Four-Velocity, Four-Acceleration, and Relativistic Force

A massive particle’s four-velocity is the unit tangent to its worldline. Its four-acceleration is automatically spatial in the particle’s instantaneous rest frame. This orthogonality is the key consistency condition for relativistic force at fixed rest mass.

Required background. Events, Intervals, and Minkowski Geometry begins the Special Relativity sequence and supplies causal geometry and four-vectors; Relativistic Energy, Momentum, and Mass Shells supplies $p^2=-m^2$. Helpful background. Calculus and Linear Algebra supplies differentiation and contractions; Classical and Variational Mechanics supplies velocity, acceleration, momentum, force, work, and power.

Normalization and orthogonality

Let $x^\mu(\tau)$ be a future-directed timelike worldline parametrized by proper time. Define

$$U^\mu=\frac{dx^\mu}{d\tau}, \qquad A^\mu=\frac{dU^\mu}{d\tau}.$$

In inertial Cartesian coordinates no connection terms are needed; the curved-spacetime generalization replaces the ordinary derivative by $U^\nu\nabla_\nu$.

Proper time gives

$$U^\mu U_\mu=-1.$$

Differentiating along the worldline yields

$$\frac{d}{d\tau}(U^\mu U_\mu)=2U_\mu A^\mu=0,$$

so

$$U_\mu A^\mu=0.$$

Therefore $A^\mu$ lies in the particle’s rest space. Its invariant magnitude

$$a=\sqrt{A^\mu A_\mu}$$

is the proper acceleration measured by an ideal accelerometer. A nonzero coordinate acceleration is neither necessary nor sufficient until the coordinates and observer are specified.

For coordinate three-velocity $\mathbf v=d\mathbf x/dt$,

$$U^\mu=\gamma(1,\mathbf v), \qquad \gamma=\frac{1}{\sqrt{1-v^2}}, \qquad v^2=\mathbf v\mathbin{\cdot}\mathbf v\lt1.$$

Differentiation gives

$$A^0=\gamma^4\,\mathbf v\cdot\mathbf a_{\rm lab}, \qquad \mathbf A =\gamma^2\mathbf a_{\rm lab} +\gamma^4(\mathbf v\cdot\mathbf a_{\rm lab})\mathbf v,$$

where $\mathbf a_{\rm lab}=d\mathbf v/dt$. Longitudinal and transverse laboratory accelerations therefore contribute differently to $a$.

For $v\ne0$, write $\mathbf a_{\rm lab}=\mathbf a_\parallel+\mathbf a_\perp$ relative to the direction of $\mathbf v$. This gives the invariant relation

$$a^2=\gamma^6 a_\parallel^2+\gamma^4 a_\perp^2.$$

This follows either by squaring the components above or by transforming to the instantaneous rest frame. It supplies two useful checks: one-dimensional motion has $a=\gamma^3|a_\parallel|$, whereas uniform-speed bending has $a=\gamma^2|a_\perp|$. The distinction concerns the acceleration measured in a specified inertial laboratory; the invariant $a$ itself is independent of that choice.

At $v=0$, the direction used for the split is undefined, but the unsplit formula has the regular limit $a^2=|\mathbf a_{\rm lab}|^2$. Nothing physical is singular there.

Four-force at fixed rest mass

For constant rest mass $m$, set

$$p^\mu=mU^\mu, \qquad f^\mu=\frac{dp^\mu}{d\tau}=mA^\mu.$$

The mass shell implies

$$f^\mu U_\mu=0.$$

Define laboratory three-force and power by

$$\mathbf F=\frac{d\mathbf p}{dt}, \qquad \mathcal P=\frac{dE}{dt}.$$

Because $d/d\tau=\gamma\,d/dt$,

$$f^\mu=\gamma(\mathcal P,\mathbf F).$$

The orthogonality check becomes

$$0=f\cdot U =\gamma^2(-\mathcal P+\mathbf F\cdot\mathbf v),$$

and hence

$$\mathcal P=\mathbf F\cdot\mathbf v.$$

This familiar work relation is not an additional postulate; it follows from the four-force being tangent to the fixed-mass shell.

The normalization, proper-acceleration interpretation, and laboratory force conversion can be checked in Hughes 2024, Lecture 13, §§13.2 and 13.4, printed pp. 104–109. Hughes retains explicit factors of $c$; they have been set to one here.

Example: electromagnetic force

For charge $q$ in a prescribed electromagnetic field,

$$f^\mu=qF^\mu{}_\nu U^\nu.$$

Antisymmetry $F_{\mu\nu}=-F_{\nu\mu}$ guarantees

$$U_\mu f^\mu=qF_{\mu\nu}U^\mu U^\nu=0,$$

so the Lorentz force preserves $p^2=-m^2$. In a chosen inertial frame, with $F^{0i}=E^i$, $F^{ij}=\epsilon^{ijk}B_k$, and $\epsilon^{123}=+1$, its three-vector form is

$$\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B), \qquad \mathcal P=q\,\mathbf E\cdot\mathbf v.$$

The magnetic term changes direction but does no laboratory work.

When rest mass changes

If $p^\mu=m(\tau)U^\mu$, then

$$f^\mu=\dot m\,U^\mu+mA^\mu, \qquad U_\mu f^\mu=-\dot m.$$

The component of $f^\mu$ parallel to $U^\mu$ changes rest energy; the orthogonal component changes four-velocity. Variable-mass systems, rockets, absorption, emission, and unstable particles therefore require a momentum-balance model for what enters or leaves. Substituting a time-dependent $m$ into $f=mA$ silently discards that flux.

The laboratory power law changes as well. Since $f^\mu=\gamma(\mathcal P,\mathbf F)$,

$$\gamma^2(-\mathcal P+\mathbf F\mathbin{\cdot}\mathbf v)=-\dot m,$$

and therefore

$$\mathcal P =\mathbf F\mathbin{\cdot}\mathbf v +\frac{\dot m}{\gamma^2}, \qquad \dot m=\frac{dm}{d\tau}.$$

The fixed-mass work relation is recovered only when $\dot m=0$.

Worked check: uniform circular motion

For constant laboratory speed $v$ on a circle of radius $R$, $\mathbf v\cdot\mathbf a_{\rm lab}=0$ and $|\mathbf a_{\rm lab}|=v^2/R$. Thus

$$A^0=0, \qquad a=\gamma^2\frac{v^2}{R}.$$

The four-acceleration is spatial in both the momentarily comoving frame and, in this special transverse case, the laboratory coordinates at that instant. The factor $\gamma^2$ shows why the proper acceleration is not simply the coordinate centripetal acceleration.

Limits and diagnostics

• As $v\to0$, $\gamma\to1$, $d\tau\to dt$, and the spatial equation reduces to Newton’s $\mathbf F=m\mathbf a$.
• A freely moving particle has $A^\mu=0$ in every inertial coordinate system, even though its coordinate velocity need not vanish.
• Four-acceleration is defined only for timelike worldlines. A photon has no proper time and no unit four-velocity.
• A coordinate-dependent “relativistic mass” obscures the invariant $m$ and is unnecessary. Energy is the observer-dependent quantity.
• In curved spacetime, $dU^\mu/d\tau$ alone is not a vector; use the covariant derivative.

Exercises

Longitudinal and transverse three-force

For fixed mass, differentiate $\mathbf p=m\gamma\mathbf v$ and show that

$$\mathbf F_\parallel=m\gamma^3\mathbf a_\parallel, \qquad \mathbf F_\perp=m\gamma\mathbf a_\perp.$$

Solution

Using $d\gamma/dt=\gamma^3\mathbf v\mathbin{\cdot}\mathbf a_{\rm lab}$ gives

$$\mathbf F =m\gamma\mathbf a_{\rm lab} +m\gamma^3(\mathbf v\mathbin{\cdot}\mathbf a_{\rm lab})\mathbf v.$$

The second term is parallel to $\mathbf v$. Because $\mathbf v\mathbin{\cdot}\mathbf a_{\rm lab}=v a_\parallel$ and $1+\gamma^2v^2=\gamma^2$, the parallel coefficient is $m\gamma^3$; the perpendicular part receives only the first term and has coefficient $m\gamma$.

Why a photon has no four-velocity

Explain why the definition $U^\mu=dx^\mu/d\tau$ cannot be applied to a photon, even though a photon has a well-defined four-momentum $k^\mu$.

Solution

Along a null worldline, $ds^2=0$, so every segment has $d\tau=0$. Dividing the tangent by proper time is therefore undefined, and no null vector can be normalized to $U^2=-1$. A photon instead carries a future-directed null four-momentum $k^2=0$; rescaling its affine parameter rescales the tangent, while a physical energy normalization fixes $k^\mu$.

For a compact alternative derivation, Siklos 2011, §§6.8.2–6.8.4, printed pp. 20–23, and §6.9.1, printed p. 25, Eq. (6.41) develops four-velocity, four-momentum, the fixed-mass work relation, and four-force using the opposite $(+,-,-,-)$ signature.

Continue to Relative Velocity and Lorentz Factors to express the motion of one observer as measured by another without choosing a special coordinate frame.

References

• Hughes, Scott A. Introduction to Relativity and Spacetime Physics. MIT 8.033, Fall 2024. Official MIT OpenCourseWare PDF.
• Siklos, Stephen. Dynamics and Relativity. University of Cambridge, Lent Term 2011. Official PDF.

Further reading

• David Tong, Lectures on Dynamics and Relativity, Chapter 7, “Kinematics” and relativistic particle dynamics, University of Cambridge, 2013.
• David Tong, Lectures on General Relativity, §1.2 “Relativistic Particles,” University of Cambridge, 2019.