Compton Effect Conceptual Problem

In summary, the Compton effect involves a collision between a photon and an electron, where the unknowns after the collision are the outgoing photon's wavelength and direction, and the speed and direction of the electron. This can be described by three equations, but the lack of knowledge about the angle of deflection after the collision is related to a lack of knowledge about the internal geometry of the objects involved. The analogy of treating the objects as hard spheres may not fully apply in this scenario, and the phase of the incoming photon may also play a role in the scattering. Additionally, the concept of potential energy between a photon and an electron may need to be reconsidered in this situation.
  • #1
Opus_723
178
3

Homework Statement



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).

Homework Equations



(3-8)

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

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.
 
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  • #2
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.
 
  • #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.
 
  • #4
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?
 
  • #5


I can understand your confusion and frustration with this conceptual problem. The Compton Effect is a complex phenomenon that involves both particle-like and wave-like properties of light. In order to fully understand it, we need to consider both aspects and think about how they interact during the collision.

First, let's address your idea about angular momentum. In this case, we are dealing with a collision between a photon (a massless particle) and an electron (a massive particle). Angular momentum is typically associated with objects that have mass, so it may not be applicable in this scenario.

Next, let's consider your idea about the phase of the wave. This is actually a very important aspect to consider. The phase of a wave is essentially the position of the wave at a given point in time. In the case of the Compton Effect, the phase of the incoming photon and the phase of the scattered photon will be different due to the interaction with the electron. This can affect the direction and wavelength of the scattered photon.

In addition to the phase of the wave, we also need to consider the polarization of the photon. This is another property of light that can change during the collision and can affect the outcome.

Overall, the key to understanding the missing piece in this problem is to think about how the wave-like properties of light (such as phase and polarization) interact with the particle-like properties of the electron. This can help us understand the relationship between the unknowns in the equations and the lack of knowledge about the electron's direction and the scattered photon's wavelength.
 

1. What is the Compton Effect?

The Compton Effect is a phenomenon in physics where a photon (a particle of light) collides with an electron, causing the electron to recoil and the photon to lose energy. This effect was first discovered by Arthur Compton in 1923 and helps to explain the particle-like behavior of light.

2. How does the Compton Effect relate to quantum mechanics?

The Compton Effect is an important concept in quantum mechanics as it shows that light, which was previously thought to only exhibit wave-like behavior, also has particle-like properties. This helped to bridge the gap between classical and quantum physics and led to the development of the theory of quantum electrodynamics.

3. What is the equation for calculating the change in wavelength in the Compton Effect?

The equation for calculating the change in wavelength (Δλ) in the Compton Effect is Δλ = h/mc * (1 - cosθ), where h is Planck's constant, m is the mass of the electron, c is the speed of light, and θ is the angle of scattering.

4. How does the Compton Effect affect the energy and momentum of the photon and electron?

In the Compton Effect, the photon loses energy and momentum while the electron gains energy and momentum. This is due to the transfer of energy and momentum from the photon to the electron during the collision.

5. What practical applications does the Compton Effect have?

The Compton Effect has several practical applications, including medical imaging techniques such as computed tomography (CT) scans and positron emission tomography (PET) scans. It is also used in X-ray diffraction studies to determine the structure of molecules and crystals. Additionally, the Compton Effect is important in understanding the behavior of radiation in nuclear reactors and in the development of new technologies such as quantum computing.

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