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Do massless particles always have momentum in the direction of travel?

  1. Jul 21, 2006 #1

    NateTG

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    I'm not sure this is the right forum, but I wanted to ask the following (possibly bizarre) question:

    Is it implicitly assumed that zero rest mass particles have momentum in the same direction as they travel?
     
  2. jcsd
  3. Jul 21, 2006 #2
    Yes, but this is not an assumption. Momentum for zero rest mass particles is given by:

    p=hbar * omega / c

    where hbar is planck's constant divided by 2*Pi and omega is the angular frequency.
     
  4. Jul 21, 2006 #3

    NateTG

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    But isn't that a scalar quantity:
    [tex]\left|\vec{p}\right|=\hbar \omega c[/tex]

    The derivation I've seen is from special relativity and starts with energy (also scalar).
     
  5. Jul 23, 2006 #4
    Yes, that's true. Also true is that momentum is always in the direction of motion. The quantum hypothesis is that [itex] \vec{p} = \hbar \vec{k} [/itex]. The k vector is in the direction of motion, always, for any particle.
     
    Last edited: Jul 23, 2006
  6. Jul 23, 2006 #5
    how can momentum NOT be in the direction of motion?
     
  7. Jul 24, 2006 #6

    NateTG

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    I don't see what the problem is. For example, it seems like it should be possible to model gravity as being propogated by virtual particles that have linear momentum opposing the direction of travel.
     
  8. Jul 24, 2006 #7
    It's not possible. If you're talking about the static GMm/r^2 force, one can't view that force as being carried by virtual particles at all. Any virtual particles carrying momentum (hence energy) must take it away from the static body, draining energy away somehow. That doesn't happen. This is true in GR as well, so long as the situation is static.

    If the situation is not static, ie gravity waves, momentum (hence energy) is carried away from the body. Particles radiated from the body must hence carry momentum in the direction of motion. The only way to get around this would be to have particles come in from infinity carrying momentum in the opposite direction of their motion, which would be the ultimate example of non-locality.
     
  9. Jul 24, 2006 #8

    you should read more about force carying particles... static EM fields do have virtual photons just like gravitation might have virtual gravitons... though i don't understand how can momentum in a different direction might exist
     
    Last edited: Jul 24, 2006
  10. Jul 25, 2006 #9

    NateTG

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    Mostly this was supposed to be a, 'has anyone tried this' sort of post more than anything else.

    Not that I'm claiming that momentum-velocity seperation is necessarily a good idea, but stranger things have been successful in physics.
     
  11. Jul 29, 2006 #10

    Meir Achuz

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    It depends on what is meant by "momentum". In an EM field, the "canonical momentum" of a charged particle is given by
    p=mv\gamma+qA/c. In that case, p and v need not be in the same direction.
     
  12. Jul 29, 2006 #11
    Its quite possiible within the domain of special relativity. If the body whose momentum you seek is under stress then the momentum of the body need not be in the direction of motion. However the total momentum of a closed system is always in the direction of its velocity.

    Pete
     
  13. Jul 30, 2006 #12
    Well, it's a tricky question.. Usually, momentum WILL be in the direction of motion unless its an alternative force that is creating the motion while pushing it away aswell... ---> O< Have you tried taking a look at EMF papers?
     
  14. Jul 30, 2006 #13
    It seems clear that the OP is thinking about linear mechanical momentum mv since this is the quantity which says that a photon has a proper mass E^2 - (pc)^2 = 0. The p in that exxpression is linear mechanical momentum. If you put in the canonical momentum then the expression looses its meaning.

    Even if the OP was refering to canonical linear momentum then the momentum is still in the direction of the velocity. Notice that the photon has zero proper mass and zero charge so that

    p=mv\gamma+qA/c.

    is invalid. Notice also that a photon has no charge so that A will be zero.


    Pete
     
    Last edited: Jul 30, 2006
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