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Relativistic Addition of Velocities

Key idea: Use the relativistic velocity addition formula to combine speeds near c, keep light speed invariant, and solve exam-style problems.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • discuss qualitatively the results of the Michelson–Morley interferometer experiment and its implications on the ether theory (knowledge of the details of the experiment is not required)
  • state the postulates of the special theory of relativity, that in all inertial frames, the laws of physics are the same and the speed of light in free space is the same regardless of the motion of the light source or observer
  • appreciate the failure of Galilean transformation equations when applied to a moving source of light
  • discuss the concept of simultaneity
  • show an understanding of the terms proper time and proper length
  • apply the Lorentz transformation equations to solve one-dimensional problems
  • Derive the time dilation formula and the length contraction formula, making use of the Lorentz factor.
  • apply the time dilation formula and the length contraction formula in related situations (e.g. the lifetime of fast-moving muons) or to solve problems
  • use the one-dimensional relativistic velocity addition formula to calculate velocities in different inertial frames or to solve problems
  • Apply the relativistic energy–momentum relation E² = (pc)² + (mc²)² to solve problems, including selecting its limiting form.
  • Show that E² = (pc)² + (mc²)² reduces to E = pc for massless particles and to E = mc² + ½mv² at low speeds.

At high speeds, velocities do not add linearly. The relativistic velocity addition formula ensures that if one observer measures something moving at the speed of light c, every inertial observer also measures c.

Use this page right after Lorentz transformations so the velocity-addition formula feels like a direct consequence, not a standalone rule.

1. Definitions (Must Know)

  • Standard setup: frame S' moves at speed v along + x relative to frame S.
  • Object speed in S (along x): u (m s⁻¹).
  • Object speed in S' (along x): u' (m s⁻¹).
  • Speed of light in vacuum: c ≈ 3.00 × 10⁸ m s⁻¹.

2. Key Ideas (What Earns Marks)

  • Relativistic velocity addition (1D): u' = (u-v)/(1-uv/c²)
  • Inverse form: u = (u' + v)/(1 + u'v/c²)
  • If u = c, then u' = c (invariance of light speed).
  • Classical limit: if u≪ c and v≪ c, then u' ≈ u-v (Galilean).

3. Detailed Explanations

A. Where it comes from (Lorentz transformations)

Start from the Lorentz transformations: x' = γ(x-vt), t' = γ(t-vx/c²) Differentiate and take the ratio: u' = dx'/dt' = (dx-vdt)/(dt-(v/c²)dx) = (u-v)/(1-uv/c²)

B. Why it matters

Galilean addition would let you exceed c by adding speeds. The relativistic formula prevents that for any |u| < c and |v| < c.

4. Common Mistakes

  • Using u' = u-v at relativistic speeds.
  • Dropping the minus sign in the denominator (1-uv/c²).
  • Mixing up which frame is moving: clearly define S' moving at + v relative to S (then the formula above applies).
  • Treating the formula as symmetric without changing the sign of v when reversing frames (use the inverse form instead).

5. Exam Tips

  • State the frame setup in one line before substituting.
  • Use a quick check:
    • if v = 0, then u' = u,
    • if u = v, then u' = 0.
  • Keep signs (directions) explicit: u and v are signed velocities in 1D.

6. Worked Examples

Modelled example 1

Two spaceships (same direction)

Core

Problem

A moves at 0.80c relative to Earth and B at 0.60c in the same direction. Find A’s velocity relative to B.
Study the worked solution

  1. Define frames

    Method

    Take Earth as S and B as S'.

    Reason

    Then S' moves at v = +0.60c and A has u = +0.80c in S.

    Working

    u' = v_(A/B)

  2. Use the forward rule

    Method

    Transform A’s Earth velocity into B’s frame.

    Reason

    The signed values match the stated standard setup.

    Working

    u' = (u-v)/(1-uv/c²)

  3. Calculate

    Method

    Substitute dimensionless speed ratios.

    Reason

    The relativistic denominator differs materially from one.

    Working

    u' = ((0.80-0.60)c)/(1-(0.80)(0.60)) = 0.38c

Guided practice 2

Light speed stays c

About 4 min

Problem

In S, light has u = c. Substitute symbolically to show what an inertial frame S' moving at speed v measures.

Try this before viewing the solution

Hints

Hint 1: factor the numerator
Write c-v = c(1-v/c).
View solution step by step

  1. Substitute

    Method

    Set u = c in the forward rule.

    Reason

    The light ray moves along + x.

    Working

    u' = (c-v)/(1-cv/c²)

  2. Simplify

    Method

    Factor the identical dimensionless terms.

    Reason

    They cancel for any permitted inertial boost.

    Working

    u' = (c(1-v/c))/(1-v/c) = c

Common misconception 3

Opposite directions (signs matter)

Find and correct the mistake

Learner claim

A particle has u = -0.50c while S' has v = +0.60c. A learner substitutes both as positive speed magnitudes. Explain the sign error and find u'.

Try this before viewing the solution

Sign of particle velocity in S

View solution step by step

  1. Set signed values

    Method

    Use u = -0.50c and v = +0.60c.

    Reason

    Velocity addition is directional.

    Working

    + x is rightward

  2. Substitute

    Method

    Keep the negative product in the denominator.

    Reason

    uv/c² = -0.30.

    Working

    u' = (-0.50c-0.60c)/(1-(-0.50)(0.60)) = -0.85c

  3. Interpret

    Method

    The particle moves left in S' at speed 0.85c.

    Reason

    The negative result encodes direction.

    Working

    u' = -0.85c

Examiner practice 4

Solve for u given u' (inverse form)

4 marks

Examination question

In S', a rocket has u' = 0.50c along + x; S' moves at 0.60c along + x relative to S. Find u in S. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Identify direction

    1 mark

    Method

    Convert from the primed frame to the unprimed frame.

    Reason

    The known object velocity is u'.

    Working

    S' → S

  2. Select equation

    1 mark

    Method

    Use the inverse addition formula.

    Reason

    The plus signs correspond to reversing the boost.

    Working

    u = (u' + v)/(1 + u'v/c²)

  3. Substitute

    1 mark

    Method

    Insert both positive velocities.

    Reason

    Both point along + x.

    Working

    u = ((0.50 + 0.60)c)/(1 + (0.50)(0.60))

  4. Calculate

    1 mark

    Method

    The rocket has u = 0.85c in S.

    Reason

    The result remains below c.

    Working

    u = 0.85c

Challenge 5

Two very fast ships (high-speed difference)

Minimal support

Independent transfer

A moves at 0.95c and B at 0.90c relative to Earth in the same direction. Define frames, find A’s velocity relative to B, and compare it with naive subtraction.

Try this before viewing the solution

Hints

Hint 1: make B the primed frame
Orient with Earth as S, B as S', u = 0.95c and v = 0.90c.
View solution step by step

  1. Define frames

    Method

    Earth is S and B is S'.

    Reason

    The required answer is A’s velocity in B’s frame.

    Working

    u' = v_(A/B)

  2. Calculate

    Method

    Use relativistic subtraction.

    Reason

    Both Earth-frame speeds are relativistic.

    Working

    u' = ((0.95-0.90)c)/(1-(0.95)(0.90)) = 0.34c

  3. Compare

    Method

    The relative speed is not the naive 0.05c.

    Reason

    The Lorentz denominator is 0.145, not approximately one.

    Working

    0.34c > 0.05c

7. Mind Stretchers

Mind stretcher 1: Why you never get u' > cExtension

For |u| < c and |v| < c, explain qualitatively why the denominator prevents the result from exceeding c.

Answer

As u and v approach c, the product uv/c² approaches 1, making the denominator small in just the right way to keep u' below c. The Lorentz transformations “couple” space and time so that boosts cannot push any physical speed past c.

Mind stretcher 2: Recover the Galilean limit (qualitative)Extension

Why does u' ≈ u-v when u≪ c and v≪ c?

Answer

When u/c and v/c are tiny, the product uv/c² is extremely small, so the denominator 1-uv/c² is essentially 1. The formula then reduces to u' ≈ u-v, matching the low-speed (Galilean) limit.

The velocity addition rule is a direct consequence of the Lorentz transformations:

Next in the maintained sequence: Relativistic Energy & Momentum.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027