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Quantum Mechanical analogy of Newton's third law

  1. Jun 16, 2009 #1
    Greetings the venerable PF society,

    I have a question that has been lingering in my mind for a while, so I thought I'd ask it here.

    Firstly, how far below does the action-reaction (Newton's third law) principle go down fundamentally? How far below can I take it with me so that I don't make gross mistakes? Or else does it not break down at all?

    Please enlighten me on the meaning as well: "To every action there is an equal and opposite reaction." (from Wikipedia)

    What does this mean? Why would you necessarily have an equal an opposite force for every acting force element?

    I know from informal discussions that it breaks down for some specific cases, but I don't know how severely that's the issue.

    I just need to connect it up to something I know better, so any help or general remark would be useful.
    BTW, the context I am going to use it is the effect of spin polarized electrons flowing through a magnetic body, namely spin-torque. It happens that the magnetic body itself feels a torque because it exerts a torque on the incoming spin-polarized electrons to align their spins. So essentially it's an action-reaction type of interaction. But I don't know if that's truly the most fundamental picture

  2. jcsd
  3. Jun 16, 2009 #2
    The principle works perfectly in quantum mechanics, due to conservation of momentum. It looks a bit like it breaks down when the location is well known and the momentum is uncertain, but momentum like energy is perfectly conserved in quantum mechanics.
  4. Jun 16, 2009 #3
    Newton's third law in terms of forces already shows signs of breaking down in electrodynamics, where we account for the finite propagation speed and when discussing the radiative backreaction no one talks about the equal and opposite force which is exerted on the EM field. Even as a student I vaguely thought of Newton's third law in terms of conservation of momentum, which is correct, and with this it seems acceptable to say that in electrodynamics the momentum imparted on a particle by the radiation backreaction has an equal and opposite counterpart momentum that is carried by the radiation itself (this is the key, its ok for EM fields to carry radiation, but its not ok to speak of a force acting on the field).

    The reason that Newton3 is equivalent to momentum conservation, in basic mechanics where both apply, is that the duration of the contact force between two objects is necessarily equal (A pushed B for as long as B pushed A, pushing is a symmetric relation), and so if we multiply the equal and opposite forces by equal durations we get equal and opposite impulses and hence momenta. It's just that force is not a good concept for physics (even in classical mechanics energy and momentum are far more important), and so Newton3 only survives in QM, classical E&M, etc in the form of momentum conservation.
  5. Jun 16, 2009 #4
    Remarkable answers, thank you very much
    Last edited: Jun 16, 2009
  6. Jun 16, 2009 #5
    Action reaction works for the Compton effect to give one basic example, where we analyze collisions using conservation of momentum albeit using special relativity.
  7. Jun 18, 2009 #6
    Dear Civilized, Is "pushing" a symmetric relation in Relativity ?
  8. Jun 18, 2009 #7
    While "pushing" is not a technical term, I meant it in the sense of a force which obeys the principle of locality i.e. it occurs at a particular point in spacetime. As I said, this notion already breaks down in classical electrodynamics. But if we posit a mathematical theory of classical mechanics of point particles in minkowski space, then for localized contact forces (delta function potentials) "pushing" is indeed a symmetric relation.
  9. Jun 19, 2009 #8
    Dear Civilized , I agree with what you said about pushing. My question was not well phrased. What I wanted to ask: "pushing as long as" is it a symmetric relation in relativity?
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