How can I eliminate phi and u to derive the Compton effect equation?

In summary, the conversation discusses deriving the Compton effect equation and the use of momentum and energy conservation equations. The approach involves solving for momentum and energy in terms of their components, and using algebraic manipulation to eliminate variables and arrive at the final result. The process can be simplified using natural units and the link provided offers a helpful approach.
  • #1
b2386
35
0
Hi everyone,

Right now, I am working on a homework problem asking me to derive the Compton effect, which is given by [tex]\lambda\prime-\lambda=\frac{h}{m_ec}(1-cos\theta)[/tex]

A diagram of the situation can be found here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compton.html#c1 (However, in my diagram, the recoil electron and scattered photon are switched with the electron and [tex]\phi[/tex] below the x-axis and the photon and [tex]\theta[/tex] above the x axis.)

To derive the equation, I use the momentum conservation equations and energy conservation equation.

There are two equations for the momentum, one for the x component and one for the y component:

x-component: [tex]\frac{h}{\lambda}=\frac{h}{\lambda\prime}cos\theta+\gamma m u cos\phi[/tex]

y-component: [tex]0=\frac{h}{\lambda\prime}sin\theta-\gamma mu sin\phi[/tex]

where [tex]\gamma=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]
m=electron mass
u=electron velocity after the collision

My equation for energy conservation is:

[tex]h\frac{c}{\lambda}+mc^2=h\frac{c}{\lambda\prime}+\gamma m c^2[/tex]


To start off, I solved the x component of the momentum for cos(phi) and the y component for sin(phi). I then added the equations together and got (since sin^2 + cos^2 = 1):

[tex]\frac{h^2}{\lambda\prime^2\gamma^2m^2u^2}+\frac{h^2}{\lambda^2\gamma^2m^2u^2}-\frac{2h^2cos\theta}{\lambda\prime\lambda\gamma^2m^2u^2}=1[/tex]

At this point, I am to now eliminate u using the resulting equation and the equation for the energy conservation as given above. However, I am stuck on this next step because it seems as though any algabraic manipulation I try makes the equation extremely complicated. Could anyone please give me some hints as to what step should be next?
 
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  • #3
The easy way to do it is (in natural units):
1. Conservation of 4-momentum is k-k'=p'-p, where p=(m;0).
2. Square each 4-vector, getting 2kk'(1-cos)=2mE'-2m^2.
3. Conservation of energy is E'-m=k-k', so kk'(1-cos)=m(k-k').
4. Divide by kk' and use lambda=2pi/k. Put in hbar/c if you want.
 
  • #4
Thanks for your replies. However, in the problem wording, it specifically tells us to start by eliminating "phi" first and then eliminate "u" from the 3equations. I have a hunch that this will be accomplished by substituting

in [tex]\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/tex] for [tex]\gamma[/tex]. This eliminates gamma and also results in more "u"s, which may allow me to cancel them out. It's just getting to that point that is the problem.
 
  • #5
b2386 said:
Thanks for your replies. However, in the problem wording, it specifically tells us to start by eliminating "phi" first and then eliminate "u" from the 3equations. I have a hunch that this will be accomplished by substituting

in [tex]\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/tex] for [tex]\gamma[/tex]. This eliminates gamma and also results in more "u"s, which may allow me to cancel them out. It's just getting to that point that is the problem.
The link I posted outlines the approach that will get you there. The key is to express the electron energy in terms of its momentum and rest energy instead of the representation you used. Solve your last equation and the energy equation for momentum squared and set them equal. With a bit of Algebra many terms will cancel and you will have the result.
 
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  • #6
b2386 said:
Thanks for your replies. However, in the problem wording, it specifically tells us to start by eliminating "phi" first and then eliminate "u" from the 3equations.
Who would tell you to do it that way?
 

Related to How can I eliminate phi and u to derive the Compton effect equation?

1. What is the Compton effect?

The Compton effect, also known as Compton scattering, is a phenomenon in which a photon (particle of light) collides with an electron, transferring some of its energy to the electron and causing it to scatter at a different angle with a longer wavelength.

2. Who discovered the Compton effect?

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

3. What is the importance of the Compton effect?

The Compton effect was one of the first pieces of evidence for the particle nature of light, as it showed that photons have energy and can transfer that energy to other particles. It also helped to confirm Einstein's theory of relativity and has many applications in fields such as X-ray imaging and nuclear physics.

4. How is the Compton effect derived mathematically?

The Compton effect can be derived using the laws of conservation of energy and momentum, as well as Einstein's equation E=mc^2. This results in a formula that relates the change in wavelength of the scattered photon to the original wavelength and the angle of scattering.

5. What are some real-world examples of the Compton effect?

The Compton effect is used in medical imaging, such as X-ray computed tomography (CT) scans, to produce detailed images of the inside of the body. It is also used in radiation therapy for cancer treatment. Additionally, the Compton effect is important in understanding the structure of atoms and the behavior of particles in particle accelerators.

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