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Doesn't wavelength require a medium?

  1. Sep 30, 2004 #1
    In audio, I've learned that sound energy are particles being compressed and rarified at specific rates, causing a "wave of pressure". Thus, sound is represented in a wavelength because of the medium it is "distorting"

    My barely-elementary understandinf of QED makes me believe that in order for an electron to have a wavelength, it is attracted to the nucleus and repelled by antiparticle collisions between the electron and the nucleus. Is this true or did I misread something? Also, how can light travelling through a vacuum have a wavelength at all? Or is my definition of wavelength convoluted?
     
  2. jcsd
  3. Oct 1, 2004 #2
    wavelength

    So here is a problem that bothered scientists for a long time - how can light travel without a medium.

    Well, its just that some waves require a medium for propagation and some don't. Those who do are called mechanical waves.

    Let us take sound waves. They require a medium because they travel by oscillations of the praticles of the medium. But in light ( and other electromagnetic waves ) the oscillations are of Electric and Magnetic fields which don't require a medium at all. Although they can travel in machanical mediums.

    I am not sure about what varies in an electron but I think it is the probability of finding it at a certain position.

    spacetime
    www.geocities.com/physics_all/index.html
     
  4. Oct 1, 2004 #3

    selfAdjoint

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    After Maxwell and before Einstein's photon, physicists thought they had a good handle on how light could propagate without a medium; there were the electric and magnetic fields (you could visualize "lines of force" a la Faraday if it helped you) and changes in one generated changes in the other, in accordance with Maxwell's equations. So the changes would propagate, and the equations said the propagation would travel at a number which, based on measurements, looked very close to the measured speed of light. This was simple and natural, and IT IS STILL AVAILABLE. Although Maxwell EM is now just an effective theory, it is still valid in its range of application, and though the electrowweak sector of the standard model has a much more difficult version of the subject, that version has to be captured at a certain level of detail by the old view. Engineers working with wave guides can "see" (very clearly measure) the electric and magnetic maxes coming down the pipe.
     
  5. Oct 1, 2004 #4

    russ_watters

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    So, does that explain how Einstein may still call them an "ether" (from another thread)? The fields are a medium for the propagation of light, its just that its self-creating/sustaining and not tied to anything else? Just trying to get my arms around the concept... I'm picturing a wave on a string that carries the string with it.
     
    Last edited: Oct 1, 2004
  6. Oct 1, 2004 #5
    I see... so the wavelength of a photon is really it's interchanging magnetic and electric fields?
     
  7. Oct 1, 2004 #6

    selfAdjoint

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    It would be the distance from an electric maximum (and magnetic minimum) through the next magnetic maximum and on to the next electric maximum agin. One full cyle of the fields.
     
  8. Oct 1, 2004 #7
    Interesting. So does this mean that when a photon with an extremely long wavelength passes a magnet, that will will not curve as sharp as a photon with an extremely short wave length?
     
  9. Oct 1, 2004 #8
    I don't think anyone said magnets curve photons. Magnets are known to curve the path of charged particle. A photon is not a charged particle, it is another field, an oscillating one. Nothing in EM theory predicts photon deflection if it crosses another field.
     
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