Special relativity — Michelson-Morley experiment: Visual walkthrough

2.3.25 · D2 · Physics › Modern Physics › Special relativity — Michelson-Morley experiment

Step 1 — "Wind" jo kisi cheez se nahi bana hota, wo hota kya hai?

KYA HAI. Kisi bhi light se pehle, ek wide nadi imagine karo. Paani left-to-right $v$ speed se beh raha hai. Ek boat still water mein speed $c$ se push kar sakti hai. Bas yahi machinery chahiye.

YE YAHAN SE KYUN SHURU KAREN. Is experiment ka har mushkil idea ek nadi mein already visible hai. 19th-century ka belief (dekho Special Relativity — Einstein's Postulates) tha ki light ek invisible medium mein chalti hai — the aether — jaise sound air mein chalti hai. Agar Earth us aether mein speed $v$ se ghus rahi hai, toh hum ek aether wind feel karte hain. Toh "boat in a river" hi "light beam through aether" hai. Hum easy version solve karte hain aur phir bas words rename karte hain.

PICTURE. Neeche, blue arrows current hain (speed $v$ ). Boat ka apna engine hamesha speed $c$ deta hai water ke relative. Do sawaal jinhe hum race karaenge: flow ke saath down-and-back vs. seedha cross karke aur waapas.

Figure — Special relativity — Michelson-Morley experiment

Step 2 — Along-the-wind trip: kyun fast aur slow cancel NAHI hote

KYA HAI. Boat ko distance $L$ downstream bhejo, phir $L$ upstream waapas lao. Hum total time chahte hain.

Downstream current help karta hai: ground speed $= c + v$ . Upstream current ladhta hai: ground speed $= c - v$ .

t_{∥} = upstream, slow \frac{L}{c - v} + downstream, fast \frac{L}{c + v}

Har piece ka same shape hai: $time = \frac{distance L}{effective speed}$ . Left term bada hai kyunki uska denominator $c - v$ chota hai. (Dhyan raho ye tabhi sense karta hai jab $v < c$ ho, taaki $c - v > 0$ ho — upar ka domain remark dekho.)

YE CANCEL KYUN NAHI HOTA. Tempting galti: "aadha trip $+ v$ hai, aadha $- v$ hai, average out ho jaata hai." Nahi. Tum slow zyada der tak hote ho than fast, isliye slow leg clock ko zyada use karta hai. Figure dekho: do coloured bars durations hain, aur slow (pink) bar visibly longer hai fast (yellow) wale se. Unka sum no-wind time $\frac{2 L}{c}$ se zyada hai.

PICTURE. Boat do baar drawn hai — ek leg flow ke saath, ek against — aur neeche do time-bars taaki imbalance literally page par longer dikhe.

Ab ise ek common denominator se clean karo:

t_{∥} = \frac{L ( c + v ) + L ( c - v )}{( c - v ) ( c + v )} = \frac{2 L c}{c ^{2} - v ^{2}} = \frac{2 L}{c} \cdot penalty > 1 \frac{1}{1 - \frac{v ^{2}}{c ^{2}}}

$2 L c$ — do legs ki distances ek single numerator mein collapse ho jaati hain.
$c^{2} - v^{2}$ — $(c - v) (c + v)$ se aata hai; wind ka fingerprint (positive kyunki $v < c$ ).
Bracketed factor 1 se bada hai har $0 < v < c$ ke liye, toh wind hamesha time cost karta hai. $v = 0$ set karo aur ye $1$ ban jaata hai: no wind, time exactly $\frac{2 L}{c}$ hai. Sahi — hamare formula ka degenerate case mein common sense se match hota hai.

Step 3 — Across-the-wind trip: kyun boat ko upstream aim karna padta hai

KYA HAI. Ab boat ko ek mark tak directly across nadi pahunchna hai aur seedha waapas aana hai. Agar wo seedha across point kare, current use downstream sweep kar dega aur wo off-target land karega. Seedha jaane ke liye, use apni nose upstream aim karni hogi taaki sideways drift exactly cancel ho.

KYU PYTHAGORAS AATA HAI. Boat ki speed water ke relative $c$ par fixed hai — yahi boat ki total velocity hai, hypotenuse ke roop mein drawn. Iska kuch hissa ( $v$ ) current se sideways ladne mein kharcha hota hai; jo bacha wo seedha across point karta hai. Teen arrows, ek right triangle:

full boat speed c^{2} = spent cancelling drift v^{2} + (cross speed)^{2} ⟹ cross speed = c^{2} - v^{2}

Hum right triangle use karte hain kyunki ye ek hi tool hai jo ek speed ko do perpendicular pieces mein split karta hai jab total fixed ho — exactly haari situation. Note karo $c^{2} - v^{2}$ tabhi real speed hai jab $v < c$ ho; agar current boat se tez hoti toh koi straight-across path hi nahi hota — ek aur reason ki poora analysis $β < 1$ mein rehta hai.

PICTURE. Velocity triangle: hypotenuse $c$ (aim), horizontal leg $v$ (current khaa jaata hai), vertical leg $c^{2} - v^{2}$ (useful cross speed).

Round trip distance $2 L$ hai is reduced cross speed par:

t_{⊥} = \frac{2 L}{c ^{2} - v ^{2}} = \frac{2 L}{c} \cdot \frac{1}{1 - \frac{v ^{2}}{c ^{2}}}

$c^{2} - v^{2}$ — triangle se vertical leg, woh speed jo boat ko cross karaati hai.
Bracket bhi 1 se bada hai — toh cross arm bhi slow ho jaata hai. Lekin note karo square root: ye penalty parallel arm ke comparison mein gentler hai. Yehi gap poora experiment hai.

Step 4 — Do penalties ko side by side rakhna

KYA HAI. Do round-trip times ko saath rakho aur subtract karo:

Δ t = t_{∥} - t_{⊥} = \frac{2 L}{c} parallel penalty \frac{1}{1 - β ^{2}} - cross penalty \frac{1}{1 - β ^{2}}, β \equiv \frac{v}{c}

$β$ KYUN DEFINE KARTE HAIN. Speeds jaise $30000$ vs $300000000$ awkward hain. Single ratio $β = v / c$ "light ke comparison mein kitna tez hai" measure karta hai, aur sab kuch isi par depend karta hai. Earth ki orbit ke liye $β \approx 1 0^{- 4}$ , toh $β^{2} \approx 1 0^{- 8}$ — absurdly chota. Ye chhotapan agla step ka key hai.

PICTURE. Do penalty curves $β$ ke against plotted. Dono $β = 0$ par 1 se start hote hain (no wind, equal times — phir se degenerate case), aur parallel curve tez chadhti hai. Unke beech ka vertical gap hai $Δ t \cdot \frac{c}{2 L}$ .

$β = 0$ par: dono penalties $= 1$ , gap $= 0$ . No wind ⇒ no time difference. Sanity confirmed.
Jaise $β \to 1$ , do denominators $1 - β^{2}$ aur $1 - β^{2}$ $0$ ki taraf shrink hote hain, toh dono penalty factors blow up ho jaate hain (parallel wala bahut tez). Ye edge sirf mathematical hai — hamare experiment ke liye $β \approx 1 0^{- 4} ≪ 1$ , toh hum curve ke far-left toe par hain. Poora plot jo hume batata hai wo hai shape near origin: gap $β^{2}$ ki tarah badhta hai, jo hum aage extract karte hain.

Step 5 — Zoom in karo: tiny gap $\propto β^{2}$ hai

KYA HAI. Kyunki $β^{2}$ minute hai, hum har penalty ko zero ke paas approximate karte hain. Do standard expansions (dono bas "curve 1 ki value se kaise nikalti hai"):

\frac{1}{1 - x} \approx 1 + x, \frac{1}{1 - x} \approx 1 + \frac{1}{2} x (tiny x = β^{2})

EXPAND KYUN KARTE HAIN. Hume exact monstrous fraction nahi chahiye — hume pehla non-zero difference chahiye. Do " $1 + kuch$ " numbers subtract karo, boring $1$ 's cancel ho jaate hain aur sirf "kuch" bachta hai:

Δ t \approx \frac{2 L}{c} [(1 + β^{2}) - (1 + \frac{1}{2} β^{2})] = \frac{2 L}{c} \cdot \frac{β ^{2}}{2} = \frac{L β ^{2}}{c} = \frac{L v ^{2}}{c ^{3}}

Parallel arm ki penalty $β^{2}$ ki tarah badhti hai; cross arm ki $\frac{1}{2} β^{2}$ ki tarah.
Unka difference $\frac{1}{2} β^{2}$ hai — surviving term.
$β = v / c$ reinstate karne se $\frac{2 L}{c} \cdot \frac{β ^{2}}{2}$ $\frac{L v ^{2}}{c ^{3}}$ ban jaata hai.

PICTURE. Pehle wale plot par $β = 0$ ke paas ek zoom box: do curves almost-parallel straight-ish lines jaisi lagti hain, aur unke beech ka shaded sliver " $\propto β^{2}$ " label hai — poora signal jo experiment dhoondh raha hai.

Step 6 — Time ko ek readable picture mein convert karna: fringes

KYA HAI. Ek stopwatch $1 0^{- 16}$ s resolve nahi kar sakta. Lekin light ki apni waves kar sakti hain. Jab do beams screen par dobara milti hain (recombination interference hai), ek time difference $Δ t$ bright/dark stripes ke roop mein dikhti hai — fringes. Poore table ko $9 0^{\circ}$ rotate karne se swap hota hai ki kaunsa arm aether wind ke parallel hai. Predicted number of fringes jo slip karte hain:

Δ N = \frac{c ( 2 Δ t )}{λ} \approx \frac{2 L v ^{2}}{λ c ^{2}}

$c (2Δ t)$ — extra path length jo light cover karti hai doubled time gap ki wajah se.
$\div λ$ — extra path ka har wavelength $λ$ ek full fringe; yahi ise countable banata hai.

Factor of 2 kyun hai (algebraic footnote). Starting orientation mein, parallel arm cross arm se $Δ t = t_{∥} - t_{⊥} = + \frac{L v ^{2}}{c ^{3}}$ aage hai. $9 0^{\circ}$ rotate karne ke baad, pehle wala cross arm ab parallel hai aur vice versa, toh labels swap ho jaate hain aur difference ban jaata hai

Δ t_{rotated} = t_{⊥} - t_{∥} = - \frac{L v ^{2}}{c ^{3}} .

Measurable cheez ye hai ki fringe pattern do orientations ke beech kaise change hota hai:

Δ t_{start} - Δ t_{rotated} = (+ \frac{L v ^{2}}{c ^{3}}) - (- \frac{L v ^{2}}{c ^{3}}) = 2 Δ t .

Sign rotation par flip ho jaata hai, toh do measurements $2Δ t$ se differ karte hain — yahi $c (2Δ t)$ mein $2$ hai. (Tum kabhi single orientation par fringe count jaanne par rely nahi karte; tum shift dekhte ho jab table ghumate ho.)

YE GENIUS KYUN HAI. Ye ek unmeasurable time ko stripes count karne mein convert karta hai — light ki wavelength ( $λ \sim 500 nm$ ) duniya ka finest ruler hai.

PICTURE. Interferometer schematic: source → beam splitter → do arms of length $L$ → mirrors → recombine → fringe pattern, $9 0^{\circ}$ rotation arrow ke saath jo arms swap karta hai (aur $Δ t$ ka sign flip karta hai).

Step 7 — Degenerate case jo actually hua: null

KYA HAI. Predicted signal: $0.44$ fringes. Observed: essentially $0$ , har orientation par, har season mein.

YE PUNCHLINE KYUN HAI. Observation ko haare formula mein daalo. $Δ t = \frac{L v ^{2}}{c ^{3}} = 0$ jab $L \neq = 0$ force karta hai $v = 0$ — lekin Earth space mein rest par nahi hai. Iska sirf ek escape hai ki poori premise "light kisi aether ke relative $c$ par move karti hai" galat hai. Balki, light $c$ par chalti hai sabke liye, har direction mein — special relativity ka second postulate. Rival patch, FitzGerald–Lorentz contraction, ne parallel arm ko itna shrink kiya ki $Δ t$ khatam ho jaaye jabki aether bacha rahe; Einstein ne instead aether ko bilkul bahar uchhal diya (aur same maths Lorentz Transformation, Time Dilation, aur Galilean velocity addition ki failure ke roop mein dobara aata hai).

PICTURE. Do fringe screens side by side: predicted (stripes visibly shifted after rotation) vs. observed (stripes nahi hilte). Unke beech ka gap relativity ka janam hai.

Ek-picture summary

Upar sab kuch compress kiya: do beams splitter se nikalte hain, ek wind ke saath race karta hai (penalty $\frac{1}{1 - β ^{2}}$ ), ek cross karta hai (penalty $\frac{1}{1 - β ^{2}}$ ), gap $\frac{L v ^{2}}{c ^{3}}$ hai, $\approx 0.44$ fringes ke roop mein padha — aur nature ne 0 jawab diya.

Recall Feynman retelling — poora walkthrough simple words mein

Ek nadi imagine karo jo sideways beh rahi hai. Ek boat nadi ke saath jaati hai aur waapas aati hai; doosri seedha cross karti hai aur waapas aati hai. Dono boats ka same engine hai, phir bhi current unki trips ko alag time leti hai — down-river boat upstream crawl karne mein zyada khoyi hai jo usne rushing down mein gain ki, aur crossing boat sirf seedha jaane ke liye current mein lean karne mein effort waste karta hai. Ab "boat" ko "beam of light" aur "river" ko invisible "aether" se replace karo jo log sochte the space ko fill karta hai, Earth moving boat hai. Humne do time penalties draw ki, dekha ki along-the-wind wali hamesha zyada loser hai, subtract kiya, aur paya ki difference ek tiny number hai — $(v / c)^{2}$ ke proportional, lagbhag ek sow-millionth. Clock se time karne ke liye too small, lekin light ki apni wavelength itni fine ruler hai: difference stripe pattern ko lagbhag 0.44 fringes slide karana chahiye tha jab tum table spin karo (aur spinning flip karti hai ki kaunsa arm quicker hai, shift ko double karti hai jo tum dhoondh rahe ho). Jab Michelson aur Morley ne actually dekha — kuch nahi hila. Nadi thi hi nahi. Light koi current nahi ride karti; wo simply sabke liye same speed par travel karti hai. Wo "kuch nahi" physics ki history ke loudest results mein se ek hai.

Kaun sa arm zyada time penalty suffer karta hai, aur leading order mein kitne factor se?

Parallel (along-wind) arm; uska

β^{2}

term cross arm ke

\frac{1}{2} β^{2}

se do guna hai.

Stopwatch seedha

Δ t

kyun measure nahi kar sakta?

Ye ~

1 0^{- 16}

s hai; iske bajaye hum ise light ki wavelength

λ

use karke countable fringe shift mein convert karte hain.

Fringe formula mein factor of 2 kyun hai?

9 0^{\circ}

rotate karne se

Δ t

ka sign flip ho jaata hai, toh do orientations

2Δ t

se differ karte hain.

v = 0

set karne se is page par har formula par kya hoga?

Dono penalties 1 ban jaati hain,

Δ t = 0

Δ N = 0

— no wind, no effect (sanity/degenerate check).

Derivation ke sense karne ke liye

v < c

kyun hona chahiye?

Taaki

c - v > 0

ho (upstream leg complete ho) aur

c^{2} - v^{2}

real ho (straight cross-path exist kare).

Predicted vs observed fringe shift?

Predicted

\approx 0.44

; observed

\approx 0

— null result.