# Four-momentum

(Redirected from 4-momentum)

In special relativity, four-momentum is the generalization of the classical three-dimensional momentum to four-dimensional spacetime. Momentum is a vector in three dimensions; similarly four-momentum is a four-vector in spacetime. The contravariant four-momentum of a particle with three-momentum p = (px, py, pz) and energy E is

$\mathbf{P} = \begin{pmatrix} P^0 \\ P^1 \\ P^2 \\ P^3 \end{pmatrix} = \begin{pmatrix} E/c \\ p_x \\ p_y \\ p_z \end{pmatrix}$

The four-momentum is useful in relativistic calculations because it is a Lorentz vector. This means that it is easy to keep track of how it transforms under Lorentz transformations.

The above definition applies under the coordinate convention that x0 = ct. Some authors use the convention x0 = t which yields a modified definition with P0 = E/c2. It is also possible to define covariant four-momentum Pμ where the sign of the 3 momentum is reversed.

## Minkowski norm

Calculating the Minkowski norm of the four-momentum gives a Lorentz invariant quantity equal (up to factors of the speed of light c) to the square of the particle's proper mass:

$-\|\mathbf{P}\|^2 = - P^\mu P_\mu = - \eta_{\mu\nu} P^\mu P^\nu = {E^2 \over c^2} - |\mathbf p|^2 = m^2c^2$

where we use the convention that

$\eta^{\mu\nu} = \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$

is the reciprocal of the metric tensor of special relativity. The magnitude ||P||2 is Lorentz invariant, meaning its value is not changed by Lorentz transformations/boosting into different frames of reference.

## Relation to four-velocity

For a massive particle, the four-momentum is given by the particle's invariant mass m multiplied by the particle's four-velocity:

$P^\mu = m \, U^\mu\!$

where the four-velocity is

$\begin{pmatrix} U^0 \\ U^1 \\ U^2 \\ U^3 \end{pmatrix} = \begin{pmatrix} \gamma c \\ \gamma v_x \\ \gamma v_y \\ \gamma v_z \end{pmatrix}$

and

$\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$

is the Lorentz factor, c is the speed of light.

## Conservation of four-momentum

The conservation of the four-momentum yields two conservation laws for "classical" quantities:

1. The total energy E = P0c is conserved.
2. The classical three-momentum p is conserved.

Note that the invariant mass of a system of particles may be more than the sum of the particles' rest masses, since kinetic energy in the system center-of-mass frame and potential energy from forces between the particles contribute to the invariant mass. As an example, two particles with four-momenta (−5 GeV/c, 4 GeV/c, 0, 0) and (−5 GeV/c, −4 GeV/c, 0, 0) each have (rest) mass 3 GeV/c2 separately, but their total mass (the system mass) is 10 GeV/c2. If these particles were to collide and stick, the mass of the composite object would be 10 GeV/c2.

One practical application from particle physics of the conservation of the invariant mass involves combining the four-momenta P(A) and P(B) of two daughter particles produced in the decay of a heavier particle with four-momentum P(C) to find the mass of the heavier particle. Conservation of four-momentum gives P(C)μ = P(A)μ + P(B)μ, while the mass M of the heavier particle is given by −||P(C)||2 = M2c2. By measuring the energies and three-momenta of the daughter particles, one can reconstruct the invariant mass of the two-particle system, which must be equal to M. This technique is used, e.g., in experimental searches for Z' bosons at high-energy particle colliders, where the Z' boson would show up as a bump in the invariant mass spectrum of electron-positron or muon-antimuon pairs.

If an object's mass does not change, the Minkowski inner product of its four-momentum and corresponding four-acceleration Aμ is zero. The four-acceleration is proportional to the proper time derivative of the four-momentum divided by the particle's mass, so

$P^{\mu} A_\mu = \eta_{\mu\nu} P^{\mu} A^\nu = \eta_{\mu\nu} P^\mu \frac{d}{d\tau} \frac{P^{\nu}}{m} = \frac{1}{2m} \frac{d}{d\tau} \|\mathbf{P}\|^2 = \frac{1}{2m} \frac{d}{d\tau} (-m^2c^2) = 0 .$

## Canonical momentum in the presence of an electromagnetic potential

For a charged particle of charge q, moving in an electromagnetic field given by the electromagnetic four-potential:

$\begin{pmatrix} A^0 \\ A^1 \\ A^2 \\ A^3 \end{pmatrix} = \begin{pmatrix} \phi / c \\ A_x \\ A_y \\ A_z \end{pmatrix}$

where φ is the scalar potential and A = (Ax, Ay, Az) the vector potential, the "canonical" momentum four-vector is

$Q^\mu = P^\mu + q A^\mu. \!$

This allows the potential energy from the charged particle in an electrostatic potential and the Lorentz force on the charged particle moving in a magnetic field to be incorporated in a compact way, in relativistic quantum mechanics.