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-space momentum $p = (p x, p y, p z) = γm v$ , where $v$ is the particle's three-space velocity and $γ$ the Lorentz factor, is

p=(p^{0},p^{1},p^{2},p^{3})=\left({E \over 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 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 $x 0 = ct$ . Some authors use the convention $x 0 = t$ , which yields a modified definition with $p 0 = E / c 2$ . It is also possible to define covariant four-momentum $p μ$ where the sign of the energy is reversed.

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

where we use the convention that

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

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.^[1] 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 $p i$ ,^[2]

p_i = \frac{\partial S}{\partial q_i}, \quad E = -\frac{\partial S}{\partial t},

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-space 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[{\frac {\partial L}{\partial {\dot q}}}\delta q\right]\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}.

Observing that

\delta S=\sum _{i}{\frac {\partial S}{\partial {q}_{i}}}\delta q_{i},

one concludes

p_{i}={\frac {\partial S}{\partial q_{i}}}.

In a similar fashion, keep endpoints fixed, but let $t 2 = t$ vary. This time, the system is allowed to move through configuration space at "arbitrary speed" or with "more or less energy", the field equations still assumed to hold and variation can be carried out on the integral, but instead observe

{\frac {dS}{dt}}=L

by the fundamental theorem of calculus. Compute using the above expression for canonical momenta,

{\frac {dS}{dt}}={\frac {\partial S}{\partial t}}+\sum _{i}{\frac {\partial S}{\partial q_{i}}}{\dot {q}}_{i}={\frac {\partial S}{\partial t}}+\sum _{i}p_{i}{\dot {q}}_{i}=L.

Now using

H=\sum _{i}p_{i}{\dot {q}}_{i}-L,

where $H$ is the Hamiltonian, leads to, since $E = H$ in the present case,

E=H=-{\frac {\partial S}{\partial t}}.

Incidentally, using $H = H (q, p, t)$ with $p = \partial S / \partial q$ in the above equation yields the Hamilton–Jacobi equations. In this context, $S$ is called Hamilton's principal function.

The action $S$ is given by

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

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

glossing over these details,

The variation of the action is

\delta S = -mc\int \delta ds.

To calculate $δds$ , observe first that $δds 2 = 2 dsδds$ and that

\delta ds^2 = \delta \eta_{\mu\nu}dx^\mu dx^\nu = \eta_{\mu\nu} (\delta (dx^\mu) dx^\nu + dx^\mu \delta (dx^\nu)) = 2\eta_{\mu\nu} \delta (dx^\mu) dx^\nu.

\delta ds = \eta_{\mu\nu} \delta dx^\mu \frac{dx^\nu}{ds} = \eta_{\mu\nu} d\delta x^\mu \frac{dx^\nu}{ds},

\delta ds = \eta_{\mu\nu} \frac{d\delta x^\mu}{d\tau} \frac{dx^\nu}{cd\tau}d\tau,

and thus

\delta S = -m\int \eta_{\mu\nu} \frac{d\delta x^\mu}{d\tau} \frac{dx^\nu}{d\tau}d\tau =-m\int \eta_{\mu\nu} \frac{d\delta x^\mu}{d\tau} u^\nu d\tau =-m\int \eta_{\mu\nu} \left[\frac{d}{d\tau}(\delta x^\mu u^\nu) - \delta x^\mu\frac{d}{d\tau}u^\nu\right] d\tau

which is just

\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

\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 $du μ / ds = 0$ , $(δx μ) t 1 = 0$ , and $(δx μ) t 2 \equiv δx μ$ as in the observations above. Now compare the last three expressions to find

p^{\mu }=-\partial ^{\mu }[S]=-{\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 $- m 2 c 2$ , and the famed result for the relativistic energy,

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

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

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

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

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

Substituting

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

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

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

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

\begin{align} \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{align}

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 four-vector.

The energy and the three-space momentum are separately conserved quantities for isolated systems in the Lagrangian framework. Hence four-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 three-space 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 four-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 three-space vector part.

Conservation of four-momentum

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

The four-momentum $p$ (either covariant or contravariant) is conserved.
The total energy $E = p 0 c$ is conserved.
The 3-space 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 $- P C \cdot P C = M 2 c 2$ . By measuring the energies and three-space 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 the mass of an object does not change, the Minkowski inner product of its four-momentum and corresponding four-acceleration $A μ$ is simply 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 }}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:

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 $A = (A x, A y, A z)$ the vector potential, the components of the canonical momentum four-vector $P$ is

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.

References

↑ Landau & Lifshitz 2002, pp. 25–29
↑ Landau & Lifshitz 1975, pp. 139
↑ Landau & Lifshitz 1975, p. 30
↑ Landau & Lifshitz 1975, pp. 15–16
↑ Sard 1970, Section 3.1
↑ Sard 1970, Section 3.2
↑ Lewis & Tolman 1909 Wikisource version

Goldstein, Herbert (1980). Classical mechanics (2nd ed.). Reading, Mass.: Addison–Wesley Pub. Co. ISBN 0201029189.
Landau, L. D.; Lifshitz, E. M. (1975) [1939]. Mechanics. Translated from Russian by J. B. Sykes and J. S. Bell. (3rd ed.). Amsterdam: Elsevier. ISBN 978-0-7506-28969.
Landau, L.D.; Lifshitz, E.M. (2000). The classical theory of fields. 4th rev. English edition, reprinted with corrections; translated from the Russian by Morton Hamermesh. Oxford: Butterworth Heinemann. ISBN 9780750627689.
Rindler, Wolfgang (1991). Introduction to Special Relativity (2nd ed.). Oxford: Oxford University Press. ISBN 0-19-853952-5.
Sard, R. D. (1970). Relativistic Mechanics - Special Relativity and Classical Particle Dynamics. New York: W. A. Benjamin. ISBN 978-0805384918.
Lewis, G. N.; Tolman, R. C. (1909). "The Principle of Relativity, and Non-Newtonian Mechanics". Phil. Mag. 6. Taylor & Francis. 18 (106): 510–523. doi:10.1080/14786441008636725. (Subscription required (help)). Wikisource version

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