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Two-body problem

Set

definiendum R2×3,Hit … Classical Hamiltonian system
postulate H(r1,r2,p1,p2)=1m112p21+1m212p22+V(|r1r2|)

Discussion

Equations of motion in suitable coordinates
range Mm1+m2
range μm1m2/M

The following choice of coordinates eliminates the singles out the center of mass, which the Hamiltonian is independent of.

range rr2r1
range R(m1 r1+m2 r2)/M
range p(m1 p2m2 p1)/M
range Pp1+p2
range r|r|
H(r,R,p,R)=1M12P2+1μ12p2+V(r)

The Hamiltonian equations of motion give p(t)=μtr(t) and the vector P(t)=MtR(t) is conserved. The conserved angular momentum L turns out to imply hat the planar angle component of p(t), denoted pϕ(t) is conserved as well. We end up with planar motion in the center of mass frame the system is reduced to only two parameters. The energy takes the form H=1μ12(p2r+(L/r)2)+V(r)+const. and we denote it's value for r(0),pr(0) by E. The equations of motion read

μtr=2μ(EV(r))(L/r)2
μtϕ=L/r2

These furthermore imply the relation

ϕ=L/r22μ(EV(r))(L/r)2dr+const.
Scattering process
range gtr
range glim
range {\bf g}_{+\infty} \equiv \lim_{t\to+\infty}{\bf g}

Geometric considerations lead to the conclusion that there is a unit vector {\bf \alpha}^V, depending on the particle interaction potential V, with

range S_{ij} \equiv \delta_{ij}-2\alpha_i^V\alpha_j^V
{\bf g}_{+\infty} = S\ {\bf g}_{-\infty}

If b is the impact parameter of a collision under consideration, for V(r)=K/r^n the angle between {\bf g}_{-\infty} and {\bf g}_{+\infty} is given by

\bar\beta+(\bar\beta/b)^n = 1
2\int_0^{\bar\beta} \frac{1}{\sqrt{1-\beta^2-(\beta/b)^n}} \mathrm d\beta

For N=1, this gives the \sin^{-4} formulas of Rutherford scattering.

Reference

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