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

Photon propogation

  1. Apr 28, 2010 #1
    As I was reading a post on 'light' I got to thinking. The propagation of the photon, is it the original photon? or a 'dasiey chain' of interactions that results in it not being the original photon?

    I'm most likely off on a number of points here so please do correct where needed. Here was the thought, as the change in an energy state creates the photon and the photon has no rest mass. Does this mean that you could look at it as that the photon is a disruption to the 'field' that connects everything? it would to me explain the no rest mass as at rest there is no 'point' existence, once 'disrupted' and the "ripple" propagates the mass is related to the energy state of that change.

    at first glance it sounded like the aether idea but that's not at all what I'm thinking of, I got an image of billard balls and the interaction of them as a way to picture the photon but then suddenly thought, wait what about looking at it as the billard balls are the changes in energy and with each contact of a ball is the possible photon creation. taking it a step further the energy from the sun for example wouldn't in reality be traveling as you'd imagine an object moving along but it would be more of a chain reaction that propagates in an interaction between atoms.

    So the above confusing description would be that the photon as described as the smallest packet of 'energy' is just that, it's the EM field itself.

    I wish I could get my thoughts out on paper with clarity they have in my mind. I know the above is disorganized and I could spend another couple hours rewriting it to fit what I'm thinking. so my apologies to few who took the time to read this. If you're wondering it's a neurological disorder that does this. Oh and this post took me 1hr to write too... but 30secs to come up with.
     
  2. jcsd
  3. Apr 29, 2010 #2

    clem

    User Avatar
    Science Advisor

    It depends on the wave length lambda of the photon. If lambda is large compared to the molecular spacing, then simple physical optics applies. This is the case for visible light.
    If lambda is smaller, then there is a sequence of collisions. Whether you consider the scattered photon as the original photon, just scattered, or a new photon after absorption is unimportant, since all photons are identical particles.
     
  4. Apr 29, 2010 #3
    I'm approaching it as that the photon is not a particle but an 'effect' of particle interaction. As the photon can be created from those interactions it would be as though the 'field' or 'fabric' is everywhere and that the photon is that 'field'.

    I see it as geometric construct. I just took a different frame of reference and approached it as though the photon isn't a particle. And this is where it gets hard to explain how I see it, the terms field, fabric etc.. only fall short as they are tied to 2~3 dimensions whereas I see it as another dimension.
    When using the term particle for photon should not the interactions follow the same pattern as other particles? and wouldn't also explain the inverse square distance of the EM field?

    I'm going to have to do some more reading to help bridge my thoughts to the language used here to try an avoid cross talk.
     
  5. Apr 30, 2010 #4

    Born2bwire

    User Avatar
    Science Advisor
    Gold Member

    The inverse square law is explained using simple geometry. If we send out a spherical wave front, then we can imagine the wave front traveling away from our isotropic source as a uniformly expanding spherical shell. If we are in a lossless medium, then the energy density (1/m in this case since we have an infinitesimally thin shell) over the entire wave front cannot change. So if the energy is evenly distributed across the wave front, then the energy density must decrease as the sphere expands so that the total energy when derived by integrating across the surface of the sphere stays the same. It turns out for this to be true, the intensity must drop off by 1/r^2 (since if we assume a constant amplitude across the wave front then the energy would be related to |A|^2*4\pi*r^2).

    As for particle, the term particle carries a different meaning in quantum mechanics than it does in classical physics. Particles in quantum mechanics behave with both classical wave properties and classical particle properties. I think clem hits on the more meaningful point though, photons are indistinguishable particles. We do not allow ourselves to say that it is a new photon or the original photon regardless of any annihilation and creation events that may have occurred between point A and B.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Photon propogation
  1. Propogation Amplitude (Replies: 2)

Loading...