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

> IB Physics HL · IB Physics HL 2025+
> Source: https://www.owlsprep.com/study/ib-physics-hl-u1-a-5-special-relativity-energy/

This AHL sub-topic extends the postulates of special relativity to energy and momentum, covering mass-energy equivalence, relativistic kinetic energy, and conservation rules for high-speed particle interactions, a frequently tested topic for IB Physics HL.

**Prerequisites:** [Special relativity fundamentals and Lorentz factor](https://www.owlsprep.com/study/ib-physics-hl-u1-a-4-special-relativity-origins/); Conservation of momentum and energy in classical mechanics

## Learning objectives

- Calculate relativistic momentum for particles moving at relativistic speeds
- Distinguish between rest energy, total energy and relativistic kinetic energy
- Apply mass-energy equivalence to calculate energy released in nuclear reactions
- Solve problems using conservation of relativistic energy and momentum

## Relativistic Momentum

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

*Notation:* $p = \gamma m_0 v$

*Example:* For $v \ll c$, $\gamma \approx 1$, 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 $m_0 = 1.67 \times 10^{-27}$ kg) moves at $v = 0.8c$ relative to a lab observer. Calculate its relativistic momentum.

1. First calculate the Lorentz factor $\gamma$:
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.

## Total Energy and Rest Energy

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

*Notation:* $E_0 = m_0 c^2$

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 ($v = 0.8c$, $m_0 = 1.67 \times 10^{-27}$ 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 $v > 0.1c$.

## Mass-Energy Equivalence

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

*Notation:* $\Delta E = \Delta m c^2$

*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 $\Delta E = \Delta m c^2$.

**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.

## Conservation of Relativistic Energy-Momentum

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. $$E_{\gamma} = \frac{1.022}{2} = 0.511 \text{ MeV}$$

> **tip**
>
> If your system starts at rest, always note that total initial momentum is zero, this simplifies most problems significantly.

## Common pitfalls

- **Wrong:** Using the classical kinetic energy formula $\frac{1}{2}m_0v^2$ for relativistic speeds
  - Why it fails: Classical kinetic energy severely underestimates kinetic energy at speeds above $0.1c$, leading to large mark losses
  - Correct: Always calculate relativistic kinetic energy as $K = \gamma m_0 c^2 - m_0 c^2 = E_{total} - E_0$
- **Wrong:** Using the relativistic mass concept in calculations
  - Why it fails: IB Physics explicitly does not use relativistic mass, and answers using this concept will be marked incorrect
  - Correct: Always work with invariant rest mass $m_0$ and use the energy-momentum invariant $E^2 = (pc)^2 + (m_0 c^2)^2$
- **Wrong:** Forgetting to convert units when calculating energy from mass defect
  - Why it fails: Using atomic mass units directly in $\Delta E = \Delta m c^2$ gives the wrong unit for energy if joules are requested
  - Correct: Convert $\Delta m$ to kg for energy in joules, or convert MeV to joules after calculating in MeV
- **Wrong:** Only conserving energy and ignoring momentum conservation
  - Why it fails: Many problems require both conservation laws to find the correct answer, especially for photon production and collisions
  - Correct: Always write equations for both total energy and total momentum conservation for the entire system
- **Wrong:** Assuming total rest mass is conserved in reactions
  - Why it fails: Rest mass can be converted to kinetic energy (and vice versa), so only total energy (not total rest mass) is conserved
  - Correct: Calculate the mass defect $\Delta m$ (change in total rest mass) to find the energy released or absorbed

## Cheatsheet

| Quantity | Formula | Key Notes |
| --- | --- | --- |
| Relativistic momentum | $p = \gamma m_0 v$ | $\gamma = 1/\sqrt{1-v^2/c^2}$ |
| Rest energy | $E_0 = m_0 c^2$ | Invariant across all frames |
| Total energy | $E = \gamma m_0 c^2$ | Rest + kinetic energy |
| Relativistic KE | $K = (\gamma - 1) m_0 c^2$ | Not $\frac{1}{2}m_0 v^2$ |
| Energy-momentum invariant | $E^2 = (pc)^2 + (m_0 c^2)^2$ | Same in all inertial frames |
| Mass-energy conversion | $\Delta E = \Delta m c^2$ | $1 u = 931.5$ MeV c$^{-2}$ |

## 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.

- [A.6 Circular motion and gravitation (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u1-a-6-circular-motion-and/)
- [Theme B: The particulate nature of matter](https://www.owlsprep.com/study/ib-physics-hl-u2-overview/)
- [B.1 Temperature and thermal energy](https://www.owlsprep.com/study/ib-physics-hl-u2-b-1-temperature-and-thermal/)

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