# Momenum and differential of momentum

by Cyrus
 P: 4,777 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.
 Sci Advisor P: 6,106 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.
 P: 483 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.
 P: 27 Momenum and differential of momentum 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.
 P: 179 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.
P: 4,777
 Quote by 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.
I think you might have misunderstood my question.
P: 27
 Quote by 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.
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.

 Related Discussions Introductory Physics Homework 11 Calculus & Beyond Homework 1 Introductory Physics Homework 8 Introductory Physics Homework 0 Introductory Physics Homework 2