F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special RelativityThe Ether and the Michelson–Morley Experiment
F1.1 F3 A1.8 A1.10About 70 minutes
Learning goals- I can explain why 19th-century physicists believed in a luminiferous ether.
- I can calculate round-trip times for a boat moving along and across a current, and connect them to light in an 'ether wind'.
- I can describe how a Michelson interferometer works and what Michelson and Morley expected to observe.
- I can explain the significance of the null result and how it changed scientific thinking.
Words to know
luminiferous ether
A hypothetical, invisible medium filling all space through which light waves were thought to travel. Its rest frame would be the one frame where light moves at exactly c.
ether wind
The apparent flow of ether past Earth that would result from Earth's motion through the ether.
interferometer
An instrument that splits a light beam into two paths and recombines them; tiny differences in travel time show up as shifts in the interference fringes.
fringe shift
The movement of the bright and dark interference bands when the path difference between two beams changes. A shift of one fringe means a change of one wavelength in path difference.
null result
An experimental result showing no effect where an effect was predicted.
Lorentz–FitzGerald contraction
An 1889–1892 hypothesis that objects shrink along their direction of motion through the ether, just enough to hide the ether wind.
Hunting for the invisible
Sound needs air. Water waves need water. In the 1800s it seemed obvious that light, also a wave, must need a medium. Physicists called it the luminiferous ether ("light-carrying ether"). It had to fill all of space, even the vacuum between stars, and it had to be extremely stiff to carry waves as fast as light, yet planets moved through it without any drag. Strange, but the idea fitted Maxwell's equations: c would be the speed of light relative to the ether.
Earth orbits the Sun at about 30 km/s. So Earth must be moving through the ether, and in the lab we should feel an ether wind. Light going upwind, downwind and crosswind should take slightly different times. This lesson is about the most famous attempt to measure that difference, and why its failure is one of the most important results in physics.
A river analogy
Picture two equally fast boats in a river. Each moves at speed c relative to the water, and the current flows at speed v. Each makes a round trip of distance L out and L back.
- Boat A goes downstream and back. Downstream its ground speed is c + v, upstream c − v. Time:
t_A = L/(c + v) + L/(c − v) = (2L/c) × 1/(1 − v²/c²)
- Boat B goes straight across and back. It must aim slightly upstream so its across-river speed is √(c² − v²) both ways. Time:
t_B = 2L/√(c² − v²) = (2L/c) × 1/√(1 − v²/c²)
Both are longer than the still-water time 2L/c, but the along-stream boat is slower: 1/(1 − v²/c²) is bigger than 1/√(1 − v²/c²). Michelson and Morley built the same race for light, with the ether wind as the current.
Worked exampleExample 1: The boat race
Each boat moves at 5.0 m/s relative to the water. The current is 1.0 m/s. Boat A goes 100 m downstream and back; boat B goes 100 m across and back. Find both times.
- Boat A: downstream speed 5.0 + 1.0 = 6.0 m/s; upstream 5.0 − 1.0 = 4.0 m/s.
- t_A = 100/6.0 + 100/4.0 = 16.7 + 25.0 = 41.7 s.
- Boat B: across speed √(5.0² − 1.0²) = √24 = 4.90 m/s each way.
- t_B = 200/4.90 = 40.8 s.
- Check with the formulas: 2L/c = 200/5.0 = 40 s; v²/c² = 0.04; t_A = 40/0.96 = 41.7 s; t_B = 40/√0.96 = 40.8 s. ✓
t_A ≈ 41.7 s, t_B ≈ 40.8 s. The trip across the current is faster by about 0.9 s.
The Michelson interferometer
In 1881 Albert A. Michelson, and in 1887 Michelson with Edward W. Morley in Cleveland, Ohio, used an interferometer:
- Light from a source hits a half-silvered mirror (a beam splitter) at 45°. Half the light goes along arm 1, half along arm 2, at right angles.
- Mirrors at the ends of the arms send each beam back. The beams recombine and travel into a telescope.
- The two beams interfere. You see a pattern of bright and dark fringes. (You will study interference in detail in the sibling unit The Wave Nature of Light.)
If the ether wind blows along arm 1, the light in arm 1 is like boat A and the light in arm 2 is like boat B. The beams return slightly out of step. Rotating the whole apparatus by 90° swaps the roles of the arms, so the fringes should shift. The apparatus floated on a pool of mercury so it could be turned smoothly without distortion.
How big should the shift be?
For v much smaller than c, use the approximations 1/(1 − x) ≈ 1 + x and 1/√(1 − x) ≈ 1 + x/2, with x = v²/c². Then the time difference between the arms is
Δt = t₁ − t₂ ≈ (2L/c)(1 + v²/c²) − (2L/c)(1 + v²/2c²) = (L/c)(v²/c²)
Rotating by 90° reverses the difference, so the total change is 2Δt. Light travels a distance c(2Δt) in that time, and every wavelength λ of path difference moves the pattern by one fringe:
Expected fringe shift: N = 2Lv²/(λc²)
Michelson and Morley made each arm effectively about 11 m long by bouncing the light back and forth with extra mirrors. With v = 3.0 × 10⁴ m/s (Earth's orbital speed) and visible light of λ = 5.5 × 10⁻⁷ m:
N = 2(11)(3.0 × 10⁴)² / [(5.5 × 10⁻⁷)(3.0 × 10⁸)²] = 0.40 fringe
Their apparatus could detect a shift far smaller than that.
Worked exampleExample 2: Predicting the fringe shift
An interferometer with arms of effective length 11 m uses light of wavelength 550 nm. (a) What time difference Δt between the arms would an ether wind of 3.0 × 10⁴ m/s produce? (b) What fringe shift is expected when the apparatus is rotated by 90°?
- (a) Δt = (L/c)(v/c)² = (11/3.0 × 10⁸)(1.0 × 10⁻⁴)² = (3.67 × 10⁻⁸ s)(1.0 × 10⁻⁸) = 3.7 × 10⁻¹⁶ s.
- (b) Rotation reverses the difference, so the change in delay is 2Δt = 7.3 × 10⁻¹⁶ s.
- Path difference: c × 2Δt = (3.0 × 10⁸)(7.33 × 10⁻¹⁶) = 2.2 × 10⁻⁷ m.
- Fringes: N = 2.2 × 10⁻⁷ / 5.5 × 10⁻⁷ = 0.40.
(a) Δt ≈ 3.7 × 10⁻¹⁶ s; (b) N ≈ 0.40 fringe. A tiny time, but an easily visible shift: that is the power of interference.
The null result
Michelson and Morley watched the fringes as they turned the apparatus slowly, at noon and in the evening, over several days in July 1887. They reported that the actual shift was certainly less than one-twentieth of the expected 0.4 fringe, and quite possibly zero. Later versions of the experiment, in different seasons (when Earth moves in different directions around the Sun), at different heights and with lasers, have pushed the limit far lower. No ether wind has ever been detected.
Physicists tried to save the ether:
- Ether drag: maybe Earth drags the nearby ether along with it. But this clashed with stellar aberration (the small, yearly tilt in the direction of starlight caused by Earth's orbital motion), which only makes sense if Earth moves through the ether.
- Lorentz–FitzGerald contraction: George FitzGerald (1889) and Hendrik Lorentz (1892) proposed that objects moving through the ether shrink along their direction of motion by exactly the factor √(1 − v²/c²). This would make the two arms' times equal and hide the ether wind. It worked, but it looked like a patch invented only to explain away the result.
In 1905 Einstein took a bolder step: there is no ether, and the speed of light is the same in every inertial frame. The next lesson, Einstein's Two Postulates and the Power of Thought Experiments, starts from there. (Historians still discuss how much the Michelson–Morley result directly influenced Einstein, who said his thinking came mainly from electromagnetic theory and his own thought experiments. But the null result became one of the strongest pieces of evidence for his theory.)
Thinking like a scientist
- A null result is still a result. A well-designed experiment that finds 'nothing' where theory predicts 'something' can overturn the theory.
- Good experiments rotate, repeat at different times and estimate the smallest effect they could detect. Michelson and Morley's predicted shift (0.4 fringe) was much larger than their detection limit, so 'no shift' was a meaningful answer.
- When a theory needs a special fix for every new result (ether drag, contraction), scientists start to look for a simpler explanation.
Just for funMichelson and Morley floated their heavy stone slab on a pool of mercury so they could spin it gently. In other words, one of the most important experiments in physics was done on a very expensive lazy Susan.
Hands-on activityFringes you can make move
This shows how interference makes tiny distances visible, the idea behind the interferometer.
You need: two clean, flat, clear plastic sheets (for example two pieces cut from a clear page protector, or two clear CD case lids), a lamp, and a pencil with an eraser. (If you use two glass microscope slides instead, ask an adult to help: glass edges can cut.)
- Wipe both sheets so they are free of dust. Press them firmly together.
- Hold them under the lamp and look at the reflection at an angle. You should see faint, coloured, wavy bands (fringes). They come from light reflecting off the two surfaces of the very thin air gap between the sheets.
- Press gently on one spot with the eraser. Watch the fringes move and form rings around the spot.
- Record: How far did you think you squeezed the gap? (Far less than a hair's width.) How far did the fringes move?
What to look for: a single fringe corresponds to a change in the air gap of only about half a wavelength of visible light (roughly 0.27 µm for green light), so fringes move when you barely touch the sheets. Michelson and Morley's experiment expected the fringes to move by 0.4 of a fringe when they rotated the apparatus. They saw almost no movement, even though their instrument was sensitive enough to detect it.
Analyse (A1.8, A1.10): Michelson and Morley estimated they could detect a shift of about 0.01 fringe. Their predicted shift was 0.40 fringe. Explain why their 'no shift' result was strong evidence against the ether wind rather than a sign that their equipment was not sensitive enough.
Your turn
1Why did physicists in the 1800s believe in the ether?
Every other known wave (sound, water waves) needs a medium. Light was known to be a wave, so they assumed it travelled through a medium too. The ether also gave a frame in which Maxwell's equations (speed c) would hold.
2A swimmer moves at 2.0 m/s relative to the water in a river with a 0.50 m/s current. How long does a round trip of 50 m upstream and 50 m back take?
t = 50/(2.0 + 0.50) + 50/(2.0 − 0.50) = 20 + 33.3 = 53.3 s.
3Same swimmer, same river: how long for 50 m straight across and back?
Across speed √(2.0² − 0.50²) = √3.75 = 1.94 m/s. t = 100/1.94 = 51.6 s.
4What is the job of the half-silvered mirror in a Michelson interferometer?
It splits the incoming light into two beams travelling along perpendicular arms, and later recombines the returning beams so they can interfere.
5Why did Michelson and Morley rotate their apparatus?
Rotating by 90° swaps which arm lies along the supposed ether wind, so the time difference would reverse and the fringes would shift. Looking for a shift during rotation avoided needing to know the exact arm lengths.
6Calculate the expected fringe shift for an interferometer with 1.2 m arms, light of wavelength 600 nm and an ether wind of 3.0 × 10⁴ m/s.
N = 2Lv²/(λc²) = 2(1.2)(9.0 × 10⁸)/[(6.0 × 10⁻⁷)(9.0 × 10¹⁶)] = 0.040 fringe.
7What did Michelson and Morley observe, and what does 'null result' mean?
They saw a fringe shift less than one-twentieth of the predicted 0.4 fringe, consistent with no shift at all. A null result is finding no effect where one was predicted.
More practice: warm-up
1About how fast does Earth orbit the Sun?
About 30 km/s (3.0 × 10⁴ m/s).
2In the river analogy, which round trip takes longer: along the current or across it?
Along the current (downstream and back).
3What does a shift of one fringe correspond to?
A change of one wavelength in the path difference between the two beams.
4In what year did Michelson and Morley perform their famous experiment?
5Name the contraction hypothesis proposed to rescue the ether.
The Lorentz–FitzGerald contraction.
More practice: on your own
1A plane flies at 200 m/s relative to the air. A wind blows at 20 m/s. Compare the round-trip times for a 400 km trip out and back (a) along the wind and (b) across the wind.
(a) 400 000/220 + 400 000/180 = 1818 + 2222 = 4040 s. (b) Cross-wind ground speed √(200² − 20²) = 199.0 m/s; 800 000/199.0 = 4020 s. Along-the-wind is about 20 s longer.
2Show that, for v ≪ c, the time difference between the arms of an interferometer is about (L/c)(v²/c²).
t₁ = (2L/c)/(1 − x) ≈ (2L/c)(1 + x) and t₂ = (2L/c)/√(1 − x) ≈ (2L/c)(1 + x/2), with x = v²/c². Subtract: Δt ≈ (2L/c)(x/2) = (L/c)(v²/c²).
3Why did Michelson and Morley repeat the experiment at different times of day (and why did later experimenters repeat it at different times of year)?
Earth's spin and orbit change the direction of the lab's velocity. Even if Earth happened to be at rest in the ether at one moment, it could not be at rest at all times of day and year. Repeating ruled out that coincidence.
4Explain why stellar aberration was a problem for the 'ether drag' idea.
Aberration (a small yearly tilt in starlight's direction) is explained by Earth moving through the medium that carries light. If Earth dragged the nearby ether along, there would be no relative motion near Earth and no aberration, yet aberration is observed.
5A modern interferometer has arms of 4.0 km (like the LIGO gravitational-wave detectors) and uses light of wavelength 1064 nm. What fringe shift would an ether wind of 3.0 × 10⁴ m/s produce?
N = 2(4000)(9.0 × 10⁸)/[(1.064 × 10⁻⁶)(9.0 × 10¹⁶)] = 7.2 × 10¹²/9.58 × 10¹⁰ ≈ 75 fringes. (This simple estimate ignores the extra mirrors that make LIGO's light bounce many times.) No such ether effect is seen.
Challenge
1Use the river formulas to show that, if the arm along the ether wind were shortened by the factor √(1 − v²/c²), the two arms would have equal round-trip times.
t₁ = (2L₁/c)/(1 − v²/c²) with L₁ = L√(1 − v²/c²) gives t₁ = (2L/c)/√(1 − v²/c²), which equals t₂ = (2L/c)/√(1 − v²/c²). The contraction hides the ether wind exactly.
2Michelson and Morley could detect a shift of about 0.01 fringe. What is the smallest ether-wind speed their 11 m, 550 nm apparatus could have detected?
N = 2Lv²/(λc²), so v = c√(Nλ/2L) = 3.0 × 10⁸ × √(0.01 × 5.5 × 10⁻⁷/22) = 3.0 × 10⁸ × √(2.5 × 10⁻¹⁰) = 3.0 × 10⁸ × 1.58 × 10⁻⁵ ≈ 4.7 × 10³ m/s (about 5 km/s), far below Earth's 30 km/s orbital speed.
3A friend says, 'The Michelson–Morley experiment failed because they saw nothing.' Write a short reply explaining why a null result can be a successful experiment, and how it changed scientific thinking.
The experiment did exactly what it was designed to do: measure a predicted effect that was well above its sensitivity. It found the effect was absent. That is a strong, reliable finding. It showed the ether model was wrong (or needed ad hoc fixes), and it later became key evidence for Einstein's postulate that light has the same speed in every inertial frame, which removed the need for an ether.
Quiz
1. In the river analogy, a boat moves at speed c relative to the water, and the current is v. How long does a round trip of L downstream and L back take?
2. What did Michelson and Morley expect to see when they rotated their interferometer?
3. What did they actually observe?
4. Why was the half-silvered mirror needed?
5. Which proposal tried to save the ether by saying moving objects shrink along their motion?
6. Which statement best describes the importance of the Michelson–Morley experiment?
Big idea (F1.1, F3): a carefully designed experiment that finds 'nothing' can overturn a theory. The Michelson–Morley null result, together with Einstein's thought experiments, led physics to abandon the ether.
How to help: The river-boat race is the key to understanding the experiment. Ask your teen to explain, with the boat numbers in Example 1, why the cross-river boat wins. Then ask: 'If the light in one arm is like boat A, why would rotating the apparatus move the fringes?' For the activity, glass slides need adult help because of sharp edges; plastic page protectors are safer.
Common difficulties:
- Forgetting that the cross-stream boat has to aim upstream, so its speed across is √(c² − v²), not c.
- Using the binomial approximations without checking that v ≪ c.
- Squaring incorrectly in N = 2Lv²/(λc²). Keep powers of ten organized: v² = 9.0 × 10⁸, c² = 9.0 × 10¹⁶.
- Thinking the experiment 'failed'. It succeeded in showing the predicted effect does not exist.