Let us say that you have a spacetime with wormholes, and a coordinate system. The two ends are created at almost the same point in spacetime. Then the one end is taken far away and time dilated; (note: this is all in reference to the coordinate system). How would coordinate time "work" in this case?

The reason I ask is that there appears be a contradiction in applying coordinate time. Consider the paths of the wormhole ends. If the end that didn't move aged a time of $t$, then the other end would age a time of $t' < t$, due to the time dilation. This is the premise of the wormhole time-machine: by time dilating the one end, you have a way to travel to a different time (according to coordinate time). However, it is not clear how coordinate time works, since the ends of the wormhole touch each other in spacetime (that is what a wormhole is). Is the inside of the wormhole coordinate time $t$ or $t'$?

How do you go about defining coordinate time in this situation?

My guess is that you can still define a differential $dt$ though of as the differential of coordinate time, but that this does not consistently define a coordinate time globally, since you can have spacetime loops with $\Delta t = \int {d t}$ that are non-zero. Or maybe each point of spacetime has more than one coordinate and therefore more than one coordinate time?

  • $\begingroup$ I'm not sure I follow you. "One end is taken far away and time dilated" – since you say "far away", are you assuming a metric is present? And what does it mean that 'time works" or "to apply coordinate time"? $\endgroup$
    – pglpm
    Dec 5, 2020 at 20:41
  • $\begingroup$ Keep in mind that physical time (the one you measure with a calibrated clock) is not a quantity we can associate with any point in spacetime (unlike in Newtonian mechanics). It's a quantity associated with a moving physical entity, a worldline. If three observers start from a spacetime event $A$ with perfectly syncronized clocks, and then meet at another event $B$, they generally find three completely different readings of their clocks. So no time can be associated with $B$. $\endgroup$
    – pglpm
    Dec 5, 2020 at 20:50
  • $\begingroup$ @pglpm far away according to the coordinate system. I mean can you assign a coordinate time to each point. $\endgroup$
    – PyRulez
    Dec 5, 2020 at 22:16
  • $\begingroup$ Not sure, but the concept of a non-coordinate basis might be part of what you're looking for. It's like $dx,\,dx,\,dy,\,dz$, but it's not constrained to be "aligned" with the coordinates. $\endgroup$ Dec 5, 2020 at 23:43

1 Answer 1


In the general case, coordinates cannot be defined globally. A coordinate space, or chart, only covers a region. To map the whole space requires an atlas, that is a collection of charts (by analogy with a geographical atlas).

  • $\begingroup$ Ah, that makes sense. It seems that Frame fields let you define $dt$ but not $t$. In general, does every spacetime at least one global frame field (which would provide a global $dt$)? $\endgroup$
    – PyRulez
    Dec 5, 2020 at 22:38
  • $\begingroup$ I don't know what you mean. A frame field covers the manifold, but each frame is local. There is no concept of a global $dt$ here. Specific spacetimes (such as Friedmann solutions) can be covered with a single coordinate system, which does define a cosmic time . $\endgroup$ Dec 6, 2020 at 8:50

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