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Momentum of massive classical free field

  1. Jul 18, 2011 #1

    atyy

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    Momentum of "massive" classical free field

    A "massive" classical free field such as that given by the http://arxiv.org/abs/0809.1003" [Broken]. If the de Broglie relations are used, and v is the group velocity, the field can have momentum given by p=γmv. Can we assign such a field p=γmv without the de Broglie relations?
     
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  3. Jul 18, 2011 #2

    bcrowell

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    Re: Momentum of "massive" classical free field

    The relations m2=E2-p2 (for all m), p=mγv (for m>0) and E=mγ (for m>0) are purely classical, so the answer is yes. For example, in the m=0 case of the Proca equation, you have Maxwell's equations, which are classical, and m2=E2-p2 holds because E=p. E=p was known 50 years before quantum mechanics. One way to get E=p from Maxwell's equations is to calculate the momentum and energy absorbed by an Ohmic surface from an electromagnetic wave. All classical.

    Of course the bare wave equation, without physical context or interpretation, doesn't necessarily have energy and momentum in the sense that we usually mean in physics. For instance, you can use wave equations to describe how disturbances propagate through a school of fish. Let's say for the sake of argument that the wave equation happens to be the Proca equation. The disturbances will have "energy" and "momentum" that are conserved, that obey the usual relativistic relations, etc., but those won't be the same as the E and p we usually talk about in physics.

    In the case of the fish, there is no quantum mechanics, no Planck's constant, no probability interpretation, no normalization of wavefunctions. We wouldn't write E=ℏω and p=ℏk, because ℏ would be irrelevant. We would probably just associate E directly with ω, p with k. I think we would still be able to talk about things like the local density of "energy," or the transport of "momentum" from one place to another, but they wouldn't be the energy and momentum we talk about in physics.
     
    Last edited: Jul 18, 2011
  4. Jul 18, 2011 #3

    atyy

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    Re: Momentum of "massive" classical free field

    How? I usually associate energy=momentum=EXB with the massless Maxwell field, so for Proca, I expect energy=EXB+mA2. Classically, I don't associate free field energy-momentum with frequency or wavelength or velocity.
     
  5. Jul 18, 2011 #4

    bcrowell

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    Re: Momentum of "massive" classical free field

    How what?
     
  6. Jul 18, 2011 #5

    atyy

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    Re: Momentum of "massive" classical free field

    How do you associate p=γmv to a "massive" classical free field?
     
  7. Jul 18, 2011 #6

    bcrowell

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    Re: Momentum of "massive" classical free field

    Because it was derived in a way that was valid regardless of the context.
     
  8. Jul 18, 2011 #7

    atyy

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    Re: Momentum of "massive" classical free field

    So v is the speed of the wave?
     
  9. Jul 18, 2011 #8

    bcrowell

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    Re: Momentum of "massive" classical free field

    Associating v with the group velocity is the only thing that works. E.g., the phase velocity is always greater than c for m>0. Information also travels at the group velocity, so that also makes it necessary to interpret v as the group velocity, since relativity is basically about cause and effect relationships between events.
     
  10. Jul 18, 2011 #9

    atyy

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    Re: Momentum of "massive" classical free field

    And are you associating energy with frequency, and momentum with wave vector?
     
    Last edited: Jul 18, 2011
  11. Jul 18, 2011 #10

    bcrowell

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    Re: Momentum of "massive" classical free field

    No, not necessarily. E.g., in the m=0 case, E=p holds even classically.

    Here's a way of looking at it that may help. These are all linear wave equations. Therefore without context, there is no way to look at a particular wavefunction Ψ(t,x,y,z) and say that wavefunction Ψ has a certain E and p. I could always define some other wavefunction Φ=2Ψ and say that Φ was the "real" function, so E and p would be quadrupled.

    One way of pinning down this scaling is if the context is quantum mechanical. Then the de Broglie relations tell you the real E and p, and you can fix some convention for normalization, so that probabilities are defined.

    But there are other ways of fixing the scaling. For example, what Maxwell did was to examine the interaction of EM waves with charged particles. Since massive particles already had a normalization defined for their E and p, that fixed the normalization for the EM waves.
     
  12. Jul 18, 2011 #11

    atyy

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    Re: Momentum of "massive" classical free field

    The question isn't whether E2=p2+m2. It does, by definition.

    However, I don't see how to get p=γmv from a "massive" classical field, except via de Broglie.

    Here's my understanding:

    Classically, E2=p2+m2 where E and p are defined via field strength F and potential A, the definition being motivated by the action of the field on a test charge.

    Let u=dE/dp.

    This gives: u2=p2/(p2+m2).

    Rearranging: p=mu/(1-u2)1/2.

    However, this doesn't seem to be p=γmv, where v is the group velocity, unless E~w and p~k, so that u=dE/dp~dw/dk=v
     
    Last edited: Jul 18, 2011
  13. Jul 18, 2011 #12

    bcrowell

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    Re: Momentum of "massive" classical free field

    Let it interact with something that has known E and p, just like Maxwell did in the m=0 case.

    You said this, but then you never spelled out the details of what it meant, never did anything with it. What is the motivation for using words like "charge?" This isn't the electromagnetic field, so electric charge isn't relevant. Did you mean "charge" in a more general sense?
     
  14. Jul 19, 2011 #13

    atyy

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    Re: Momentum of "massive" classical free field

    Working out Lorentz force laws with Proca is too hairy for me. Let's try your school of fish approach in post #2. Isn't that a form of wave-particle duality? Let the fish move with uniform 4-velocity U. The wave propagating through the fish has wave 4-vector K. To assign "energy-momentum" P to the fish, let's define P=mU where m2 is K2. So U is a particle property, but the mass m to be thought of abstractly as a particle property is physically the dispersion relation of the wave-vector.
     
    Last edited: Jul 19, 2011
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