Compton Scattering of a photon by a moving electron

[athrun200]

Homework Statement

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 wavelength after scattering
m is the mass of electron

Homework Equations

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
this cannot get the answer

 

[vela]

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

 

[athrun200]

they are both along x axis, even after collision

The origin question should be

Consider Comton scattering of a photon by a moving electron. Before the collision the photon has wave length λ and is moving in the +x direction, and the electron is moving in the -x direction with total energy E. The photon and electron collide head-on. After the collision both are moving in the -x direction.
(a) Drive an expression for the wavelength λ' of the scattered photon.
(b) Show that for the case E>>mc^2, the result reduces to $\lambda '$=$\frac{hc}{E}$(1+$\frac{m^{2}c^{4}\lambda}{4hcE}$

 

[vela]

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 $E\approx Pc$.

 

[athrun200]

OK, you have
$$\lambda' = \frac{\lambda(E-Pc)+2hc}{E+Pc} = \frac{2hc}{E+Pc} + \lambda\frac{E-Pc}{E+Pc}$$

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

 

[vela]

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

 

[athrun200]

thx!