16.5: Vectorial Derivation of the Scattering Angle

The essential result of the above analysis was the scattering angle as a function of impact parameter, for a given incoming energy. It's worth noting that this can be found more directly by vectorial methods from Hamilton's equation.

Recall from the last lecture Hamilton’s equation

\begin{equation}
\vec{L} \times m \ddot{\vec{r}}=-m r^{2} f(r) \dfrac{d \widehat{\vec{r}}}{d t}
\end{equation}

and the integral for an inverse square force \(f(r)=k / r^{2}\) (changing the sign of \(\vec{A}\) for later convenience)

\begin{equation}
\vec{L} \times m \dot{\vec{r}}=k m\widehat{\vec{r}}+\vec{A}
\end{equation}

As previously discussed, multiplying by \(\vec{L}\). establishes that \(\vec{A}\) is in the plane of the orbit, and multiplying by \(\vec{r}\) gives

\begin{equation}
-L^{2}=k m r+A r \cos \theta
\end{equation}

This corresponds to the equation

\begin{equation}
1 / r=-e \cos \theta-1
\end{equation}

(the left-hand branch with the right-hand focus as origin, note from diagram above that \(\cos \theta\) is negative throughout) and

\begin{equation}\dfrac{L^{2}}{k m r}=-1-\dfrac{A}{k m} \cos \theta\end{equation}

To find the scattering angle, suppose the unit vector pointing parallel to the asymptote is \(\widehat{\vec{r}}_{\infty}\), so the asymptotic velocity is \(v_{\infty} \widehat{\vec{r}}_{\infty}\).

Note that as before, \(\vec{A}\) is along the major axis (to give the correct form for the \((r, \theta)\) equation), and \(r=\infty\) gives the asymptotic angles from

\begin{equation}
\cos \theta_{r=\infty}=-k m / A
\end{equation}

We’re not rotating the hyperbola as we did in the alternative treatment above: here we keep it symmetric about the \(x\)-axis, and find its asymptotic angle to that axis, which is one-half the scattering angle.

Now take Hamilton’s equation in the asymptotic limit, where the velocity is parallel to the displacement:

the vector product of Hamilton’s equation \(\times \widehat{\vec{r}}_{\infty}\) yields

\begin{equation}
\vec{A} \times \widehat{\vec{r}}_{\infty}=\left(\vec{L} \times m v_{\infty} \widehat{\vec{r}}_{\infty}\right) \times \widehat{\vec{r}}_{\infty}=-\vec{L}(L / b)
\end{equation}

It follows that

\begin{equation}
\sin \theta_{r=\infty}=-L^{2} / A b
\end{equation}

And together with \(\cos \theta_{r=\infty}=-k m / A\), we find

\begin{equation}
\tan \theta_{r=\infty}=\dfrac{L^{2}}{k m b}=\dfrac{m b v_{\infty}^{2}}{k}
\end{equation}

This is the angle between the asymptote and the major axis: the scattering angle

\begin{equation}
\chi=\pi-2 \theta_{r=\infty}=2\left(\dfrac{\pi}{2}-\theta_{r=\infty}\right)=2 \tan ^{-1}\left(\dfrac{k}{m b v_{\infty}^{2}}\right)
\end{equation}

agreeing with the previous result.