A.5 Special relativity: energy and momentum (AHL)

IB Physics HL· Theme A: Space, time and motion, A.5 AHL· 45 min read

1. Relativistic Momentum★★★☆☆HL only⏱ 10 min

📘 Definition

Relativistic momentum

$p = γ m_{0} v$

The corrected expression for momentum of a particle moving at speed $v$ that preserves conservation of momentum in all inertial reference frames.

Example:

For $v ≪ c$ , $γ \approx 1$ , so $p \approx m_{0} v$ , matching the classical definition.

The classical definition of momentum fails to conserve in relativistic particle collisions. Adding the Lorentz factor to the definition fixes this, and correctly shows that momentum approaches infinity as $v$ approaches $c$ , making the speed of light unreachable for massive particles.

📐 Worked Example

A proton (rest mass $m_{0} = 1.67 \times 1 0^{- 27}$ kg) moves at $v = 0.8 c$ relative to a lab observer. Calculate its relativistic momentum.

1
First calculate the Lorentz factor $γ$ :
2
$\gamma = \frac{1}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{0.6} \approx 1.67$
3
Substitute into the relativistic momentum formula:
4
$p = \gamma m_0 v = 1.67 \times (1.67 \times 10^{-27}) \times (0.8 \times 3 \times 10^8) \approx 6.7 \times 10^{-19} \text{ kg m s}^{-1}$

Exam tip:

IB does not use the relativistic mass convention; always use rest mass $m_{0}$ in all formulas.

2. Total Energy and Rest Energy★★★☆☆HL only⏱ 10 min

📘 Definition

Rest energy

$E_{0} = m_{0} c^{2}$

The inherent energy a particle has due to its rest mass, even when it is not moving relative to the observer.

The total relativistic energy of a moving particle is the sum of its rest energy and relativistic kinetic energy. The key invariant relation between total energy, momentum and rest mass holds in all inertial frames, making it extremely useful for problem solving.

E^2 = (p c)^2 + (m_0 c^2)^2

📐 Worked Example

Find the relativistic kinetic energy of the proton from the previous example ( $v = 0.8 c$ , $m_{0} = 1.67 \times 1 0^{- 27}$ kg), and compare to the classical prediction.

1
Calculate total relativistic energy:
2
$E = \gamma m_0 c^2 = 1.67 \times 1.67 \times 10^{-27} \times (3 \times 10^8)^2 \approx 2.51 \times 10^{-10} \text{ J}$
3
Calculate rest energy:
4
$E_0 = m_0 c^2 = 1.67 \times 10^{-27} \times 9 \times 10^{16} \approx 1.50 \times 10^{-10} \text{ J}$
5
Relativistic kinetic energy is $K = E - E_{0}$ :
6
$K = 2.51 \times 10^{-10} - 1.50 \times 10^{-10} \approx 1.01 \times 10^{-10} \text{ J}$
7
Compare to classical kinetic energy $\frac{1}{2} m_{0} v^{2}$ :
8
$K_{classical} = 0.5 \times 1.67 \times 10^{-27} \times (0.8 \times 3 \times 10^8)^2 \approx 4.81 \times 10^{-11} \text{ J}$
9
The classical prediction underestimates kinetic energy by over 50% at this relativistic speed, showing the large error of classical mechanics at high speed.

Exam tip:

Kinetic energy is always total energy minus rest energy; never use the classical formula for $v > 0.1 c$ .

3. Mass-Energy Equivalence★★☆☆☆HL only⏱ 8 min

📘 Definition

Mass-energy equivalence

$Δ E = Δ m c^{2}$

The principle that mass and energy are interchangeable; the total energy of a system is proportional to its total mass, even at rest.

Example:

This explains the energy released in nuclear fission and fusion, where rest mass is converted to kinetic energy.

In nuclear reactions, the total rest mass of the products is different from the total rest mass of the reactants. This difference (the mass defect) corresponds to the energy released or absorbed in the reaction, calculated directly from $Δ E = Δ m c^{2}$ .

📐 Worked Example

The mass defect for a single U-235 fission event is approximately $0.18$ u. Calculate the energy released in MeV, given $1 u = 931.5$ MeV c $^{- 2}$ .

1
Substitute directly into the mass-energy relation:
2
$\Delta E = \Delta m c^2 = 0.18 \text{ u} \times 931.5 \text{ MeV c}^{-2} \times c^2$
3
Cancel $c^{2}$ and calculate:
4
$\Delta E = 0.18 \times 931.5 \approx 170 \text{ MeV}$

Exam tip:

IB usually gives $1 u = 931.5$ MeV c $^{- 2}$ , which simplifies calculations by canceling the $c^{2}$ term automatically.

4. Conservation of Relativistic Energy-Momentum★★★★★HL only⏱ 15 min

For any closed system, total relativistic energy and total relativistic momentum are both conserved in all interactions, including particle decays, collisions, annihilation and pair production. The energy-momentum invariant can be used to simplify problems where particles are created or destroyed.

📐 Worked Example

An electron and positron (each rest mass $0.511$ MeV c $^{- 2}$ ) annihilate at rest to produce two identical gamma rays. Find the energy of each gamma ray.

1
Calculate total initial energy; both particles are at rest so total energy equals the sum of their rest energies:
2
$E_{total} = 2 \times m_0 c^2 = 2 \times 0.511 \text{ MeV} = 1.022 \text{ MeV}$
3
Total initial momentum is zero (both particles at rest), so the two gamma rays must have equal and opposite momentum, hence equal energy to conserve momentum.
4
Total energy is conserved, so each gamma carries half the total energy:
5
$E_{\gamma} = \frac{1.022}{2} = 0.511 \text{ MeV}$

5. Common Pitfalls

Wrong move:

Using the classical kinetic energy formula $\frac{1}{2} m_{0} v^{2}$ for relativistic speeds

Why:

Classical kinetic energy severely underestimates kinetic energy at speeds above $0.1 c$ , leading to large mark losses

Correct move:

Always calculate relativistic kinetic energy as $K = γ m_{0} c^{2} - m_{0} c^{2} = E_{t o t a l} - E_{0}$

Wrong move:

Using the relativistic mass concept in calculations

Why:

IB Physics explicitly does not use relativistic mass, and answers using this concept will be marked incorrect

Correct move:

Always work with invariant rest mass $m_{0}$ and use the energy-momentum invariant $E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}$

Wrong move:

Forgetting to convert units when calculating energy from mass defect

Why:

Using atomic mass units directly in $D e l t a E = D e l t am c^{2}$ gives the wrong unit for energy if joules are requested

Correct move:

Convert $Δ m$ to kg for energy in joules, or convert MeV to joules after calculating in MeV

Wrong move:

Only conserving energy and ignoring momentum conservation

Why:

Many problems require both conservation laws to find the correct answer, especially for photon production and collisions

Correct move:

Always write equations for both total energy and total momentum conservation for the entire system

Wrong move:

Assuming total rest mass is conserved in reactions

Why:

Rest mass can be converted to kinetic energy (and vice versa), so only total energy (not total rest mass) is conserved

Correct move:

Calculate the mass defect $Δ m$ (change in total rest mass) to find the energy released or absorbed

6. Quick Reference Cheatsheet

Quantity	Formula	Key Notes
Relativistic momentum	$p = γ m_{0} v$	$γ = 1/ 1 - v^{2} / c^{2}$
Rest energy	$E_{0} = m_{0} c^{2}$	Invariant across all frames
Total energy	$E = γ m_{0} c^{2}$	Rest + kinetic energy
Relativistic KE	$K = (γ - 1) m_{0} c^{2}$	Not $\frac{1}{2} m_{0} v^{2}$
Energy-momentum invariant	$E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}$	Same in all inertial frames
Mass-energy conversion	$Δ E = Δ m c^{2}$	$1 u = 931.5$ MeV c $^{- 2}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 2
Calculate energy of accelerated proton
2022 · 1
Mass defect in alpha decay
2021 · 2
Relativistic momentum collision problem

Going deeper

ib syllabusIB Physics HL 2025 SyllabusOfficial syllabus for reference

What's Next

The concepts of relativistic energy and momentum you learned here are foundational for almost all advanced topics in IB Physics HL. Mass-energy equivalence is critical for nuclear physics topics, where you will use it to calculate binding energy and energy released in fission, fusion, and radioactive decay. Relativistic energy-momentum conservation is used constantly in particle physics, to solve problems involving particle decays and collisions. This topic also forms the base for understanding general relativity, where energy and momentum curve spacetime to produce gravity. Mastering the invariant relation here will pay off across many other topics in the syllabus.