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Transport defining a nonrotating frame along an accelerated timelike curve. With four-velocity normalization $u^{\mu }{u}_{\mu }=-c^2$ and four-acceleration $a^{\mu }=D{u}^{\mu }/d\tau$, the displayed equation preserves inner products and keeps a spatial spin orthogonal to $u$. Along a geodesic, $a^{\mu }=0$ and it reduces to parallel transport.

The four-acceleration is the covariant derivative of four-velocity along a massive observer's worldline: $A^{\mu }=d{U}^{\mu }/d\tau +{\Gamma }_{\nu \rho }^{\mu }{U}^{\nu }{U}^{\rho }$. In Cartesian Minkowski spacetime coordinates the Christoffel symbols vanish, leaving the ordinary proper time derivative. Differentiating the constant velocity norm gives $U\cdot A=0$; in the instantaneous rest frame, $A=(0,\mathbf{a})$, so $A\cdot A=-a^2$ with $+---$ metric signature. The magnitude $a$ is the proper acceleration.

For uniform circular motion in an inertial laboratory, the Lorentz factor is constant. With metric signature $+---$, four-acceleration is $A^{\mu }=(0,{\gamma }^2\mathbf{a})$ and proper acceleration has magnitude ${\gamma }^2{v}^2/r$. The instantaneous spatial acceleration is perpendicular to velocity, even though the particle has nonzero proper acceleration.

With metric signature $(+---)$, normalize a timelike Killing vector field by $U^a=K^a/N$, $N^2=K^a{K}_a$. The Killing equation gives $K^b{\nabla }_b{K}_a=-\frac{1}{2}{\nabla }_a{N}^2$, while $K^b{\nabla }_{b}N=0$. Thus the four-acceleration of its normalized flow is the displayed gradient. With metric signature $(-+++)$ the corresponding lower-index formula has the opposite sign.

In general coordinates the basis changes along the world line. Differentiating only the components of a vector omits that basis change, and does not transform as a vector under a position-dependent coordinate transformation. The appropriate rate is the covariant derivative along a curve:For the metric-compatible torsion-free connection, the Christoffel symbols are determined by the metric tensor and its first derivatives:Thus the ship's four-acceleration isIn Cartesian coordinates of an inertial frame in flat spacetime, the connection coefficients vanish and the ordinary component derivative happens to suffice. In curved spacetime it suffices at a chosen event in a local inertial frame, but not generally along an extended trajectory.

Write $D=u^a{\nabla }_a$, $\theta ={\nabla }_a{u}^a$ for the expansion scalar, $a_b=u^a{\nabla }_a{u}_b$ for the four-acceleration, and $P_b{}^c={\delta }_b{}^c+u_b{u}^c$ for the spatial projector of a timelike congruence. The normalization of the four-velocity gives $u^b{a}_b=0$ and $u^b{\nabla }_a{u}_b=0$. The product rule and metric compatibility give the covariant derivative of the perfect fluid in general relativity stress-energy tensor:Its tensor contraction with the four-velocity isand its projection with the spatial projector of a timelike congruence isThus stress-energy conservation implies the relativistic perfect-fluid energy equation and the relativistic Euler equation:Conversely, the two displayed equations set both projections of $F_b$ to zero. Since $F_b=P_b{}^c{F}_c-u_b{u}^c{F}_c$, they imply $F_b=0$. This establishes equivalence, including points where $\rho +p=0$, without dividing by that quantity.

In the prescribed synchronous coordinates, the Christoffel symbols satisfyConsequently the four-acceleration of $u={\partial }_t$ vanishes. For a spatial index $i$, $u_i=0$, so the spatial component of the relativistic Euler equation becomesOn a connected coordinate neighbourhood,A nonzero pressure is allowed, but its spatial gradient must vanish if the fluid's four-velocity follows these freely falling coordinate lines. No division by $p$ or $\rho +p$ is needed.

For relativistic dust, $p=0$, the relativistic Euler equation gives $\rho a_b=0$. The nowhere-vanishing density therefore makes the four-acceleration zero: $u$ is a timelike geodesic vector field. Let $\alpha =u^{♭}$ be the differential one-form obtained by the musical isomorphism. Its exterior derivative satisfiesHere ${\iota }_u$ is the interior product. By hypersurface orthogonality and the Frobenius theorem, $\alpha \wedge d\alpha =0$. Taking the interior product and using $\alpha (u)=-1$ givesThus $\boxed{d{u}^{♭}=0}$. By the Poincaré lemma, locally $\alpha =-dt$, and then $u(t)=-\alpha (u)=1$. Choose spatial coordinate functions $x^i$ on one level set of $t$ and extend them constantly along the integral curves of a vector field $u$, using its smooth flow. This constructs $u={\partial }_t$. Since $dt({\partial }_i)=0$,The metric on the orthogonal complement of a timelike vector is positive-definite, so the synchronous coordinates for irrotational dust haveThis construction is local; crossing geodesics can prevent these synchronous coordinates from covering a larger region.

Use metric signature $(+---)$ throughout this solution. Stress-energy conservation for the perfect fluid is ${\nabla }_b{T}^{ab}=0$. Expanding it and using metric compatibility givesThe normalization of the four-velocity implies $U_a{U}^b{\nabla }_b{U}^a=0$. Contracting with $U_a$ therefore gives the relativistic perfect-fluid energy equation,while subtracting the component parallel to $U$ gives the relativistic Euler equation,Equivalently, apply the orthogonal projection ${\delta }_a{}^c-U_a{U}^c$ to the conservation equation. This form makes clear that a pressure gradient orthogonal to the four-velocity supplies the fluid's four-acceleration.

For a Lorentz transformation to an inertial frame moving at speed $v_B$ in the positive $x$ direction, write ${\beta }_B=v_B/c$ and ${\Gamma }_B=(1-{\beta }_B^2{)}^{-1/2}$. Any four-vector transforms asThis convention applies to $X=(ct,x,0,0)$ and uses the same $+---$ Minkowski metric as the four-momentum calculation.

The definition of proper time gives $d\tau =dt\sqrt{1-u^2/c^2}=dt/\gamma$ for $|u|<c$. Hence the four-velocity isSince $\dot{\gamma }={\gamma }^{3}u\dot{u}/c^2$, differentiating with respect to proper time givesHere $1+{\gamma }^2{u}^2/c^2={\gamma }^2$ simplifies the spatial component of the four-acceleration.

At each event, boost with $v_B=u$ to the instantaneous rest inertial frame. Then $A^{\prime 0}=0$ and $A^{\prime 1}=\gamma {\gamma }^4\dot{u}(1-u^2/c^2)={\gamma }^3\dot{u}$. Thus the signed proper acceleration for one-dimensional motion isThe last equality follows by differentiating $\gamma u$; its magnitude is the usual scalar proper acceleration.

If this signed proper acceleration is constant and the particle starts from rest, integration gives $\gamma u=at$, soIntegrating once more and imposing $x(0)=0$ yields the hyperbolic motion worldlineFor $a=0$ the continuous limiting solution is $x=0$. Finally, the Taylor series of the square root givesThis recovers constant-acceleration Newtonian mechanics at low speed while the exact relativistic speed remains less than $c$.

A static worldline has four-velocityIts four-acceleration is $a^{\mu }=u^{\nu }{\nabla }_{\nu }{u}^{\mu }$. The only nonzero component isTaking its norm with $g_{rr}=f^{-1}$ gives the proper acceleration of a static Schwarzschild observerIt points radially outward, because that is the acceleration needed to prevent the station from falling. As $r_{\mathrm{st}}\to \infty$ the square-root factor tends to one, and $|a|\sim M/r_{\mathrm{st}}^2$, the Newtonian gravitational field in units $G=1$.

Write the four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$ and four-acceleration $A^{\mu }=d{U}^{\mu }/d\tau$. Since $U_{\mu }{A}^{\mu }=0$, spatial orthogonality $\mathbf{U}\cdot \mathbf{A}=0$ implies $U^0{A}^0=0$. Thus $A^0=cd\gamma /d\tau =0$ and the speed is constant. If “acceleration” is read as the laboratory three-acceleration instead, $\gamma \mathbf{v}\cdot d\mathbf{v}/dt=0$ immediately gives the same conclusion.

For four-acceleration in uniform circular motion, the speed is $r|\omega |<c$, so $\gamma =(1-r^2{\omega }^2/c^2{)}^{-1/2}$ is constant. Differentiating the four-velocity using $d/d\tau =\gamma d/dt$ givesThe proper acceleration is $\sqrt{-A_{\mu }{A}^{\mu }}={\gamma }^{2}r{\omega }^2$. The vanishing time component reflects constant energy; the spatial components describe the changing momentum direction.

tidal forces measure differences in gravitational acceleration across an extended region. Neighboring freely falling particles each have zero four-acceleration, but their separation need not accelerate uniformly. Let $u^{\mu }$ be the tangent to a reference timelike geodesic and ${\xi }^{\mu }$ its infinitesimal separation from a neighboring one, chosen from an affinely parametrized geodesic congruence. Using the Riemann curvature tensor convention specified in Solution 3,Changing the overall curvature convention changes this displayed sign. The covariant derivatives compare vectors by parallel transport, avoiding a coordinate-dependent subtraction of accelerations.

In the Newtonian limit, expand the force field at two nearby points:For $\Phi =-GM/r$, a local orthonormal frame aligned with the radius givesThere is radial stretching and transverse compression; the trace vanishes outside the mass. Two freely falling test masses and a measurement of their changing separation implement gravity gradiometry. Their common acceleration drops out, while a relative acceleration proportional to separation remains. A local inertial frame removes Christoffel symbols at a point, but cannot remove nonzero curvature over a finite region.

The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.

Throughout this solution $c=1$ and $a>0$.

Null metric tensor and boost. Inverting the coordinate definitions gives $t=(p+q)/2$, $x=(q-p)/2$, and henceFor a frame moving at speed $v$ along positive $x$, use $t^{\prime }=\gamma (t-vx)$, $x^{\prime }=\gamma (x-vt)$. Their difference and sum giveThe reciprocal scale factors preserve $dpdq$. They are $e^{\pm {\eta }_v}$ in rapidity notation.

acceleration invariants. Proper-time normalization gives ${\dot{t}}^2-{\dot{x}}^2=1$, so $\dot{p}\dot{q}=1$. The four-acceleration has squared norm ${\ddot{t}}^2-{\ddot{x}}^2=-a^2$; thereforeFor future motion write $\dot{p}=e^{-\eta }$ and $\dot{q}=e^{\eta }$. Then the second equation is $-(\dot{\eta }{)}^2=-a^2$. acceleration toward positive $x$ selects $\eta ={\eta }_0+a\tau$.

hyperbolic motion from an initial event and rapidity. The stated initial event fixes $p(0)=-1/a$, $q(0)=1/a$, but not the initial rapidity ${\eta }_0$. Integrating with positive signed acceleration gives the general familyThe initial event alone does not fix the initial velocity or signed acceleration. The trajectory explicitly requested in the next paragraph selects the additional conditions ${\eta }_0=0$ and positive-$x$ acceleration. With that convention,and consequentlyWith arbitrary ${\eta }_0$ the equivalent expressions are $t=[\sinh ({\eta }_0+a\tau )-\sinh {\eta }_0]/a$ and $x=1/a+[\cosh ({\eta }_0+a\tau )-\cosh {\eta }_0]/a$. Thus position data alone do not select the displayed special hyperbola. The following calculations use that intended hyperbola, which is also explicitly given by the paper.

Diagram and horizon. The selected world line is the right-hand branch $x^2-t^2=a^{-2}$. Its future asymptote $x=t$ is a reception horizon: rightmoving signals have constant $p=t-x$, whereas the observer has $p<0$ and approaches zero only as $\tau \to \infty$. Events with $p\ge 0$ cannot send a future signal to this observer if the acceleration continues indefinitely. This is an observer-dependent causal restriction in flat spacetime, not a curvature singularity.

photon reception by a uniformly accelerated observer. A rightmoving photon from the stationary source has four-momentum $P^{\mu }=h{\nu }_0(1,1)$. Contracting with the observer's four-velocity givesThe same result follows by differentiating its emission-time relation. A photon emitted at $x=L$ and time $t_e$ has $p=t_e-L$, so intersection requiresThus $t_e=L$ is the supremum of receivable emission times, reached only in the limit of infinite receiver proper time. A photon emitted exactly then lies on the horizon and is never received at finite time; later photons cannot catch the ship. Restoring dimensions gives the limiting emission time $L/c$ for this chosen origin and trajectory.

Rindler coordinates. Substitution of the selected trajectory into the coordinate transformation givesDifferentiating the transformation also gives $d{s}^2=e^{2a\zeta }(d{\xi }^2-d{\zeta }^2)$. Curves of fixed $\zeta$ are hyperbolae, while curves of fixed $\xi$ are straight rays $t/x=\tanh (a\xi )$ within the right wedge. The coordinate $\xi$ is the selected ship's proper clock; at another fixed $\zeta$, proper time obeys $d{\tau }_{\zeta }=e^{a\zeta }d\xi$. The spatial coordinate is logarithmic in $\rho =\sqrt{x^2-t^2}=e^{a\zeta }/a$, rather than being proper distance itself. Fixed-$\zeta$ observers have proper acceleration $a{e}^{-a\zeta }$.

Let $\mathbf{a}=d\mathbf{u}/dt$ denote coordinate acceleration and use $A^{\mu }$ for four-acceleration, to distinguish these two objects. Applying the chain rule to the Lorentz factor givesSince $dt/d\tau ={\gamma }_u$, differentiation of the four-velocity $U^{\mu }={\gamma }_u(c,\mathbf{u})$ with respect to proper time yieldsThis is also ${\gamma }_u^2[{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})/c,\mathbf{a}+{\gamma }_u^2(\mathbf{u}\cdot \mathbf{a})\mathbf{u}/c^2]$, as required.

The invariant Minkowski norm of the four-velocity is $U^{\mu }{U}_{\mu }=c^2$. Differentiating this identity gives $U^{\mu }{A}_{\mu }=0$. In the instantaneous rest frame, $U^{\mu }=(c,0)$, so $A^0=0$ and the spatial components equal the physically experienced proper acceleration $\mathbf{\alpha }$. Therefore ${\alpha }^2=-A^{\mu }{A}_{\mu }$ in the $+---$ metric signature.

For a direct calculation, put $q=\mathbf{u}\cdot \mathbf{a}$ and expand the temporal and spatial squares:The last step uses $1-u^2/c^2={\gamma }_u^{-2}$. This proper acceleration from coordinate velocity and acceleration also reads ${\alpha }^2={\gamma }_u^6{a}_{\parallel }^2+{\gamma }_u^4{a}_{\perp }^2$, providing the familiar one-dimensional limit $\alpha ={\gamma }_u^3|a_{\parallel }|$. Each expression has units ${\mathrm{m}}^2{\mathrm{s}}^{-4}$.

For motion along one spatial axis, introduce the rapidity $\eta$ so that the four-velocity is $u^{\mu }=c(\cosh \eta ,\sinh \eta ,0,0)$. Its four-acceleration isA constant signed proper acceleration $g$ therefore gives $d\eta /d\tau =g/c$, hence $\eta ={\eta }_0+g\tau /c$. Differentiating the four-velocity twice yieldsThe zero transverse components also satisfy this equation. The constant acceleration measured on board is the proper acceleration, rather than a constant coordinate acceleration in the laboratory.

At any event on a massive observer's worldline, choose an orthonormal tetrad with timelike basis vector $e_{(0)}=U/c$. Its components obey $g(e_{(a)},e_{(b)})={\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, so locally $d{s}^2={\eta }_{ab}d{X}^{(a)}d{X}^{(b)}$: the Minkowski metric. A local inertial frame can also remove the Christoffel symbols at that event. Curvature cannot be removed on a finite region, and a frame following an accelerated observer need not remain inertial in its neighbourhood.

In general coordinates, $X^{\mu }(\tau )$ labels the trajectory, while the four-velocity and four-acceleration areCoordinate values $X^{\mu }$ are not themselves a vector under arbitrary nonlinear coordinate changes; their tangent $U$ is. For fixed rest mass $m$, the four-momentum is $P^{\mu }=m{U}^{\mu }$ and the four-force is $f^{\mu }=D{P}^{\mu }/d\tau =m{A}^{\mu }$. With the $+---$ metric signature,where $a$ is the proper acceleration. The middle identity follows by differentiating the constant velocity norm; in the instantaneous rest frame, $U=(c,0)$ and $A=(0,\mathbf{a})$, proving the last identity.

The proper time normalization of the astronaut's four-velocity gives ${\dot{t}}^2-{\dot{x}}^2=1$. In the light-cone coordinates this isThe four-acceleration has squared norm ${\ddot{t}}^2-{\ddot{x}}^2=-a^2$, because $a$ is the constant proper acceleration. The same coordinate change therefore givesEquivalently write $\dot{q}=e^{\eta }$, $\dot{p}=e^{-\eta }$. Their second derivatives give $\ddot{p}\ddot{q}=-{\dot{\eta }}^2$, so the rapidity form of one-dimensional proper acceleration makes $\dot{\eta }=\pm a$, with sign fixing the acceleration direction.

Differentiate each segment and express the four-velocity using a continuous rapidity:At $T$ both rapidities are $\omega T$; at $3T$ both are $-\omega T$. Hence the four-velocity is continuous at both junctions. This expression also gives $g(U,U)=c^2$ everywhere, so the parameter remains proper time on the later segments.

Away from the junctions,with ${\theta }^{\prime }$ successively $+\omega ,-\omega ,+\omega$. The rapidity is continuous but its slope reverses. Thus the four-acceleration has a finite jump to its negative at each junction, while its magnitude stays $c\omega$ on either side. There is no velocity jump; the idealized path is continuously differentiable but not twice differentiable at the switching events.