# Compton scattering of photons

A quick question regarding compton scattering: if we consider the situation in which a photon incident on a free electron is scattered through an angle of 180 degrees, its energy essentially does not change. Since energy must be conserved, I assume this means there is no effect whatsoever on the energy, momentum, or velocity of the electron? Seems counter-intuitive somehow...

EDIT: This is only for the case that $$E_{\gamma} = 2m_e$$

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phyzguy
The photon's energy certainly does change. Compton scattering is essentially elastic scattering, like billiard balls. So if a photon is back-scattered off a stationary electron, it imparts energy and momentum to the electron. So the photon loses energy, and its frequency decreases.

tom.stoer
... a photon ... is scattered through an angle of 180 degrees, its energy essentially does not change
But that's wrong; look at the change of the wavelength as a function of the photon scattering angle

$$\Delta\lambda = \lambda_C(1-\cos\theta_\gamma)$$

So its wavelength and therefore its energy changes!

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But that's wrong; look at the change of the wavelength as a function of the photon scattering angle

$$\Delta\lambda = \lambda_C(1-\cos\theta_\gamma)$$

So its wavelength and therefore its energy changes!
Sorry, you're absolutely right. However, I'm talking about the specific case in which the photon energy $$E_{gamma} = 2m_e$$. I should have mentioned that earlier.

tom.stoer
Can you please explain? Wavelength and energy always change.

phyzguy
Sorry, you're absolutely right. However, I'm talking about the specific case in which the photon energy$$E_\gamma=2m_e$$. I should have mentioned that earlier.
Why do you think that special case is "special"? If I look at the Klein-Nishina formula, I find:

$$\frac{Eout}{Ein}=\frac{1}{1+\frac{E_\gamma}{m_ec^2}(1-cos(\theta))}$$

This says that in the specific case you mentioned the outgoing photon has 1/5 the energy of the incoming, no?

(Not sure why the tex in you're quote is messed up.)

Can you please explain? Wavelength and energy always change.
Sure.

Initial photon energy: $$E = 2m_e$$

So $$\lambda = hc/2m_e$$.

Using the compton formula, $$\lambda ' = h/m_e c(1-\cos\theta_\gamma) + \lambda$$, with \theta = 180:

lambda' = 2h/m_e c + hc/2m_e = h/m_e (2/c + c/2) = 2h/m_e c

EDIT: This is ridiculous. Latex refuses to work.

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tom.stoer
Fractions (and units)?

$$\frac{2}{1} + \frac{1}{2} = \frac{3}{2}$$

tom.stoer
I think we agree that for 180 degrees scattering angle we have

$$\lambda^\prime = \lambda + \lambda_C(1-\cos\theta_\gamma) = \lambda + 2\lambda_c$$

Now we use

$$\lambda = \frac{c}{\nu} = \frac{hc}{E_\gamma} =\frac{hc}{xm_ec^2} = \frac{1}{x}\lambda_C$$

where x means the fraction of the electron's rest energy. In your case x=2.

Then we get

$$\lambda^\prime = \frac{1}{x}\lambda_C + 2\lambda_c = \left(\frac{1}{x} + 2\right)\lambda_C$$

So again: the energy always changes, even for backward scattering and your special choice of energy

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A quick question regarding compton scattering: if we consider the situation in which a photon incident on a free electron is scattered through an angle of 180 degrees, its energy essentially does not change.
If you shoot a photon (or laser beam) of energy Elaser head-on at a high-energy electron beam of energy

Ee = γmec2

the backscattered Compton photon energy is roughly

Ebackscatter = 4γ2 Elaser. See Eqn (2) in

http://physics.princeton.edu/~mcdonald/examples/accel/aoki_nim_a516_228_04.pdf [Broken]

This is because there are two Lorentz transformations from the lab to center-of-mass coordinates and back. See Sections 37.1 and 37.2 in

http://pdg.lbl.gov/2002/kinemarpp.pdf

Bob S

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tom.stoer