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Relativistic Momentum and Energy

Key idea: Use relativistic momentum p=γmv and the energy–momentum relation to solve problems for massive particles and photons, with worked examples.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • 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 relativistic speeds, momentum and energy take forms that keep the laws of physics consistent across inertial frames. The key H3 tool is the energy–momentum relation: E² = (pc)² + (mc²)²

This is the maintained dynamics capstone before interpretation checks like the twin paradox and full-sequence confirmation.

1. Definitions (Must Know)

  • Lorentz factor: γ = 1/(square root of (1-v²/c²))
  • Rest mass, m (kg): invariant mass of the particle (do not use “relativistic mass”).
  • Relativistic momentum (1D magnitude): p = γ mv
  • Total energy: E = γ mc²
  • Rest energy: E₀ = mc²
  • Kinetic energy: K = E-E₀ = (γ-1)mc²
  • Energy–momentum relation (H3): E² = (pc)² + (mc²)²
  • Symbols used in this lesson: m (kg), v (m s⁻¹), c (≈ 3.00 × 10⁸ m s⁻¹), γ (dimensionless), p (kg m s⁻¹), E,E₀,K (J).

2. Key Ideas (What Earns Marks)

  • In an isolated system, total energy and total momentum are conserved.
  • The energy–momentum relation lets you solve problems without needing v explicitly.
  • Required limits (H3 LO):
    • massless particle (m = 0): E = pc (e.g. photon),
    • low speed (v≪ c): E ≈ mc² + 1/2 mv².

Quick reference:

QuantityRelativistic formNotes
Momentump = γ mvUse m as rest mass
Total energyE = γ mc²Includes rest energy
Kinetic energyK = (γ-1)mc²Reduces to 1/2 mv² at low speed
Photon energyE = pcFor m = 0

3. Detailed Explanations

A. Using the energy–momentum relation

If a question gives any two of (E,p,m), you can find the third from: E² = (pc)² + (mc²)²

This is especially useful for:

  • high-speed particles where Newtonian K = 1/2 mv² is not valid, and
  • photons (where m = 0).

B. The two required limits

Massless (m = 0): E² = (pc)² ⇒ E = pc

Low-speed (v≪ c): using the binomial approximation γ ≈ 1 + (1/2)(v²/c²), E = γ mc² ≈ mc² + 1/2 mv²

4. Common Mistakes

  • Saying “protons are massless” (false). A photon is an example of a massless particle; a proton has non-zero rest mass.
  • Using K = 1/2 mv² at relativistic speeds without checking v≪ c.
  • Dropping the mc² term and treating total energy as just kinetic energy.
  • Forgetting that conservation laws apply to an isolated system (external work/impulse breaks simple conservation statements).

5. Exam Tips

  • If you see “photon” or “massless”, immediately use E = pc.
  • If you see “show it reduces to…”, write the limit explicitly (m = 0 or v≪ c).
  • Keep track of whether you’re asked for total energy E or kinetic energy K.

6. Worked Examples

Modelled example 1

Photon momentum from energy

Core

Problem

A photon has energy 3.2 × 10⁻¹⁹ J. Find its momentum.
Study the worked solution

  1. Select the limit

    Method

    Set rest mass to zero.

    Reason

    A photon is massless.

    Working

    E² = (pc)² ⇒ E = pc

  2. Rearrange

    Method

    Divide energy by c.

    Reason

    Momentum has units J divided by m s⁻¹.

    Working

    p = E/c

  3. Calculate

    Method

    Substitute c = 3.00 × 10⁸ m s⁻¹.

    Reason

    Keep the supplied significant figures.

    Working

    p = 1.1 × 10⁻²⁷ kg m s⁻¹

Guided practice 2

Total energy from momentum and rest mass

About 6 min

Problem

A particle has m = 1.0 × 10⁻²⁷ kg and p = 4.0 × 10⁻¹⁹ kg m s⁻¹. Find its total energy.

Try this before viewing the solution

Hints

Hint 1: calculate comparable energy terms
Find pc and mc² in joules before combining them.
Hint 2: combine squares, not values
The relation is a quadrature: take the square root only after adding the two squared terms.
View solution step by step

  1. Momentum term

    Method

    Convert p to the energy scale pc.

    Reason

    Both terms inside the relation must have energy units.

    Working

    pc = 1.20 × 10⁻¹⁰ J

  2. Rest term

    Method

    Calculate rest energy.

    Reason

    c² = 9.00 × 10¹⁶ m² s⁻².

    Working

    mc² = 9.0 × 10⁻¹¹ J

  3. Combine

    Method

    Add squares and take the positive root.

    Reason

    Total energy is positive.

    Working

    E = square root of ((1.20 × 10⁻¹⁰)² + (9.0 × 10⁻¹¹)²) = 1.50 × 10⁻¹⁰ J

Common misconception 3

Low-speed check

Find and correct the mistake

Learner claim

A learner says K = (γ-1)mc² cannot agree with K = (1/2)mv² because it contains rest-energy factors. Use the low-speed expansion to correct the claim.

Try this before viewing the solution

Low-speed gamma expansion

View solution step by step

  1. Expand gamma

    Method

    For v≪ c, retain the leading correction.

    Reason

    Higher powers of v/c are negligible.

    Working

    γ ≈ 1 + (1/2)v²/c²

  2. Subtract rest contribution

    Method

    γ-1 removes the leading one.

    Reason

    Kinetic energy excludes rest energy.

    Working

    γ-1 ≈ (1/2)v²/c²

  3. Cancel factors

    Method

    The c² factors cancel.

    Reason

    The classical kinetic-energy scale remains.

    Working

    K ≈ ((1/2)v²/c²)mc² = (1/2)mv²

Examiner practice 4

Kinetic energy at 0.80c

4 marks

Examination question

A particle has m = 2.0 × 10⁻²⁷ kg and v = 0.80c. Find its relativistic kinetic energy. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Calculate gamma

    1 mark

    Method

    At 0.80c, γ = 1.67.

    Reason

    1/square root of (1-0.80²) = 1/0.60.

    Working

    γ = 1.67

  2. Select energy

    1 mark

    Method

    Use K = (γ-1)mc².

    Reason

    The question asks for kinetic, not total, energy.

    Working

    K = (0.67)mc²

  3. Substitute

    1 mark

    Method

    Insert rest mass and c².

    Reason

    c² = 9.00 × 10¹⁶ in SI.

    Working

    K = (0.67)(2.0 × 10⁻²⁷)(9.00 × 10¹⁶)

  4. Calculate

    1 mark

    Method

    Give the result in joules.

    Reason

    Two significant figures suit the data.

    Working

    K = 1.2 × 10⁻¹⁰ J

Challenge 5

When kinetic energy equals rest energy

Minimal support

Independent transfer

For a massive particle, find v/c when kinetic energy equals rest energy: K = mc².

Try this before viewing the solution

Hints

Hint 1: cancel the common rest-energy scale
Orient with (γ-1)mc² = mc² before solving for speed.
View solution step by step

  1. Find gamma

    Method

    Cancel mc² from the non-zero-mass equation.

    Reason

    The energy equality fixes a dimensionless factor.

    Working

    γ-1 = 1 ⇒ γ = 2

  2. Invert gamma

    Method

    Square the reciprocal relation.

    Reason

    This isolates the squared speed ratio.

    Working

    1-v²/c² = 1/4

  3. Calculate speed

    Method

    Take the positive speed magnitude.

    Reason

    The result must be below c.

    Working

    v/c = square root of (3/4) = 0.866

7. Mind Stretchers

Mind stretcher 1: Why use E² = (pc)² + (mc²)² at all?Extension

Why is the energy–momentum relation often easier than working with v directly?

Answer

Because v can be hard to find from partial information, and many problems give p, E, or m directly. The relation links them algebraically and works cleanly for both massive and massless particles.

Mind stretcher 2: Why momentum must change from p = mvExtension

Why isn’t it enough to keep Newtonian momentum p = mv and just “add” relativistic energy corrections?

Answer

Because momentum conservation must hold consistently across inertial frames at high speeds, and the Newtonian p = mv does not transform correctly under Lorentz transformations. The relativistic momentum p = γ mv is needed so that conservation laws and collision outcomes remain consistent between frames.

8. Optional/Enrichment: Useful Unit Conversions

  • Electronvolt: 1 eV = 1.602 × 10⁻¹⁹ J (energy gained by charge e across 1 V).
  • Atomic mass unit: 1 u = 1.66 × 10⁻²⁷ kg.

Next in the maintained sequence: The Twin Paradox.

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