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Momenum and differential of momentum

by Cyrus
Tags: differential, momentum, momenum
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Cyrus
#1
Dec27-07, 12:59 PM
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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.
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mathman
#2
Dec27-07, 06:04 PM
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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
#3
Dec27-07, 06:59 PM
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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
#4
Dec27-07, 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.
gabee
#5
Dec27-07, 08:04 PM
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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, 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
#6
Dec27-07, 11:02 PM
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P: 4,780
Quote Quote by mathman View Post
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
#7
Dec27-07, 11:32 PM
P: 27
Quote Quote by gabee View Post
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, 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 [tex] dV \neq 0[/tex]) 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.


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