3

In Wikipedia, spacetime intervals are presented explicitly under the heading "Spacetime intervals in flat space"; apparently including a presentation of spacetime intervals for (all) pairs of events in flat spacetime.

But is it correct and understood that spacetime interval values, $s^2$, can also be attributed (unambiguously, up to a common non-zero factor) to (all) pairs of events in a (any) spacetime which is not flat?

user12262
  • 4,258
  • 17
  • 40

1 Answers1

0

is it correct and understood that spacetime interval values, $s^2$, can also be attributed (unambiguously, up to a common non-zero factor) to (all) pairs of events in a (any) spacetime which is not flat?

No. There can be multiple paths between the same events. And while each path can be broken into small sections that have a well defined interval and you could take the square root of each section of them and add them all up (getting a proper length of the path) and then take the square you will get different answers for the different paths.

And if you try to limit your paths to geodesics, some events have no geodesics between them and others have multiple geodesics between them with different proper lengths for different geodesics. (The proper length is the sum of the square roots of the $s^2$ for each little piece of path.)

for all all-timelike paths I'd evaluate the plain extremum

Consider the set $$\{(v,w,x,y,z)\in\mathbb R^5: v^2+w^2=1\}$$ and give the larger space the metric $ds^2=dz^2+dy^2+dx^2-dw^2-dv^2.$ This induces an obvious metric on the 4d spacetime as an isometrically embedded surface. Then there are arbitrarily long timelike curves and arbitrarily short timelike curves between any two events, I don't even know what you could mean by a plain extremum.

Timaeus
  • 25,523
  • isn't geodesic the shortest path? how can there be multiple geodesics of different length? – Ali Moh Sep 09 '15 at 22:32
  • @AliMoh The geodesic is not the shortest path. There is a unique geodesic between two events if they are sufficiently close (topologically). But two paths can both be geodesics as long as they aren't too close. Consider the most well known case of gravitational lensing. Two paths start out in different directions curve about a mass and get bent towards each other and cross. Or more at home notice that every line of longitude is a geodesic from the north pole to the south pole. – Timaeus Sep 09 '15 at 23:26
  • @AliMoh Consider, e.g. the sphere. You can go the "long way" or the "short way" around a great circle. The "short way" is a minimum, the "long way" is a saddle point in curve space. – Ryan Unger Sep 10 '15 at 02:29
  • Timaeus: "[...] sections that have a well defined interval and take the square root [...]" -- Do you know how to name the result of this operation, in general (then please consider contributing to http://physics.stackexchange.com/q/186745). "And if you try to limit your paths to geodesics" -- I wouldn't get involved with that before having evaluated intervals between all event pairs. Null intervals seem obvious anyways; for all all-timelike paths I'd evaluate the plain extremum; the all-spacelike paths would first have to be preselected. – user12262 Sep 10 '15 at 05:17
  • @user12262 If you have too many paths each with different length, there is no obvious interval. Limiting to geodesics was an attempt to have fewer paths, so that you might get an obvious interval. It fails because it can still have too many and can sometimes have none. – Timaeus Sep 10 '15 at 05:24
  • Timaeus: "If you have too many paths each with different length [...]" -- That should be good enough to evaluate a bound (if there is any), shouldn't it? "[...] can sometimes have none." -- Are you thinking of spacetimes in which there are pairs of events which don't have any path connecting them that is exclusively all-Null, or all-timelike, or all-spacelike? Well, if you don't expand/modify your answer accordingly then I may have to edit/limit my OP question to discount such cases ... – user12262 Sep 10 '15 at 05:34
  • @user12262 It is not good form to edit a question to invalidate questions that correctly and appropriately answered the question as written. There is nothing wrong with asking a new question if you have a different question. That said I see this often for these kind of let's say ... Mathematical Relativity questions that seem to endlessly restrict spacetimes and change meanings of words over and over again. And it isn't good. – Timaeus Sep 10 '15 at 05:44
  • Timaeus: "It is not good form to edit a question to invalidate [...]" -- Certainly. I had been thinking about asking for separate treatment (as necessary). (Besides thinking about invalidating. Of course. Always.) Now, I noted your recent edit ("Consider the set [...]"). What exactly does the symbol $ds^2$ denote, which you introduced? Does it require values of intervals, $s^2$, having been introduced or admitted first, and then (perhaps) considering certain limits? – user12262 Sep 10 '15 at 05:57
  • @user12262 Have you done basic stuff yet, like learned to compute arc length of a curve in polar coordinates for a flat 2d Euclidean plane? You can compute the arc length of any (suitably rectifiable) curve by breaking it into a series of straight lines and finding the lengths of those and adding them up. If instead of adding you integrate then you only need infinitesimal line segments and $ds^2$ gives you the square of that length. Much as $ds^2=dr^2+r^2d\theta^2$ does for polar coordinates. – Timaeus Sep 10 '15 at 06:04
  • Timaeus: "Have you done basic stuff yet, [...]" -- Apparently only, if at all, on the basis of given metric spaces (or suitable generalizations such as these). (Got to go now, sorry; back perhaps in 12 hours.) – user12262 Sep 10 '15 at 06:13
  • Timaeus: Returning to your edit (based on my comment, which was terse due to that cumbersome 600 char. limit): If your example does indeed describe a set of events with arbitrarily long timelike curves ("everywhere timelike"; not only "piecewise") between any pair of events then, sticking to my proposal, I'd evaluate $s^2 = -\infty$ for each pair. (Similar to how "extended metric space" extends "plain metric space".) That's ugly. But is it false? Do you have another suggestion how to evaluate intervals in your example?? – user12262 Sep 10 '15 at 18:39
  • @user12262 I have no advice because your project seems pointless, general relativity is a local theory. The example I gave, is a spacetime that looks just like regular flat spacetime ... for a while. All general relativity does is piece together little regions of spacetime. Which is fine since experiments happen in little regions of spacetime. – Timaeus Sep 10 '15 at 18:53
  • Timaeus: "All general relativity does is piece together little regions of spacetime." -- That begs for the notion "little region" to be defined in the first place. (Presumably it's not equivalent to "strictly flat region". Therefore my question still stands, if restricted to suitable "regions" of spacetime. (I plan on submitting this as a new question. Hopefully it won't be deemed "too localized" by the community ...)) "general relativity is a local theory." -- MTW Box 13.1 seems to suggest otherwise; and your own referring to integration in order to explain the symbol $ds^2$ does, too. – user12262 Sep 10 '15 at 19:24