# What is the physical meaning of the integral of momentum with respect to time?

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. $$f(s) = \int_{s}^{s+\Delta s} \sum_im_i\mathbf{v}_i(t)dt= \sum_i\int_{s}^{s+\Delta s} m_i\mathbf{v}_i(t)dt$$

• The time-integral of momentum sounds like some sort of "total momentum over a time period". – Steeven Feb 8 '18 at 12:05
• 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 – Stefan Reinhardt Feb 8 '18 at 12:10
• 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. – Steeven Feb 8 '18 at 12:45
• 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? – Kyle Kanos Feb 10 '18 at 20:37
• -1 Not useful to the broader community. – sammy gerbil Feb 12 '18 at 0:14