Special Relativity — Student Workbook

1. The postulates of special relativity

Galilean relativity assumes that if you add the speed of a moving source to the speed of whatever it fires out, you get the total speed as seen by someone standing still. Fire a laser from a spaceship moving at half the speed of light, and Galilean relativity predicts the light should now travel at 1.5 times its normal speed. But when this was tested — most famously in the Michelson–Morley experiment of 1887 — no such change was ever detected. The speed of light stubbornly refused to add up the way Newtonian physics said it should.

In 1905, Einstein resolved this by starting from two simple statements, or postulates, and working out what must logically follow from them.

Einstein's postulates of special relativity:

The laws of physics are the same in all inertial reference frames.
The speed of light in a vacuum is measured to be the same by all observers in inertial reference frames, regardless of the motion of the light source or the observer.

The first postulate is really just Galilean relativity again — no inertial observer's frame is any more "correct" than any other. The second is the strange, new idea: however fast you are moving, and however fast the thing that emitted the light was moving, you will always measure light in a vacuum to travel at exactly c = 3.00 × 10⁸ m s⁻¹.

Fig. 1.1 Ship A and ship B move away from the Sun at different speeds. Each is an equally valid inertial reference frame, so an observer on either ship — and an observer who stayed behind at the Sun — all measure the speed of any light beam they observe as exactly c.

Worth remembering Postulate 2 is not really a separate, independent assumption — it follows logically from postulate 1. Maxwell's equations, which describe how electric and magnetic fields behave, predict the speed of light from two physical constants (the permittivity and permeability of free space).

c = 1 / √(ε₀μ₀) where ε₀ is the permittivity of free space and μ₀ is the permeability of free space

Since these are the same constants everywhere, Maxwell's equations are a "law of physics" — so if postulate 1 is true and the laws of physics really are the same in every inertial frame, the speed of light they predict, c, must be measured as the same in every inertial frame too.

Check your understanding

1State Einstein's two postulates of special relativity, in your own words.

Model answer 1. The laws of physics are the same in every inertial reference frame. 2. The speed of light in a vacuum is measured as the same by every observer in an inertial reference frame, no matter how the light source or the observer are moving.

2Using Figure 1.1, explain why an observer on ship A and an observer on ship B both measure the same beam of sunlight to be travelling at c, even though A and B are moving away from the Sun (and from each other) at different speeds.

Think about postulate 2 directly — it doesn't matter how fast the observer, or the source of the light, is moving.

Model answer By Einstein's second postulate, the speed of light in a vacuum is measured as c by every observer in an inertial reference frame, regardless of their own velocity or the velocity of the source. So although A and B are moving at different speeds relative to the Sun and to each other, each of them individually measures the same beam of light to be travelling at exactly c.

3Explain briefly why the Michelson–Morley experiment was significant for the development of special relativity.

Model answer The experiment tried to detect a difference in the speed of light depending on the direction it was measured in, as Newtonian physics predicted there should be. No such difference was found: the speed of light appeared constant regardless of the observer's motion. This result could not be explained by Galilean relativity, and it was evidence that led Einstein towards his second postulate.

2. The Lorentz transformation equations

Recall the Galilean transformation equations from the last workbook: x′ = x − vt and t′ = t. These work because Newtonian physics assumes every observer agrees on time. But Einstein's second postulate breaks that assumption — if every observer measures the same speed of light, then their measurements of distance and time travelled by that light cannot both be the ordinary, Galilean kind. Something has to give.

Working through the consequences of the two postulates, Einstein arrived at a new pair of transformation equations — the Lorentz transformation equations — which replace the Galilean ones at any speed, and reduce to the familiar Galilean equations when v is small compared with c.

x′ = γ(x − vt) t′ = γ(t − vx/c²) where v is the velocity of frame S′ relative to frame S, c is the speed of light, and γ (gamma) is the Lorentz factor below

γ = 1 / √(1 − v²/c²) the Lorentz factor — always greater than or equal to 1, and equal to 1 only when v = 0

You are not expected to derive these equations — only to use them, and to understand what γ tells you. Notice that when v is very small compared with c, v²/c² is almost zero, γ is almost exactly 1, and the Lorentz equations collapse back into the Galilean equations x′ = x − vt and t′ = t. This is why Galilean relativity works so well for everyday speeds: it is simply the v ≪ c approximation of the full, Lorentz picture.

Fig. 2.1 The Lorentz factor γ stays close to 1 until v/c gets close to 1, then rises very steeply. At everyday speeds (v/c well below 0.1), γ is indistinguishable from 1 and relativistic effects are negligible.

Worked example 2.1

Calculate the Lorentz factor γ for a spacecraft travelling at v = 2.4 × 10⁸ m s⁻¹ (c = 3.00 × 10⁸ m s⁻¹).

γ = 1 / √(1 − v²/c²)

γ = 1 / √(1 − (2.4×10⁸)²/(3.00×10⁸)²)

γ = 1 / √(1 − 0.64) = 1 / √0.36 = complete this step yourself

Worked example 2.2

In frame S, an event occurs at x = 240 m and t = 2.0 × 10⁻⁶ s. Frame S′ moves at v = 1.8 × 10⁸ m s⁻¹ relative to S, with their origins coinciding at t = t′ = 0. Given γ = 1.25 for this speed, find x′.

x′ = γ(x − vt)

x′ = 1.25 × (240 − (1.8×10⁸ × 2.0×10⁻⁶))

x′ = 1.25 × (240 − 360) = 1.25 × (−120) = complete this step yourself

Check your understanding

4Explain why the Galilean transformation equations are only an approximation, and state the condition under which they become accurate.

Model answer The Galilean equations assume every observer agrees on time (t′ = t exactly), which is not true once v²/c² becomes significant. They are the v ≪ c approximation of the full Lorentz transformation — accurate whenever the relative velocity v is very small compared with the speed of light c, so that γ ≈ 1.

5Calculate the Lorentz factor γ for a particle travelling at v = 0.90c.

Since v is already given as a fraction of c, substitute v/c = 0.90 directly — the c's cancel and you don't need the value of c at all.

Model answer

γ = 1 / √(1 − v²/c²) = 1 / √(1 − 0.90²)

γ = 1 / √(1 − 0.81) = 1 / √0.19 = (finish evaluating this yourself)

6In frame S, an event occurs at x = 500 m at t = 1.0 × 10⁻⁶ s. Frame S′ moves at v = 1.5 × 10⁸ m s⁻¹ relative to S (γ = 1.15 for this speed). Calculate x′.

Model answer

x′ = γ(x − vt)

x′ = 1.15 × (500 − (1.5×10⁸ × 1.0×10⁻⁶))

x′ = 1.15 × (500 − 150) = 1.15 × 350 = (finish evaluating this yourself)

3. Proper time and proper length

Because different inertial observers can measure different values for the time between two events, and for the length of an object, physicists give a special name to the measurement made by the one observer for whom the measurement is as simple as possible.

Proper time interval, Δt₀: the time interval between two events as measured by an observer for whom both events happen at the same position — that is, an observer who does not need to move between the two events to witness them both.

Proper length, L₀: the length of an object as measured by an observer who is at rest relative to that object — that is, in the same reference frame as the object itself.

Every other inertial observer, moving relative to the events or the object, will measure a longer time interval than the proper time, and a shorter length than the proper length. The proper time is always the smallest possible time interval between two events, and the proper length is always the largest possible length of an object.

Common mistake "Proper" here does not mean "correct" or "official" — every inertial observer's measurement is equally valid. It simply identifies which observer is in the most convenient position: at rest relative to the two events (for proper time), or at rest relative to the object (for proper length).

Check your understanding

7A muon is created high in the atmosphere and decays a short time later, closer to the ground. One observer travels down with the muon; a second observer stays fixed on the ground and watches the muon pass. Which observer measures the proper time interval between the muon's creation and its decay? Explain your answer.

The proper time is measured by whoever sees both events (creation and decay) happen at the same position as themselves.

Model answer The observer travelling down with the muon measures the proper time, because both events — the muon's creation and its decay — happen at the same position relative to that observer (right where the muon is). The ground observer sees the muon created at one position and decaying at a different position, so they measure a longer, non-proper time interval.

8A metre rule is carried on board a spaceship. Which observer measures its proper length: an astronaut on the spaceship, or a scientist watching the spaceship fly past from a planet? Explain your answer.

Model answer The astronaut on the spaceship measures the proper length, because the astronaut is at rest relative to the metre rule (they are in the same reference frame). The scientist on the planet is moving relative to the rule, so they measure a shorter, contracted length.

9Two firework shells explode at different points along a straight firing range, one after the other. An engineer stands exactly halfway between the two explosion points and does not move. A drone flies at constant velocity from directly above the first explosion point towards the second, arriving just as the second shell explodes. Which of these two observers could, in principle, measure the proper time interval between the two explosions? Explain your answer.

Model answer Neither observer measures the proper time interval, because neither is present at the same position as both explosions: the engineer is at a fixed point between the two, not at either explosion, and the drone is at a different position for each explosion too. The proper time interval could only be measured by an observer who is at the position of the first explosion when it happens, and also at the position of the second explosion when it happens.

4. Time dilation

We can now combine the second postulate (light always travels at c) with the definition of proper time to work out exactly how much longer a non-proper time interval is. A neat way to see this is with a thought experiment called a light clock: two mirrors, facing each other, with a single pulse of light bouncing between them. Every time the light completes one bounce, the clock "ticks".

Rachel is on board a rocket, holding a light clock oriented so the light bounces straight up and down. To Rachel, the light simply travels straight up to the top mirror and straight back down — a distance she calls 2L, taking a proper time Δt₀ (because, to Rachel, both the "up" event and the "down" event happen at the same position: right in front of her).

Mateo is floating in his own inertial frame, watching Rachel's rocket fly past at speed v. From Mateo's point of view, the light does not simply go straight up and down — because the whole clock (and its mirrors) are moving sideways with the rocket. So Mateo sees the light trace out a longer, zig-zag path.

Simulation — the light clock

Rocket velocity, v = 0.60c

Proper time (Rachel, on the rocket), Δt₀ = 1.00 tick-units

Time Mateo measures, Δt = γΔt₀ = 1.25 tick-units

Lorentz factor, γ = 1.25

Interactive simulation — open the online version of this workbook to use it.

Fig. 4.1 The light clock: Rachel (top path, moving with the rocket) always sees the light travel straight up and down. Mateo (bottom path, watching from outside) sees the same light trace a longer zig-zag, because the mirrors themselves are moving. Since both must measure the same speed of light, c, but Mateo's path is longer, Mateo's clock reading must be longer too.

Because both Rachel and Mateo must measure the light travelling at exactly the same speed, c (postulate 2), and Mateo's path is geometrically longer, Mateo must measure a longer time for one tick of the clock. Using Pythagoras' theorem on the zig-zag triangle gives exactly the Lorentz factor, γ, as the ratio between the two:

Δt = γΔt₀ where Δt₀ is the proper time interval (measured by an observer for whom both events happen at the same place) and Δt is the (longer) time interval measured by any other inertial observer

Time dilation: the effect by which a moving clock is measured to run slow — that is, any observer who is moving relative to a clock measures a longer time between its ticks than an observer at rest relative to the clock does.

Exam hint Δt₀ (proper time) is always the smaller of the two values, and Δt is always the larger. If a question gives you a speed as a fraction of c (for example 0.80c), it is usually easiest to substitute 0.80c directly for v in the equation for γ — the two c's cancel, so you never need to look up the value of c at all.

Want to try a different, independently built version of the light clock? This one lets you set the rocket's velocity directly and watch the light bounce in real time.

Launch the light clock simulation ↗

Simulation by Michael Fowler, University of Virginia (galileoandeinstein.phys.virginia.edu).

Worked example 4.1

A spacecraft passes the Earth at v = 0.80c. A clock on board the spacecraft measures a proper time interval of Δt₀ = 5.0 s between two events that happen at the same place on the spacecraft. Calculate the time interval Δt measured by an observer on Earth.

γ = 1 / √(1 − v²/c²) = 1 / √(1 − 0.80²) = 1 / √0.36 = 1.67

Δt = γΔt₀

Δt = 1.67 × 5.0 = complete this step yourself

Check your understanding

10A muon, created in the upper atmosphere, has a proper lifetime of Δt₀ = 2.2 μs before it decays. It travels towards the Earth at v = 0.995c. Calculate the lifetime Δt of the muon as measured by an observer on the ground.

First find γ using v/c = 0.995, then use Δt = γΔt₀.

Model answer

γ = 1 / √(1 − 0.995²) = 1 / √(1 − 0.990025) = 1 / √0.009975 ≈ 10.0

Δt = γΔt₀ = 10.0 × 2.2×10⁻⁶ = (finish evaluating this yourself)

11A spaceship passes the Earth at 2.7 × 10⁸ m s⁻¹. A signal light on Earth is switched on for a proper time of 1.4 × 10⁻⁵ s, as measured on the Earth. For how long is the light on, as measured by an observer on the spaceship? (c = 3.00 × 10⁸ m s⁻¹)

Be careful — here it is the Earth-based observer who is at rest relative to the two events (the light switching on and off), so it is the Earth time that is the proper time, Δt₀.

Model answer

v/c = 2.7×10⁸ / 3.00×10⁸ = 0.90

γ = 1 / √(1 − 0.90²) = 1 / √0.19 ≈ 2.29

Δt = γΔt₀ = 2.29 × 1.4×10⁻⁵ = (finish evaluating this yourself)

12The average lifetime of muons at rest in a laboratory is measured as 2.2 μs. A separate beam of fast-moving muons, produced in the same laboratory, is measured to have an average lifetime of 6.4 μs. Calculate the speed of these muons as a fraction of c.

Find γ first (γ = Δt / Δt₀), then rearrange γ = 1/√(1 − v²/c²) for v. Remember to square both γ and the ratio correctly when you rearrange — it is easy to forget to square one of the terms.

Model answer

γ = Δt / Δt₀ = 6.4 / 2.2 = 2.91

γ = 1/√(1 − v²/c²) → v²/c² = 1 − 1/γ² = 1 − 1/2.91²

v/c = √(1 − 1/2.91²) = (finish evaluating this yourself)

13Using Figure 4.1 and the light clock simulation, explain in your own words why Mateo (the outside observer) must measure a longer time for one tick of the clock than Rachel does, without using any equations.

Model answer Mateo sees the light travel further for each tick, because the mirrors are moving sideways with the rocket while the light also moves up and down, so the light traces a longer, diagonal path. Both Rachel and Mateo must measure the light travelling at the same speed, c (postulate 2). If the distance Mateo measures is longer but the speed is the same for both of them, then the time Mateo measures for one tick must also be longer.

5. Length contraction

Time dilation has a natural partner. If a moving clock runs slow, then a moving ruler must also measure shorter — otherwise observers in different frames would disagree about basic quantities like speed itself. Using the Lorentz transformation equations to compare how two observers measure the length of the same object leads directly to the length contraction equation.

L = L₀ / γ where L₀ is the proper length of the object (measured at rest relative to it) and L is the (shorter) length measured by an observer moving relative to the object, at relative speed v

Length contraction: the effect by which a moving object is measured to be shorter, along its direction of travel, than its proper length. There is no contraction in directions perpendicular to the motion.

Notice that L is always smaller than L₀, since γ ≥ 1 — exactly the opposite pattern to time dilation, where the non-proper time Δt is always larger than Δt₀. It can help to remember: moving clocks run slow (time gets longer), moving rulers shrink (length gets shorter).

Simulation — length contraction

Spaceship velocity, v = 0.60c

Proper length, L₀ = 100 m

Lorentz factor, γ = 1.25

Observed length, L = L₀/γ = 80.0 m

Interactive simulation — open the online version of this workbook to use it.

Fig. 5.1 The top ship is shown at rest, with its full proper length L₀. The bottom ship is the same ship, travelling at speed v relative to the observer — its measured length, L, is contracted along the direction of travel.

Worked example 5.1

A spaceship has a proper length of L₀ = 100 m. It moves at v = 0.80c relative to an observer on a space station. Calculate the length of the spaceship as measured by the observer.

γ = 1 / √(1 − 0.80²) = 1 / √0.36 = 1.67

L = L₀ / γ

L = 100 / 1.67 = complete this step yourself

Check your understanding

14A train carriage has a proper length of 15 m. Observer B, standing on the platform, watches the train pass at a uniform speed of 0.80c. Calculate the length of the carriage as measured by observer B.

The 15 m given is the proper length, L₀ — it's measured by an observer (on the train) at rest relative to the carriage.

Model answer

γ = 1 / √(1 − 0.80²) = 1 / √0.36 = 1.67

L = L₀ / γ = 15 / 1.67 = (finish evaluating this yourself)

15A spaceship of length 60.0 m (as measured by its own crew) passes the Earth at a speed of 0.980c. Calculate its length as measured by an observer on Earth.

Model answer

γ = 1 / √(1 − 0.980²) = 1 / √(1 − 0.9604) = 1 / √0.0396 ≈ 5.03

L = L₀ / γ = 60.0 / 5.03 = (finish evaluating this yourself)

16A UFO is spherical when at rest. Describe, without calculation, what shape it would appear to an observer that it passed at a relative speed of 0.50c.

Length contraction only shortens an object along its direction of travel — nothing happens to the dimensions at right angles to the motion.

Model answer It would appear flattened into an oval (ellipsoid) shape: contracted along the direction of travel, but unchanged in the two directions perpendicular to the motion.

17An observer measures the length of a metre rule, orientated along the direction of travel, to be 80 cm. Calculate the speed of the rule relative to the observer.

Rearrange L = L₀/γ to find γ first (γ = L₀/L), then rearrange γ = 1/√(1 − v²/c²) for v, remembering to square both γ and the length ratio.

Model answer

γ = L₀ / L = 100 / 80 = 1.25

v/c = √(1 − 1/γ²) = √(1 − 1/1.25²)

v/c = √(1 − 0.64) = (finish evaluating this yourself, then multiply by c)

6. The relativity of simultaneity

Two events are simultaneous if they happen at the same time. In everyday life we take this for granted as an absolute fact — either two things happened at the same moment, or they didn't. But because different inertial observers can measure different time intervals (time dilation), it turns out they can also disagree about whether two events happened at the same time at all.

Imagine two bolts of lightning strike a fast-moving train at the same instant, at two points: A (the front of the train) and B (the back). An observer standing on the embankment, exactly halfway between A and B, sees both flashes of light arrive at the same moment — for this observer, the two strikes were simultaneous.

Now consider Carla, sitting exactly halfway between A and B — but on board the moving train. By the time the light from each strike reaches her, the train has moved. Because the train is travelling towards the position where the strike at A happened, the light from A has less distance left to travel to reach her; because the train is moving away from where B happened, the light from B has further to travel. So Carla sees the flash from A arrive first, and the flash from B arrive some time later.

Fig. 6.1 Lightning strikes A and B at the same instant, as measured by an observer on the embankment. The train (and Carla, sitting at its centre) is moving towards A, so light from A reaches her sooner than light from B — to Carla, the two strikes are not simultaneous.

The relativity of simultaneity: two events that are simultaneous in one inertial reference frame are, in general, not simultaneous in a different inertial reference frame that is moving relative to the first. Simultaneity is only absolute if the two events happen at exactly the same place.

Worth remembering This is not a trick of light delay that could be "corrected for" — it is a genuine disagreement about whether the two events happened at the same time. Both the embankment observer and Carla are correct, in their own reference frames. There is no absolute answer to "did A and B really happen at the same time?"

Check your understanding

18Using Figure 6.1, explain why the embankment observer (standing still, exactly halfway between A and B) measures the two lightning strikes as simultaneous.

Model answer The embankment observer is exactly halfway between A and B and is not moving, so the light from each strike has to travel exactly the same distance to reach them, at exactly the same speed, c. Since the strikes happened at the same instant and the light from each takes the same time to arrive, the observer sees both flashes arrive together — simultaneously.

19Suppose the train in Figure 6.1 were travelling in the opposite direction (towards B instead of A). State which strike Carla would now observe first, and explain why.

Think about which strike's light now has less distance to travel to reach Carla, given the new direction of travel.

Model answer Carla would now see strike B first. Since the train is moving towards B, the light from B has less distance to travel to reach her, while the light from A has to catch up to the train, taking longer to arrive.

20Explain why two events that happen at exactly the same position are simultaneous for every inertial observer, with no disagreement possible.

Model answer If two events happen at the same position, there is no distance for light (or any signal) from one event to travel further than light from the other to reach any given observer — there is no asymmetry in arrival time for any observer to disagree about, regardless of how that observer is moving. The relativity of simultaneity only arises when two events happen at different positions.

7. Evidence from muon decay

Time dilation and length contraction are not just abstract mathematics — they are tested every day by cosmic ray muons. Muons are unstable particles, produced when cosmic rays strike the upper atmosphere, roughly 10 km above the Earth's surface. They travel down towards the ground at speeds close to c, but they are unstable, decaying with a half-life of only about 1.5 μs.

Using ordinary (non-relativistic) physics, we can calculate roughly how long the journey to the ground should take, and therefore how many half-lives pass, and therefore what fraction of the original muons should survive to reach a detector at ground level. That classical calculation predicts that only a tiny fraction — often quoted as around one in a million — should survive the trip. In reality, a much larger fraction, often around 20%, is detected at the ground. Special relativity resolves the mystery, and it does so in two different but equally valid ways, depending on whose reference frame you choose to work in.

Simulation — muon survival

Muon speed, v = 0.995c

Show it from the muon's frame (length contraction) instead of the Earth's frame (time dilation)

Earth frame: distance fixed at 10 000 m

Lorentz factor, γ = 10.0

Muon half-life, as measured on Earth: 15.0 μs

Half-lives elapsed: 1.02

Fraction surviving to the ground: 49%

Muons remaining (of 5 shown): 2 / 5

Interactive simulation — open the online version of this workbook to use it.

Fig. 7.1 Muons falling from the upper atmosphere. Tick the box to switch between the Earth's explanation (the muon's decay clock is time-dilated, so it survives longer) and the muon's own explanation (the 10 km of atmosphere is length-contracted, so there is less distance to cross).

Explanation 1 — from the Earth's reference frame (time dilation): the Earth observer sees the muon's internal "clock" (its decay) running slow, because the muon is moving at high speed. Its half-life, measured from Earth, is dilated to γ times its proper half-life. Because each half-life lasts longer as measured from Earth, fewer half-lives pass during the (roughly fixed) 10 km journey, so far more muons survive to reach the ground than the non-relativistic calculation predicts.

Explanation 2 — from the muon's reference frame (length contraction): in the muon's own frame, its half-life is simply its normal, proper half-life — nothing is dilated. Instead, it is the 10 km thickness of the atmosphere that is contracted, to just 10 km / γ. Since the muon has much less distance to cross, the journey takes far less proper time, so once again, far more muons survive.

Both explanations are correct descriptions of the same physical outcome — the number of muons that reach the ground — because they are simply the same situation, described from two different, equally valid inertial reference frames.

Worked example 7.1

Muons are created 10 000 m above the ground, travelling at v = 0.995c. Their proper half-life is 1.5 μs. Using the Earth's frame, calculate (a) the time taken for the journey, as measured from Earth, and (b) how many half-lives this represents.

(a) Non-relativistically, time ≈ distance / speed = 10 000 / (0.995 × 3.00×10⁸)

time = 10 000 / 2.985×10⁸ = complete this step yourself s

(b) Number of half-lives = time ÷ 1.5×10⁻⁶ s. (Note: this is the time as measured on Earth — but because of time dilation, the muon's own decay clock ticks through fewer proper half-lives than this calculation alone would suggest for a stationary muon; that is exactly why more muons survive than classical physics predicts.)

Check your understanding

21Explain, in terms of time dilation, why more muons reach the Earth's surface than a non-relativistic calculation predicts.

Model answer From the Earth's frame, the muon is moving at high speed, so its internal decay "clock" is time-dilated — it runs slow compared with a muon at rest. This means the muon's half-life, as measured from Earth, is longer than its proper half-life. Since each half-life takes longer, fewer half-lives occur during the journey to the ground, so a larger fraction of the original muons survive to be detected than a non-relativistic calculation (which ignores time dilation) would predict.

22Explain, in terms of length contraction, why more muons reach the Earth's surface than a non-relativistic calculation predicts. Do not refer to time dilation in your answer.

Think about the problem entirely from the muon's own point of view — in the muon's frame, its own half-life is not dilated at all.

Model answer In the muon's own reference frame, its half-life is simply its ordinary, proper half-life. However, the thickness of the atmosphere the muon must cross is length-contracted in the muon's frame, since the atmosphere is moving relative to the muon at high speed. With a much shorter distance to cross, the journey takes less proper time, so fewer half-lives pass, and a larger fraction of muons survive to reach the ground.

23A muon is created 2000 m above a detector and travels at 0.996c. Its proper half-life is 1.5 μs. Using the Earth's frame, calculate (a) the time taken for the journey as measured on Earth, and (b) the number of half-lives this represents.

For part (a), use time = distance / speed with speed ≈ 0.996 × 3.00×10⁸ m s⁻¹. For part (b), divide your answer to (a) by the half-life.

Model answer

(a) t = distance / speed = 2000 / (0.996 × 3.00×10⁸) = 2000 / 2.988×10⁸ ≈ 6.7×10⁻⁶ s

(b) number of half-lives = 6.7×10⁻⁶ / 1.5×10⁻⁶ = (finish evaluating this yourself)

24Explain why muon decay is considered strong experimental evidence for special relativity, rather than just a mathematical curiosity.

Model answer Muon decay is a real, measurable phenomenon: detectors on the ground genuinely count far more muons arriving than non-relativistic physics predicts. Special relativity is the only theory that correctly predicts the observed survival rate, whether the calculation is carried out from the Earth's frame (using time dilation) or from the muon's frame (using length contraction) — both give the same, correct answer, which matches experiment. This makes it direct, repeatable experimental evidence, not just a theoretical prediction.

Glossary

Postulate: A basic statement taken as the starting point for a theory, from which other results are worked out logically.
Lorentz transformation equations: The equations x′ = γ(x − vt) and t′ = γ(t − vx/c²), which relate the position and time of an event in two inertial reference frames at any relative speed, and which reduce to the Galilean transformation equations when v ≪ c.
Lorentz factor, γ: γ = 1 / √(1 − v²/c²). Always ≥ 1, and equal to 1 only when v = 0.
Proper time interval, Δt₀: The time interval between two events as measured by an observer for whom both events happen at the same position. Always the smallest possible measured time interval.
Proper length, L₀: The length of an object as measured by an observer at rest relative to the object. Always the largest possible measured length.
Time dilation: The effect by which a clock moving relative to an observer is measured to run slow: Δt = γΔt₀.
Length contraction: The effect by which an object moving relative to an observer is measured to be shorter, along its direction of travel, than its proper length: L = L₀ / γ.
Relativity of simultaneity: Two events that are simultaneous in one inertial reference frame are, in general, not simultaneous in a different inertial reference frame moving relative to the first.