Mass-energy equivalence and relativistic energy: VCE Physics Unit 4

Syllabus dot point

“Use the relationship between mass and energy, total energy as the sum of rest energy and kinetic energy, and relativistic kinetic energy, and apply it to mass conversion in the Sun, electron-positron annihilation and particle accelerators”

10 min readUpdated 7 Oct 2026

Quick answer

An object of mass $m$ has rest energy $E_0 = mc^2$ . Its total energy is $E_{tot} = \gamma mc^2$ , so its kinetic energy is $E_k = (\gamma - 1)mc^2$ . Mass can be converted into energy and energy into mass, as in the Sun, in electron-positron annihilation and in accelerators.

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What this dot point is asking
Rest energy
Total energy and kinetic energy
Worked examples
Mass and energy in particle accelerators
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What this dot point is asking

VCAA expects you to interpret Einstein's prediction that the total "mass-energy" of an object is

E_{tot} = E_k + E_0 = \gamma mc^2, \qquad E_0 = mc^2, \qquad E_k = (\gamma - 1)mc^2

and to apply it to mass conversion in the Sun, positron-electron annihilation and nuclear transformations in particle accelerators. The details of particular nuclear processes are not required.

Rest energy

Rest energy is the energy an object has just because it has mass, even when at rest:

E_0 = mc^2

Because $c^2 = 9.00 \times 10^{16}$ m $^2$ s $^{-2}$ is enormous, a small mass corresponds to a very large amount of energy. One gram of mass is equivalent to $1.0 \times 10^{-3} \times 9.00 \times 10^{16} = 9.0 \times 10^{13}$ J.

Mass and energy are two forms of one thing. When mass is converted to other forms of energy, $\Delta E = \Delta m\, c^2$ .

Total energy and kinetic energy

For an object moving with Lorentz factor $\gamma$ , the total energy is

E_{tot} = \gamma mc^2

and since $E_{tot} = E_k + E_0$ ,

E_k = \gamma mc^2 - mc^2 = (\gamma - 1)mc^2

At low speeds, $(\gamma - 1) \approx \tfrac{1}{2}v^2/c^2$ and this reduces to the familiar $E_k = \tfrac{1}{2}mv^2$ . At high speed the relativistic kinetic energy is larger than $\tfrac{1}{2}mv^2$ , and it grows without limit as $v \to c$ . That is why no object with mass can be accelerated all the way to $c$ : it would need infinite energy.

Useful constants: $c = 3.00 \times 10^8$ m s $^{-1}$ , electron mass $9.11 \times 10^{-31}$ kg, $1\ \text{eV} = 1.60 \times 10^{-19}$ J.

Worked examples

Worked example

Example 1. Electron at 0.80c

Find the rest energy, total energy and kinetic energy of an electron moving at $0.80c$ , in joules and in MeV.

$\gamma = 1/\sqrt{1 - 0.80^2} = 1.667$

Rest energy: $E_0 = (9.11 \times 10^{-31})(3.00 \times 10^8)^2 = 8.20 \times 10^{-14}$ J $= 0.512$ MeV

Total energy: $E_{tot} = \gamma E_0 = 1.667 \times 8.20 \times 10^{-14} = 1.37 \times 10^{-13}$ J $= 0.854$ MeV

Kinetic energy: $E_k = (\gamma - 1)E_0 = 0.667 \times 8.20 \times 10^{-14} = 5.47 \times 10^{-14}$ J $= 0.342$ MeV

Check: $E_{tot} - E_0 = 0.854 - 0.512 = 0.342$ MeV. The classical estimate $\tfrac{1}{2}mv^2$ gives only $0.164$ MeV, less than half the correct value.

Example 2. Speed from kinetic energy

An electron has kinetic energy $1.0$ MeV. How fast is it moving?

$\gamma = 1 + E_k/E_0 = 1 + 1.0/0.512 = 2.95$

$v/c = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - 0.115} = 0.941$ , so $v = 0.94c$ .

Example 3. The Sun

The Sun radiates energy at a rate of about $3.8 \times 10^{26}$ W, supplied by nuclear fusion in the core, where mass is converted to energy. What mass is converted per second?

$\Delta m = \Delta E/c^2 = (3.8 \times 10^{26})/(9.00 \times 10^{16}) = 4.2 \times 10^{9}$ kg each second.

That is over four million tonnes of mass every second. The Sun is so massive that this loss is a tiny fraction of its mass over human timescales.

Example 4. Positron-electron annihilation

An electron and a positron (same mass, opposite charge) at rest meet and annihilate into two identical photons. Find the energy and wavelength of each photon.

Total energy released $= 2 \times 8.20 \times 10^{-14} = 1.64 \times 10^{-13}$ J. Each photon carries half: $8.20 \times 10^{-14}$ J $= 0.512$ MeV.

$\lambda = hc/E = (6.63 \times 10^{-34})(3.00 \times 10^8)/(8.20 \times 10^{-14}) = 2.43 \times 10^{-12}$ m (a gamma ray).

Two photons, travelling in opposite directions, are needed so momentum is conserved. The reverse process, energy becoming matter, also happens.

Mass and energy in particle accelerators

In an accelerator, particles are given a very large kinetic energy. When they collide, kinetic energy can be converted into the mass of new particles, using $\Delta m = \Delta E/c^2$ . For example, $1.0$ GeV $= 1.60 \times 10^{-10}$ J of available energy corresponds to $1.6 \times 10^{-10}/(9.00 \times 10^{16}) = 1.8 \times 10^{-27}$ kg, a little more than the mass of a proton. This is why high-energy colliders can create particles heavier than the ones that went in.

Relativistic energy is also why accelerators are designed around $\gamma$ , as in the accelerator example on the muons, GPS and accelerators page: at high energy, adding more energy raises $\gamma$ but the particle's speed barely changes.

Key fact

In the Sun, annihilation and accelerators the total energy is conserved. What changes is the split between rest energy (mass) and other forms of energy. When a question gives a mass change, use $\Delta E = \Delta m\,c^2$ , with the mass in kilograms for an answer in joules.

Common errors

Using $\tfrac{1}{2}mv^2$ at high speed: Above roughly $0.1c$ the classical kinetic energy starts to be noticeably wrong. Use $(\gamma - 1)mc^2$ .
Confusing $E_{tot}$ and $E_k$: $E_{tot} = \gamma mc^2$ includes the rest energy. Kinetic energy is $E_{tot} - E_0$ .
Mixing units: $mc^2$ gives joules when $m$ is in kg. Convert to eV only at the end, dividing by $1.60 \times 10^{-19}$ .
Forgetting the second photon: A single photon cannot carry away zero momentum, so annihilation of a particle pair at rest gives two photons.

Try this

Q1. Calculate the rest energy of a proton of mass $1.67 \times 10^{-27}$ kg in joules and in MeV. [2 marks]

Cue. $E_0 = 1.67 \times 10^{-27} \times 9.00 \times 10^{16} = 1.50 \times 10^{-10}$ J. Dividing by $1.60 \times 10^{-19}$ gives $9.39 \times 10^{8}$ eV $= 939$ MeV.

Q2. A proton moves at $0.60c$ . Calculate its kinetic energy in MeV. [3 marks]

Cue. $\gamma = 1.25$ , $E_k = 0.25 \times 939 = 235$ MeV.

Q3. A power station converts $1.0$ g of mass to energy. How much energy is released, and how long would this supply a $1.0$ GW station? [3 marks]

Cue. $E = 1.0 \times 10^{-3} \times 9.00 \times 10^{16} = 9.0 \times 10^{13}$ J. Time $= 9.0 \times 10^{13}/(1.0 \times 10^{9}) = 9.0 \times 10^{4}$ s, about $25$ hours.

Q4. Explain why an electron and positron annihilating at rest produce two photons rather than one. [2 marks]

Cue. Initial momentum is zero, so the final momentum must be zero. Two photons of equal energy travelling in opposite directions have zero total momentum. One photon would carry momentum.