Special relativity - inverse Compton scattering

In summary, the inverse Compton scattering involves a particle with energy ##E## and mass ##m## colliding head on with a photon with energy ##E_\gamma##. The maximum energy the photon can have after being scattered is given by the formula: $$E'_\gamma=E_\gamma\frac{E+\sqrt{E^2-m^2}}{E_\gamma+E-\cos{\theta}(E_\gamma-\sqrt{E^2-m^2})}$$ The condition for maximum energy is on ##\cos{\theta}##, with the extreme cases being when ##E_{\gamma} = \sqrt{E^2-m^2}## and ##E_{\gamma} < \sqrt{
  • #1
Aleolomorfo
73
4

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.
 
Physics news on Phys.org
  • #2
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}##?
 
Last edited:
  • #3
TSny said:
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?
 
  • #4
Aleolomorfo said:
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.
 
  • Like
Likes Aleolomorfo
  • #5
Thank you very much indeed for your help!
 

Related to Special relativity - inverse Compton scattering

1. What is special relativity?

Special relativity is a theory developed by Albert Einstein in 1905 that describes the relationship between space and time. It explains that the laws of physics are the same for all observers in uniform motion, and that the speed of light is constant for all observers.

2. What is inverse Compton scattering?

Inverse Compton scattering is a phenomenon in which a high-energy photon collides with a low-energy electron, resulting in the transfer of energy from the photon to the electron. This process is important in understanding the production of high-energy radiation in astrophysical sources such as pulsars and active galactic nuclei.

3. How does special relativity relate to inverse Compton scattering?

Special relativity plays a crucial role in understanding inverse Compton scattering because it explains how energy and momentum are conserved in these interactions. The theory also helps to explain how the scattering angle of the photon changes in relation to the velocity of the electron.

4. What are some real-world applications of inverse Compton scattering?

Inverse Compton scattering has many practical applications, such as in medical imaging and radiation therapy. It is also used in particle accelerators to produce high-energy photons for research purposes. In astrophysics, it is used to study the properties of high-energy radiation emitted from various celestial sources.

5. How does inverse Compton scattering impact our understanding of the universe?

Inverse Compton scattering plays a crucial role in our understanding of the universe because it helps us to study the properties of high-energy radiation emitted from celestial sources. This process allows us to gather information about the composition, structure, and evolution of these objects, which in turn helps us to better understand the universe as a whole.

Similar threads

  • Advanced Physics Homework Help
Replies
4
Views
720
  • Advanced Physics Homework Help
Replies
0
Views
381
  • Advanced Physics Homework Help
Replies
2
Views
900
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
547
  • Advanced Physics Homework Help
Replies
1
Views
3K
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
2K
Replies
1
Views
941
  • Advanced Physics Homework Help
Replies
7
Views
2K
Back
Top