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

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

 

[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!

 

[leonidas24]

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.)

 

[leonidas24]

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.

 

[tom.stoer]

Fractions (and units)?

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

Please check you calculations carefully.

 

[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

 

[Bob S]

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 E_laser head-on at a high-energy electron beam of energy

E_e = γm_ec²

the backscattered Compton photon energy is roughly

E_backscatter = 4γ² E_laser. 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

 

[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.