Compton scattering of photons

  • Thread starter 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 [tex]E_{\gamma} = 2m_e[/tex]
 
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  • #2
phyzguy
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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.
 
  • #3
tom.stoer
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... 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

[tex]\Delta\lambda = \lambda_C(1-\cos\theta_\gamma)[/tex]

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

[tex]\Delta\lambda = \lambda_C(1-\cos\theta_\gamma)[/tex]

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 [tex]E_{gamma} = 2m_e[/tex]. I should have mentioned that earlier.
 
  • #5
tom.stoer
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Can you please explain? Wavelength and energy always change.
 
  • #6
phyzguy
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Sorry, you're absolutely right. However, I'm talking about the specific case in which the photon energy[tex]E_\gamma=2m_e[/tex]. 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:

[tex]\frac{Eout}{Ein}=\frac{1}{1+\frac{E_\gamma}{m_ec^2}(1-cos(\theta))}[/tex]

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.)
 
  • #7
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Can you please explain? Wavelength and energy always change.
Sure.

Initial photon energy: [tex]E = 2m_e[/tex]

So [tex]\lambda = hc/2m_e[/tex].

Using the compton formula, [tex]\lambda ' = h/m_e c(1-\cos\theta_\gamma) + \lambda [/tex], 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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  • #8
tom.stoer
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Fractions (and units)?

[tex]\frac{2}{1} + \frac{1}{2} = \frac{3}{2}[/tex]

Please check you calculations carefully.
 
  • #9
tom.stoer
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I think we agree that for 180 degrees scattering angle we have

[tex]\lambda^\prime = \lambda + \lambda_C(1-\cos\theta_\gamma) = \lambda + 2\lambda_c[/tex]

Now we use

[tex]\lambda = \frac{c}{\nu} = \frac{hc}{E_\gamma} =\frac{hc}{xm_ec^2} = \frac{1}{x}\lambda_C[/tex]

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

Then we get

[tex]\lambda^\prime = \frac{1}{x}\lambda_C + 2\lambda_c = \left(\frac{1}{x} + 2\right)\lambda_C[/tex]

So again: the energy always changes, even for backward scattering and your special choice of energy
 
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  • #10
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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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  • #11
tom.stoer
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@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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