Special relativity - inverse Compton scattering

[Aleolomorfo]

Homework Statement

In the inverse Compton scattering there is a particle, with energy $E$ in the laboratory frame and mass at rest $m$, which collide head on with a photon with energy $E_\gamma$. Finding the maximum energy the photon can have after being scattered.

The Attempt at a Solution

My result for the energy of the scattered photon is:
$$E'_\gamma=E_\gamma\frac{E+\sqrt{E^2-m^2}}{E_\gamma+E-\cos{\theta}(E_\gamma-\sqrt{E^2-m^2})}$$
$\theta$ is the scattering angle of the photon (I hope the formula is correct without calcus mistakes)
My doubt is about the condition to impose in order to get the maximum energy. The energy is maximum when the denominator is minimum and so the condition is on $\cos{\theta}$. I would impose $\cos{\theta}=1$ and so $\theta=0$, but my sixth sense is telling me that this is not the correct answer.

 

[TSny]

Your formula looks correct to me.

What happens if $E_{\gamma} = \sqrt{E^2-m^2}\,\,$? (What's special about the lab frame in this case?)
What happens if $E_{\gamma} < \sqrt{E^2-m^2}\,\,$?

What do you get for the maximum final photon energy for the extreme case where $E \gg m$ and $E > E_{\gamma}$?

 

[Aleolomorfo]

Your formula looks correct to me.

What happens if $E_{\gamma} = \sqrt{E^2-m^2}\,\,$? (What's special about the lab frame in this case?)
What happens if $E_{\gamma} < \sqrt{E^2-m^2}\,\,$?

What do you get for the maximum final photon energy for the extreme case where $E \gg m$ and $E \gg E_{\gamma}$?

If $E_\gamma = \sqrt{E^2-m^2}$ the total momentum of the system is zero and so the lab frame coincides with the CM frame. In this case the energy is indipendent of the angle and so I can say that the maximum energy is $E'_\gamma=E_\gamma\frac{E+\sqrt{E^2-m^2}}{E_\gamma+E}$

If $E_\gamma < \sqrt{E^2-m^2}$ the denominator is minimum when $\cos{\theta} = -1$ and so when $\theta=\pi$ (back-scattering).

To sum up, I can't say anything about the maximun energy without knowing the relation between $E_\gamma$ and the momentum of the particle, can I?

In the extreme case I've found that if $E_\gamma \ge \sqrt{E^2-m^2}$ the maximum energy is $E'_\gamma\simeq E_\gamma$, on the other hand if $E_\gamma < \sqrt{E^2-m^2}$ the maximum energy is $E'_\gamma\simeq E$.

Is it right what I've done?

 

[TSny]

If $E_\gamma = \sqrt{E^2-m^2}$ the total momentum of the system is zero and so the lab frame coincides with the CM frame. In this case the energy is indipendent of the angle and so I can say that the maximum energy is $E'_\gamma=E_\gamma\frac{E+\sqrt{E^2-m^2}}{E_\gamma+E}$

Yes. This expression will simplify nicely.

If $E_\gamma < \sqrt{E^2-m^2}$ the denominator is minimum when $\cos{\theta} = -1$ and so when $\theta=\pi$ (back-scattering).

Yes

To sum up, I can't say anything about the maximun energy without knowing the relation between $E_\gamma$ and the momentum of the particle, can I?

Right. You have the three cases where the initial 3-momentum, $p_{\gamma}$, of the photon is greater than, less than, or equal to the initial 3-momentum, $p$, of the particle. Check to see if the following statements are true:

1. For $p_{\gamma} > p$, the photon will lose energy if it is scattered at any nonzero angle.
2. For $p_{\gamma} < p$, the photon gains energy if it is scattered at any nonzero angle and it gains the most energy for back-scattering.
3. For $p_{\gamma} = p$, the photon maintains its initial energy for any angle of scattering.

In the extreme case I've found that if $E_\gamma \ge \sqrt{E^2-m^2}$ the maximum energy is $E'_\gamma\simeq E_\gamma$, on the other hand if $E_\gamma < \sqrt{E^2-m^2}$ the maximum energy is $E'_\gamma\simeq E$.

That looks good.

 

[Aleolomorfo]

Thank you very much indeed for your help!