0
$\begingroup$

The time-ordering of two space-like separated events is undefined in special relativity. Is there an analogous concept in GR?

$\endgroup$
2
$\begingroup$

Yes. For us to assign a time-value to every point, we need to define a time function $t$ on our spacetime (this is possible if the spacetime is stably causal). This is a function $t : M \to \mathbb{R}$ such that $t$ is strictly increasing along every future-directed geodesics.

Once we have a time function, we can define a timelike time coordinate out of it, with the timelike vector defined by $\nabla t$. From the properties of the time function, it does define an order for causally-separated points. ie, if $p$ and $q$ are two points, and there exists a causal curve between $p$ and $q$, then either $t(p) < t(q)$ or $t(q) < t(p)$. This property is independent of the time function we defined, as it is not unique.

On the other hand, the same is not true if the two points are not causally-separated. It's not too hard to show this, at least locally. Take a convex normal neighbourhood $U$ around $p$. The tangent space $T_p M$ has the same structure as special relativity, and also mapping the time coordinate of $T_p M$ to $M$ defines a time function on $M$ (ie if your vector $v = (v_t, \vec{v}) \in T_p M$ has time coordinate $v_t$, and $\exp_p(v) = q$, then define the time function is $t(q) = \langle (\exp_p^{-1}(q)), e_t \rangle$).

Then you can check, using the same arguments as special relativity, that after a change of basis by a Lorentz boost, the time ordering of two spacelike-separated point can change, and it is therefore not a coordinate-independent notion.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! If you have the chance, I'd love to read how to do that last part in more detail, i.e. the basis change. Where did you use a basis here? $\endgroup$ – Eric David Kramer Feb 18 at 12:30
  • $\begingroup$ The basis is the basis of the tangent space, ie $\mathrm{Span}(e_i) = T_pM$. As per usual in a Lorentz vector space, a boost applied to a timelike vector will remain a timelike vector, but this will cause an inversion on the order of some spacelike-related events (cf. Tolman) $\endgroup$ – Slereah Feb 18 at 12:32
  • $\begingroup$ So if in one frame, my "time function" increases locally along a spacelike curve, in another Lorentz frame this time function would become locally decreasing? $\endgroup$ – Eric David Kramer Feb 18 at 12:42
  • $\begingroup$ Yes, it's easy to see by looking at a Lorentz transform diagram : i.stack.imgur.com/5NiiK.gif An increasing spacelike line below the line $x'$ would be increasing with respect to the original coordinates (ie $t$ would increase), but would be decreasing with respect to the new coordinates ($t'$ decreases) $\endgroup$ – Slereah Feb 18 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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