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The chronological future $I^{+}(S)$ of a subset $S$ of a spacetime is the set of points reachable from $S$ by future-directed timelike curves. The chronological future is an open set.

The chronological past $I^{-}(S)$ is the set of points from which a future-directed timelike curve reaches $S$. Equivalently, it is the time reverse of the chronological future.

The clock hypothesis says that an ideal clock following a timelike curve measures its proper time: its instantaneous rate depends on velocity through $d\tau =dt\sqrt{1-v^2/c^2}$, with no separate acceleration term. Acceleration still changes the world line and hence the accumulated proper time. A physical clock may suffer disturbances under forces or environmental changes, so experiments must distinguish its mechanism's errors from the geometric prediction.

With the mostly-plus metric signature $(-,+,+,+)$ and coordinates $X^{\mu }=(ct,x,y,z)$, the Minkowski metric is the nondegenerate bilinear formIt is a symmetric bilinear form, but not a positive-definite bilinear form. For example, $U=(1,1,0,0)$ is nonzero yet $U\cdot U=0$, making it a null vector. This does not make the Minkowski metric degenerate: some other vectors have nonzero pairing with $U$.

Applied to infinitesimal displacements, the Minkowski metric givesFor a timelike curve, a clock records $d\tau =\sqrt{-d{s}^2}/c$. For a null curve, $d{s}^2=0$. For spacelike separation, the positive Minkowski interval measures a different kind of separation. A Lorentz transformation preserves this form, expressed in matrix notation by ${\Lambda }^T\eta \Lambda =\eta$, although it need not preserve ordinary Euclidean distance in the four-component coordinate list.

A spacetime is geodesically complete if every maximal geodesic has an affine parameter ranging over all of $\mathbb{R}$. For timelike geodesics this is equivalent to unbounded proper time in both directions. An extendible geodesic segment can be prolonged in the same spacetime; an inextendible geodesic cannot. A finite coordinate endpoint need not imply finite affine parameter.

For the Kruskal spacetime, use $U,V$ withIn the right exterior $U<0<V$. A truncated ray $U=U_0<0$, $V_1<V<V_2$ is an extendible geodesic of radial null type: neither artificial endpoint is a spacetime boundary. A future ray $U=U_0>0$ in the black hole reaches $UV=1$, hence $r=0$, and is inextendible geodesic and future null-geodesically incomplete. Its Killing energy $E\ne 0$ gives $dr/d\lambda =-|E|$, so the Schwarzschild singularity occurs at finite affine parameter. The maximal continuation toward the past supplies the other half of this same null geodesic.

There is no inextendible, complete radial timelike geodesic in positive-mass Kruskal spacetime. This requested example is impossible as printed. For a radial timelike geodesic, the conserved Killing energy and normalization giveIf $E^2<1$, $r$ has at most one turning point, a maximum $r_{\max }=2M/(1-E^2)$; a maximal trajectory runs from the white hole Schwarzschild singularity to the Schwarzschild singularity. If $E^2\ge 1$, there is no finite turning point; one end can lie at infinity but the other reaches $r=0$. The exceptional $E=0$ trajectory through the bifurcation surface also reaches $r=0$ in both time directions. Near $r=0$,whose integral is finite. Constant-$r$ radial timelike curves are accelerated, not geodesics.

Two plausible repairs have different meanings. Removing “radial” permits a complete circular timelike geodesic at $r=R>6M$, with nonzero angular momentum and proper time ranging over $\mathbb{R}$. Replacing “timelike” by “null” permits a complete horizon null geodesic: $U=0$, $V\in \mathbb{R}$, with $V$ an affine parameter. The Penrose diagram shows both repairs explicitly, together with the two valid requested examples; the circular trajectory is only a radial projection and is labelled as nonradial.

The horizontal boundaries are the Schwarzschild singularities, diagonal dashed lines are the Killing horizons, and outer diagonal edges are null infinity.

The proper time elapsed along the smooth timelike curve isThe expression is invariant under every orientation-preserving reparametrization of the curve.

Proper time is the time measured by a clock following a timelike worldline. With metric signature $(-,+,+,+)$ and units $c=1$,along that worldline.

Restoring the speed of light, the same metric signature gives $d\tau =\sqrt{-d{s}^2}/c=dt\sqrt{1-|\mathbf{v}{|}^2/c^2}$ for a future-directed timelike curve. Integrating along a clock's world line gives its elapsed proper time, even when its velocity varies. For uniform motion this becomes the time dilation formula; it is not generally the square root of the interval between endpoints of a curved world line.

A timelike curve has a nonzero tangent vector with negative Minkowski norm in metric signature $(-,+,+,+)$ at every regular point. A physical massive observer's world line is also future-directed. Parametrizing it by coordinate time gives $-c^2+|\mathbf{v}{|}^2<0$, or $|\mathbf{v}|<c$: the local tangent condition is exactly the subluminal speed condition. It is stronger than checking only that two endpoints have timelike separation.

Two distinct spacetime events have timelike separation when their displacement has negative Minkowski interval in metric signature $(-,+,+,+)$, equivalently $c|\Delta t|>|\Delta \mathbf{x}|$. A massive observer can connect them along a straight timelike curve. If the later spacetime event has $\Delta t>0$, every Lorentz transformation in the orthochronous Lorentz group keeps it later. The elapsed proper time on the straight inertial segment is $\sqrt{-s^2}/c$.

In the light-second units of the Minkowski spacetime example, $O$ to $A$ has $s^2=(-4^2+3^2){\mathrm{ls}}^2=-7{\mathrm{ls}}^2$. Its straight moving clock therefore records $\sqrt{7}\mathrm{s}$, approximately $2.646\mathrm{s}$, while the coordinate time difference is $4\mathrm{s}$. Coordinate time and proper time answer different questions.

Work in Minkowski spacetime, neglect gravity, and choose the stay-at-home clock's inertial frame. Let departure be $O=(t,x)=(0,0)$ and reunion be $R=(T,0)$. The home clock follows $x_H(t)=0$. A traveller moves outward with velocity $+v$ and returns with velocity $-v$, where $0<v<c$ and $c$ is the speed of light. The ideal world line has a sharp turnaround at $P=(T/2,vT/2)$:Each straight leg lies in an inertial frame, but no single traveller inertial frame covers both legs.

For any differentiable timelike curve in this home inertial frame, its clock accumulates proper time through the integralApply it separately to the traveller's two legs. With the Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For $v=0.6c$ and $T=10\mathrm{yr}$, the Lorentz factor is $\gamma =1.25$. The turnaround is $3$ light-years away at home time $5\mathrm{yr}$. The traveller records $4\mathrm{yr}$ outward and $4\mathrm{yr}$ returning: the reunited clocks read $10\mathrm{yr}$ and $8\mathrm{yr}$. These are illustrative model values, not experimental measurements.

The same integral gives a short geometric explanation: $\sqrt{1-v(t{)}^2/c^2}\le 1$, so every such competing timelike curve between these fixed spacetime events has $\Delta \tau \le T$. Equality requires zero velocity throughout, apart from sets of zero duration. Thus the straight home world line maximizes proper time in this Minkowski spacetime problem. This statement concerns fixed endpoints and an inertial home clock; in a gravitational field, compare paths using the appropriate metric tensor.

A world line is an object's history as a curve of spacetime events. In an inertial frame, a particle at rest has constant spatial coordinates, so its world line is vertical in a spacetime diagram with $ct$ upward. A particle at constant velocity $v$ along one axis has $x=x_0+vt$, a straight sloping world line. The coordinate representation changes with the inertial frame, while the physical history is the same.

A massive observer follows a future-directed timelike curve. Along a differentiable world line with velocity $\mathbf{v}(t)$, its clock accumulatesThe Minkowski metric therefore measures clock time along the curve, not just the difference between two coordinate readings. In the straight $O$-to-$A$ example, $v=0.75c$ for $4\mathrm{s}$, giving $4\sqrt{1-0.75^2}\mathrm{s}=\sqrt{7}\mathrm{s}$. If the observer changes velocity along the path, use the integral rather than assigning the straight-segment formula to the whole path.