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Since $s$ is proper time, the four-acceleration isIts Minkowski norm squared is $g(A,A)=-c^2{\omega }^2$, so the invariant magnitude of proper acceleration isThe vector is spacelike, which accounts for the minus sign under the square root.

Use the metric $\eta =\mathrm{diag}(-1,1,1,1)$ and four-velocity $u^{\mu }=\gamma (c,\mathbf{v})$. Its derivative with respect to proper time is the four-acceleration. Sincea purely transverse laboratory acceleration has $d\gamma /dt=0$. With $d/d\tau =\gamma d/dt$,The invariant proper acceleration is the magnitude measured in the instantaneous rest frame, $a_0^2=A^{\mu }{A}_{\mu }$. Consequently,For longitudinal acceleration the factor is instead ${\gamma }^3$ multiplying the laboratory acceleration. The distinction matters for circular storage-ring motion, where the acceleration is centripetal and transverse even though its magnitude is nonzero.

For a massive particle on a timelike world line, proper time satisfies $d\tau =dt\sqrt{1-v^2/c^2}=dt/\gamma$. Its four-velocity and four-acceleration are $U=dX/d\tau$ and $A=dU/d\tau$. Differentiating $U=\gamma (c,\mathbf{v})$ givesThe invariant Minkowski inner products areFor the explicitly requested contraction,Thus four-acceleration is orthogonal to four-velocity. Equivalently this follows by differentiating the constant Lorentz scalar $U^2=c^2$ with respect to proper time.

Along a massive particle's world line, proper time satisfies $d\tau =dt\sqrt{1-v^2/c^2}$. A pure force preserves its rest mass. Differentiating $P=mU$ and $U^2=c^2$ givesso $\boxed{F\cdot U=0}$ for a pure force. This is a restriction on rest-mass transfer, not a statement that its work in every frame vanishes.

Write $\mathbf{a}=d\mathbf{v}/dt$. Since $\dot{\gamma }={\gamma }^3\mathbf{v}\cdot \mathbf{a}/c^2$, the four-acceleration isIf $\mathbf{v}\cdot \mathbf{a}=0$, this reduces to $A=(0,{\gamma }^2\mathbf{a})$, whenceHere $\alpha$ is the proper acceleration. The positive expression printed in the source is $A^2$ in the opposite metric signature, or the squared acceleration magnitude in ours. For example, at rest with nonzero acceleration, $A=(0,\mathbf{a})$ has $A^2=-a^2$ in our convention. Thus one must not combine the printed positive sign with a mostly-minus norm.

Use coordinates $X^{\mu }=(ct,\mathbf{x})$ and the mostly-plus Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(-1,1,1,1)$. Along a future-directed timelike world line, proper time satisfiesThe four-velocity and four-acceleration areThe Minkowski norm is fixed: $U^{\mu }{U}_{\mu }=-c^2$. Differentiating with respect to proper time, using the constant Minkowski metric, givesIn the instantaneous rest reference frame, $U^{\mu }=(c,0,0,0)$, so the orthogonality says $A^0=0$ there. This construction uses a timelike world line; a massless particle has no rest-frame proper time parameter.

Let $z^{\mu }(\tau )$ be the point electric charge's world line, with proper time $\tau$ and four-velocity $U^{\mu }=d{z}^{\mu }/d\tau =\gamma (c,\mathbf{u})$. For the field event $x^{\mu }$, choose the retarded source event: $R^{\mu }=x^{\mu }-z^{\mu }({\tau }_{\mathrm{ret}})$ is future-directed and null. Write $R^{\mu }=(R,R\mathbf{n})$, where $R=|\mathbf{x}-\mathbf{z}(t_{\mathrm{ret}})|=c(t-t_{\mathrm{ret}})$. In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian $1/(1-\mathbf{n}\cdot \mathbf{\beta })$ from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.^[ref]

In any inertial frame, put $\mathbf{\beta }=\mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot \mathbf{\beta }$. Then $-R\cdot U=\gamma cR\kappa$, and the Liénard–Wiechert potential isDifferentiate using $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, including the position and time dependence of the retarded time. For example, $\partial t_{\mathrm{ret}}/\partial t=1/\kappa$ and $\nabla t_{\mathrm{ret}}=-\mathbf{n}/(c\kappa )$. The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as $R^{-2}$ and exists even without acceleration. The second is the acceleration or radiation field, falling as $R^{-1}$ and transverse to $\mathbf{n}$. Its Poynting flux falls as $R^{-2}$, so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.

For uniform circular motion, the transverse Lorentz force supplies the change in direction of three-momentum:The magnetic rigidity relation gives $pc=0.299792458B[\mathrm{T}]r[\mathrm{m}]\mathrm{GeV}=2.99792458\mathrm{GeV}$. Using $m_e{c}^2=0.511\mathrm{MeV}$,The coordinate centripetal acceleration is $a=u^2/r$. For transverse acceleration, the magnitude of the four-acceleration, or proper acceleration, is $a_0={\gamma }^{2}a$. The supplied Larmor formula, in its proper acceleration form, therefore givesThe emitted power here is also the laboratory total emitted power: in the instantaneous rest inertial frame the dipole radiation pattern has zero net radiated three-momentum, so both emitted energy and elapsed coordinate time gain the same Lorentz factor on transformation. This does not assert that a directional detector flux is invariant.

Numerically $\mathcal{T}\simeq 1.048\times 10^{-5}\mathrm{s}$, $P\simeq 2.18\times 10^{-10}\mathrm{W}$ per electron, andThe synchrotron's radiofrequency accelerating system must replace this lost energy to keep $r$ and magnetic rigidity fixed over many revolutions. The fraction lost per turn is about $4.77\times 10^{-6}$, so a fixed-orbit calculation over one turn is self-consistent.^[ref]

For motion on a fixed axis, the signed spatial four-acceleration in the instantaneous rest inertial frame is $a={\gamma }^3\dot{u}$. Its magnitude is the scalar proper acceleration. Constant signed $a$ from rest gives $\gamma u=at$ and the hyperbolic motion worldline $x=(c^2/a)(\sqrt{1+a^2{t}^2/c^2}-1)$.

For $d{s}^2=Ad{t}^2-d{r}^2/A-r^{2}d{\Omega }^2$, $A=1-2GM/r-\Lambda r^2/3$, and $c=1$, a radial timelike geodesic has ${\dot{r}}^2=k^2-A$. The radial geodesic equation gives $\ddot{r}=-A^{\prime }/2=-GM/r^2+\Lambda r/3$, including at a turning point. This is a second derivative of the areal radius with respect to proper time; the geodesic's four-acceleration still vanishes. In the slow-motion weak-field limit, restoring units gives $d^{2}r/d{t}^2=-GM/r^2+\Lambda c^{2}r/3$.

For relativistic dust with nowhere-zero density, stress-energy conservation implies zero four-acceleration. If its normalized four-velocity $u$ has hypersurface orthogonality, then $u^{♭}\wedge d{u}^{♭}=0$ and ${\iota }_{u}d{u}^{♭}=0$. The interior product of the first identity gives $d{u}^{♭}=0$. The Poincaré lemma supplies $u^{♭}=-dt$ locally; spatial coordinate functions constant along the smooth flow of $u$ give $u={\partial }_t$, $g_{tt}=-1$ and $g_{ti}=0$. The resulting synchronous coordinates describe the dust until the local coordinate construction fails.