# Momenum and differential of momentum

1. Dec 27, 2007

### Cyrus

This was writen on the board this last semester, and I cant seem to figure it out and its been bothering me to no end (mentally).

Momentum is defined as:

$$p=mv$$

therefore, if you want to find the differential momentum, it should be, mathematically speaking:

$$d(p)=d(mv)=dp=dm*V+m*dV$$

But differetial momentum is always writen as:

$$dp=dm*V$$

I cant make sense out of what happened to the second term on the right side.

In fluid mechanics we have:

$$d(\rho VA)=\frac{d \rho}{\rho}+\frac{dV}{V} +\frac{DA}{A}$$

So d(p) for momentum should follow just the same using the product rule.

It makes sense conceptually, as each particle dm has a velocity V, and if you sum it over the body you get the total momentum, but it seems totally wrong mathematically.

2. Dec 27, 2007

### mathman

classically, dm=0 in most cases. However, when special relativity is involved, mass does change. Another case where mass changes involves selfpropelled objects (rockets, airplanes) where fuel is burned off changing the mass.

3. Dec 27, 2007

### z-component

Cyrus is asking what happened to the second term, m * dV. If the differential moment definition is correct as written above, then dp would always be zero according to you. Good question.

4. Dec 27, 2007

### ok123jump

I suspect that the answer to your question only makes sense in a relativistic framework.

If our frame of reference is moving with the same acceleration as a given particle, then

$$m\cdot dV=0$$ since $$dV=0$$.

Then, in this setting,

$$dp=dm \cdot V+m \cdot dV=dm \cdot V$$.

However, in this framework, the only useful hint is that momentum affects mass.

Last edited: Dec 27, 2007
5. Dec 27, 2007

### gabee

Hmm.

Using the equation

$$dp = dm*V$$

must implicitly assume that V is constant wrt the variable by which you differentiate.

If that variable is length x, say, then m=m(x) and V=V(x). Then

$$\frac{dp}{dx} = V\frac{dm}{dx} + m\frac{dV}{dx}$$.

First, it's important to note immediately that this is only valid in one dimension.
The second term on the right is zero if dV/dx=0, which implies that V is a constant with respect to x. Note that V could actually be a function of some other variable (like time), but our consideration is only with respect to one variable, x. So it appears that dp=dm*V is only valid if you're talking about rigid, non-rotating? bodies in which velocity does not change with respect to the integration variable. The full formula would apply in more general cases.

I could be wrong, tell me what you think.

Last edited: Dec 27, 2007
6. Dec 27, 2007

### Cyrus

I think you might have misunderstood my question.

7. Dec 27, 2007

### ok123jump

I think that you're mostly correct (there is one trivial case which is important to remember).

The body must be rigid (otherwise at some point along the body there might exist a place where $$dV \neq 0$$) and must be either non-rotating or rotational velocity is neglected. I suspect that for most cases that the body will be non-rotating.

Your formula is somewhat limiting because it doesn't generalize to higher dimensions, but I think that you have the right idea.

Last edited: Dec 27, 2007