Mass-Energy Equivalence
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You must be comfortable with the definition of the Lorentz factor \(\gamma\) and the relativistic momentum \(p = \gamma mv\). Familiarity with the work-energy theorem in classical mechanics is also assumed.
Why this matters
Mass is not merely a measure of inertia; it is a concentrated form of energy.
The discovery that mass and energy are equivalent fundamentally altered our understanding of conservation laws. In classical physics, mass and energy are conserved separately. In special relativity, they are two aspects of the same physical quantity. This insight explains how stars shine, how nuclear reactors operate, and why the total energy of a system includes a term that depends only on its mass, even when it is at rest. It resolves the apparent paradoxes of high-speed particle collisions where mass appears to change, by showing that mass is simply a measure of the total energy content in the rest frame.
The Idea in Plain Language
Imagine a box containing a photon bouncing between two mirrors. In the box's rest frame, the photon has energy but no mass. However, if you observe this box from a frame where it is moving, the photon's energy changes due to the Doppler effect, and the box itself must have momentum to conserve the total momentum of the system. Working this through shows that the trapped energy \(E\) adds \(E/c^2\) to the inertia of the box. More profoundly, even when the box is at rest, the energy of the photon contributes to the total mass of the system. This suggests that energy itself has inertia. Einstein realized that, for both to be conserved in every inertial frame, energy and momentum must transform together between frames. The result is that any object with mass \(m\) possesses an intrinsic energy \(mc^2\) simply by existing. This is not energy of motion; it is the energy of being.
Relativistic Energy Definition
To maintain the conservation of momentum in all inertial frames, we must define a corresponding conserved quantity for energy. If momentum is \(p = \gamma mv\), dimensional analysis and the requirement for a Lorentz invariant structure suggest that energy must be proportional to \(\gamma mc^2\). The constant of proportionality is determined by matching the classical limit. At low speeds, where \(v \ll c\), the energy must reduce to the classical kinetic energy plus a constant. The term \(\gamma mc^2\) expands to \(mc^2 + \frac{1}{2}mv^2 + \dots\). The term \(\frac{1}{2}mv^2\) is the familiar kinetic energy. The remaining term, \(mc^2\), is constant for a given particle. We define the total relativistic energy \(E\) as:
Here, \(m\) is the invariant mass (rest mass), which is the same in all inertial frames. The factor \(\gamma = (1 - v^2/c^2)^{-1/2}\) accounts for the relativistic effects. This definition ensures that energy and momentum transform together under Lorentz transformations, preserving the laws of physics across frames.
Rest Energy and Kinetic Energy
We can decompose the total energy \(E\) into two parts. The first part is the rest energy \(E_0\), which is the energy the particle possesses when it is at rest (\(v=0\), so \(\gamma=1\)).
The second part is the kinetic energy \(K\), which is the energy associated with motion. It is defined as the difference between the total energy and the rest energy:
In the non-relativistic limit, where \(v \ll c\), we can use the binomial expansion for \(\gamma\):
Substituting this into the expression for kinetic energy:
This recovers the classical expression for kinetic energy, confirming that our relativistic definition is consistent with everyday experience at low speeds. The rest energy \(mc^2\) is often huge because \(c^2\) is a very large number (\(9 \times 10^{16} \ \mathrm{m^2/s^2}\)). For a 1 kg mass, the rest energy is approximately \(9 \times 10^{16} \ \mathrm{J}\), which is equivalent to the energy released by a large nuclear weapon. This explains why nuclear reactions, which convert a small fraction of mass into energy, are so powerful.
The Energy-Momentum Relation
A crucial consequence of the definitions of relativistic energy and momentum is the invariant relationship between them. By squaring the energy equation and using the identity \(\gamma^2 v^2 = \gamma^2 c^2 - c^2\), we can eliminate the velocity \(v\) and the Lorentz factor \(\gamma\). Starting from \(E = \gamma mc^2\) and \(p = \gamma mv\):
Consider the difference \(E^2 - (pc)^2\):
Since \(\gamma^2 = \frac{1}{1 - v^2/c^2} = \frac{c^2}{c^2 - v^2}\), we have \(\gamma^2 (c^2 - v^2) = c^2\). Therefore:
Rearranging this gives the famous energy-momentum relation:
This equation is Lorentz invariant, meaning it holds true in all inertial frames. It connects the three fundamental quantities: total energy, momentum, and invariant mass. For a particle at rest, \(p=0\), and we recover \(E = mc^2\). For a massless particle like a photon, \(m=0\), and we get \(E = pc\), which is consistent with quantum mechanics where \(E = hf\) and \(p = h/\lambda\), and \(c = f\lambda\).
The derivation
Mass-Energy Equivalence E = mc^2
For a body of invariant (rest) mass m moving with speed v in an inertial frame, requiring that the relativistic momentum p = γmv be conserved in every inertial frame forces the existence of a conserved scalar E = γmc2, where γ = (1 − v2/c2)−1/2. In the rest frame this reduces to E0 = mc2, the rest energy, so that a body possesses energy purely by virtue of its mass.
A common misconception is that mass is "converted" into energy. In modern physics, mass is an invariant property of a system. In a nuclear reaction, the mass of the final products is less than the mass of the initial reactants. This "missing" mass corresponds to the kinetic energy of the products and the energy of emitted radiation. The total energy (including rest energy and kinetic energy) is conserved, and the total invariant mass of the isolated system is also conserved. The change in mass refers to the change in the rest mass of the specific particles involved, not a destruction of mass itself. C
Worked example
Problem. An electron has a rest mass of \(9.11 \times 10^{-31} \ \mathrm{kg}\). Calculate its rest energy in joules and in electron-volts (eV). Given \(c = 3.00 \times 10^8 \ \mathrm{m/s}\) and \(1 \ \mathrm{eV} = 1.60 \times 10^{-19} \ \mathrm{J}\).
- Calculate the rest energy in joules using \(E_0 = mc^2\).
- Convert the result from joules to electron-volts by dividing by the charge of an electron.
Units check. \(\mathrm{kg \cdot (m/s)^2} = \mathrm{kg \cdot m^2/s^2} = \mathrm{J}\). The conversion to eV is dimensionless as it is a ratio of energy units.
Three quick tests
- What is the rest energy of a proton with mass \(1.67 \times 10^{-27} \ \mathrm{kg}\)?
Answer
Approximately \(1.50 \times 10^{-10} \ \mathrm{J}\) or \(938 \ \mathrm{MeV}\). - If a particle has zero rest mass, what is the relationship between its energy and momentum?
Answer
\(E = pc\). - Why does the classical kinetic energy formula \(K = \frac{1}{2}mv^2\) fail at high speeds?
Answer
It is only the low-speed limit of the exact result \(K = (\gamma - 1)mc^2\); the true kinetic energy grows without limit as \(v \to c\), so \(\tfrac12 mv^2\) badly underestimates the work needed at high speed.
Problem set for PU-105 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.