A.5 Galilean and special relativity flashcards: IB Physics HL

All 68 A.5 Galilean and special relativity cards

1. DefinitionHL only

   Define a reference frame and an inertial reference frame.

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   Reference frame: a coordinate system (with clocks) used to measure positions and times of events. Inertial frame: one in which Newton's first law holds—a frame that is not accelerating (moves at constant velocity relative to other inertial frames).

2. DefinitionHL only

   State the principle of Galilean relativity.

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   The laws of mechanics are the same in all inertial reference frames; no mechanical experiment can detect absolute uniform motion. Time and lengths are absolute (same for all observers); velocities add: u' = u − v.

3. DefinitionHL only

   State the two postulates of special relativity.

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   (1) The laws of physics are the same in all inertial reference frames. (2) The speed of light in vacuum is the same (c) for all inertial observers, independent of the motion of the source or observer.

4. DefinitionHL only

   Define proper time interval.

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   Time interval between two events measured in the frame in which both events occur at the SAME POSITION (measured by a single clock present at both events). It is the shortest time interval any observer measures.

5. DefinitionHL only

   Define proper length.

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   Length of an object measured in the frame in which the object is AT REST (the rest frame). It is the longest length any observer measures; moving observers measure a contracted length along the direction of motion only.

6. DefinitionHL only

   Define an event and simultaneity in special relativity.

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   Event: something that happens at a specific point in space and instant in time (x, t). Two events are simultaneous in a frame if they have the same t in that frame. Simultaneity is relative: events simultaneous in one frame are generally not simultaneous in another moving frame (unless they occur at the same place).

7. DefinitionHL only

   Define the spacetime interval and state its key property.

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   (Δs)² = (cΔt)² − (Δx)². It is INVARIANT: all inertial observers obtain the same value for the same pair of events, even though Δt and Δx individually differ. For events on a light ray the interval is zero.

8. EquationHL onlyData booklet: Yes

   Write the Galilean transformations and the Galilean velocity addition.

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   x' = x − vt; t' = t; u' = u − v. Frame S' moves at velocity v along +x relative to S; u = velocity of object in S, u' in S'. Valid only for v ≪ c. Misuse: applying to light or relativistic particles.

9. EquationHL onlyData booklet: Yes

   Write the Lorentz factor and describe its behaviour.

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   γ = 1/√(1 − v²/c²). No unit; γ ≥ 1; γ = 1 at v = 0; γ → ∞ as v → c. Useful values: 0.6c → 1.25; 0.8c → 1.67; 0.9c → 2.29; 0.99c → 7.1. Misuse: forgetting the square root, or using v in km h⁻¹ with c in m s⁻¹.

10. EquationHL onlyData booklet: Yes

    Write the Lorentz transformations for coordinates and for intervals.

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    x' = γ(x − vt); t' = γ(t − vx/c²). For intervals: Δx' = γ(Δx − vΔt); Δt' = γ(Δt − vΔx/c²). S' moves at +v along x relative to S. Inverse: swap primes and replace v by −v. Misuse: dropping the vx/c² term in the time transformation.

11. EquationHL onlyData booklet: Yes

    Write the relativistic velocity addition formula.

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    u' = (u − v)/(1 − uv/c²). u = velocity of object in S; v = velocity of S' relative to S; u' = velocity in S'. Result never exceeds c; for u = c, u' = c. Misuse: sign errors—if object and frame move in opposite directions, v is negative.

12. EquationHL onlyData booklet: Yes

    Write the time dilation equation and identify the proper time.

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    Δt = γΔt₀. Δt₀ = proper time (clock at rest relative to the events); Δt = time interval measured by an observer moving relative to that clock (always longer: 'moving clocks run slow'). Misuse: assigning Δt₀ to the wrong frame.

13. EquationHL onlyData booklet: Yes

    Write the length contraction equation and identify the proper length.

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    L = L₀/γ. L₀ = proper length (object's rest frame); L = length measured by an observer for whom the object moves (shorter). Contraction only along the direction of motion; perpendicular lengths unchanged. Misuse: multiplying by γ instead of dividing.

14. EquationHL onlyData booklet: Yes

    Write the invariance of the spacetime interval.

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    (ct')² − (x')² = (ct)² − (x)². Same for all inertial observers. Use to find a quantity in one frame given the other three, or to calibrate axes on a spacetime diagram (hyperbolae). Misuse: adding instead of subtracting the space term.

15. Concept/explainHL only

    Explain how muon decay provides evidence for special relativity.

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    • Muons created ~10 km up in atmosphere, half-life ≈ 1.5 μs (proper)
    • At ~0.99c, non-relativistic range ≈ 450 m, so almost none should reach ground
    • Many are detected at sea level
    • Earth frame: muon lifetime dilated by γ, so they travel further
    • Muon frame: atmosphere length-contracted, so shorter distance
    • Both descriptions agree with observed count.
16. Concept/explainHL only

    Explain, using a light source in the middle of a moving train, why simultaneity is relative.

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    • Source flashes; light travels at c in both directions in every frame
    • Train frame: front and back equidistant, light arrives simultaneously
    • Platform frame: back wall moves towards the light, front wall moves away
    • Light reaches back first
    • Same two events, different order/timing → simultaneity depends on frame.
17. Concept/explainHL only

    Use a light clock to explain why moving clocks run slow.

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    • Light clock: pulse bounces between mirrors separated by d; tick = 2d/c in its rest frame
    • For a moving clock, observer sees the pulse travel a longer diagonal path
    • Speed of light is still c (postulate 2)
    • Longer path at same speed → longer tick
    • So the moving clock's tick interval is dilated by γ.
18. Concept/explainHL only

    Explain why no object with mass can reach the speed of light.

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    • γ → ∞ as v → c
    • Time dilation and length contraction become infinite
    • Velocity addition: adding any velocities below c always gives a result below c
    • Infinite energy would be required (not required in syllabus detail, but valid)
    • Only massless particles (photons) travel at exactly c.
19. Concept/explainHL only

    Explain why Galilean relativity fails for light and how Einstein resolved it.

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    • Galilean addition predicts light speed c ± v for moving observer
    • Experiment/Maxwell's equations give c for all observers
    • Contradiction
    • Einstein kept the relativity principle and made c invariant (postulate 2)
    • Consequence: space and time intervals are frame-dependent (Lorentz transformations) rather than absolute.
20. Worked problemHL onlyData booklet: Yes

    Calculate γ for v = 0.80c and for v = 0.60c.

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    0.80c: γ = 1/√(1 − 0.64) = 1/√0.36 = 1/0.60 = 1.67. 0.60c: γ = 1/√(1 − 0.36) = 1/0.80 = 1.25.

21. Worked problemHL onlyData booklet: Yes

    A muon has a proper mean lifetime of 2.2 μs and moves at 0.98c. Find its lifetime and distance travelled in the Earth frame.

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    γ = 1/√(1 − 0.9604) = 1/0.199 = 5.0. Δt = γΔt₀ = 5.0 × 2.2 = 11 μs. Distance = vΔt = 0.98 × 3.0 × 10⁸ × 11 × 10⁻⁶ = 3.2 × 10³ m (3.2 km). Muon frame: distance = 3.2 km/5.0 = 0.65 km in 2.2 μs.

22. Worked problemHL onlyData booklet: Yes

    A spacecraft of proper length 100 m passes Earth at 0.60c. Find its length as measured from Earth.

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    γ = 1.25. L = L₀/γ = 100/1.25 = 80 m.

23. Worked problemHL onlyData booklet: Yes

    A rocket moves at 0.60c relative to Earth and launches a probe forward at 0.70c relative to the rocket. Find the probe's speed relative to Earth.

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    Use inverse addition: u = (u' + v)/(1 + u'v/c²) = (0.70 + 0.60)c/(1 + 0.42) = 1.30c/1.42 = 0.92c (< c, as required; Galilean would wrongly give 1.30c).

24. Worked problemHL onlyData booklet: Yes

    In frame S an event occurs at x = 300 m, t = 1.0 μs. Frame S' moves at 0.60c along +x. Find x' and t'.

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    γ = 1.25; v = 1.8 × 10⁸ m s⁻¹. x' = 1.25(300 − 1.8 × 10⁸ × 1.0 × 10⁻⁶) = 1.25(300 − 180) = 150 m. t' = 1.25(1.0 × 10⁻⁶ − 1.8 × 10⁸ × 300/9.0 × 10¹⁶) = 1.25(1.0 − 0.60) × 10⁻⁶ = 0.50 μs.

25. Worked problemHL onlyData booklet: Yes

    Two events in S are separated by Δx = 600 m and Δt = 1.0 μs. Find their separation in time in a frame where they occur at the same place, using the invariant interval.

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    (cΔt)² − Δx² = (300)² − (600)² = 90 000 − 360 000 < 0. Negative interval → space-like: NO frame exists where events occur at the same place (they cannot be causally connected). If Δx were 200 m: (cΔt₀)² = 90 000 − 40 000 → Δt₀ = 0.75 μs.

26. Graph/diagramHL only

    Describe the axes and basic features of a spacetime (Minkowski) diagram.

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    Vertical axis ct (m), horizontal axis x (m). Worldline of a stationary object: vertical line. Object at constant v: straight line with gradient c/v (steeper = slower). Light: lines at 45° (gradient ±1). Massive objects have worldlines steeper than 45°.

27. Graph/diagramHL only

    How are the axes of a moving frame S' drawn on the spacetime diagram of S?

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    ct' axis: worldline of S' origin, line ct = (c/v)x. x' axis: line ct = (v/c)x. Both tilt symmetrically towards the 45° light line by angle θ where tanθ = v/c. As v → c both axes close onto the light line. Scales on tilted axes are stretched (calibrate with invariant hyperbolae).

28. Graph/diagramHL only

    How do you read simultaneity from a spacetime diagram?

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    Events simultaneous in S lie on a horizontal line (parallel to x axis). Events simultaneous in S' lie on a line parallel to the tilted x' axis. Two events on a horizontal line are therefore NOT on a line parallel to x'—showing relativity of simultaneity.

29. Graph/diagramHL only

    How are time dilation and length contraction shown on a spacetime diagram?

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    Time dilation: one tick of the S' clock is a unit along the ct' axis; projecting onto ct (along a line parallel to x) gives a longer interval γ. Length contraction: rod at rest in S' has worldlines of its ends parallel to ct'; its length in S is the horizontal separation at fixed t, shorter than L₀. Calibration hyperbola (ct)² − x² = const marks equal intervals on all axes.

30. Exam technique/trapHL only

    Trap: how do you decide which observer measures the proper time?

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    Ask: 'In which frame do both events happen at the same place (one clock present at both)?' That frame measures Δt₀. E.g. a moving spaceship's clock measures proper time for events aboard the ship; Earth measures the dilated interval. For lengths, the proper length belongs to the frame where the object is at rest.

31. Exam technique/trapHL only

    Trap: check your answer in velocity addition and length/time transformation.

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    Velocity addition: result must be < c; relativistic answer smaller than Galilean sum. Time dilation: moving observer always measures a LONGER time (Δt > Δt₀). Length contraction: moving observer always measures a SHORTER length (L < L₀). If your numbers go the other way, you have swapped frames.

32. Exam technique/trapHL only

    Trap: Lorentz transformation of intervals versus coordinates.

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    Transform DIFFERENCES: Δx' = γ(Δx − vΔt), Δt' = γ(Δt − vΔx/c²). Setting Δx = 0 (events at same place in S) gives Δt' = γΔt (time dilation). Setting Δt' = 0 (simultaneous measurement of ends in S') gives length contraction. Do not apply time dilation formula blindly when events occur at different places.

33. Exam technique/trapHL only

    NOS: give two experimental confirmations of special relativity that the mark scheme expects.

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    (1) Muon decay: far more cosmic-ray muons reach sea level than classical physics allows, matching γ-dilated lifetimes. (2) Atomic clocks flown on aircraft/satellites (GPS) run slow relative to ground clocks by the predicted amount; GPS must correct for this or accumulate km-scale errors daily. Also particle accelerators: particles never exceed c despite increasing energy.

34. DefinitionHL only

    State Newton's postulates of absolute time and absolute space, and state what special relativity replaces them with.

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    Newton assumed time flows at the same rate for all observers (a single universal clock) and that the distance between two points is the same in every reference frame; both are therefore absolute and independent of the observer's motion. Special relativity replaces them: time intervals and lengths are frame-dependent, and only the speed of light in vacuum and the spacetime interval are invariant. Exam tip: students lose the mark by writing only "time is the same everywhere" without linking it to the observer/reference frame. Say explicitly "the same for all inertial observers". Time and length are scalars; both are measured in s and m respectively. Newtonian results are recovered when v ≪ c (correspondence principle).

35. DefinitionHL only

    Define a worldline and state what its gradient represents on a spacetime diagram.

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    A worldline is the path traced by an object through spacetime, i.e. the line joining all the events (x, ct) that the object occupies. On a diagram with ct vertical and x horizontal, gradient = c/v, so a stationary object has a vertical worldline, light has gradient 1 (45°) and no massive object can have a worldline tilted more than 45° from the ct axis. Exam tip: the commonest error is quoting the gradient as v/c — that is the gradient measured from the ct axis (and the tilt of the ct′ axis), not the gradient of the worldline on ct-versus-x axes. State which axis is which before quoting a gradient. A worldline is a locus, not a vector.

36. DefinitionHL only

    Define the light cone and describe the three regions it divides spacetime into.

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    The light cone at an event is the surface traced by light worldlines (45° lines) through that event. It separates spacetime into: the future light cone (events that can be reached by a signal from the event), the past light cone (events that could have sent a signal to it) and "elsewhere", the region outside the cone that has no causal connection with the event. Exam tip: answers must link the cone to causality, not just to geometry — the mark is for "only events inside or on the cone can be causally connected, because a signal would need v > c to reach elsewhere". Note the ordering in time of two events in "elsewhere" is frame-dependent.

37. DefinitionHL only

    Distinguish between timelike, spacelike and lightlike (null) separations of two events.

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    Using Δs² = (cΔt)² − Δx²: timelike if Δs² > 0 — the events can be causally connected, a frame exists in which they occur at the same place, and √(Δs²)/c is the proper time between them; spacelike if Δs² < 0 — no causal connection, a frame exists in which they are simultaneous, and √(−Δs²) is the proper distance; lightlike/null if Δs² = 0 — only a light signal can connect them. Exam tip: state the sign convention you are using before classifying, and remember Δs² is the same number in every inertial frame, so the classification is absolute. Δs² has units m² (scalar, invariant).

38. DefinitionHL only

    Explain what is meant by an invariant quantity in special relativity and give three examples.

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    An invariant quantity has the same numerical value when measured in any inertial reference frame, unlike a quantity that merely happens to be conserved within one frame. Examples: the speed of light in vacuum c; the spacetime interval Δs² = (cΔt)² − Δx²; proper time interval Δt₀ and proper length L₀ (each is defined in a specified frame and so is agreed on by all observers); and the rest mass of a particle. Exam tip: students confuse "invariant" with "constant" or "conserved" — momentum is conserved but not invariant, and c is both. Say "same value for all inertial observers" to earn the mark. All the quantities listed are scalars.

39. DefinitionHL only

    Distinguish a coordinate time interval from a proper time interval.

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    A proper time interval Δt₀ is the time between two events measured by a single clock present at both events, i.e. in the frame where the events occur at the same place. A coordinate time interval Δt is the time between the same two events measured in a frame where they occur at different places, and therefore requires two synchronised clocks; Δt = γΔt₀ ≥ Δt₀. Exam tip: the mark is for "same place / single clock", not for "the clock at rest" alone — a clock can be at rest and still not be present at both events. Never call the longer (dilated) time the proper time. Both are scalars measured in s; only Δt₀ is invariant.

40. DefinitionHL only

    Distinguish the rest frame of a particle from the laboratory frame, using a decaying particle as an example.

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    The rest frame is the inertial frame in which the particle is at rest, so it measures the particle's proper lifetime Δt₀ (e.g. 2.2 μs for a muon) and any distance travelled by the ground in that frame is contracted. The laboratory (or Earth) frame is the frame in which the particle moves at speed v, so it measures the dilated lifetime γΔt₀ and the uncontracted proper distance. Exam tip: the mark scheme awards marks for consistently attaching each quantity to a named frame — write "in the muon frame" or "in the Earth frame" beside every number. Mixing a contracted length with a dilated time in the same calculation is the single most penalised error in this sub-topic.

41. DefinitionHL only

    State the units and scaling used on the axes of a spacetime diagram and explain the choice.

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    The vertical axis is ct (not t) and the horizontal axis is x, so both axes carry the same unit — metres (or both in light-seconds/seconds, or light-years/years). With this choice the worldline of light has gradient 1 and lies at 45° to both axes, and the light cone is symmetric. Exam tip: a very common lost mark is labelling the vertical axis "t" while drawing light at 45°; label it ct or t with the horizontal in light-seconds. Speeds are read as fractions of c, so the tilt of a worldline from the ct axis is v/c, a dimensionless ratio between 0 and 1.

42. EquationHL onlyData booklet: Yes

    Write the relationship between the tilt angle of the axes of a moving frame on a spacetime diagram and its speed.

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    tan θ = v/c, where θ is the angle (in degrees or rad) between the ct′ axis and the ct axis, and equally between the x′ axis and the x axis, v is the speed of frame S′ along +x relative to S (m s⁻¹) and c = 3.00 × 10⁸ m s⁻¹. Valid for uniform relative motion along a common x axis with axes drawn at a common origin. Data booklet: Yes. Common misuse: writing tan θ = c/v, or tilting both axes the same way — the ct′ and x′ axes rotate towards the 45° light line by equal angles in opposite senses (they close like scissors). Sanity check: v = 0.50c gives θ = 27°; v → c gives θ → 45°.

43. EquationHL onlyData booklet: No – derive

    Write an expression for the speed of a frame in terms of its Lorentz factor, and use it.

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    From γ = 1/√(1 − v²/c²), rearranging gives v = c√(1 − 1/γ²), with v in m s⁻¹, c = 3.00 × 10⁸ m s⁻¹ and γ dimensionless (γ ≥ 1). Valid for any inertial relative speed v < c. Data booklet: No – derive (only γ itself is printed). Common misuse: writing v = c(1 − 1/γ) or forgetting to square γ. Sanity checks: γ = 2 gives v = 0.866c; γ = 5 gives v = 0.980c; γ = 10 gives v = 0.995c. Because v depends so weakly on γ near c, quote v to at least 3 significant figures when γ > 5.

44. EquationHL onlyData booklet: Yes

    Write the inverse Lorentz transformations and state when to use them.

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    x = γ(x′ + vt′) and t = γ(t′ + vx′/c²), with x, x′ in m, t, t′ in s, v the speed of S′ relative to S along +x in m s⁻¹, c = 3.00 × 10⁸ m s⁻¹ and γ = 1/√(1 − v²/c²). Use them to convert coordinates measured in the moving frame S′ back to the ground frame S. Data booklet: Yes (obtained from the printed forms by swapping primes and reversing the sign of v). Common misuse: keeping the minus sign when transforming from S′ to S, which double-counts the motion. Sanity check: setting v = 0 gives γ = 1 and x = x′, t = t′.

45. EquationHL onlyData booklet: Yes

    Write the equation used to predict the surviving fraction of muons reaching sea level and identify the relativistic step.

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    N = N₀e^(−λt) with λ = ln2/T½ (or N = N₀e^(−t/τ) using mean lifetime τ), where N is the number surviving, N₀ the initial number, t the time elapsed in the frame used, and λ in s⁻¹. The relativistic step: in the Earth frame the half-life is dilated to γT½, so the exponent is much smaller than the classical prediction. Data booklet: Yes (decay law). Common misuse: dividing the Earth-frame flight time by the proper half-life — the two must be in the same frame. Sanity check: for v = 0.98c, γ ≈ 5.0, so a flight lasting 3 proper half-lives lasts only 0.6 dilated half-lives.

46. EquationHL onlyData booklet: No – derive

    Show, using the Galilean transformation, that acceleration is invariant, and state its significance.

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    With x′ = x − vt and t′ = t (v constant, m s⁻¹), differentiating gives u′ = u − v and differentiating again gives a′ = a (m s⁻²), because dv/dt = 0 for uniform relative motion. Significance: since mass was assumed absolute, F = ma has the same form in all inertial frames, so no mechanical experiment can identify an absolute rest frame — this is Galilean relativity. Data booklet: No – derive. Common misuse: applying u′ = u − v to light, which contradicts Maxwell's prediction of a single value for c. Sanity check: a ball dropped in a uniformly moving train falls with the same 9.8 m s⁻² measured by both observers.

47. Concept/explainHL only

    Explain the train-and-tunnel (ladder-and-barn) paradox and its resolution.

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    • A train of proper length 200 m travels at high speed through a tunnel of proper length 150 m
    • In the tunnel frame the train is contracted to less than 150 m, so both tunnel doors can be closed simultaneously with the train entirely inside
    • In the train frame the tunnel is contracted to less than 200 m, so the train cannot possibly fit
    • The resolution is the relativity of simultaneity: the two door-closing events are simultaneous only in the tunnel frame
    • In the train frame the far door closes and reopens before the near door closes, so the train is never trapped
    • Both observers agree on every local event (no collision occurs), so there is no physical contradiction. Exam tip: incomplete answers stop at "the train is contracted" and never mention simultaneity — the marks are for the ordering of the two door events.
48. Concept/explainHL only

    Explain qualitatively why the twin paradox is not a genuine paradox.

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    • Special relativity says each inertial observer sees the other's clock running slow, which appears to make the situation symmetric
    • The situation is not symmetric: the Earth twin remains in a single inertial frame throughout, while the travelling twin must decelerate and accelerate to turn around
    • The travelling twin therefore changes inertial frame, and only the Earth twin's worldline is a straight line between the two meeting events
    • The travelling twin measures the proper time between departure and return only if the journey is treated in segments; the straight worldline always records the greatest proper time
    • Hence the traveller returns younger, by a factor γ
    • Confirmed by atomic clocks flown around the Earth (Hafele–Keating). Exam tip: answers that say "acceleration causes the ageing" score poorly — say the acceleration breaks the symmetry by changing frames; the ageing difference comes from the different worldline lengths.
49. Concept/explainHL only

    Explain how the light cone preserves causality even though the time order of some events is frame-dependent.

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    • Two events separated by a timelike or lightlike interval lie inside or on each other's light cone and can be connected by a signal travelling at v ≤ c
    • For such events every inertial observer agrees on which came first, so cause always precedes effect
    • Two events separated by a spacelike interval lie in each other's "elsewhere" and cannot be connected by any signal
    • For these events different observers can disagree on the time order, and a frame exists in which they are simultaneous
    • This disagreement is harmless because no information passes between them, so neither can influence the other
    • Signalling faster than light would connect spacelike events and allow an effect to precede its cause in some frame. Exam tip: students often claim "relativity means anything can happen in any order" — restrict the reordering to spacelike-separated events.
50. Concept/explainHL only

    Explain why length contraction occurs only along the direction of relative motion.

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    • Length contraction follows from the Lorentz transformation, which mixes only x and t; y′ = y and z′ = z are unchanged
    • A thought experiment: if transverse lengths changed, a rod moving through a ring of equal diameter would both pass through and fail to pass through, depending on the frame, which is a physical contradiction
    • Physical events such as "the rod scratches the ring" must be agreed on by all observers, so transverse dimensions must be invariant
    • The relativity principle also requires the effect to be symmetric between the two frames, which is only consistent if transverse lengths are unchanged
    • Hence a moving cube appears shortened only in the direction of travel, not in cross-section. Exam tip: many students state that a moving object "shrinks in all directions" or that it "looks smaller" — contraction is a measurement of length, made at a single instant in the observing frame.
51. Concept/explainHL only

    Explain why relativistic effects are not noticed in everyday life.

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    • γ = 1/√(1 − v²/c²) differs measurably from 1 only when v is a substantial fraction of c
    • For an aircraft at 250 m s⁻¹, v/c ≈ 8 × 10⁻⁷, so γ − 1 ≈ 3 × 10⁻¹³, far below the resolution of everyday instruments
    • As v/c → 0 the Lorentz transformations reduce to the Galilean transformations and relativistic velocity addition reduces to u′ = u − v
    • This is the correspondence principle: the new theory must reproduce the well-tested old theory in its domain of validity
    • Effects are detectable at low speeds only with extremely precise clocks (atomic clocks, GPS satellites) or with particles moving near c. Exam tip: a bare statement "speeds are too small" gains little; quote a value of v/c and the resulting γ, and name the correspondence principle.
52. Concept/explainHL only

    Explain how the muon experiment is consistent when analysed in the muon's own frame rather than the Earth frame.

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    • In the Earth frame the muon's mean lifetime is dilated from 2.2 μs to γ × 2.2 μs, so it survives long enough to cross the full atmospheric distance (about 4800 m)
    • In the muon's frame the lifetime is only the proper 2.2 μs and no dilation occurs, because the muon is at rest in that frame
    • Instead the atmosphere rushes past the muon, so the distance to the ground is length contracted to 4800/γ
    • Dividing the contracted distance by the same relative speed gives the same flight time as the proper lifetime allows
    • Both frames therefore predict the same physical outcome — the same fraction of muons detected at sea level
    • The two effects are two views of one phenomenon, never applied together. Exam tip: applying both dilation and contraction in one calculation is the classic error and scores zero for the physics.
53. Concept/explainHL only

    Explain why the concept of a luminiferous ether was abandoned.

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    • Nineteenth-century physics assumed light, like sound, needed a medium, so Maxwell's speed c was taken to be relative to a stationary ether
    • Earth's orbital motion should then produce a measurable "ether wind", making the speed of light direction-dependent
    • The Michelson–Morley interferometer was sensitive enough to detect this but repeatedly gave a null result, with no fringe shift as the apparatus was rotated
    • Attempts to rescue the ether (ether drag, the Lorentz–FitzGerald contraction as an ad hoc fix) added assumptions without new predictions
    • Einstein instead elevated the invariance of c to a postulate, removing the need for any ether and explaining the null result naturally
    • Nature of Science: a simpler theory with greater predictive power replaced a patched one. Exam tip: students often say Michelson–Morley "proved" relativity — it falsified the ether hypothesis; relativity explains the result.
54. Concept/explainHL only

    Explain what relativistic velocity addition predicts when one of the speeds is c, and why this matters.

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    • The formula u′ = (u − v)/(1 − uv/c²) replaces the Galilean u′ = u − v
    • Substituting u = c gives u′ = (c − v)/(1 − v/c) = c(c − v)/(c − v) = c for any v < c
    • So light measured at c in one inertial frame is measured at c in every inertial frame, exactly as the second postulate requires
    • For any u, v < c the formula always returns a value less than c, so no combination of sub-light speeds can produce a speed of c or greater
    • When u and v are both much less than c the denominator → 1 and the Galilean result is recovered
    • This makes c a limiting speed for the transmission of energy and information. Exam tip: incomplete answers assert the result rather than substituting u = c and showing the algebraic cancellation, which is what "show that" demands.
55. Concept/explainHL only

    Explain why two clocks synchronised in one inertial frame are not synchronised in another.

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    • Clocks are synchronised in a frame by sending light signals from the midpoint and setting both clocks when the light arrives
    • In a frame moving relative to that one, the light still travels at c but the clocks are moving, so the light reaches the rear clock before the front clock
    • Therefore the two setting events, simultaneous in the first frame, are not simultaneous in the second
    • The Lorentz transformation quantifies this: for events with Δt = 0 in S, Δt′ = −γvΔx/c², proportional to their spatial separation Δx
    • Clocks separated along the direction of motion are therefore offset, with the leading clock reading behind (the "leading clocks lag" rule)
    • Clocks at the same place (Δx = 0) remain synchronised in all frames. Exam tip: state that the offset depends on separation along the motion — answers omitting Δx cannot explain the paradoxes.
56. Worked problemHL onlyData booklet: Yes

    The ct′ axis of a moving frame is drawn at 31° to the ct axis of a spacetime diagram. Determine the speed of the frame and the angle between the x and x′ axes.

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    Principle: on a spacetime diagram with ct and x axes, the axes of the moving frame satisfy tan θ = v/c (data booklet). Substituting: tan 31° = 0.601, so v = 0.601 × 3.00 × 10⁸ m s⁻¹ = 1.80 × 10⁸ m s⁻¹, i.e. v = 0.60c. The x′ axis is tilted from the x axis by the same angle, 31°, but in the opposite rotational sense, so that both axes close symmetrically about the 45° light worldline. The angle between the ct′ and x′ axes is therefore 90° − 2(31°) = 28°. Answer: v = 1.8 × 10⁸ m s⁻¹ (0.60c); angle between x and x′ = 31°. Check/Trap: θ can never exceed 45°, since that would require v > c; check tan θ < 1 before quoting a speed.

57. Worked problemHL onlyData booklet: Yes

    A train of proper length 200 m passes through a tunnel of proper length 150 m at 0.80c. Determine the train's length in the tunnel frame and the tunnel's length in the train frame, and state whether the train is ever fully enclosed.

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    Principle: L = L₀/γ, with the contracted length measured in the frame in which the object moves. γ = 1/√(1 − 0.80²) = 1/√0.36 = 1/0.600 = 1.67. Tunnel frame: the train moves, so L = 200/1.67 = 120 m. Since 120 m < 150 m, in this frame the train fits entirely inside and both doors could be shut at the same instant. Train frame: the tunnel moves, so L = 150/1.67 = 90 m, which is much shorter than the 200 m train, so the train cannot fit. Answers: 1.2 × 10² m and 90 m. Check/Trap: the two frames are not contradictory — the door-closing events are simultaneous only in the tunnel frame; in the train frame the far door opens before the near door closes.

58. Worked problemHL onlyData booklet: Yes

    A twin travels to a star 6.0 ly away at 0.80c and returns immediately at the same speed. Determine the ageing of each twin.

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    γ = 1/√(1 − 0.64) = 1.67. Earth frame: the distance is the proper distance 6.0 ly, so each leg takes t = d/v = 6.0/0.80 = 7.5 y, giving a round trip of 15 y for the Earth twin. Traveller: the clock carried on the ship measures the proper time for each leg, Δt₀ = Δt/γ = 15/1.67 = 9.0 y. Cross-check in the ship frame: the distance is contracted to 6.0/1.67 = 3.6 ly, so each leg takes 3.6/0.80 = 4.5 y, total 9.0 y — the same. Answer: the Earth twin ages 15 y, the traveller ages 9.0 y, a difference of 6.0 y. Check/Trap: use light-years with speeds in units of c to avoid unit conversion; never apply time dilation and length contraction in the same leg.

59. Worked problemHL onlyData booklet: Yes

    Two events in the Earth frame are separated by Δx = 1200 m and Δt = 3.0 μs. Classify their separation and determine the speed of the frame in which they are simultaneous.

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    Principle: Δs² = (cΔt)² − Δx² is invariant. cΔt = 3.00 × 10⁸ × 3.0 × 10⁻⁶ = 900 m, so Δs² = 900² − 1200² = 8.10 × 10⁵ − 1.44 × 10⁶ = −6.3 × 10⁵ m². Since Δs² < 0 the separation is spacelike: the events cannot be causally connected, and the proper distance between them is √(6.3 × 10⁵) = 7.9 × 10² m. For simultaneity, set Δt′ = γ(Δt − vΔx/c²) = 0, giving v = c²Δt/Δx = c(cΔt/Δx) = c(900/1200) = 0.75c = 2.3 × 10⁸ m s⁻¹. Answer: spacelike; v = 0.75c. Check/Trap: v must come out less than c for a spacelike pair — if it does not, the separation was timelike and you should have solved Δx′ = 0 instead.

60. Worked problemHL onlyData booklet: Yes

    Two spacecraft approach each other, each moving at 0.90c relative to Earth. Determine the speed of one spacecraft as measured from the other.

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    Principle: relativistic velocity addition u′ = (u − v)/(1 − uv/c²). Work in the frame of spacecraft A, which moves at v = −0.90c relative to Earth; spacecraft B moves at u = +0.90c in the Earth frame. Substituting: u′ = (0.90c − (−0.90c))/(1 − (0.90c)(−0.90c)/c²) = 1.80c/(1 + 0.81) = 1.80c/1.81 = 0.9945c. Answer: 0.99c, or 2.98 × 10⁸ m s⁻¹ (3 s.f.). Check/Trap: the Galilean answer of 1.8c is impossible; the relativistic result is always less than c. Watch the signs — approaching craft have velocities of opposite sign, which makes the denominator 1 + 0.81 rather than 1 − 0.81; using the wrong sign gives the absurd 9.5c.

61. Worked problemHL onlyData booklet: Yes

    Muons are created 4800 m above sea level and travel down at 0.98c with a proper mean lifetime of 2.2 μs. Determine the fraction reaching sea level, with and without time dilation.

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    Earth-frame flight time: t = 4800/(0.98 × 3.00 × 10⁸) = 4800/2.94 × 10⁸ = 1.63 × 10⁻⁵ s = 16.3 μs. Classical prediction: N/N₀ = e^(−t/τ) = e^(−16.3/2.2) = e^(−7.42) = 6.0 × 10⁻⁴, i.e. about 0.06%. Relativistic: γ = 1/√(1 − 0.98²) = 1/√0.0396 = 5.03, so the mean lifetime in the Earth frame is τ′ = 5.03 × 2.2 = 11.1 μs, giving N/N₀ = e^(−16.3/11.1) = e^(−1.48) = 0.23, i.e. about 23%. Answers: 6.0 × 10⁻⁴ classically, 0.23 relativistically — a factor of about 380. Check/Trap: the large measured muon flux at sea level matches the relativistic figure. Use mean lifetime with e^(−t/τ) and half-life with e^(−λt), λ = ln2/T½ — never mix them.

62. Worked problemHL onlyData booklet: Yes

    Determine the speed at which a rod's measured length is half its proper length, and state the corresponding time dilation factor.

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    Principle: L = L₀/γ, so L/L₀ = 1/γ = 0.500 gives γ = 2.00. Then from γ = 1/√(1 − v²/c²): 1 − v²/c² = 1/γ² = 0.250, so v²/c² = 0.750 and v = c√0.750 = 0.866c = 2.60 × 10⁸ m s⁻¹ (3 s.f.). Because γ appears as a factor in both effects, a clock moving at this speed also runs slow by the same factor: a proper time interval of 1.00 s is measured as 2.00 s in the frame in which the clock moves. Answer: v = 0.866c; time dilation factor 2.00. Check/Trap: halving the length needs 87% of c, not 50% — γ rises very slowly at first, so students who guess v = 0.5c (γ = 1.15) are far out. Carry at least 4 figures inside the square root.

63. Graph/diagramHL onlyData booklet: Yes

    Describe how worldlines of objects with different speeds appear on a spacetime diagram, and how to extract a speed from one.

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    Axes: ct (vertical, in metres or light-seconds) against x (horizontal, same unit). Shapes: an object at rest in the frame has a vertical straight worldline; a light pulse has a straight worldline at 45°; a uniformly moving object has a straight worldline tilted between these, and an accelerating object has a curved worldline. To extract a speed, measure the tilt angle θ from the ct axis and use v = c tan θ, or read the gradient of the line, which equals c/v, so v = c/gradient. As the speed increases the worldline tilts further towards the 45° light line but can never reach or cross it, since that would require v > c. A worldline crossing another means the two objects are at the same place at the same time — a collision or meeting event.

64. Graph/diagramHL onlyData booklet: No – derive

    Describe how the twin paradox appears on a spacetime diagram.

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    Axes: ct (vertical) against x (horizontal), drawn in the Earth frame. The Earth twin's worldline is a straight vertical line from the departure event to the reunion event. The traveller's worldline is a bent (kinked) path: one straight segment tilted at tan θ = v/c out to the turning event, then a segment tilted the same angle the other way back to the reunion. Both worldlines join the same two events, so the two twins can compare clocks directly. The proper time along each worldline is found from Δs²/c², and because Δs² = (cΔt)² − Δx² subtracts the spatial term, the bent path always accumulates less proper time than the straight one. Increasing v tilts the segments closer to 45° and reduces the traveller's ageing further; the kink shows the change of inertial frame that breaks the symmetry.

65. Graph/diagramHL onlyData booklet: No – derive

    Describe the shape of a graph of the Lorentz factor γ against v/c and how to use it.

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    Axes: γ (dimensionless, vertical, starting at 1) against v/c (horizontal, 0 to 1). Shape: a curve starting at (0, 1) with zero initial gradient, almost flat up to about v = 0.3c (γ < 1.05), then rising increasingly steeply and asymptotic to the vertical line v/c = 1, where γ → ∞. There is no gradient with a simple physical meaning, but the intercept γ = 1 at v = 0 expresses the correspondence principle — Newtonian physics is recovered at low speed. Use it by reading γ at a given v/c to scale a proper time (Δt = γΔt₀) or a proper length (L = L₀/γ), or by reading backwards from a required γ to the necessary speed. The steepness near v = c shows why quoting v to 3 or 4 significant figures matters at high γ.

66. Exam technique/trapHL only

    Outline how a school-level muon detection investigation tests time dilation, including method, controls, limitations and improvements.

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    Apparatus: a scintillator with photomultiplier (or coincidence Geiger tubes) and a timer/counter. Method: record the muon count rate for a fixed long interval at a high altitude (mountain, aircraft) and then at sea level; use absorber thickness or coincidence between two detectors to select downward-going muons and reject other radiation. Controls: identical detector, threshold and counting time at both sites; measure and subtract the background rate. Analysis: compare the measured ratio of rates with the classical prediction e^(−t/τ) and the relativistic prediction e^(−t/γτ). Uncertainty: counting is random, so absolute uncertainty is √N and fractional uncertainty 1/√N — count for long times to reduce it, and use error bars. Limitations: a spread of muon speeds and altitudes of production, plus systematic error from detector efficiency. Improvements: repeat at several altitudes and plot ln(rate) against time of flight to linearise.

67. Exam technique/trapHL only

    Trap: what precision errors arise when calculating γ and speeds close to c, and how are they avoided?

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    The trap: rounding γ early. Because γ = 1/√(1 − v²/c²), the bracket is a small difference of numbers close to 1, so premature rounding destroys precision — writing γ = 5 instead of 5.025 for v = 0.98c changes a lifetime by 0.5%. Students fall for it because they treat γ like an ordinary factor. Correct approach: keep v/c exact as a fraction where possible, carry at least 4 significant figures through the square root, and round only at the final line, quoting to the number of significant figures in the data (usually 2 or 3). Going the other way, v = c√(1 − 1/γ²) is very insensitive to γ, so speeds near c need 3–4 significant figures (0.9945c, not 0.99c) to be meaningful. Command terms: "determine" and "calculate" require working plus a unit; "estimate" allows 1 significant figure.

68. Exam technique/trapHL only

    Trap: distinguish what a relativistic observer measures from what they would see, and identify what is not contracted.

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    The trap: assuming that length contraction describes the visual appearance of a fast object, and that everything about it shrinks. Students fall for it because "observer" suggests looking. What relativity gives is a measurement — the positions of the two ends recorded simultaneously in the observer's frame — not an image. A photograph also involves the different light travel times from near and far parts of the object, which makes a fast cube appear rotated rather than flattened; this optical effect is not required by the syllabus but the distinction is. Also not contracted: dimensions perpendicular to the motion, the proper length itself (a frame-fixed number), the object's rest mass, and c. Correct approach: for every quantity, name the frame, ask which frame the object is at rest in, and apply L = L₀/γ only along the direction of motion.

Practise this topic with exam-style questions: A.5 Galilean and special relativity questions (HL) · all flashcards

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