-2
$\begingroup$

The geometric relations between pairs of events of a spacetime $\mathcal S$ can generally be characterized by values of Lorentzian distance (cmp. J.K.Beem, P. Ehrlich, K.Easley, "Global Lorentzian Geometry"), $$\ell : \mathcal S \times \mathcal S \rightarrow [0 ... \infty];$$ especially also in cases where values of spacetime intervals, $s^2$, cannot be obtained (i.e. if the spacetime under consideration is not flat).

For any two distinct events $\varepsilon_{A J}, \varepsilon_{A K} \in \mathcal S$ which were timelike separated from each other, such that some participant ("material point", "observer") $A$ had taken part in both, holds

$$\ell[ \, \varepsilon_{A J}, \varepsilon_{A K} \, ] + \ell[ \, \varepsilon_{A K}, \varepsilon_{A J} \, ] \gt 0;$$

where in particular

  • either $\varepsilon_{A J}$ chronologically preceded $\varepsilon_{A K}$, and therefore
    $\ell[ \, \varepsilon_{A J}, \varepsilon_{A K} \, ] \gt 0, \qquad \ell[ \, \varepsilon_{A K}, \varepsilon_{A J} \, ] = 0$,

  • or the other way around;

while any pair of two distinct events which were not timelike seprated from each other (but instead lightlike, or instead spacelike), $\varepsilon_{A B}, \varepsilon_{J K} \in \mathcal S$, satisfy

$$\ell[ \, \varepsilon_{A B}, \varepsilon_{J K} \, ] = \ell[ \, \varepsilon_{J K}, \varepsilon_{A B} \, ] = 0.$$

My question:
Given these $\ell$-values of all pairs of events in $\mathcal S$, is it possible to distinguish and to determine which pairs were spacelike separated from each other (and not lightlike, nor timelike) ?


(The characterization of two events as "lightlike separated from each other" means, in terms of their causal relations, that one of them strictly causally preceded the other, without explicitly prescribing which one "came first", and that neither preceded the other chronologically.)

$\endgroup$
4

2 Answers 2

0
$\begingroup$

Why is $\ell$ defined such that it's always nonnegative? I'm not sure I see the physical intuition or significance of such a definition.

That said, since the future/past of a point $I^\pm(p)$ is open, $S-I^\pm(p)$, the set of spacelike or null separated points from p, is closed. The interior of $S-I^\pm(p)$ corresponds to spacelike separated, and the boundary is null separated. In the context of this Lorentzian distance $\ell$, a point $q$ is spacelike separated from $p$ if $$q\in \text{int(}\ell^{-1}_p(0)).$$

This should in principle be known if knowledge of $\ell$-values of all pairs of points in S are known.

EDIT: It's been brought to my attention in the comments that the assignment of boundary $\rightarrow$ null separated, interior $\rightarrow$ spacelike separated doesn't (necessarily) hold for general spacetimes. That being the case, the problem seems much harder to me because it depends on global considerations (existence of causal curves) rather than local ones (values of $\ell_p$ in an open set).

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Mar 23, 2017 at 20:42
-1
$\begingroup$

In order to assert that an event pair consisting of $\varepsilon_{A B}, \varepsilon_{J K} \in \mathcal S$, for which $$\ell[ \, \varepsilon_{A B}, \varepsilon_{J K} \, ] = \ell[ \, \varepsilon_{J K}, \varepsilon_{A B} \, ] = 0$$ is known to begin with, is specifically spacelike separated, and not lightlike,

the existence of a suitable "$\ell = 0$"-boundary is required, and must be established, which separates

  • the set of all timelike ("$\ell \gt 0$") event pairs to which either event $\varepsilon_{A B}$ or $\varepsilon_{J K}$ belong

  • from the "$\ell = 0$"-set of event pairs including the pair $(\varepsilon_{A B}, \varepsilon_{J K})$.

This is accomplished by requiring explicitly that:

$\forall \varepsilon_{A Q} \in \mathcal S {\Large \, | \, } \ell[ \, \varepsilon_{A B}, \varepsilon_{A Q} \, ] + \ell[ \, \varepsilon_{A Q}, \varepsilon_{A B} \, ] \gt 0 :$

$ \qquad \exists \, \varepsilon_{A P} \in \mathcal S {\Large \, | \, } \ell[ \, \varepsilon_{A B}, \varepsilon_{A P} \, ] + \ell[ \, \varepsilon_{A P}, \varepsilon_{A B} \, ] \gt 0 \text{ and } \ell[ \, \varepsilon_{A P}, \varepsilon_{A Q} \, ] + \ell[ \, \varepsilon_{A Q}, \varepsilon_{A P} \, ] \gt 0 \text{ and } $
$ \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{A B}, \varepsilon_{A P} \, ] + \ell[ \, \varepsilon_{A P}, \varepsilon_{A B} \, ] + \ell[ \, \varepsilon_{A P}, \varepsilon_{A Q} \, ] + \ell[ \, \varepsilon_{A Q}, \varepsilon_{A P} \, ] \le $
$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{A B}, \varepsilon_{A Q} \, ] + \ell[ \, \varepsilon_{A Q}, \varepsilon_{A B} \, ] $
$ \qquad \qquad \qquad \qquad \text{ and } \, \ell[ \, \varepsilon_{A P}, \varepsilon_{J K} \, ] = \ell[ \, \varepsilon_{J K}, \varepsilon_{A P} \, ] = 0 \, \text{ and } $

$ \qquad \forall \varepsilon_{A O} \in \mathcal S {\Large \, | \, } \ell[ \, \varepsilon_{A B}, \varepsilon_{A O} \, ] + \ell[ \, \varepsilon_{A O}, \varepsilon_{A B} \, ] \gt 0 \text{ and } \ell[ \, \varepsilon_{A O}, \varepsilon_{A P} \, ] + \ell[ \, \varepsilon_{A P}, \varepsilon_{A O} \, ] \gt 0 \text{ and } $
$ \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{A B}, \varepsilon_{A O} \, ] + \ell[ \, \varepsilon_{A O}, \varepsilon_{A B} \, ] + \ell[ \, \varepsilon_{A O}, \varepsilon_{A P} \, ] + \ell[ \, \varepsilon_{A P}, \varepsilon_{A O} \, ] \le $
$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{A B}, \varepsilon_{A P} \, ] + \ell[ \, \varepsilon_{A P}, \varepsilon_{A B} \, ] : $

$ \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{A O}, \varepsilon_{J K} \, ] = \ell[ \, \varepsilon_{J K}, \varepsilon_{A O} \, ] = 0,$

and correspondingly (interchanging the roles of $\varepsilon_{A B}$ and $\varepsilon_{J K}$):

$\forall \varepsilon_{J F} \in \mathcal S {\Large \, | \, } \ell[ \, \varepsilon_{J K}, \varepsilon_{J F} \, ] + \ell[ \, \varepsilon_{J F}, \varepsilon_{J K} \, ] \gt 0 :$

$ \qquad \exists \, \varepsilon_{J G} \in \mathcal S {\Large \, | \, } \ell[ \, \varepsilon_{J K}, \varepsilon_{J G} \, ] + \ell[ \, \varepsilon_{J G}, \varepsilon_{J K} \, ] \gt 0 \text{ and } \ell[ \, \varepsilon_{J G}, \varepsilon_{J F} \, ] + \ell[ \, \varepsilon_{J F}, \varepsilon_{J G} \, ] \gt 0 \text{ and } $
$ \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{J K}, \varepsilon_{J G} \, ] + \ell[ \, \varepsilon_{J G}, \varepsilon_{J K} \, ] + \ell[ \, \varepsilon_{J G}, \varepsilon_{J F} \, ] + \ell[ \, \varepsilon_{J F}, \varepsilon_{J G} \, ] \le $
$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{J K}, \varepsilon_{J F} \, ] + \ell[ \, \varepsilon_{J F}, \varepsilon_{J K} \, ] $
$ \qquad \qquad \qquad \qquad \text{ and } \, \ell[ \, \varepsilon_{A B}, \varepsilon_{J G} \, ] = \ell[ \, \varepsilon_{J G}, \varepsilon_{A P} \, ] = 0 \, \text{ and } $

$ \qquad \forall \varepsilon_{J H} \in \mathcal S {\Large \, | \, } \ell[ \, \varepsilon_{J K}, \varepsilon_{J H} \, ] + \ell[ \, \varepsilon_{J H}, \varepsilon_{J K} \, ] \gt 0 \text{ and } \ell[ \, \varepsilon_{J H}, \varepsilon_{J G} \, ] + \ell[ \, \varepsilon_{J G}, \varepsilon_{J H} \, ] \gt 0 \text{ and } $
$ \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{J K}, \varepsilon_{J H} \, ] + \ell[ \, \varepsilon_{J H}, \varepsilon_{J K} \, ] + \ell[ \, \varepsilon_{J H}, \varepsilon_{J G} \, ] + \ell[ \, \varepsilon_{J G}, \varepsilon_{J H} \, ] \le $
$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{J K}, \varepsilon_{J G} \, ] + \ell[ \, \varepsilon_{J G}, \varepsilon_{J K} \, ] : $

$ \qquad \qquad \qquad \qquad \ell[ \, \varepsilon_{A B}, \varepsilon_{J H} \, ] = \ell[ \, \varepsilon_{J H}, \varepsilon_{A B} \, ] = 0.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.