Worked examples — Special relativity — Michelson-Morley experiment

This page drills the Michelson–Morley formulas until no case can surprise you. We start by listing every kind of situation the topic can throw at you, then hit each one with a fully worked example.

Before we begin, one reminder of the symbols we already earned in the parent note (nothing new is used before it is stated). The picture below shows what $L$, $c$, $v$ and the two arms actually are:

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The scenario matrix

Every problem in this topic is one of these cells. The worked examples below are tagged with the cell they cover.

Cell	What varies	Degenerate / limit to watch	Example
A. Baseline arm times	ordinary $v\ll c$	—	Ex 1
B. Zero wind	$v=0$	both arms give the same time	Ex 2
C. Extreme wind	$v\to c$ ($\beta \to 1$)	parallel time $\to \infty$	Ex 3
D. Fringe-shift prediction	plug numbers into $\Delta N$	—	Ex 4
E. Exact vs. approximate	does ${\beta }^2$ shortcut hold?	small vs. large $\beta$	Ex 5
F. Inverse / design problem	solve for $L$ given target $\Delta N$	—	Ex 6
G. Real-world word problem	swimmer-in-river numbers	current $>$ swimmer?	Ex 7
H. Exam twist	which arm is faster, sign of $\Delta t$	$t_{\parallel }>t_{\perp }$ always	Ex 8
I. Geometry of the cross-arm	why $\sqrt{c^2-v^2}$	$v=0$ and $v\to c$ corners	Ex 9

We will use these master formulas throughout (all derived in the parent note). The exact forms hold for any $\beta$; the approximate forms ($\approx$) only when $\beta \ll 1$:

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Example 1 — Cell A: the baseline round-trip times

Forecast: Both times are near $2L/c$. Guess: are they equal, or is one bigger? By how many parts in a million?

1. Compute $\beta$ and ${\beta }^2$. $\beta =\frac{3\times 10^4}{3\times 10^8}=10^{-4}$, so ${\beta }^2=10^{-8}$. Why this step? Every formula is built from ${\beta }^2$; get it first so we never re-derive it.

2. The no-wind baseline time $\frac{2L}{c}=\frac{2\times 11}{3\times 10^8}=7.333\times 10^{-8}\text{ s}$. Why this step? Both arms are just this baseline multiplied by a factor barely above 1, so it's the anchor.

3. Parallel: $t_{\parallel }=7.333\times 10^{-8}\times \frac{1}{1-10^{-8}}\approx 7.333\times 10^{-8}(1+10^{-8})\text{ s}$. Why this step? $\frac{1}{1-x}\approx 1+x$ because $x=10^{-8}$ is minuscule.

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

5. Difference: $t_{\parallel }-t_{\perp }\approx 7.333\times 10^{-8}\times \frac{1}{2}\times 10^{-8}\approx 3.67\times 10^{-16}\text{ s}$. Why this step? Subtracting the two $(1+\dots )$ factors leaves only the difference of the tiny corrections, $10^{-8}-\frac{1}{2}\times 10^{-8}=\frac{1}{2}\times 10^{-8}$.

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

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Example 2 — Cell B: the zero-wind degenerate case

Forecast: With no current, the "river" is a still pond. Guess: do the two arms tie exactly?

1. Set $\beta =0$. Then $1-{\beta }^2=1$ and $\sqrt{1-{\beta }^2}=1$. Why this step? $v=0\Rightarrow \beta =0$; both correction factors collapse to $1$.

2. Both times equal the baseline: $t_{\parallel }=t_{\perp }=\frac{2L}{c}$. Why this step? Multiplying the baseline by $1$ leaves it unchanged — with no wind, "along" and "across" are physically identical.

3. Difference: $\Delta t=t_{\parallel }-t_{\perp }=\frac{2L}{c}-\frac{2L}{c}=0$. Why this step? Two equal times subtract to zero; there is literally nothing for the experiment to detect.

Verify: $\Delta t\approx L{v}^2/c^3=L\cdot 0/c^3=0$. ✓ Physical sense: no wind means no preferred direction, so no arm can be favoured — exactly the situation a null result looks like. This is why Michelson–Morley's zero shift "felt like" $v=0$ everywhere.

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

Forecast: The shortcut $L{v}^2/c^3$ dies here — $\beta$ is not small. Guess: which time blows up faster?

1. Baseline $\frac{2L}{c}=7.333\times 10^{-8}\text{ s}$ (same as Ex 1). Why this step? Both exact times are still this baseline times a wind-factor, so we reuse it rather than recomputing $2L/c$.

2. Factors: $1-{\beta }^2=1-0.81=0.19$; $\sqrt{1-{\beta }^2}=\sqrt{0.19}=0.43589$. Why this step? At large $\beta$ we must use the exact denominators — the Taylor shortcut is invalid.

3. Parallel: $t_{\parallel }=7.333\times 10^{-8}/0.19=3.860\times 10^{-7}\text{ s}$. Why this step? Dividing the baseline by the small factor $0.19$ blows the time up fivefold — the along-wind trip is badly slowed.

4. Perpendicular: $t_{\perp }=7.333\times 10^{-8}/0.43589=1.682\times 10^{-7}\text{ s}$. Why this step? The cross factor $0.436$ is larger than $0.19$, so the cross-arm is slowed less than the parallel arm.

5. Limit $\beta \to 1$: $1-{\beta }^2\to 0$ makes $t_{\parallel }\to \infty$ (light going straight against a near-$c$ wind barely advances), while $t_{\perp }\to \infty$ too but more slowly. $t_{\parallel }$ diverges faster. Why this step? $1/(1-{\beta }^2)$ has a stronger singularity than $1/\sqrt{1-{\beta }^2}$.

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: the predicted fringe shift

Forecast: The instrument can see $0.01$ fringes. Guess: will the prediction be far above that threshold?

1. Use $\Delta N\approx \frac{2L{v}^2}{\lambda c^2}$. Why this step? This is the rotation formula (the $2$ is the "there-and-back" $2\Delta t$ swap explained in the master-formula box); dividing by $\lambda$ converts a path-length into a countable number of fringes.

2. Numerator: $2\times 11\times (3\times 10^4{)}^2=22\times 9\times 10^8=1.98\times 10^{10}$. Why this step? Group constants first — squaring $3\times 10^4$ gives $9\times 10^8$ — to dodge exponent slips.

3. Denominator: $\lambda c^2=(5\times 10^{-7})(9\times 10^{16})=4.5\times 10^{10}$. Why this step? Same tactic: square $c$ once ($9\times 10^{16}$) then multiply by $\lambda$, so the two big powers combine cleanly.

4. Divide: $\Delta N\approx \frac{1.98\times 10^{10}}{4.5\times 10^{10}}=0.44$ fringes. Why this step? The matching $10^{10}$ powers cancel, leaving a plain ratio $1.98/4.5$ — the final pure fringe count.

Verify: $0.44\gg 0.01$, so a real shift would be unmissable. The observed shift was $\approx 0$. Units: $\frac{\text{m}\cdot (\text{m/s}{)}^2}{\text{m}\cdot (\text{m/s}{)}^2}$ = dimensionless. ✓ Matches the parent note's $0.44$. ✓

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Example 5 — Cell E: exact vs. approximate, does the shortcut hold?

Forecast: The shortcut is a Taylor approximation. Guess: it's near-perfect for $10^{-4}$, but how badly does it fail at $0.5$?

1. Rescale to $2L/c=1$. Then the exact difference is $\Delta t=\frac{1}{1-{\beta }^2}-\frac{1}{\sqrt{1-{\beta }^2}}$ and the shortcut is $\frac{{\beta }^2}{2}$. Why this step? Dividing both formulas by the common factor $2L/c$ leaves pure numbers, so we compare the shapes of the two expressions without any metres or seconds getting in the way.

2. At $\beta =10^{-4}$ (${\beta }^2=10^{-8}$): exact bracket $=\frac{1}{1-10^{-8}}-\frac{1}{\sqrt{1-10^{-8}}}\approx 5.0\times 10^{-9}$; shortcut $=\frac{1}{2}\times 10^{-8}=5.0\times 10^{-9}$. Why this step? Confirms the shortcut is essentially exact when $\beta$ is tiny.

3. At $\beta =0.5$ (${\beta }^2=0.25$): exact bracket $=\frac{1}{0.75}-\frac{1}{\sqrt{0.75}}=1.3333-1.1547=0.1786$; shortcut $=0.125$. Why this step? Shows the shortcut under-predicts by $\sim 30\%$ once $\beta$ is large — the higher Taylor terms we threw away are no longer negligible.

Verify: Relative error at $\beta =0.5$: $(0.1786-0.125)/0.1786=0.30$, i.e. $30\%$ off. At $\beta =10^{-4}$: error $<10^{-8}$. ✓ Lesson: the $L{v}^2/c^3$ shortcut is only for $\beta \ll 1$.

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Example 6 — Cell F: inverse design problem

Forecast: From Ex 4, $L=11\text{ m}$ gave $0.44$. Guess: you'll need a bit more than double that.

1. Rearrange $\Delta N=\frac{2L{v}^2}{\lambda c^2}$ for $L$: $L=\frac{\Delta N\lambda c^2}{2v^2}$. Why this step? We know everything except $L$, so isolate it algebraically by multiplying both sides by $\lambda c^2$ and dividing by $2v^2$.

2. Numerator: $\Delta N\lambda c^2=1.0\times (5\times 10^{-7})\times (9\times 10^{16})=4.5\times 10^{10}$. Why this step? Build the top of the fraction first; squaring $c$ once ($9\times 10^{16}$) keeps the exponents tidy.

3. Denominator: $2v^2=2\times (3\times 10^4{)}^2=1.8\times 10^9$. Why this step? Square $v$ ($9\times 10^8$) then double it — this is the bottom of the fraction, kept separate to avoid mixing powers.

4. Divide: $L=\frac{4.5\times 10^{10}}{1.8\times 10^9}=25\text{ m}$. Why this step? Dividing the two powers ($10^{10}/10^9=10$) and the mantissas ($4.5/1.8=2.5$) gives $25$ — the required arm length.

Verify: Plug $L=25$ back into Ex 4's ratio: $\Delta N=0.44\times (25/11)=1.0$. ✓ Units: $\frac{1\cdot \text{m}\cdot (\text{m/s}{)}^2}{(\text{m/s}{)}^2}=\text{m}$. ✓ (This is why Michelson folded the beam with many mirrors to reach effective $L\sim 11\text{ m}$.)

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

Forecast: Guess: does the across route beat the downstream-and-back route? And can the swimmer even go straight across if the current outpaces them?

1. Define the ratio for THIS problem. Because the "swimmer speed" here is $s$, not light-speed $c$, the wind fraction is $\beta \equiv \frac{v}{s}=\frac{3}{5}=0.6$ (careful: in every optics example $\beta =v/c$; only in this swimmer analogy does $c$ get replaced by $s$). Not small, so the exact forms are mandatory. Why this step? $\beta$ controls whether the Taylor shortcut is allowed; $0.6$ is far from $0$, so we must use raw speeds.

2. Parallel (exact, two legs): $t_{\parallel }=\frac{L}{s-v}+\frac{L}{s+v}=\frac{100}{2}+\frac{100}{8}=50+12.5=62.5\text{ s}$. Why this step? Downstream the ground speed is $s+v$, upstream it is $s-v$; each leg is distance $÷$ its own speed.

3. Perpendicular: cross-speed $=\sqrt{s^2-v^2}=\sqrt{25-9}=4\text{ m/s}$, so $t_{\perp }=\frac{2L}{\sqrt{s^2-v^2}}=\frac{200}{4}=50\text{ s}$. Why this step? The swimmer aims upstream so drift cancels; Pythagoras gives the useful cross-speed.

4. Difference: $\Delta t=t_{\parallel }-t_{\perp }=62.5-50=12.5\text{ s}$. The downstream route is slower — matching Cell H's rule $t_{\parallel }>t_{\perp }$. Why this step? The whole point of the analogy is that unequal-speed legs cost more time than the straight-across route.

5. Degenerate case $v=6>s=5$: now $s-v=-1$, so $t_{\parallel }$ is nonsensical (the swimmer is swept backwards — can never return upstream), and $\sqrt{s^2-v^2}=\sqrt{25-36}=\sqrt{-11}$ is imaginary — the swimmer cannot go straight across at all. Why this step? It shows the formulas honestly break when the current exceeds the swimmer, the analogue of $v>c$ which relativity forbids for light.

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: the exam twist (sign of $\Delta t$)

Forecast: Guess: the parallel (along-wind) arm is always the slow one.

1. Form the ratio $R=\frac{t_{\parallel }}{t_{\perp }}=\frac{1/(1-{\beta }^2)}{1/\sqrt{1-{\beta }^2}}=\frac{\sqrt{1-{\beta }^2}}{1-{\beta }^2}=\frac{1}{\sqrt{1-{\beta }^2}}$. Why this step? A ratio removes the shared $2L/c$ so only the wind-factors remain.

2. Bound it: for $0<\beta <1$, $\sqrt{1-{\beta }^2}<1$, hence $R=1/\sqrt{1-{\beta }^2}>1$. Why this step? Anything less than $1$ has a reciprocal greater than $1$.

3. Conclusion: $t_{\parallel }>t_{\perp }$ always. So the correct signed prediction (matching our toolkit definition $\Delta t\equiv t_{\parallel }-t_{\perp }$) is positive. The student's negative sign just means they subtracted in the wrong order. Why this step? It pins down the sign convention so the student never mislabels which arm leads.

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

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Example 9 — Cell I: the geometry of the cross-arm

Forecast: Guess: the light's own speed $c$ is the hypotenuse, not a leg.

1. Read the triangle. The light always moves at speed $c$ through the aether — that is the hypotenuse. The wind sweeps it sideways at speed $v$ — the horizontal leg. To land straight across, the useful (vertical) component is the remaining leg. Why this step? Velocities add as vectors (see Galilean Relativity & Velocity Addition); the beam must aim upstream so the sideways wind exactly cancels.

2. Apply Pythagoras on the right triangle: $c^2=v^2+(\text{cross-speed}{)}^2$, so cross-speed $=\sqrt{c^2-v^2}$. Why this step? The hypotenuse squared equals the sum of the leg squares — the only relation that ties $c$, $v$, and the useful speed together.

3. Corner $v=0$: cross-speed $=\sqrt{c^2}=c$. Sensible — no wind, full speed across. Why this step? Checks the formula against the still-water limit where nothing should be lost.

4. Corner $v\to c$: cross-speed $=\sqrt{c^2-c^2}=0$. The beam must aim entirely into the wind just to hold position and never crosses — matching Ex 7's imaginary blow-up. Why this step? Checks the opposite extreme, confirming the formula stays physical right up to the breaking point.

Verify: At $v=3\times 10^4,c=3\times 10^8$: $\sqrt{c^2-v^2}=c\sqrt{1-{\beta }^2}=3\times 10^8\times \sqrt{1-10^{-8}}\approx 2.99999999985\times 10^8\text{ m/s}$, barely below $c$. ✓

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Recall Quick self-test

Which arm is always slower, parallel or perpendicular? ::: Parallel (along the wind), because $1/(1-{\beta }^2)>1/\sqrt{1-{\beta }^2}$. When does the $\Delta t\approx L{v}^2/c^3$ shortcut fail? ::: When $\beta$ is not $\ll 1$; at $\beta =0.5$ it is already $\sim 30\%$ off. What arm length gives a $1.0$-fringe prediction at Earth speed and $\lambda =500$ nm? ::: About $25$ m. What is $\Delta t$ when $v=0$? ::: Exactly zero — both arms tie; this is the "null-result" look. Why is there a factor of $2$ in $\Delta N$ on rotation? ::: Rotating $90^{\circ }$ swaps the two arms, so the fringe pattern moves by $\Delta t$ each way — a total change of $2\Delta t$. Cross-arm effective speed and the reason? ::: $\sqrt{c^2-v^2}$; $c$ is the hypotenuse, $v$ a leg, useful speed the other leg (Pythagoras).

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