Understanding the Compton Effect: Derivation of the Equation

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Discussion Overview

The discussion revolves around the derivation of the Compton effect equation, focusing on the principles of conservation of energy and momentum. Participants explore the mathematical relationships involved in the derivation, particularly the origins of specific terms in the equation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the derivation of the Compton effect equation, specifically the term \(\sqrt{p^2+m^2}\) and its expansion to \((E-E' \cos \theta)^2 + m^2\).
  • Another participant explains that the term \(\sqrt{p^2+m^2}\) relates to the relativistic relationship between mass, momentum, and energy, suggesting that the source may use units where \(c=1\).
  • A different participant notes that the expansion likely arises from conservation of momentum, indicating that the direction of the incoming photon is typically aligned with the x-axis, leading to conservation of both x- and y-components of momentum.
  • This participant also clarifies that for a photon, the relationship \(E = pc\) simplifies to \(E = p\) in the specified units, linking the outgoing photon's momentum to its energy.
  • One participant expresses gratitude for the clarifications but still seeks further understanding of the origin of the radical term.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the origin of the radical term in the equation, and multiple interpretations of the derivation process are presented.

Contextual Notes

The discussion highlights potential variations in textbook approaches to deriving the Compton effect equation, which may lead to different interpretations of the steps involved.

The_ArtofScience
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The Compton effect does make sense conceptually but I'm having trouble with the actual derivation.

The equation starts out from the principles of conservation of energy and mass: [tex]E_{\gamma}[/tex]+[tex]E_{e}[/tex]=[tex]E_{\gamma'}[/tex]+[tex]E_{e'}[/tex]
which strangely proceeds into the form: [tex]\ E-E'}+m[/tex]=[tex]\sqrt{p^2+m^2}}[/tex] The [tex]\sqrt{p^2+m^2}}[/tex] actually expands into [tex](E-E'cos\theta)^2{}+m^2[/tex] when the RHS is squared

So my question is where does the [tex]\sqrt{p^2+m^2}}[/tex] come from and why does it expand to [tex](E-E'cos\theta)^2{}+m^2[/tex]?
 
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The_ArtofScience said:
where does the [tex]\sqrt{p^2+m^2}}[/tex] come from

The relativistic relationship between (rest) mass, momentum and energy is [itex]E^2 = (pc)^2 + (mc^2)^2[/itex]. Your source must be using units that make [itex]c = 1[/itex]. I assume this refers to the energy of the electron after the interaction (because the electron is usually assumed to be stationary before the interaction) so it really should read

[tex]E_{\gamma} - E^{\prime}_{\gamma} + m = \sqrt {(p^{\prime}_e)^2 + m^2}[/tex]


and why does it expand to [tex](E-E'cos\theta)^2{}+m^2[/tex]?

I assume the E and the E' refer to the outgoing photon (gamma). I'm sorry, different textbooks use different routes to derive the Compton-scattering equation and I'm too tired to try to guess which route your source uses. Does your source give any more details?
 
Looking at your second question again, that step probably comes from conservation of momentum. Usually in this derivation one let's the direction of the incoming photon be along the x-axis. Then the outgoing electron and photon both have x- and y-components of momentum. The x- and y-components are each conserved.

Also, for a photon, E = pc which in your units (with c = 1) becomes E = p. So that [itex]E^{\prime} \cos \theta[/itex] is probably the x-component of the outgoing photon momentum: [itex]p_x^{\prime} = p^{\prime} \cos \theta = E^{\prime} \cos \theta[/itex].
 
Thank you jtbell for clearing that up. I'm still curious where that radical came from, but for now I'm very satisfied with the explanations. =D
 

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