Compton Scattering

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Homework Statement


The energy E and momentum P of a relativistic electron and energy ε and momentum p of a photon are related by:
[tex]E^2 - P^2c^2 = m_{e}^2c^4 and \epsilon = pc[/tex]

let [tex]E_{i}, P_{i} and E_{f}, P_{f}[/tex] denote the inital and final energies and momenta of the electron, and let [tex]\epsilon_{i}, p_{i} and \epsilon_{f}, p_{f}[/tex] denote the initial and final energies and momenta of the photon. Assume the electron is initially at rest, so that [tex]E_{i} = m_{e}c^2[/tex] and [tex]P_{i} = 0[/tex], and assume that the photon is scattered through the angle [tex]\theta[/tex].

By considering the conservation of momentum show that:
[tex]\epsilon_{i}^2 - 2\epsilon_{i}\epsilon_{f}cos\theta + \epsilon_{f}^2 = E_{f}^2 - m_{e}^2c^4[/tex]


Homework Equations


momentum before = momentum after


The Attempt at a Solution



initial momentum = [tex]P_{i} + p_{i} = \frac{\epsilon_{i}}{c}[/tex]
final momentum = [tex]P_{f} + p_{f} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

therefore: [tex]\frac{\epsilon_{i}}{c} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

After a bit of rearranging I can get to:

[tex]\epsilon_{i}^2 - \frac{\epsilon_{f}^2}{cos^2\theta} = E_{f}^2 - m_{e}^2c^4[/tex], where the RHS is as required, but I can't rearrange the LHS to get what they want. I get the feeling I may have left a cosθ out somewhere, or that I've forgotten quite an important (yet simple) technique in here somewhere, to rearrange the LHS. I can see that it could be written as the difference of 2 squares, but I cannot see how that would help me.

Any hints appreciated.
 
Last edited:

Answers and Replies

  • #2
3,763
9

Homework Statement


The energy E and momentum P of a relativistic electron and energy ε and momentum p of a photon are related by:
[tex]E^2 - P^2c^2 = m_{e}^2c^4 and \epsilon = pc[/tex]

let [tex]E_{i}, P_{i} and E_{f}, P_{f}[/tex] denote the inital and final energies and momenta of the electron, and let [tex]\epsilon_{i}, p_{i} and \epsilon_{f}, p_{f}[/tex] denote the initial and final energies and momenta of the photon. Assume the electron is initially at rest, so that [tex]E_{i} = m_{e}c^2[/tex] and [tex]P_{i} = 0[/tex], and assume that the photon is scattered through the angle [tex]\theta[/tex].

By considering the conservation of momentum show that:
[tex]\epsilon_{i}^2 - 2\epsilon_{i}\epsilon_{f}cos\theta + \epsilon_{f}^2 = E_{f}^2 - m_{e}^2c^4[/tex]


Homework Equations


momentum before = momentum after


The Attempt at a Solution



initial momentum = [tex]P_{i} + p_{i} = \frac{\epsilon_{i}}{c}[/tex]
final momentum = [tex]P_{f} + p_{f} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

therefore: [tex]\frac{\epsilon_{i}}{c} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

After a bit of rearranging I can get to:

[tex]\epsilon_{i}^2 - \frac{\epsilon_{f}^2}{cos^2\theta} = E_{f}^2 - m_{e}^2c^4[/tex], where the RHS is as required, but I can't rearrange the LHS to get what they want. I get the feeling I may have left a cosθ out somewhere, or that I've forgotten quite an important (yet simple) technique in here somewhere, to rearrange the LHS. I can see that it could be written as the difference of 2 squares, but I cannot see how that would help me.

Any hints appreciated.

I can directly see a mistake in the way you got rid off the square root in :

[tex]\frac{\epsilon_{i}}{c} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

you need to do SQUARE THIS :

[tex]\frac{\epsilon_{i}}{c} - \frac{\epsilon_{f}}{ccos\theta}= \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}}[/tex]

But still, there is something wrong with the position of the cosine.

marlon

edit : beware, i did not check the entire algebra or the application of the conservation laws. I just denoted an obvious mistake :)
 
Last edited:
  • #3
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I can directly see a mistake in the way you got rid off the square root in :

[tex]\frac{\epsilon_{i}}{c} = \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}} + \frac{\epsilon_{f}}{ccos\theta}[/tex]

you need to do SQUARE THIS :

[tex]\frac{\epsilon_{i}}{c} - \frac{\epsilon_{f}}{ccos\theta}= \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}}[/tex]

marlon

edit : beware, i did not check the entire algebra or the application of the conservation laws. I just denoted an obvious mistake :)

Oh, ok so I can't just square all the terms then? I'll give it a go and see what happens.
 
  • #4
3,763
9
Oh, ok so I can't just square all the terms then?
Yes you can, only you need to do it correctly :)

you did (A+B)²=C² --> A² + B² = C²

This is, ofcourse, wrong


marlon
 
  • #5
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So the cos is in the wrong place? Well I'll work through and see where it should go. On the top of the fraction maybe?
 
  • #6
Doc Al
Mentor
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Remember that the initial and final momenta are vectors.
 
  • #7
3,763
9
So the cos is in the wrong place? Well I'll work through and see where it should go. On the top of the fraction maybe?

Let's start all over.

1) apply conservation of energy

2) apply conservation of momentum

From 1) write that equation as [tex](P_f c)^2[/tex] =

From 2) we get : [tex]\vec {P_f} = \vec {p_i} - \vec {p_f} [/tex]

The trick is to write 2) as [tex](P_f c)^2[/tex] = ; so that we can get rid off the electron parameters [tex](P_f c)^2[/tex] . To do that (AND THIS IS WHERE YOU MADE YOUR MISTAKE WITH THE COSINE) you need to be aware of the fact that the momenta are VECTORS.

So [tex] (P_f c)^2 = c^2 \vec{P_f} \cdot \vec{P_f} = c^2 ( \vec {p_i} - \vec {p_f}) \cdot (\vec {p_i} - \vec {p_f}) [/tex]

THIS IS A SCALAR PRODUCT !!!

marlon
 
Last edited:

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