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Homework Help: Compton Effect Conceptual Problem

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    This isn't an assigned problem, but it's the on conceptual question in the chapter that's bugging me, regarding the Compton effect.

    In the Compton Effect....(other questions)....The incoming photon's wavelength [itex]\lambda[/itex] is assumed to be known. The unknowns after the collision are the outgoing photon's wavelength and direction, [itex]\lambda'[/itex] and [itex]\theta[/itex], and the speed and direction of the electron, u[itex]_{e}[/itex] and [itex]\phi[/itex]. With only three equations- two components of momentum conservation and one of energy- we can't find all four. Equation (3-8) [Given in the book, see below] gives [itex]\lambda'[/itex] in terms of [itex]\theta[/itex]. Our lack of knowledge of [itex]\theta[/itex] after the collision (without and experiment) is directly related to a lock of knowledge of something before the collision. What is it? (Imagine the two objects as hard spheres).

    2. Relevant equations


    [itex]\lambda'[/itex]-[itex]\lambda[/itex] = [itex]\frac{h}{m_{e}c}[/itex](1-cos([itex]\theta[/itex])

    3. The attempt at a solution

    When you have two hard spheres colliding, the angle is determined by whether it's a head-on collision or a glancing blow. Basically, they're not quite lined up the way we usually treat them mathematically (point masses on the x-axis). Or, you could have objects that aren't spheres, in which case the deflection angle is determined by their geometry (imagine a collision between a sphere and a triangle). It could get even more complicated if the spheres were spinning. So maybe angular momentum plays a role?

    None of these ideas seems helpful though. I can't think of what the "internal geometry" of a photon would be. Angular momentum... maybe, but I wouldn't know the first thing about what that would even mean in this case. Probably stretching the analogy too far.

    The only other idea I had wasn't connected to the spheres analogy at all. It just occurred to me, that, if we treated the incoming light as a wave, the only property we haven't talked about is the phase of the wave at the point and moment of the collision. Now, I have no idea how that would translate to a photon, or if it's even meaningful for a photon. But it seems to be the only thing you could really change or measure in any way.

    Mostly, I'm clueless.
  2. jcsd
  3. Feb 9, 2013 #2


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    Homework Helper

    you got the right idea with the imagining two objects as hard spheres. The angle after collision depends on whether it was a glancing blow or head - on.

    And of course, as you mentioned, the electron and photon are not actually hard spheres. If we take the photon to be a point particle, what happens to the potential energy of the photon as it gets very close to the electron? So now, what does the 'hard sphere' around the electron actually represent?

    About the phase of the incoming photon - uh that's an interesting idea. I am not that familiar with the QM theory of scattering (I only have had a short introduction to the subject). So I am not certain, but I think the phase of the incoming photon would not matter. Remember that QM is not deterministic (i.e. even if we know the initial state exactly, we will not be able to predict the outcome generally). So don't be surprised that even if we know the initial state exactly, the photon can still scatter in a random direction.
  4. Feb 10, 2013 #3
    Not sure what you mean by "potential energy of the photon as it gets very close to the electron." A photon doesn't have charge, so I'm not sure how you would define that.
  5. Feb 11, 2013 #4


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    well, for the scattering to occur, there must be some kind of short-range 'force' between the photon and electron. The scattering does not have to be caused by the electromagnetic 'force'. Hard-sphere scattering is really a particular model for the interaction. What does 'hard-sphere' scattering imply?
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