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# Four-momentum

In special relativity, four-momentum is a four-vector that replaces classical momentum; the four-momentum of a particle is defined as the particle's mass times the particle's four-velocity.

$\mathbf{p} = m \left( \gamma c , \gamma v_x , \gamma v_y ,\gamma v_z \right) = \left( \gamma m c^2 /c , \gamma m v_x , \gamma m v_y ,\gamma m v_z \right) = \left( {E \over c} , \gamma p_x , \gamma p_y ,\gamma p_z \right)$

where

$\gamma m c^2 = E \,\!$

is the energy of the moving body.

Calculating the norm of the momentum-energy quad-vector we obtain:

$\| \mathbf{p} \| = \sqrt{{E^2 \over c^2} - {\gamma}^2 m^2 v^2} = mc$

and since c is a constant we may say that the norm of the four-momentum vector is equal to the body's mass; although, when computing values, it is really only equal to the mass if we choose to work in units of measurement in which the speed of light is simply c = 1.

The conservation of the four-momentum yields 3 laws of "classical" conservation:

1. The energy (p0) is conserved.
2. The classical momentum is conserved.
3. The norm of the four-momentum is conserved.

In reactions between an isolated handful of particles, four-momentum is conserved. The mass of a system of particles may be more than the sum of the particle's masses, since kinetic energy counts as mass. As an example, two particles with the four-momentums {5, 4, 0, 0} and {5, -4, 0, 0} both have the mass 3, but their total mass is 10. Note that the length of the four-vector {t, x, y, z} is $\sqrt{t^2-x^2-y^2-z^2}$

The scalar product of a four-momentum and the corresponding four-acceleration is always 0.