Visual walkthrough — Special relativity — Michelson-Morley experiment

Step 1 — What is a "wind" made of nothing?

WHAT. Before any light, picture a wide river. The water flows left-to-right at speed $v$. A boat can push through still water at speed $c$. That is all the machinery we need.

WHY start here. Every hard idea in this experiment is already visible in a river. The 19th-century belief (see Special Relativity — Einstein's Postulates) was that light rides an invisible medium — the aether — the way sound rides air. If Earth ploughs through that aether at speed $v$, we feel an aether wind. So "boat in a river" is "light beam through aether." We solve the easy version and then just rename the words.

PICTURE. Below, the blue arrows are the current (speed $v$). The boat's own engine always gives speed $c$ relative to the water. The two questions we will race: down-and-back along the flow vs. straight across and back.

Step 2 — The along-the-wind trip: why fast and slow do NOT cancel

WHAT. Send the boat a distance $L$ downstream, then bring it $L$ back upstream. We want the total time.

Downstream the current helps: ground speed $=c+v$. Upstream the current fights: ground speed $=c-v$.

Each piece is the same shape: $\text{time}=\frac{\text{distance }L}{\text{effective speed}}$. The left term is bigger because its denominator $c-v$ is smaller. (Note this only makes sense while $v<c$, so that $c-v>0$ — see the domain remark above.)

WHY it does not cancel. The tempting mistake: "half the trip is $+v$, half is $-v$, they average out." No. You are slow for longer than you are fast, so the slow leg hogs the clock. Look at the figure: the two coloured bars are the durations, and the slow (pink) bar is visibly longer than the fast (yellow) one. Their sum beats the no-wind time $\frac{2L}{c}$.

PICTURE. The boat drawn twice — one leg with the flow, one against — and the two time-bars underneath so the imbalance is literally longer on the page.

Now clean it up with one common denominator:

• $2Lc$ — the two legs' distances collapse to a single numerator.
• $c^2-v^2$ — comes from $(c-v)(c+v)$; the wind's fingerprint (positive since $v<c$).
• The bracketed factor is bigger than 1 for every $0<v<c$, so wind always costs time. Set $v=0$ and it becomes $1$: no wind, time is exactly $\frac{2L}{c}$. Good — our formula agrees with common sense at the degenerate case.

Step 3 — The across-the-wind trip: why the boat must aim upstream

WHAT. Now the boat must reach a mark directly across the river and come straight back. If it points straight across, the current sweeps it downstream and it lands off-target. To go straight, it must aim its nose upstream so the sideways drift is exactly cancelled.

WHY Pythagoras enters. The boat's speed relative to the water is fixed at $c$ — that is the boat's total velocity, drawn as the hypotenuse. Part of it ($v$) is spent fighting the current sideways; what is left over points straight across. Three arrows, one right triangle:

We use the right triangle because it is the only tool that splits one speed into two perpendicular pieces with the total held fixed — exactly our situation. Note $\sqrt{c^2-v^2}$ is only a real speed while $v<c$; if the current were faster than the boat there would be no straight-across path at all — another reason the whole analysis lives in $\beta <1$.

PICTURE. The velocity triangle: hypotenuse $c$ (the aim), horizontal leg $v$ (eaten by the current), vertical leg $\sqrt{c^2-v^2}$ (the useful cross speed).

The round trip is distance $2L$ at this reduced cross speed:

• $\sqrt{c^2-v^2}$ — the vertical leg from the triangle, the only speed that carries the boat across.
• The bracket is bigger than 1 too — so the cross arm is also slowed. But note the square root: this penalty is gentler than the parallel arm's. That gap is the whole experiment.

Step 4 — Stacking the two penalties side by side

WHAT. Put the two round-trip times together and subtract:

WHY define $\beta$. Speeds like $30000$ vs $300000000$ are awkward. The single ratio $\beta =v/c$ measures "how fast compared to light," and everything depends on it. For Earth's orbit $\beta \approx 10^{-4}$, so ${\beta }^2\approx 10^{-8}$ — absurdly small. That smallness is the key to the next step.

PICTURE. Two penalty curves plotted against $\beta$. Both start at 1 when $\beta =0$ (no wind, equal times — the degenerate case again), and the parallel curve climbs faster. The vertical gap between them is $\Delta t\cdot \frac{c}{2L}$.

• At $\beta =0$: both penalties $=1$, gap $=0$. No wind ⇒ no time difference. Sanity confirmed.
• As $\beta \to 1$ the two denominators $1-{\beta }^2$ and $\sqrt{1-{\beta }^2}$ shrink toward $0$, so both penalty factors blow up (the parallel one far faster). This edge is only mathematical — for our experiment $\beta \approx 10^{-4}\ll 1$, so we sit at the far-left toe of the curve. What the full plot does tell us is the shape near the origin: the gap grows like ${\beta }^2$, which we extract next.

Step 5 — Zooming in: the tiny gap is $\propto {\beta }^2$

WHAT. Because ${\beta }^2$ is minuscule, we approximate each penalty near zero. Two standard expansions (each just "how the curve leaves the value 1"):

WHY expand at all. We do not need the exact monstrous fraction — we need the first non-zero difference. Subtracting two "$1+\text{something}$" numbers, the boring $1$'s cancel and only the "something" survives:

• The parallel arm's penalty grows like ${\beta }^2$; the cross arm's like $\frac{1}{2}{\beta }^2$.
• Their difference is $\frac{1}{2}{\beta }^2$ — the surviving term.
• Reinstating $\beta =v/c$ turns $\frac{2L}{c}\cdot \frac{{\beta }^2}{2}$ into $\frac{L{v}^2}{c^3}$.

PICTURE. A zoom box on the previous plot near $\beta =0$: the two curves look like nearly-parallel straight-ish lines, and the shaded sliver between them is labelled "$\propto {\beta }^2$" — the entire signal the experiment hunts.

Step 6 — Turning time into a picture you can read: fringes

WHAT. A stopwatch cannot resolve $10^{-16}$ s. But light's own waves can. When the two beams recombine on a screen (the recombination is interference), a time difference $\Delta t$ shows up as bright/dark stripes — fringes. Rotating the whole table by $90^{\circ }$ swaps which arm is parallel to the aether wind. The predicted number of fringes that slide past:

• $c(2\Delta t)$ — the extra path length light covers because of the doubled time gap.
• $÷\lambda$ — one full fringe per wavelength $\lambda$ of extra path; this is what makes it countable.

Why the factor of 2 (algebraic footnote). In the starting orientation, the parallel arm leads the cross arm by $\Delta t=t_{\parallel }-t_{\perp }=+\frac{L{v}^2}{c^3}$. After rotating $90^{\circ }$, the former cross arm is now parallel and vice versa, so the labels swap and the difference becomes

The measurable thing is how the fringe pattern changes between the two orientations:

The sign flips on rotation, so the two measurements differ by twice $\Delta t$ — that is the $2$ in $c(2\Delta t)$. (You never rely on knowing the fringe count at a single orientation; you watch the shift as you turn the table.)

WHY this is genius. It converts an unmeasurable time into counting stripes — the wavelength of light ($\lambda \sim 500\text{ nm}$) is the world's finest ruler.

PICTURE. The interferometer schematic: source → beam splitter → two arms of length $L$ → mirrors → recombine → fringe pattern, with the $90^{\circ }$ rotation arrow that swaps the arms (and flips the sign of $\Delta t$).

Step 7 — The degenerate case that actually happened: null

WHAT. Predicted signal: $0.44$ fringes. Observed: essentially $0$, at every orientation, every season.

WHY this is the punchline. Feed the observation back into our formula. $\Delta t=\frac{L{v}^2}{c^3}=0$ with $L\ne 0$ forces $v=0$ — but Earth is not at rest in space. The only escape is that the whole premise "light moves at $c$ relative to some aether" is wrong. Instead, light moves at $c$ for everyone, in every direction — the second postulate of special relativity. The rival patch, FitzGerald–Lorentz contraction, shrank the parallel arm just enough to kill $\Delta t$ while keeping the aether; Einstein instead threw the aether out entirely (and the same maths reappears as the Lorentz Transformation, Time Dilation, and the failure of Galilean velocity addition).

PICTURE. Two fringe screens side by side: predicted (stripes visibly shifted after rotation) vs. observed (stripes don't budge). The gap between them is the birth of relativity.

The one-picture summary

Everything above, compressed: two beams leave the splitter, one races the wind (penalty $\frac{1}{1-{\beta }^2}$), one crosses it (penalty $\frac{1}{\sqrt{1-{\beta }^2}}$), the gap is $\frac{L{v}^2}{c^3}$, read as $\approx 0.44$ fringes — and nature answered 0.