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My textbook says roughly the following in then context of the invariant spacetime interval:

The proper time $\Delta \tau$ between two events is the time interval as observed in a coordinate system where the two events take place at the same position. This is not always possible. The invariant interval between two events equals c times the proper time, or $c\Delta \tau= \Delta S^2$. The proper time is the time interval measured by an observer who doesn't move relative to this position. The proper time is the time as for the observer himself; it is his own time. If the interval is positive, there will always be a coordinate system where the position of the two events is the same. This is true because a positive interval implies $|c\Delta t|>|\Delta r|$, so a reference frame that moves with velocity $\vec{v} = (\Delta \vec{r})/(\Delta t)$ will transform the events to the same position. And this speed is always smaller than the speed of light.

Alright, I understand that a positive spacetime interval implies that we could find an observer that measures two events at the same position, because even thought $\Delta x= 0$, we still have a positive time-interval $c\Delta t$. It makes sense that a time-interval is positive, so this seems to be about right to me.

However, I don't see how they come up with $\vec{v} = (\Delta \vec{r})/(\Delta t)$. It kind of makes sense intuitively, because I imagine following travelling from the first event to the second, so that they happen where you're at... but I can't connect the mathematics behind this intuition.

Can someone help me? Thanks!

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  • $\begingroup$ Isn't that just a definition? If so, there's nothing else to explain. $\endgroup$ – probably_someone Jan 6 '17 at 0:28
  • $\begingroup$ I just don't have a feel for it, so if someone could elaborate a little bit on this concept, that'd be great. I'm really stuck on this part. $\endgroup$ – Sha Vuklia Jan 6 '17 at 0:29
  • $\begingroup$ Answer attempt below. Let me know if that's not what you're asking! $\endgroup$ – probably_someone Jan 6 '17 at 0:37
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The reason we define the proper time this way is to eliminate any effect of the travel time of light on the measurement of coordinates of events. If you assume that light travels at a finite speed, then the best way to eliminate any ambiguity is to simply travel in such a way that you happen to be at exactly the site of each event when it happens.

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  • $\begingroup$ Yea, got it! The equation says that you're travelling along with the two events. If $\Delta S>0$, your speed will be smaller than the speed of light. If $\Delta S<0$, your speed would have to be greater (contradiction). I did not interpret $\Delta r$ as being the 'spatial interval' of the two events, because I'm trying to see space and time as being part of a continuous system (i.e., spacetime) - however they still seem to make a distinction every now and then. Anyhow, thanks a lot! Your answers helped me clear up a lot of confusion. $\endgroup$ – Sha Vuklia Jan 6 '17 at 0:41
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    $\begingroup$ Yeah, the reason they sometimes resort to "classical" (i.e. 3+1 dimensional instead of 4-dimensional) quantities is because those quantities are easily measurable, and it's intuitively obvious how to go about measuring them. Of course, if you set $c=1$, so that velocity is unitless and distance and time have the same units, then you get rid of this distinction relatively easily. $\endgroup$ – probably_someone Jan 6 '17 at 0:46

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