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Why does F=dp/dt?

  1. Nov 20, 2005 #1
    Why does F=dp/dt?
     
  2. jcsd
  3. Nov 21, 2005 #2
    I'm not sure that is a question you're posing, but..

    With [itex]\vec{p}=m\vec{v}[/tex] the chain rule gives

    [tex]\frac{d\vec{p}}{dt}=\frac{dm}{dt}\vec{v}+\frac{d\vec{v}}{dt}m=m\vec{a}+\frac{d\vec{v}}{dt}m[/tex]

    For objects of constant mass the last term vanishes and you remain with [itex]\vec{F}=m\vec{a}[/tex]...
     
    Last edited: Nov 21, 2005
  4. Nov 21, 2005 #3

    Pengwuino

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    Well I'll give the non-mathematical proof for it.

    momentum = mass * velocity right?

    Force = mass * acceleration as well.

    Well if you have a 1kg mass initially moving at 20m/s being pushed by a 10N force, it's going to accelerate at 10m/s right? Well lets say it only acts for 1 second. That means it should accelerate 10m/s and end at 30m/s. It's initial momentum would be 1*20 = 20kg*m/s. Its final momentum at 30m/s would be 1*30=30kg *m/s. So through this, you can see that the momentum changed 10kg*m/s in that 1 thus you can see the change in momentum over the change in time is (10kg*m/s)/1s which is equal to 1kg *m/s^2 which is equal to 10 N which is equal to your original force!
     
  5. Nov 21, 2005 #4
    I don't know if this is what you call it but I think it is the product rule.

    Also, it would appear that you have answered John Doe's question. He seemed to want to know why the derivative of momentum is equal to ma. That is would you have found.

    The Bob (2004 ©)
     
    Last edited: Nov 21, 2005
  6. Nov 21, 2005 #5

    dx

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    Right now, theres no answer to why [tex]F[/tex] equals [tex]\dot{p}[/tex]. It is a basic law deduced from experiment. In fact, this is the way newton originally gave his law, sometimes also
    [tex]\frac{F}{m}\delta{T}=\delta{v}[/tex]
    All these are equivalent to
    [tex]F = ma[/tex]
    when the mass is constant. Differentiating [tex]p[/tex] with respect to time assuming mass is constant gives [tex]ma[/tex]
     
    Last edited: Nov 21, 2005
  7. Nov 21, 2005 #6

    ahrkron

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    The expression you quote is a definition for force. It has the nice feature of sharing some properties with our intuitive idea of "force", but the quantity dp/dt is what you have to think about when using the concept in the context of physics.
     
  8. Nov 21, 2005 #7

    arildno

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    As ahrkon says, F=dp/dt is pretty much a definitional equation in MODERN physics, rather than a derived one.

    Quite differently, from a historical point of view, the definitional equations within classical mechanics (for the closed system) was F=ma, and mass conservation.

    The subtle shift to the view in modern physics consists of replacing the velocity vector&mass with the momentum vector&energy as the primary dynamical quantities, whereas the velocity vector&mass were the primary dynamical quantities in classical mechanics.
     
  9. Nov 21, 2005 #8

    Physics Monkey

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    I would just point out that force has always been
    [tex]
    \vec{F} = \frac{d \vec{p}}{dt},
    [/tex]
    this is the way Newton wrote it in the Principia. The simplification [tex] \vec{F} = m \vec{a} [/tex] can be made for most systems, but Newton did have it right from the beginning.

    Also, the equation
    [tex]
    \vec{F} = \frac{d \vec{p}}{d t}
    [/tex]
    cannot be just differentiated to obtain
    [tex]
    \vec{F} = \frac{dm}{dt} \vec{v} + m \frac{d \vec{v}}{dt},
    [/tex]
    one simple way to see this is that the right hand side of the above equation isn't Galilean invariant! The error comes because the above equation has an implicit redefinition of the system which is invalid, in plain words you have to think about the momentum change of the little mass element [tex] dm [/tex] because it makes a finite contribution to the right hand side. A more careful analysis indicates that what multiplies the [tex] \frac{d m}{dt} [/tex] term is basically the speed of the mass [tex] dm [/tex] relative to the main body. Clearly this modification produces a Galilean invariant result since it is the relative velocity that enters. This result is quite general in non-relativistic classical mechanics since bodies lose mass by shedding some of their material (think a rocket) and gain mass by acquiring extra bulk from their surroundings (think a snowball).

    Note that these considerations are modified in a relativstic setting where the "relativistic mass" can change without the body shedding any material, and the desired invariance is Lorentz invariance. In this case, it turns out that the naive differentiation is essentially ok.
     
    Last edited: Nov 21, 2005
  10. Nov 21, 2005 #9

    arildno

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    Which is why the basic equations in classical mechanics PROPERLY stated (for the material system) should be be F=ma and conservation of mass (from which F=dp/dt trivially follows).
     
  11. Nov 21, 2005 #10

    Physics Monkey

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    arildno,

    I have to disagree, the basic equation as formulated by Newton is properly stated as
    [tex]
    \vec{F} = \frac{d \vec{p}}{dt},
    [/tex]
    not [tex] \vec{F} = m \vec{a} [/tex]. Why are you arguing with this statement?
     
  12. Nov 21, 2005 #11

    arildno

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    Because:
    1. It is readily verified that, within classical limits, mass does not vary with velocity.
    2. Therefore, F=ma is NECESSARILY Galilean invariant (whether or not we have verified that a closed system has conserved mass) , whereas F=dp/dt requires the ADDITIONAL assumption for the closed system that mass is conserved in order to be Galilean invariant.
    3. Thus, to regard F=ma and mass conservation as the basic equations for the closed /material system is simpler than F=dp/dt and mass conservation.
     
    Last edited: Nov 21, 2005
  13. Nov 21, 2005 #12

    Physics Monkey

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    The assumption of Galilean invariance and the equation F = dp/dt together require that mass be conserved and that mass not vary with velocity. I do not have to assume that mass is conserved. Moreover, in both the Lagrangian and Hamiltonian frameworks it is dp/dt that enters (this is especially important when considering angular momentum). Furthermore, in relativity the correct expression is dp/dt which only reduces to ma in the limit of small velocities. Clearly the expression dp/dt is more fundamental.
     
  14. Nov 21, 2005 #13
    To add another way of looking at it is from the point of view of the action: [tex]S[/tex]
    The action is the generating function from taking the system from one moment in time to another or the time integral of the Lagrangian.
    [tex]\vec{F} = \frac{d \vec{p}}{dt}[/tex], is the only value which allows [tex]\delta{S}[/tex] to be zero.
    In other words it is the only value for which the action is stationary, which allows classical physics to be same through out time.*
    *I'm only an undergraduate, so I could be incorrect in my assessment of this.
     
  15. Nov 21, 2005 #14

    D H

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    I disagree. F=dp/dt describes the basic equations of motion. F=ma is valid for only constant mass system only. F=dp/dt is the general form. Newtons Laws apply to much more than simple closed systems.

    While mass does not vary with velocity in classical mechanics, it does vary with time in some systems. Watch the next liftoff of the Space Shuttle to see such a system in action. One had better use F=dp/dt to model rocket dynamics. F=ma leads to invalid results.
     
  16. Nov 22, 2005 #15
    I would like to see a solid proof for this statement:

    F=dp/dt

    Modern physics is dependant on this premise. This cannot be a definition given that a definition still requires justification for it's usage, and it's validity as a scientific statement is not readily obvious. I accept the fact that science is based upon experiment, however any theory should be logically consistent.

    Could someone please explain?
     
  17. Nov 22, 2005 #16
    F=dp/dt is the definition of force. It's usefulness stems from the observation that in nature there exist all sorts of phenomena that can elegantly described by this definition: [itex]-GMm/r^2, -kx, q_1q_2/4\pi\epsilon[/tex]...
     
  18. Nov 22, 2005 #17

    robphy

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    It should be noted that F=dp/dt (or F=ma) is NOT a definition of force. The left hand side is the NET FORCE (i.e. the vector sum of the external forces on the object in the question)... and the equality is not a definition, but a physical law.
     
  19. Nov 22, 2005 #18
    The relation F=dp/dt is not an equality but an identity. I.e. one does not derive this relation. It is a definition of F in terms of p. It used to be that it was a postulate and referred to as Newton's second law of motion but physicists have a difficult time when its referred to as a law since it is unclear how to measure F unless it is first defined. However this in itself opens up a whole can of worms since different people have very strong feelings about what I just said on both sides of the issue.

    Pete
     
  20. Nov 22, 2005 #19
    The relation
    [tex]F = \frac{dp}{dt}[/tex]
    can be integrated and rewritten as
    [tex]p = \int{Fdt}[/tex]
    does this help?
     
  21. Nov 22, 2005 #20
    hi

    As most people here has said, F=\dot p is a definition of force(theoretical) and a physical law (fact, empirical). This Newton's Second Law is a statement from the theoretical point of view (one of the 3 Newton defined) that can be verified experimentally. Of course i'm talking in the reing of classical mechanics.


    If make no sense to ask about a proof that F=\dot p. p=int(\F dt) dt is the same but expressed in different language.

    Furthermore, Asking ' prove that F=\dot p ' is like asking: Prove that t is time and that R^3 is space. Those are also basic definitions in the classical mechanics reign, since The Principia, they're taken as 'granted' from our 'expirience'. They are the analogs of point, line , plane, etc. of Euclid's geometry. Abstract but 'well known' (supposely) objects (in the practical perspective), whose existence and properties are axioms (in the theoretical sense).

    Regards
     
    Last edited: Nov 22, 2005
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