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MathematicalPhysicist

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**Problem statement**

A photon of energy E0 head on with a free electron of rest mass m0 and speed v.

the photon is scattered at at 90 degrees.

find the energy E of the scattered photon.

**attemtp at solution**

the answer in the book is [tex] E=\frac{E_0(1+

\frac{v}{c})}{1+\frac{E_0}{E_i}}[/tex]

where [tex] E_i=\gamma*m_0c^2[/tex]

now to answer this i move to an inertial frame of the electron prior the collision.

which means the photon's energy in this system is [tex] E'_{ph}=\gamma*E_0(1+\frac{v}{c})[/tex]

now from conservation of energy and momentum and from the angle of 90 degrees between the electron and the photon after the collision, we get:

[tex]p'_{ph}^2=p_e^2+p_{ph,after}^2[/tex]

[tex]E'_{ph}+m_0c^2=E_{ph,after}+E_e[/tex]

and we can eliminate the energy of the electron after the collision, by writing

[tex]E'_{ph}^2/c^2-E_{ph,after}^2/c^2=p_e^2[/tex]

and then entering it the second equation:

[tex]E'_{ph}+m_0c^2=E_{ph,after}+\sqrt{E'_{ph}^2-E_{ph,after}^2+(m_0c^2)^2}[/tex]

from here after some algebraic manipulations, i get a quadratic equation for [tex]E_{ph,after}[/tex]

and solve for it, by taking the positive value of the two roots.

i just want to see if i got this correct, havent tried to solve the equation yet, seems a bit long, i just want to see if i got the physics correct?

thanks in advance.

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