Calculating Final Photon Energy in Inverse Compton Scattering

[erok81]

Homework Statement

Suppose that an X ray has initial energy E_γ=100keV, and the incident (relativistic) electron has energy E_e=100GeV. Compute the final energy of the photon E'_γ assuming the final direction of the photon makes an angle θ=Π with the initial direction.

For solving this problem, use the conservation of energy and each component of the momentum.

Homework Equations

None yet - still setting up problem.

The Attempt at a Solution

I am familiar with normal Compton scattering where a photon interacts with a rest electron, but not the inverse.

Maybe this works a certain way, but here is how I see it and what I want to verify.

I picture relativistic electron moving along the x-axis and getting rear ended by the x ray. The resulting gamma ray goes the other direction, still along x, and the electron continues moving along the x-axis with much less energy and momentum. At least that's what I hope because it makes the problem easier - although I am pretty sure that is wrong.

So a more realistic view is the electron is coming in at an angle and leave at an angle with the resulting photon going the opposite direction but still along the x-axis.

Is there a relationship between the initial angle and the resulting angle of the electron? I have it half setup but end up with way too many unknowns.

 

[erok81]

I worked a little more on this one and came up with the following. Since it is a sketch I just threw the equations on the attachment as well.

This diagram makes sense to me.

Any ideas?

 

[erok81]

I've narrowed the question down some and now it's more conceptual. I am missing something.

So I have the energy of the x-ray, and the energy of the relativistic electron. I don't know the final energy of the produced gamma ray. My equation is this for conservation of energy.

$$E_{\gamma} + E_{e} = E^{'}_{\gamma} +E^{'}_{e}$$

$$h\frac{c}{\lambda}+\gamma m_{e} c^{2} = h\frac{c}{\lambda'}+m_{e} c^{2}$$

I also know m_ec² since those are both known values. Since energy is conserved couldn't I just do this?


$$h\frac{c}{\lambda}+\gamma m_{e} c^{2} - m_{e} c^{2} = h\frac{c}{\lambda'}$$

The question states I need to use both energy and both components of momentum to solve.

What am I missing?