# Four-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 relativistic energy E and three-momentum p = (px, py, pz) = γmv, where v is the particle's 3-velocity and γ the Lorentz factor, is

${\displaystyle p=(p^{0},p^{1},p^{2},p^{3})=\left({E \over c},p_{x},p_{y},p_{z}\right).}$

The quantity mv of above is ordinary non-relativistic momentum of the particle and m its rest mass. 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 energy 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:

${\displaystyle p\cdot p=p^{\mu }p_{\mu }=\eta _{\mu \nu }p^{\mu }p^{\nu }=-{E^{2} \over c^{2}}+|\mathbf {p} |^{2}=-m^{2}c^{2}}$

where we use the convention that

${\displaystyle \eta _{\mu \nu }=\left({\begin{matrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{matrix}}\right)}$

is the metric tensor of special relativity. That the norm is negative reflects that the momentum is a timelike 4-vector for massive particles.

The Minkowski norm is Lorentz invariant, meaning its value is not changed by Lorentz transformations/boosting into different frames of reference. More generally, for any two 4-momenta p and q, the quantity pq is invariant.

## 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,

${\displaystyle p^{\mu }=mu^{\mu },}$

where the four-velocity u is

${\displaystyle u=(u^{0},u^{1},u^{2},u^{3})=\gamma (c,v_{x},v_{y},v_{z}),}$

and

${\displaystyle \gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$

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

## Derivation

There are several ways to arrive at the correct expression for 4-momentum. One way is to first define the 4-velocity u = dx/ and simply define p = mu, being content that it is a 4-vector with the correct units and correct behavior. Another, more satisfactory, approach is to begin with the principle of least action and use the Lagrangian framework to derive the 4-momentum, including the expression for the energy.[1] One may at once, using the observations detailed below, define 4-momentum from the action S. Given that in general for a closed system with generalized coordinates qi and canonical momenta pi,[2]

${\displaystyle p_{i}={\frac {\partial S}{\partial q_{i}}},\quad E=-{\frac {\partial S}{\partial t}},}$

it is immediate (recalling x0 = ct, x1 = x, x2 = y, x3 = z and x0 = −x0, x1 = x1, x2 = x2, x3 = x3 in the present metric convention) that

${\displaystyle p_{\mu }=-{\frac {\partial S}{\partial x^{\mu }}}=\left({E \over c},-\mathbf {p} \right)}$

is a covariant 4-vector with the 3-vector part being the (negative of) canonical momentum.

The action S is given by

${\displaystyle S=-mc\int ds=\int Lds,\quad L=-mc^{2}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}},}$

where L is the relativistic Lagrangian for a free particle. From this,

${\displaystyle \delta S=\left[-mu_{\mu }\delta x^{\mu }\right]_{t_{1}}^{t_{2}}+m\int _{t_{1}}^{t_{2}}\delta x^{\mu }{\frac {du_{\mu }}{ds}}ds=-mu_{\mu }\delta x^{\mu }={\frac {\partial S}{\partial x^{\mu }}}\delta x^{\mu }=-p_{\mu }\delta x^{\mu },}$

where the second step employs the field equations duμ/ds = 0, (δxμ)t1 = 0, and (δxμ)t2δxμ as in the observations above. Now compare the last three expressions to find

${\displaystyle p^{\mu }=-{\frac {\partial S}{\partial x_{\mu }}}=mu^{\mu }=m\left({\frac {c}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}},{\frac {v_{x}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}},{\frac {v_{y}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}},{\frac {v_{z}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\right),}$

with norm m2c2, and the famed result for the relativistic energy,

 ${\displaystyle E={\frac {mc^{2}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}=m_{r}c^{2},}$

where mr is the now unfashionable relativistic mass, follows. By comparing the expressions for momentum and energy directly, one has

 ${\displaystyle \mathbf {p} =E{\frac {\mathbf {v} }{c^{2}}},}$

that holds for massless particles as well. Squaring the expressions for energy and 3-momentum and relating them gives the energy–momentum relation,

 ${\displaystyle {\frac {E^{2}}{c^{2}}}=\mathbf {p} \cdot \mathbf {p} +m^{2}c^{2}.}$

Substituting

${\displaystyle p_{\mu }\leftrightarrow -{\frac {\partial S}{\partial x^{\mu }}}}$

in the equation for the norm gives the relativistic Hamilton–Jacobi equation,[3]

 ${\displaystyle \eta ^{\mu \nu }{\frac {\partial S}{\partial x^{\mu }}}{\frac {\partial S}{\partial x^{\nu }}}=-m^{2}c^{2}.}$

It is also possible to derive the results from the Lagrangian directly. By definition,[4]

{\displaystyle {\begin{aligned}\mathbf {p} &={\frac {\partial L}{\partial \mathbf {v} }}=\left({\partial L \over \partial {\dot {x}}},{\partial L \over \partial {\dot {y}}},{\partial L \over \partial {\dot {z}}}\right)=m(\gamma v_{x},\gamma v_{y},\gamma v_{z})=m\gamma \mathbf {v} =m\mathbf {u} ,\\E&=\mathbf {p} \cdot \mathbf {v} -L={\frac {mc^{2}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}},\end{aligned}}}

which constitute the standard formulae for canonical momentum and energy of a closed (time-independent Lagrangian) system. With this approach it is less clear that the energy and momentum are parts of a 4-vector.

The energy and the 3-momentum are separately conserved quantities for isolated systems in the Lagrangian framework. Hence 4-momentum is conserved as well. More on this below.

More pedestrian approaches include expected behavior in electrodynamics.[5] In this approach, the starting point is application of Lorentz force law and Newton's second law in the rest frame of the particle. The transformation properties of the electromagnetic field tensor, including invariance of electric charge, are then used to transform to the lab frame, and the resulting expression (again Lorentz force law) is interpreted in the spirit of Newton's second law, leading to the correct expression for the relativistic 3-momentum. The disadvantage, of course, is that it isn't immediately clear that the result applies to all particles, whether charged or not, and that it doesn't yield the complete 4-vector.

It is also possible to avoid electromagnetism and use well tuned experiments of thought involving well-trained physicists throwing billiard balls, utilizing knowledge of the velocity addition formula and assuming conservation of momentum.[6][7] This too gives only the 3-vector part.

## Conservation of 4-momentum

As shown above, there are three (not independent, the last two implies the first) conservation laws:

• The 4-momentum p (either covariant or contravariant) is conserved.
• The total energy E = p0c is conserved.
• The 3-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 pA and pB of two daughter particles produced in the decay of a heavier particle with four-momentum pC to find the mass of the heavier particle. Conservation of four-momentum gives pCμ = pAμ + pBμ, while the mass M of the heavier particle is given by PCPC = 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 electronpositron or muon–antimuon pairs.

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

${\displaystyle p^{\mu }A_{\mu }=\eta _{\mu \nu }p^{\mu }A^{\nu }=\eta _{\mu \nu }p^{\mu }{\frac {d}{d\tau }}{\frac {p^{\nu }}{m}}={\frac {1}{2m}}{\frac {d}{d\tau }}p\cdot p={\frac {1}{2m}}{\frac {d}{d\tau }}(-m^{2}c^{2})=0.}$

## 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:

${\displaystyle A=(A^{0},A^{1},A^{2},A^{3})=\left({\phi \over c},A_{x},A_{y},A_{z}\right)}$

where Φ is the scalar potential and A = (Ax, Ay, Az) the vector potential, the components of the canonical momentum four-vector P is

${\displaystyle P^{\mu }=p^{\mu }+qA^{\mu }.\!}$

This, in turn, 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.