0.7 c$ se upar guess karo.

Step 1 — Ratio form likho. $\frac{L}{L _{0}} = \frac{1}{γ} = 1 - \frac{v ^{2}}{c ^{2}} \Rightarrow 0.60 = 1 - \frac{v ^{2}}{c ^{2}} .$ Yeh step kyun? Hume ratio $L / L_{0}$ diya gaya hai, aur ratio exactly square-root factor ke barabar hai. $γ$ alag se compute karne ki zaroorat nahi.

Step 2 — Dono sides square karo root hatane ke liye. $0.36 = 1 - \frac{v ^{2}}{c ^{2}} \Rightarrow \frac{v ^{2}}{c ^{2}} = 0.64.$ Square kyun? Unknown $v$ ek square root ke andar phansa hua hai; squaring woh tool hai jo ise undo karta hai.

Step 3 — Root lo. $\frac{v}{c} = 0.80 \Rightarrow v = 0.8 c .$ Positive root kyun? Speed ek magnitude hai — kabhi negative nahi.

Verify: Plug back karo: $γ = 1/ 1 - 0.64 = 5/3$ , toh $L = 1.00 \times 0.6 = 0.60$ m ✔. Aur $0.8 c > 0.7 c$ , forecast se match karta hai.

Cells C & D — dono extremes

Worked example Example 3 — Zero speed aur light speed (Cells C, D)

Same $L_{0} = 10$ m rod lo. Lab kya length measure karegi (a) jab $v = 0$ ho, aur (b) jab $v \to c$ ho?

Forecast: Rest mein, kuch nahi hilta — bilkul shrink nahi. Light speed ke paas, shrink wild ho jaani chahiye — guess karo yeh zero ki taraf jaati hai.

Part (a) — $v = 0$ (Cell C, degenerate input). $γ = \frac{1}{1 - 0} = 1, L = \frac{10}{1} = 10 m .$ Yeh kyun matter karta hai: Jab rod lab mein bhi rest mein ho, toh lab frame hi rest frame hai, isliye $L = L_{0}$ . Formula sahi tarike se proper length wapas deta hai — kisi bhi relativity formula ke liye yeh ek vital sanity check hai.

Part (b) — $v \to c$ (Cell D, limiting behaviour). Jab $v \to c$ , tab term $v^{2} / c^{2} \to 1$ , isliye $1 - \frac{v ^{2}}{c ^{2}} \to 0 = 0 \Rightarrow L \to 0.$ Yeh kyun matter karta hai: Light speed ke paas pahunchte-pahunchte, rod lab mein zero length ki taraf squash ho jaati hai. Figure mein red curve dekho — choti $v$ ke liye flat aur lazy, phir $c$ ke paas ek cliff se neeche gir jaati hai. Isliye "everyday" motion mein koi effect nahi dikhta, lekin light-speed motion dramatic hoti hai.

Verify (numerically): $v = 0.99 c$ pe, $γ = 1/ 1 - 0.9801 \approx 7.09$ , $L \approx 10/7.09 \approx 1.41$ m — pehle se bahut chota, dive confirm karta hai. ✔

Cell E — everyday speeds

Worked example Example 4 — Yeh tumhe kyun nahi dikhta (Cell E)

$100$ m lambi ek train $v = 300$ km/h $\approx 83.3$ m/s se chalti hai. Lab isko kitna contracted dekhti hai? ( $c = 3 \times 1 0^{8}$ m/s use karo.)

Forecast: Trainen clearly choti nahi dikhtin. Toh yeh effect bahut hi tiny honi chahiye — guess karo ek atom ki width se kaafi kam.

Step 1 — $v / c$ compute karo. $\frac{v}{c} = \frac{83.3}{3 \times 1 0 ^{8}} \approx 2.78 \times 1 0^{- 7} .$ Yeh step kyun? Contraction sirf ratio $v / c$ pe depend karti hai; train ki speed light ke compare mein ek whisper hai.

Step 2 — Tiny- $x$ shortcut use karo. Bahut chhote number $x = v^{2} / c^{2}$ ke liye, exact $1 - x$ almost $1 - \frac{1}{2} x$ hota hai. Toh kitna loss hua yeh hai: $L_{0} - L \approx L_{0} \cdot \frac{1}{2} \frac{v ^{2}}{c ^{2}} = 100 \times \frac{1}{2} (2.78 \times 1 0^{- 7})^{2} \approx 3.86 \times 1 0^{- 12} m .$ Yeh tool (approximation) kyun? Calculator ke square root mein $2.78 \times 1 0^{- 7}$ daalo toh sirf $1.000000$ return karta hai — change iske precision ke neeche chhupta hai. Linear approximation hi ek maatra tarika hai yeh dekhne ka ki loss actually kitna chota hai.

Verify: $3.86 \times 1 0^{- 12}$ m roughly ek atom ki width ( $\sim 1 0^{- 10}$ m) ka ek forty-thousandth hai — bilkul unmeasurable ✔. Isliye length contraction tab tak chhupa raha jab tak hum $c$ ke paas move hone wali cheezein nahi study karne lage.

Cell F — symmetry, "who sees whom"

Worked example Example 5 — Dono ek doosre ko chota dekhte hain (Cell F)

Do identical rods, dono apne apne frame mein $L_{0} = 12$ m lambi, $v = 0.6 c$ speed se ek doosre ke past guzarti hain. Ann rod A pe sawaar hai; Bob rod B pe. Ann Bob ki rod kitni lambi measure karti hai, aur Bob Ann ki rod kitni lambi? Kya yeh contradiction hai?

Forecast: Relativity ke principle se, koi bhi "specially moving" nahi hai. Toh ek symmetric answer expect karo — har koi doosre ko equally chota dekhta hai.

Step 1 — Shared $γ$ compute karo. $γ = \frac{1}{1 - 0.36} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25.$ Ek $γ$ kyun? Frames ke beech relative speed dono taraf same $0.6 c$ hai, toh dono identical factor use karte hain.

Step 2 — Ann, Bob ki moving rod measure karti hai. $L_{Bob, by Ann} = \frac{12}{1.25} = 9.6 m .$

Step 3 — Bob, Ann ki moving rod measure karta hai. $L_{Ann, by Bob} = \frac{12}{1.25} = 9.6 m .$ Identical kyun? Ann ki seat se, Bob ki rod move karti hai; Bob ki seat se, Ann ki rod move karti hai. Har koi doosre ki rod pe same formula apply karta hai.

Verify — koi contradiction nahi: Woh same pair of events compare nahi kar rahe. Ann, Bob ke dono ends ko Ann ke time mein simultaneously mark karti hai; Bob apni khud ki "same time" use karta hai. Kyunki simultaneity frame-dependent hai, woh literally alag-alag measurements hain. Dono ka $9.6$ m hona perfectly consistent hai ✔.

Cell G — woh direction jo shrink nahi hoti

Worked example Example 6 — Perpendicular dimension (Cell G)

Ek rectangular flag $2.0$ m lambi (motion ke along) aur $1.0$ m tall (perpendicular) hai. Yeh apni length ke along $v = 0.8 c$ se fly karti hai. Lab mein dono measured dimensions batao.

Forecast: Sirf $v$ ke aligned length ko shrink hona chahiye. Height waisi hi rehni chahiye. Guess: length $\to 1.2$ m, height $\to 1.0$ m.

Step 1 — Along-motion length contract karo. $γ = 5/3$ ke saath: $L_{∥} = \frac{2.0}{1.667} = 1.2 m .$ Sirf yahi kyun? Derivation ne $x^{'} = γ (x - v t)$ use kiya — yeh equation sirf $x$ -axis ke baare mein hai, motion ki direction ke baare mein. Isme $y$ ya $z$ kuch nahi hai.

Step 2 — Height ko waise hi chhodo. $L_{⊥} = 1.0 m (unchanged) .$ Unchanged kyun — ring argument: Socho ek ring ek hole se hokar guzar rahi hai jisme woh rest mein exactly fit hoti hai. Agar perpendicular dimension shrink hoti, toh ek frame kehta "ring freely slip ho jaati hai" aur doosra "ring jam ho jaati hai" — yeh impossible physical disagreement hoga. Isliye perpendicular lengths frame-independent honee chahiye.

Verify: Final flag = $1.2$ m $\times$ $1.0$ m. Sirf red (motion-aligned) side badi ✔.

Cell H — real-world word problem

Worked example Example 7 — Muon ki patli atmosphere (Cell H)

Ek muon $v = 0.98 c$ se ground ki taraf streak karta hai. Earth ke frame mein woh atmosphere jo ise cross karni hai $L_{0} = 10$ km thick hai. Muon ke apne frame mein woh layer kitni thick hai?

Forecast: Muon ki seat se, atmosphere iske past rush karti hai. Toh muon ek contracted, thinner layer dekhta hai — guess karo kuch km.

Step 1 — Identify karo kiska proper length kiska hai. $10$ km Earth ne measure kiya, jo atmosphere ke relative rest mein hai — toh $10$ km atmosphere ki proper length $L_{0}$ hai. Kyun dhyan dena chahiye? Proper vs. contracted ulta karna hi #1 tarika hai galat number ke saath divide karne ka.

Step 2 — $γ$ compute karo. $γ = \frac{1}{1 - 0.9 8 ^{2}} = \frac{1}{1 - 0.9604} = \frac{1}{0.0396} \approx 5.025.$

Step 3 — Muon ke frame mein contract karo. $L = \frac{10}{5.025} \approx 1.99 km .$ Divide kyun? Muon ke frame mein atmosphere move karti hai, isliye muon contracted value dekhta hai.

Verify: $1.99$ km $< 10$ km ✔. Yahi thinner layer exactly isliye hai ki muons — jo jaldi decay ho jaate hain — phir bhi ground tak pahunchte hain: unhe apne frame mein sirf $\sim 2$ km cross karna hota hai. Dekho Muon decay experiment. ✔

Cell I — exam twist (time / distance ke saath combine karo)

Worked example Example 8 — Contracted distance vs. dilated time (Cell I)

Ek spaceship Earth se ek star tak $v = 0.8 c$ speed se jaati hai jo $L_{0} = 8.0$ light-years door hai (Earth-frame distance). (a) Ship ke frame mein trip kitni door hai? (b) Trip mein ship ki apni clock se kitna time lagta hai?

Forecast: Ship choti distance dekhti hai (contraction) aur chota time (iska clock kam tick karta hai). Dono chote.

Step 1 — $γ$ phir se. $γ = 5/3 \approx 1.667$ (Ex 1 jaisa hi, kyunki $v = 0.8 c$ ).

Step 2 — (a) Distance contract karo. $8.0$ ly Earth ka proper distance hai: $L = \frac{8.0}{1.667} = 4.8 ly .$ Divide kyun? Ship ke frame mein star iske taraf rush karta hai, toh gap contracted hai.

Step 3 — (b) Ship ka travel time. Ship ke frame mein yeh contracted distance $4.8$ ly speed $0.8 c$ se cover karti hai: $t_{ship} = \frac{L}{v} = \frac{4.8 ly}{0.8 c} = 6.0 years .$ Contracted $L$ yahan kyun use karein? Ship ko distances jaise woh unhe measure karti hai apni clock ke saath use karni chahiye — Earth ki distance ko ship ki clock ke saath mix karna classic trap hai.

Verify (cross-check via time dilation): Earth-frame travel time $= 8.0 ly /0.8 c = 10$ yr. Ship ki clock (proper time) $10/ γ = 10/1.667 = 6.0$ yr ✔ padhni chahiye — dono methods agree karte hain, confirm karta hai ki length contraction aur Time dilation ek hi Lorentz transformation ke do faces hain.

Cell J — diagonal object

Worked example Example 9 — Motion ke saath tilted rod (Cell J)

Apne rest frame mein ek rod ki length $L_{0} = 5.0$ m hai aur yeh us direction se $3 7^{\circ}$ upar point karti hai jis direction mein baad mein move karegi. Phir yeh $v = 0.8 c$ se $x$ -axis ke along fly karti hai. Lab mein iska measured length kya hai? ( $cos 3 7^{\circ} = 0.8$ , $sin 3 7^{\circ} = 0.6$ .)

Forecast: Sirf along-motion part shrink hoga; upar wala part bachega. Toh rod choti bhi hogi aur steeper bhi dikhegi. Guess karo total thodi 5 m se kam hogi.

Step 1 — Rod ko $x$ aur $y$ pieces mein split karo. $L_{0 x} = 5.0 cos 3 7^{\circ} = 4.0 m, L_{0 y} = 5.0 sin 3 7^{\circ} = 3.0 m .$ Split kyun? Contraction sirf $x$ ke along kaam karta hai. Rod ko ek along-part aur ek across-part mein todne se hum rule ko alag-alag apply kar sakte hain.

Step 2 — Sirf $x$ -piece contract karo ( $γ = 5/3$ ): $L_{x} = \frac{4.0}{1.667} = 2.4 m, L_{y} = 3.0 m (unchanged) .$ Sirf $x$ kyun? Cell G mein prove kiye gaye rule se hi, perpendicular ( $y$ ) part untouched rehta hai; sirf along-motion ( $x$ ) part $γ$ feel karta hai.

Step 3 — Length rebuild karo Pythagorean theorem se: $L = L_{x}^{2} + L_{y}^{2} = 2. 4^{2} + 3. 0^{2} = 5.76 + 9.0 = 14.76 \approx 3.84 m .$ Pythagoras kyun? Do surviving pieces $L_{x}$ aur $L_{y}$ right angle pe milte hain, toh true rod length us chhote triangle ka hypotenuse hai — Pythagoras exactly woh tool hai jo do perpendicular sides ko ek straight length mein turn karta hai.

Verify: $3.84 < 5.0$ ✔ (chota, forecast se match karta hai). Naya angle $arctan (3.0/2.4) = arctan (1.25) \approx 51. 3^{\circ}$ hai — $3 7^{\circ}$ se steeper, exactly jaise forecast kiya, kyunki sirf horizontal side shrink hua. ✔

Active recall

Recall Cells pe self-test

$L_{0}$ aur $v$ diye hain, $γ$ se multiply karte hain ya divide? ::: Divide — lab contracted (choti) length dekhti hai. Ek rod apni rest length ki $0.60$ tak shrink ho jaati hai; $v$ nikalo. ::: $0.6 = 1 - v^{2} / c^{2} \Rightarrow v = 0.8 c$ . $v = 0$ pe formula kya deta hai? ::: $L = L_{0}$ — koi contraction nahi, sanity check. Jab $v \to c$ tab $L$ ka kya hota hai? ::: $L \to 0$ — zero length ki taraf squash ho jaata hai. Kya moving flag ki height change hoti hai? ::: Nahi — sirf motion ke along dimension contract hoti hai. Agar dono ships ek doosre ki rod ko choti measure karein, kya yeh paradox hai? ::: Nahi — woh "same time" ke alag notions use karte hain (relativity of simultaneity). Tilted rod ke liye, kaunsa component contract karta hai? ::: Sirf motion ki direction ke along component; Pythagoras se dobara combine karo. Ship 8 ly gap $0.8 c$ pe cross karti hai — uski apni clock kya padhti hai? ::: $6.0$ years (contracted $4.8$ ly ÷ $0.8 c$ use karta hai).

Connections

Length contraction — derivation — woh parent jise yeh page drill karta hai.
Lorentz transformation — jahan se ultimately har step aati hai.
Time dilation — partner effect, Ex 8 mein cross-check kiya.
Relativity of simultaneity — Ex 5 ka "paradox" resolve karta hai.
Lorentz factor gamma — har example mein use hone wala $γ$ dial.
Muon decay experiment — real-world Ex 7.
Spacetime interval — Ex 8 agreement ke peeche ka invariant.

Worked examples — Length contraction — derivation