In general relativity, time-like geodesics are the trajectories of free-falling test particles, parametrized by proper time. Thus, they are easy to interpret in physical terms and are easy to measure (at least in principle).

Is there a similar interpretation/measurement for space-like geodesics? For example, how do I measure the shortest path between two space-like separated points? What is the interpretation of such a path (apart from it being a geodesic)?

My first guess is that the answer will involve a static picture (with no change in time), e.g. how elastic springs stretch between two fixed points. The problem with this picture is that the curves the springs bend along depend on the stiffness of the spring. (I believe one gets a geodesic when the stiffness is infinite, but this is no effect one would be able to measure for everyday gravitational fields.)


As VM9 pointed out below, I should fix something like a space (a space-like hypersurface) before talking about the geodesic distance.

So, let me define space-like geodesic for my purpose as follows, which will be a local notion: Take an observer, which simply shall be a time-like vector $u$ at an event $P$ of space-time $M$. Let $V$ be the orthogonal complement of $u$ (a three-dimensional space of space-like directions at $P$). Let $\Sigma$ be the image of a small neighborhood of $0$ in $V$ under the exponential mapping (that is the set of events that are connected to $P$ by small geodesics that are orthogonal to $u$ at $P$).

It is that space-like submanifold my observer is interested in. How can one physically measure lengthes in $\Sigma$? While every measurement process takes time as VM9 pointed out as well, let's assume that time flies rather slowly ($c \to \infty$) so that one could work, for example, in Gaussian coordinates (synchronous coordinates) with respect to $\Sigma$.

  • $\begingroup$ One may imagine that space-like geodesics are the trajectories of free-falling tachyons ($m^2<0$), that is with $u^\mu u_\mu = -1$ [in a metrics being locally $g = Diag(1,-1,-1,-1$) in some inertial frame] $\endgroup$ – Trimok Dec 2 '13 at 22:50
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    $\begingroup$ That sounds like a mere mathematical answer, not a physical one. $\endgroup$ – Marc Dec 3 '13 at 14:33
  • $\begingroup$ Physically, I always imagined that you pictured spacelike seperations as the distances between timelike geodesics. Their spacelike extent can then be inferred from laser-ranging experiments, where you use the time gap between sending the signal and recieving it to infer a spacelike distance. I'm not sure that there is a great way to do a direct measurement. To my knowledge, no one has even directly detected length contraction. $\endgroup$ – Jerry Schirmer Dec 3 '13 at 17:15
  • $\begingroup$ Does this describe space-like distances well? I am thinking of the deflection of light by a star. As it is well known, general relativity gives twice the deflection angle than Newtonian gravity. One half is directly through the acceleration by the mass (which is also in Newton's theory), while the other half is due to the curvature of the space hyper-surface. My question is more or less about this second half. Is this linked in a simple way to the space-like distances you measure with the laser-ranging experiment you are suggesting? $\endgroup$ – Marc Dec 3 '13 at 18:38

If you have two points $p,q$ spacelike separated in a spacetime $M$ there is not anything like the shortest spacelike curve joining them! Any spacelike curve joining them can be continuously deformed closer and closer to a lightlike curve joining the same points. So the $inf$ of the set of the lengths of spacelike curves joining the points is always zero and this value is attained for a lightlike curve.

To answer your question we have to fix a reference frame. So, first of all we have to fix a family of spacelike 3-surfaces $\{\Sigma_t\}_{t\in R}$ whose union is the spacetime $\cup_{t\in R} \Sigma_t =M$ and pairwise disjoint $\Sigma_t \cap \Sigma_{t'}=\emptyset$ for $t\neq t'$. Each $\Sigma_t$ equipped with the positive metric induced by the one of the spacetime is a three dimensional rest space.

If you consider one of them, say $\Sigma_{0}$ and fix $p,q \in \Sigma_{0}$ (supposed to be connected), the shortest (obviously spacelike) curve belonging to $\Sigma_{0}$ and joining them exists, if $p$ and $q$ are sufficiently close to each other, in view of a well known result of Riemannian geometry.

Very unfortunately all that I said above is theoretical, in the sense that it cannot be realized in practice. This is because "tempus fugit". I mean that you have to take time evolution into account, since an experiments is not an instantaneous procedure. So $p$ and $q$ have to better think of as the intersections with $\Sigma_0$ of a pair of world-lines $\gamma_p$, $\gamma_q$ describing the stories of material points. In view of time evolution, while you are performing experiments (searching for the shortest curve joining the points) in the interval of time $[t_1,t_2]$, you are actually dealing with the whole subclass of rest spaces $\Sigma_t$ with $t\in [t_1,t_2]$.

Here a pair of problems arise.

(1) First of all the geometry of the $\Sigma_t$, that induced by the spacetime metric, can be different for every instant $t$.

(2) There is no trivial way to identify points belonging to different $\Sigma_t$ in order to define a notion of point at rest (at least during the interval of time $[t_1,t_2]$) with the considered reference frame.

The simplest way to get rid of both problems, without supposing unphysical instantaneous measurement procedures, is assuming that the spacetime admits a timelike Killing symmetry and that the $\Sigma_t$ are compatible with that symmetry. This means that there is a family of disjoint timelike curves $\gamma_r = \gamma_r(u)$ -- $r$ varying in some set -- filling the universe and that, moving along one of them, the metrical properties of the spacetime remain fixed. Moreover the parameter $t$ labelling the surfaces $\Sigma_t$ coincides with the parameter $u$ of each $\gamma_r$, so that $\Sigma_u$ is nothing but the evolution of $\Sigma_0$ along the curves $\gamma_u$. $t$ is the time parameter of the reference frame. The set of the indices $r$ labelling the Killing curves $\gamma_r$ can be identified to the points of $\Sigma_0$ and, redefining the origin of $t$ on each curve, we can arrange things in order that every curve $\gamma_r$ intersects $\Sigma_0$ exactly for $t=0$.

Within this picture, to come to your issue, we assume that $\gamma_p$ and $\gamma_q$ are two curves in the said family, and we can say that the material point whose $\gamma_p$ and $\gamma_q$ represent the stories are at rest with the reference frame $(\{\gamma_r\}_{r\in \Sigma_0},\{\Sigma_t\}_{t\in R} )$. Since the curves $\gamma_r$ represent isometries, $\Sigma_{t_1}$ and $\Sigma_{t_2}$ have the same geometry that does not depend on time $t$. In other words the spacetime splits into the product $R \times \Sigma$, where $\Sigma$ is anyone of the $\Sigma_t$ equipped with the induced (positive Riemannian) metric from the spacetime which, by construction, does not depend on time $t$. In particular, the distances of $\gamma_p(t)$ and $\gamma_q(t)$, measured in each $\Sigma_t$ do not depend on $t$.

Referring to the rest space $\Sigma$ equipped with a static geometry we can safely answer your question. The shortest curve joining the two points (now at rest in the reference frame!) is the geodesic of the natural geometry on $\Sigma$. In practice, it can be constructed as the chain joining the points with the smallest number of links. Or you could adopt the parallel transport viewpoint: you have to construct an uninterrupted sequence of identical rigid rulers parallelly transported (the (n+1)th ruler is moved while remaining in contact with the nth) joining $p$ and $q$. The length of the geodesic is the number of links in the first case or the number of rulers in the second case.

  • $\begingroup$ You are right, my question isn't formulated in a very precise manner when it comes to the definition of space and space-like. In fact, I was thinking locally. I am going to amend my question accordingly. $\endgroup$ – Marc Dec 11 '13 at 20:33
  • $\begingroup$ It was nice instead, my efforts were just to constraint your question into a more precise mathematical form in order to be able to give an answer! I do not know if it is the only one. $\endgroup$ – Valter Moretti Dec 11 '13 at 20:38
  • $\begingroup$ Added an extra paragraph. Please let me know in case you still think it is not well-defined. $\endgroup$ – Marc Dec 11 '13 at 21:03
  • $\begingroup$ OK, now everything is well defined, but now I do not understand well what type of answer you wish. A sort of practical recipe to define geodesics and measure lengths along them? You must assume that a certain "microscopic" level geometry is the Euclidean one (deviations are negligible at those scales), so you know what a practical geodesic segment is (a ruler) at those scales. Using the parallel transport of rulers you construct "macroscopic" geodesics and you can see (if any) deviations from Euclidean geometry. $\endgroup$ – Valter Moretti Dec 11 '13 at 21:26
  • $\begingroup$ Normal neighbourhoods in the actual space-time seem to be quite large, so I am hoping that measurements are possible on a macroscopic enough scale. $\endgroup$ – Marc Dec 11 '13 at 21:32

I do not buy into the view that spacelike geodesics are not physical entities or, at best, can only be understood in tachyon terms. Mathematically spacelike geodesic paths are well understood in 3+1 spacetime. They can be computed in principle given a metric. Given 2 events in a local (but not infinitesimal) part of manifold they can be joined by a unique geodesic curve. If this is timelike we have an interpretation of a test particle in free fall, and its length is the time elapsed on the body. If spacelike I agree that for null observers the distance along these geodesics vanishes and in general the length is observer dependent. However, there is a maximum distance along the spacelike geodesic once the curve is defined. I know this intuitively because the distance from Earth to the moon as observed by us, 384,400 km, cannot be measured as longer than this by any observer. A complication comes when the spacetime is not static. Then all we have is two particles on two trajectories which emit a flash each to define 2 events. If we know everything about the spacetime in this neighbourhood then we can once again compute the geodesic and then find its unique maximum length. So in principle it is possible to find the geodesic and length if we assume it is possible to know the metric of the neighbourhood and coordinates of the 2 events. In our solar system we do know the metric and we can give coordinates to events and so we can compute geodesics between events and thus its (maximum) length. Clearly there are pathological cases where there are problems, i.e. at singularities, but until persuaded otherwise I hold the view that spacelike geodesics are physical and they have an invariant length that all observers can agree on.

  • $\begingroup$ geodesics need not be maxima or minima, but in general are only extrema of the interval. You wrote "there is a maximum distance along the spacelike geodesic once the curve is defined." This is misleading, as once a curve is chosen, the distance / interval along it is determined. Perhaps you have a sort of 3-dimensional view in mind. Or maybe a radial line in a spherically symmetric spacetime, which one could take under various time slices (so really this is a 2-dimensional submanifold spanned by one spacelike vector and one timelike vector at each point). I also recommend Wald 1984 $\S3.3$ $\endgroup$ – Colin MacLaurin May 9 at 6:09

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