Special Case of Compton Scattering

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Summary:

Is there anything special about a Compton scattering event in which the incident photon has the same energy (512 keV) as the electron at rest ?

Main Question or Discussion Point

Was just wondering if there's anything special about the physics of Compton scattering where the incident photon has the same energy (512 keV) as the electron at rest. Then: $$1−cosθ_d=\frac{E_0ΔE}{Ei Er}=\frac{E_0(E_0−Er)}{E_0E_r}=\frac{E_0}{E_r}−1=\frac{λ_r}{λ_0}−1 $$ $$⇒\frac{1}{E_r}\propto -\cos\theta_d$$ $$⇒\lambda_r\propto -\cos\theta_d$$ where ##E_0## and ##\lambda_0## are the energy and (Compton) wavelength of the stationary electron and ##E_r##,##\lambda_r## are the energy and wavelength of the scattered photon.

Might be relevant if the gamma ray laser ever becomes a reality.
 

Answers and Replies

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Suppose the gamma ray laser becomes a reality. Could it be used for imaging purposes and if so then wouldn't the simplified Compton scattering relationships prove useful ? Just speculating admittedly but Compton effect imaging is already 'real' science:

https://www.nature.com/articles/s41598-019-49130-z
 
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There's a result which shows that if the scattered photon has energy equal to the rest mass of the electron, then the trajectories of the scattered photon and the electron are at right angles. Scattering angle is given by ##sin\theta=v/c## where v is the post collision velocity of the electron.

As I imagine it, the 'reverse time' version of this collision is an electron (beam of electrons perhaps) colliding with a gamma ray photon (beam of gamma rays) at right angles. The gamma ray photons have energy equal to rest mass of electrons. The electrons are "stopped" (if that is possible ??) and the gamma photons pick up the energy/momentum and are deflected through an angle ##\theta## where ##sin\theta## = v/c and v is the pre-collision velocity of the electron(s).

Does that sound plausible in 'real' time ? I can't quite imagine how a "stopped" electron would behave!
 
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The electrons are "stopped" (if that is possible ??)
Why is this a problem? For all scatters there is a frame where the outgoing electron is at rest.
 
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Ok - so per scenario above, you fire beams of electrons and gamma ray photons (512 keV) at right angles to each other. What happens - does the electron 'beam' simply disappear ? One can imagine stationary electrons within the context of being bound to some atom but not suspended in mid air / mid vacuum as it were. Do they just fall to the ground under gravity or what ?

Alternatively if you want to carry out Compton scattering on 'free' rather than 'bound' electrons, is it possible to have 'free but stationary' electrons ? Stationary in the same way as bound electrons are 'stationary' ?
 
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What happens - does the electron 'beam' simply disappear ?
Why would it? If you see a beam of electrons moving past you at v=non-zero what is the problem with moving along the beam so v=0 relative to you?
 
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Why would it? If you see a beam of electrons moving past you at v=non-zero what is the problem with moving along the beam so v=0 relative to you?
In principle, no problem. Except that in the scenario described above the electrons are 'stopped' in the observer reference frame - he/she is not moving alongside the beam because the beam has been 'stopped' relative to him/her. The electrons are 'stationary' in a fashion identical to that of loosely bound electrons in an atom - the original target of Compton scattering experiments.

From: Libre Texts

Compton scattering refers to the scattering of light off of free electrons. Experimentally, it’s impossible to create a target of completely free electrons. However, if the incident photons have energy much greater than the typical binding energies of electrons to atoms, the electrons will be “knocked off” of the atoms by the photons and act as free particles. Therefore, Compton scattering typically refers to scattering of high energy photons off of atomic targets.
 
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Let's go step by step.

OK, so you don't have a problem with a stopped electron.
In a collision with billiard balls, do you have a problem with the final velocity of a billiard ball being zero?
What about a photon-billiard ball collision?
 

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