Exercises — Special relativity — Michelson-Morley experiment

(a) Luminiferous aether wo hypothetical invisible medium hai jo poori space mein bhari hogi, jisme light waves ripple karti hain — woh "nadi" jisme light tairti hai. (b) Experiment ne Earth ki speed $v$ aether mein measure karne ki koshish ki, do arms ke beech light-speed difference detect karke (jo fringe shift ke roop mein padha jaata). (c) Null result ka matlab hai koi fringe shift observe nahi hua — predicted effect simply wahan tha hi nahi.

Parallel arm slower hai. Uski round-trip time hai: $t_{\parallel }=\frac{2L}{c}\cdot \frac{1}{1-{\beta }^2},t_{\perp }=\frac{2L}{c}\cdot \frac{1}{\sqrt{1-{\beta }^2}}.$ Kyunki $0<{\beta }^2<1$ hai, humein milta hai $1-{\beta }^2<\sqrt{1-{\beta }^2}$ (1 se chhota number apne square root se bhi chhota hota hai). Chhota denominator $t_{\parallel }$ ko bada banata hai. Toh $t_{\parallel }>t_{\perp }$: parallel arm zyada time leti hai.

Hum kya karenge: seedha numbers daalo. Ye formula kyun: yeh parent note mein derive kiya gaya leading-order result hai — tiny ${\beta }^2$ term bachta hai, usse chhota sab drop kar diya jaata hai. $\Delta N\approx \frac{2(11)(3\times 10^4{)}^2}{(5\times 10^{-7})(3\times 10^8{)}^2}.$ Numerator: $2\cdot 11=22$; $(3\times 10^4{)}^2=9\times 10^8$; toh $22\times 9\times 10^8=1.98\times 10^{10}$. Denominator: $(5\times 10^{-7})(9\times 10^{16})=4.5\times 10^{10}$. $\Delta N\approx \frac{1.98\times 10^{10}}{4.5\times 10^{10}}\approx \mathrm{0.44.}$ Apparatus $0.01$ fringes resolve karta hai, toh $0.44$ obvious hota — phir bhi zero dekha gaya.

$\beta =\frac{v}{c}=\frac{3\times 10^4}{3\times 10^8}=10^{-4},{\beta }^2=10^{-8}.$ Expansion mein agla term order ${\beta }^4=10^{-16}$ ka hai — ${\beta }^2$ se ek-sau-million guna chhota. Yeh $\Delta N$ ko $0.01$-fringe resolution se kaafi neeche change karta hai, toh isko drop karne se kuch measurable nahi bigadta.

Hum kya karenge: $t_{\parallel }$ ko $t_0$ se exactly compare karo. Kyun: yeh prove karne ke liye ki "fast aur slow legs cancel ho jaate hain" wali intuition galat hai, bina approximation ke crutch ke. $t_{\parallel }=\frac{L}{c-v}+\frac{L}{c+v}=\frac{L(c+v)+L(c-v)}{(c-v)(c+v)}=\frac{2Lc}{c^2-v^2}.$ $t_0=2L/c$ se divide karo: $\frac{t_{\parallel }}{t_0}=\frac{2Lc}{c^2-v^2}\cdot \frac{c}{2L}=\frac{c^2}{c^2-v^2}=\frac{1}{1-{\beta }^2}.$ Kyunki $0<{\beta }^2<1$ hai, denominator $1-{\beta }^2$ zero aur 1 ke beech hai, toh ratio 1 se bada hai. Parallel round trip hamesha no-wind time se zyada hoti hai — legs kabhi cancel nahi hote.

Triangle dekho. Magenta arrow beam ki true speed aether ke relative hai, length $c$. Orange arrow wind hai jo use sideways push kar raha hai, length $v$. Beam ko actually seedha across travel karne ke liye (violet arrow), use upstream aim karna hoga taaki sideways component wind ko cancel kare. Us right triangle par Pythagoras se: $(\text{cross speed}{)}^2+v^2=c^2\Rightarrow \text{cross speed}=\sqrt{c^2-v^2}.$ Round trip $2L$ distance cover karta hai is speed par: $t_{\perp }=\frac{2L}{\sqrt{c^2-v^2}}=\frac{2L}{c}\cdot \frac{1}{\sqrt{1-{\beta }^2}}.$ $t_0=2L/c$ se divide karo: $\frac{t_{\perp }}{t_0}=\frac{1}{\sqrt{1-{\beta }^2}}.$ Yeh 1 se bada hai lekin $t_{\parallel }/t_0=1/(1-{\beta }^2)$ se chhota hai, jo confirm karta hai ki perpendicular arm bhi slow hoti hai — bas thodi kam.

Exact se shuru karo: $\Delta t=\frac{2L}{c}\left[\frac{1}{1-{\beta }^2}-\frac{1}{\sqrt{1-{\beta }^2}}\right].$ Expand kyun karein: ${\beta }^2\sim 10^{-8}$ tiny hai, toh hum sirf ${\beta }^2$ ka pehla surviving power rakhte hain. Standard small-$x$ approximations use karo (with $x={\beta }^2$): $\frac{1}{1-x}\approx 1+x,\frac{1}{\sqrt{1-x}}=(1-x{)}^{-1/2}\approx 1+\frac{1}{2}x.$ Substitute karo: $\Delta t\approx \frac{2L}{c}[(1+{\beta }^2)-(1+\frac{1}{2}{\beta }^2)]=\frac{2L}{c}\cdot \frac{{\beta }^2}{2}=\frac{L{\beta }^2}{c}.$ Kyunki ${\beta }^2=v^2/c^2$ hai: $\Delta t\approx \frac{L}{c}\cdot \frac{v^2}{c^2}=\frac{L{v}^2}{c^3}.$ Do "$+1$" terms cancel ho gaye; difference poori tarah mismatched ${\beta }^2$ coefficients ($1$ vs $\frac{1}{2}$) mein rehta hai.

Factor of 2 kyun: Rotate karne se pehle, arm A parallel hai aur arm B perpendicular hai, toh path-time difference $\Delta t$ hai. $90^{\circ }$ turn ke baad, roles swap ho jaate hain — ab B parallel hai aur A perpendicular — toh difference $-\Delta t$ ho jaata hai. Observer jo change actually dekhta hai woh hai $\Delta t-(-\Delta t)=2\Delta t$. $c$ se multiply karne par time extra path length ban jaata hai, aur $\lambda$ se divide karne par count hota hai ki kitne wavelengths hain: $\Delta N=\frac{c(2\Delta t)}{\lambda }=\frac{2c}{\lambda }\cdot \frac{L{v}^2}{c^3}=\frac{2L{v}^2}{\lambda c^2}.$ Values daalo ($L=11$, $v=3\times 10^4$, $c=3\times 10^8$, $\lambda =5\times 10^{-7}$): $\Delta t=\frac{11(3\times 10^4{)}^2}{(3\times 10^8{)}^3}=\frac{11\cdot 9\times 10^8}{2.7\times 10^{25}}\approx 3.67\times 10^{-16}\text{ s},$ $\Delta N=\frac{2(3\times 10^8)(3.67\times 10^{-16})}{5\times 10^{-7}}\approx \mathrm{0.44.}$ Same $0.44$ pehle jaisa — dono routes agree karte hain.

Hum kya karenge: $\Delta N=2L{v}^2/(\lambda c^2)$ ko $v$ ke liye invert karo, $\Delta N=0.004$ ko detection ceiling maanke. Kyun: null result $v=0$ prove nahi karta; yeh cap karta hai ki $v$ kitna bada ho sakta hai abhi bhi chhupa hua. $v=c\sqrt{\frac{\lambda c\Delta N}{2L}}\text{— cleanly rederive karte hain.}$ $\Delta N=\frac{2L{v}^2}{\lambda c^2}$ se: $v^2=\frac{\Delta N\lambda c^2}{2L}=\frac{(0.004)(5\times 10^{-7})(3\times 10^8{)}^2}{2(22)}.$ Numerator: $(0.004)(5\times 10^{-7})=2\times 10^{-9}$; times $(9\times 10^{16})=1.8\times 10^8$. Denominator: $44$. $v^2=\frac{1.8\times 10^8}{44}\approx 4.09\times 10^6{\text{m}}^2/{\text{s}}^2,$ $v\approx 2.02\times 10^3\text{m/s}\approx 2.0\text{km/s},\beta =\frac{v}{c}\approx 6.7\times 10^{-6}.$ Earth $\sim 30$ km/s par orbit karti hai, toh yeh $\sim 2$ km/s ki ceiling full orbital aether wind ko ~15 ke factor se rule out karti hai — aether, agar real hota, to almost perfectly "dragged along" hona padta, jo doosre experiments forbid karte hain.

(a) $\beta \to 0$ (koi wind nahi). Dono brackets $1/1-1/1=0$ par jaate hain, toh $\Delta t\to 0$. Physical meaning: koi aether wind nahi toh dono arms identically behave karti hain — exactly wahi jo observe hua. Yeh sanity check hai: classical formula correctly predict karta hai "no wind ⇒ no shift." (b) $\beta \to 1^{-}$ (apparatus almost light speed par aether mein). Factor $1/(1-{\beta }^2)\to +\infty$ faster jaata hai compared to $1/\sqrt{1-{\beta }^2}\to +\infty$ (ek number apni square root se faster blow up karta hai). Toh bracket $\to +\infty$ aur $\Delta t\to +\infty$. Physical meaning: parallel arm ke "against the wind" leg ki speed $c-v\to 0$ hai — light barely upstream fight kar sakti hai, toh uska round trip diverge ho jaata hai. Classical model ek enormous, unmissable shift predict karta hai. Ki real experiment ne kuch bhi nahi dekha Earth ki modest speed par bhi — exactly wohi hai jisne aether picture ko demolish kiya.

Null result poori theory ka launchpad hai: koi preferred frame nahi (Galilean Relativity & Velocity Addition light ke liye fail karta hai), ek constant $c$ (Maxwell's Equations and the Speed of Light), aur woh machinery jo aether patch ki jagah aayi — Lorentz Transformation, Length Contraction, aur Time Dilation. Fringe reading khud Interference of Light & Fringes par rely karti thi.