3.12.3.5 - Mass and Energy

3.12.3.5 - Mass and Energy

Separate rest energy, total energy and kinetic energy before calculating at relativistic speeds. The AQA mass-versus-speed convention and Bertozzi's independent measurements show why increasing energy drives a massive particle towards, but never to, the vacuum speed of light.

Rest energy and changes of system mass

An object has rest energy even when its centre of mass is stationary. Its rest mass m0m_0 and rest energy E0E_0 are related by

Rest energy

E0=m0c2,c=3.00×108 m s−1.E_0=m_0c^2, \qquad c=3.00\times10^8\,\mathrm{m\,s^{-1}}.

This is an equivalence of mass and energy, with c2c^2 supplying the conversion between kilograms and joules. It does not mean that every object's entire rest energy is available in an ordinary reaction.

If a stationary system loses energy to its surroundings, its mass decreases by Δm=ΔE/c2\Delta m=\Delta E/c^2, where the deltas here denote positive amounts lost. Energy retained as internal energy contributes to the mass of the system. Heating therefore increases a sealed system's rest mass slightly; choosing a frame in which the whole system moves faster is a different change and does not change its rest mass.

For example, energy 9.00×1010 J9.00\times10^{10}\,\mathrm{J} leaving a stationary system corresponds to a mass loss 9.00×1010/(3.00×108)2=1.00×10−6 kg9.00\times10^{10}/(3.00\times10^8)^2=1.00\times10^{-6}\,\mathrm{kg}. The energy and mass account must refer to the same system. A reaction's lower product rest masses can supply kinetic energy and radiation; total energy is conserved when all products and transfers are included.

AQA's relativistic-mass convention

The current AQA specification uses E=mc2E=mc^2, E=m0c2/1−v2/c2E=m_0c^2/\sqrt{1-v^2/c^2}, and graphs of mass against speed. In this convention mm is relativistic mass, defined by total energy divided by c2c^2; m0m_0 is the rest mass.

Relativistic mass

m=γm0=m01−v2/c2,γ=11−v2/c2.m=\gamma m_0=\frac{m_0}{\sqrt{1-v^2/c^2}}, \qquad \gamma=\frac{1}{\sqrt{1-v^2/c^2}}.

For the same unchanged particle, rest mass is invariant: observers in relative motion agree on m0m_0. They can disagree about its energy and therefore about mm in the AQA convention. Many modern accounts use “mass” to mean invariant rest mass throughout; the physical predictions are the same once the notation is identified. Do not say the particle's rest mass increases just because it is accelerated.

At v=0.60cv=0.60c, γ=1.25\gamma=1.25, so m=1.25m0m=1.25m_0. At 0.990c0.990c, γ=7.089\gamma=7.089: a proton with m0=1.67×10−27 kgm_0=1.67\times10^{-27}\,\mathrm{kg} has m=1.18×10−26 kgm=1.18\times10^{-26}\,\mathrm{kg} in this convention, while its rest mass stays 1.67×10−27 kg1.67\times10^{-27}\,\mathrm{kg}.

Avoid inserting this speed-dependent mass into F=maF=ma as though all classical dynamics remained unchanged. The energy equations give the required conclusions directly.

Separating total and kinetic energy

The total energy of a free massive particle is its rest energy plus its kinetic energy, all referred to the same inertial frame.

Total and kinetic energy

E=γm0c2=mc2,Ek=E−E0=(γ−1)m0c2.E=\gamma m_0c^2=mc^2, \qquad E_k=E-E_0=(\gamma-1)m_0c^2.

The subtraction matters: at rest, γ=1\gamma=1, so Ek=0E_k=0 but E=m0c2E=m_0c^2. Calling γm0c2\gamma m_0c^2 the kinetic energy would wrongly give a stationary particle a non-zero kinetic energy.

An electron at 0.800c0.800c

With m0=9.11×10−31 kgm_0=9.11\times10^{-31}\,\mathrm{kg}, E0=8.199×10−14 JE_0=8.199\times10^{-14}\,\mathrm{J}. Here γ=1/0.600=1.667\gamma=1/0.600=1.667, giving

E=1.37×10−13 J,Ek=5.47×10−14 J.E=1.37\times10^{-13}\,\mathrm{J}, \qquad E_k=5.47\times10^{-14}\,\mathrm{J}.

The classical calculation gives 12m0v2=2.62×10−14 J\tfrac12m_0v^2=2.62\times10^{-14}\,\mathrm{J}. Its percentage underestimate is measured relative to the relativistic value:

Ek,rel−Ek,classEk,rel×100=52.0%.\frac{E_{k,\mathrm{rel}}-E_{k,\mathrm{class}}}{E_{k,\mathrm{rel}}}\times100=52.0\%.

At v≪cv\ll c, the approximation γ≈1+v2/(2c2)\gamma\approx1+v^2/(2c^2) yields Ek≈12m0v2E_k\approx\tfrac12m_0v^2. Classical mechanics is an excellent low-speed approximation, with accuracy determined by the measurement precision required.

If kinetic energy is given instead, first find γ=1+Ek/(m0c2)\gamma=1+E_k/(m_0c^2), then v/c=1−1/γ2v/c=\sqrt{1-1/\gamma^2}. For example, Ek=3E0E_k=3E_0 gives γ=4\gamma=4, not 3, and v=0.968cv=0.968c.

Reading the graphs and the speed limit

Graphs distinguish the non-zero rest energy from the zero kinetic energy at rest and show how rapidly relativistic effects grow near cc. Normalised axes allow the same curves to represent particles with different rest masses.

[DIAGRAM: asset_name: D.3.5 - Mass and Energy - Diagram 1; asset_slug: D.3.5 - Mass and Energy - Diagram 1; recommended_method: matplotlib; caption: Relativistic mass rises from its rest value as speed increases; description: Exact graph of relativistic mass ratio m/m0 against speed ratio v/c, with rest-mass reference at 1 and vertical asymptote at v/c=1.]
Relativistic mass rises from its rest value as speed increases

Relativistic mass starts at m/m0=1m/m_0=1, remains close to 1 at low speed and rises without bound as v/c→1v/c\to1. The dashed horizontal line represents invariant rest mass.

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