# What is the definition of momentum when a mass distribution $\rho(r,t)$ is given?

This question is Edited after recieving comments.

What is the definition of momentum when a mass distribution $\rho(r,t)$ is given?
Assuming a particle as a point mass we know the definition of momentum as $p = mv$.
I need a definition where it is assumed that point masses are not present.

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I don't know that they have other, more descriptive names.   In what context are you asking your question?   I mean, are you trying to derive the Navier Stokes equations or something like that? – sigoldberg1 Nov 29 '10 at 6:58
@sigoldberg1: I want to know what is meant by momentum and momentum distribution, in this context. – Rajesh D Nov 29 '10 at 7:04
maybe you'd be better off asking that. Or at least expand on your question to indicate why you're asking what you're asking and how momentum relates to it. – David Zaslavsky♦ Nov 29 '10 at 7:12
@David Zaslavsky: please answer the edited part in the question. – Rajesh D Nov 29 '10 at 8:39

## 3 Answers

If you have multiple masses the total momentum is $$p = \sum_i m_i v_i$$

If you have a continuum distribution then you can proceed like follows. You divide the continuum into small boxes where each part of the box has approximately same mass and velocity (this assumes some kind of smoothness of the distribution). Then you can obtain whole momentum with the above formula. Now letting the box sizes go to zero you obtain integral

$$p(t) = \int \rho(r, t) v(r, t) dr$$

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@Marek: what do you call the quantity $\rho(r, t) v(r, t)$. What is $v(r, t)$ called ? – Rajesh D Nov 29 '10 at 11:17
@Marek: How are $\rho(r, t), v(r, t)$ coupled if we were to describe the newtonian gravitation in a set of differential equations. – Rajesh D Nov 29 '10 at 11:20

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@Rajesh If it is a rigid body, this does not need a name since it is just constant velocity (rotation will cancel up). If it is fluid dynamics, this is just the flow. If it is elasticity, this is deformation velocity field or something like this, I don't recall it too well. – mbq♦ Nov 29 '10 at 12:09

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@Rajesh: I would call it unphysical. And probably also mathematically inconsistent. – Marek Nov 29 '10 at 13:22

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@Rajesh: For any fluid that is at all reasonable, $\rho$ and $\vec v$ have to satisfy the continuity equation $\frac{\partial \rho}{\partial t} + \nabla\cdot\left(\vec v \rho\right) = 0$, so they are not completely independent degrees of freedom anyway. – Jerry Schirmer Nov 30 '10 at 3:16
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Counter-question: What is the definition of mass, when a mass distribution given?

Strange question, isn't it?

Since you have the mass distribution you got to have the momentum distribution as well.

It is simply ρ*v.

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Motion of a non-point object can be described as a combination of translation (which only depends on the total mass of the object) and rotation (which depends on the mass distribution). Rotational momentum is defined as $\mathbf{L}= I \boldsymbol{\omega}$ where $I$ is a moment of inertia $I = \int_V \rho(\mathbf{r})\,d(\mathbf{r})^2 \, \mathrm{d}V\!(\mathbf{r})$

As you can see, the mass distribution defines the rotational momentum of a system.

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