What is Four-momentum and invariant mass? — Physics

Definition 5.9 University Physics — Year 3 · Chapter 5 — Relativistic Dynamics

Bundle a particle’s energy and momentum into its four-momentum $P = (E/c, \vect p)$ . For a system of particles, sum componentwise: $E_{\text{tot}} = \sum E_i$ , $\vect p_{\text{tot}} = \sum\vect p_i$ . The system’s invariant mass $M$ is defined by

M^2c^4 = E_{\text{tot}}^2 - \|\vect p_{\text{tot}}\|^2c^2 :

the same number in every frame, equal to the total energy (over $c^2$ ) in the centre-of-momentum frame where $\vect p_{\text{tot}} = \vect 0$ . Note that $M$ exceeds the sum of the parts’ masses whenever they move relative to each other — two photons flying apart have $M > 0$ though each is massless.

Examples

Example 5.11 (The pion’s fingerprint)

A charged pion at rest decays, $\pi^+ \to \mu^+ + \nu$ (the neutrino effectively massless). Momentum conservation makes the products back-to-back with equal $p$ ; energy conservation reads $m_\pi c^2 = \sqrt{p^2c^2 + m_\mu^2c^4} + pc$ . Solving:

pc = \frac{(m_\pi^2 - m_\mu^2)c^2}{2m_\pi} = 29.8\,\mathrm{MeV} ,

so every muon from a pion decaying at rest is born with exactly $4.1\,\mathrm{MeV}$ of kinetic energy — a monoenergetic line, and its observation (Powell, 1947) is how the pion’s mass was first weighed. Two-body decays always produce such lines; three-body decays smear them into spectra, which is precisely how the neutrino was first suspected in nuclear $\beta$ decay.

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