Two-body problem
Set
definiendum | ⟨R2×3,H⟩∈it … Classical Hamiltonian system |
postulate | H(r1,r2,p1,p2)=1m112p21+1m212p22+V(|r1−r2|) |
Discussion
Equations of motion in suitable coordinates
range | M≡m1+m2 |
range | μ≡m1m2/M |
The following choice of coordinates eliminates the singles out the center of mass, which the Hamiltonian is independent of.
range | r≡r2−r1 |
range | R≡(m1 r1+m2 r2)/M |
range | p≡(m1 p2−m2 p1)/M |
range | P≡p1+p2 |
H(r,R,p,R)=1M12P2+1μ12p2+V(r) |
The Hamiltonian equations of motion give p(t)=μ∂∂tr(t) and the vector P(t)=M∂∂tR(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
μ∂∂tr=√2μ(E−V(r))−(L/r)2 |
μ∂∂tϕ=L/r2 |
These furthermore imply the relation
ϕ=∫L/r2√2μ(E−V(r))−(L/r)2dr+const. |
Scattering process
range | g≡∂∂tr |
range | g−∞≡lim |
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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