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Is there any fundamental reason why (at least mathematically) momentum is the integral of mass wrt velocity and kinetic energy the integral of momentum also wrt velocity? ie $$p= \int m \ dv = mv$$ $$E = \int p \ dv = \int mv \ dv = \frac 12 mv^2$$ If so, does this work for higher order integrals and what would be the associated physical parameters?

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  • $\begingroup$ It doesn't work with the relativistic equations of momentum ($\gamma m u$) and kinetic energy ($(\gamma -1) mc^2$), so it isn't generalized. $\endgroup$
    – Cross
    Commented Jul 15, 2022 at 10:09
  • $\begingroup$ The first integral is not the momentum but the change in momentum. If v is constant the momentum is still mv but your integral is zero. The premise is false so there is no question actually. $\endgroup$
    – nasu
    Commented Jul 15, 2022 at 10:43
  • $\begingroup$ @Cross - since you are free to change inertial frames, momentum is always relative. There is no absolute momentum. What you define as momentum it is the change in momentum between rest to the current state in the reference frame used. $\endgroup$
    – JAlex
    Commented Jul 15, 2022 at 13:31

1 Answer 1

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On the relation between momentum and kinetic energy:

Momentum

I regard momentum $mv$ as the integral of $ma$ (mass times acceleration) with respect to time, from starting time coordinate $t_0$ to final time coordinate $t$.

$$ \int_{t_0}^t m \ a \ dt \tag{1} $$

For the purpose of evaluating (1) we can restate $a=\tfrac{dv}{dt}$ as: $a \ dt = dv$

Using that to change from $dt$ to $dv$, with corresponding change of limits:

$$ \int_{v_0}^v m \ dv = mv - mv_0 \tag{2} $$

Which of course simplifies to $mv$ when the initial velocity is zero.


Kinetic energy

Kinetic energy is the integral of $ma$ with respect to position coordinate, from starting position $s_0$ to final position $s$.

$$ \int_{s_0}^s m \ a \ ds \tag{3} $$

Use $ds = v \ dt$, with corresponding change of limits:

$$ \int_{t_0}^t m \ a \ v \ dt \tag{4} $$

Use $a \ dt = dv$, with corresponding change of limits:

$$ \int_{v_0}^v m \ v \ dv = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{5} $$

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  • $\begingroup$ Thanks - it makes sense to frame the question as the integral of ma wrt to a fundamental quantity (time and distance) rather than m wrt a derived quantity (velocity) $\endgroup$
    – PetGriffin
    Commented Jul 15, 2022 at 13:35
  • $\begingroup$ @PetGriffin Well, mathematically $ma$ is a derived quantity; it is the second derivative of position. The thing that makes acceleration fundamental is that it is the same for all members of the equivalence class of inertial coordinate systems. In that sense acceleration is not relative. Incidentally, I find it odd that $ma$ does not have a name of its own. I propose the name 'Newtonum', as it is the counterpart of the SI unit of force: 'Newton'. $\endgroup$
    – Cleonis
    Commented Jul 15, 2022 at 16:12
  • $\begingroup$ How about 'Hannah'? That, after all, was Newton's ma. $\endgroup$
    – Simon
    Commented Jul 15, 2022 at 19:07

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