0
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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 '20 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 '20 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 '20 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. $\endgroup$ – Chiral Anomaly Dec 5 '20 at 23:43
3
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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 '20 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 . $\endgroup$ – Charles Francis Dec 6 '20 at 8:50

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.