---
title: Four-Momentum, Four-Force, and Accelerated Motion
module: Relativistic Dynamics
moduleNumber: 3
lessonNumber: 1
order: 301
summary: >
  The four-momentum packages energy and momentum into a single vector whose
  invariant length is the rest mass. Its proper-time derivative is the four-force,
  always orthogonal to the four-velocity, and a constant orthogonal four-force
  produces hyperbolic motion. Constant proper acceleration gives rapidity linear in
  proper time, the relativistic rocket equation, and the Rindler horizon behind an
  eternally accelerating observer.
topics: [Relativistic Dynamics]
draft: false
sources:
  - book: Schutz
    ref: "A First Course in GR, Ch. 2 — Vector Analysis in Special Relativity; §2.4 Four-momentum; problems on uniform acceleration"
  - book: Hartle
    ref: "Gravity, Ch. 5 — Special Relativistic Mechanics; §5.3 Four-momentum"
---

The kinematics of [four-vectors and index
notation](/relativity/spacetime-and-the-lorentz-group/four-vectors-and-index-notation)
supplies the objects; dynamics tells them how to move. A particle carries a
four-velocity tangent to its worldline, and multiplying by the rest mass gives
the central object of relativistic mechanics: the four-momentum, whose time
component is energy and whose space components are momentum. Newton's second law
becomes a statement about the proper-time derivative of this vector, and the
constraint that the rest mass is fixed forces the four-force to be orthogonal to
the four-velocity in the Minkowski sense. Applying a constant such force produces
not the parabola of Galilean free fall but a hyperbola in spacetime, the
worldline of constant proper acceleration that underlies the relativistic rocket
and the accelerated observer's horizon.

Throughout, the metric signature is $(-,+,+,+)$, so the Minkowski inner product
of two four-vectors $A$ and $B$ with components $A^\mu = (A^0, \vec A)$ is
$A\cdot B = \eta_{\mu\nu}A^\mu B^\nu = -A^0 B^0 + \vec A\cdot\vec B$.

## The four-momentum

For a particle of rest mass $m$ moving with velocity $\vec u$ relative to an
inertial frame, the four-velocity is the derivative of position with respect to
proper time $\tau$,

$$
U^\mu = \frac{\d x^\mu}{\d\tau} = \gamma_u\,(c,\ \vec u),
\qquad
\gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}},
$$

where the factor $\gamma_u = \d t/\d\tau$ converts coordinate time to proper
time along the worldline. Its Minkowski square is fixed,

$$
U\cdot U = \gamma_u^2\!\left(-c^2 + u^2\right) = -c^2,
$$

independent of the particle's speed. The four-momentum is rest mass times
four-velocity:

$$
p^\mu = m\,U^\mu = \bigl(\gamma_u m c,\ \gamma_u m\vec u\bigr)
      = \left(\frac{E}{c},\ \vec p\right).
$$

The time component reproduces the relativistic energy $E = \gamma_u m c^2$ and
the space components reproduce the relativistic momentum $\vec p = \gamma_u
m\vec u$, so the split of energy and momentum that looked frame-dependent in
[the foundations](/relativity/foundations/relativistic-momentum-energy) is the
split of one four-vector into time and space parts.

> **Definition (Four-momentum).** For a particle of rest mass $m$ and
> four-velocity $U^\mu$, the four-momentum is $p^\mu = mU^\mu = (E/c,\ \vec p)$
> with $E = \gamma_u mc^2$ and $\vec p = \gamma_u m\vec u$. Total four-momentum is
> conserved: for an isolated system $\sum_i p_i^\mu$ is the same before and after
> any interaction, a single equation that carries both energy and momentum
> conservation.

Its invariant length follows from $U\cdot U = -c^2$:

$$
p\cdot p = m^2\,U\cdot U = -m^2c^2.
$$

Writing the inner product in components, $p\cdot p = -E^2/c^2 + |\vec p|^2$, and
equating to $-m^2c^2$ gives the energy–momentum relation

$$
E^2 = (|\vec p|\,c)^2 + (mc^2)^2.
$$

The rest mass is the Minkowski length of the four-momentum, an invariant every
observer agrees on, while $E$ and $\vec p$ separately are frame-dependent
projections. A massless particle has $p\cdot p = 0$: its four-momentum is null,
$E = |\vec p|c$, and it moves along the light cone.

$$
% caption: The energy-momentum four-vector on a hyperbola of fixed invariant
% length m c; a boost slides E and the momentum along the curve without changing
% the rest mass, which is the Minkowski distance from the origin.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (5.4,0) node[right] {momentum $|\vec p|c$};
  \draw[->, black] (0,0) -- (0,4.4) node[above] {energy $E$};
  % hyperbola E^2 = (pc)^2 + (mc^2)^2 ; vertex at E = mc^2 = 1.6
  \draw[acc, very thick, domain=0:1.15, samples=60, variable=\t]
    plot ({1.6*sinh(\t)},{1.6*cosh(\t)});
  % asymptote E = pc (massless)
  \draw[black, dashed] (0,0) -- (3.2,3.2) node[right, black] {$E=|\vec p|c$};
  \fill (0,1.6) circle (1.6pt);
  \node[left, black!70] at (0,1.6) {rest};
  \fill (2.2,2.72) circle (1.6pt);
  \draw[->, black] (0.15,1.6) to[bend left=14] (2.05,2.66);
  \node[right] at (2.24,2.72) {boosted};
  \node[black!70] at (0.95,0.55) {length $= mc$};
  \draw[<->, black] (0.06,0.06) -- (0.66,0.66);
\end{tikzpicture}
$$

## The four-force and the constraint of fixed mass

The relativistic equation of motion replaces the coordinate-time derivative of
Newton's law with a proper-time derivative, giving the four-force (Minkowski
force)

$$
F^\mu = \frac{\d p^\mu}{\d\tau}.
$$

Its space part relates to the ordinary three-force $\vec f = \d\vec p/\d t$ by
the chain rule $\d/\d\tau = \gamma_u\,\d/\d t$, so $\vec F = \gamma_u\vec f$, and
its time part is the rate at which the force does work: with $E = \gamma_u mc^2$,

$$
F^0 = \frac{1}{c}\frac{\d E}{\d\tau} = \frac{\gamma_u}{c}\,\frac{\d E}{\d t}
     = \frac{\gamma_u}{c}\,\vec f\cdot\vec u.
$$

When the rest mass is constant, $p\cdot p = -m^2c^2$ is constant along the
worldline. Differentiating with respect to proper time,

$$
\frac{\d}{\d\tau}\,(p\cdot p) = 2\,p\cdot\frac{\d p}{\d\tau}
   = 2\,p\cdot F = 0,
$$

so $F\cdot U = 0$: **the four-force is Minkowski-orthogonal to the
four-velocity.** This is the covariant statement that a force which only steers a
particle, without changing its rest mass, cannot have a component along its own
four-velocity. Equivalently the four-acceleration $A^\mu = \d U^\mu/\d\tau =
F^\mu/m$ satisfies $A\cdot U = 0$; differentiating $U\cdot U = -c^2$ gives the
same relation directly.

> **Theorem (Orthogonality of four-force and four-velocity).** For a particle of
> constant rest mass, $F\cdot U = 0$ and equivalently $A\cdot U = 0$. The
> four-acceleration is a spacelike vector, $A\cdot A \ge 0$, and its invariant
> length is the square of the proper acceleration $\alpha$ measured in the
> particle's instantaneous rest frame: $A\cdot A = \alpha^2$.

The last identity is read off in the instantaneous rest frame (IRF), the
inertial frame momentarily comoving with the particle. There $U^\mu = (c, \vec
0)$, so $A\cdot U = 0$ forces $A^0 = 0$, and the four-acceleration is purely
spatial, $A^\mu = (0, \vec a_{\text{IRF}})$, with $A\cdot A = |\vec
a_{\text{IRF}}|^2 = \alpha^2$. The proper acceleration $\alpha$ is what an
accelerometer carried by the particle reads.

$$
% caption: In the instantaneous rest frame the four-velocity points along the
% time axis and the orthogonal four-acceleration lies in the space direction;
% Minkowski orthogonality tilts the pair away from a Euclidean right angle.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.3,0) -- (5.0,0) node[right] {space $x$};
  \draw[->, black] (0,-0.3) -- (0,4.3) node[above] {time $ct$};
  \draw[black, dashed] (0,0) -- (3.6,3.6);
  \draw[black, dashed] (0,0) -- (-0.3,-0.3);
  \node[black, right] at (3.4,3.65) {light cone};
  % worldline (mild curve)
  \draw[very thick] (0.4,0) .. controls (1.1,1.6) and (1.9,2.6) .. (2.7,4.0);
  % event point
  \fill[black!75] (1.5,2.2) circle (1.6pt);
  % four-velocity U tangent
  \draw[->, acc, very thick] (1.5,2.2) -- (2.15,3.2) node[right] {four-velocity};
  % four-acceleration A, spacelike, orthogonal
  \draw[->, black!70, very thick] (1.5,2.2) -- (2.5,1.75) node[right] {four-acceleration};
  \node[left] at (0.35,0.05) {worldline};
\end{tikzpicture}
$$

## The three-force and longitudinal versus transverse response

The ordinary three-force $\vec f = \d\vec p/\d t$ relates to the acceleration
$\vec a = \d\vec u/\d t$ in a way that mixes the direction of motion. Differentiate
$\vec p = \gamma_u m\vec u$,

$$
\vec f = m\,\frac{\d(\gamma_u\vec u)}{\d t}
       = m\bigl(\dot\gamma_u\,\vec u + \gamma_u\,\vec a\bigr),
\qquad
\dot\gamma_u = \gamma_u^3\,\frac{\vec u\cdot\vec a}{c^2},
$$

so

$$
\vec f = \gamma_u m\,\vec a
       + \gamma_u^3\,m\,\frac{(\vec u\cdot\vec a)}{c^2}\,\vec u .
$$

The force is not parallel to the acceleration unless $\vec a$ is purely along or
purely across $\vec u$. For acceleration **along** the motion ($\vec a \parallel
\vec u$), the identity $1 + \gamma_u^2 u^2/c^2 = \gamma_u^2$ collapses the two
terms; for acceleration **across** the motion ($\vec a \perp \vec u$) the second
term drops out. The two responses are

$$
f_\parallel = \gamma_u^3\, m\, a_\parallel,
\qquad
f_\perp = \gamma_u\, m\, a_\perp .
$$

A given force produces $\gamma_u^2$ times less acceleration along the motion than
across it. The coefficients $\gamma_u^3 m$ and $\gamma_u m$ were historically
called the **longitudinal** and **transverse mass**; the modern reading is that
there is one rest mass $m$ and the direction dependence lives in the kinematic
factors. The longitudinal stiffness $\gamma_u^3 m$ is the reason a linear
accelerator finds it progressively harder to speed a particle up as $u \to c$:
the response to a forward push falls as $\gamma_u^{-3}$, while a transverse
magnetic bending force is comparatively cheap, which is why circular machines
steer high-energy beams with magnets.

## Constant proper acceleration and hyperbolic motion

Set a particle moving along the $x$-axis under a constant proper acceleration
$\alpha$, its accelerometer reading a fixed value forever. In the IRF the
four-acceleration is $(0,\alpha,0,0)$. Boosting to the lab frame in which the
particle has instantaneous rapidity $\phi$ (with $\beta = \tanh\phi$, $\gamma =
\cosh\phi$), the four-velocity and four-acceleration are

$$
U^\mu = c\,(\cosh\phi,\ \sinh\phi,\ 0,\ 0),
\qquad
A^\mu = \frac{\d U^\mu}{\d\tau}
      = c\,\dot\phi\,(\sinh\phi,\ \cosh\phi,\ 0,\ 0),
$$

where the dot is $\d/\d\tau$. Then $A\cdot A = c^2\dot\phi^2 = \alpha^2$, so the
rapidity accumulates linearly with proper time:

$$
\frac{\d\phi}{\d\tau} = \frac{\alpha}{c}
\quad\Longrightarrow\quad
\phi(\tau) = \frac{\alpha\tau}{c}.
$$

Rapidity is the natural "angle" of the motion, advancing at a constant rate
under constant proper acceleration exactly as an ordinary angle advances under
constant angular velocity. The lab-frame velocity is then

$$
\beta(\tau) = \tanh\!\left(\frac{\alpha\tau}{c}\right),
\qquad
\gamma(\tau) = \cosh\!\left(\frac{\alpha\tau}{c}\right),
$$

which approaches but never reaches $c$: the hyperbolic tangent saturates at
unity. Integrating $U^\mu = \d x^\mu/\d\tau$ with $\phi = \alpha\tau/c$ and the
particle starting from rest at $x = c^2/\alpha$ at $\tau = 0$ gives the
worldline

$$
ct(\tau) = \frac{c^2}{\alpha}\,\sinh\!\left(\frac{\alpha\tau}{c}\right),
\qquad
x(\tau)  = \frac{c^2}{\alpha}\,\cosh\!\left(\frac{\alpha\tau}{c}\right).
$$

Eliminating $\tau$ with $\cosh^2 - \sinh^2 = 1$ yields the trajectory in
spacetime,

$$
x^2 - (ct)^2 = \left(\frac{c^2}{\alpha}\right)^2,
$$

a hyperbola with asymptotes $x = \pm ct$. This is why constant proper
acceleration is called **hyperbolic motion**: the worldline is a hyperbola whose
asymptotes are the light cone, in contrast to the Galilean parabola $x =
x_0 + \tfrac12 a t^2$ that the same equations reduce to for $\alpha\tau \ll c$.

$$
% caption: Constant proper acceleration traces a hyperbola asymptotic to the
% light cone; equal proper-time ticks crowd together in lab time as the speed
% approaches c, and the tangent four-velocity tips toward the cone.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,-3.2) -- (0,3.5) node[above] {time $ct$};
  \draw[->, black] (-0.2,0) -- (5.2,0) node[right] {space $x$};
  % light cone asymptotes through origin
  \draw[black, dashed] (0,0) -- (3.4,3.4);
  \draw[black, dashed] (0,0) -- (3.4,-3.4);
  \node[black, right] at (3.2,3.5) {horizon};
  % hyperbola x = S cosh t, ct = S sinh t, S = 1.4
  \draw[acc, very thick, domain=-1.15:1.15, samples=80, variable=\s]
    plot ({1.4*cosh(\s)},{1.4*sinh(\s)});
  % proper-time ticks
  \foreach \s in {-1,-0.5,0,0.5,1}
    \fill[black!75] ({1.4*cosh(\s)},{1.4*sinh(\s)}) circle (1.3pt);
  % four-velocity tangent at s=0.5
  \draw[->, black!70, very thick] ({1.4*cosh(0.5)},{1.4*sinh(0.5)})
     -- ({1.4*cosh(0.5)+0.55*sinh(0.5)},{1.4*sinh(0.5)+0.55*cosh(0.5)});
  \node[right] at ({1.4*cosh(0.9)},{1.4*sinh(0.9)}) {worldline};
  \node[black!70, below right] at (1.5,-0.05) {vertex};
\end{tikzpicture}
$$

> **Worked example.** A ship holds a proper acceleration of one Earth gravity,
> $\alpha = 9.81\ \text{m/s}^2$, comfortable for the crew. The characteristic
> length and time are $c^2/\alpha = 9.5\times10^{15}\ \text{m} \approx 1.0\
> \text{ly}$ and $c/\alpha = 3.1\times10^7\ \text{s} \approx 0.97\ \text{yr}$.
> After one year of proper time the rapidity is $\phi = \alpha\tau/c \approx
> 1.03$, so $\beta = \tanh(1.03) \approx 0.77$ and $\gamma = \cosh(1.03) \approx
> 1.58$. After two proper years $\phi \approx 2.06$, $\beta \approx 0.968$; after
> five, $\phi \approx 5.15$, $\beta \approx 0.99994$. Proper time runs far slower
> than lab time once $\gamma$ is large, which is what lets a crew cross galactic
> distances within a lifetime while the home clock records millennia.

## The relativistic rocket

A rocket accelerates by expelling exhaust; its own accelerometer reads the
proper acceleration $\alpha$. Because rapidity is additive under collinear boosts
and accumulates linearly in proper time, the velocity, distance, and coordinate
time of a constant-$\alpha$ rocket are the hyperbolic-motion results above:

$$
\beta = \tanh\!\left(\frac{\alpha\tau}{c}\right),
\qquad
t = \frac{c}{\alpha}\,\sinh\!\left(\frac{\alpha\tau}{c}\right),
\qquad
x = \frac{c^2}{\alpha}\!\left[\cosh\!\left(\frac{\alpha\tau}{c}\right) - 1\right],
$$

the last shifted so the rocket starts at the origin. The fuel cost follows from
momentum conservation in the exhaust. If the engine ejects propellant at exhaust
speed $u_e$ (relative to the rocket, with rapidity $\phi_e$ set by $\tanh\phi_e =
u_e/c$), conservation of four-momentum in the instantaneous rest frame gives the
relativistic rocket equation for the rest mass $M$,

$$
\frac{\d M}{M} = -\frac{\d\phi}{\tanh\phi_e},
$$

whose integral relates the mass ratio to the total rapidity gained,

$$
\frac{M_0}{M} = \exp\!\left(\frac{\phi}{\tanh\phi_e}\right)
             = \left(\frac{1+\beta}{1-\beta}\right)^{\!c/(2u_e)},
$$

using $\phi = \tfrac12\ln[(1+\beta)/(1-\beta)]$. For a photon rocket, $u_e = c$
and $\tanh\phi_e = 1$, giving the most efficient possible drive $M_0/M =
e^{\phi} = \gamma(1+\beta)$. The additive rapidity turns the awkward
velocity-addition law into simple exponential bookkeeping: each increment of
proper acceleration adds a fixed slice of rapidity, and the fuel penalty is
exponential in the total.

$$
% caption: Under constant proper acceleration rapidity rises linearly with
% proper time while the lab velocity beta saturates below c; the fuel mass ratio
% grows exponentially in the accumulated rapidity.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (5.6,0) node[right] {proper time $\alpha\tau/c$};
  \draw[->, black] (0,0) -- (0,3.9) node[above] {value};
  % axis ticks
  \foreach \x/\l in {1.3/1, 2.6/2, 3.9/3, 5.2/4} \draw[black] (\x,0.05)--(\x,-0.05) node[below]{\l};
  \draw[black, densely dotted] (0,3.2) -- (5.4,3.2) node[right, black] {$\beta=1$};
  % rapidity phi = x (linear), scaled: phi value maps 1 -> 0.8 height
  \draw[acc, very thick] (0,0) -- (5.2,{5.2/1.3*0.8});
  \node[acc, above, rotate=32] at (3.9,{3.9/1.3*0.8+0.05}) {rapidity};
  % beta = tanh(phi), phi = x/1.3 ; height scale 3.2
  \draw[black, very thick, dashed, domain=0:5.4, samples=80, variable=\x]
    plot ({\x},{3.2*tanh(\x/1.3)});
  \node[black, above] at (4.5,{3.2*tanh(4.5/1.3)+0.06}) {velocity $\beta$};
\end{tikzpicture}
$$

## The Rindler horizon

The hyperbola $x^2 - (ct)^2 = (c^2/\alpha)^2$ never crosses its asymptote $x =
ct$. An observer riding it therefore has a one-way causal boundary. Any light
signal emitted from a point to the past-left of the asymptote — from the region
$x < ct$ — never catches the eternally accelerating observer, because the
observer's worldline stays forever to the right of the line $x = ct$ while
approaching it. The null line $x = ct$ is the **Rindler horizon**: events beyond
it can never send a signal to the accelerated observer, even though nothing about
the local spacetime there is unusual.

The horizon is observer-dependent. It exists because of the observer's eternal
acceleration, not because of any feature of spacetime, and an observer who stops
accelerating destroys it: the moment the worldline goes inertial, past-hidden
signals begin to arrive. This is the flat-spacetime rehearsal for the black-hole
event horizon, where the causal boundary becomes a feature of the geometry
itself rather than of the observer's motion.

$$
% caption: Family of constant-proper-acceleration hyperbolae; the null line x =
% ct is the Rindler horizon that no eternally accelerating observer can receive
% signals from below, and each hyperbola carries a different proper acceleration.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,-2.6) -- (0,3.2) node[above] {time $ct$};
  \draw[->, black] (-0.2,0) -- (5.3,0) node[right] {space $x$};
  % horizon lines
  \draw[acc, very thick] (0,0) -- (3.0,3.0) node[above right] {horizon $x=ct$};
  \draw[black, dashed] (0,0) -- (2.6,-2.6);
  % family of hyperbolae, S = 1.2, 2.3
  \foreach \S in {1.2,2.3}
    \draw[black, domain=-0.95:0.95, samples=64, variable=\s]
      plot ({\S*cosh(\s)},{\S*sinh(\s)});
  \node[black!70, right] at (3.25,1.95) {accelerating};
  \node[black!70, right] at (3.25,1.55) {observers};
  % a blocked signal
  \draw[->, black, densely dotted] (0.5,-1.5) -- (1.75,0.05);
  \node[black, left] at (0.5,-1.5) {signal from below};
  \node[black] at (3.7,-0.7) {never received};
\end{tikzpicture}
$$

The four objects of this lesson — four-momentum, its invariant length, the
four-force orthogonal to the four-velocity, and the hyperbolic worldline of
constant proper acceleration — are the mechanics that the next lessons apply to
decays and collisions. There the conservation law $\sum_i p_i^\mu =
\text{const}$ and the invariant $p\cdot p = -m^2c^2$ do all the work, first for a
single parent breaking into daughters, then for colliding beams.[^schutz][^hartle]

[^schutz]: Schutz, _A First Course in General Relativity_, 2nd ed., §2.4
(four-momentum) and the Ch. 2 problems on uniformly accelerated motion.
[^hartle]: Hartle, _Gravity: An Introduction to Einstein's General Relativity_,
§5.3–§5.6 (four-momentum, uniform acceleration, and the accelerated observer).