Four-Momentum, Four-Force, and Accelerated Motion

The kinematics of 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 $A^{μ} = (A^{0}, A)$ is $A \cdot B = η_{μν} A^{μ} B^{ν} = - A^{0} B^{0} + A \cdot B$ .

The four-momentum

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

U^{μ} = \frac{d x ^{μ}}{d τ} = γ_{u} (c, u), γ_{u} = \frac{1}{1 - u ^{2} / c ^{2}},

where the factor $γ_{u} = d t / d τ$ converts coordinate time to proper time along the worldline. Its Minkowski square is fixed,

U \cdot U = γ_{u}^{2} (- c^{2} + u^{2}) = - c^{2},

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

p^{μ} = m U^{μ} = (γ_{u} m c, γ_{u} m u) = (\frac{E}{c}, p) .

The time component reproduces the relativistic energy $E = γ_{u} m c^{2}$ and the space components reproduce the relativistic momentum $p = γ_{u} m u$ , so the split of energy and momentum that looked frame-dependent in the foundations is the split of one four-vector into time and space parts.

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

p \cdot p = m^{2} U \cdot U = - m^{2} c^{2} .

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

E^{2} = (∣ p ∣ c)^{2} + (m c^{2})^{2} .

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

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.

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^{μ} = \frac{d p ^{μ}}{d τ} .

Its space part relates to the ordinary three-force $f = d p / d t$ by the chain rule $d / d τ = γ_{u} d / d t$ , so $F = γ_{u} f$ , and its time part is the rate at which the force does work: with $E = γ_{u} m c^{2}$ ,

F^{0} = \frac{1}{c} \frac{d E}{d τ} = \frac{γ _{u}}{c} \frac{d E}{d t} = \frac{γ _{u}}{c} f \cdot u .

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

\frac{d}{d τ} (p \cdot p) = 2 p \cdot \frac{d p}{d τ} = 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 $A^{μ} = d U^{μ} / d τ = F^{μ} / m$ satisfies $A \cdot U = 0$ ; differentiating $U \cdot U = - c^{2}$ gives the same relation directly.

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

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.

The three-force and longitudinal versus transverse response

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

f = m \frac{d ( γ _{u} u )}{d t} = m (\overset{γ}{˙}_{u} u + γ_{u} a), \overset{γ}{˙}_{u} = γ_{u}^{3} \frac{u \cdot a}{c ^{2}},

f = γ_{u} m a + γ_{u}^{3} m \frac{( u \cdot a )}{c ^{2}} u .

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

f_{∥} = γ_{u}^{3} m a_{∥}, f_{⊥} = γ_{u} m a_{⊥} .

A given force produces $γ_{u}^{2}$ times less acceleration along the motion than across it. The coefficients $γ_{u}^{3} m$ and $γ_{u} m$ 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 $γ_{u}^{3} m$ 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 $γ_{u}^{- 3}$ , 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 $α$ , its accelerometer reading a fixed value forever. In the IRF the four-acceleration is $(0, α, 0, 0)$ . Boosting to the lab frame in which the particle has instantaneous rapidity $ϕ$ (with $β = tanh ϕ$ , $γ = cosh ϕ$ ), the four-velocity and four-acceleration are

U^{μ} = c (cosh ϕ, sinh ϕ, 0, 0), A^{μ} = \frac{d U ^{μ}}{d τ} = c \dot{ϕ} (sinh ϕ, cosh ϕ, 0, 0),

where the dot is $d / d τ$ . Then $A \cdot A = c^{2} \dot{ϕ}^{2} = α^{2}$ , so the rapidity accumulates linearly with proper time:

\frac{d ϕ}{d τ} = \frac{α}{c} ⟹ ϕ (τ) = \frac{α τ}{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

β (τ) = tanh (\frac{α τ}{c}), γ (τ) = cosh (\frac{α τ}{c}),

which approaches but never reaches $c$ : the hyperbolic tangent saturates at unity. Integrating $U^{μ} = d x^{μ} / d τ$ with $ϕ = α τ / c$ and the particle starting from rest at $x = c^{2} / α$ at $τ = 0$ gives the worldline

c t (τ) = \frac{c ^{2}}{α} sinh (\frac{α τ}{c}), x (τ) = \frac{c ^{2}}{α} cosh (\frac{α τ}{c}) .

Eliminating $τ$ with $cosh^{2} - sinh^{2} = 1$ yields the trajectory in spacetime,

x^{2} - (c t)^{2} = (\frac{c ^{2}}{α})^{2},

a hyperbola with asymptotes $x = \pm c t$ . 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 $x = x_{0} + \frac{1}{2} a t^{2}$ that the same equations reduce to for $α τ ≪ c$ .

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.

The relativistic rocket

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

β = tanh (\frac{α τ}{c}), t = \frac{c}{α} sinh (\frac{α τ}{c}), x = \frac{c ^{2}}{α} [cosh (\frac{α τ}{c}) - 1],

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 $ϕ_{e}$ set by $tanh ϕ_{e} = u_{e} / c$ ), 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 ϕ}{tanh ϕ _{e}},

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

\frac{M _{0}}{M} = exp (\frac{ϕ}{tanh ϕ _{e}}) = (\frac{1 + β}{1 - β})^{c / (2 u_{e})},

using $ϕ = \frac{1}{2} ln [(1 + β) / (1 - β)]$ . For a photon rocket, $u_{e} = c$ and $tanh ϕ_{e} = 1$ , giving the most efficient possible drive $M_{0} / M = e^{ϕ} = γ (1 + β)$ . 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.

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.

The Rindler horizon

The hyperbola $x^{2} - (c t)^{2} = (c^{2} / α)^{2}$ never crosses its asymptote $x = c t$ . 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 $x < c t$ — never catches the eternally accelerating observer, because the observer's worldline stays forever to the right of the line $x = c t$ while approaching it. The null line $x = c t$ 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.

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.

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 $\sum_{i} p_{i}^{μ} = const$ and the invariant $p \cdot p = - m^{2} c^{2}$ do all the work, first for a single parent breaking into daughters, then for colliding beams.¹²

Schutz, A First Course in General Relativity, 2nd ed., §2.4 (four-momentum) and the Ch. 2 problems on uniformly accelerated motion. ↩
Hartle, Gravity: An Introduction to Einstein's General Relativity, §5.3–§5.6 (four-momentum, uniform acceleration, and the accelerated observer). ↩