---
title: Relativistic Momentum and Energy
module: Foundations of Relativity
moduleNumber: 1
lessonNumber: 4
order: 104
summary: >
  Conserving momentum in every inertial frame forces the redefinition p = gamma
  m u, which diverges as the speed approaches c. Integrating the corresponding
  force gives the total energy E = gamma m c-squared, whose rest term m c-squared
  is Einstein's mass-energy equivalence. Energy and momentum join into a
  four-vector whose invariant length is the rest energy, giving E-squared =
  (pc)-squared + (m c-squared)-squared, massless particles, and nuclear binding
  energy.
topics: [Foundations of Relativity]
draft: false
sources:
  - book: Tipler & Llewellyn
    ref: "Ch. 2 — Relativity II; §2-1 Relativistic Momentum; §2-2 Relativistic Energy"
  - book: Tipler & Llewellyn
    ref: "§2-3 Mass/Energy Conversion and Binding Energy; §2-4 Invariant Mass"
---

The [kinematics of special
relativity](/relativity/foundations/time-dilation-length-contraction)
forces a matching change in dynamics. Classical momentum $\vec{p} =
m\vec{u}$ is not conserved in every inertial frame once velocities transform
by the relativistic rule, and Newton's $\vec{F} = m\vec{a}$ cannot be
right if it leads to conservation of $m\vec{u}$. Repairing conservation
gives new definitions of momentum and energy, both reducing to the classical
forms at low speed, and it ties mass to energy.[^tl-mom]

## Relativistic momentum

Consider an elastic glancing collision of two identical balls, symmetric between
frames $S$ and $S'$ that move relative to each other at speed $v$. Requiring the
$y$-momentum to balance in both frames, in the limit where each ball is nearly at
rest in its home frame, forces the momentum of a ball moving at speed $u$ to be

$$
\vec{p} = \frac{m\vec{u}}{\sqrt{1 - u^2/c^2}} = \gamma_u\, m\vec{u},
\qquad \gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}}.
$$

Here $m$ is the **rest mass**, the mass measured in the ball's own frame, and
$\gamma_u$ uses the particle's speed $u$ relative to the observer (distinct from
the $\gamma$ of a frame's relative velocity). The definition meets the two
requirements: momentum stays conserved across frames, and for $u \ll c$ the
denominator is $\approx 1$ and $\vec{p} \to m\vec{u}$.

> **Definition (Relativistic momentum).** For a particle of rest mass $m$ moving
> at velocity $\vec{u}$, $\vec{p} = \gamma_u m \vec{u}$ with $\gamma_u =
> (1 - u^2/c^2)^{-1/2}$. The rest mass $m$ is a fixed property of the particle;
> the factor $\gamma_u$ carries all the speed dependence and makes $p$ diverge
> as $u \to c$.

Because $\gamma_u \to \infty$ as $u \to c$, the momentum grows without bound as
the speed approaches $c$, while the classical $mu$ would level off at $mc$. No
finite momentum transfer brings a massive particle to the speed of light.

$$
% caption: Relativistic momentum in units of m c versus u over c (solid),
% diverging at the speed of light, against the classical straight line m u
% (dashed), which the relativistic curve tracks only at low speed.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.6,0) node[right] {speed (u over c)};
  \draw[->, black] (0,0) -- (0,4.4) node[above] {momentum (p over mc)};
  \foreach \x/\l in {0/0, 3/0.5, 6/1} \draw[black] (\x,0.05) -- (\x,-0.05) node[below] {\l};
  \foreach \y/\l in {1.2/1, 2.4/2, 3.6/3} \draw[black] (-0.05,\y) -- (0.05,\y) node[left] {\l};
  % classical p = mu -> line to (6,1.2)
  \draw[black, dashed] (0,0) -- (6,1.2) node[right] {classical};
  % relativistic p/mc = (u/c)/sqrt(1-(u/c)^2), x in 0..6 maps u/c 0..1, y scaled 1.2
  \draw[acc, very thick]
    (0,0) .. controls (2.0,0.5) and (3.4,0.95) ..
    (4.2,1.45) .. controls (4.9,2.0) and (5.4,2.7) ..
    (5.7,3.4) .. controls (5.85,3.85) and (5.93,4.15) .. (5.97,4.3);
  \node[acc, anchor=east] at (5.6,3.0) {relativistic};
\end{tikzpicture}
$$

The relativistically correct force keeps Newton's second law in the form
$\vec{F} = \d\vec{p}/\d t$, now with the relativistic $\vec{p}$:

$$
\vec{F} = \frac{\d(\gamma_u m \vec{u})}{\d t}.
$$

## Relativistic energy

Kinetic energy is the work a net force does in accelerating a particle from rest
to speed $u$. In one dimension,

$$
E_k = \int_0^u F\,\d x = \int_0^u u\,\d(\gamma_u m u).
$$

Using $\d(\gamma_u m u) = m(1 - u^2/c^2)^{-3/2}\,\d u$ and integrating gives

$$
E_k = \gamma_u m c^2 - m c^2 = (\gamma_u - 1)\,m c^2.
$$

For $u \ll c$, the binomial expansion $\gamma_u \approx 1 + \tfrac{1}{2}u^2/c^2$
recovers $E_k \approx \tfrac{1}{2}m u^2$. The exact expression diverges as $u \to
c$: bringing a massive particle to the speed of light would take infinite
energy, the dynamical statement of the same speed limit the momentum shows.

$$
% caption: Relativistic kinetic energy in units of m c-squared versus u over c
% (solid) diverges at the speed of light; the classical one-half m u-squared
% (dashed) agrees only at low speed.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.6,0) node[right] {speed (u over c)};
  \draw[->, black] (0,0) -- (0,4.4) node[above] {kinetic energy};
  \foreach \x/\l in {0/0, 3/0.5, 6/1} \draw[black] (\x,0.05) -- (\x,-0.05) node[below] {\l};
  \foreach \y/\l in {1.4/1, 2.8/2} \draw[black] (-0.05,\y) -- (0.05,\y) node[left] {\l};
  % classical 0.5 (u/c)^2, at u/c=1 -> 0.5 -> y=0.7
  \draw[black, dashed] (0,0) .. controls (2.5,0.12) and (4.5,0.35) .. (6,0.7)
    node[right] {classical};
  % relativistic (gamma-1): rises slowly then diverges
  \draw[acc, very thick]
    (0,0) .. controls (2.2,0.25) and (3.6,0.65) ..
    (4.4,1.2) .. controls (5.0,1.8) and (5.4,2.6) ..
    (5.7,3.4) .. controls (5.85,3.9) and (5.93,4.2) .. (5.97,4.35);
  \node[acc, anchor=east] at (5.5,2.9) {relativistic};
\end{tikzpicture}
$$

The kinetic energy splits into a speed-dependent piece $\gamma_u m c^2$ and a
constant $m c^2$. The constant is the **rest energy**, and the **total energy**
is their sum:

$$
E = E_k + m c^2 = \gamma_u m c^2 = \frac{m c^2}{\sqrt{1 - u^2/c^2}}.
$$

> **Theorem (Mass-energy equivalence).** A particle at rest has energy $E = m
> c^2$. Any energy $\Delta E$ stored in a body in any form increases its mass by
> $\Delta E / c^2$; conversely, a mass loss $\Delta m$ releases energy $\Delta m
> \,c^2$. Mass and energy are the same quantity in different units.

Total energy $E$, not $E_k$ and not $mc^2$ separately, is the conserved quantity
for an isolated system in each frame. Conservation and invariance are different
properties: a conserved quantity is unchanged before and after an interaction
within one frame, while an invariant quantity has the same value in all frames.
Total energy is conserved but not invariant; different frames measure different
$E$.

> **Example (Solar mass loss).** The Sun radiates $P = 4\pi R^2 \times
> 1.36 \times 10^3\ \mathrm{W/m^2} \approx 3.85 \times 10^{26}\ \mathrm{W}$ at
> Earth's orbital radius $R = 1.50 \times 10^{11}\ \mathrm{m}$. By $E = mc^2$ the
> corresponding mass loss rate is
> $$
> \frac{\d m}{\d t} = \frac{P}{c^2} = \frac{3.85 \times 10^{26}}{(3.00 \times 10^8)^2}
> \approx 4.3 \times 10^9\ \mathrm{kg/s},
> $$
> about four million tonnes of mass converted to radiant energy every second.

## The energy-momentum four-vector

Energy and momentum transform between frames together, exactly as time and space
do. Writing out the transformation for a particle's $E$ and $\vec{p}$ between
$S$ and $S'$ moving at relative speed $v$,

$$
p'_x = \gamma\parens{p_x - \frac{vE}{c^2}}, \qquad
p'_y = p_y, \qquad p'_z = p_z, \qquad
E' = \gamma\,(E - v p_x).
$$

The momentum $\vec{p}$ transforms like the position $\vec{r}$, and the
energy $E$ transforms like the time $t$. Momentum and energy are the components
of a **four-vector**, and like the invariant interval $(\Delta s)^2 = (c\Delta
t)^2 - (\Delta x)^2$, this four-vector has an invariant length. Squaring the
definitions of $E$ and $p$ and subtracting gives

$$
E^2 - (pc)^2 = (m c^2)^2, \qquad\text{i.e.}\qquad E^2 = (pc)^2 + (m c^2)^2.
$$

> **Definition (Invariant mass).** The magnitude of the energy-momentum
> four-vector, $(m c^2)^2 = E^2 - (pc)^2$, is the rest energy and is the same in
> every inertial frame. Only in the rest frame, where $p = 0$, do the total
> energy and rest energy coincide, $E = m c^2$.

The relation is a right triangle in $(E, pc, mc^2)$: the total energy is the
hypotenuse, momentum times $c$ and rest energy the legs. At low speed $pc \ll
mc^2$ and $E \approx mc^2 + p^2/2m$; at high speed $pc \gg mc^2$ and $E \approx
pc$.

$$
% caption: The energy-momentum triangle. Total energy E is the hypotenuse of a
% right triangle whose legs are momentum times c and rest energy m c-squared, so
% E-squared equals (pc)-squared plus (m c-squared)-squared.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \coordinate (o) at (0,0);
  \coordinate (a) at (4.2,0);
  \coordinate (b) at (4.2,2.6);
  \draw[black, thick] (o) -- (a) node[midway, below] {$pc$};
  \draw[black, thick] (a) -- (b) node[midway, right] {$mc^2$};
  \draw[acc, very thick] (o) -- (b) node[midway, above, sloped] {$E$};
  \draw[black] (3.95,0) -- (3.95,0.25) -- (4.2,0.25);
  \fill[black] (o) circle (1.6pt);
\end{tikzpicture}
$$

> **Example (Speed of a fast electron).** An electron has total energy $E =
> 2.40\ \mathrm{MeV}$ and rest energy $m c^2 = 0.511\ \mathrm{MeV}$. Its momentum
> follows from the invariant relation, $pc = \sqrt{E^2 - (mc^2)^2} =
> \sqrt{2.40^2 - 0.511^2} = 2.34\ \mathrm{MeV}$. Its speed comes from $u/c =
> pc/E = 2.34/2.40 = 0.975$, so $u = 0.975c$.

## Massless particles

The relation $E^2 = (pc)^2 + (mc^2)^2$ allows $m = 0$, which has no classical
analog. Setting $mc^2 = 0$,

$$
E = pc \qquad (m = 0),
$$

and combined with $u/c = pc/E$ this gives $u = c$. A massless particle moves at
the speed of light in every frame, carrying energy and momentum despite zero
rest mass; the product $\gamma_u m$ stays finite as $m \to 0$ and $\gamma_u \to
\infty$. The photon is the established example, with the gluon and the
hypothetical graviton the others. For the photon the interval and the rest mass
both vanish, and "rest frame" has no meaning, since light moves at $c$ relative
to all frames.

The invariant mass of a _system_ is not the sum of its parts' masses when the
parts move relative to one another. Two photons of energies $5\ \mathrm{MeV}$ and
$2\ \mathrm{MeV}$ approaching each other have total energy $7\ \mathrm{MeV}$ and
net momentum $(5 - 2)/c = 3\ \mathrm{MeV}/c$, so the system rest energy is
$\sqrt{7^2 - 3^2} = 6.3\ \mathrm{MeV}$, nonzero even though each photon is
massless.

## Binding energy

When particles are bound together, energy $E_b$ must be supplied to separate
them, so the bound system's mass is _less_ than the sum of its parts by $E_b/c^2$.

> **Theorem (Mass and binding energy).** The mass of a bound system is less than
> the total mass of its separated constituents by $E_b / c^2$, where $E_b$ is the
> binding energy, the energy needed to pull the system apart.

The deuteron makes the effect visible. It is a proton and neutron bound together.
The rest energies of the free proton and neutron sum to $938.272 + 939.565 =
1877.837\ \mathrm{MeV}$, while the deuteron's rest energy is $1875.613\
\mathrm{MeV}$. The difference,

$$
E_b = 1877.837 - 1875.613 = 2.224\ \mathrm{MeV},
$$

is the deuteron's binding energy: $2.224\ \mathrm{MeV}$ must be added to split it,
and the same energy is released when a proton and neutron fuse into a deuteron.

$$
% caption: Mass-energy bookkeeping for the deuteron. The free proton plus
% neutron outweigh the bound deuteron by the binding energy over c-squared,
% about 2.224 MeV over c-squared.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (0,4.0) node[above] {rest energy (MeV)};
  \draw[black] (0,0) -- (6.4,0);
  % free p+n bar
  \fill[black] (0.6,0) rectangle (2.2,3.5);
  \draw[black] (0.6,0) rectangle (2.2,3.5);
  \node[anchor=north, align=center] at (1.4,-0.1) {free p + n\\1877.837};
  % deuteron bar
  \fill[acc!14] (3.4,0) rectangle (5.0,3.24);
  \draw[acc] (3.4,0) rectangle (5.0,3.24);
  \node[anchor=north, align=center] at (4.2,-0.1) {deuteron\\1875.613};
  % gap indicator
  \draw[black] (2.2,3.5) -- (5.4,3.5);
  \draw[black] (5.0,3.24) -- (5.4,3.24);
  \draw[<->, black] (5.35,3.24) -- (5.35,3.5);
  \node[anchor=west, align=center] at (5.45,3.37) {$\frac{E_b}{c^2}$\\2.224};
\end{tikzpicture}
$$

The same accounting drives every energy release from the nucleus. Fission of a
heavy nucleus and fusion of light ones both convert a mass defect into kinetic
energy of the products, and the amounts are large because nuclear binding
energies are millions of times the electron-volt binding energies of atoms. The
[nuclear physics](/nuclear-physics) module follows that thread; the mass-energy
relation established here is its foundation. The last lesson leaves inertial
frames behind and asks what happens under acceleration and gravity, in
[general relativity](/relativity/foundations/general-relativity).

[^tl-mom]: **Tipler & Llewellyn**, _Modern Physics_, §2-1 to §2-4 — Relativistic Momentum ($p = \gamma m u$ from momentum conservation), Relativistic Energy ($E_k = (\gamma - 1)mc^2$, $E = \gamma m c^2$, $E = mc^2$), Mass/Energy Conversion and Binding Energy, and Invariant Mass with the energy-momentum four-vector, $E^2 = (pc)^2 + (mc^2)^2$, and massless particles.