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
$$\require{cancel} \begin{aligned} \left(p \, – p’ + mc \right)^{2} \, &= m^{2}c^{2} + \left(p^{2} + p’^{2} – 2pp’ \, \text{cos} \, \theta \right) \\ \cancel{p^{2}} + \cancel{p’^{2}} + \cancel{m^{2}c^{2}} + 2 \left( – pp’ + pmc \, – p’mc \right) \, &= \cancel{m^{2}c^{2}} + \cancel{p^{2}} + \cancel{p’^{2}} – 2pp’ \, \text{cos} \, \theta \\ pmc – p’mc \, &= pp’ \left( 1 \, – \, \text{cos} \, \theta \right) \\ \frac{mc}{p’} – \frac{mc}{p} \, &= 1 \, – \, \text{cos} \, \theta \\ \frac{mc \lambda’}{h} – \frac{mc \lambda}{h} \, &= 1 – \, \text{cos} \, \theta \\ \lambda’ – \lambda \, &= \frac{h}{mc} \left( 1 \, – \, \text{cos} \, \theta \right) \end{aligned}$$
Alternative Method:
(p-p’+mc)² how it is solved ? Please help me .
(P-P’ + MC)²
[(P-P’)² + 2(P-P’)(MC) +(MC)²]
[{(P)²-2(P)(P’)+(P’)²} +2PMC -2P’MC +M²C²]
[{P² -2PP’ +P’²} +2PMC -2P’MC + M²C²]
REARRANGING
[ P² +P’² +M²C² -2PP’ +2PMC +2P’MC]
TAKING 2 AS COMMON
[P² +P’² + M²C² +2(-PP’ +PMC +P’MC)]
Its a late reply ikr.
Thankyou so much it helps me alot
IT IS A EASIEST METHOD HELP FOR STUDENTS VERY NICE
very easy and clean derivation
Thank you. 🙂 I’m planning to get a latex version out sometime in the future.