Does Compton Scattering Have No Effect on Electron Energy and Velocity?

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Compton scattering involves the interaction of a photon with a free electron, where the photon loses energy and momentum upon scattering. In the specific case of a photon with energy equal to twice the electron rest mass (Eγ = 2me), the energy and wavelength of the photon change, even when scattered at a 180-degree angle. The Klein-Nishina formula indicates that the outgoing photon has reduced energy compared to the incoming photon. Discussions highlight that energy conservation applies, and the scattering process is akin to elastic collisions. Overall, the energy of the photon and the electron is affected during Compton scattering, contradicting the initial assumption that there is no change.
leonidas24
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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. 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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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.
 
leonidas24 said:
... 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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tom.stoer said:
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.
 
Can you please explain? Wavelength and energy always change.
 
leonidas24 said:
Sorry, you're absolutely right. However, I'm talking about the specific case in which the photon energyE_\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.)
 
tom.stoer said:
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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Fractions (and units)?

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

Please check you calculations carefully.
 
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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  • #10
leonidas24 said:
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

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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  • #11
@leonidas24: note that the usual Compton formaula is valid only for scattering in the electron's rest frame. Bob_S' experimental is different.
 

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