Convex hull algorithms

In classical mechanics, a Liouville dynamical system is an exactly soluble dynamical system in which the kinetic energy T and potential energy V can be expressed in terms of the s generalized coordinates q as follows:

$T=\frac{1}{2}\{u_1(q_1)+u_2(q_2)+\cdots +u_s(q_s)\}\{v_1(q_1){\dot{q}}_1^2+v_2(q_2){\dot{q}}_2^2+\cdots +v_s(q_s){\dot{q}}_s^2\}$

$V=\frac{w_1(q_1)+w_2(q_2)+\cdots +w_s(q_s)}{u_1(q_1)+u_2(q_2)+\cdots +u_s(q_s)}$

The solution of this system consists of a set of separably integrable equations

$\frac{\sqrt{2}}{Y}dt=\frac{d{\phi }_1}{\sqrt{E{\chi }_1-{\omega }_1+{\gamma }_1}}=\frac{d{\phi }_2}{\sqrt{E{\chi }_2-{\omega }_2+{\gamma }_2}}=\cdots =\frac{d{\phi }_s}{\sqrt{E{\chi }_s-{\omega }_s+{\gamma }_s}}$

where E = T + V is the conserved energy and the ${\gamma }_s$ are constants. As described below, the variables have been changed from q_s to φ_s, and the functions u_s and w_s substituted by their counterparts χ_s and ω_s. This solution has numerous applications, such as the orbit of a small planet about two fixed stars under the influence of Newtonian gravity. The Liouville dynamical system is one of several things named after Joseph Liouville, an eminent French mathematician.

Example of bicentric orbits

In classical mechanics, Euler's three-body problem describes the motion of a particle in a plane under the influence of two fixed centers, each of which attract the particle with an inverse-square force such as Newtonian gravity or Coulomb's law. Examples of the bicenter problem include a planet moving around two slowly moving stars, or an electron moving in the electric field of two positively charged nuclei, such as the first ion of the hydrogen molecule H₂, namely the hydrogen molecular ion or H₂⁺. The strength of the two attractions need not be equal; thus, the two stars may have different masses or the nuclei two different charges.

Solution

Let the fixed centers of attraction be located along the x-axis at ±a. The potential energy of the moving particle is given by

$V(x,y)=\frac{-{\mu }_1}{\sqrt{{(x-a)}^2+y^2}}-\frac{{\mu }_2}{\sqrt{{(x+a)}^2+y^2}}.$

The two centers of attraction can be considered as the foci of a set of ellipses. If either center were absent, the particle would move on one of these ellipses, as a solution of the Kepler problem. Therefore, according to Bonnet's theorem, the same ellipses are the solutions for the bicenter problem.

Introducing elliptic coordinates,

$x=a\cosh \xi \cos \eta ,$

$y=a\sinh \xi \sin \eta ,$

the potential energy can be written as

$V(\xi ,\eta )=\frac{-{\mu }_1}{a(\cosh \xi -\cos \eta )}-\frac{{\mu }_2}{a(\cosh \xi +\cos \eta )}=\frac{-{\mu }_1(\cosh \xi +\cos \eta )-{\mu }_2(\cosh \xi -\cos \eta )}{a({\cosh }^2\xi -{\cos }^2\eta )},$

and the kinetic energy as

$T=\frac{m{a}^2}{2}({\cosh }^2\xi -{\cos }^2\eta )({\dot{\xi }}^2+{\dot{\eta }}^2).$

This is a Liouville dynamical system if ξ and η are taken as φ₁ and φ₂, respectively; thus, the function Y equals

$Y={\cosh }^2\xi -{\cos }^2\eta$

and the function W equals

$W=-{\mu }_1(\cosh \xi +\cos \eta )-{\mu }_2(\cosh \xi -\cos \eta )$

Using the general solution for a Liouville dynamical system below, one obtains

$\frac{m{a}^2}{2}{({\cosh }^2\xi -{\cos }^2\eta )}^2{\dot{\xi }}^2=E{\cosh }^2\xi +\left(\frac{{\mu }_1+{\mu }_2}{a}\right)\cosh \xi -\gamma$

$\frac{m{a}^2}{2}{({\cosh }^2\xi -{\cos }^2\eta )}^2{\dot{\eta }}^2=-E{\cos }^2\eta +\left(\frac{{\mu }_1-{\mu }_2}{a}\right)\cos \eta +\gamma$

Introducing a parameter u by the formula

$du=\frac{d\xi }{\sqrt{E{\cosh }^2\xi +\left(\frac{{\mu }_1+{\mu }_2}{a}\right)\cosh \xi -\gamma }}=\frac{d\eta }{\sqrt{-E{\cos }^2\eta +\left(\frac{{\mu }_1-{\mu }_2}{a}\right)\cos \eta +\gamma }},$

gives the parametric solution

$u=\int \frac{d\xi }{\sqrt{E{\cosh }^2\xi +\left(\frac{{\mu }_1+{\mu }_2}{a}\right)\cosh \xi -\gamma }}=\int \frac{d\eta }{\sqrt{-E{\cos }^2\eta +\left(\frac{{\mu }_1-{\mu }_2}{a}\right)\cos \eta +\gamma }}.$

Since these are elliptic integrals, the coordinates ξ and η can be expressed as elliptic functions of u.

Constant of motion

The bicentric problem has a constant of motion, namely,

$r_1^2{r}_2^2\left(\frac{d{\theta }_1}{dt}\right)\left(\frac{d{\theta }_2}{dt}\right)-2c[{\mu }_1\cos {\theta }_1+{\mu }_2\cos {\theta }_2],$

from which the problem can be solved using the method of the last multiplier.

Derivation

New variables

To eliminate the v functions, the variables are changed to an equivalent set

${\phi }_r=\int d{q}_r\sqrt{v_r(q_r)},$

giving the relation

$v_1(q_1){\dot{q}}_1^2+v_2(q_2){\dot{q}}_2^2+\cdots +v_s(q_s){\dot{q}}_s^2={\dot{\phi }}_1^2+{\dot{\phi }}_2^2+\cdots +{\dot{\phi }}_s^2=F,$

which defines a new variable F. Using the new variables, the u and w functions can be expressed by equivalent functions χ and ω. Denoting the sum of the χ functions by Y,

$Y={\chi }_1({\phi }_1)+{\chi }_2({\phi }_2)+\cdots +{\chi }_s({\phi }_s),$

the kinetic energy can be written as

$T=\frac{1}{2}YF.$

Similarly, denoting the sum of the ω functions by W

$W={\omega }_1({\phi }_1)+{\omega }_2({\phi }_2)+\cdots +{\omega }_s({\phi }_s),$

the potential energy V can be written as

$V=\frac{W}{Y}.$

Lagrange equation

The Lagrange equation for the r^th variable ${\phi }_r$ is

$\frac{d}{dt}\left(\frac{\partial T}{\partial {\dot{\phi }}_r}\right)=\frac{d}{dt}\left(Y{\dot{\phi }}_r\right)=\frac{1}{2}F\frac{\partial Y}{\partial {\phi }_r}-\frac{\partial V}{\partial {\phi }_r}.$

Multiplying both sides by $2Y{\dot{\phi }}_r$, re-arranging, and exploiting the relation 2T = YF yields the equation

$2Y{\dot{\phi }}_r\frac{d}{dt}\left(Y{\dot{\phi }}_r\right)=2T{\dot{\phi }}_r\frac{\partial Y}{\partial {\phi }_r}-2Y{\dot{\phi }}_r\frac{\partial V}{\partial {\phi }_r}=2{\dot{\phi }}_r\frac{\partial }{\partial {\phi }_r}[(E-V)Y],$

which may be written as

$\frac{d}{dt}\left(Y^2{\dot{\phi }}_r^2\right)=2E{\dot{\phi }}_r\frac{\partial Y}{\partial {\phi }_r}-2{\dot{\phi }}_r\frac{\partial W}{\partial {\phi }_r}=2E{\dot{\phi }}_r\frac{d{\chi }_r}{d{\phi }_r}-2{\dot{\phi }}_r\frac{d{\omega }_r}{d{\phi }_r},$

where E = T + V is the (conserved) total energy. It follows that

$\frac{d}{dt}\left(Y^2{\dot{\phi }}_r^2\right)=2\frac{d}{dt}(E{\chi }_r-{\omega }_r),$

which may be integrated once to yield

$\frac{1}{2}{Y}^2{\dot{\phi }}_r^2=E{\chi }_r-{\omega }_r+{\gamma }_r,$

where the ${\gamma }_r$ are constants of integration subject to the energy conservation

$\sum _{r=1}^s{\gamma }_r=0.$

Inverting, taking the square root and separating the variables yields a set of separably integrable equations:

$\frac{\sqrt{2}}{Y}dt=\frac{d{\phi }_1}{\sqrt{E{\chi }_1-{\omega }_1+{\gamma }_1}}=\frac{d{\phi }_2}{\sqrt{E{\chi }_2-{\omega }_2+{\gamma }_2}}=\cdots =\frac{d{\phi }_s}{\sqrt{E{\chi }_s-{\omega }_s+{\gamma }_s}}.$

References

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