The photon and Electromagnetic field

  1. How does an EM field, with both an electric and a magnetic component, each of which "disappears" produce an apparently steady photon?

    A point charge has an electromagnetic field it creates...A stationary observer with respect to the charge sees and electric field; an observer in motion sees a magnetic field. But photons, presumably bumps in the EM field, don't wink in and out of existence depending on our motion or the appearance of one field component or the other. .
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
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    Hi Naty1! :smile:

    I'm not sure I'm getting the point you're making …

    if you're saying that the observer in motion will see only a magnetic field, and no electric field, then that's wrong …

    the field he sees will still be mostly electric (E2 - B2 = E02) …

    there is no observer for whom the electric field disappears. :smile:
     
  4. Mentz114

    Mentz114 4,191
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    "Photons" only exist at the moment they are emitted or absorbed i.e. when they interact with matter. There is no evidence ( nor any way of getting any ) that photons exist in the EM field when it is not interacting with matter.

    Speculations about 'free' photons usually lead to apparent contradictions, as evidenced by your question.
     
  5. atyy

    atyy 10,100
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  6. Well that's contrary to another thread here...I don't remember which.....I'll see if I can find it... I clearly remember the (apparent) conclusion because prior to that my understanding was the same as yours...even so the magnetic field seems to appear and disappear...right??

    Well thats novel.... and difficult for messenger particles to comply with I should think! From Fabric of the Cosmos, Brian Greene:
    And what about the polarization of photons, (their spin directions), wave particle duality....photoelectric effect, etc,etc,etc...

    Had someone answered I was mixing quantum theory with classicial field theory I would have immediately understood....because there are always anomalies when trying to do that...

    Also, the answer may be related to the following:

    PARALLEL WORLDS, Michio Kaku, Pg 216
    (Without monopoles the equations look almost the same.)
     
  7. I found a quote, but it was used in another thread..still looking for the thread...

     
  8. tiny-tim

    tiny-tim 26,054
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    Hi Naty1! :smile:

    E2 - B2 is a relativistic invariant.

    (btw, so is E.B)

    So you can't have E = 0 in one frame and B = 0 in another. :wink:

    EDIT:
    The second half of that is correct. The first half should say "Also keep in mind that electric field in one reference frame has an additional magnetic field in another"
     
  9. I am confused. Do u say photon doesn't exist when they are not interacting with matter?
    Supposing photon is emitted from sun and reaches earth.So, you say photon exist when it is emitted by Sun and when it interacts with earth.In the mean time, photon doesn't exist between sun and earth.
     
  10. Jonathan Scott

    Jonathan Scott 1,342
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    A photon is a way of describing a quantized transfer of electromagnetic energy and angular momentum. However, while the energy is in flight, it isn't really a collection of independent particles but rather a flow of waves containing the total amount of energy. When those waves interact with something, they will transfer quantized amounts of energy of the appropriate frequency. If the energy density is very low, then there may be a one to one relationship between photons being emitted and received, but various physical effects such as interference demonstrate that one cannot assume that the emitted and received photons were actually in some sense the same particle.

    However, it's quite common to describe the flow of electromagnetic waves as a stream of photons anyway in cases where the distinction doesn't really matter.
     
  11. Jonathan posted..

    I'm not sure I agree entirely with the first sentence; however it's largely immaterial as we are into the wave (classical field theory) versus particle (quantum mechanical) viewpoint....The difficulty is akin to the double slit experiment...electromegnatic waves in some situations appear to be continuous and wavelike, in other situations discrete quantum like particle effects are manifested....it's a good concept to remember, but has been debated for about 90 years...
     
  12. DrGreg

    DrGreg 1,929
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    It's true that we can measure photons only by emitting them or absorbing them. But to conclude that they "do not exist" between those events seems a very perverse use of the English language.

    You say that there is no evidence that they exist, but the evidence is that they are emitted (i.e. come into existence!) and are absorbed (i.e. cease to exist!).

    I think what you meant to say is that we can't tell exactly where a photon is between its creation and destruction. Quantum theory tells us the photon takes all possible routes -- so it can be in many places at once -- and the best we can do is describe it via a wavefunction.
     
  13. Now that I can buy into!!!
     
  14. Here's a quote from wikipedia,
    http://en.wikipedia.org/wiki/Electromagnetic_field

    which supports Tiny tim's post....so I'm back to my original understanding...thanks tim...

    (my boldface added)
     
  15. So nobody has answered my original question yet...If a characteristic of the EM field is observer dependent, why not the photon?

    Rather odd...but maybe because the photon only moves at lightspeed, that locks in the characteristic which might otherwise change....
     
  16. tiny-tim

    tiny-tim 26,054
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    Hi Naty1! :smile:

    The EM fields which produce photons have E = B (and E.B = 0) (I think: :redface:) …

    and if E = B in one frame, then E = B in every frame …

    so, unlike an electron, a "photon-field" will look the same in any frame. :smile:
     
  17. Mentz114

    Mentz114 4,191
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    (my bolding).

    I think that strengthens my case. The wave function for a photon is not a Schroedinger equation in the normal sense. I quote from (1)

    So the wave function does not describe a particle, even in the loosest sense.

    (1)"The Maxwell wave function of the photon" M. G. Raymer and Brian J. Smith.
    In proc. SPIE conference Optics and Photonics, The Nature of Light: What is a Photon?" (San Diego, Aug. 2005)
     
    Last edited: Nov 4, 2008
  18. jtbell

    Staff: Mentor

     
  19. tiny-tim

    tiny-tim 26,054
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    oops!

    Hi jtbell! :smile:

    As you've probably guessed, I learnt my electromagnetism with c = 1, and I keep forgetting to put the c back in. :redface:

    hmm … then I suppose it's E2 - c2B2 which is invariant? :smile:
     
  20. DrGreg

    DrGreg 1,929
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    I'm out of my depth with the fine details of quantum theory, but if I understand you correctly you are saying that a photon behaves like a particle when it is emitted and absorbed, but not in between. Phrased like that, that makes sense, but I wouldn't say a photon "doesn't exist" between those events, just that it isn't very particle-like. (To my way of thinking, "wavefunction" and "photon" mean pretty much the same thing.)
     
  21. Sorry if I come into your discussion.
    If the photon is defined as the object representing the particle-like behaviour of light, and if that particle-like behaviour doesn't exist between source and detector, then the photon...shouldn't exist there.
    (I have put two "if" and a "should" because in the past I have almost been banished from this forum for having wrote exactly that statement (without "if" and with "does" instead of "should") :smile:).
     
    Last edited: Nov 5, 2008
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