Four-momentum

In special relativity, four-momentum (also called momentum–energy or momenergy^[1]) 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 $p = (p x, p y, p z) = γm v$ , where $v$ is the particle's three-velocity and $γ$ the Lorentz factor, is

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

The quantity $m v$ 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 covariant vector. This means that it is easy to keep track of how it transforms under Lorentz transformations.

Minkowski norm

Calculating the Minkowski norm squared 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:

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

where

\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 metric tensor of special relativity with metric signature for definiteness chosen to be

(-1, 1, 1, 1)

. The negativity of the norm reflects that the momentum is a timelike four-vector for massive particles. The other choice of signature would flip signs in certain formulas (like for the norm here). This choice is not important, but once made it must for consistency be kept throughout.

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 four-momenta $p$ and $q$ , the quantity $p \cdot q$ 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,

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

where the four-velocity

u

is

u=\left(u^{0},u^{1},u^{2},u^{3}\right)=\gamma _{v}\left(c,v_{x},v_{y},v_{z}\right),

and

\gamma _{v}:={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}

is the Lorentz factor (associated with the speed

v

),

c

is the speed of light.

Derivation

There are several ways to arrive at the correct expression for four-momentum. One way is to first define the four-velocity $u = dx / dτ$ and simply define $p = mu$ , being content that it is a four-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 four-momentum, including the expression for the energy.^[2] One may at once, using the observations detailed below, define four-momentum from the action $S$ . Given that in general for a closed system with generalized coordinates $q i$ and canonical momenta $p i$ ,^[3]

p_{i}={\frac {\partial S}{\partial q_{i}}}={\frac {\partial S}{\partial x_{i}}},\quad E=-{\frac {\partial S}{\partial t}}=-c\cdot {\frac {\partial S}{\partial x_{0}}},

it is immediate (recalling

x 0 = ct

,

x 1 = x

,

x 2 = y

,

x 3 = z

and

x 0 = - x 0

,

x 1 = x 1

,

x 2 = x 2

,

x 3 = x 3

in the present metric convention) that

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

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

Observations

Consider initially a system of one degree of freedom $q$ . In the derivation of the equations of motion from the action using Hamilton's principle, one finds (generally) in an intermediate stage for the variation of the action,

\delta S=\left.\left\right|_{t_{1}}^{t_{2}}+\int _{t_{1}}^{t_{2}}\left({\frac {\partial L}{\partial q}}-{\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {q}}}}\right)\delta qdt.

The assumption is then that the varied paths satisfy $δq (t 1) = δq (t 2) = 0$ , from which Lagrange's equations follow at once. When the equations of motion are known (or simply assumed to be satisfied), one may let go of the requirement $δq (t 2) = 0$ . In this case the path is assumed to satisfy the equations of motion, and the action is a function of the upper integration limit $δq (t 2)$ , but $t 2$ is still fixed. The above equation becomes with $S = S (q)$ , and defining $δq (t 2) = δq$ , and letting in more degrees of freedom,

\delta S=\sum _{i}{\frac {\partial L}{\partial {\dot {q}}_{i}}}\delta q_{i}=\sum _{i}p_{i}\delta q_{i}.

[1]

[2]

[3]