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Origin of Newton's Third

  1. Nov 5, 2008 #1

    cepheid

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    Hey, quick question. Is there any way that we can sort of "explain" Newton's Third Law as a corollary of Newton's 2nd (consv. of momentum)? I mean, if object A is interacting with object B, and initially the total momentum is p_A + p_B, then

    change in p_A = -change in p_B

    for momentum to be conserved. This means that the integral of F_AB over the period of interaction is the negative of the integral of F_BA. But can we go from saying that those two integrals are negatives of each other to saying that the two forces *themselves* are negatives of each other? Or does that not follow in general?

    Also, is it sensible to try to "explain" Newton's Third Law, or is something best just explained as an experimental fact?

    P.S. I am aware that in more advanced formulations of classical physics, fundamental laws such as the conservation laws for energy, angular momentum, and momentum follow from certain symmetries. And I seem to recall that using the Lagrangian formalism and the principle of least action, you can *arrive* at Newton's 2nd law (cons. of momentum) in terms of generalized coordinates. I need to go and brush up on that stuff, but *that's* why I'm wondering whether there is some sort of more underlying and fundamental theoretical explanation for Newton's Third, or whether I should just accept it as being true.
     
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  3. Nov 5, 2008 #2

    atyy

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    Conservation of momentum requires Newton's 2nd and 3rd, so you've answered your own question.
     
  4. Nov 5, 2008 #3

    cepheid

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    I thought cons. of momentum followed only from Newton's 2nd. Can you explain?
     
  5. Nov 5, 2008 #4

    atyy

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    Let's see, I understand it something like this.

    Say:
    F12=force exerted on object 1 by object 2
    p1=momentum of object 1

    F12=dp1/dt (Newton's 2nd law)
    F21=dp2/dt (Newton's 2nd law)
    F12=-F21 (Newton's 3rd)

    0=F12-F12=F12+F21=dp1/dt+dp2/dt=d(p1+p2)/dt

    Conservation of momentum is the equality of the first and last terms.
     
  6. Nov 5, 2008 #5

    cepheid

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    what's stopping you from saying that:

    ptotal = p1+p2 = (momentum of our system)

    Fext = (force on our system due to anything that is not in it) = 0

    ==> dptotal / dt = 0 (by Newton's 2nd)

    ==> ptotal = const. (conservation of momentum).

    ...without referencing Newton's Third law?
     
  7. Nov 5, 2008 #6

    atyy

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    You have to include the internal forces when you use Newton's 2nd law.
     
  8. Nov 6, 2008 #7

    atyy

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    Newton's law of gravitation is:

    F12=-F21=-Gm1m2ro/r2, where ro is a unit vector.

    But without Newton's 3rd law, you could imagine some funny force like:

    F12=q1q2/r2
    F21=0
    (I never get my signs right, let's say F12 represents magnitude and is attractive.)

    If that were the case, then object 2 would not accelerate. So a set of cartesian coordinates attached to object 2 would be inertial. Object 1 would be attracted to object 2 and gain momemtum without any change in momentum of object 1. We often use this approximation when we model a ball bouncing off the ground, but actually bouncing a ball also bounces the earth (though I don't know what experiment verifies this).
     
  9. Nov 6, 2008 #8

    cepheid

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    Okay...I see what you're saying. What I should have done was, for a system of N particles with total momentum P,

    [tex] P = \sum_i p_i [/tex]

    therefore

    [tex] \frac{dP}{dt} = \sum_i \frac{dp_i}{dt} [/tex]

    which, by Newton's 2nd :

    [tex] = \sum_i F_i [/tex]

    where Fi is the force on the ith particle due to *anything*, not just due to the external potential it may or may not be in. In other words:

    [tex] F_i = F_{ext,i} + \sum_{j \not= i} F_{ij} [/tex]

    or something like that, where Fij is the internal force of the jth particle on the ith one. And if we didn't have Newton's third law for the pairwise interactions, then the sum on the right hand side could be non-zero even if Fext was zero on all particles. Momentum would not be conserved.

    EDIT: this was in response to post 6.

    EDIT 2: I forgot vector arrows on everything.

    EDIT 3: So what I was trying to say with this post was that I realized that my post #5 was quite silly, what with me treating the whole system as though it were single particle that had some sort of single external force on it that entirely determined its rate of change of momentum.
     
    Last edited: Nov 6, 2008
  10. Nov 6, 2008 #9

    atyy

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    Yes, I think that's right. The confusing that is that in bouncing a ball we often use a frame attached to the ground. Actually since the earth is undergoing acceleration due to the attractive force of the ball etc, we should use a centre of mass frame (which according to the Newton's 2nd and 3rd laws will be an inertial frame if no external forces are present). We can write the equations in the centre of mass frame and then do a coordinate transformation to a frame attached to the earth. Then we will find a law that looks like Newton's 2nd law in standard form, but isn't since it involves a "reduced mass". Since the earth is so much bigger than the ball, we don't usually have to bother with these things.
     
    Last edited: Nov 6, 2008
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