# Compton Scattering of a photon by a moving electron

1. Dec 22, 2011

### athrun200

1. The problem statement, all variables and given/known data
If the total energy of the electron E>>m$c^{2}$
Prove that the formula of wavelength would be
$\lambda '$=$\frac{hc}{E}$(1+$\frac{m^{2}c^{4}\lambda}{4hcE}$)
where $\lambda '$ is the wavelenght after scattering
m is the mass of electron

2. Relevant equations

3. The attempt at a solution
I have proved that $\lambda '$=$\frac{\lambda(E-Pc)+2hc}{E+Pc}$
In my equation, P is the initial momentum of the electron.
I haven't use the given assumption E>>m$c^{2}$. So we must use it to eliminate P.

But if E>>m$c^{2}$$\Rightarrow$ E$\approx$Pc

2. Dec 22, 2011

### vela

Staff Emeritus
There's no angular dependence? The problem seems a bit ambiguous to me. What are the directions of the photon and electron initially?

3. Dec 23, 2011

### athrun200

they are both along x axis, even after collision

The origin question should be

Last edited: Dec 23, 2011
4. Dec 23, 2011

### vela

Staff Emeritus
OK, you have
$$\lambda' = \frac{\lambda(E-Pc)+2hc}{E+Pc} = \frac{2hc}{E+Pc} + \lambda\frac{E-Pc}{E+Pc}$$Now multiply the second term by (E+Pc)/(E+Pc) and then use the approximation $E\approx Pc$.

5. Dec 23, 2011

### athrun200

The second term will become zero after using E=Pc, since the second term has the factor E-Pc.

6. Dec 23, 2011

### vela

Staff Emeritus
That's not quite true. You need to multiply by (E+Pc)/(E+Pc) and simplify the numerator and then let E=pC.

Last edited: Dec 23, 2011
7. Dec 23, 2011

thx!