Applications: Particle Decays and Collisions
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You must be comfortable with the energy-momentum four-vector \( p^\mu = (E/c, \mathbf{p}) \) and the invariant mass relation \( E^2 = (pc)^2 + (mc^2)^2 \). Familiarity with conservation of energy and momentum in classical mechanics is assumed.
Why this matters
Relativistic kinematics provides the only consistent framework for predicting outcomes in high-energy particle physics.
In particle physics, we rarely track trajectories; instead, we analyze initial and final states. Because the speed of light is invariant, classical Galilean transformations fail to preserve the laws of electromagnetism and mechanics simultaneously. Special relativity resolves this by treating energy and momentum as components of a single geometric object. This allows us to solve complex collision and decay problems using invariant quantities, such as the Mandelstam variables, which remain constant across all inertial frames. This geometric approach simplifies calculations significantly, particularly when dealing with particles moving at speeds close to \( c \).
The Geometry of Conservation
Imagine a collision not as a dynamic process in time, but as a static geometric relationship between vectors in four-dimensional spacetime. In classical mechanics, we conserve energy and momentum separately. In relativity, these are linked. The total four-momentum of an isolated system is conserved. This means the sum of the four-momenta of the initial particles equals the sum of the four-momenta of the final particles. Because the "length" (Minkowski norm) of a four-vector is invariant, we can calculate properties of the system in the most convenient reference frame, often the center-of-momentum frame, where the total spatial momentum is zero. This frame-independent nature is the key power of relativistic kinematics.
Two-Body Decays
Consider a particle of mass \( M \) at rest decaying into two particles of masses \( m_1 \) and \( m_2 \). In the rest frame of the parent particle, the total momentum is zero. By conservation of momentum, the daughter particles must have equal and opposite momenta: \( \mathbf{p}_1 = -\mathbf{p}_2 \). Let the magnitude of this momentum be \( p \).
Using the energy-momentum relation for each daughter particle:
Energy conservation requires \( E_1 + E_2 = Mc^2 \). Solving this system for \( p \) yields the unique momentum of the decay products. This result is crucial because it shows that for a two-body decay, the energies of the products are fixed (monochromatic), unlike in three-body decays where a continuous spectrum is observed.
Collisions and the Center-of-Momentum Frame
In a collision, we often know the energy and momentum of the incoming particles in the laboratory frame. To find the maximum possible energy available for creating new particles, we transform to the center-of-momentum (CM) frame. The Mandelstam invariant \( s \) is the squared total four-momentum, written in energy units:
In the CM frame \( \mathbf{P}_{\text{total}} = 0 \), so \( s = E_{\text{CM}}^2 \) and \( \sqrt{s} = E_{\text{CM}} \) is the total energy available in the CM frame. For a fixed-target experiment where a particle of mass \( m \) and total energy \( E \) hits a stationary target of mass \( M \):
This formula highlights the inefficiency of fixed-target experiments: most energy goes into the forward motion of the CM frame rather than into internal excitation. In colliders, where two particles of equal energy \( E \) collide head-on, \( s = 4E^2 \), making far more energy available for particle production.
Threshold Energy for Pair Production
To produce new particles, the total energy in the CM frame must be at least the sum of their rest energies. Consider the reaction \( p + p \to p + p + p + \bar{p} \). The final state has four protons (one is the antiproton, but mass is the same). The minimum CM energy required is \( E_{\text{CM, min}} = 4 m_p c^2 \). Therefore, the threshold invariant mass squared is \( s_{\text{min}} = (4 m_p c^2)^2 = 16 (m_p c^2)^2 \).
If this reaction occurs in a fixed-target setup (proton on stationary proton), we equate the invariant \( s \) calculated from the lab frame to \( s_{\text{min}} \):
Solving for \( E_{\text{lab}} \):
Since \( E_{\text{lab}} = \gamma m_p c^2 \), the kinetic energy \( K = E_{\text{lab}} - m_p c^2 = 6 m_p c^2 \). This demonstrates that a significant portion of the beam energy is "wasted" in the motion of the center of mass.
The derivations
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.
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 error is assuming that the kinetic energy in the lab frame equals the energy available for particle creation. In fixed-target collisions, the CM frame moves at relativistic speeds relative to the lab. The invariant mass \( \sqrt{s} \) is the correct measure of available energy, not the lab kinetic energy. Confusing these leads to underestimating the required beam energy by orders of magnitude. C
Worked example
Problem. A pion of mass \( m_\pi = 139.6 \ \mathrm{MeV/c^2} \) decays at rest into a muon of mass \( m_\mu = 105.7 \ \mathrm{MeV/c^2} \) and a massless neutrino (\( m_\nu = 0 \)). Calculate the energy of the muon in the pion rest frame.
- Identify the system: Two-body decay \( \pi \to \mu + \nu \). The pion is at rest, so \( E_\pi = m_\pi c^2 \) and \( \mathbf{p}_\pi = 0 \).
- Apply conservation of momentum: \( \mathbf{p}_\mu = -\mathbf{p}_\nu \). Let \( p \) be the magnitude of the momentum for both particles.
- Write energy conservation: \( E_\mu + E_\nu = m_\pi c^2 \).
- Express energies in terms of \( p \): For the muon, \( E_\mu = \sqrt{(pc)^2 + (m_\mu c^2)^2} \). For the massless neutrino, \( E_\nu = pc \).
- Substitute into the conservation equation: \( \sqrt{(pc)^2 + (m_\mu c^2)^2} + pc = m_\pi c^2 \).
- Rearrange to solve for \( pc \): \( \sqrt{(pc)^2 + (m_\mu c^2)^2} = m_\pi c^2 - pc \).
- Square both sides: \( (pc)^2 + (m_\mu c^2)^2 = (m_\pi c^2)^2 - 2 m_\pi c^2 (pc) + (pc)^2 \).
- Simplify: \( (m_\mu c^2)^2 = (m_\pi c^2)^2 - 2 m_\pi c^2 (pc) \).
- Solve for \( pc \): \( pc = \frac{(m_\pi c^2)^2 - (m_\mu c^2)^2}{2 m_\pi c^2} \).
- Calculate the muon energy: \( E_\mu = m_\pi c^2 - pc = m_\pi c^2 - \frac{(m_\pi c^2)^2 - (m_\mu c^2)^2}{2 m_\pi c^2} \).
- Substitute numerical values: \( m_\pi c^2 = 139.6 \ \mathrm{MeV} \), \( m_\mu c^2 = 105.7 \ \mathrm{MeV} \).
- Compute \( pc \): \( pc = \frac{139.6^2 - 105.7^2}{2 \times 139.6} = \frac{19488.16 - 11172.49}{279.2} = \frac{8315.67}{279.2} \approx 29.78 \ \mathrm{MeV} \).
- Compute \( E_\mu \): \( E_\mu = 139.6 - 29.78 = 109.82 \ \mathrm{MeV} \).
Units check. Masses are in \( \mathrm{MeV/c^2} \), so \( mc^2 \) is in \( \mathrm{MeV} \). The formula yields energy in \( \mathrm{MeV} \), which is consistent.
Three quick tests
- Why is the center-of-momentum frame preferred for analyzing particle collisions?
Answer
It maximizes the available energy for particle creation because the total spatial momentum is zero, meaning no energy is "locked" in the bulk motion of the system. - If a particle decays into three massless particles, is the energy of each daughter particle fixed?
Answer
No. In a three-body decay, the energy distribution is continuous, unlike the monochromatic energies in a two-body decay. - What is the invariant mass \( \sqrt{s} \) for two photons of energy \( E \) colliding head-on?
Answer
\( \sqrt{s} = 2E \). Since photons are massless, their four-momenta are lightlike. For head-on collision, \( \mathbf{p}_1 = -\mathbf{p}_2 \), so \( \mathbf{P}_{\text{total}} = 0 \), and \( E_{\text{total}} = 2E \). Thus \( s = (2E/c)^2 \) and \( \sqrt{s} = 2E/c \) (in energy units, \( 2E \)).
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.
Derivations cited: D-100 · D-099 · Threads: force, energy, light