Compton Scattering Energy Equation?

[x(x-y)]

Could anybody point me in the right direction as to how one would derive the equation below?

 

[latex]E_{\gamma}'=\frac{E_{\gamma}}{1+\frac{E_{\gamma}}{m_0 c^2} \left(1- \cos \theta \right)}[/latex]

 

I read it in my lab manual along with the Compton Wavelength equation (which I know how to derive) and I'm wondering where it comes from.

 

Thanks.

Edited February 26, 2013 by x(x-y)

[mathematic]

Check "Compton scattering" using Google. The expression results from conservation of momentum and energy.

[swansont]

The expression results from conservation of momentum and energy.

 

Remembering that both the x and y components of momentum have to be conserved.

[x(x-y)]

I see, thanks. So after searching for specific terms I can find the derivations just by searching for "compton scattering" generically. Great.

[imatfaal]

Take your derivation of the compton scattering (which you say you are happy with) - at one point you will have

 

[latex]\lambda_{f} - \lambda_{i}= \frac{h}{m_{e}c}(1-\cos \theta)[/latex]

 

replace wavelength with hc/E ie planck relation

 

divide both sides by hc

 

subtract E_i from both sides

 

take reciprocals of each side

 

multiple top and bottom of RHS by E_i

 

clean up - and you have your equation. It's just a simple algebraic mix up of the standard equation and use of planck. Let me know if it doesnt work out when you try

 

 

 

 

[x(x-y)]

Yep that works - but you can skip a few of those last steps actually and get to the answer easily from

 

[latex]\frac{1}{E_2} - \frac{1}{E_1} = \frac{1}{m_e c^2} \left(1-\cos \theta \right)[/latex]

 

Anyway, thanks for the help.