Spring 2006 13 Em

This only covers part of number 13 in Spring 2006, but it is the full answer to the stated problem below.

An electron is incident with impact parameter $\rho$ and speed vo on a proton at rest. Calculate the energy radiated during the collision assuming the ordering

(1)
\begin{align} \frac{e^2}{\rho} \ll mv^2_o \ll mc^2 \end{align}

Hint: this simplifies the orbit approximation.

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(2)
\begin{align} F=ma=\frac{e^2}{r^2} \end{align}
(3)
\begin{align} \left | \vec{a}\right |=\frac{e^2}{mr^2}\\ \end{align}
(4)
\begin{align} \left | \vec{a}\right |^2=\frac{e^4}{m^2r^4}\\ \end{align}
(5)
\begin{align} \ddot{\vec{p}}=e\ddot{\vec{r}} \end{align}
(6)
\begin{align} \vec{a}=\ddot{\vec{r}} \end{align}
(7)
\begin{align} P=\frac{2}{3}\frac{\left(\ddot{\vec{p}}\right)^2}{c^3} \end{align}
(8)
\begin{align} P=\frac{2}{3}\frac{e^2}{c^3}\left | \vec{a}\right |^2=\frac{2}{3}\frac{e^6}{m^2r^4c^3} \end{align}

Energy radiated$= \int P \,dt$

(9)
\begin{align} r^2=\rho^2+\left(v_o t\right)^2 \end{align}

since

(10)
\begin{align} mv_o^2 \gg \frac{e^2}{\rho^2} \end{align}
(11)
\begin{align} E=\frac{2}{3}\frac{e^6}{m^2c^3}\int_{- \infty }^{ \infty } \frac{1}{\left (\rho^2+\left (v_o t \right )^2\right)^2}\,dt \end{align}
(12)
\begin{align} E=\frac{2}{3}\frac{e^6}{m^2c^3}\frac{1}{v_o\rho^4}\underbrace{\int_{- \infty }^{ \infty } \frac{1}{\left (1+x^2\right)^2}\,dx}_{\frac{\pi}{2}} \end{align}