Chapter 2
Below Decks
No experiment can tell you that you’re moving, not even with light.
What this chapter does
The prologue ended with a suspect: rule 3, “every clock ticks at the same rate, moving or not.” This chapter finds the measured fact that puts that rule in danger. It has three parts.
- A sealed cabin. Galileo’s principle: inside a cabin moving steadily in a straight line, no experiment shows that you’re moving. You’ll test it with a ball, a pendulum and a flash of light.
- Speed relative to what? Why the light result is so strange: light’s speed doesn’t add up the way a ball’s does. Then, which rooms count as “steady”.
- Ask Nature. Two real experiments showing that light’s speed is the same for everyone, however fast the lamp or the lab moves.
Then practice, a small function to build, and a checkpoint.
Warm-up
Three quick ones from the prologue and the surveyors, to wake up the parts of your brain this chapter needs.
A right triangle has legs 8 and 15. How long is its hypotenuse? Need a refresher?
That’s 8 + 15. Pythagoras adds the squares of the legs. What is 8² + 15²?
Hint
Square each leg: \(64\) and \(225\). Add them, then take the square root.
Solution
\(\sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\). It’s one of the friendly triangles, 8-15-17.
How many meters of time is one nanosecond? (Light covers 299 792 458 m in one second.)
That’s how many nanoseconds are in one meter of time: the conversion run backwards. In one billionth of a second, does light go more or less than a meter?
Hint
A nanosecond is \(10^{-9}\) s. Multiply by the number of meters light covers per second.
Solution
\(299\,792\,458 \times 10^{-9} \approx 0.2998\) m. Light covers about 30 cm per nanosecond: about a foot.
The Daytimers and the Nighttimers survey the same town with grids turned 10° apart. What do they agree on about the mill?
Hint
Turn one grid by 10°. Which of the three numbers would you not have to recompute?
Solution
The distance. North and east offsets depend on the grid; \(\sqrt{x^2 + y^2}\) does not. Keep that idea close: this chapter is about another thing that refuses to depend on who measures it.
The puzzle
Shut yourself up with a friend in the main cabin below decks of a large ship. This isn’t a new idea: Galileo suggested it in 1632, and he meant it as an experiment. Bring some flies, a bowl of fish, and a bottle that drips into a jar below. Watch everything while the ship sits in harbor. Then let the ship sail off, smoothly, with no rocking, and watch again.
Two ship words you’ll need. The front of a ship is the bow, so the cabin’s front wall is the fore wall. The back of a ship is the stern, so the back wall is the aft wall.
Here’s the first test. A ball hangs from the ceiling right above a chalk mark on the floor. The ship sails smoothly at a steady 10 m/s. You let go.
Predict
In the smoothly sailing cabin, the dropped ball lands…
Experiment
Galileo only had flies and fish. We have flashlights. Light has a speed: about 30 centimeters every nanosecond. Here’s the question that will end up costing us our common-sense idea of time: speed relative to what? If light moved at its speed relative to the water, or to some invisible stuff filling space, then inside a moving cabin you’d clock light slower one way and faster the other, and you’d know you were moving. Galileo’s promise would be broken.
Why would light need “something” to move through? Light is a wave, and every other wave anyone knew moves through some stuff: sound through air, ripples through water. So for most of the 1800s physicists were sure light moved through an invisible stuff filling space. They called it the ether.
The lab below lets you test both possibilities. In the ether universe, light moves at speed 1 relative to the ether, and the ship moves through the ether the way it moves through still air. The other universe is ours. Before you flash the lamp, one more prediction.
Predict
In our universe, the ship coasts fast. The lamp in the middle of the cabin flashes. As the crew’s wall clocks measure it, the light reaches…
Here is the cabin, sealed tight. You can run three experiments in it, in either universe, and set the ship’s speed. Everything else stays the same.
A ship’s cabin with a hanging ball over a chalk mark, a drip over a jar, a pendulum, and a lamp in the middle with a clock on each end wall. At every steady ship speed the ball lands on the mark, the drip lands in the jar, and one full swing of the pendulum, there and back, takes 2.46 s. In the imaginary ether universe a flash reaches the aft wall at 2 and the fore wall at 6 (meters of time) when the ship moves at 0.5 through the ether; in our universe both clocks read 3, at any speed.
Describe this lab
A cutaway drawing of a cabin 6 m long. A ball hangs 2.28 m above a chalk mark; a bottle drips into a jar; a 1.5 m pendulum hangs near the fore wall; a lamp stands in the middle, with a clock on the aft wall and one on the fore wall. Controls: the experiment (ball and drip, pendulum, light flash), the universe (ether or ours), where you watch from (the cabin or the dock, for the ball), the ship speed (0 to 30 m/s for the ball and pendulum; 0 to 0.5 of light speed for the light test), a run button, and a button that hides the speed so you can try to work it out. The light flash waits until you have made the light prediction above the lab, or chosen to skip it.
Every run is written into the lab notebook below the drawing: the ship speed, the universe and the readings (where the ball landed, how long one full swing took, or what the two wall clocks read in meters of time). The Describe button reads out the latest result. Keys, with focus inside the lab: Space runs, U switches universe, V switches view, E changes experiment.
Try it
(a) With Ball & drip, drop the ball at ship speeds 0, 5 and 10 m/s. Where does it land each time? Now watch from the dock and drop it again at 10 m/s. What shape is the path, and where does the ball land? While you’re here, time one full swing of the pendulum at 0 and at 30 m/s.
(b) Choose Light flash and the Ether universe. Set u to 0.5, flash, and note both clock readings. Then press Hide the speed, flash again, and work out the secret speed from the two clocks alone. Get within 10%.
(c) Switch to Our universe, hide the speed again, and try once more.
Stuck on (b)?
Start from what you saw at u = 0.5. The aft clock read 2, which is 3 ÷ 1.5: the light and the aft wall move toward each other, so the gap between them shrinks at 1 + u = 1.5. So for a secret flash, 1 + u = 3 ÷ (the aft reading). If the aft clock reads 2.7, then 1 + u = 3 ÷ 2.7 ≈ 1.11, so u ≈ 0.11. Check it with the fore clock, which should read 3 ÷ (1 − u) ≈ 3.37. The worked example under “Light, in an ether universe” shows a shortcut that uses both clocks at once.
Then play. Is there any experiment in this cabin, in either universe, that tells you the ship is moving? Which one?
What just happened?
At 0, 5, 10 and even 30 m/s the ball lands on the chalk mark, the drip finds the jar, and one full swing of the pendulum takes 2.46 s.
The flies don’t pile up at the back of the cabin. The drips don’t miss the jar. The fish don’t get tired swimming toward the bow. Nothing you can do inside that cabin tells you the ship is moving. Not “it’s hard to tell”. Nothing.
Watched from the dock, the ball’s path is an arc. It was already moving forward with the ship when you let go, and it keeps that forward speed all the way down. The mark moves forward by exactly the same amount, so the ball still lands on it.
Both. A path is a description, and a description needs a ruler. The cabin’s ruler and the dock’s ruler slide past each other, so they describe the fall differently. What they agree on is the thing that happened: the ball hit the mark. Hold on to that idea; Chapter 3 gives it a name.
Drop the ball in the cabin lab three times, at two or more different speeds. This card opens by itself when you have.
In our universe both wall clocks read 3.00 meters of time, at every speed you tried. Only the imaginary ether universe gave the speed away.
So a flash of light can’t tell the crew the ship is moving, any more than a ball can. Light obeys Galileo’s principle too: inside a cabin moving steadily in a straight line, no experiment shows that you’re moving.
Now imagine the same flash seen from the dock. Everyday speeds add. Throw a ball toward the bow at 15 m/s on a ship doing 10 m/s, and from the dock it moves at \(10 + 15 = 25\) m/s. Let’s call that the everyday adding rule. Say the ship coasts at half of light’s speed, 0.5. On board, the forward flash moves at 1: that’s what the tie told you. So the everyday adding rule says that from the dock, it moves at \(1 + 0.5 = 1.5\). Measured from the dock, it moves at 1. (A team at CERN checked exactly this in 1964, with lamps far faster than any ship. You’ll see it below.)
So Nature kept Galileo’s principle and broke the everyday adding rule, for light. That has a price. The crew and the dock time the same flash, each with their own rulers and clocks, and both get 1, although one is moving past the other. With everyday distance and time, that can’t happen. Something in how they measure has to give.
The prime suspect is one you’ve met. The prologue’s rule 3 said every clock ticks at the same rate, moving or not. Chapter 5 shows how light’s fixed speed forces that rule to break, and Chapter 8 brings the muons back to check.
Flash the lamp in our universe twice, at two different speeds. This card opens by itself when you have.
Quick check. The ship coasts past the dock at 0.5, and the lamp flashes toward the bow. One of these ideas broke. Which one?
Hint
Which idea makes a prediction that a measurement contradicts? The crew measures 1 on board. What does the dock measure?
Solution
The everyday adding rule. Galileo’s principle held: on board the flash moves at 1 both ways. The adding rule then predicts \(1 + 0.5 = 1.5\) from the dock, but the dock measures 1 (CERN, 1964).
Speed relative to what?
The principle of relativity
Here is Galileo’s principle in one sentence. In a sealed room that moves at a steady speed in a straight line, every experiment comes out exactly as it would at rest. A room like that, with its rulers and clocks, is a frame of reference, and the sentence is called the principle of relativity. We’ll also call it Galileo’s principle.
Two words carry the weight: steady and straight. When the ship turns, the coffee sloshes. When it brakes, you lurch forward. You can feel changes in motion. You can’t feel motion.
Myth
“You can feel how fast you’re going.” You feel bumps, turns, speeding up and slowing down. Right now Earth carries you around the Sun at about 30 kilometers every second, and nothing in your room notices.
Adding speeds, the everyday way
Rae, who always rides whatever is moving (today, the ship), walks toward the bow at 2 m/s on a ship doing 10 m/s. From the dock she moves at 12 m/s. The ball she throws forward at 15 m/s, measured on board, does 25 from the dock. That’s the everyday adding rule:
\[ \lab{w} = \lab{u} + \rkt{v'} \]
Here \(\lab{u}\) is the ship’s speed and \(\lab{w}\) the ball’s speed, both measured from the dock (blue), and \(\rkt{v'}\) is the ball’s speed measured on board (vermilion; the prime marks an on-board measurement). It has worked for every ball, arrow and train anyone has ever timed. Light is the exception.
Sound plays by this rule too, with a twist: it moves at 340 m/s relative to the air. Inside the sealed cabin the air sails along with you, so a shout crosses the cabin equally fast both ways. No clue there. But stand on the open deck in a wind.
Quick check. On the open deck you feel a 10 m/s headwind: measured on board, the air streams from bow to stern at 10 m/s. You shout toward the bow. Measured on board, your shout moves forward at…
Hint
Sound moves at 340 relative to the air. Measured on board, which way is the air moving, and how fast?
Solution
On board, the air moves aft at 10 m/s, and sound moves at 340 relative to it: forward \(340 - 10 = 330\) m/s, backward \(340 + 10 = 350\) m/s. The wind gives your motion away, because sound has something to move relative to.
Light, in an ether universe
Physicists of the 1800s were sure the ether was real (the deep dive below says why). If it were, the light in a cabin moving through the ether at speed u would behave like your shout in the wind: slower toward the bow, faster toward the stern.
Let’s do the ether arithmetic once, slowly, with friendly numbers. The cabin is 6 m long, so each wall is 3 m from the lamp. Clocks are in meters of time, so light in still ether covers 1 m per meter of time.
- The ship moves through the ether at \(u = 0.5\). Light heading for the fore wall moves at 1 through the ether, but the wall runs away at 0.5. Relative to the cabin, the light gains at \(1 - 0.5 = 0.5\). This is the everyday adding rule \(w = u + v'\), solved for \(v'\). Here \(w = 1\) is the light’s speed measured from the ether and \(u = 0.5\) is the cabin’s, so \(v' = w - u = 0.5\) is the light’s speed measured from the cabin.
- Time to cover 3 m at speed 0.5: \(3 \div 0.5 = 6\). The fore clock reads 6. Time is distance divided by speed.
- The aft wall rushes toward the light, so the closing speed is \(1 + 0.5 = 1.5\), and the time is \(3 \div 1.5 = 2\). The aft clock reads 2. Same rule, wall moving the other way.
- Now run it backwards, as the crew would. Divide the difference of the readings by their sum: \(\dfrac{6 - 2}{6 + 2} = 0.5\). That’s \(u\). The readings are \(3/(1-u)\) and \(3/(1+u)\). Over the common bottom \((1-u)(1+u) = 1 - u^2\), their difference is \(\dfrac{3 \cdot 2u}{1-u^2}\) and their sum is \(\dfrac{3 \cdot 2}{1-u^2}\). Divide one by the other: the \(3\), the \(2\) and the \(1-u^2\) all cancel, and what’s left is \(u\).
So in an ether universe a crew with one lamp and two clocks could measure their speed, and Galileo’s principle would hold for balls and fail for light. That universe makes sense. It just isn’t ours.
Pitfall
Light speeds don’t add like ball speeds. A rocket coasting at 0.5 switches on a headlight. On the rocket the light moves at 1: that’s Galileo’s principle, and it’s what the cabin’s tied clocks showed. The everyday adding rule then predicts \(1 + 0.5 = 1.5\) in the lab. The lab measures 1 (CERN, 1964, below).
What Nature says
Every test so far, in labs and in particle accelerators, gives the same answer. You’ll meet two of them in a moment. Here it is, in words this course uses from now on:
\[ \begin{gathered} \text{speed of light} = \inv{1} \\ \text{in every free-float frame} \end{gathered} \]
A free-float frame is a room, with its rulers and clocks, in which anything you let go stays put or coasts in a straight line at a steady speed. Our sailing cabin counts for sideways motion: a ball rolled across its floor coasts straight. Up and down, it doesn’t count, because a dropped ball still falls. The falling room below shows which rooms count fully.
It doesn’t matter how fast the lamp moves, or the lab. This is not an assumption we picked because it looked pretty. It is a measured fact, as solid as the ball landing on the chalk mark. Teal is the color this course uses for anything every observer agrees on: an invariant. In 1905 Einstein made it one of his two starting points, alongside Galileo’s principle. Many of the most direct tests came later, and every one has backed him. Everything else in this course, from slow clocks to \(E = mc^2\), follows from those two. The first casualty is the prologue’s rule 3: if two observers moving past each other both time the same flash at 1, their clocks can’t be ticking at the same rate. Chapter 5 shows exactly how they differ.
Quick check. A rocket coasting past the lab at 0.5 flashes a lamp backward, toward its tail. What speed does Lin, who always stays in the lab, measure for that light?
Hint
Does light’s speed depend on how fast its lamp moves, or which way it points?
Solution
1. The everyday adding rule would give \(1 - 0.5 = 0.5\), but light’s speed is 1 in every free-float frame, whatever its source is doing.
Real numbers
A speed of 1 means 299 792 458 m/s, exactly, by the definition of the meter. A ship at 10 m/s has \(u = 10 / 299\,792\,458 \approx 3.3 \times 10^{-8}\). While a flash crosses half our cabin (3 m of time, about 10 ns), that ship moves about 0.1 micrometers: a thousandth of the width of a hair.
Why, really?
Why should light refuse to add like everything else? I wish I could tell you. Nobody has a deeper reason. The best answer anyone has is a description, not an explanation: the universe has a speed built into its geometry, the way a sheet of paper has right angles built into it, and anything without mass, like light, travels at that speed. From Chapter 6 on you’ll see that geometry drawn out. Why the universe has that shape, nobody can say. We only know, very precisely, that it does.
Free-float frames: the falling room
Which rooms count as “steady”? Our ship is good enough for anything that happens sideways. But drop a ball and it falls: gravity picks out a direction, “down”. Now picture a room very high above the ground that falls freely, like an elevator whose cable has snapped (a thought experiment; nobody gets hurt). Everything in it falls together: the floor, you, and the ball you let go. So the ball stays beside your hand, and your feet no longer press on the floor: the gentlest push would send you drifting. Inside that room, no experiment detects gravity at all.
Such a room is a free-float frame. (Textbooks also say “inertial frame”.) The principle of relativity is really a statement about free-float frames: the laws of physics, and the speed of light, are the same in all of them.
A room seen from the side, 12 m wide. Resting on Earth, a released ball falls to the floor. Falling freely, the room, the person and the ball fall together, so the ball floats beside the hand and the person no longer presses on the floor. With two balls 10 m apart, each falls toward the center of the Earth, so their paths converge: they drift together by 7.7 micrometers in the first second and by about 28 mm in a minute. To an accuracy of 1 mm, over those 10 m, the room stays free-float for about 11 seconds.
Describe this lab
Left: a room seen from the side, with a person and a ball at shoulder height. Beside it, a strip of the shaft wall outside, marked every 5 m. Controls: the room (falling freely or resting on Earth), a button that releases the ball, and a switch for two balls 10 m apart. Resting, the ball falls to the floor and the wall stands still. In the falling room, the button lets go of the room and the ball together: the wall’s marks rush upward, the person no longer presses on the floor, and the ball stays beside the hand.
With two balls, a second panel shows the room from far away: two lines run from the balls to the center of the Earth, 6371 km below, and meet there (the angle is hugely exaggerated). A time slider runs from 0 to 60 seconds of falling, with a mark at 11.4 s, where the drift reaches 1 mm. Readouts give the time, how much closer the balls have drifted at real size, and whether the room is still free-float to an accuracy of 1 mm. In the drawing the drift is 100 times bigger than it is. Keys, with focus inside the lab: Space releases, F switches the room, T switches two balls on or off.
Try it
(a) Release the ball in the falling room, then in the room resting on Earth. In which room does the ball stay beside the hand? That room is free-float.
(b) Back in the falling room, switch on Two balls, 10 m apart and release them. Watch the far view: where is each ball heading? Read how far they have drifted together after 1 s, and after 60 s.
(c) Drag the time until the drift passes 1 mm. Up to then, the room counts as free-float to 1 mm accuracy. After that, it doesn’t.
Why do the balls drift together? Each one falls toward the center of the Earth, along its own line, and two lines aimed at one point aren’t parallel. This drift is called a tidal effect: gravity pulls a little differently at different places. (The Moon’s version of the same effect raises the ocean tides.) For balls 10 m apart the drift is 7.7 micrometers after 1 s and about 28 mm after a minute. The rule is \(\tfrac12 (g\,d/R)\,t^2\), where \(g = 9.8\un{m/s^2}\), \(d = 10\un{m}\) is the distance between the balls, and \(R = 6371\un{km}\) is Earth’s radius. A spacecraft in orbit is a falling room that never lands.
So is the falling room free-float? For two balls 10 m apart, to 1 mm accuracy, yes, for about 11 seconds. To 1 micrometer, for under half a second. Physicists say free-float frames are local: every claim that a room is free-float comes with a stated accuracy, over a stated region and time.
Deep dive: why everyone believed in the ether
Every wave anyone had studied was a wave in something: water waves in water, sound in air, a twang in a guitar string. In the 1860s James Clerk Maxwell found that his equations for electricity and magnetism allowed waves that travel at one fixed speed, and that speed matched the measured speed of light. Light was an electromagnetic wave.
But a fixed speed relative to what? To physicists the answer seemed obvious. They had pictured light as a wave in an invisible “ether” since the early 1800s, and Maxwell himself built his theory on one. So that one speed had to be the speed relative to the ether. It had to be strange stuff. It had to be stiff enough to carry waves faster than anything else, yet thin enough that planets move through it without slowing down. Strange, but people believed it the way you believe sound needs air. Until somebody went looking for the wind.
So far, two facts: balls can’t tell you your speed, and in our universe neither can light. Let’s see the evidence for the second one.
Ask Nature
If there is an ether, Earth can’t be at rest in it all year: it circles the Sun at about 30 km/s, which is \(u \approx 10^{-4}\). In 1887, in Cleveland, Albert Michelson and Edward Morley set out to feel that wind.
Their trick was a race. Split one beam of light in two. Send one half along the wind and back, the other across it and back, then bring them together. In an ether universe the along trip takes longer, by about \(L u^2\) (Build it, below, computes it exactly). With arms of 11 m that’s \(11 \times (10^{-4})^2 = 1.1 \times 10^{-7}\) m of time, less than a millionth of a nanosecond. No clock can time that.
But light can. In \(1.1 \times 10^{-7}\) m of time, light goes 110 nanometers, so the slow half arrives 110 nm behind the other. Yellow sodium light has waves 589 nm long: the lag is about a fifth of a wave. When the halves meet, their crests and troughs make bright and dark stripes, called fringes. Turn the apparatus by 90° and the arms swap jobs, so the stripes should slide by twice that: about 0.37 of a fringe. That is easy to see.
A Michelson interferometer on a stone slab. If an ether wind of 10⁻⁴ blew across it, turning the slab by 90° would slide the fringes by 0.37 of a fringe (with 11 m arms, counting in wavelengths of 589 nm sodium light). Michelson and Morley saw a slide of less than 0.02.
Describe this lab
Left: the apparatus seen from above. A lamp sends light to a half-silvered mirror in the middle of a round stone slab; one half goes along arm 1 to a mirror and back, the other along arm 2, at right angles, to a mirror and back. They recombine in an eyepiece. Dashed arrows show the ether wind that would blow if there were an ether. A button turns the slab by 90°, and then back to 0°.
Right: the eyepiece, showing vertical bright and dark fringes, a red crosshair and a dashed outline of where the fringes started; under it, a scale from 0 to 1 fringe with a marker that shows how far they have slid; and a gauge comparing the predicted slide of the fringes for a 90° turn with the measured one. Controls: arm length (1 to 11 m), wind speed (0 to 2 × 10⁻⁴), and whether the eyepiece shows the ether prediction or Nature’s result. Readouts show the predicted slide (0.37 of a fringe for the default settings), the observed slide (less than 0.02) and the difference between the two round-trip times, in meters of time; the Describe table lists both trips. After your etherRoundTrips passes its tests in the Build it section, the prediction is computed by your code. Keys, with focus inside the lab: Space turns the slab, N switches the fringe view.
Try it
Turn the slab with the ether prediction showing. How far do the fringes slide past the crosshair? Now switch to Nature, 1887 and turn again. Then halve the arms to 5.5 m, and separately double the wind: what does each do to the predicted slide?
Ask Nature
Michelson and Morley mounted their mirrors on a sandstone slab resting on a wooden float in a trough of mercury, so it turned smoothly, and folded each arm with mirrors to about 11 m. The ether predicted a slide of 0.37 of a fringe (their paper, rounding, says about 0.4). They saw less than about 0.02, consistent with nothing at all. A. A. Michelson and E. W. Morley, “On the Relative Motion of the Earth and the Luminiferous Ether”, American Journal of Science 34, 333 (1887)
It has been repeated many times since, in every season, with lasers and far better precision. The answer has never changed.
A skeptic could still argue that light keeps its speed relative to the lamp that sent it, like a bullet from a gun. Then a moving lamp would send faster light. In 1964 a team at CERN tested exactly that, with the fastest lamps available.
Ask Nature
Neutral pions are particles that decay into two flashes of gamma light. The CERN team made pions moving at \(v \approx 0.99975\) and timed the gamma flashes over a measured distance of a few tens of meters. The everyday adding rule predicts nearly 2. They measured the light’s speed to be 1, to within about 1 part in 10 000. A lamp moving at 99.975% of light speed sends light at light speed. T. Alväger, F. J. M. Farley, J. Kjellman and I. Wallin, Physics Letters 12, 260 (1964)
Myth
“The Michelson–Morley experiment is what led Einstein to relativity.” The honest answer: historians still argue about it. Einstein’s 1905 paper names no ether experiment; it mentions only, in passing, failed attempts to detect Earth’s motion through the light-carrying medium. Later, Einstein gave different accounts of how much this experiment mattered to him. What nobody disputes is the evidence: the ether wind isn’t there.
Practice
Three tiers: ◇ back of the envelope, ◆ pencil and paper, ★ head-scratcher. Speeds are fractions of light speed unless a unit says otherwise.
A passenger on a ferry gliding smoothly at a steady 8 m/s drops her keys from 1.5 m above the deck. Where do they land?
Hint
Before she lets go, how fast are the keys moving forward, as measured from the shore?
Solution
Right below her hand. The keys share the ferry’s 8 m/s forward speed and keep it as they fall; the deck keeps it too. On the ferry nothing differs from a ferry tied up at the dock.
A plane flies at 250 m/s through still air, and sound moves at 340 m/s relative to the air. Using the everyday adding rule, how fast does the plane’s own engine noise move ahead of the plane, measured from the plane?
That adds the speeds. The sound moves at 340 relative to the air, and the plane is chasing it at 250. How fast does the gap grow?
Hint
Measured from the ground (the air is still), the sound moves forward at 340 m/s and the plane at 250 m/s.
Solution
From the ground: sound at 340, plane at 250. From the plane, the sound pulls ahead at \(340 - 250 = 90\) m/s. A pilot could measure 90 ahead and \(340 + 250 = 590\) behind, and so learn the plane’s speed through the air. Sound has a medium; that’s why it gives you away.
A rocket coasting at 0.5 shines a lamp forward. What speed does the everyday adding rule predict for that light, measured in the lab?
That’s what Nature says, and well spotted. But the question asks what the everyday adding rule predicts: what is \(u + v'\)?
That’s the rocket’s own speed. The light leaves the rocket at 1; what does adding the rocket’s speed give?
Hint
The everyday adding rule: \(w = u + v'\), with \(u\) the rocket’s speed and \(v'\) the light’s speed measured on the rocket.
Solution
\(w = 0.5 + 1 = 1.5\). Nature says 1: the 1964 CERN pions, moving at 0.99975, sent light at 1, not nearly 2. The everyday adding rule fails for light, and Chapter 13 shows how to fix it for everything.
Ether universe. An interferometer arm is \(L = 11\) m, and Earth’s orbital speed is \(u = 10^{-4}\). The round trip along the wind exceeds the trip across it by about \(L u^2\). Find this in meters of time. Need a refresher?
That’s \(L u\), not \(L u^2\). Square the \(10^{-4}\) first.
Hint
\(u^2 = (10^{-4})^2 = 10^{-8}\).
Hint
Multiply by 11. You can type 11*1e-8.
Solution
\(L u^2 = 11 \times 10^{-8} = 1.1 \times 10^{-7}\) m of time. Exactly, the along trip is \(2L/(1-u^2) = 22.000\,000\,22\) and the across trip \(2L/\sqrt{1-u^2} = 22.000\,000\,11\), so the difference is \(1.1 \times 10^{-7}\). (The shortcut uses \(1/(1-x) \approx 1 + x\) and \(1/\sqrt{1-x} \approx 1 + x/2\) for tiny \(x\).) Your code task computes it exactly.
A sealed cabin on a ship that sails at a steady speed in a straight line, except where a card says otherwise. Can each experiment show the crew, from inside and with ordinary instruments, that the ship is moving or changing how it moves?
- The period of a pendulum
- Where a dropped ball lands
- The round-trip time of a light flash
- Coffee sloshing as the ship turns
- The cabin thermometer
Hint
Only one card changes how the ship moves. Which one involves a change of direction?
Solution
Only the sloshing coffee, and that’s because turning is not steady straight-line motion: it’s a change of direction, which you can feel. The pendulum, the ball, the light and the thermometer all behave as they do in harbor.
Two balls 10 m apart, side by side, are released in a freely falling room near Earth. In 1 s they drift together by about 8 μm, because of the tidal effect: \(\tfrac12 (g\,d/R)\,t^2 = 7.7\) μm. Is the room free-float to an accuracy of 1 mm over 1 s?
Hint
1 mm = 1000 μm.
Solution
Yes. The drift, about 8 μm, is less than a hundredth of the 1 mm accuracy. Over a minute the drift grows by \(60^2 = 3600\) times, to about 28 mm, and then the answer would be no. Free-float is always local.
Why is a freely falling elevator a better free-float frame than a room on the ground? Answer in two or three sentences.
Model answer
In a falling elevator everything falls together, so a released ball floats and no experiment inside detects gravity: it behaves like a room far out in space. A room on the ground isn’t free-float, because things in it fall to the floor. The falling elevator only works locally, though: over a large region or a long time, the tidal effect makes free things drift together.
- Everything inside falls together
- No experiment inside detects gravity locally
- It holds only over limited space and time
Finish the worked example. Ether universe, the same 6 m cabin with the lamp in the middle, but now \(u = 0.2\).
- Light toward the fore wall gains at \(1 - 0.2 = 0.8\), so the fore clock reads \(3 \div 0.8 = 3.75\).
- Light toward the aft wall closes at ____, so the aft clock reads ____.
What does the aft clock read?
That’s the fore clock. The aft wall moves toward the light: is the closing speed bigger or smaller than 1?
That’s our universe. In the ether universe the aft wall rushes toward the light at 0.2.
Hint
Closing speed \(1 + u\).
Solution
Closing speed \(1 + 0.2 = 1.2\); time \(3 \div 1.2 = 2.5\). Check with the crew’s rule: \(\dfrac{3.75 - 2.5}{3.75 + 2.5} = \dfrac{1.25}{6.25} = 0.2\). ✓
In an ether universe, a crew flashes a lamp in the middle of their cabin. The aft clock reads 2.5 and the fore clock 3.75 (meters of time). How fast does the ship move through the ether, and how long is the cabin? Need a refresher?
That’s the distance from the lamp to each wall: half the cabin.
That’s the difference of the readings. Divide it by their sum.
Hint
\(u = \dfrac{\text{fore} - \text{aft}}{\text{fore} + \text{aft}}\), from the worked example.
Hint
Once you know \(u\): the fore wall is a distance \(d\) away and the light gains on it at \(1 - u\), so \(d = 3.75 \times (1 - u)\).
Solution
\(u = (3.75 - 2.5)/(3.75 + 2.5) = 1.25/6.25 = 0.2\). Then \(d = 3.75 \times 0.8 = 3\) m from lamp to wall (check: \(2.5 \times 1.2 = 3\)), so the cabin is 6 m long.
A ship sails at 10 m/s. What is its speed \(u\) as a fraction of light speed? Scientific notation is fine, like 5e-7.
Close, but that’s rounded to one digit, which is 10% off. Give two or three significant figures.
That’s light speed divided by the ship’s. Which is the fraction of which?
Hint
Divide the ship’s speed by light’s speed, both in m/s.
Solution
\(u = 10 / 299\,792\,458 \approx 3.34 \times 10^{-8}\). Squared, that’s about \(10^{-15}\): any ether effect on a ship would be a millionth of a billionth. That’s why Michelson needed a whole planet’s speed, and waves of light as his ruler.
Which of these are measured facts, not assumptions?
Hint
Ask of each one: who measured it, and with what?
Solution
The first, second and fourth: everyday experience on smooth ships and planes, the 1964 CERN pions, and Michelson and Morley (and everyone since). The ether was an assumption that Nature declined, and 1.5 is the everyday adding rule applied where it fails.
★ The predicted slide grows like \(u^2\): 0.37 of a fringe for \(u = 10^{-4}\) (about 30 km/s). Michelson and Morley saw less than 0.02. What is the fastest ether wind, in km/s, that their result still allows? (Light speed is 299 792 km/s.)
That scales the speed by 0.02/0.37, as if the slide grew like \(u\). It grows like \(u^2\): you need a square root.
Hint
If the slide is proportional to \(u^2\), then \(u\) is proportional to the square root of the slide.
Hint
\(u_{\max} = 10^{-4}\times\sqrt{0.02/0.37}\). Then multiply by 299 792 km/s.
Solution
\(u_{\max} = 10^{-4}\sqrt{0.02/0.3735} \approx 10^{-4} \times 0.231 = 2.31 \times 10^{-5}\), which is \(2.31 \times 10^{-5} \times 299\,792 \approx 6.9\) km/s. Their 1887 runs took only a few days in summer. Could the ether have happened to drift along with Earth just then? Earth’s orbital velocity reverses every six months, a change of about 60 km/s, so half a year later the wind would have to be far above 7 km/s. Michelson and Morley never made those later runs, but repetitions since have, in every season, and modern ones push the limit far lower.
For the formulas in one place, print the formula card; the small-number shortcuts live in the math toolkit.
Build it
Time to build the ether universe’s stopwatch in code. Your function will predict Michelson’s fringe slide, and once it passes its tests, the interferometer above runs on your numbers, with Nature’s data point sitting right beside them.
Code you need
A function can return two answers at once, packed in an object with named parts: { area: …, perimeter: … }. And it can refuse a bad input with throw:
function rectangle(w, h) {
if (w < 0 || h < 0) throw new RangeError('sides must not be negative');
return { area: w * h, perimeter: 2 * (w + h) };
}
rectangle(3, 4) returns { area: 12, perimeter: 14 }, and rectangle(-1, 4) stops with the message. || means “or”, and Math.abs(x) is the size of x without its sign.
Here’s the physics, with friendly numbers: arm \(L = 1\), wind \(u = 0.6\).
- Along the wind. Out at \(1 - 0.6 = 0.4\): that takes \(1 \div 0.4 = 2.5\). Back at \(1 + 0.6 = 1.6\): \(1 \div 1.6 = 0.625\). Total \(3.125\).
- Across the wind. The light has to aim upstream to beat the drift, the way a swimmer crosses a river. Of its speed 1, it spends 0.6 cancelling the wind, which leaves \(\sqrt{1 - 0.6^2} = 0.8\) for going across. Each way takes \(1 \div 0.8 = 1.25\), so the total is \(2.5\).
For a general arm \(L\) and wind \(u\), the along trip is \(L/(1-u) + L/(1+u) = 2L/(1-u^2)\), and the across trip is \(2L/\sqrt{1-u^2}\). If \(|u| \ge 1\), the light heading upstream could never catch its mirror, so the trip would never end: your function should refuse.
The starter code’s comments repeat these steps: compute the along trip, then the across trip, and refuse \(|u| \ge 1\) with a RangeError.
Build it: etherRoundTrips(L, u)
Return { along, across }, the two round-trip times in meters of time for an arm of length L in an ether wind u. The tests check no wind (both 2L), the friendly 0.6 wind (3.125 and 2.5), an arm of 2 in a 0.8 wind (11.11 and 6.667), Michelson’s 11 m arm at 10⁻⁴, a refusal for u = 1, and that along is never shorter than across.
When your function passes, scroll back up to the interferometer: the chip in its toolbar says it’s running on your code, and the purple bar on the gauge is your prediction. Nature’s bar doesn’t move. That’s the point.
Checkpoint
Five scored items, then one to explain in your own words. Feedback waits until you submit, and you can retake it as often as you like.
A rocket coasting past the lab at 0.6 fires a laser pulse forward. Lin, in the lab, measures the pulse’s speed as…
Solution
1. Light’s speed is the same in every free-float frame, however fast its source moves (CERN, 1964).
A windowless cabin sails smoothly at a steady speed in a straight line. Which experiment inside can reveal that it’s moving?
Solution
None. Steady straight-line motion can’t be detected from inside a sealed room, with balls, pendulums or light.
Ether universe: a 6 m cabin moves through the ether at \(u = 0.25\), with the lamp in the middle. What does the fore clock read when the flash arrives?
That’s the aft clock: closing speed 1.25. The fore wall runs away from the light.
Solution
The light gains on the fore wall at \(1 - 0.25 = 0.75\), so it takes \(3 \div 0.75 = 4\).
Which statement about the Michelson–Morley experiment is accurate?
Solution
About 0.37 predicted (their paper rounds it to 0.4), less than 0.02 observed: no ether wind. Its role in Einstein’s thinking is historically contested.
Same in every free-float frame, or different from frame to frame?
- The speed of light
- The speed of a thrown ball
- The time a weight on a spring takes for one bounce, measured by a clock riding with it
- The shape of a dropped ball’s path
- The laws of motion
Solution
Same: light’s speed, the spring’s own bounce time and the laws of physics. (A spring works anywhere; a pendulum would need gravity, and in a free-float room it wouldn’t swing at all.) Depends on the frame: a ball’s speed (\(w = u + v'\)) and the shape of its path (straight down in the cabin, an arc from the dock).
In one or two sentences: what makes a room “free-float”, and why only locally?
Model answer
A free-float room is one in which free objects stay at rest or move steadily in straight lines, so no experiment inside detects gravity or motion; a freely falling room is one. It’s only local, because the tidal effect makes free objects drift together over large distances or long times.
- Free objects float or move steadily
- Gives an example (falling room, coasting spaceship)
- Mentions the tidal effect or a limited region and time
Explain it to Sagredo
Answer Sagredo, in a few sentences of your own.
Model answer
It’s moving relative to the dock, and the dock is moving relative to the Sun. But “moving” on its own, with nothing to compare against, has no meaning: no experiment can find an absolute rest, not with balls and not with light, because the laws of physics and the speed of light are the same in every free-float frame. So the honest answer is: moving relative to what?
- Says motion is always relative to something
- Mentions that no experiment, including light, detects absolute motion
- Names what is the same in every free-float frame
What’s invariant here?
- The speed of light, and the laws of physics, are the same in every free-float frame. \(\inv{c = 1}\) in the dock’s frame, the cabin’s or a planet’s, whether the lamp sits still or is a pion flying at 0.99975.
- Frame-dependent: speeds of balls (\(\lab{w} = \lab{u} + \rkt{v'}\), the everyday adding rule) and the shapes of paths.
- Free-float frames are local: to a stated accuracy, over a stated region and time.
Still strange?
It should be. A beam of light shot past you by a rocket at half its speed still passes you at full speed; something about our ideas of distance and time has to give, and over the next chapters you’ll find out exactly what.
Going further
Deep dive: could the ether hide?
After 1887 physicists tried to save the ether. One idea: maybe the arm moving along the wind shrinks by exactly the amount needed to cancel the delay. That happens to be the right amount of shrinking, and you’ll meet it as a real effect in Chapter 9, for a very different reason. But that trick leaves a loophole. If the arms are unequal, the difference between the two light-travel times still depends on how fast the lab moves, so as Earth’s speed changes through the day and the year, the fringes should slowly drift. In 1932 Kennedy and Thorndike built an interferometer with deliberately unequal arms, watched it for months, and saw no such drift. Their null result rules out the ether-plus-shrinking story and points to a second effect, slow clocks, which you’ll meet in Chapters 5 and 8.
Deep dive: how would an ether crew set their clocks?
In the cabin lab the two wall clocks started together when the lamp flashed. How did the crew make sure of that? In the ether universe time is the same for everyone, so they could set both clocks side by side in the middle and carry them slowly to the walls. In our universe that question turns out to be deep, but it doesn’t undo the lab’s result: clocks carried slowly from the middle to the walls agree with clocks set using light, so the tie is a real measurement, not a choice of how to set the clocks. Chapter 3 shows how to synchronize clocks with light, and Chapter 7 shows that two frames can’t agree on which clocks are in step.
★★ Ether universe, \(u = 0.6\). A lamp sits 1 m behind the fore wall of a cabin. A flash goes to a mirror on the fore wall and comes back to the lamp. How long does the round trip take, by the cabin’s clocks? Compare with etherRoundTrips(1, 0.6).
That’s the across trip. This flash goes along the wind: out against it, back with it.
That’s our universe. In the ether universe the outbound light gains on the mirror at only 0.4.
Hint
Out: the light gains on the mirror at \(1 - 0.6\). Back: it closes on the lamp at \(1 + 0.6\).
Solution
\(1 \div 0.4 + 1 \div 1.6 = 2.5 + 0.625 = 3.125\), which is etherRoundTrips(1, 0.6).along. In our universe it would be 2.
Reading: Galileo’s own cabin, on the Second Day of his Dialogue Concerning the Two Chief World Systems (1632), is still a pleasure to read. Taylor and Wheeler’s Spacetime Physics (2nd edition, 1992) gives the principle of relativity a whole chapter.