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A clue for de Broglie?

  1. Dec 24, 2008 #1
    If a ball with mass [tex]m_1[/tex] moving at speed [tex]v[/tex] strikes another ball of mass [tex]m_2[/tex] that is initially at rest in a head-on elastic collision, the speeds of the two balls after the collision are

    [tex]
    v_1 = v (\frac {m_1 - m_2}{m_1 + m_2})[/tex], [tex]v_2 = v (\frac {2 m_1}{m_1 + m_2})[/tex] respectively.

    If a wave with amplitude A moving in a medium with speed [tex]v_1[/tex] hits (head-on) the boundary of another medium in which it moves at speed [tex]v_2[/tex], the amplitudes of the reflected and transmitted waves are

    [tex]
    A_1 = A (\frac {v_2 - v_1}{v_2 + v_1})[/tex], [tex]A_2 = A (\frac {2 v_1}{v_2 + v_1})[/tex] respectively.

    And if instead of speed one defines a variable that is inversely proportional to it, as in

    let [tex]k_1 = 1/v_1[/tex] and [tex]k_2 = 1/v_2[/tex] then for the waves we get

    [tex]
    A_1 = A (\frac {k_1 - k_2}{k_1 + k_2})[/tex], [tex]A_2 = A (\frac {2 k_1}{k_1 + k_2})[/tex]

    which look exactly like the equations for the colliding balls.

    Since balls are not waves and mass is not speed (or the inverse of speed), why are these formulae isomorphic? I know that wave-particle duality is a central idea in quantum mechanics, but these relationships are derived using classical assumptions, and rather different-looking assumptions at that; conservation of momentum and kinetic energy in the first case and continuity and differentiability of waves at a boundary in the second. Is it a coincidence or is there something deeper at work here?
     
    Last edited: Dec 24, 2008
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  3. Dec 24, 2008 #2

    tiny-tim

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    Emmy Noether

    Hi snoopies622! :smile:

    I think it's all Emmy Noether's fault …

    she proved that whenever there's a geometric symmetry, there's an associated conserved current …

    for balls, the current associated with translational symmetry (of space-time) is energy-momentum …

    for waves, it has to be something, and it will obey the same rules. :smile:
     
  4. Dec 24, 2008 #3

    HallsofIvy

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    Well said, tiny tim.
     
  5. Dec 24, 2008 #4
    Interesting. You know, I keep hearing about Noether's theorem but I have yet to make a serious attempt to understand it. I guess this is the time. The word "theorem" is one that I associate with pure mathematics and not science (one does not think of Kepler's laws of planetary motion - for example - as theorems even though they can be derived from Newton's laws) so my first question is, is Noether's theorem derived from physical laws (which are observable and falsifiable) or purely mathematical propositions, like Euclid's axioms?
     
  6. Dec 24, 2008 #5

    Vanadium 50

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    Are you sure mass is not the inverse of speed? Both equations rely on momentum conservation, and m = p/v.
     
  7. Dec 25, 2008 #6
    Actually the wave equations are derived from two assumptions:

    1. the displacement at the boundary is the same on both sides (the curve is continuous there)
    2. the derivative of the displacement at the boundary with respect position is also the same on both sides (the curve is differentiable there).

    -- conservation of momentum is not an assumption.

    In any case, I have never quite understood the idea that waves carry momentum. I know that a wave carries energy and has a speed, and that dividing energy by speed produces something with the same dimensions as momentum, but dividing again by speed produces mass. Would one say that (for instance) a transverse mechanical wave carries an associated mass along with it, too?
     
    Last edited: Dec 25, 2008
  8. Dec 25, 2008 #7

    Vanadium 50

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    A wave most certainly carries momentum. Look at a surfer - where does their momentum come from?
     
  9. Dec 25, 2008 #8
    Doesn't a surfer begin by paddling ahead of an oncoming wave? If not he would simply bob up and down (or move in a circle) like everything else in the water.

    If a mechanical wave carries momentum, what the formula for it and how is it derived?
     
  10. Dec 25, 2008 #9

    Vanadium 50

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    A surfer certainly doesn't get his speed by paddling. These guys go, what 20 mph? The fastest swimmers can go maybe 5 mph, and paddling on a surfboard is slower still.

    Any text with a treatment of waves beyond Halliday and Resnick will have the derivation of the momentum carried by a wave: e.g. Section 1.11 of Elmore and Heald.
     
  11. Dec 25, 2008 #10
    Thanks for the reference. I just checked - the UNH physics library has that book, but they're closed for at least a couple weeks for the holiday/between-semester break. Can you tell me anything about this subject (how much momentum a mechanical wave carries and how one arrives at this relationship) in the meantime?
     
    Last edited: Dec 26, 2008
  12. Dec 26, 2008 #11

    tiny-tim

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    Noether's theorem

    Hi snoopies622! :smile:

    I gave the impression that Noether's theorem is derived purely from geometric symmetry, but that's not entirely true …

    it also relies on the physical assumption that motion is governed by an action (the integral of a Lagrangian) …

    whenever there's a symmetry of the action, there's an associated conserved current :smile:

    the symmetry of the action is is usually blindingly obviously derived from a geometric symmetry (including things like charge as part of geometry), so we tend to say, eg, that conservation of energy-momentum comes from translational symmetry of space-time when we really should say translational symmetry of the action. :wink:

    I think it follows that if you impose a physical restraint on a system so that it has an extra, non-geometric, symmetry, then that restraint will create a new conserved current.

    For example, the physical requirements that two particles be indistinguishable, or be Fermi-Dirac antisymmetric, are non-geometric symmetries which must have associated conserved currents … but i can't think what they are :redface:
     
  13. Dec 26, 2008 #12
    Thanks tiny-tim. This seems like a deep subject. I guess when my local physics library re-opens I'll be taking out at least TWO books (one on waves and one on Noether's theorem).

    Do Lagrangians apply to waves? So far I've only see them used for particles, or for systems of particles.

    Another question if anyone out there knows the answer: is the momentum of a mechanical wave directly proportional to its energy divided by its speed? If so, what is the constant of proportionality?
     
    Last edited: Dec 26, 2008
  14. Jan 15, 2009 #13
    A couple notes for posterity:

    1. In entry #1, [tex]A_2 = A (\frac {2 v_1}{v_2 + v_1})[/tex] should be [tex]A_2 = A (\frac {2 v_2}{v_2 + v_1})[/tex]

    2. This thread effectively continues under a different name ("wave momentum") here https://www.physicsforums.com/showthread.php?t=281995
     
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