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Consider two infinitesimally close, timelike separated but otherwise arbitrary events $P$ and $Q$ with coordinates $(t,\vec{x})$ and $(t+dt,\vec{x}+d\vec{x})$. For example, imagine event $P$ is "a bulb turns on in a room" and the later event $Q$ is "someone sneezes in the room". The invariant spacetime interval between them is $$(ds)^2=-c^2(dt)^2+|\vec{dx}|^2.$$

Is it possible to write the timelike interval between any two such random events as $$(ds)^2=- c^2(d\tau)^2?$$ If so, can we interpret this $d\tau$ as some proper time interval? But proper time interval of what? Proper time is usually linked with a moving object/observer; it is the time recorded by a moving object/observer in its rest frame.

In this scenario, however, we do not have a moving object to start with. But we can imagine a moving observer who has the correct velocity so that at time $\tau$ in her clock, she is at the spatial location of $P$ (location of the bulb) and at the time $\tau+d\tau$, at the spatial location of $Q$ (location of the sneeze). Thus, the $d\tau$ of this case, can be thought of as the proper time interval of a special observer who has an appropriate velocity to be at the location of the events when they occur. Am I correct?

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  • $\begingroup$ The time between the 2 events is dt, the distance of the bulb and the W"sneeze is measured in the room. So it has nothing to do with some other moving observer. $\endgroup$
    – trula
    Mar 22 at 18:33
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    $\begingroup$ The question is whether you can write $ds^2=-c^2d\tau^2$. If yes, then the question is how will you interpret this $d\tau$. It's not about the meaning of $dt$ or $\vec{dx}$. Of course, they're measured in the room. $\endgroup$ Mar 22 at 18:38
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    $\begingroup$ Why wouldn't you be able to write $ds^2=-c^2d\tau^2$ for any timelike interval? $\endgroup$
    – PM 2Ring
    Mar 22 at 18:43
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    $\begingroup$ Yes, you are correct $\endgroup$ Mar 22 at 18:46
  • $\begingroup$ @PM2Ring Not saying can't be done. Surely, it can be. Maybe my wording of that line was not very clear. I'm asking how to interpret this $d\tau$. Proper time? Proper time of whom/what? $\endgroup$ Mar 22 at 20:37

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Yes, this is possible. All that is required is that one be able to find a frame where the four-vector pointing form $P$ to $Q$ is rotated into the temporal dimension in the new frame, since the Lorentz group consists of the rotations and boosts, then there is no reason why this couldn't be done. Let the primed frame be the rest frame of the special observer, and the unprimed frame the frame where the events are separated both spatially and temporally. One should interpret $d\tau$ as the proper time for an observer that has just the right velocity with respect to the unprimed frame to be at the location of the events when they occur. For the unprimed frame, the events are at different times and places, however in the primed frame, the events are at the same location as the observer but at different times.

To my mind, at least, observers such as the one in the example have proper times, however, events just have a time-like or space-like interval between then, however, as the case may be. The proper time of an observer is unique, however, the space-time interval between two events is not. Thus, why should one choose one particular interval over anyone of the others?

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$P$ is "a bulb turns on in a room" and the later event $Q$ is "someone sneezes in the room"

If the "room" is regarded as a mathematical-point in space, then these are acceptable as events. Then, $P$ and $Q$ are two events on the worldline of the "room".

A "proper time from $P$ to $Q$" is best thought of as the
"elapsed time along a worldline from $P$ to $Q$"
as read off of the wristwatch of an observer on that worldline (assuming that $Q$ is in the chronological future of $P$). Thus, different worldlines from $P$ to $Q$ will have different elapsed proper times... which is precisely the Clock Effect. (The Euclidean geometric analogue is the length of a path from point $R$ to point $S$.)

(Note: each observer along a worldline from $P$ to $Q$ will regard those events to happen "at the same place".)

The "Spacetime Interval from $P$ to $Q$" is best thought of as the
"magnitude of the spacetime-displacement vector $\vec s_{PQ}$ from $P$ to $Q$"
... in analogy to the [straight-line] distance between points.
( For me, in this context, the "squared-interval" is the quadratic quantity: the Minkowski dot product of $\vec s_{PQ}\cdot \vec s_{PQ}$.)

For chronologically-related events $P$ and $Q$, the spacetime-interval is associated with the timelike displacement $\vec s_{PQ}$, which could be interpreted as the "elapsed time along the inertial worldline from $P$ to $Q$". If your "room" is inertial, then this is the "elapsed time of the room's wristwatch from $P$ to $Q$".

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  • $\begingroup$ Why is the room a mathematical point in space? In the room frame, events P and Q have a spatial separation of $d\vec{x}$ $\endgroup$
    – PM 2Ring
    Mar 23 at 4:22
  • $\begingroup$ @PM2Ring In that case, I would say "the room" is superfluous. P is an event on the bulb worldline and Q is an event on the sneezer worldline. The spacetime-interval is associated with the spacetime-displacement vector from P to Q, which we are told is timelike-related. Thus, the spacetime-displacement vector is associated with the inertial worldline from P to Q. $\endgroup$
    – robphy
    Mar 23 at 5:46

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