# Inverse compton effect.

Gold Member
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 $$E=\frac{E_0(1+ \frac{v}{c})}{1+\frac{E_0}{E_i}}$$
where $$E_i=\gamma*m_0c^2$$
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 $$E'_{ph}=\gamma*E_0(1+\frac{v}{c})$$
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:
$$p'_{ph}^2=p_e^2+p_{ph,after}^2$$
$$E'_{ph}+m_0c^2=E_{ph,after}+E_e$$
and we can eliminate the energy of the electron after the collision, by writing
$$E'_{ph}^2/c^2-E_{ph,after}^2/c^2=p_e^2$$
and then entering it the second equation:
$$E'_{ph}+m_0c^2=E_{ph,after}+\sqrt{E'_{ph}^2-E_{ph,after}^2+(m_0c^2)^2}$$
from here after some algebraic manipulations, i get a quadratic equation for $$E_{ph,after}$$
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?

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Gold Member
anyone?

Astronuc
Staff Emeritus
I think the approach is correct, but I haven't done such a problem in a such a long time, I'd have to think about it.

The electron is heading toward the photon with a speed v, so in the electrons rest frame, it would see a slightly higher frequency photon by virtue of the Doppler effect.

Are you sure about the 90° angle between electron and photon? The problem statement indicates that the scattered photon is scattered 90° from the original photon if I'm reading it correctly, so does this translate to 90° angle between electron and photon.

Have you worked it out yet?

Another approach would be to look at px and py before and after, since the resulting photon travels at 90° (y direction) with respect to the original direction (x direction), so the py of the photon must be equal in magnitude to the py of the electron.

This might be of interest - http://venables.asu.edu/quant/proj/compton.html

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Gold Member
yes you are ofcourse correct it should be 90 degrees between the direction of the initial photon.

anyway here's what i got:
equations of energy and momentum:
E_gamma-energy of the initial photon in the rest frame of the initial electron
E-the energy of the photon after the collision.
P-the momentum of the photon after the collision.
P_e-the momentum of the electron after the collision.
E_e-the energy of the electron after the collision.
$$E_e=\gamma(1+\frac{v}{c})E_0+m_0c^2-E$$
$$P_e^2=P^2+P_\gamma^2$$
$$(m_0c^2)^2=E_e^2-(P_ec)^2=(\gamma(1+\frac{v}{c})E_0+m_0c^2-E)^2-E^2-(\gamma(1+\frac{v}{c})E_0)^2$$
after rearranging i get:
$$E=\frac{E_0\gamma(1+\frac{v}{c})}{1+\frac{E_0\gamma(1+\frac{v}{c})}{m_0c^2}}$$
which is different from the book, this i have a mistake somewhere in reasoning?

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Astronuc
Staff Emeritus
Just a thought. In the rest frame of the electron, the problem is simply one of Compton scattering, and so the results of that should be consistent. Then one has to transform that result back to the initial reference frame.

Gold Member
well, even this doesnt seem to work.
cause presumably we should get:
$$E=\gamma*(E'-v/c*E')=\gamma*E'(1-v/c)$$
after rearranging i still don't get the answer in the book, i.e i get something like this
$$E'=\frac{E_0(1+\frac{v}{c})}{1-\frac{v}{c}+\frac{E_0}{\gamma m_0c^2}}$$

Gold Member
eventually i solved it by doing all the calculations in the lab system, quite straightforward.

thanks.

Astronuc
Staff Emeritus