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Does it mean anything physically relevant if you take the arclength of a position time graph or a velocity time graph? Does it actually tell you the total position traveled along the entire curve, or is it some completely arcane result that has no meaning, like the integral of a position/time graph ?

For instance, the theory of relativity uses arc-length for worldlines to calculate proper time, but what is an arclength operator doing on any distance/time interval that accounts not only for worldlines, but for other physical interpretations as well?

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  • $\begingroup$ Do you think that's the same circumstance as the arclength of a function like y=ax^2+bx+c? $\endgroup$ – user165197 Aug 2 '17 at 23:40
  • $\begingroup$ +1 My apologies that I mislead you, but can I suggest, and not just because I was asleep at the wheel here, that you include the relativity tag (and possibly more tags ) and change references to distance to intervals where necesary. Best of luck with it. $\endgroup$ – user163104 Aug 3 '17 at 3:00
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It doesn't mean anything, because the axes have different units, so the computation of arc length doesn't make sense. The arc length would be $\sqrt{\Delta x^2+\Delta t^2}$, but the two terms have different units, so you can't add them.

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  • $\begingroup$ But on the other hand, you special relativity uses arclength en.wikipedia.org/wiki/World_line $\endgroup$ – user165197 Aug 2 '17 at 23:41
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    $\begingroup$ In theory of relativity, arc length of timelike curves (world lines) is the proper time elapsed along the world line en.wikipedia.org/wiki/Arc_length. One axis is space, the other is time, and it seems to make sense, and it seems to work. $\endgroup$ – user165197 Aug 2 '17 at 23:44
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    $\begingroup$ @DaneJoe Sure, in Relativity space and time use the same dimension (in the dimension analysis sense): spacetime; so they can be added together. In classical physics they are different dimensions. $\endgroup$ – rodrigo Aug 3 '17 at 0:47
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    $\begingroup$ @DaneJoe $\Delta x^2+\Delta t^2$ makes sense in relativity because time is being measured in meters (or space is being measured in seconds). $\endgroup$ – user12029 Aug 3 '17 at 4:23
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    $\begingroup$ @DaneJoe: Sorry, but I don't understand what cross multiplication has to do with anything. When you compute the length of a position/time curve you are adding position plus time, as this answer clearly states. And dimensional analysis says that you cannot do that unless your model treats these two dimensions as equivalent. Relativity does; classical mechanics does not. $\endgroup$ – rodrigo Aug 3 '17 at 21:48
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One must specify a metric to measure an arc-length.

If you use the Galilean metric, $$ds^2=dt^2$$, to measure the arc-length, then you have the Galilean proper time along that worldline. So, when various worldlines start at event A and end at event B, you find that all worldline segments from A to B have the same Galilean proper time. This is "absolute time."

(For special relativity, you of course use a different metric, e.g. $$ds^2=dt^2-(dx/c)^2$$, to get the special-relativistic proper time.... which then displays the clock effect.)

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  • $\begingroup$ Okay but you realize this is not limited to wordlines right? That's just one single example of where it happens to work. How does it work in general? Specifically, what transformation or unit is it physically representing on a graph not only to make it work for a wordline but for other things as well? $\endgroup$ – user165197 Aug 3 '17 at 5:26
  • $\begingroup$ This metric [sometimes called the temporal metric] applies to all curves in a Galilean spacetime. It adds up the time-components of all segments of the curve, yielding the elapsed-time between its endpoints [and which is "absolute", independent of the spacetime-path between the events and independent of the observer making the measurement]. For a segment that is "spacelike", that segment has zero length with this metric. One can assign a length to such a segment using the spatial metric of Galilean spacetime $dL^2=dx^2$... two metrics are needed because the Galilean metric is degenerate. $\endgroup$ – robphy Aug 3 '17 at 13:56
  • $\begingroup$ The "Galilean spacetime" is developed, for example, in a book by I.M. Yaglom "A Simple Noneuclidean Geometry and its Physical Basis". archive.org/details/… $\endgroup$ – robphy Aug 3 '17 at 13:59
  • $\begingroup$ A Galilean metric was replaced with the more accurate model used in the theory of relativity. What im looking for is the commonalities across different interpretations. So by inference of what you're suggesting, because you didn't explain further, all arclengths on all position time graphs always yield time, just like how all integrals on all velocity time graphs yield position. $\endgroup$ – user165197 Aug 3 '17 at 21:18
  • $\begingroup$ Spacetime arc-length $ds^2=dt^2-k(dx/c)^2$ applied to timeline curves gives elapsed proper time... k=1 for special relativity and k=0 for Galilean. Note that arc-length $\int ds$ is not the same kind of quantity as as the "area under the curve" $\int v dt$. $\endgroup$ – robphy Aug 3 '17 at 21:41

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