Worked examples — Special relativity — Michelson-Morley experiment

2.3.25 · D3 · Physics › Modern Physics › Special relativity — Michelson-Morley experiment

Is page par Michelson–Morley ke formulas itne baar drill kiye jayenge ki koi bhi case aapko surprise nahi kar sake. Pehle hum har tarah ki situation list karenge jo is topic mein aa sakti hai, phir har ek ka fully worked example dekhenge.

Shuru karne se pehle, ek reminder — woh symbols jo hum parent note mein pehle se earn kar chuke hain (koi bhi naya symbol use hone se pehle bataya jayega). Neeche ki figure mein $L$, $c$, $v$ aur dono arms actually kaisa dikhte hain, yeh show kiya gaya hai:

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Scenario matrix

Is topic ka har problem in cells mein se ek hai. Neeche ke worked examples mein us cell ka tag laga hua hai jise woh cover karta hai.

Cell	Kya vary karta hai	Degenerate / limit dhyaan rakhne wali	Example
A. Baseline arm times	ordinary $v\ll c$	—	Ex 1
B. Zero wind	$v=0$	dono arms same time deti hain	Ex 2
C. Extreme wind	$v\to c$ ($\beta \to 1$)	parallel time $\to \infty$	Ex 3
D. Fringe-shift prediction	$\Delta N$ mein numbers plug karo	—	Ex 4
E. Exact vs. approximate	kya ${\beta }^2$ shortcut kaam karta hai?	small vs. large $\beta$	Ex 5
F. Inverse / design problem	target $\Delta N$ diya ho, $L$ nikalo	—	Ex 6
G. Real-world word problem	swimmer-in-river numbers	current $>$ swimmer?	Ex 7
H. Exam twist	kaun sa arm faster hai, $\Delta t$ ka sign	$t_{\parallel }>t_{\perp }$ hamesha	Ex 8
I. Geometry of the cross-arm	$\sqrt{c^2-v^2}$ kyun	$v=0$ aur $v\to c$ corners	Ex 9

Hum poore note mein yeh master formulas use karenge (sab parent note mein derive kiye gaye hain). Exact forms kisi bhi $\beta$ ke liye valid hain; approximate forms ($\approx$) sirf tab jab $\beta \ll 1$:

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Example 1 — Cell A: baseline round-trip times

Forecast: Dono times $2L/c$ ke aas-paas hongi. Guess karo: kya woh equal hain, ya ek doosre se kitne parts per million zyada hai?

1. $\beta$ aur ${\beta }^2$ compute karo. $\beta =\frac{3\times 10^4}{3\times 10^8}=10^{-4}$, isliye ${\beta }^2=10^{-8}$. Yeh step kyun? Har formula ${\beta }^2$ se bana hai; pehle ise nikaal lo taaki hum ise dobara derive na karein.

2. No-wind baseline time $\frac{2L}{c}=\frac{2\times 11}{3\times 10^8}=7.333\times 10^{-8}\text{ s}$. Yeh step kyun? Dono arms bas yahi baseline hain, $1$ se thoda upar ke factor se multiply ki gayi, isliye yeh anchor hai.

3. Parallel: $t_{\parallel }=7.333\times 10^{-8}\times \frac{1}{1-10^{-8}}\approx 7.333\times 10^{-8}(1+10^{-8})\text{ s}$. Yeh step kyun? $\frac{1}{1-x}\approx 1+x$ kyunki $x=10^{-8}$ bahut chhota hai.

4. Perpendicular: $t_{\perp }=7.333\times 10^{-8}\times \frac{1}{\sqrt{1-10^{-8}}}\approx 7.333\times 10^{-8}(1+\frac{1}{2}\times 10^{-8})\text{ s}$. Yeh step kyun? $\frac{1}{\sqrt{1-x}}\approx 1+\frac{x}{2}$.

5. Difference: $t_{\parallel }-t_{\perp }\approx 7.333\times 10^{-8}\times \frac{1}{2}\times 10^{-8}\approx 3.67\times 10^{-16}\text{ s}$. Yeh step kyun? Dono $(1+\dots )$ factors subtract karne par sirf tiny corrections ka difference bachta hai, $10^{-8}-\frac{1}{2}\times 10^{-8}=\frac{1}{2}\times 10^{-8}$.

Verify: $\Delta t\approx L{v}^2/c^3=\frac{11\times (3\times 10^4{)}^2}{(3\times 10^8{)}^3}=3.67\times 10^{-16}\text{ s}$ use karo. ✓ Same. Units: $\text{m}\cdot (\text{m/s}{)}^2/(\text{m/s}{)}^3=\text{s}$. ✓ Aur $t_{\parallel }>t_{\perp }$ kyunki $\frac{1}{1-x}>\frac{1}{\sqrt{1-x}}$ for $0<x<1$. ✓

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Example 2 — Cell B: zero-wind degenerate case

Forecast: Koi current nahi, toh "river" ek still pond hai. Guess karo: kya dono arms exactly tie karti hain?

1. $\beta =0$ set karo. Tab $1-{\beta }^2=1$ aur $\sqrt{1-{\beta }^2}=1$. Yeh step kyun? $v=0\Rightarrow \beta =0$; dono correction factors $1$ par collapse ho jaate hain.

2. Dono times baseline ke equal: $t_{\parallel }=t_{\perp }=\frac{2L}{c}$. Yeh step kyun? Baseline ko $1$ se multiply karne par woh unchanged rehti hai — koi wind nahi, toh "along" aur "across" physically identical hain.

3. Difference: $\Delta t=t_{\parallel }-t_{\perp }=\frac{2L}{c}-\frac{2L}{c}=0$. Yeh step kyun? Do equal times subtract hone par zero aata hai; experiment ke liye literally kuch detect karne ko nahi.

Verify: $\Delta t\approx L{v}^2/c^3=L\cdot 0/c^3=0$. ✓ Physical sense: koi wind nahi matlab koi preferred direction nahi, isliye koi arm favoured nahi ho sakti — exactly wahi situation jaise null result dikhta hai. Isliye Michelson–Morley ka zero shift "lagta tha" ki $v=0$ hai har jagah.

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Example 3 — Cell C: extreme-wind limit $\beta \to 1$

Forecast: Yahan shortcut $L{v}^2/c^3$ kaam nahi karega — $\beta$ chhota nahi hai. Guess karo: kaun sa time zyada tezi se blow up karta hai?

1. Baseline $\frac{2L}{c}=7.333\times 10^{-8}\text{ s}$ (Ex 1 jaise hi). Yeh step kyun? Dono exact times abhi bhi yahi baseline hain times a wind-factor, isliye $2L/c$ dobara compute karne ki jagah reuse karte hain.

2. Factors: $1-{\beta }^2=1-0.81=0.19$; $\sqrt{1-{\beta }^2}=\sqrt{0.19}=0.43589$. Yeh step kyun? Bade $\beta$ par hume exact denominators use karne honge — Taylor shortcut invalid hai.

3. Parallel: $t_{\parallel }=7.333\times 10^{-8}/0.19=3.860\times 10^{-7}\text{ s}$. Yeh step kyun? Baseline ko chhote factor $0.19$ se divide karne par time paanch guna blow up ho jaata hai — along-wind trip bahut slow ho jaati hai.

4. Perpendicular: $t_{\perp }=7.333\times 10^{-8}/0.43589=1.682\times 10^{-7}\text{ s}$. Yeh step kyun? Cross factor $0.436$, $0.19$ se bada hai, isliye cross-arm, parallel arm se kam slow hoti hai.

5. Limit $\beta \to 1$: $1-{\beta }^2\to 0$ se $t_{\parallel }\to \infty$ (near-$c$ wind ke against seedha jaati light barely aage badhti hai), jabki $t_{\perp }\to \infty$ bhi hota hai lekin dhire. $t_{\parallel }$ zyada tezi se diverge karta hai. Yeh step kyun? $1/(1-{\beta }^2)$ mein $1/\sqrt{1-{\beta }^2}$ se stronger singularity hai.

Verify: Ratio $t_{\parallel }/t_{\perp }=\sqrt{1-{\beta }^2}/(1-{\beta }^2)=1/\sqrt{1-{\beta }^2}=1/0.43589=2.294$. Check: $3.860/1.682=2.295$. ✓

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Example 4 — Cell D: predicted fringe shift

Forecast: Instrument $0.01$ fringes dekh sakta hai. Guess karo: kya prediction us threshold se kaafi upar hogi?

1. Use $\Delta N\approx \frac{2L{v}^2}{\lambda c^2}$. Yeh step kyun? Yeh rotation formula hai (mein $2$ woh "there-and-back" $2\Delta t$ swap hai jo master-formula box mein explain kiya gaya hai); $\lambda$ se divide karne par path-length, countable fringes ki sankhya mein convert ho jaati hai.

2. Numerator: $2\times 11\times (3\times 10^4{)}^2=22\times 9\times 10^8=1.98\times 10^{10}$. Yeh step kyun? Pehle constants group karo — $3\times 10^4$ ko square karne par $9\times 10^8$ milta hai — exponent slips se bachne ke liye.

3. Denominator: $\lambda c^2=(5\times 10^{-7})(9\times 10^{16})=4.5\times 10^{10}$. Yeh step kyun? Same tactic: $c$ ko ek baar square karo ($9\times 10^{16}$) phir $\lambda$ se multiply karo, taaki dono bade powers cleanly combine ho jayein.

4. Divide: $\Delta N\approx \frac{1.98\times 10^{10}}{4.5\times 10^{10}}=0.44$ fringes. Yeh step kyun? $10^{10}$ powers cancel ho jaate hain, sirf plain ratio $1.98/4.5$ bachta hai — final pure fringe count.

Verify: $0.44\gg 0.01$, isliye ek real shift unmissable hoti. Observed shift $\approx 0$ thi. Units: $\frac{\text{m}\cdot (\text{m/s}{)}^2}{\text{m}\cdot (\text{m/s}{)}^2}$ = dimensionless. ✓ Parent note ke $0.44$ se match karta hai. ✓

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Example 5 — Cell E: exact vs. approximate, kya shortcut kaam karta hai?

Forecast: Shortcut ek Taylor approximation hai. Guess karo: $10^{-4}$ ke liye yeh near-perfect hai, lekin $0.5$ par yeh kitna badly fail karta hai?

1. $2L/c=1$ par rescale karo. Tab exact difference hai $\Delta t=\frac{1}{1-{\beta }^2}-\frac{1}{\sqrt{1-{\beta }^2}}$ aur shortcut hai $\frac{{\beta }^2}{2}$. Yeh step kyun? Dono formulas ko common factor $2L/c$ se divide karne par pure numbers bacha lete hain, isliye hum dono expressions ki shapes compare kar sakte hain bina kisi metres ya seconds ke.

2. $\beta =10^{-4}$ par (${\beta }^2=10^{-8}$): exact bracket $=\frac{1}{1-10^{-8}}-\frac{1}{\sqrt{1-10^{-8}}}\approx 5.0\times 10^{-9}$; shortcut $=\frac{1}{2}\times 10^{-8}=5.0\times 10^{-9}$. Yeh step kyun? Confirm karta hai ki shortcut essentially exact hai jab $\beta$ tiny ho.

3. $\beta =0.5$ par (${\beta }^2=0.25$): exact bracket $=\frac{1}{0.75}-\frac{1}{\sqrt{0.75}}=1.3333-1.1547=0.1786$; shortcut $=0.125$. Yeh step kyun? Dikhata hai ki shortcut $\sim 30\%$ under-predict karta hai jab $\beta$ large ho — woh higher Taylor terms jo hum ne throw away kiye the ab negligible nahi hain.

Verify: $\beta =0.5$ par relative error: $(0.1786-0.125)/0.1786=0.30$, yaani $30\%$ off. $\beta =10^{-4}$ par: error $<10^{-8}$. ✓ Lesson: $L{v}^2/c^3$ shortcut sirf $\beta \ll 1$ ke liye hai.

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Example 6 — Cell F: inverse design problem

Forecast: Ex 4 se, $L=11\text{ m}$ ne $0.44$ diya. Guess karo: tumhe us se thoda zyada double chahiye hoga.

1. Rearrange $\Delta N=\frac{2L{v}^2}{\lambda c^2}$ ko $L$ ke liye: $L=\frac{\Delta N\lambda c^2}{2v^2}$. Yeh step kyun? Hume $L$ ke alawa sab kuch pata hai, isliye dono sides ko $\lambda c^2$ se multiply karke aur $2v^2$ se divide karke algebraically isolate karo.

2. Numerator: $\Delta N\lambda c^2=1.0\times (5\times 10^{-7})\times (9\times 10^{16})=4.5\times 10^{10}$. Yeh step kyun? Fraction ka top pehle banao; $c$ ko ek baar square karo ($9\times 10^{16}$) exponents tidy rakhne ke liye.

3. Denominator: $2v^2=2\times (3\times 10^4{)}^2=1.8\times 10^9$. Yeh step kyun? $v$ ko square karo ($9\times 10^8$) phir double karo — yeh fraction ka bottom hai, powers mix hone se bachne ke liye alag rakha.

4. Divide: $L=\frac{4.5\times 10^{10}}{1.8\times 10^9}=25\text{ m}$. Yeh step kyun? Dono powers divide karne par ($10^{10}/10^9=10$) aur mantissas ($4.5/1.8=2.5$) se $25$ milta hai — required arm length.

Verify: $L=25$ ko Ex 4 ke ratio mein plug karo: $\Delta N=0.44\times (25/11)=1.0$. ✓ Units: $\frac{1\cdot \text{m}\cdot (\text{m/s}{)}^2}{(\text{m/s}{)}^2}=\text{m}$. ✓ (Isliye Michelson ne beam ko kai mirrors se fold karke effective $L\sim 11\text{ m}$ reach ki.)

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Example 7 — Cell G: real-world swimmer word problem

Forecast: Guess karo: kya across route, downstream-and-back route ko beat karta hai? Aur kya swimmer seedha across ja sakta hai agar current unhe outpace kare?

1. IS problem ke liye ratio define karo. Kyunki yahan "swimmer speed" $s$ hai, light-speed $c$ nahi, wind fraction hai $\beta \equiv \frac{v}{s}=\frac{3}{5}=0.6$ (dhyan raho: har optics example mein $\beta =v/c$; sirf is swimmer analogy mein $c$ ki jagah $s$ aata hai). Chhota nahi, isliye exact forms mandatory hain. Yeh step kyun? $\beta$ control karta hai ki Taylor shortcut allowed hai ya nahi; $0.6$, $0$ se bahut door hai, isliye hume raw speeds use karne honge.

2. Parallel (exact, two legs): $t_{\parallel }=\frac{L}{s-v}+\frac{L}{s+v}=\frac{100}{2}+\frac{100}{8}=50+12.5=62.5\text{ s}$. Yeh step kyun? Downstream ground speed $s+v$ hai, upstream $s-v$ hai; har leg distance $÷$ apni speed hai.

3. Perpendicular: cross-speed $=\sqrt{s^2-v^2}=\sqrt{25-9}=4\text{ m/s}$, isliye $t_{\perp }=\frac{2L}{\sqrt{s^2-v^2}}=\frac{200}{4}=50\text{ s}$. Yeh step kyun? Swimmer upstream aim karta hai taaki drift cancel ho; Pythagoras useful cross-speed deta hai.

4. Difference: $\Delta t=t_{\parallel }-t_{\perp }=62.5-50=12.5\text{ s}$. Downstream route slower hai — Cell H ke rule $t_{\parallel }>t_{\perp }$ se match karta hai. Yeh step kyun? Analogy ka poora point yahi hai ki unequal-speed legs, straight-across route se zyada time lete hain.

5. Degenerate case $v=6>s=5$: ab $s-v=-1$, isliye $t_{\parallel }$ nonsensical hai (swimmer backwards sweep ho jaata hai — kabhi upstream wapas nahi aa sakta), aur $\sqrt{s^2-v^2}=\sqrt{25-36}=\sqrt{-11}$ imaginary hai — swimmer seedha across ja hi nahi sakta. Yeh step kyun? Yeh dikhata hai ki formulas honestly break ho jaate hain jab current swimmer se zyada ho, jo $v>c$ ka analogue hai jise relativity light ke liye forbid karti hai.

Verify: $t_{\parallel }/t_{\perp }=62.5/50=1.25$; predicted $1/\sqrt{1-{\beta }^2}=1/\sqrt{1-0.36}=1/0.8=1.25$. ✓

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Example 8 — Cell H: exam twist ($\Delta t$ ka sign)

Forecast: Guess karo: parallel (along-wind) arm hamesha slow wali hoti hai.

1. Ratio form karo $R=\frac{t_{\parallel }}{t_{\perp }}=\frac{1/(1-{\beta }^2)}{1/\sqrt{1-{\beta }^2}}=\frac{\sqrt{1-{\beta }^2}}{1-{\beta }^2}=\frac{1}{\sqrt{1-{\beta }^2}}$. Yeh step kyun? Ratio shared $2L/c$ hatata hai taaki sirf wind-factors bacho.

2. Bound karo: $0<\beta <1$ ke liye, $\sqrt{1-{\beta }^2}<1$, isliye $R=1/\sqrt{1-{\beta }^2}>1$. Yeh step kyun? Jo bhi $1$ se chhota hota hai uska reciprocal $1$ se bada hota hai.

3. Conclusion: $t_{\parallel }>t_{\perp }$ hamesha. Isliye correct signed prediction (hamare toolkit definition $\Delta t\equiv t_{\parallel }-t_{\perp }$ se match karta hua) positive hai. Student ka negative sign bas iska matlab hai ki unhone wrong order mein subtract kiya. Yeh step kyun? Sign convention pin down karta hai taaki student kabhi mislabel na kare ki kaun sa arm lead karta hai.

Verify: $\beta =0.5$ par, $R=1/\sqrt{0.75}=1.1547>1$. ✓ $\beta =0.9$ par (Ex 3), $R=2.294>1$. ✓ Rule holds.

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Example 9 — Cell I: cross-arm ki geometry

Forecast: Guess karo: light ki apni speed $c$ hypotenuse hai, leg nahi.

1. Triangle padho. Light hamesha aether ke through speed $c$ se chalti hai — yeh hypotenuse hai. Wind ise sideways $v$ speed se sweep karti hai — horizontal leg. Seedha across land karne ke liye, useful (vertical) component remaining leg hai. Yeh step kyun? Velocities vectors ki tarah add hoti hain (dekho Galilean Relativity & Velocity Addition); beam ko upstream aim karna padta hai taaki sideways wind exactly cancel ho jaaye.

2. Pythagoras apply karo right triangle par: $c^2=v^2+(\text{cross-speed}{)}^2$, isliye cross-speed $=\sqrt{c^2-v^2}$. Yeh step kyun? Hypotenuse squared, leg squares ke sum ke barabar hota hai — yahi woh ek relation hai jo $c$, $v$, aur useful speed ko tie karta hai.

3. Corner $v=0$: cross-speed $=\sqrt{c^2}=c$. Sensible — koi wind nahi, full speed across. Yeh step kyun? Formula ko still-water limit ke against check karta hai jahan kuch bhi lost nahi hona chahiye.

4. Corner $v\to c$: cross-speed $=\sqrt{c^2-c^2}=0$. Beam ko poori tarah wind mein aim karna padta hai sirf position hold karne ke liye aur kabhi cross nahi karta — Ex 7 ke imaginary blow-up se match karta hai. Yeh step kyun? Opposite extreme check karta hai, confirm karta hai ki formula physical rehta hai breaking point tak.

Verify: $v=3\times 10^4,c=3\times 10^8$ par: $\sqrt{c^2-v^2}=c\sqrt{1-{\beta }^2}=3\times 10^8\times \sqrt{1-10^{-8}}\approx 2.99999999985\times 10^8\text{ m/s}$, $c$ se thoda hi neeche. ✓

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Recall Quick self-test

Kaun sa arm hamesha slower hota hai, parallel ya perpendicular? ::: Parallel (wind ke saath), kyunki $1/(1-{\beta }^2)>1/\sqrt{1-{\beta }^2}$. $\Delta t\approx L{v}^2/c^3$ shortcut kab fail karta hai? ::: Jab $\beta$, $\ll 1$ na ho; $\beta =0.5$ par yeh already $\sim 30\%$ off hai. Earth speed aur $\lambda =500$ nm par $1.0$-fringe prediction ke liye kitna arm length chahiye? ::: Lagbhag $25$ m. Jab $v=0$ ho toh $\Delta t$ kya hai? ::: Exactly zero — dono arms tie karti hain; yeh "null-result" jaisa dikhta hai. $\Delta N$ rotation mein $2$ ka factor kyun hai? ::: $90^{\circ }$ rotate karne par dono arms swap ho jaate hain, isliye fringe pattern $\Delta t$ ek taraf aur $\Delta t$ doosri taraf move karta hai — total change $2\Delta t$ hota hai. Cross-arm effective speed aur reason? ::: $\sqrt{c^2-v^2}$; $c$ hypotenuse hai, $v$ ek leg hai, useful speed doosri leg hai (Pythagoras).

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Yeh bhi dekho: Special Relativity — Einstein's Postulates, Length Contraction, Time Dilation, Lorentz Transformation, Maxwell's Equations and the Speed of Light.