Two-body problem


definiendum $\langle \mathbb R^{2\times 3}, H\rangle \in \mathrm{it} $ … Classical Hamiltonian system
postulate $ H({\bf r}_1,{\bf r}_2,{\bf p}_1,{\bf p}_2) = \frac{1}{m_1}\frac{1}{2}{\bf p}_1^2 + \frac{1}{m_2}\frac{1}{2}{\bf p}_2^2 + V(|{\bf r}_1-{\bf r}_2|) $


Equations of motion in suitable coordinates
range $ M \equiv m_1+m_2 $
range $ \mu \equiv m_1 m_2/M $

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

range $ {\bf r} \equiv {\bf r}_2 - {\bf r}_1 $
range $ {\bf R} \equiv (m_1\ {\bf r}_1 + m_2\ {\bf r}_2)/M $
range $ {\bf p} \equiv (m_1\ {\bf p}_2 - m_2\ {\bf p}_1)/M $
range $ {\bf P} \equiv {\bf p}_1 + {\bf p}_2 $
range $ r \equiv |{\bf r}| $
$ H({\bf r},{\bf R},{\bf p},{\bf R}) = \frac{1}{M}\frac{1}{2}{\bf P}^2 + \frac{1}{\mu}\frac{1}{2}{\bf p}^2 + V(r) $

The Hamiltonian equations of motion give ${\bf p}(t)=\mu\frac{\partial}{\partial t}{\bf r}(t)$ and the vector ${\bf P}(t) = M\frac{\partial}{\partial t} {\bf R}(t)$ is conserved. The conserved angular momentum $L$ turns out to imply hat the planar angle component of ${\bf p}(t)$, denoted $p_{\phi}(t)$ 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 = \frac{1}{\mu}\frac{1}{2}\left(p_r^2+(L/r)^2\right) + V(r) + \text{const.} $ and we denote it's value for $r(0), p_r(0)$ by $E$. The equations of motion read

$ \mu\frac{\partial}{\partial t} r = \sqrt{2\mu(E-V(r))-(L/r)^2}$
$ \mu\frac{\partial}{\partial t} \phi = L/r^2 $

These furthermore imply the relation

$\phi = \int \frac{L/r^2}{\sqrt{2\mu(E-V(r))-(L/r)^2}} \mathrm dr + \text{const.}$
Scattering process
range $ {\bf g} \equiv \frac{\partial}{\partial t}{\bf r} $
range $ {\bf g}_{-\infty} \equiv \lim_{t\to-\infty}{\bf g} $
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.


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