Math 2.0

Relativity and gravitational theory Definition

Invariant mass

Builds on Minkowski spaceFour-momentum

Definition

Let (ℝ4,η)(\R^4, \eta) be Minkowski space with η=diag⁡(−1,1,1,1)\eta = \operatorname{diag}(-1, 1, 1, 1), and let cc be the speed of light. Let a system of particles have four-momenta p1,…,pnp_1, \ldots, p_n, each future-directed timelike or null, and let P=p1+⋯+pnP = p_1 + \cdots + p_n be the total four-momentum. The invariant mass of the system is M=1c−η(P,P).M = \frac{1}{c} \sqrt{-\eta(P, P)}. In an inertial frame where P=(E/c,P⃗)P = (E/c, \vec P), this reads M2c4=E2−∣P⃗∣2c2M^2 c^4 = E^2 - \abs{\vec P}^2 c^2.

Remarks

A sum of future-directed timelike or null vectors is future-directed timelike or null, so −η(P,P)≥0-\eta(P, P) \ge 0 and MM is a real number ≥0\ge 0. Since η\eta is preserved by Lorentz transformations, MM does not depend on the inertial frame. For one particle of mass mm, P=muP = m u and M=mM = m. Example: two photons with four-momenta (E/c)(1,1,0,0)(E/c)(1, 1, 0, 0) and (E/c)(1,−1,0,0)(E/c)(1, -1, 0, 0) give P=(2E/c,0,0,0)P = (2E/c, 0, 0, 0), so M=2E/c2>0M = 2E/c^2 > 0 although each photon has mass 00. Non-example: the sum ∑iγimi\sum_i \gamma_i m_i of the relativistic masses of the particles is not the invariant mass: it equals E/c2E/c^2, which depends on the frame. When M>0M > 0 there is a frame where P⃗=0\vec P = 0, the centre-of-momentum frame, and there M=E/c2M = E/c^2.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Invariant_mass
About this page
TagQXACW5TYAF
URIm2://QXACW5TYAF
Kinddefinition
WritingW0 stub
MSC83A05, 70H40
DefinesInvariant mass
UsesMinkowski space, Four-momentum
Changed3 November 2026

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