Teaching

1. 1

   Betting question

   5 min

   Lightning strikes both ends of a moving train. Seen from the embankment, the two strikes happen at the same time. A passenger sits exactly in the middle of the train. What does she see?

   • ABoth flashes at the same moment.
   • BThe flash from the front of the train first.
   • CThe flash from the back of the train first.

   Solution for the teacher

   Correct answer: B

   She moves towards the light from the front and away from the light from the back, and the light reaches her at the same speed c from both sides. So the front flash arrives first. Because she sits in the middle, she concludes that it must also have struck first: for her the strikes are not simultaneous.

2. 2

   Research task

   20 min

   On their own device, alone or in pairs. The same questions are on the worksheet.

   Open the experiment: Train and lightning The light-clockYour own twin journey

   1. Set the speed to 0 % c and press Start: when does M′ in the middle of the train see the two flashes? Then set 50 % c and compare.

      Answer

      At 0 % c both flashes arrive at the same moment (after 500 ns). At 50 % c, M′ sees the front flash B after about 334 ns and the rear flash A only after about 1,001 ns.

   2. Switch to “View from the train” and read the gap between the two strikes at 40 % c, then at 80 % c. Is the gap twice as big, less, or more?

      Answer

      At 40 % c about 437 ns, at 80 % c about 1,334 ns: a little over three times as big, not just twice. The gap grows faster than the speed because of the factor γ.

   3. Open the light clock and set 99.9 % c. How many years pass on Earth for every year on board?

      Answer

      About 22.4 years (γ = 22.4). At 50 % c it is only 1.15 years.

3. 3

   Consolidation

   15 min

   Together at the board. Key statements:

   • The speed of light is the same for every observer. Experiments such as Michelson–Morley support this; it is the starting point of the theory.
   • “Simultaneous” is not absolute: two events at different places can be simultaneous for one observer and not for another. What we see is not what we measure, because the light travel time must be taken into account.
   • A moving clock ticks slower by the factor γ = 1/√(1 − v²/c²). In everyday life γ − 1 is tiny (at 100 km/h about 4 · 10⁻¹⁵); at 99.9 % of c it is 22.4.
   • Time dilation has been measured many times: fast muons from the upper atmosphere, atomic clocks on aeroplanes (Hafele and Keating, 1971), clocks in satellites. The relativity of simultaneity belongs to the same theory and enters all of these calculations.

   For the curious: the calculation

   Light clock: a light pulse bounces between two mirrors a distance d apart. At rest one tick lasts Δt₀ = 2d/c. If the clock moves sideways at speed v, the pulse takes a slanted path. In the time Δt of one tick it covers c·Δt while the clock moves v·Δt. Pythagoras for one leg gives (c·Δt/2)² = d² + (v·Δt/2)². With d = c·Δt₀/2 this gives Δt = Δt₀ / √(1 − v²/c²) = γ·Δt₀.

4. 4

   Transfer

   5 min

   One question for the end, or for homework.

   You fly to Sirius (8.6 light-years away) and back at 99.9 % of the speed of light. How long does the trip take for your twin on Earth, and how long for you? (Use the twin-journey experiment.)

   Answer

   About 17.2 years on Earth and about 0.77 years (roughly 9 months) on board. Simplified: instant turn-around, acceleration left out.

To think furtherLearning journey: Journey 1: What does “at the same time” mean?The Bern patent office

1. 1

   Betting question

   5 min

   You are in a windowless cabin. Is it standing on Earth, or is a rocket in empty space pushing it upwards at exactly 9.81 m/s²? Which statement is correct?

   • AA ball-drop experiment inside the cabin shows which situation it is.
   • BNo experiment inside a small cabin can tell the two apart.
   • COnly a beam of light would tell them apart.

   Solution for the teacher

   Correct answer: B

   In both situations a dropped ball reaches the floor with 9.81 m/s², and light behaves the same way too. This was Einstein’s idea of 1907: locally, gravity and acceleration cannot be distinguished. (A very large cabin would show tidal differences, which is why the statement applies to small cabins.)

2. 2

   Research task

   20 min

   On their own device, alone or in pairs. The same questions are on the worksheet.

   Open the experiment: Two cabins, one result The dent in the sheetStars at the edge of the Sun

   1. Press “Drop the ball”. What happens in cabin A (on Earth) and in cabin B (rocket)? Compare the two accelerations.

      Answer

      In both cabins the ball falls to the floor at 9.81 m/s². From inside there is no difference.

   2. Switch on the light. What does the beam do in each cabin? Why does the drawing say “exaggerated”?

      Answer

      In both cabins the beam bends down equally. In a 2 m cabin at 1 g it really sinks only about 2 · 10⁻¹⁶ m, because light needs just about 7 ns for the distance. The drawing exaggerates it.

   3. Press “Open the windows”. Which cabin stands on Earth and which is accelerated in space? Could you have found out from inside?

      Answer

      Cabin A stands on Earth, cabin B is accelerated by a rocket at 1 g in empty space. With the windows closed, nothing inside a small cabin tells them apart.

3. 3

   Consolidation

   15 min

   Together at the board. Key statements:

   • Heavy and light bodies fall equally fast when air resistance is absent. On the Moon in 1971 (Apollo 15) a hammer and a feather hit the ground together.
   • This works because the “heavy” mass that gravity pulls on and the “inert” mass that resists acceleration are equal. This has been tested to a precision of about 10⁻¹⁵ (MICROSCOPE satellite, final report 2022).
   • Einstein concluded: in a small region, free fall removes gravity, and standing on the ground is like being accelerated upwards. Later he called this thought the happiest of his life.
   • Consequence: if light bends in the accelerated cabin, it must bend in a gravitational field, too. In 1919 star positions near the Sun showed this (see the eclipse experiment). The rubber-sheet picture shows only curved space; for slow bodies the curvature of time matters even more.

   For the curious: the calculation

   How much does the beam sink? Cabin 2 m wide: the light needs t = 2 m / c = 6.7 ns. In the accelerated cabin the floor moves up by s = ½ · g · t² = ½ · 9.81 m/s² · (6.7 · 10⁻⁹ s)² ≈ 2.2 · 10⁻¹⁶ m, less than the size of an atomic nucleus. Bending of light is therefore tiny near the Earth and measurable only near very large masses such as the Sun.

4. 4

   Transfer

   5 min

   One question for the end, or for homework.

   Astronauts on the space station float, yet the station is only about 400 km above the ground. Is there no gravity up there? What does “floating” really mean?

   Answer

   Gravity there is still about 8.7 m/s², roughly 89 % of the value on the ground. The station and the astronauts are in free fall around the Earth, together. In free fall, as in the falling lift, you feel no weight.

To think furtherLearning journey: Journey 2: Why does everything fall alike?The Bern patent office

1. 1

   Betting question

   5 min

   A very bright red lamp frees no electrons from a metal plate. You swap it for a very dim violet lamp. What happens?

   • AStill nothing: the lamp is far too weak.
   • BA few electrons leave immediately, but only a few.
   • CElectrons leave only after a delay, once the metal has collected enough energy.

   Solution for the teacher

   Correct answer: B

   Violet light carries more energy in each portion (about 3.1 eV) than red light (about 1.8 eV). For a metal like sodium, which needs about 2.3 eV per electron, one violet portion is enough and one red portion never is. Brightness only decides how many portions arrive, so a dim violet lamp frees few electrons, but at once.

2. 2

   Research task

   20 min

   On their own device, alone or in pairs. The same questions are on the worksheet.

   Open the experiment: Light vs. electrons

   1. Move the frequency slider from low to high. At which frequency do electrons leave the metal for the first time?

      Answer

      Between 5.5 and 5.6 × 10¹⁴ Hz. At 5.6 the display changes to “Electrons ejected” and shows 0.02 eV.

   2. Read the kinetic energy of the electrons at 8 × 10¹⁴ Hz and at 12 × 10¹⁴ Hz. How does it change with the frequency?

      Answer

      1.01 eV and 2.66 eV. The higher the frequency, the more energy the electrons get: about 0.41 eV for every additional 10¹⁴ Hz.

   3. Calculate the energy of one light portion at 8 × 10¹⁴ Hz with E = h · f (h = 4.14 · 10⁻¹⁵ eV·s). Subtract the work function of 2.3 eV. Does your result match the display?

      Answer

      E = 4.14 · 10⁻¹⁵ eV·s · 8 · 10¹⁴ Hz = 3.31 eV. 3.31 eV − 2.3 eV = 1.01 eV, exactly what the display shows.

3. 3

   Consolidation

   15 min

   Together at the board. Key statements:

   • The wave picture predicts: brighter light carries more energy, so any colour should free electrons if only it is bright enough, possibly after a delay. The measurements show something else: below a threshold frequency nothing happens, however bright the light (at least at ordinary intensities); above it, electrons leave at once.
   • Brightness changes only the number of electrons (the current), not their maximum energy. The maximum energy depends only on the frequency.
   • Einstein’s explanation (1905): light arrives in portions, each with the energy E = h · f. One portion frees at most one electron: E(kin, max) = h · f − W, where W is the work function. Below f₀ = W / h every portion is too small.
   • Millikan tested this equation in experiments (about 1914 to 1916). Einstein received the Nobel Prize in Physics of 1921 chiefly for the law of the photoelectric effect, not for relativity. Light still shows wave behaviour in interference and diffraction: it shows both.

   For the curious: the calculation

   E = h · f with h = 6.626 · 10⁻³⁴ J·s = 4.136 · 10⁻¹⁵ eV·s (1 eV = 1.602 · 10⁻¹⁹ J). Threshold for W = 2.3 eV: f₀ = W / h = 2.3 eV / 4.136 · 10⁻¹⁵ eV·s ≈ 5.6 · 10¹⁴ Hz, wavelength λ = c / f₀ ≈ 540 nm (green). A shortcut: E in eV = 1240 / λ in nm. Red light (700 nm) has 1.8 eV, violet (400 nm) 3.1 eV, UV-B (300 nm) 4.1 eV.

4. 4

   Transfer

   5 min

   One question for the end, or for homework.

   Why can ultraviolet light damage skin cells chemically, while the light of a red lamp cannot? (Hint: the photoelectric idea.)

   Answer

   Breaking a chemical bond takes a minimum energy per portion, a few eV. A UV-B portion has about 4.1 eV, a red portion only about 1.8 eV, and with ordinary light intensities red portions do not add up to one break. Strong red light can warm the skin, but that is a different effect.

To think furtherLearning journey: Journey 3: What is light made of?

1. 1

   Betting question

   5 min

   A paperclip has a mass of 1 gram. Suppose its whole mass could be turned into electrical energy. How much would that be?

   • AAbout 25 kilowatt-hours.
   • BAbout 25,000 kilowatt-hours.
   • CAbout 25 million kilowatt-hours.

   Solution for the teacher

   Correct answer: C

   E = m · c² = 0.001 kg · (3 · 10⁸ m/s)² ≈ 9 · 10¹³ J. With 1 kWh = 3.6 · 10⁶ J that is about 25 million kWh. The factor c² is so large that even a tiny mass stands for a gigantic amount of energy. In practice no one can convert a paperclip completely.

2. 2

   Research task

   20 min

   On their own device, alone or in pairs. The same questions are on the worksheet.

   Open the experiment: Energy calculator

   1. Press the chip “Paperclip (1 g)”. Write down the energy in joules and in kilowatt-hours.

      Answer

      About 8.99 · 10¹³ J, which is about 25 million kWh (the display says 24,965,422 kWh).

   2. Now press “A person (62 kg)”. By what factor is the energy larger than for the paperclip? Why exactly this factor?

      Answer

      By the factor 62,000 (5.57 · 10¹⁸ J), because the energy is directly proportional to the mass, and 62 kg are 62,000 g.

   3. The experiment compares with the Hiroshima bomb (about 6.3 · 10¹³ J). Change the mass until the comparison reads about “1×”. How much mass is that?

      Answer

      About 0.7 g: m = E / c² = 6.3 · 10¹³ J / 9.0 · 10¹⁶ m²/s² ≈ 0.7 g. The mass of the material before and after the explosion differed by less than a gram.

3. 3

   Consolidation

   15 min

   Together at the board. Key statements:

   • Every body at rest has a rest energy E₀ = m · c². Because c² is huge (9 · 10¹⁶ m²/s²), a little mass corresponds to a lot of energy.
   • The equation does not say that mass “disappears”. It says: if a system gives off energy, its mass becomes smaller by Δm = ΔE / c², and what is given off (light, heat) carries that share away.
   • Burning coal changes the mass by about one part in three billion, far too little to weigh. In nuclear reactions it is measurable: about 0.1 % when uranium splits, about 0.7 % when hydrogen fuses to helium in the Sun.
   • The Sun radiates 3.8 · 10²⁶ W. That costs it about 4.3 million tonnes of mass every second, and it has done so for billions of years.

   For the curious: the calculation

   E = m · c² with c = 299,792,458 m/s, so c² = 8.99 · 10¹⁶ m²/s². For 1 g = 0.001 kg: E = 8.99 · 10¹³ J. One kilowatt-hour is 3.6 · 10⁶ J, so E ≈ 2.5 · 10⁷ kWh. For the Hiroshima comparison: 15 kilotons of TNT are about 6.3 · 10¹³ J (1 kt = 4.184 · 10¹² J), so m = 6.3 · 10¹³ J / 8.99 · 10¹⁶ m²/s² ≈ 7.0 · 10⁻⁴ kg = 0.70 g.

4. 4

   Transfer

   5 min

   One question for the end, or for homework.

   The Sun radiates 3.8 · 10²⁶ watts. How many tonnes of mass does it lose through radiation every second? (Rearrange E = m · c² for the mass.)

   Answer

   Δm per second = 3.8 · 10²⁶ W / 9.0 · 10¹⁶ m²/s² ≈ 4.2 · 10⁹ kg, a little over 4 million tonnes per second (with the more precise values 3.83 · 10²⁶ W and 8.99 · 10¹⁶ m²/s² it is 4.26 · 10⁹ kg). Compared with the Sun’s mass of about 2 · 10³⁰ kg that is very little, so it can keep shining for billions of years.

To think furtherLearning journey: Journey 3: What is light made of?

1. 1

   Betting question

   5 min

   GPS satellites circle at about 20,200 km height and 14,000 km/h. Because of relativity their clocks run differently from clocks on the ground. By how much do they gain per day?

   • AAbout 38 nanoseconds.
   • BAbout 38 microseconds (millionths of a second).
   • CAbout 38 milliseconds (thousandths of a second).

   Solution for the teacher

   Correct answer: B

   Net about +38.6 microseconds per day. That looks tiny, but light covers about 11.6 km in this time, and GPS works by timing signals. Two effects add up: the satellite’s speed makes its clock slower (−7.2 µs per day), the weaker gravity at that height makes it faster (+45.8 µs per day).

2. 2

   Research task

   20 min

   On their own device, alone or in pairs. The same questions are on the worksheet.

   Open the experiment: GPS without the relativity correction The light-clock

   1. Leave both switches on and set 1 day. Read the clock difference and the distance below it.

      Answer

      +38.6 µs and 11.6 km (the distance light travels in that time).

   2. Switch off “weaker gravity” so that only the speed effect is left. What sign does the difference have? What does it mean for the satellite clocks?

      Answer

      −7.2 µs per day (about 2.2 km of light path): through its speed alone the satellite clock would run slow, as a moving clock does.

   3. Switch both effects on again and drag the days to 30. How does the difference grow with the days?

      Answer

      In proportion: after 30 days +1,158 µs and about 347 km. The clocks keep drifting apart by the same amount every day.

3. 3

   Consolidation

   15 min

   Together at the board. Key statements:

   • A receiver measures how long the signals from several satellites travel. Light covers about 30 cm in one nanosecond, so the clocks must be accurate to a few nanoseconds.
   • Two effects of opposite sign: a moving clock ticks slower (special relativity, −7.2 µs per day); a clock in weaker gravity ticks faster (general relativity, +45.8 µs per day). Net: the satellite clock gains 38.6 µs per day.
   • That is why engineers set the satellite clocks before launch to tick slightly slower: 10.23 MHz becomes 10.22999999543 MHz (Ashby, 2003). On the ground they then agree with Earth time.
   • GPS is thus an everyday case in which both of Einstein’s theories are used, and every navigation app relies on both.

   For the curious: the calculation

   Light path of the clock difference: c · Δt = 299,792 km/s · 38.6 · 10⁻⁶ s ≈ 11.6 km per day. Frequency offset: 38.6 µs / 86,400 s ≈ 4.47 · 10⁻¹⁰; 10.23 MHz · (1 − 4.47 · 10⁻¹⁰) = 10.22999999543 MHz. Note: 11.6 km is the light path belonging to the clock error, not an exact position error, because a receiver removes an error that all its satellites share. It shows the order of magnitude and why the correction is built in.

4. 4

   Transfer

   5 min

   One question for the end, or for homework.

   A GPS signal travels 20,200 km from a satellite straight overhead. How long does it take? How far does light go in one nanosecond? What does that mean for the clocks if you want your position to within a few metres?

   Answer

   20,200 km / 299,792 km/s ≈ 0.067 s. In 1 ns light covers about 0.3 m, so a position within a few metres needs timing good to about ten nanoseconds.

To think furtherLearning journey: Journey 1: What does “at the same time” mean?

1. 1

   Betting question

   5 min

   In August 1939 a letter signed by Albert Einstein reached US President Roosevelt. Which statement is correct?

   • AEinstein worked for years on the American atomic bomb project.
   • BLeo Szilárd drafted the letter and Einstein signed it. Einstein was not part of the later bomb project.
   • CEinstein wrote the letter alone and led the bomb project.

   Solution for the teacher

   Correct answer: B

   The letter of 2 August 1939 was drafted largely by the physicist Leo Szilárd and signed by Einstein. It warned that bombs of a new type might become possible and that Germany might be working on them. Einstein himself did not work on the Manhattan Project; the US Army regarded the well-known pacifist as a security risk.

2. 2

   Research task

   20 min

   On their own device, alone or in pairs. The same questions are on the worksheet.

   Open the experiment: The letter in full

   1. Read the first paragraph of the letter. Whose work does Einstein name as the reason for writing, and which use of uranium does he mention first?

      Answer

      He names recent work by E. Fermi and L. Szilárd. The first use is a new and important source of energy. Bombs appear only in the third paragraph.

   2. Find the paragraph about bombs. How certain does the letter sound? How does it expect such a bomb to be transported?

      Answer

      It calls such bombs “conceivable”, though “much less certain” than the chain reaction itself. It expects a bomb carried by boat and exploded in a port, and says such bombs might be too heavy to carry by air. In 1945 the bombs were dropped from aircraft, so the letter was wrong on this point: the knowledge of 1939 was uncertain.

   3. Open the note on Germany on page 2. Which two things does the letter report about Germany?

      Answer

      Germany had stopped the sale of uranium from the Czechoslovak mines it had taken over. And the son of the German state secretary, von Weizsäcker, worked at the Kaiser Wilhelm Institute in Berlin, where some of the American uranium work was being repeated.

3. 3

   Consolidation

   15 min

   Together at the board. Key statements:

   • Spring 1933: after Hitler came to power, Einstein, who was in the USA, did not return to Germany. On 28 March 1933 he landed in Antwerp and resigned from the Prussian Academy. In October 1933 he arrived in Princeton. A law of 7 April 1933 allowed civil servants, including university teachers, to be dismissed for being Jewish or politically unwanted.
   • Nuclear fission was discovered in December 1938 (Hahn and Strassmann) and explained soon after (Meitner and Frisch). In summer 1939 Szilárd feared that Germany might be ahead and asked Einstein to lend his name and his standing.
   • The letter shows how uncertain the knowledge was: it speaks of what is “conceivable” and was wrong about the transport. In 1945 atomic bombs destroyed Hiroshima and Nagasaki.
   • Einstein later spoke out against nuclear weapons and signed the Russell–Einstein Manifesto days before his death in April 1955. Whether and when scientists must warn, stay silent or refuse is a question without one correct answer; it can be argued from sources.
4. 4

   Transfer

   5 min

   One question for the end, or for homework.

   Researchers rarely know what others will later do with their results. Pick a present-day example (for instance artificial intelligence, gene editing or climate research). Which two questions should researchers ask before they publish?

   Answer

   There is no single correct answer. Sensible questions: Who could use this, and for what? Who could be harmed, and how likely is it? Are there ways to reduce the harm? Who decides, and who is asked? Accept any answer that gives reasons, and compare with 1939: Szilárd chose to warn the government.

To think furtherExile & PrincetonThe Letter & the Bomb