16.5: Vectorial Derivation of the Scattering Angle
( \newcommand{\kernel}{\mathrm{null}\,}\)
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
→L×m¨→r=−mr2f(r)dˆ→rdt
and the integral for an inverse square force f(r)=k/r2 (changing the sign of →A for later convenience)
→L×m˙→r=kmˆ→r+→A
As previously discussed, multiplying by →L. establishes that →A is in the plane of the orbit, and multiplying by →r gives
−L2=kmr+Arcosθ
This corresponds to the equation
1/r=−ecosθ−1
(the left-hand branch with the right-hand focus as origin, note from diagram above that cosθ is negative throughout) and
L2kmr=−1−Akmcosθ
To find the scattering angle, suppose the unit vector pointing parallel to the asymptote is ˆ→r∞, so the asymptotic velocity is v∞ˆ→r∞.
Note that as before, →A is along the major axis (to give the correct form for the (r,θ) equation), and r=∞ gives the asymptotic angles from
cosθr=∞=−km/A
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 ׈→r∞ yields
→A׈→r∞=(→L×mv∞ˆ→r∞)׈→r∞=−→L(L/b)
It follows that
sinθr=∞=−L2/Ab
And together with cosθr=∞=−km/A, we find
tanθr=∞=L2kmb=mbv2∞k
This is the angle between the asymptote and the major axis: the scattering angle
χ=π−2θr=∞=2(π2−θr=∞)=2tan−1(kmbv2∞)
agreeing with the previous result.