Special relativity — Michelson-Morley experiment: Question bank

Prerequisite ideas worth having open: Galilean Relativity & Velocity Addition (the "wrong" velocity-adding that this experiment tests), Maxwell's Equations and the Speed of Light (where $c$ comes from), Interference of Light & Fringes (how the answer is read), and where it all leads: Special Relativity — Einstein's Postulates, Length Contraction, Time Dilation, Lorentz Transformation.

Definition The notation you'll see on this page

So you never have to hunt in the parent note, here is every symbol used below, in plain words:

$c$ — the speed of light relative to the aether (the assumed invisible medium). About $3 \times 1 0^{8}$ m/s.
$v$ — the speed of the apparatus (Earth) drifting through that aether: the "aether wind". About $3 \times 1 0^{4}$ m/s.
$L$ — the length of one arm of the interferometer: the distance from the beam splitter out to each mirror. The beam covers $L$ out and $L$ back, so $2 L$ per round trip.
$t_{∥}$ — round-trip time for the beam in the arm lying along the wind (down-wind then up-wind).
$t_{⊥}$ — round-trip time for the beam in the arm lying across the wind.
$Δ t = t_{∥} - t_{⊥}$ — the time difference between the two beams; this is the quantity the aether predicts should be non-zero.
$β = v / c$ — the wind speed as a fraction of light speed. For Earth $β \sim 1 0^{- 4}$ , so $β^{2} \sim 1 0^{- 8}$ — tiny.
$λ$ — the wavelength of the light (about $500$ nm), the ruler that converts a path difference into fringes.
$Δ N$ — the predicted fringe shift: how many bright-to-dark cycles the interference pattern should slide by when the apparatus is rotated.

True or false — justify

Every "true/false" answer must state why — a bare verdict scores zero.

The Michelson–Morley experiment measured the speed of light directly.

False. It never timed a beam. It compared the arrival of two beams via interference fringes, detecting a tiny time difference

Δ t

between arms — a difference so small no clock could catch it, but fringes can.

A null result means the experiment failed.

False. A precise null is a discovery: it ruled out a detectable aether wind and forced physics toward the constancy of

c

. Negative results constrain theory hard.

If the aether existed and Earth were momentarily at rest in it, the experiment would still show a shift.

False. With

v = 0

there is no aether wind, so both effective speeds equal

c

, giving

t_{∥} = t_{⊥} = 2 L / c

and

Δ N = 0

— the very case that looks identical to the real null result. That is why they repeated it across seasons, hoping to catch Earth moving through the aether at some point.

Rotating the apparatus by

9 0^{\circ}

is what produces the measurable fringe shift.

True. A single orientation's

Δ t

is baked into a fringe pattern whose "zero" you don't know, so you can't read it. Rotating by

9 0^{\circ}

swaps which arm is parallel: the arm that was slow (

t_{∥}

) becomes fast (

t_{⊥}

) and vice-versa, so the path difference flips from

+ c Δ t

- c Δ t

— a change of

2 c Δ t

, i.e. a change of

2Δ t

in time. That change is what the fringes register, and it equals

Δ N = 2 c Δ t / λ

cycles.

The parallel arm's downstream and upstream legs cancel out.

False. The two legs each cover distance

L

: the up-wind leg takes

L / (c - v)

(slow), the down-wind leg

L / (c + v)

(fast). You spend longer in the slow leg, so they add to more, not less, than the no-wind time:

\frac{L}{c - v} + \frac{L}{c + v} = \frac{2 L c}{c ^{2} - v ^{2}} > \frac{2 L}{c}

. The excess is what the wind would reveal.

The perpendicular arm is completely unaffected by the aether wind.

False. To travel a straight cross-path the beam must aim slightly upstream, so its useful cross-speed drops to

c^{2} - v^{2}

(see the cross-arm "why" below). Its round trip is

t_{⊥} = 2 L / c^{2} - v^{2}

, which is longer than

2 L / c

— just less longer than the parallel arm, which is exactly why a difference

Δ t

survives.

The experiment directly proved Einstein's special relativity.

False. It killed the detectable aether and was consistent with SR, but Lorentz–FitzGerald contraction also fit the null result. SR earned its status from a much broader web of predictions, not this one experiment alone.

To leading order the expected time difference grows linearly with the aether speed

v

False. It scales as

v^{2}

Δ t \approx L v^{2} / c^{3}

. The linear-in-

v

terms of the two arms are equal and cancel in the subtraction (see the

v^{2}

"why" below), leaving

β^{2} = (v / c)^{2}

as the leading effect — tiny, but the interferometer was built to see it.

Spot the error

Each statement below contains a flaw. Name it and explain the mechanism.

"Light needs no medium, so the aether hypothesis was obviously silly in 1887."

The flaw is hindsight. In 1887 every known wave (sound, water, string) needed a medium, and Maxwell's equations gave a definite

c

that begged the question "relative to what?". An aether was the reasonable default answer, not a silly one — this very experiment is what forced physicists to abandon it.

"Since light travels at

c

in the aether, on Earth we'd measure

c + v

heading into the wind."

The flaw is a sign/direction error. Heading into the wind, the medium streams against the beam, so the ground speed is

c - v

(slower); only when moving with the wind is it

c + v

. Swapping these would flip the sign of the whole prediction, because the aether effect lives entirely in the difference between the slow leg

L / (c - v)

and the fast leg

L / (c + v)

"The cross-arm light travels at

c - v

because it fights the wind head-on."

The flaw is wrong geometry. The cross beam does not move against the wind head-on; it moves perpendicular to it. Its speed

c

is a vector: to cancel the sideways drift it aims upstream, so part of

c

is "used up" fighting the current. By Pythagoras,

c^{2} = v^{2} + (cross-speed)^{2}

, giving cross-speed

c^{2} - v^{2}

— not

c - v

. The head-on value

c - v

belongs to the parallel arm, not this one.

"Because both arms are length

L

, both round-trip times must be equal."

The flaw is confusing equal distance with equal time. Time is distance divided by effective speed, and the two arms have different effective speeds: the parallel arm gives

t_{∥} = \frac{2 L c}{c ^{2} - v ^{2}}

, the perpendicular arm

t_{⊥} = \frac{2 L}{c ^{2} - v ^{2}}

. Same

L

, different denominators, so

t_{∥} > t_{⊥}

whenever

v \neq = 0

"A 0.4-fringe prediction is too small to matter, so the null result proves nothing."

The flaw is a mis-scaled sense of "small". The apparatus resolved shifts down to about

0.01

fringe, so a predicted

Δ N \approx 0.4

is forty times its sensitivity — glaringly visible. Seeing essentially zero against a prediction that large is a strong constraint, not a wash.

"If they'd just used a longer arm

L

, the wind would eventually cancel and give zero anyway."

The flaw is a backwards dependence. Since

Δ N = 2 L v^{2} / (λ c^{2})

, the predicted shift is proportional to

L

: longer arms make the wind easier, not harder, to detect. Michelson deliberately folded the light path with extra mirrors to boost the effective

L

for exactly this reason.

Why questions

Answer with the mechanism, not just a label.

Why does the expected effect depend on

v^{2}

and not

v

Write each time as

t_{∥} = \frac{2 L}{c} \frac{1}{1 - β ^{2}}

and

t_{⊥} = \frac{2 L}{c} \frac{1}{1 - β ^{2}}

. For small

β

expand:

\frac{1}{1 - β ^{2}} \approx 1 + β^{2}

and

\frac{1}{1 - β ^{2}} \approx 1 + \frac{1}{2} β^{2}

. Both have the same constant

1

(which cancels in

t_{∥} - t_{⊥}

) and no term linear in

v

at all — the arms are symmetric to first order. Subtracting leaves

\frac{2 L}{c} (β^{2} - \frac{1}{2} β^{2}) = \frac{2 L}{c} \cdot \frac{β ^{2}}{2} = \frac{L v ^{2}}{c ^{3}}

, so the leading survivor is

\propto v^{2}

Why must the cross-beam be aimed upstream rather than straight across?

If aimed straight, the wind of speed

v

would drift it sideways during the trip and it would miss the mirror. Aiming upstream lets the drift cancel that sideways push exactly. But then, of the total speed

c

, a component

v

is spent counter-drifting, and by Pythagoras only

c^{2} - v^{2}

remains for the useful straight-across motion.

Why did they rotate the apparatus and watch across the whole year?

Rotating by

9 0^{\circ}

swaps which arm is parallel to the wind, flipping the path difference by

2 c Δ t

and turning the hidden

Δ t

into a moving fringe pattern you can measure. Watching all year covers Earth's ever-changing orbital direction, so even if the aether wind vanished on one date it could not stay hidden on every date.

Why was interference (fringes) the chosen readout instead of a stopwatch?

The predicted

Δ t \sim 1 0^{- 16}

s is far below any clock's reach, but it corresponds to a path difference of a fraction of the wavelength

λ

. Interference converts a sub-wavelength path difference directly into visible movement of bright/dark fringes, amplifying an unmeasurable time into a countable shift

Δ N = 2 c Δ t / λ

Why does the swimmer-in-a-river analogy work for light beams?

In the aether picture, light's speed is fixed relative to the aether just as a swimmer's speed is fixed relative to the water; the lab frame plays the role of the riverbank. So the same round-trip asymmetry — down/up-stream being slower overall than straight across — carries over term for term.

Why does the null result point toward "

c

is the same for all observers"?

If no arm orientation or season ever shows a shift, then the light speed measured in the lab is independent of the lab's own motion

v

. That is precisely the statement that

c

is frame-independent — later enshrined as Einstein's second postulate.

Why did Lorentz's length contraction "save" the aether, and why did Einstein abandon it anyway?

Shrinking the parallel arm

L

by just the right factor makes

t_{∥} = t_{⊥}

, so

Δ t = 0

and the wind hides — a patch that fits the data. Einstein dropped the aether because the same equations follow more simply from two postulates, with no undetectable medium to invent.

Edge cases

Boundary and degenerate inputs the reader must not be surprised by.

What is the predicted fringe shift if

v = 0

(Earth at rest in the aether)?

Zero. With no wind both effective speeds are

c

, so

t_{∥} = t_{⊥} = 2 L / c

and

Δ N = 0

— indistinguishable from the actual observed null, which is why one measurement alone can't decide anything.

What happens to the predicted

Δ t

v \to c

(a hypothetical extreme wind)?

t_{∥} = \frac{2 L c}{c ^{2} - v ^{2}}

blows up because

c^{2} - v^{2} \to 0

(the up-wind leg

L / (c - v)

can never make headway), so the naive aether prediction diverges — a red flag that the classical velocity-addition it assumes has broken down.

If the two arms had unequal lengths

L_{1} \neq = L_{2}

, does the experiment still work?

Yes — the rotation comparison still isolates the aether-dependent term because rotating swaps the arms' roles; unequal lengths only add a fixed path offset that stays put under rotation and cancels out, which is why exact arm-length matching was not the critical requirement.

What if there were literally no wave medium and no relativity — could the null still occur by luck?

Only if

v

happened to be exactly zero at every single measurement, which the year-round repetition rules out. A persistent null across all directions and dates demands a principle (constant

c

), not a lucky coincidence.

In the limit

β = v / c ≪ 1

, why is keeping only the

β^{2}

term legitimate?

For Earth

β \sim 1 0^{- 4}

, so the next correction

β^{4} \sim 1 0^{- 16}

is utterly negligible beside

β^{2} \sim 1 0^{- 8}

; the truncated expansion is accurate to far better than the

0.01

-fringe resolution.

If light did obey Galilean velocity addition, what would this experiment have shown?

A clear, orientation-dependent fringe shift of order

0.4

fringes — see Galilean Relativity & Velocity Addition. The observed absence of this shift is the direct evidence that light does not add velocities the Galilean way.