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Momenum and differential of momentum 
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#1
Dec2707, 12:59 PM

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: [tex]p=mv[/tex] therefore, if you want to find the differential momentum, it should be, mathematically speaking: [tex] d(p)=d(mv)=dp=dm*V+m*dV[/tex] But differetial momentum is always writen as: [tex] dp=dm*V[/tex] I cant make sense out of what happened to the second term on the right side. In fluid mechanics we have: [tex]d(\rho VA)=\frac{d \rho}{\rho}+\frac{dV}{V} +\frac{DA}{A}[/tex] 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
Dec2707, 06:04 PM

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.



#3
Dec2707, 06:59 PM

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.



#4
Dec2707, 07:19 PM

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 [tex] m\cdot dV=0 [/tex] since [tex] dV=0 [/tex]. Then, in this setting, [tex] dp=dm \cdot V+m \cdot dV=dm \cdot V [/tex]. However, in this framework, the only useful hint is that momentum affects mass. 


#5
Dec2707, 08:04 PM

P: 179

Hmm.
Using the equation [tex]dp = dm*V[/tex] 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 [tex]\frac{dp}{dx} = V\frac{dm}{dx} + m\frac{dV}{dx}[/tex]. 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, nonrotating? 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. 


#7
Dec2707, 11:32 PM

P: 27

The body must be rigid (otherwise at some point along the body there might exist a place where [tex] dV \neq 0[/tex]) and must be either nonrotating or rotational velocity is neglected. I suspect that for most cases that the body will be nonrotating. Your formula is somewhat limiting because it doesn't generalize to higher dimensions, but I think that you have the right idea. 


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