# Compton Scattering

1. Dec 27, 2006

### Brewer

1. The problem statement, all variables and given/known data
The energy E and momentum P of a relativistic electron and energy ε and momentum p of a photon are related by:
$$E^2 - P^2c^2 = m_{e}^2c^4 and \epsilon = pc$$

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

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

2. Relevant equations
momentum before = momentum after

3. The attempt at a solution

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

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

After a bit of rearranging I can get to:

$$\epsilon_{i}^2 - \frac{\epsilon_{f}^2}{cos^2\theta} = E_{f}^2 - m_{e}^2c^4$$, 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: Dec 27, 2006
2. Dec 27, 2006

### marlon

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

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

you need to do SQUARE THIS :

$$\frac{\epsilon_{i}}{c} - \frac{\epsilon_{f}}{ccos\theta}= \sqrt{\frac{E_{f}^2-m_{e}^2c^4}{c^2}}$$

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: Dec 27, 2006
3. Dec 27, 2006

### Brewer

Oh, ok so I can't just square all the terms then? I'll give it a go and see what happens.

4. Dec 27, 2006

### marlon

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. Dec 27, 2006

### Brewer

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. Dec 27, 2006

### Staff: Mentor

Remember that the initial and final momenta are vectors.

7. Dec 27, 2006

### marlon

Let's start all over.

1) apply conservation of energy

2) apply conservation of momentum

From 1) write that equation as $$(P_f c)^2$$ =

From 2) we get : $$\vec {P_f} = \vec {p_i} - \vec {p_f}$$

The trick is to write 2) as $$(P_f c)^2$$ = ; so that we can get rid off the electron parameters $$(P_f c)^2$$ . 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 $$(P_f c)^2 = c^2 \vec{P_f} \cdot \vec{P_f} = c^2 ( \vec {p_i} - \vec {p_f}) \cdot (\vec {p_i} - \vec {p_f})$$

THIS IS A SCALAR PRODUCT !!!

marlon

Last edited: Dec 27, 2006