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rpardo
Jan30-10, 08:45 PM
Hey guys,

Im trying to derive the following equation

(mc^2) [(1/E2)-(1/E1)]+cos(theta)-[((E1-E2)^2)/(2E1E2)]=1

E1 = Incident Photon's energy
E2 = Scattered Photon's energy
theta= scattering angle
m = mass of electron
c = s

Using conservation of energy,conservation of mass, and the cosine law

I've derived the relativistic equation, but I am really stumped on how to derive this...its been a couple days
Any help at all would be appreciated. I feel like I'm really close but the algebra is the problem

Thanks in advance for your help guys and gals,

rpardo
Jan31-10, 02:29 PM
\[
\begin{gathered}
{\text{Conservation of momentum: }}p_1 + p_{e_1 } = p_2 + p_{e_2 } \hfill \\
{\text{assume electron starts from rest }}p_1 = p_2 + p_{e_2 } \hfill \\
{\text{Conservation of energy: }}E_1 + E_{e_1 } = E_2 + E_{e_2 } \hfill \\
E_1 = p_1 c \hfill \\
E_2 = p_2 c \hfill \\
E_{e_1 } = 0 \hfill \\
E_{e_2 } = \frac{1}
{2}m_0 v^2 = \frac{1}
{{2m_0 }}p_e ^2 \hfill \\
\hfill \\
p_1 c = p_2 c + \frac{1}
{{2m_0 }}p_e ^2 {\text{ }} \hfill \\
\Leftrightarrow 2m_0 c(p_1 - p_2 ) = p_e ^2 \hfill \\
{\text{Sub in }}p_e ^2 = p_1 ^2 + p_2 ^2 - 2p_1 p_2 \cos \Theta {\text{ from cosine law}} \hfill \\
\Rightarrow 2m_0 c(p_1 - p_2 ) = p_1 ^2 + p_2 ^2 - 2p_1 p_2 \cos \Theta \hfill \\
\Leftrightarrow (2m_0 c(p_1 - p_2 ))^2 = (p_1 ^2 + p_2 ^2 - 2p_1 p_2 \cos \Theta )^2 \hfill \\
\Leftrightarrow 4m_0 ^2 c^2 (p_1 ^2 - 2p_1 p_2 + p_2 ^2 ) = p_1 ^4 + 2p_1 ^2 p_2 ^2 - 4p_1 ^3 p_2 \cos \Theta + p_2 ^4 - 4p_2 ^3 p_1 \cos \Theta + 4p_1 ^2 p_2 ^2 \cos ^2 \Theta \hfill \\
\Leftrightarrow 4m_0 ^2 c^2 p_1 ^2 - 8m_0 ^2 c^2 p_1 p_2 + 4m_0 ^2 c^2 p_2 = p_1 ^4 + 2p_1 ^2 p_2 ^2 - 4p_1 ^3 p_2 \cos \Theta + p_2 ^4 - 4p_2 ^3 p_1 \cos \Theta + 4p_1 ^2 p_2 ^2 \cos ^2 \Theta \hfill \\
\end{gathered}
\]

torquil
Jan31-10, 03:19 PM
\[
\begin{gathered}
{\text{Conservation of momentum: }}p_1 + p_{e_1 } = p_2 + p_{e_2 } \hfill \\
{\text{assume electron starts from rest }}p_1 = p_2 + p_{e_2 } \hfill \\
{\text{Conservation of energy: }}E_1 + E_{e_1 } = E_2 + E_{e_2 } \hfill \\
E_1 = p_1 c \hfill \\
E_2 = p_2 c \hfill \\
E_{e_1 } = 0 \hfill \\
E_{e_2 } = \frac{1}
{2}m_0 v^2 = \frac{1}
{{2m_0 }}p_e ^2 \hfill \\
\hfill \\
p_1 c = p_2 c + \frac{1}
{{2m_0 }}p_e ^2 {\text{ }} \hfill \\
\Leftrightarrow 2m_0 c(p_1 - p_2 ) = p_e ^2 \hfill \\
{\text{Sub in }}p_e ^2 = p_1 ^2 + p_2 ^2 - 2p_1 p_2 \cos \Theta {\text{ from cosine law}} \hfill \\
\Rightarrow 2m_0 c(p_1 - p_2 ) = p_1 ^2 + p_2 ^2 - 2p_1 p_2 \cos \Theta \hfill \\
\Leftrightarrow (2m_0 c(p_1 - p_2 ))^2 = (p_1 ^2 + p_2 ^2 - 2p_1 p_2 \cos \Theta )^2 \hfill \\
\Leftrightarrow 4m_0 ^2 c^2 (p_1 ^2 - 2p_1 p_2 + p_2 ^2 ) = p_1 ^4 + 2p_1 ^2 p_2 ^2 - 4p_1 ^3 p_2 \cos \Theta + p_2 ^4 - 4p_2 ^3 p_1 \cos \Theta + 4p_1 ^2 p_2 ^2 \cos ^2 \Theta \hfill \\
\Leftrightarrow 4m_0 ^2 c^2 p_1 ^2 - 8m_0 ^2 c^2 p_1 p_2 + 4m_0 ^2 c^2 p_2 = p_1 ^4 + 2p_1 ^2 p_2 ^2 - 4p_1 ^3 p_2 \cos \Theta + p_2 ^4 - 4p_2 ^3 p_1 \cos \Theta + 4p_1 ^2 p_2 ^2 \cos ^2 \Theta \hfill \\
\end{gathered}
\]



Fixed your latex!

vela
Jan31-10, 07:25 PM
Once you get to

2m_0c(p_1-p_2) = p_1^2+p_2^2-2p_1p_2\cos\theta

don't square the equation. Instead, divide by 2p_1p_2. Throw in factors of c here and there, and you'll have the cosine term and the mc2(1/E2-1/E1) terms that you want. You just have to then figure out how the rest works out.