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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.

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