Momenum and differential of momentum

[Cyrus]

This was written on the board this last semester, and I can't 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 written as:

dp=dm*V

I can't 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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[Cyrus]

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.

I think you might have misunderstood my question.

 

[ok123jump]

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.

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.