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Potential energy of a system of particles

  1. Nov 1, 2004 #1

    quasar987

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    Hi.

    In trying to convince myself that the total potential energy of a system of two particles was just the potential energy of one particle computed while considering the other particle fixed, I imagined two identical particles a distance 2r apart and attracting each other, and concluded that their total kinetic energy when they meet would be the same if both particles were free as if one was fixed.

    But I don't know how to prove the thing in the most general case possible. Can anyone help?

    Thanks.
     
  2. jcsd
  3. Nov 1, 2004 #2

    arildno

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    Sure enough:
    This is because the interaction potential must depend (as in the cases of gravitational/electrostatic potentials) on the particles RELATIVE position.
    Let us state Newton's 2.laws on 2 particles, along with Newton's 3.law:
    [tex]\vec{F}_{12}=m_{1}\vec{a}_{1} (1)[/tex]
    [tex]\vec{F}_{21}=m_{2}\vec{a}_{2}(2)[/tex]
    [tex]\vec{F}_{12}=-\vec{F}_{21}[/tex]
    (I hope the notations speak for themselves..)
    Now, form the dot products with the respective velocities, and add them together:
    [tex]\vec{F}_{12}\cdot\vec{v}_{1}+\vec{F}_{21}\cdot\vec{v}_{2}=\frac{d}{dt}(\frac{1}{2}m_{1}\vec{v}_{1}^{2}+\frac{1}{2}m_{2}\vec{v}_{2}^{2})[/tex]
    Or, identifying the kinetic energy of the system [tex]\mathcal{K}[/tex] along with use of 3.law:
    [tex]\vec{F}_{12}\cdot\vec{v}_{12}=\frac{d\mathcal{K}}{dt}(3)[/tex]
    where I have introduced the relative velocity of particle 1, with respect to particle 2.

    Now, let [tex]\vec{F}_{12}=-\nabla{V}_{12}, V_{12}=V_{12}(\vec{r}_{12})[/tex]
    that is, the potential is a function of the relative velocity.
    This is required in order that we may integrate (3) with respect to time to gain conservation of mechanical energy:
    [tex]\mathcal{K}+V_{12}=C[/tex]
     
    Last edited: Nov 1, 2004
  4. Nov 1, 2004 #3

    quasar987

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    Fantastic! Thanks arildno!
     
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