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Are there useful physical quantities with dimensions of distance-time?

The background for this question is a differential equation which I try to make sense of:
https://math.stackexchange.com/questions/1095477/connections-between-the-solution-of-simple-ordinary-equation-normal-distributio

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    $\begingroup$ I feel like I've seen this question before on this site but I can't find an earlier version. Separately, a related question: physics.stackexchange.com/q/32096 $\endgroup$
    – David Z
    Commented Jan 9, 2015 at 7:52
  • $\begingroup$ @DavidZ: I have also tried to find a similar question on this site before posting but haven't found one either. $\endgroup$
    – vonjd
    Commented Jan 9, 2015 at 7:54
  • $\begingroup$ Can you give us some background? Has something with dimensions of $LT$ cropped up in a calculation, or is this just curiousity? $\endgroup$
    – John Rennie
    Commented Jan 9, 2015 at 7:56
  • $\begingroup$ @JohnRennie: I edited the question accordingly. $\endgroup$
    – vonjd
    Commented Jan 9, 2015 at 8:29
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    $\begingroup$ @AlfredCentauri: Well, I think the answer is what Carl has already mentioned: That you have a constant (which is in these cases $1$) that has a unit so that both sides are unitwise equal again. $\endgroup$
    – vonjd
    Commented Jan 9, 2015 at 12:40

3 Answers 3

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According to Wikipedia, the integral of position with respect to time is called the "Absement" which is a portmanteau of "absence" and "displacement." It seems the quantity is useful for measuring the net/average displacement, as you would divide this integral by the total time interval. It would also be useful for systems in which accumulated distance is dependent on a displacement and the length of a time of a given displacement. Examples include systems which expel power as a function of displacement, i.e. any kind of valve opening.

https://en.wikipedia.org/wiki/Absement

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The quantity LxT is of interest because it is invariant between inertial reference frames, even as velocity approaches c. LT = LoTo Doc

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Distance is rate of change of position. Now let's play around with physics formulas:

Acceleration is velocity over time, or distance over time over time, d/t/t. Multiply by time, and you get:

Velocity is distance over time, or distance over time, d/t. Multiply by time, and you get:

Displacement is distance, or position over time, d. Multiply by time, and you get:

Position is distance into time, or position, dt. Multiply by time...

Please give credit if this turns out to be an amazing physics insight.

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    $\begingroup$ Distance and position are both quantities with dimensions of length. $\endgroup$
    – J. Murray
    Commented Oct 11, 2018 at 19:43
  • $\begingroup$ Was the first premise that distance is rate of change of position correct? If that is true, then how is the rest of my logic wrong? $\endgroup$
    – Abdul Moiz Qureshi
    Commented Oct 12, 2018 at 12:26
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    $\begingroup$ No, it was not. $\endgroup$
    – J. Murray
    Commented Oct 12, 2018 at 15:40
  • $\begingroup$ Ah, I see my mistake now. Distance is not a kind of motion, and hence can not have a time term in it. $\endgroup$
    – Abdul Moiz Qureshi
    Commented Oct 12, 2018 at 15:50
  • $\begingroup$ However, I still insist that position and displacement do not have the same dimensions. Positions are points, not lengths, and are scaler rather than vector quantities. $\endgroup$
    – Abdul Moiz Qureshi
    Commented Oct 12, 2018 at 15:57

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