# Finding the 'proper time' frame of reference

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!

• Isn't that just a definition? If so, there's nothing else to explain. Jan 6, 2017 at 0:28
• 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. Jan 6, 2017 at 0:29
• Answer attempt below. Let me know if that's not what you're asking! Jan 6, 2017 at 0:37

• 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. Jan 6, 2017 at 0:41
• 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. Jan 6, 2017 at 0:46