Solving Compton Scattering Homework Statement

In summary: 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?No, it is not in the wrong place. It is just that you are dealing with vectors, not scalars.marlonNo, it is not in the wrong place. It is just that you are dealing with vectors, not scalars.Right, so that would be (P_f c)^2 = c^2 (p_i^2 + p_f^2 - 2 p_i p_f cos\theta)? So I can then use that to get my answer? Thanks for the help.Right, so that would be (P_f c)^2 = c^2
  • #1
Brewer
212
0

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:
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  • #2
Brewer said:

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
marlon said:
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
Brewer said:
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
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
Remember that the initial and final momenta are vectors.
 
  • #7
Brewer said:
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
 
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1. What is Compton scattering?

Compton scattering is a phenomenon in which a photon of high energy collides with a free electron, resulting in a decrease in the energy of the photon and the emission of an electron of lower energy.

2. What is the purpose of solving Compton scattering homework?

The purpose of solving Compton scattering homework is to better understand the principles and applications of Compton scattering, which is an important concept in quantum mechanics and photon-electron interactions.

3. How do you solve Compton scattering problems?

To solve Compton scattering problems, you need to use the Compton scattering formula, which calculates the change in wavelength of a photon after scattering off an electron. This formula takes into account the initial and final energies and angles of the photon and electron.

4. What are some real-world applications of Compton scattering?

Compton scattering has many applications in various fields, such as medical imaging, nuclear medicine, and astronomy. In medical imaging, it is used to produce images of bones and tissues, while in nuclear medicine, it is used to detect and diagnose diseases. In astronomy, it is used to study the composition and structure of celestial objects.

5. How does Compton scattering relate to other concepts in physics?

Compton scattering is related to other concepts in physics, such as the wave-particle duality of light, the photoelectric effect, and the Heisenberg uncertainty principle. It also has connections to other phenomena, such as X-ray diffraction and Compton edge in gamma-ray spectroscopy.

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