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PU-105 / L16 · lecture

The Energy-Momentum Four-Vector and E^2 = (pc)^2 + (mc^2)^2

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BEFORE YOU START

You should be comfortable with the Lorentz transformation, the definition of the Minkowski metric, and the concept of a four-vector from previous lectures.

Why this matters

Relativistic mechanics requires a unified description of energy and momentum that remains consistent across all inertial frames.

In classical mechanics, energy and momentum are separate conserved quantities. In special relativity, they are inextricably linked components of a single geometric object. Understanding this unification is essential for predicting particle behavior in accelerators and for interpreting high-energy physics experiments where kinetic energies vastly exceed rest-mass energies.

The Geometric Unification

Imagine energy and momentum not as two separate lists of numbers, but as the time and space coordinates of a single event in a higher-dimensional space. Just as the distance between two points in Euclidean space is invariant under rotation, the "length" of this energy-momentum object is invariant under changes of reference frame. This invariant length is directly tied to the particle's rest mass. This geometric view resolves the apparent paradoxes of relativistic dynamics by treating energy and momentum as projections of a single entity onto different axes of spacetime.

Constructing the Four-Vector

To form a four-vector, we must combine quantities that transform in the same way under Lorentz transformations. The three-momentum \(\mathbf{p}\) transforms as the spatial part of a four-vector. To complete the set, we need a temporal component that scales with energy. We define the energy-momentum four-vector \(p^\mu\) as:

\[ p^\mu = \left( \frac{E}{c}, \mathbf{p} \right) = \left( \frac{E}{c}, p_x, p_y, p_z \right) \]

Here, \(E\) is the total relativistic energy and \(\mathbf{p}\) is the relativistic three-momentum. The factor of \(c\) in the time component ensures that all components have the same physical dimensions (momentum). This construction guarantees that \(p^\mu\) transforms as a four-vector, meaning its Minkowski norm is invariant across all inertial frames.

The Invariant Norm

The Minkowski norm of any four-vector is calculated using the metric signature \((+,-,-,-)\). For the energy-momentum four-vector, the norm is:

\[ p^\mu p_\mu = \left( \frac{E}{c} \right)^2 - |\mathbf{p}|^2 \]

Because this quantity is a Lorentz invariant, it must have the same value in every inertial frame. The most convenient frame to evaluate this invariant is the particle's rest frame, where the momentum \(\mathbf{p}' = 0\). In this frame, the energy is simply the rest energy \(E' = mc^2\). Substituting these values into the norm expression gives:

\[ p^\mu p_\mu = \left( \frac{mc^2}{c} \right)^2 - 0 = m^2 c^2 \]

Since the norm is invariant, this result holds in any frame. Equating the general expression to the rest-frame value yields the fundamental dispersion relation.

Deriving the Dispersion Relation

By equating the general expression for the norm with the invariant value derived from the rest frame, we obtain the exact relationship between total energy, momentum, and rest mass:

\[ \left( \frac{E}{c} \right)^2 - p^2 = m^2 c^2 \]

Multiplying through by \(c^2\) rearranges this into the standard form:

\[ E^2 = (pc)^2 + (mc^2)^2 \]

This equation is valid for all particles, whether massive or massless. It replaces the classical kinetic energy formula \(K = \frac{1}{2}mv^2\) and the separate definition of momentum \(p = mv\). It shows that total energy \(E\) is composed of two orthogonal contributions: the rest energy \(mc^2\) and the momentum-dependent term \(pc\). This relation is the cornerstone of relativistic kinematics.

The derivation

◆ DERIVATION EMBED · D-100 · not re-derived here

The Energy-Momentum Four-Vector

The energy E and three-momentum p of a particle of rest mass m are the timelike and spacelike parts of a single four-vector pμ = (E/c, p), whose Minkowski norm is a Lorentz invariant fixed by the mass. Evaluating that invariant in two frames — the particle's rest frame and a general frame — yields the exact dispersion relation E2 = (pc)2 + (mc2)2.

E2 = (pc)2 + (mc2)2
Open full derivation →
Tier 3 · watch out

A common error is to assume that \(E = mc^2\) is the general energy equation. This is only true in the rest frame where \(p=0\). In any other frame, \(E\) is always greater than \(mc^2\). Furthermore, the term \(pc\) is not kinetic energy; kinetic energy is defined as \(K = E - mc^2\). Confusing total energy with kinetic energy leads to significant calculation errors in high-speed scenarios. C

Worked example

Problem. An electron (rest mass \(m = 9.11 \times 10^{-31}\ \mathrm{kg}\)) has momentum \(p = 1.50 \times 10^{-22}\ \mathrm{kg \cdot m/s}\). Find its total energy in joules and in electronvolts, and its kinetic energy. Use \(c = 3.00 \times 10^8\ \mathrm{m/s}\) and \(1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}\).

  1. Rest energy: \(mc^2 = (9.11 \times 10^{-31})(3.00 \times 10^8)^2 = 8.20 \times 10^{-14}\ \mathrm{J}\).
  2. Momentum term: \(pc = (1.50 \times 10^{-22})(3.00 \times 10^8) = 4.50 \times 10^{-14}\ \mathrm{J}\), comparable with \(mc^2\), so the electron is relativistic.
  3. Energy-momentum relation: \(E^2 = (pc)^2 + (mc^2)^2 = 2.03 \times 10^{-27} + 6.72 \times 10^{-27} = 8.75 \times 10^{-27}\ \mathrm{J^2}\), so \(E = 9.35 \times 10^{-14}\ \mathrm{J}\).
  4. In electronvolts: \(E = 9.35 \times 10^{-14}/1.602 \times 10^{-19} \approx 5.84 \times 10^{5}\ \mathrm{eV} = 584\ \mathrm{keV}\); the kinetic energy is \(K = E - mc^2 \approx 1.15 \times 10^{-14}\ \mathrm{J} \approx 72\ \mathrm{keV}\).
\[ E \approx 9.35 \times 10^{-14}\ \mathrm{J}\ (584\ \mathrm{keV}), \qquad K \approx 72\ \mathrm{keV} \]

Units check. \(pc\) and \(mc^2\) are both in \(\mathrm{kg\,m^2/s^2} = \mathrm{J}\), so \(E = \sqrt{(pc)^2 + (mc^2)^2}\) is in joules.

CHECK YOURSELF

Three quick tests

  1. What is the value of the invariant norm \(p^\mu p_\mu\) for a photon?
    AnswerZero. Since \(m=0\), \(p^\mu p_\mu = 0\), implying \(E = pc\).
  2. If a particle's kinetic energy is equal to its rest energy, what is the ratio \(pc/E\)?
    Answer\(\sqrt{3}/2\). If \(K = mc^2\), then \(E = 2mc^2\), so \((pc)^2 = E^2 - (mc^2)^2 = 3(mc^2)^2\), \(pc = \sqrt{3}\,mc^2\) and \(pc/E = \sqrt{3}/2\) (equivalently \(v/c = \sqrt{3}/2\)).
  3. Why is the factor \(1/c\) necessary in the time component of the four-vector?
    AnswerTo ensure dimensional consistency. Energy has units of Joules (\(\mathrm{kg \cdot m^2/s^2}\)), while momentum has units of \(\mathrm{kg \cdot m/s}\). Dividing energy by \(c\) (\(\mathrm{m/s}\)) gives \(\mathrm{kg \cdot m/s}\), matching the units of momentum.
PROBLEMS

Problem set for PU-105 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-100 · Threads: force, energy, light