KAHIBARO
Table of Contents
A Single Object in Relativistic Motion
In relativistic particle physics, ordinary momentum and energy are not best treated as separate quantities. They belong together in one spacetime object called the four-momentum. This is extremely useful because different observers may disagree on the separate values of energy and momentum, but the four-momentum transforms in a clean and consistent way between inertial frames.
For a particle of rest mass $m$, moving with velocity $\vec{v}$, we define the Lorentz factor as
$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
where $c$ is the speed of light.
The relativistic momentum is
$$
\vec{p} = \gamma m \vec{v}
$$
and the total energy is
$$
E = \gamma mc^2
$$
These combine into the four-momentum.
Definition of Four-Momentum
The four-momentum of a particle is written as
$$
P^\mu = \left(\frac{E}{c}, \vec{p}\right)
$$
In component form,
$$
P^\mu = \left(\frac{E}{c}, p_x, p_y, p_z\right)
$$
This looks similar to the spacetime position four-vector, where time and space are grouped together. Here, energy and momentum are grouped together.
The time-like component is $E/c$, and the space-like components are the three momentum components.
The four-momentum is
$$
P^\mu = \left(\frac{E}{c}, \vec{p}\right)
$$
with
$$
E = \gamma mc^2, \qquad \vec{p} = \gamma m \vec{v}
$$
Energy and momentum are not independent in relativity, they are parts of the same four-vector.
Why It Matters
In particle physics, reactions and decays are analyzed using conservation of four-momentum. Instead of separately tracking energy and each component of momentum in an ad hoc way, we use one unified object.
This is especially important at speeds close to $c$, where classical formulas fail.
Relation to Four-Velocity
The four-momentum is closely related to the four-velocity. If $U^\mu$ is the four-velocity, then
$$
P^\mu = m U^\mu
$$
For a particle of mass $m$,
$$
U^\mu = \gamma (c, \vec{v})
$$
so
$$
P^\mu = m \gamma (c, \vec{v}) = \left(\frac{E}{c}, \vec{p}\right)
$$
This shows that four-momentum is the natural relativistic extension of ordinary momentum.
Invariant Magnitude
A central feature of four-vectors is that their spacetime magnitude is invariant. For four-momentum, this invariant is related to the particle's rest mass.
Using the Minkowski spacetime relation,
$$
P^\mu P_\mu = \left(\frac{E}{c}\right)^2 - p^2
$$
where
$$
p^2 = p_x^2 + p_y^2 + p_z^2
$$
For a particle of rest mass $m$,
$$
P^\mu P_\mu = m^2 c^2
$$
Multiplying by $c^2$ gives the famous energy-momentum relation,
$$
E^2 = p^2 c^2 + m^2 c^4
$$
This equation is one of the most important results in relativistic physics.
For any particle,
$$
E^2 = p^2 c^2 + m^2 c^4
$$
This relation connects total energy, momentum, and rest mass.
Massive and Massless Particles
For a massive particle, $m \ne 0$, so both the energy and momentum satisfy
$$
E^2 = p^2 c^2 + m^2 c^4
$$
If the particle is at rest, then $\vec{p} = 0$, and the equation becomes
$$
E = mc^2
$$
which is the rest energy.
For a massless particle, such as a photon, $m = 0$. Then the relation becomes
$$
E^2 = p^2 c^2
$$
so
$$
E = pc
$$
This is why photons can carry momentum even though they have no rest mass.
Conservation of Four-Momentum
In particle interactions, four-momentum is conserved. This means the total four-momentum before an event equals the total four-momentum after it.
If particles 1 and 2 collide and produce particles 3 and 4, then
$$
P_1^\mu + P_2^\mu = P_3^\mu + P_4^\mu
$$
This single four-vector equation contains both energy conservation and conservation of the three momentum components.
In component form,
$$
E_1 + E_2 = E_3 + E_4
$$
and
$$
\vec{p}_1 + \vec{p}_2 = \vec{p}_3 + \vec{p}_4
$$
provided the same frame is used throughout.
In every allowed particle decay or collision,
$$
\sum P^\mu_{\text{before}} = \sum P^\mu_{\text{after}}
$$
This means both total energy and total momentum are conserved together.
Example, Particle at Rest
Consider a particle of rest mass $m$ sitting still.
Then
$$
\vec{v} = 0, \qquad \gamma = 1
$$
so
$$
\vec{p} = 0
$$
and
$$
E = mc^2
$$
Its four-momentum is
$$
P^\mu = (mc, 0, 0, 0)
$$
This is the simplest form of four-momentum, and it is especially useful in the particle's own rest frame.
Example, Moving Particle
Suppose a particle moves in the $x$ direction with speed $v$. Then
$$
\vec{p} = (\gamma mv, 0, 0)
$$
and
$$
E = \gamma mc^2
$$
so the four-momentum is
$$
P^\mu = \left(\frac{\gamma mc^2}{c}, \gamma mv, 0, 0\right)
= (\gamma mc, \gamma mv, 0, 0)
$$
Its invariant magnitude is still
$$
P^\mu P_\mu = (\gamma mc)^2 - (\gamma mv)^2 = m^2 c^2
$$
as required.
Geometric Picture
Four-momentum can be pictured as an arrow in spacetime, with one energy-like component and three momentum-like components. Different observers divide that arrow differently into time-like and space-like parts, but the invariant combination stays the same.
This drawing shows only one spatial momentum component for simplicity. In reality, the momentum part has three components.
Useful Special Cases
The most common forms are summarized below.
| Particle type | Momentum | Energy | Relation |
|---|---|---|---|
| Massive particle at rest | $0$ | $mc^2$ | $E = mc^2$ |
| Massive moving particle | $\gamma m \vec{v}$ | $\gamma mc^2$ | $E^2 = p^2 c^2 + m^2 c^4$ |
| Massless particle | nonzero | nonzero | $E = pc$ |
Units
The momentum components have units of momentum, and the time component is written as $E/c$ so that all four components have the same units.
| Quantity | SI unit |
|---|---|
| Energy $E$ | joule, J |
| Momentum $p$ | kg m/s |
| $E/c$ | kg m/s |
This matching of units is one reason the form $P^\mu = (E/c, \vec{p})$ is so natural.
Rest Frame Importance
For a massive particle, there exists a frame where the particle is at rest. In that frame,
$$
P^\mu = (mc, 0, 0, 0)
$$
This makes the rest frame especially useful for calculations in decays and collisions. The invariant mass of a system can often be found most easily by first forming the total four-momentum and then taking its invariant magnitude.
Total Four-Momentum of a System
For several particles, the total four-momentum is the sum of the individual four-momenta:
$$
P^\mu_{\text{tot}} = \sum_i P^\mu_i
$$
So for two particles,
$$
P^\mu_{\text{tot}} = \left(\frac{E_1 + E_2}{c}, \vec{p}_1 + \vec{p}_2\right)
$$
This idea is fundamental in particle physics because experiments usually involve systems of particles, not just one particle.
For a system of particles,
$$
P^\mu_{\text{tot}} = \sum_i P^\mu_i
$$
The invariant built from the total four-momentum gives the invariant mass of the whole system.
Final Perspective
Four-momentum is the relativistic quantity that unifies energy and momentum into one object. Its components change from one inertial frame to another, but its invariant magnitude is tied to mass. Because it is conserved in all particle interactions, it is one of the most powerful tools in nuclear and particle physics.