Compton scattering of photons

1. May 14, 2010

leonidas24

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 $$E_{\gamma} = 2m_e$$

Last edited: May 14, 2010
2. May 14, 2010

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.

3. May 14, 2010

tom.stoer

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!

Last edited: May 14, 2010
4. May 14, 2010

leonidas24

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.

5. May 14, 2010

tom.stoer

Can you please explain? Wavelength and energy always change.

6. May 14, 2010

phyzguy

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.)

7. May 14, 2010

leonidas24

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.

Last edited: May 14, 2010
8. May 14, 2010

tom.stoer

Fractions (and units)?

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

9. May 14, 2010

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

Last edited: May 14, 2010
10. May 14, 2010

Bob S

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

Last edited by a moderator: May 4, 2017
11. May 15, 2010

tom.stoer

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