KAHIBARO
Table of Contents
Why energy and momentum must be linked
In special relativity, energy and momentum are not separate ideas in the same way they are in classical mechanics. They are deeply connected. A moving particle has both energy and momentum, and relativity shows that these quantities satisfy one universal relation.
This relation is important because it works for particles with mass, particles without mass, slow motion, and motion close to the speed of light. It is one of the most useful formulas in modern physics.
The relativistic energy and momentum formulas
For a particle of rest mass $m$ moving with speed $v$, the relativistic momentum is
$$
p = \gamma m v
$$
and the total energy is
$$
E = \gamma m c^2
$$
where
$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
and $c$ is the speed of light.
These two expressions contain the same factor $\gamma$, which hints that energy and momentum are closely related.
Deriving the energy-momentum relation
Start from
$$
E = \gamma m c^2
$$
and
$$
p = \gamma m v
$$
Square both equations:
$$
E^2 = \gamma^2 m^2 c^4
$$
$$
p^2 = \gamma^2 m^2 v^2
$$
Now multiply the momentum equation by $c^2$:
$$
p^2 c^2 = \gamma^2 m^2 v^2 c^2
$$
Subtract this from $E^2$:
$$
E^2 - p^2 c^2 = \gamma^2 m^2 c^4 - \gamma^2 m^2 v^2 c^2
$$
Factor the right side:
$$
E^2 - p^2 c^2 = \gamma^2 m^2 c^2 (c^2 - v^2)
$$
Since
$$
\gamma^2 = \frac{1}{1 - \frac{v^2}{c^2}}
$$
we get
$$
\gamma^2 (c^2 - v^2) = c^2
$$
so the result becomes
$$
E^2 - p^2 c^2 = m^2 c^4
$$
or, more commonly,
$$
E^2 = p^2 c^2 + m^2 c^4
$$
The relativistic energy-momentum relation is
$$
E^2 = p^2 c^2 + m^2 c^4
$$
This formula is valid for any particle.
Meaning of the terms
In the equation
$$
E^2 = p^2 c^2 + m^2 c^4
$$
$E$ is the total energy of the particle, not just kinetic energy.
$p$ is the relativistic momentum.
$m$ is the rest mass, sometimes called invariant mass.
The term $m c^2$ is connected to the particle's rest energy. When the particle is at rest, $p = 0$, so the equation becomes
$$
E = mc^2
$$
This is the famous mass-energy equivalence for a particle at rest.
Special case, particle at rest
If the particle is not moving, then $v = 0$, so momentum is zero:
$$
p = 0
$$
Then the energy-momentum relation gives
$$
E^2 = m^2 c^4
$$
so
$$
E = mc^2
$$
This is the rest energy,
$$
E_0 = mc^2
$$
where $E_0$ means energy when the particle is at rest.
For a particle at rest,
$$
E_0 = mc^2
$$
This is called the rest energy.
Special case, massless particles
Some particles, such as photons, have zero rest mass:
$$
m = 0
$$
Then the energy-momentum relation becomes
$$
E^2 = p^2 c^2
$$
so
$$
E = pc
$$
This is the energy formula for a massless particle.
A photon always moves at speed $c$, and its energy is determined entirely by its momentum.
For a massless particle,
$$
E = pc
$$
This is especially important for photons.
Special case, low-speed limit
When the speed is much smaller than $c$, relativity must agree with classical physics. In this limit, the total energy can be written approximately as
$$
E \approx mc^2 + \frac{1}{2}mv^2
$$
The first term is the rest energy, and the second term is the classical kinetic energy.
So classical kinetic energy appears as an approximation inside the relativistic formula.
Total energy, rest energy, and kinetic energy
The total energy can be separated into rest energy and kinetic energy:
$$
E = E_0 + K
$$
where
$$
E_0 = mc^2
$$
and $K$ is the relativistic kinetic energy.
So,
$$
K = E - mc^2
$$
Using the total energy formula,
$$
K = \gamma mc^2 - mc^2 = (\gamma - 1)mc^2
$$
This shows that kinetic energy grows very rapidly when $v$ becomes close to $c$.
Comparing classical and relativistic cases
The following table helps show the difference.
| Case | Momentum | Energy |
|---|---|---|
| Classical particle | $p = mv$ | $E = \frac{1}{2}mv^2$ for kinetic energy only |
| Relativistic particle | $p = \gamma mv$ | $E = \gamma mc^2$ total energy |
| Particle at rest | $0$ | $mc^2$ |
| Massless particle | not zero in general | $E = pc$ |
Geometric view of the relation
The equation
$$
E^2 = p^2 c^2 + m^2 c^4
$$
has a form similar to the Pythagorean theorem:
$$
a^2 = b^2 + c^2
$$
This does not mean energy and momentum are ordinary sides of a triangle in space, but it helps show that they are linked in a precise mathematical way.
This drawing is only a mathematical picture to help visualize the formula.
Solving for different quantities
The energy-momentum relation can be rearranged depending on what is known.
If energy and mass are known, momentum is
$$
p = \frac{1}{c}\sqrt{E^2 - m^2 c^4}
$$
If momentum and mass are known, total energy is
$$
E = \sqrt{p^2 c^2 + m^2 c^4}
$$
If energy and momentum are known, rest mass is
$$
m = \frac{1}{c^2}\sqrt{E^2 - p^2 c^2}
$$
A physically allowed particle must satisfy
$$
E^2 - p^2 c^2 = m^2 c^4 \ge 0
$$
This means $E \ge pc$ for ordinary particles.
Example with a particle at rest
Suppose an electron is at rest. Its momentum is zero, so the total energy is just its rest energy:
$$
E = mc^2
$$
If the electron mass is $m$, then no motion is needed for it to possess energy. This is one of the central ideas of relativity, mass itself stores energy.
Example with a photon
For a photon,
$$
m = 0
$$
so
$$
E = pc
$$
If the photon energy is known, its momentum is
$$
p = \frac{E}{c}
$$
Even though a photon has no rest mass, it still carries momentum and can transfer it to matter.
Why this relation matters
The energy-momentum relation is used everywhere in modern physics. It is essential in particle physics, nuclear physics, astrophysics, and quantum theory. It allows us to analyze collisions, decays, radiation, and high-speed particles in a single framework.
It also unifies two famous ideas, rest energy and relativistic momentum, into one compact equation.
The single most important result of this chapter is
$$
E^2 = p^2 c^2 + m^2 c^4
$$
At rest, this becomes $E = mc^2$.
For massless particles, this becomes $E = pc$.
Final perspective
In classical mechanics, mass, energy, and momentum are treated as mostly separate concepts. In special relativity, they are tied together by a deeper structure. The energy-momentum relation expresses that connection clearly.
Once this formula is understood, many results in modern physics become much easier to recognize and use.