Exercises — Special relativity — Michelson-Morley experiment

(a) The luminiferous aether is the hypothetical invisible medium filling all space, through which light waves were thought to ripple — the "river" light swims in. (b) The experiment tried to measure the Earth's speed $v$ through the aether, by detecting a light-speed difference between the two arms (read off as a fringe shift). (c) A null result means no fringe shift was observed — the predicted effect simply was not there.

The parallel arm is slower. Its round-trip time is $t_{\parallel }=\frac{2L}{c}\cdot \frac{1}{1-{\beta }^2},t_{\perp }=\frac{2L}{c}\cdot \frac{1}{\sqrt{1-{\beta }^2}}.$ Since $0<{\beta }^2<1$, we have $1-{\beta }^2<\sqrt{1-{\beta }^2}$ (a number below 1 is smaller than its own square root). A smaller denominator makes $t_{\parallel }$ larger. So $t_{\parallel }>t_{\perp }$: the parallel arm takes longer.

What we do: plug numbers straight in. Why this formula: it is the leading-order result derived in the parent note — the tiny ${\beta }^2$ term survives, everything smaller is dropped. $\Delta N\approx \frac{2(11)(3\times 10^4{)}^2}{(5\times 10^{-7})(3\times 10^8{)}^2}.$ Numerator: $2\cdot 11=22$; $(3\times 10^4{)}^2=9\times 10^8$; so $22\times 9\times 10^8=1.98\times 10^{10}$. Denominator: $(5\times 10^{-7})(9\times 10^{16})=4.5\times 10^{10}$. $\Delta N\approx \frac{1.98\times 10^{10}}{4.5\times 10^{10}}\approx \mathrm{0.44.}$ The apparatus resolves $0.01$ fringes, so $0.44$ would be obvious — yet zero was seen.

$\beta =\frac{v}{c}=\frac{3\times 10^4}{3\times 10^8}=10^{-4},{\beta }^2=10^{-8}.$ The next term in the expansion is of order ${\beta }^4=10^{-16}$ — a hundred-million times smaller than ${\beta }^2$. It changes $\Delta N$ far below the $0.01$-fringe resolution, so dropping it costs nothing measurable.

What we do: compare $t_{\parallel }$ to $t_0$ exactly. Why: to prove the "the fast and slow legs cancel" intuition is wrong with no approximation crutch. $t_{\parallel }=\frac{L}{c-v}+\frac{L}{c+v}=\frac{L(c+v)+L(c-v)}{(c-v)(c+v)}=\frac{2Lc}{c^2-v^2}.$ Divide by $t_0=2L/c$: $\frac{t_{\parallel }}{t_0}=\frac{2Lc}{c^2-v^2}\cdot \frac{c}{2L}=\frac{c^2}{c^2-v^2}=\frac{1}{1-{\beta }^2}.$ Since $0<{\beta }^2<1$, the denominator $1-{\beta }^2$ is between 0 and 1, so the ratio is greater than 1. The parallel round trip always exceeds the no-wind time — the legs never cancel.

Look at the triangle. The magenta arrow is the beam's true speed relative to the aether, length $c$. The orange arrow is the wind pushing it sideways, length $v$. For the beam to actually travel straight across (violet arrow), it must aim upstream so the sideways component cancels the wind. By Pythagoras on that right triangle: $(\text{cross speed}{)}^2+v^2=c^2\Rightarrow \text{cross speed}=\sqrt{c^2-v^2}.$ The round trip covers distance $2L$ at this speed: $t_{\perp }=\frac{2L}{\sqrt{c^2-v^2}}=\frac{2L}{c}\cdot \frac{1}{\sqrt{1-{\beta }^2}}.$ Dividing by $t_0=2L/c$: $\frac{t_{\perp }}{t_0}=\frac{1}{\sqrt{1-{\beta }^2}}.$ This is bigger than 1 but smaller than $t_{\parallel }/t_0=1/(1-{\beta }^2)$, confirming the perpendicular arm is slowed too — just less.

Start exact: $\Delta t=\frac{2L}{c}\left[\frac{1}{1-{\beta }^2}-\frac{1}{\sqrt{1-{\beta }^2}}\right].$ Why expand: ${\beta }^2\sim 10^{-8}$ is tiny, so we keep only the first surviving power of ${\beta }^2$. Use the standard small-$x$ approximations (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: $\Delta t\approx \frac{2L}{c}[(1+{\beta }^2)-(1+\frac{1}{2}{\beta }^2)]=\frac{2L}{c}\cdot \frac{{\beta }^2}{2}=\frac{L{\beta }^2}{c}.$ Since ${\beta }^2=v^2/c^2$: $\Delta t\approx \frac{L}{c}\cdot \frac{v^2}{c^2}=\frac{L{v}^2}{c^3}.$ The two "$+1$" terms cancelled; the difference lives entirely in the mismatched ${\beta }^2$ coefficients ($1$ vs $\frac{1}{2}$).

Why the factor of 2: Before rotating, arm A is parallel and arm B is perpendicular, so the path-time difference is $\Delta t$. After a $90^{\circ }$ turn, the roles swap — now B is parallel and A is perpendicular — so the difference becomes $-\Delta t$. The change the observer actually sees is $\Delta t-(-\Delta t)=2\Delta t$. Multiplying by $c$ turns time into extra path length, and dividing by $\lambda$ counts how many wavelengths that is: $\Delta N=\frac{c(2\Delta t)}{\lambda }=\frac{2c}{\lambda }\cdot \frac{L{v}^2}{c^3}=\frac{2L{v}^2}{\lambda c^2}.$ Plugging in ($L=11$, $v=3\times 10^4$, $c=3\times 10^8$, $\lambda =5\times 10^{-7}$): $\Delta t=\frac{11(3\times 10^4{)}^2}{(3\times 10^8{)}^3}=\frac{11\cdot 9\times 10^8}{2.7\times 10^{25}}\approx 3.67\times 10^{-16}\text{ s},$ $\Delta N=\frac{2(3\times 10^8)(3.67\times 10^{-16})}{5\times 10^{-7}}\approx \mathrm{0.44.}$ Same $0.44$ as before — the two routes agree.

What we do: invert $\Delta N=2L{v}^2/(\lambda c^2)$ for $v$, treating $\Delta N=0.004$ as the detection ceiling. Why: a null result doesn't prove $v=0$; it caps how large $v$ could be while still hiding. $v=c\sqrt{\frac{\lambda c\Delta N}{2L}}\text{— let’s rederive cleanly.}$ From $\Delta N=\frac{2L{v}^2}{\lambda c^2}$: $v^2=\frac{\Delta N\lambda c^2}{2L}=\frac{(0.004)(5\times 10^{-7})(3\times 10^8{)}^2}{2(22)}.$ Numerator: $(0.004)(5\times 10^{-7})=2\times 10^{-9}$; times $(9\times 10^{16})=1.8\times 10^8$. Denominator: $44$. $v^2=\frac{1.8\times 10^8}{44}\approx 4.09\times 10^6{\text{m}}^2/{\text{s}}^2,$ $v\approx 2.02\times 10^3\text{m/s}\approx 2.0\text{km/s},\beta =\frac{v}{c}\approx 6.7\times 10^{-6}.$ The Earth orbits at $\sim 30$ km/s, so this ceiling of $\sim 2$ km/s rules out the full orbital aether wind by a factor of ~15 — the aether, if real, would have to be almost perfectly "dragged along," which other experiments forbid.

(a) $\beta \to 0$ (no wind). Both brackets go to $1/1-1/1=0$, so $\Delta t\to 0$. Physical meaning: with no aether wind the two arms behave identically — exactly the observed situation. This is the sanity check: the classical formula correctly predicts "no wind ⇒ no shift." (b) $\beta \to 1^{-}$ (apparatus nearly at light speed through the aether). The factor $1/(1-{\beta }^2)\to +\infty$ faster than $1/\sqrt{1-{\beta }^2}\to +\infty$ (a number blows up faster than its square root). So the bracket $\to +\infty$ and $\Delta t\to +\infty$. Physical meaning: the parallel arm's "against the wind" leg has speed $c-v\to 0$ — light can barely fight upstream, so its round trip diverges. The classical model predicts an enormous, unmissable shift. That the real experiment saw nothing even at Earth's modest speed is exactly what demolished the aether picture.

The null result is the launchpad for the whole theory: no preferred frame (Galilean Relativity & Velocity Addition fails for light), a constant $c$ (Maxwell's Equations and the Speed of Light), and the machinery that replaced the aether patch — the Lorentz Transformation, Length Contraction, and Time Dilation. The fringe reading itself relied on Interference of Light & Fringes.