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16.5: Vectorial Derivation of the Scattering Angle

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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)dˆrdt

and the integral for an inverse square force f(r)=k/r2 (changing the sign of A for later convenience)

L×m˙r=kmˆr+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=ecosθ1

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

L2kmr=1Akmcosθ

To find the scattering angle, suppose the unit vector pointing parallel to the asymptote is ˆr, so the asymptotic velocity is vˆr.

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×mvˆr)׈r=L(L/b)

It follows that

sinθr==L2/Ab

And together with cosθr==km/A, we find

tanθr==L2kmb=mbv2k

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

χ=π2θr==2(π2θr=)=2tan1(kmbv2)

agreeing with the previous result.


This page titled 16.5: Vectorial Derivation of the Scattering Angle is shared under a not declared license and was authored, remixed, and/or curated by Michael Fowler.

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