Understanding the Compton Effect: Derivation of the Equation

In summary, the Compton effect can be derived from the principles of conservation of energy and mass, using the relativistic relationship between mass, momentum, and energy. The equation involves the expansion of the square root of the sum of p^2 and m^2, which can be explained by conservation of momentum and the equivalence of energy and momentum for photons. Different textbooks may use different routes to derive the Compton-scattering equation, but they all ultimately lead to the same result.
  • #1
The_ArtofScience
83
0
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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  • #2
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?
 
  • #3
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].
 
  • #4
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
 

1. What is the Compton Effect?

The Compton Effect, also known as the Compton Scattering, is a phenomenon in which a photon (electromagnetic radiation) collides with a free electron and loses some of its energy, causing its wavelength to increase.

2. Who discovered the Compton Effect?

The Compton Effect was discovered by American physicist Arthur H. Compton in 1923, for which he was awarded the Nobel Prize in Physics in 1927.

3. How is the Compton Effect derived mathematically?

The Compton Effect can be derived using the principles of conservation of energy and momentum. The equation, derived by Compton, is Δλ = (h/mc)(1-cosθ), where Δλ is the change in wavelength, h is Planck's constant, m is the mass of the electron, c is the speed of light, and θ is the angle of scattering.

4. What is the significance of the Compton Effect?

The Compton Effect is significant because it provides evidence for the particle nature of light, as opposed to the previously accepted wave-like nature. It also helped to confirm Einstein's theory of relativity and has many practical applications in fields such as medical imaging.

5. How does the Compton Effect differ from the photoelectric effect?

The Compton Effect and the photoelectric effect are both phenomena that involve the interaction of photons and electrons. However, the Compton Effect results in the scattering of the photon and the emission of a lower energy photon, while the photoelectric effect results in the emission of an electron from a surface when a photon of sufficient energy is absorbed. Additionally, the Compton Effect can occur with any type of electromagnetic radiation, while the photoelectric effect specifically involves visible or ultraviolet light.

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