A.5 Galilean and special relativity HL only

1. A car going round a circular roundabout at a steady 30 m s⁻¹ — A student who equates constant speed with constant velocity picks this. The car's direction is changing, so it has a centripetal acceleration and its frame is not inertial.
2. A lift at the instant it is at rest while reversing direction — A student who thinks zero velocity means zero acceleration picks this. The lift's velocity is changing through zero, so it is accelerating at that instant and its frame is not inertial.
3. A rocket whose engine gives it a uniform acceleration of 3 m s⁻² — A student who reads 'uniform' as 'steady motion' picks this. A uniform acceleration is still an acceleration, and an inertial frame must be non-accelerating.
4. A train on a straight, level track at a steady 30 m s⁻¹ — The train's velocity (speed and direction) is constant, so it is not accelerating. A non-accelerating frame is an inertial frame: in it a ball placed on a table stays at rest, just as it would on the ground.

Working Inertial frame = non-accelerating. Train: straight track, constant speed → constant velocity, a = 0. Roundabout: direction changes, a = v²/r ≠ 0. Reversing lift: v = 0 momentarily but Δv/Δt ≠ 0. Rocket: a = 3 m s⁻² ≠ 0.

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1. Behind the point of release, as the floor moves on while the ball is falling. — A student who thinks a released object is 'left behind' picks this. The ball already has the train's horizontal velocity and nothing changes it, so it keeps pace with the floor.
2. Directly below the point of release, as the air in the carriage carries it. — A student who credits the moving air with keeping objects up with the train picks this. The ball keeps the train's velocity by Newton's first law. The carriage air moves with the train and exerts no horizontal push; the result is the same in a vacuum.
3. Directly below the point of release, as it keeps the train's horizontal velocity. — Before release the ball moves with the train. No horizontal force acts after release, so it keeps that horizontal velocity and stays level with the release point. Galilean relativity: the result is the same as in a stationary train.
4. Directly below the point of release, and it falls vertically for the platform. — A student who thinks the ball's motion is the same in every frame picks this. On the platform the ball has the train's horizontal velocity as it falls, so it follows a curved (parabolic) path, not a vertical line.

Working Train frame: horizontal velocity 0, falls vertically, lands below release point. Platform frame: horizontal velocity = train's velocity, parabolic path; lands at the point that is then below the release point.

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1. +57 m s⁻¹ — A student who calculates v − u = 35 − (−22) = +57 m s⁻¹ has found the velocity of A in B's frame. The velocity of B relative to A is u − v, which is negative.
2. −57 m s⁻¹ — In A's frame, v = +35 m s⁻¹ and B's velocity in the road frame is u = −22 m s⁻¹. u′ = u − v = −22 − 35 = −57 m s⁻¹: B approaches A at 57 m s⁻¹, moving west.
3. −13 m s⁻¹ — A student who ignores directions and subtracts the speeds gets 22 − 35 = −13 m s⁻¹. B moves west, so u = −22 m s⁻¹, and the magnitudes combine.
4. −22 m s⁻¹ — A student who thinks B's velocity is the same for every observer gives its road-frame value. In A's frame the relative velocity of the frames must be subtracted: u′ = u − v.

Working u = −22 m s⁻¹ (B, road frame), v = +35 m s⁻¹ (A relative to road). u′ = u − v = −22 − 35 = −57 m s⁻¹.

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1. No object with mass can reach or exceed the speed of light in a vacuum. — A student who thinks the speed limit is a starting assumption picks this. That no massive object can reach c is a consequence of the postulates, not one of them.
2. The speed of light in a vacuum is the same for all inertial observers. — This is Einstein's second postulate: c is independent of the motion of source and observer, which contradicts Galilean velocity addition. The first postulate is that the laws of physics are the same in all inertial frames.
3. Light travels at the same speed in every medium it passes through. — A student who has dropped 'in a vacuum' from the second postulate picks this. Light travels more slowly in glass or water than in a vacuum; the postulate concerns c in a vacuum.
4. The laws of physics are the same only in frames that are at rest. — A student who thinks an inertial frame must be at rest picks this. The first postulate applies to every inertial frame, including frames moving at constant velocity; rest is just the special case of zero velocity.

Working Postulate 1: the laws of physics are the same in all inertial frames. Postulate 2: the speed of light in a vacuum is the same for all inertial observers.

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1. −0.85c — u = −0.50c, v = +0.60c. u′ = (−0.50c − 0.60c)/(1 − (−0.50)(0.60)) = −1.10c/1.30 = −0.846c ≈ −0.85c. B approaches A at 0.85c, less than c as it must be.
2. −1.10c — A student who uses Galilean addition, u − v = −0.50c − 0.60c, gets −1.10c. That exceeds c; the relativistic denominator 1 − uv/c² = 1.30 is needed.
3. −1.57c — A student who drops the signs in the denominator uses 1 − 0.30 = 0.70 and gets −1.10c/0.70 = −1.57c. uv = (−0.50c)(0.60c) is negative, so the denominator is 1 + 0.30 = 1.30.
4. −0.14c — A student who treats B's velocity as +0.50c (ignoring its direction) gets (0.50 − 0.60)/(1 − 0.30) = −0.14c. B moves in the negative direction, so u = −0.50c.

Working u = −0.50c, v = +0.60c. uv/c² = (−0.50)(0.60) = −0.30. u′ = (−0.50 − 0.60)c/(1 − (−0.30)) = −1.10c/1.30 = −0.846c ≈ −0.85c.

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1. The time interval Δt between the events, as time passes at an equal rate for every observer — A student who assumes absolute time picks this. In special relativity Δt depends on the frame (for example time dilation); only the combination (cΔt)² − (Δx)² is invariant.
2. The quantity (cΔt)² − (Δx)², the square of the space–time interval between the events — The space–time interval is invariant: every inertial observer calculates the same (Δs)² = (cΔt)² − (Δx)², even though they measure different Δt and Δx.
3. The distance Δx between the two events, as distances do not depend on the frame — A student who thinks distances are the same in every frame picks this. Δx differs between frames even in Galilean relativity when the events are not simultaneous.
4. The quantity (cΔt)² + (Δx)², the square of the straight-line distance on a space–time diagram — A student who applies Pythagoras' theorem to space–time picks this. The invariant has a minus sign: (Δs)² = (cΔt)² − (Δx)².

Working Invariant: (Δs)² = (cΔt)² − (Δx)². Δt and Δx individually depend on the frame.

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1. 3.60 min — A student who thinks the Earth measures the shorter interval divides by γ: 6.00/1.667 = 3.60 min. The proper time, measured on the spacecraft, is the shortest interval; the Earth measures γΔt₀.
2. 10.0 min — The spacecraft's clock is present at every flash, so 6.00 min is the proper time Δt₀. γ = 1/√(1 − 0.80²) = 1.667, so Δt = γΔt₀ = 1.667 × 6.00 = 10.0 min.
3. 16.7 min — A student who omits the square root uses γ = 1/(1 − 0.64) = 2.78, giving 16.7 min. γ = 1/√0.36 = 1.667.
4. 13.4 min — A student who does not square v/c uses γ = 1/√(1 − 0.80) = 2.24, giving 13.4 min. v²/c² = 0.64, so γ = 1.667.

Working Δt₀ = 6.00 min (spacecraft clock at both flashes). γ = 1/√(1 − 0.80²) = 1/√0.36 = 1.667. Δt = γΔt₀ = 1.667 × 6.00 min = 10.0 min.

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1. 75 m — A student who multiplies by γ, as in time dilation, gets 60.0 × 1.25 = 75 m. A moving object is measured shorter: L = L₀/γ.
2. 38 m — A student who omits the square root in γ uses 1/(1 − 0.36) = 1.5625 and gets 38 m. γ = 1/√0.64 = 1.25.
3. 60 m — A student who thinks contraction is only an optical appearance gives the proper length. The station observer really measures L = L₀/γ = 48 m.
4. 48 m — γ = 1/√(1 − 0.60²) = 1.25. L = L₀/γ = 60 m/1.25 = 48 m. The spacecraft is shorter along its direction of motion.

Working γ = 1/√(1 − 0.36) = 1/0.80 = 1.25. L = L₀/γ = 60 m/1.25 = 48 m.

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1. Particle: a single point on the x axis of the diagram. Light pulse: a line at 45° to both axes. — A student who reads the diagram as a picture of the path in space picks this. The particle exists at every instant, so it has an event at every ct: a vertical line, not a point.
2. Particle: a line parallel to the x axis. Light pulse: a line at 45° to both axes. — A student who reads the diagram like a distance–time graph picks this. Here time (ct) is vertical, so a horizontal line joins events at one instant; a particle at rest has a vertical world line.
3. Particle: a line parallel to the ct axis. Light pulse: a line at 45° to both axes. — A particle at rest keeps the same x as ct increases, so its world line is vertical. Light travels a distance cΔt in time Δt, so Δx = Δ(ct) and its world line is at 45°.
4. Particle: a line parallel to the ct axis. Light pulse: a line along the x axis. — A student who treats light as instantaneous picks this. With ct on the time axis, light moves one unit of x for each unit of ct, so its world line is at 45°, not horizontal.

Working At rest: x constant → line parallel to ct axis. Light: Δx = cΔt = Δ(ct) → 45°.

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1. 0.42c — A student who uses sin θ gets sin 25° = 0.42. The world line moves Δx for each Δ(ct), which is opposite over adjacent: tan θ = v/c.
2. 0.47c — tan θ = v/c, with θ measured from the ct axis. v = c tan 25° = 0.466c ≈ 0.47c.
3. 2.14c — A student who measures the angle from the x axis uses tan 65° = 2.14 and gets a speed greater than c. θ is measured from the ct axis: v = c tan 25°.
4. 0.56c — A student who assumes speed is proportional to angle, with light at 45°, calculates (25/45)c = 0.56c. tan θ is not proportional to θ: v = c tan 25° = 0.47c.

Working v = c tan θ = c tan 25° = 0.466c ≈ 0.47c.

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1. Only the platform observer is correct, because the ground is the frame truly at rest. — A student who believes the ground is the one frame that is really at rest picks this. The ground itself moves relative to the Sun; no inertial frame is preferred.
2. Both are correct: each describes the motion relative to their own inertial frame. — Motion is always described relative to a reference frame. The train and the platform are both inertial frames, and neither is preferred, so both descriptions are equally valid.
3. Only the platform observer is correct, as a moving frame cannot be inertial. — A student who thinks an inertial frame must be at rest picks this. A frame moving at constant velocity is inertial, so the passenger's frame is as valid as the platform's.
4. Only one can be correct, as velocity belongs to the object, not to the observer. — A student who thinks each object has a single true velocity picks this. Velocity is always measured relative to a frame, so the platform has one velocity in the train's frame and the train another in the platform's frame, and both are valid.

Working Train frame: platform velocity −20 m s⁻¹. Platform frame: train velocity +20 m s⁻¹. Both frames are inertial (constant velocity), so both descriptions are valid.

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1. The acceleration of the puck at any instant — u′ = u − v with v constant, so Δu′ = Δu and the accelerations are equal. Both observers therefore find the same resultant force, F = ma: Newton's laws are the same in both frames.
2. The velocity of the puck at any given instant — A student who thinks an object has one velocity for all observers picks this. The velocities differ by the relative velocity of the frames, u′ = u − v.
3. The linear momentum of the puck at any instant — A student who confuses 'conserved' with 'the same in every frame' picks this. Momentum mu depends on the velocity, which differs between the frames.
4. The position of the puck at any instant — A student who thinks positions are the same in every frame picks this. Positions are measured from each frame's own origin, and the origins move apart: x′ = x − vt.

Working Galilean: u′ = u − v, v constant → a′ = Δu′/Δt = Δu/Δt = a. Velocity, momentum and position differ between the frames.

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1. 750 m — A student who adds vt, thinking positions in a forward-moving frame are larger, calculates 450 + 300 = 750 m. The origin of S′ has moved 300 m towards the ball, so vt is subtracted.
2. 390 m — A student who uses the ball's speed in the transformation calculates 450 − 5.0 × 12 = 390 m. The v in x′ = x − vt is the velocity of frame S′ relative to S, 25 m s⁻¹.
3. 450 m — A student who thinks an event has the same position in every frame gives 450 m. The origin of S′ is 300 m from the origin of S at t = 12 s, so the positions differ.
4. 150 m — x′ = x − vt, where v is the velocity of S′ relative to S: x′ = 450 m − (25 m s⁻¹)(12 s) = 450 m − 300 m = 150 m.

Working x′ = x − vt = 450 m − (25 m s⁻¹ × 12 s) = 450 m − 300 m = 150 m. (The ball's own speed of 5.0 m s⁻¹ is not needed.)

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1. Both observers give each event the same time, whatever their relative motion. — t′ = t is the Galilean assumption that time is absolute: all inertial observers assign the same time to every event, so they agree on time intervals and on simultaneity.
2. Each observer gives a later time to an event the further away from them it happens. — A student who dates an event by when its light arrives picks this. The time of an event is read on a clock at the event; t′ = t says the two frames' clocks agree for every event, wherever it is.
3. The moving observer's clock runs slow, so it gives a smaller time to each event. — A student who carries time dilation into Galilean relativity picks this. Galilean relativity has no time dilation: t′ = t exactly. Slow-running clocks belong to special relativity.
4. Only the observer at rest gives the true time; the moving one must correct it. — A student who thinks one frame is really at rest picks this. t′ = t says both observers give the same time, so there is nothing to correct and neither frame is preferred.

Working t′ = t for all events → identical time coordinates, intervals and simultaneity in every inertial frame (absolute time).

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1. The experiments must contain an error, since u′ = u − v is common sense. — A student who treats Galilean addition as exact common sense at all speeds blames the measurement. Many independent experiments, with different methods, all find c for every inertial observer; the equation, not the measurement, is at fault.
2. Light is an exception as it is massless; u′ = u − v holds for massive bodies. — A student who thinks light is an exception because it is massless picks this. The Galilean rule also fails for fast electrons and muons; the flaw is the assumption t′ = t, which applies to everything.
3. u′ = u − v is only an approximation, valid when speeds are much less than c. — The Galilean result follows from the assumption t′ = t. It matches experiment at everyday speeds but fails for light and for any object moving at a speed comparable to c. It is replaced by the relativistic u′ = (u − v)/(1 − uv/c²).
4. The ship only appears to measure c; relative to it the light really passes at 0.50c. — A student who thinks the laser's frame gives the "true" speed, so the ship's measurement is only apparent, picks this. The ship's frame is as valid as any other, and the ship's observer really does measure c; the Galilean prediction of 0.50c is simply wrong.

Working Galilean prediction in spacecraft frame: u′ = c − 0.50c = 0.50c. Measured: c. The Galilean equation is therefore only an approximation; special relativity replaces it.

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1. 1.6c, as the speed of the spacecraft towards the star adds to the speed of light — A student who applies Galilean velocity addition to light adds the speeds. Experiment shows light has speed c for all inertial observers, so u′ = u − v does not apply to light.
2. c, but only as the star is at rest; light from a moving star would be faster — A student who thinks a moving source changes the speed of light picks this. The second postulate states that c is independent of the motion of the source as well as of the observer.
3. 0.40c, as the spacecraft's speed is subtracted from the speed of light — A student who finds a relative speed by subtracting the smaller speed from the larger, ignoring that the spacecraft and the light are moving towards each other, gets c − 0.60c = 0.40c. By the second postulate the speed of light in a vacuum is c for all inertial observers, so the astronaut measures c.
4. c, as the speed of light in a vacuum does not depend on the motion of the observer — By the second postulate every inertial observer measures the speed of light in a vacuum to be c, whatever the motion of source or observer. The spacecraft's speed changes the light's frequency and wavelength, not its speed.

Working Second postulate: the speed of light in a vacuum = c for all inertial observers. Neither Galilean addition (c + 0.60c = 1.6c) nor subtraction of speeds (c − 0.60c = 0.40c) applies; the astronaut measures c.

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1. 6.6 × 10² m — A student who leaves out γ calculates x − vt = 900 − 240 = 660 m, which is the Galilean value. At 0.80c, γ = 1.67 multiplies the bracket.
2. 4.0 × 10² m — A student who divides by γ calculates 660/1.67 = 396 m. The Lorentz transformation multiplies (x − vt) by γ.
3. 1.1 × 10³ m — γ = 1/√(1 − 0.80²) = 1/0.60 = 1.67. vt = 0.80 × 3.00 × 10⁸ × 1.0 × 10⁻⁶ = 240 m. x′ = γ(x − vt) = 1.67 × (900 − 240) = 1.67 × 660 = 1100 m.
4. 1.9 × 10³ m — A student who adds vt calculates 1.67 × (900 + 240) = 1900 m. The origin of S′ has moved 240 m towards the event, so vt is subtracted.

Working γ = 1/√(1 − 0.64) = 1/0.600 = 1.667. vt = (0.80)(3.00 × 10⁸ m s⁻¹)(1.0 × 10⁻⁶ s) = 240 m. x′ = γ(x − vt) = 1.667 × (900 − 240) m = 1.667 × 660 m = 1100 m = 1.1 × 10³ m.

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1. −2.3 μs — γ = 1.67. vx/c² = (0.80 × 3.00 × 10⁸ × 900)/(3.00 × 10⁸)² = 2.4 × 10⁻⁶ s. t′ = γ(t − vx/c²) = 1.67 × (1.0 − 2.4) μs = 1.67 × (−1.4 μs) = −2.3 μs. The event happens before the origins pass, according to S′.
2. +1.0 μs — A student who assumes time is the same in every frame (t′ = t) gives 1.0 μs. In special relativity t′ depends on the position of the event through the vx/c² term.
3. −1.4 μs — A student who leaves out γ calculates t − vx/c² = −1.4 μs. The bracket must be multiplied by γ = 1.67.
4. +1.7 μs — A student who applies time dilation to the event's time coordinate calculates γt = 1.67 × 1.0 = 1.7 μs. The full transformation t′ = γ(t − vx/c²) includes the position-dependent term.

Working γ = 1.667. vx/c² = (0.80 × 3.00 × 10⁸ m s⁻¹ × 900 m)/(3.00 × 10⁸ m s⁻¹)² = 2.4 × 10⁻⁶ s. t′ = γ(t − vx/c²) = 1.667 × (1.0 × 10⁻⁶ − 2.4 × 10⁻⁶) s = 1.667 × (−1.4 × 10⁻⁶ s) = −2.33 × 10⁻⁶ s ≈ −2.3 μs.

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1. It is never greater than 1, and it decreases towards zero as v approaches c. — A student who takes γ to be √(1 − v²/c²) itself, rather than its reciprocal, picks this. √(1 − v²/c²) is at most 1, so its reciprocal γ is at least 1.
2. It increases in direct proportion to v, so it equals 2 when v is half of c. — A student who assumes a linear relationship picks this. At 0.50c, γ = 1/√0.75 = 1.15; γ reaches 2 only at v = 0.87c.
3. It is larger for a frame approaching the observer than for one moving away at the same speed. — A student who expects γ to behave like the Doppler shift picks this. γ depends on v², so approach and recession at the same speed give the same γ.
4. It is never less than 1, and it increases without limit as v approaches c. — γ = 1 at v = 0, is greater than 1 for any non-zero v, and 1 − v²/c² tends to zero as v approaches c, so γ grows without limit. It has the same value for either direction of motion.

Working γ(0) = 1; γ(0.50c) = 1/√0.75 = 1.15; γ(0.87c) ≈ 2.0; γ → ∞ as v → c. Depends on v², so direction does not matter.

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1. The relativistic equation cannot be used, as relativity applies only to speeds near c. — A student who thinks relativity switches on only near c picks this. The relativistic equation is valid at every speed; at 800 m s⁻¹ it simply gives a result extremely close to the Galilean one.
2. u − v is exact at these speeds, so the relativistic equation gives a slightly wrong value. — A student who treats Galilean addition as exact picks this. The relativistic equation is the exact one; the Galilean result is the approximation, though the difference is negligible here.
3. uv/c² ≈ 2 × 10⁻¹², so the Galilean value differs from the relativistic one by a negligible amount. — uv/c² = (800 × 250)/(3.00 × 10⁸)² = 2.2 × 10⁻¹². The relativistic value differs from 550 m s⁻¹ by about 1 × 10⁻⁹ m s⁻¹, so u′ = u − v is an excellent approximation at everyday speeds.
4. uv/c² ≈ 7 × 10⁻⁴, so the Galilean value is in error by about 0.07 %, which is a clearly measurable amount. — A student who divides by c instead of c² gets 800 × 250/(3.00 × 10⁸) = 6.7 × 10⁻⁴. That quantity has a unit (m s⁻¹) and cannot be subtracted from 1; with c² the term is 2.2 × 10⁻¹².

Working uv/c² = (800 m s⁻¹)(250 m s⁻¹)/(3.00 × 10⁸ m s⁻¹)² = 2.0 × 10⁵/9.0 × 10¹⁶ = 2.2 × 10⁻¹². Relativistic u′ = 550/(1 − 2.2 × 10⁻¹²) m s⁻¹ = 550 m s⁻¹ + 1.2 × 10⁻⁹ m s⁻¹. (Dividing by c only: 2.0 × 10⁵/3.00 × 10⁸ = 6.7 × 10⁻⁴.)

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1. 3.0 μs — A student who takes (Δs)² = (cΔt)² + (Δx)² writes 1500² = (cΔt′)² + 1200² and gets cΔt′ = 900 m, Δt′ = 3.0 μs. With the correct minus sign, (cΔt′)² = 1500² + 1200².
2. 9.0 μs — A student who uses Δs = cΔt − Δx without squares writes 1500 = cΔt′ − 1200, giving cΔt′ = 2700 m and 9.0 μs. Only the squared combination is invariant.
3. 5.0 μs — A student who assumes time intervals are the same in every frame gives 5.0 μs. Δt changes between frames; it is (Δs)² that is invariant.
4. 6.4 μs — In S: cΔt = 3.00 × 10⁸ × 5.0 × 10⁻⁶ = 1500 m and Δx = 0, so (Δs)² = 1500² m². In S′: (cΔt′)² = (Δs)² + (Δx′)² = 1500² + 1200², so cΔt′ = 1921 m and Δt′ = 1921/(3.00 × 10⁸) = 6.4 μs.

Working S: cΔt = (3.00 × 10⁸)(5.0 × 10⁻⁶) = 1500 m, Δx = 0 → (Δs)² = 2.25 × 10⁶ m². S′: (cΔt′)² = (Δs)² + (Δx′)² = 2.25 × 10⁶ + 1.44 × 10⁶ = 3.69 × 10⁶ m² → cΔt′ = 1921 m → Δt′ = 1921/(3.00 × 10⁸) s = 6.4 × 10⁻⁶ s. (5.0 μs is the proper time, the shortest.)

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1. An observer moving with the muon, as both events occur at the same place in that frame — The proper time interval is measured in the frame in which the two events occur at the same place. In the muon's rest frame it is created and decays at the same point, so a single clock there measures Δt₀.
2. An observer at rest on the Earth, as the Earth is the frame that is not moving — A student who treats the Earth as truly at rest picks this. In the Earth frame the creation and decay happen kilometres apart, so the Earth measures a dilated interval, not the proper time.
3. Any observer with an accurate clock, as proper time means the correct time — A student who reads 'proper' as 'correct' picks this. Every inertial observer's measurement is correct for their frame; the proper time is the one measured where both events happen at the same place.
4. An observer midway between the events, as light from each reaches them equally fast — A student who thinks event times depend on when light arrives picks this. Signal delay is always corrected for; the proper time is defined by the two events being at the same place, not by the observer's position.

Working Proper time: frame in which both events occur at the same place → muon rest frame.

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1. The astronaut measures the proper length, the shortest length found by any inertial observer. — A student who reasons that proper length must be the shortest, like proper time, picks this. Proper time is the shortest interval but proper length is the longest length: L = L₀/γ ≤ L₀.
2. The Earth observer measures the proper length, as the Earth is the frame that is truly at rest. — A student who treats the Earth as truly at rest picks this. The proper length is measured in the frame in which the object is at rest, which here is the astronaut's frame.
3. The astronaut measures the proper length, the longest length found by any inertial observer. — The proper length is measured in the object's rest frame, the astronaut's. Every observer moving along its length measures L = L₀/γ, which is shorter.
4. Both measure the same length; the Earth observer merely sees it shorter because light takes time. — A student who thinks length contraction is an optical effect picks this. After allowing for light travel time the Earth observer still measures L = L₀/γ, a real difference.

Working Proper length L₀: measured in the rest frame of the object (astronaut). Earth: L = L₀/γ = L₀/1.25 = 0.80L₀.

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1. Only one can be correct: the clock that is really moving is the one that runs slow. — A student who believes one spacecraft is 'really' moving picks this. There is no absolute rest; each astronaut can regard their own craft as at rest, and both measurements are valid.
2. Neither is correct: the clocks only appear slow because light from them takes time to arrive. — A student who treats time dilation as a signal-delay illusion picks this. Each observer's result remains after correcting for light travel time; it is a real measured effect.
3. Both are correct: each measures the clock moving relative to them to run slow, by γ = 1.25. — By the first postulate neither inertial frame is preferred, so time dilation is reciprocal. Each observer measures the other's clock to run slow by γ = 1/√(1 − 0.60²) = 1.25. The relativity of simultaneity makes the two results consistent.
4. Neither is correct: the two effects cancel, so each measures the other's clock at the normal rate. — A student who sees reciprocal time dilation as a contradiction picks this. The two measurements compare different pairs of events in different frames, so they do not cancel; each is valid.

Working γ = 1/√(1 − 0.60²) = 1/0.80 = 1.25. Each frame measures the other's clock interval as Δt = 1.25Δt₀.

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1. The light reaches the rear door first, as the rear door moves towards the light. — In the platform frame both pulses travel at c (second postulate). The rear door moves towards its pulse and the front door away from its pulse, so the rear door is reached first. Events simultaneous in the carriage frame are not simultaneous in the platform frame.
2. Both at once, as light going forwards moves at c + v and backwards at c − v. — A student who applies Galilean velocity addition to light picks this. In the platform frame both pulses travel at c, so the rear door, moving towards its pulse, is reached first.
3. Both at once, as events simultaneous in one frame are simultaneous in every frame. — A student who assumes absolute simultaneity picks this. Simultaneity is relative: events at different places that are simultaneous in one frame are not simultaneous in a frame moving along the line joining them.
4. The light arrives at both doors together; she only sees one first because it is nearer to her. — A student who thinks the disagreement is a light-delay effect picks this. Even after correcting for light travel time to the observer, the platform frame assigns the rear-door event the earlier time.

Working Platform frame: both pulses at c; rear door approaches its pulse, front door recedes from its pulse → rear door reached first.

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1. 2.4 μs — A student who leaves out γ gives vΔx/c² = 2.4 μs. The Lorentz transformation multiplies the bracket by γ = 1.25.
2. 3.0 μs — Δt = 0, so Δt′ = γ(Δt − vΔx/c²) = −γvΔx/c². γ = 1.25; vΔx/c² = 0.60 × 1200/(3.00 × 10⁸) = 2.4 × 10⁻⁶ s. |Δt′| = 1.25 × 2.4 μs = 3.0 μs.
3. 1.9 μs — A student who divides by γ gets 2.4/1.25 = 1.9 μs. In t′ = γ(t − vx/c²) the bracket is multiplied by γ.
4. 4.0 μs — A student who thinks the time difference is the light travel time between the events calculates Δx/c = 1200/(3.00 × 10⁸) = 4.0 μs. The relativity of simultaneity comes from the vx/c² term, not from signal delay.

Working Δt′ = γ(Δt − vΔx/c²), Δt = 0. γ = 1/√(1 − 0.36) = 1.25. vΔx/c² = (0.60 × 3.00 × 10⁸ m s⁻¹)(1200 m)/(3.00 × 10⁸ m s⁻¹)² = 2.4 × 10⁻⁶ s. |Δt′| = 1.25 × 2.4 × 10⁻⁶ s = 3.0 × 10⁻⁶ s.

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1. They are set by lines of constant space–time interval, so their units differ from those on ct and x. — The guide states that the scales are not the same and are defined by lines of constant space–time interval. The curve (ct)² − x² = 1 unit² crosses ct′ at ct′ = 1; on S's diagram that point lies further from the origin than ct = 1.
2. They are identical to the scales on ct and x, because all four axes are measured in metres. — A student who assumes axes with the same unit share one scale picks this. Same unit does not mean same drawn scale; the scales are fixed by lines of constant interval.
3. They are shrunk by γ, because moving clocks run slow and moving lengths are contracted. — A student who builds the headline results into the drawing picks this. As drawn on S's diagram, a unit on a primed axis is longer, and it is found from a line of constant interval, not by dividing by γ.
4. They are stretched by γ on ct′ and shrunk by γ on x′, by time dilation and length contraction. — A student who maps 'time dilation' onto a stretched time axis and 'length contraction' onto a shrunk space axis picks this. Both primed units are set by the invariant hyperbolae, and on S's diagram both appear longer than the unprimed units by the same factor, √((1 + v²/c²)/(1 − v²/c²)); neither axis is shrunk.

Working Scale on ct′: where (ct)² − x² = 1 crosses the ct′ axis (x = βct), ct = γ and x = βγ, so the distance from the origin on S's diagram is γ√(1 + β²) = √((1 + β²)/(1 − β²)) > 1. By symmetry, x² − (ct)² = 1 crosses the x′ axis at the same distance. Both primed units are longer than the unprimed units by the same factor.

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1. P and Q are simultaneous, because they lie on one line parallel to the x axis of S. — A student who assumes simultaneity is the same in every frame picks this. A line parallel to the x axis joins events simultaneous in S only; simultaneity in S′ is read along lines parallel to the x′ axis.
2. P occurs before Q, because Q lies further along the x′ axis, which tilts upwards. — A student who reads the tilt the wrong way picks this. The upward tilt means the S′ simultaneity line through P passes above Q, so Q happens first in S′, consistent with t′ = γ(t − vx/c²).
3. P occurs before Q, because light from Q takes longer to reach an observer standing at x = 0. — A student who dates events by when light arrives picks this. The order of events in a frame is set by its time coordinates, not by signal arrival; in S′, Q has the earlier time coordinate.
4. Q occurs before P, as Q lies below the line through P parallel to the x′ axis. — Lines parallel to the x′ axis join events simultaneous in S′. The line through P rises 0.50 ly per ly, so at x = 4.0 ly it is at ct = 4.0 ly. Q, at ct = 2.0 ly, lies below it, so Q is earlier in S′. Check: cΔt′ = γ(cΔt − (v/c)Δx) = 1.15 × (0 − 2.0 ly) < 0.

Working x′-axis slope: Δ(ct)/Δx = v/c = 0.50. Line through P parallel to x′: ct = 2.0 + 0.50x → at x = 4.0 ly, ct = 4.0 ly. Q (ct = 2.0 ly) lies below → earlier in S′. Lorentz: γ = 1.155; cΔt′ = γ(0 − 0.50 × 4.0 ly) = −2.3 ly.

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1. It cannot be correct: tan 50° ≈ 1.2, so the electron would have to move faster than light. — tan θ = v/c gives v = 1.19c. No particle with mass can reach c, so its world line must make less than 45° with the ct axis.
2. It cannot be correct: a line at more than 45° would mean that the electron moves back in time. — A student who links faster-than-light motion with time travel picks this. The line at 50° still rises in ct, so it does not go back in time; it is impossible because it means v > c.
3. It is possible: the angle to the x axis is 40°, so its speed is c tan 40° ≈ 0.84c. — A student who measures the angle from the x axis picks this. θ in tan θ = v/c is measured from the ct axis; at 50°, v = 1.19c, which is impossible.
4. It is possible: sin 50° ≈ 0.77, so the electron is moving at a speed of about 0.77c. — A student who uses sin θ picks this. The relation is tan θ = v/c, and tan 50° = 1.19 > 1 gives a speed above c.

Working v/c = tan 50° = 1.19 > 1 → impossible for a massive particle. (tan 40° = 0.84; sin 50° = 0.77 are the errors.)

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1. 4.5 × 10⁴ m — A student who multiplies by γ gets 4500 × 10.0 = 45 km. A moving length is contracted: L = L₀/γ.
2. 4.5 × 10³ m — A student who thinks only the muon, as the 'moving' object, is affected gives the Earth-frame value. In the muons' frame the atmosphere is moving, so it is contracted.
3. 4.5 × 10¹ m — A student who omits the square root uses γ = 1/(1 − 0.995²) = 100 and gets 45 m. γ = 1/√0.009975 = 10.0.
4. 4.5 × 10² m — In the muons' frame the atmosphere moves at 0.995c, so it is contracted. γ = 1/√(1 − 0.995²) = 10.0. L = L₀/γ = 4500 m/10.0 = 450 m.

Working γ = 1/√(1 − 0.995²) = 1/√0.009975 = 10.0. Proper length (Earth frame) L₀ = 4500 m. Muon frame: L = L₀/γ = 4500 m/10.0 = 449 m ≈ 4.5 × 10² m.

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1. Time dilation: in the muons' own frame their half-life is dilated to 15 μs, so most survive. — A student who thinks a moving particle's time slows in its own frame picks this. In its own frame a muon is at rest with a half-life of 1.5 μs; it survives there because the atmosphere is contracted.
2. Length contraction: Earth observers measure the 2.25 km drop contracted to about 225 m. — A student who assigns length contraction to the wrong frame picks this. The mountain is at rest in the Earth frame, so Earth observers measure 2.25 km; the contraction to 225 m is measured in the muons' frame.
3. Time dilation: in the Earth frame the half-life is about 15 μs, so about 700 per hour should arrive. — Earth frame: half-life = γ × 1.5 μs = 15 μs, so the 7.5 μs descent is 0.5 half-lives and 1000 × 2^(−0.5) ≈ 710 per hour survive, as observed. Without time dilation the descent is 5 half-lives and only about 31 per hour would arrive.
4. Neither effect: the 1.5 μs rest half-life must be wrong, as half-life is the same in all frames. — A student who assumes time intervals are the same in every frame concludes the rest-frame value must be wrong. The measured half-life at rest is reliable; the data match the relativistic prediction of a dilated half-life.

Working Earth frame: descent time = 2250 m/(0.995 × 3.00 × 10⁸ m s⁻¹) = 7.5 μs. Dilated half-life = 10.0 × 1.5 μs = 15 μs → 0.50 half-lives → 1000 × 2^(−0.50) = 7.1 × 10² per hour ≈ observed 700. No dilation: 7.5/1.5 = 5.0 half-lives → 1000 × 2^(−5.0) ≈ 31 per hour. Muon frame: drop contracted to 2250/10.0 = 225 m, crossed in 0.75 μs = 0.50 half-lives → same prediction.

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1. Its edge along the motion is 3.3 m; its other two edges are still 2.0 m each. — A student who multiplies by γ, as in time dilation, gets 2.0 × 1.67 = 3.3 m. Lengths are divided by γ: the moving edge is measured as L = L₀/γ = 1.2 m, shorter than its proper length.
2. Its edge along the motion is 1.2 m; its other two edges are 2.0 m each. — γ = 1/√(1 − 0.80²) = 1.67, so the edge parallel to the motion is measured as L = L₀/γ = 2.0/1.67 = 1.2 m. Length contraction acts only along the direction of relative motion; the two perpendicular edges keep their proper length of 2.0 m.
3. All three of its edges are 1.2 m, because every dimension contracts by γ. — A student who applies the contraction to every dimension picks this. Only the length parallel to the relative velocity is contracted; lengths perpendicular to the motion are the same in every inertial frame, so the other two edges are measured as 2.0 m.
4. All three of its edges are 2.0 m, because a moving object merely looks shorter. — A student who treats length contraction as an appearance caused by light travel time picks this. The contracted length is what the observer measures with rulers at rest in their own frame, after any signal delay is allowed for: the edge along the motion is genuinely measured as 1.2 m.

Working γ = 1/√(1 − 0.80²) = 1/√0.36 = 1/0.60 = 1.67. Only the edge parallel to the relative velocity is contracted: L = L₀/γ = 2.0 m × 0.60 = 1.2 m. The two edges perpendicular to the motion are unchanged at 2.0 m. The observer measures a block 1.2 m × 2.0 m × 2.0 m.

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