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i am asking myself what the physical meaning of $$\int \mathbf{p}(t)dt$$ would be.

I want to provide a bit more context why i am asking myself that:
I use a particle based simulation system, where the particles $i$ can move completely asynchronous over time using individual timesteps $\Delta t_i$. Hence, i do not have a consistent state of the simulation and i cannot compute the total momentum, to check if it is conserved and maybe to correct it. I only know when they reached a certain timestamp (e.g. $1s$)
So my idea was: If the momentum is conserved then the following function $f$ must have a constant value: $$ f(s) = \int_{s}^{s+\Delta s} \mathbf{p}(t)dt = \int_{s}^{s+\Delta s} \sum_i\mathbf{p}_i(t)dt = \int_{s}^{s+\Delta s} \sum_im_i\mathbf{v}_i(t)dt , $$ where $s$ moves over time and $\Delta s$ is an arbitrary incrementing step (e.g. $\Delta s = 0.1s$)

So my questions are:
Is my asumption right at all?
And has $f$ any physical meaning or not?

Update
Currently, i only consider forces between the particles and no external ones (such as gravity or interaction forces other systems, etc...).

Update 2 To clarify why i want to do that is that i can change the sum and the integral and track the momentum of each particle individually and then sum it up. $$ f(s) = \int_{s}^{s+\Delta s} \sum_im_i\mathbf{v}_i(t)dt= \sum_i\int_{s}^{s+\Delta s} m_i\mathbf{v}_i(t)dt $$

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  • $\begingroup$ The time-integral of momentum sounds like some sort of "total momentum over a time period". $\endgroup$ – Steeven Feb 8 '18 at 12:05
  • $\begingroup$ Okay, i was just curious if my brain was fried in that point and if there would be some equivalent like some weird kind of energy or work or so on $\endgroup$ – Stefan Reinhardt Feb 8 '18 at 12:10
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    $\begingroup$ It is an interesting question, that I only just thought about now. For position there do exist definitions of several higher- and lower-order time-derivatives/-integrals . But while I have seen higher-order time-derivatives of force (yank, tug, shake etc.), I have never seen it's higher-order integrals used or mentioned. $\endgroup$ – Steeven Feb 8 '18 at 12:45
  • $\begingroup$ i cannot compute the total momentum, to check if it is conserved Well I imagine that evolving the momentum using the conservation of momentum equation for SPH would probably guarantee it (at least to some numerical precision), no? $\endgroup$ – Kyle Kanos Feb 10 '18 at 20:37
  • $\begingroup$ -1 Not useful to the broader community. $\endgroup$ – sammy gerbil Feb 12 '18 at 0:14
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I think that f cannot have constant value because you denoted that velocity v is a function of time. Thus the velocity changes with time and momentum also changes with time. Also if the velocity changes then force must have acted thus momentum is not conserved for the particle i.

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    $\begingroup$ This is not correct. Provided the only forces present are between the different particles then, while the momentum of individual particles may not be conserved, the total momentum of all particles should be. $\endgroup$ – By Symmetry Feb 8 '18 at 11:38
  • $\begingroup$ I agree with you that the momentum of one particle is not conserved over time, but it should be over all due to the law of momentum conservation, or not? I should add that i currently do not consider external forces to the system such as gravity... $\endgroup$ – Stefan Reinhardt Feb 8 '18 at 11:40
  • $\begingroup$ If no external forces and collison only cause changes then momentum must be conserved. $\endgroup$ – Sri vishnu Bharat Feb 8 '18 at 12:05
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You cannot change the momentum of particle only as a function of time. Thus will lead to wrong values. The total momentum will have wrong values if each particle has its own mass and change of momentun between particles is not according rule m1v1+m2u1=m1v2+m2u2. Check whether the function v(t) that the cause change in momentum of particle , i ,obeys above rule for every particles. If it is not correct change function according to above eqution for perfect results.

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  • $\begingroup$ Okay i should mention that all particles have the same and constant mass. Then the total momentum is only a function over time, or did i miss someting? $\endgroup$ – Stefan Reinhardt Feb 8 '18 at 12:16
  • $\begingroup$ Check change in relative velocity that is sum if velocities before collision must be equal to sum of velocities after collision. V1+u1=v2+u2. If this is also correct then i dont know where is the error. $\endgroup$ – Sri vishnu Bharat Feb 8 '18 at 12:28
  • $\begingroup$ This is kind of difficult, as i do have a SPH based system and do not handle the collisions explicitly. But i could go for overall relative velocities over time. $\endgroup$ – Stefan Reinhardt Feb 8 '18 at 12:33

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