# Can you recover the values of spacetime intervals $s^2$ from given causal relations between events?

Given a suitable set $\mathcal S$ of events together with their (pairwise) causal relations, i.e. for each pair of distinct events $\mathsf A, \mathsf B \in \mathcal S$ the assignment

• whether $\mathsf A$ chronologically precedes $\mathsf B$, or
whether $\mathsf B$ chronologically precedes $\mathsf A$, or

• whether $\mathsf A$ strictly causally precedes $\mathsf B$ but $\mathsf A$ does not chronologically precedes $\mathsf B$, or
whether $\mathsf B$ strictly causally precedes $\mathsf A$ but $\mathsf B$ does not chronologically precedes $\mathsf A$, or

• neither of the above,

is it then possible to determine the values of intervals $s^2$ for each pair of events, up to some (non-zero) constant?

And if so, how would one go about doing this?

• As that is knowing all the causal structure of spacetime, the answer is the same as physics.stackexchange.com/q/196496 Jul 28, 2015 at 21:43
• @Slereah: "As that is knowing all the causal structure of spacetime, the answer is the same as physics.stackexchange.com/q/196496" -- Perhaps you're referring specificly to the (your) answer physics.stackexchange.com/a/196500 given to PSE/q/196496 ? If so, note that the answer given there apparently does not at all mention values of intervals ($s^2$), or at least their ratios (between interval values which are not "null"); nor does it seem to address explicitly "how to go about" obtaining the requested determinations. Jul 28, 2015 at 21:57
• As the causal relationships will be identical for two conformally related metrics, the interval will also be conformal, since it is ~ g. Jul 28, 2015 at 22:10
• @Slereah: "As the causal relationships will be identical for two conformally related metrics" ... perhaps related to "$\Omega^2(x) \gt 0$" in the other answer mentioned above ... "the interval will also be conformal, since it is ~ g". -- Can you prove that (in the sense of my question, the corresponding interval values due to either metric tensor are "scaled isometric", with a non-zero proportionality constant) even allowing that neither $g$ nor $\Omega$ are necessarily constant? If so, please consider submitting that as an answer; and please don't forget the "how to go about". Jul 28, 2015 at 22:37

No you can’t.

Hawking Malament Theory stated that you can only determine space time metric up to a conformal factor by its causal structure and that is what motivated causal set theory.

Another approach so called the Ehlers-Prani-Schild formalism suggest that we should take causal structure (light cone/ massless particle)and projective structure (connections/massive particle) together as the fundamental degrees of freedom instead of spacetime. Spacetime metric should be reconstructed from them and thus a emergence object.

It is not the movement of particles through spacetime, but the movement of particles that creates the illusion of spacetime.

• jacktang1996: "No you can’t. [... Malament, EPS ...]" -- Those are relevant key words, so: +1. But I remain unconvinced ... Why should the conformal factor variability of the metric tensor $g$ render ratios of non-zero spacetime inervals (Synge world functions, Lorentzian distances ...) indeterminate, and not just present intrinsic indeterminacy of $g$ ?? Also [contd.] Dec 30, 2021 at 1:14
• Also we have "Einstein's principal proclamation": "All our space-time verifications invariably amount to a determination of space-time coincidences {... such as ...} meetings of two or more {...} material points.". Consequently causal relations is all we can work with. In particular, the EPS notion of "free particle" must be defined in terms of causal relations. Dec 30, 2021 at 1:15

Trivially no.

Consider two such sets in a flat space-time with the same number of events. Each event in the second set is associated with an event in the first event such that taking the earliest event in each set to define the origin of a frame of reference the 4-vectors of position of events in the second set are twice the four vectors of position of the related even in the first set.

All sequence and causal relationships are maintained, but the intervals in the second set are 4 times as large.

Second counter example.

Two set consisting of Three points each. Also in a flat space time.

Three of the points in each set are associated with the four vectors \begin{align*} (0, 0, 0, 0)&\\ (1, 1, 0, 0)& \end{align*}

They differ in the third event the first having $(2, 0, 2, 0)$ and the second having (2, 0, -2, 0). So now you can insist on an arbitrary scale and a reflection. Then the third counter example has four points in each set. As before they share \begin{align*} (0, 0, 0, 0)&\\ (1, 1, 0, 0)& \,, \end{align*} but now they differ in the last two, with one set having \begin{align*} (1, 0, 1, 0)&\\ (1, 0, 0, 1)& \,, \end{align*} and the other \begin{align*} (1, 0, -1, 0)&\\ (1, 0, 0, -1)& \,. \end{align*} Then you throw out cases of scaling, reflection and rotation. Then I suggest that the sets that start \begin{align*} (0, 0, 0, 0)&\\ (1, 1, 0, 0)& \,, \end{align*} but follow up with either(2,2,0,0)$or$(3,3,0,0)$both have all their intervals equal to zero. • dmckee: "[...] but the intervals in the second set are 4 times as large." -- Please note the phrase "up to some (non-zero) constant" in the OP question statement. The two example cases described in your answer are in so far to be considered equivalent (with the applicable constant of value 4, or 1/4); and your answer is therefore (trivially) incorrect. Jul 28, 2015 at 22:05 • dmckee: Re your recent edit, with several "counter examples": Yes, these seem indeed examples of events whose causal relations do not imply specific interval ratios. (I'd foremost think of "arbitrarily many events which are pairwise timelike to each other".) But may I remind you: I had been asking in the OP about a suitable set of events$\mathcal S\$, i.e. if such a set may be thought of at all. Taking (again) a hint from Synge's more or less well-known "five point curvature detector" (GR, p. 408), I'd guess that such a set should have a whole lot more than 4 elements, suitably related. Jul 29, 2015 at 4:54