Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confusion on Meaning of Momentum

  1. Jan 11, 2010 #1
    I do ask pardon if my question is stupid. I am but a mere graduate student.

    How does the definition of momentum "hang together coherently" between the de Broglie wavelength definition and the SR definition?

    My confusion comes as follows. p = (gamma) m v is one definition of momentum, where gamma is the 1 / root ( 1-(v/c)**2), m is mass, v is velocity in some reference frame. This is completely reasonable to me based on SR. It's also true that momentum can be described in terms of the de Broglie wavelength: lambda = h / p . This is satisfying in that it is consistent for both light and matter. How does the SR definition of momentum relate to the de Broglie theory?

    It seems that SR has no statement about the momentum of mass-less particles. Is the de Broglie statement more general, since it applies to both light and matter?

    Given this, what is momentum? A photon has zero mass and constant velocity; does momentum have any "meaning" in this realm? It seems like numerous experiments with regard to photons are explained in terms of photons "colliding" with different energies based different momentum, but this seems anthrapromorphic. It seems that the energy of a photon is 1-1 with momentum, such that any collision notion is silly.

    And I have no idea how momentum hangs together with Noether's theorem now; momentum is a conserved quantity due to the symmetry of the system there. Is Noether's definition better than all of these?

    What is momentum?
  2. jcsd
  3. Jan 11, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That definition of momentum is useful for massive particles only. For massless particles, like photons, you need a different relationship.
    The momentum used to calculate the de Broglie wavelength of a particle is the same momentum as defined in SR.

    Sure it does. For massless particles, the relationship between energy and momentum is E = pc. The general relationship for any particle is:

    [tex]E^2 = m^2c^4 + p^2c^2[/tex]
  4. Jan 11, 2010 #3
    "The momentum used to calculate the de Broglie wavelength of a particle is the same momentum as defined in SR. "

    What is the definition of momentum in SR for mass-less particles? I know of none.

    "Sure it does. For massless particles, the relationship between energy and momentum is E = pc. The general relationship for any particle is: E^2 = m^2c^4 + p^2c^2 "

    That's not a definition of momentum, unless you are using circular logic of momentum being energy. That just gives the energy in terms of momentum, but not what momentum is.
  5. Jan 11, 2010 #4

    Doc Al

    User Avatar

    Staff: Mentor

    The Einstein relation between energy and momentum can be found using the usual definition of relativistic momentum (p = γmv) which you didn't seem to have a problem with. While you can't use that definition of relativistic momentum directly to get the momentum of a massless particle (since plugging in m = 0 and v = c gives p = 0/0), you can rearrange things to get the Einstein relation that I gave. (You'll need to combine it with E = γmc².)
  6. Jan 11, 2010 #5
    Perhaps I am missing something; it seems like we're not talking on the same point.

    E**2 = m**2 c**4 + p**2 c**2

    I am not asking how to derive this equation, which seems to be what you wrote about. I tried to say that this doesn't answer the question. You seemed to answer that the momentum of a mass-less particle is given by this equation, which gives the energy. OK, then how would you get the energy in the first place in SR for a mass-less particle? The only relation of which I know is the de Broglie relationship, which does not come out of SR. Is there an SR relationship for energy of mass-less particles that doesn't involve momentum? Or is there an SR relationship for mass-less particles that doesn't involve energy? You need energy or momentum from something measurable to get started.
  7. Jan 11, 2010 #6
    [tex] E^2 = m^2c^4 + p^2c^2 [/tex] is part of the geometry of space-time.

    Bring out QM. Plug in m=0 and the above relation shows that

    [tex]p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}.[/tex]

    For a photon the general energy expression of SR is in fact equivalent to the de Broglie relationship. The photon can't have rest energy because it is moving at c, always, never at rest.
    Last edited: Jan 11, 2010
  8. Jan 11, 2010 #7
    "Bring out QM" I assume that means that invoke a QM observation, E=hw. This is the very point about which I am confused: is there no way using SR alone to describe mass-less particles? SR is otherwise completely stand-alone from QM, as far as I know, doesn't require QM at all...is this the one area in which it is incomplete in its breadth? Or is there an SR way to derive the de Broglie relationship?
  9. Jan 11, 2010 #8
    I see your confusion now. The relationship between E and f is in de Broglie's own PhD thesis, and leads to the de Broglie relation by SR. De Broglie even offers a line of reasoning to the proportionality within SR. However, it is easier thought of as a fact of nature which Planck observed in the flux of blackbody radiation and which Einstein observed in the photoelectric effect (and subsequently earned a Nobel Prize for this work); not unlike that other fact of nature, F=ma. We call it the Planck hypothesis, and it IS the quantum revolution. If you are still confused about momentum (of zero rest-mass particles), read Einstein's box experiment, or learn Special Relativity.
    Last edited: Jan 12, 2010
  10. Jan 12, 2010 #9
    I would like to confirm the following.
    E^2 = m^2c^4 + p^2 c^2 dividing by hbar^2, we get ω-k relation of de Broglie wave,
    ω^2 = ω0^2 + k^2 c^2 where ω0= mc^2/hbar "rest frequency"
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook