Derivation Of Compton Shift Equation

Using the above diagram, derive the Compton Shift Equation.

Conservation of momentum gives:

$$\begin{aligned}p \, &= p’ \, \text{cos} \, \theta + P_{e} \, \text{cos} \, \phi \\ P_{e} \, \text{cos} \, \phi \, &= p \, – p’ \, \text{cos} \, \theta \end{aligned} $$

AND

$$\begin{aligned} 0 \, &= p’ \, \text{sin} \, \theta \, – P_{e} \, \text{sin} \, \phi \\ P_{e} \, \text{sin} \, \phi \, &= p’ \, \text{sin} \, \theta \end{aligned}$$

Combining the above two equations gives equation 1:  (Use $\text{sin}^{2} \, \phi + \text{cos}^{2} \, \phi = 1$)

$$\begin{aligned} P^{2}_{e} \, &= \left( p \, – p’ \, \text{cos} \, \theta \right)^{2} + p’^{2} \, \text{sin}^{2} \, \theta \\ P^{2}_{e} \, &= p^{2} + p’^{2} – 2pp’ \, \text{cos} \, \theta \end{aligned}$$

Conservation of energy gives:

$$ E + E_{o} \, = E’ + E_{e} $$

Using $E^{2} = m^{2} c^{4} + p^{2} c^{2}$ and $E = pc$ for photon, the above equation becomes:

$$\begin{aligned} pc + mc^{2} \, &= p’c + \left(m^{2} c^{4} + P^{2}_{e} c^{2} \right)^{\frac{1}{2}} \\ \left(p \, – p’ + mc \right)^{2} \, &= m^{2}c^{2} + P^{2}_{e} \end{aligned}$$

Using equation 1 to get rid of $P^{2}_{e}$ in the above equation, we obtain

$$\require{cancel} \begin{aligned} \left(p \, – p’ + mc \right)^{2} \, &= m^{2}c^{2} + \left(p^{2} + p’^{2} – 2pp’ \, \text{cos} \, \theta \right) \\ \cancel{p^{2}} + \cancel{p’^{2}} + \cancel{m^{2}c^{2}} + 2 \left( – pp’ + pmc \, – p’mc \right) \, &= \cancel{m^{2}c^{2}} + \cancel{p^{2}} + \cancel{p’^{2}} – 2pp’ \, \text{cos} \, \theta \\ pmc – p’mc \, &= pp’ \left( 1 \, – \, \text{cos} \, \theta \right) \\ \frac{mc}{p’} – \frac{mc}{p} \, &= 1 \, – \, \text{cos} \, \theta \\ \frac{mc \lambda’}{h} – \frac{mc \lambda}{h} \, &= 1 – \, \text{cos} \, \theta \\ \lambda’ – \lambda \, &= \frac{h}{mc} \left( 1 \, – \, \text{cos} \, \theta \right) \end{aligned}$$

 

 

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Alternative Method:

 

 

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