Special relativity — Michelson-Morley experiment: Question bank

2.3.25 · D5 · Physics › Modern Physics › Special relativity — Michelson-Morley experiment

Prerequisite ideas jo khuli rakhni chahiye: Galilean Relativity & Velocity Addition (woh "galat" velocity-addition jo ye experiment test karta hai), Maxwell's Equations and the Speed of Light (jahan se $c$ aata hai), Interference of Light & Fringes (answer padha kaise jaata hai), aur jahan ye sab le jaata hai: Special Relativity — Einstein's Postulates, Length Contraction, Time Dilation, Lorentz Transformation.

Definition Is page par jo notation dikhegi

Taaki tumhe parent note mein dhundhna na pade, yahan har ek symbol plain words mein hai:

$c$ — aether ke relative light ki speed (assumed invisible medium). Lagbhag $3 \times 1 0^{8}$ m/s.
$v$ — apparatus (Earth) ki speed us aether mein drift karte hue: "aether wind". Lagbhag $3 \times 1 0^{4}$ m/s.
$L$ — interferometer ki ek arm ki length: beam splitter se har mirror tak ki doori. Beam $L$ jaati hai aur $L$ wapas aati hai, isliye ek round trip mein $2 L$ .
$t_{∥}$ — wind ke saath wali arm (down-wind phir up-wind) mein beam ka round-trip time.
$t_{⊥}$ — wind ke aarpaar wali arm mein beam ka round-trip time.
$Δ t = t_{∥} - t_{⊥}$ — dono beams ke beech ka time difference; ye woh quantity hai jo aether ke hisaab se non-zero honi chahiye.
$β = v / c$ — wind speed light speed ke fraction ke roop mein. Earth ke liye $β \sim 1 0^{- 4}$ , isliye $β^{2} \sim 1 0^{- 8}$ — bahut chhota.
$λ$ — light ki wavelength (lagbhag $500$ nm), woh ruler jo path difference ko fringes mein convert karta hai.
$Δ N$ — predicted fringe shift: apparatus rotate karne par interference pattern kitne bright-to-dark cycles khisake.

True or false — justify

Har "true/false" answer mein kyun batana zaroori hai — sirf verdict dene par zero milega.

Michelson–Morley experiment ne light ki speed directly measure ki.

False. Isne kisi beam ko kabhi time nahi kiya. Isne do beams ki arrival ko interference fringes ke zariye compare kiya, arms ke beech ek tiny time difference

Δ t

detect karke — itna chhota ki koi clock pakad nahi sakta, lekin fringes pakad sakti hain.

Null result ka matlab hai experiment fail ho gaya.

False. Ek precise null apne aap mein ek discovery hai: isne detectable aether wind ko rule out kar diya aur physics ko

c

ki constancy ki taraf moddiya. Negative results theory ko strongly constrain karte hain.

Agar aether exist karta aur Earth ek pal ke liye usme rest mein hoti, tab bhi experiment ek shift dikhata.

False.

v = 0

ke saath koi aether wind nahi hogi, isliye dono effective speeds

c

ke barabar honge,

t_{∥} = t_{⊥} = 2 L / c

milega aur

Δ N = 0

— bilkul wahi case jo real null result jaisa dikhta hai. Isliye unhone ise seasons ke across repeat kiya, umeed mein ki kabhi na kabhi Earth aether mein move karte hue pakdi jayegi.

Apparatus ko

9 0^{\circ}

rotate karna hi measurable fringe shift produce karta hai.

True. Ek single orientation ka

Δ t

ek fringe pattern mein embedded hota hai jiska "zero" tumhe pata nahi, isliye tum use read nahi kar sakte.

9 0^{\circ}

rotate karne se swap hota hai ki kaun si arm parallel hai: jo arm slow thi (

t_{∥}

) woh fast (

t_{⊥}

) ban jaati hai aur vice-versa, isliye path difference

+ c Δ t

- c Δ t

flip ho jaata hai — yani

2 c Δ t

ka change, matlab time mein

2Δ t

ka change. Fringes yahi change register karti hain, aur ye

Δ N = 2 c Δ t / λ

cycles ke barabar hai.

Parallel arm ke downstream aur upstream legs cancel ho jaate hain.

False. Dono legs distance

L

cover karti hain: up-wind leg mein

L / (c - v)

lagta hai (slow), down-wind leg mein

L / (c + v)

(fast). Tum zyada waqt slow leg mein guzaarte ho, isliye ye add hokar no-wind time se zyada ban jaata hai, kam nahi:

\frac{L}{c - v} + \frac{L}{c + v} = \frac{2 L c}{c ^{2} - v ^{2}} > \frac{2 L}{c}

. Yahi excess hai jo wind ko reveal karta.

Perpendicular arm aether wind se bilkul unaffected hai.

False. Ek seedha cross-path travel karne ke liye beam ko thoda upstream aim karna padta hai, isliye uski useful cross-speed

c^{2} - v^{2}

tak girr jaati hai (neeche cross-arm "why" dekho). Uska round trip hai

t_{⊥} = 2 L / c^{2} - v^{2}

, jo

2 L / c

se zyada hai — bas parallel arm se kam zyada, aur yahi wajah hai ki ek difference

Δ t

survive karta hai.

Experiment ne directly Einstein ki special relativity prove ki.

False. Isne detectable aether ko khatam kiya aur SR ke saath consistent tha, lekin Lorentz–FitzGerald contraction bhi null result ko fit karta tha. SR ne apna status predictions ke ek bahut bade web se kamaya, sirf is ek experiment se nahi.

Leading order mein expected time difference aether speed

v

ke saath linearly badhta hai.

False. Ye

v^{2}

ke scale par hai:

Δ t \approx L v^{2} / c^{3}

. Dono arms ke linear-in-

v

terms subtraction mein equal aur cancel ho jaate hain (neeche

v^{2}

wala "why" dekho),

β^{2} = (v / c)^{2}

ko leading effect ke roop mein chodke — tiny, lekin interferometer isi ko dekhne ke liye banaya gaya tha.

Spot the error

Neeche har statement mein ek flaw hai. Use naam do aur mechanism explain karo.

"Light ko koi medium chahiye hi nahi, isliye 1887 mein aether hypothesis clearly silly tha."

Flaw hai hindsight. 1887 mein har known wave (sound, water, string) ko ek medium chahiye tha, aur Maxwell's equations ne ek definite

c

diya jisme sawaal naturally uthta tha "kis ke relative?". Aether ek reasonable default answer tha, silly nahi — yahi experiment tha jisne physicists ko ise abandon karne par majboor kiya.

"Kyunki aether mein light

c

par travel karti hai, Earth par hum wind mein jaate hue

c + v

measure karte."

Flaw hai sign/direction error. Wind mein jaate hue, medium beam ke against stream karta hai, isliye ground speed

c - v

hai (slower); sirf wind ke saath jaane par ye

c + v

hai. Inhe swap karne se pure prediction ka sign flip ho jaata, kyunki aether effect poori tarah se slow leg

L / (c - v)

aur fast leg

L / (c + v)

ke beech ke difference mein rehta hai.

"Cross-arm light

c - v

par travel karti hai kyunki ye wind se head-on ladti hai."

Flaw hai wrong geometry. Cross beam wind se head-on nahi ladta; ye uske perpendicular move karta hai. Uski speed

c

ek vector hai: sideways drift cancel karne ke liye ye upstream aim karta hai, isliye

c

ka ek part current se ladne mein "kharach" ho jaata hai. Pythagoras se,

c^{2} = v^{2} + (cross-speed)^{2}

, cross-speed

c^{2} - v^{2}

milta hai —

c - v

nahi. Head-on value

c - v

parallel arm ka hai, is arm ka nahi.

"Kyunki dono arms length

L

ki hain, dono round-trip times barabar hone chahiye."

Flaw hai equal distance ko equal time samajhna. Time distance divided by effective speed hota hai, aur dono arms ki effective speeds alag hain: parallel arm deta hai

t_{∥} = \frac{2 L c}{c ^{2} - v ^{2}}

, perpendicular arm

t_{⊥} = \frac{2 L}{c ^{2} - v ^{2}}

. Same

L

, alag denominators, isliye

t_{∥} > t_{⊥}

jab bhi

v \neq = 0

"0.4-fringe prediction bahut chhoti hai matter karne ke liye, isliye null result kuch prove nahi karta."

Flaw hai "small" ka galat andaza. Apparatus lagbhag

0.01

fringe tak shifts resolve kar sakta tha, isliye predicted

Δ N \approx 0.4

uski sensitivity se chalees guna hai — bilkul saf dikhne wala. Itni badi prediction ke against essentially zero dekhna ek strong constraint hai, koi draw nahi.

"Agar unhone sirf longer arm

L

use ki hoti, to wind eventually cancel hokar waise bhi zero de deta."

Flaw hai backwards dependence. Kyunki

Δ N = 2 L v^{2} / (λ c^{2})

, predicted shift

L

ke proportional hai: longer arms wind ko detect karna aasaan banate hain, mushkil nahi. Michelson ne deliberately light path ko extra mirrors se fold kiya effective

L

badhane ke liye, bilkul isi wajah se.

Why questions

Mechanism ke saath jawab do, sirf label nahi.

Expected effect

v^{2}

par kyun depend karta hai,

v

par nahi?

Har time ko likho

t_{∥} = \frac{2 L}{c} \frac{1}{1 - β ^{2}}

aur

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

. Chhote

β

ke liye expand karo:

\frac{1}{1 - β ^{2}} \approx 1 + β^{2}

aur

\frac{1}{1 - β ^{2}} \approx 1 + \frac{1}{2} β^{2}

. Dono mein same constant

1

hai (jo

t_{∥} - t_{⊥}

mein cancel ho jaata hai) aur

v

mein koi bhi linear term nahi hai — arms first order mein symmetric hain. Subtract karne par milta hai

\frac{2 L}{c} (β^{2} - \frac{1}{2} β^{2}) = \frac{2 L}{c} \cdot \frac{β ^{2}}{2} = \frac{L v ^{2}}{c ^{3}}

, isliye leading survivor

\propto v^{2}

hai.

Cross-beam ko seedha across ki bajaye upstream aim kyun karna padta hai?

Agar seedha aim kiya jaaye, to

v

speed ki wind trip ke dauran use sideways drift kar degi aur mirror miss ho jaayega. Upstream aim karna drift ko us sideways push ko exactly cancel karne deta hai. Lekin tab, total speed

c

mein se, ek component

v

counter-drifting mein kharach hota hai, aur Pythagoras se sirf

c^{2} - v^{2}

useful straight-across motion ke liye bachta hai.

Unhone apparatus rotate kyun kiya aur poore saal kyun dekha?

9 0^{\circ}

rotate karne se swap hota hai ki kaun si arm wind ke parallel hai, path difference ko

2 c Δ t

se flip karke hidden

Δ t

ko ek moving fringe pattern mein baadal deta hai jise measure kiya ja sake. Saal bhar dekhne se Earth ki hamesha badalti orbital direction cover hoti hai, isliye aether wind ek date par bhi gayab ho jaaye to har date par chupi nahi reh sakti.

Readout ke liye stopwatch ki bajaye interference (fringes) kyun choose ki gayi?

Predicted

Δ t \sim 1 0^{- 16}

s kisi bhi clock ki reach se bahut door hai, lekin ye wavelength

λ

ke fraction ke path difference ke barabar hai. Interference ek sub-wavelength path difference ko directly bright/dark fringes ki visible movement mein convert kar deta hai, ek unmeasurable time ko countable shift

Δ N = 2 c Δ t / λ

mein amplify karke.

Swimmer-in-a-river analogy light beams ke liye kyun kaam karti hai?

Aether picture mein, light ki speed aether ke relative fixed hai bilkul waise jaise swimmer ki speed paani ke relative fixed hoti hai; lab frame nadi ke kinare ka role play karta hai. Isliye wahi round-trip asymmetry — down/up-stream overall straight across se slower hona — term-by-term carry over hoti hai.

Null result "

c

sabhi observers ke liye same hai" ki taraf kyun point karta hai?

Agar koi bhi arm orientation ya season kabhi shift nahi dikhata, to lab mein measured light speed lab ki apni motion

v

se independent hai. Ye precisely wahi statement hai ki

c

frame-independent hai — baad mein Einstein ke second postulate ke roop mein enshrined.

Lorentz ki length contraction ne aether ko kyun "bachaya", aur Einstein ne use kyun chhoda?

Parallel arm

L

ko bilkul sahi factor se shrink karne par

t_{∥} = t_{⊥}

milta hai, isliye

Δ t = 0

aur wind chupp ho jaati hai — ek aise patch jo data fit karta hai. Einstein ne aether isliye chhodda kyunki same equations do postulates se zyada simply follow hoti hain, bina kisi undetectable medium ko invent kiye.

Edge cases

Boundary aur degenerate inputs jo reader ko surprise nahi karni chahiye.

Agar

v = 0

ho (Earth aether mein rest mein), to predicted fringe shift kya hoga?

Zero. Bina wind ke dono effective speeds

c

hain, isliye

t_{∥} = t_{⊥} = 2 L / c

aur

Δ N = 0

— jo actual observed null se bilkul indistinguishable hai, isliye ek akela measurement akele kuch decide nahi kar sakta.

Ek hypothetical extreme wind ke liye

v \to c

par predicted

Δ t

kya hoga?

t_{∥} = \frac{2 L c}{c ^{2} - v ^{2}}

blow up kar jaata hai kyunki

c^{2} - v^{2} \to 0

(up-wind leg

L / (c - v)

kabhi aage nahi badh sakta), isliye naive aether prediction diverge kar jaata hai — ek red flag ki jis classical velocity-addition ko ye assume karta hai woh toot gayi hai.

Agar dono arms ki unequal lengths

L_{1} \neq = L_{2}

hon, tab bhi experiment kaam karta hai?

Haan — rotation comparison tab bhi aether-dependent term ko isolate karta hai kyunki rotate karne se arms ke roles swap hote hain; unequal lengths sirf ek fixed path offset add karti hain jo rotation ke under wahi rehta hai aur cancel ho jaata hai, isliye exact arm-length matching critical requirement nahi thi.

Agar literally koi wave medium bhi nahi hota aur relativity bhi nahi — kya null result phir bhi luck se ho sakta tha?

Tabhi agar

v

har ek single measurement par exactly zero hota, jo saal bhar ki repetition rule out kar deti hai. Sabhi directions aur dates par persistent null ek principle (constant

c

) maangta hai, lucky coincidence nahi.

β = v / c ≪ 1

ki limit mein, sirf

β^{2}

term rakhna legitimate kyun hai?

Earth ke liye

β \sim 1 0^{- 4}

, isliye agla correction

β^{4} \sim 1 0^{- 16}

β^{2} \sim 1 0^{- 8}

ke beside bilkul negligible hai; truncated expansion

0.01

-fringe resolution se kahin behtar accurate hai.

Agar light sach mein Galilean velocity addition follow karti, to ye experiment kya dikhata?

Ek clear, orientation-dependent fringe shift of order

0.4

fringes — dekho Galilean Relativity & Velocity Addition. Is shift ki observed absence direct evidence hai ki light velocities Galilean tarike se add nahi karti.