Convex hull algorithms: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
Improved flow of first sentence.
 
Line 1: Line 1:
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 coordinate]]s ''q'' as follows:<ref name="liouville_1849">{{cite journal | last = Liouville | year = 1849 | title = Mémoire sur l'intégration des équations différentielles du mouvement d'un nombre quelconque de points matériels | journal = Journal de Mathématiques Pures et Appliquées | volume = 14 | pages = 257&ndash;299 | url = http://visualiseur.bnf.fr/ConsulterElementNum?O=NUMM-16393&Deb=263&Fin=305&E=PDF}}</ref>
I would like to introduce myself to you, I am Jayson Simcox but I don't like when people use my complete name. Distributing manufacturing has been his occupation for some time. Mississippi is where his home is. To perform lacross is one of the issues she loves most.<br><br>Review my blog: telephone psychic ([http://checkmates.co.za/index.php?do=/profile-56347/info/ just click the up coming internet page])
 
:<math>
T = \frac{1}{2} \left\{ u_{1}(q_{1}) + u_{2}(q_{2}) + \cdots + u_{s}(q_{s}) \right\}
\left\{ 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} \right\}
</math>
 
:<math>
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}) }
</math>
 
The solution of this system consists of a set of separably integrable equations
 
:<math>
\frac{\sqrt{2}}{Y}\, dt = \frac{d\varphi_{1}}{\sqrt{E \chi_{1} - \omega_{1} + \gamma_{1}}} =
\frac{d\varphi_{2}}{\sqrt{E \chi_{2} - \omega_{2} + \gamma_{2}}} = \cdots =
\frac{d\varphi_{s}}{\sqrt{E \chi_{s} - \omega_{s} + \gamma_{s}}}
</math>
 
where ''E = T + V'' is the conserved energy and the <math>\gamma_{s}</math> are constants. As described below, the variables have been changed from ''q<sub>s</sub>'' to φ<sub>s</sub>, and the functions ''u<sub>s</sub>'' and ''w<sub>s</sub>'' substituted by their counterparts ''χ<sub>s</sub>'' and ''ω<sub>s</sub>''.  This solution has numerous applications, such as the orbit of a small planet about two fixed stars under the influence of [[gravitation|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 law|inverse-square force]] such as [[gravitation|Newtonian gravity]] or [[Coulomb's law]]. Examples of the bicenter problem include a [[planet]] moving around two slowly moving [[star]]s, or an [[electron]] moving in the [[electric field]] of two positively charged [[atomic nucleus|nuclei]], such as the first [[ion]] of the hydrogen molecule H<sub>2</sub>, namely the [[Dihydrogen cation|hydrogen molecular ion]] or H<sub>2</sub><sup>+</sup>.  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
 
:<math>
V(x, y) = \frac{-\mu_{1}}{\sqrt{\left( x - a \right)^{2} + y^{2}}} - \frac{\mu_{2}}{\sqrt{\left( x + a \right)^{2} + y^{2}}} .
</math>
 
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]],
 
:<math>
x = a \cosh \xi \cos \eta,
</math>
 
:<math>
y = a \sinh \xi \sin \eta,
</math>
 
the potential energy can be written as
 
:<math>
V(\xi, \eta) = \frac{-\mu_{1}}{a\left( \cosh \xi - \cos \eta \right)} - \frac{\mu_{2}}{a\left( \cosh \xi + \cos \eta \right)}
= \frac{-\mu_{1} \left( \cosh \xi + \cos \eta \right) - \mu_{2} \left( \cosh \xi - \cos \eta \right)}{a\left( \cosh^{2} \xi - \cos^{2} \eta \right)},</math>
 
and the kinetic energy as
 
:<math>
T = \frac{ma^{2}}{2} \left( \cosh^{2} \xi - \cos^{2} \eta \right) \left( \dot{\xi}^{2} + \dot{\eta}^{2} \right).
</math>
 
This is a Liouville dynamical system if ξ and η are taken as φ<sub>1</sub> and φ<sub>2</sub>, respectively; thus, the function ''Y'' equals
 
:<math>
Y = \cosh^{2} \xi - \cos^{2} \eta
</math>
 
and the function ''W'' equals
 
:<math>
W = -\mu_{1} \left( \cosh \xi + \cos \eta \right) - \mu_{2} \left( \cosh \xi - \cos \eta \right)
</math>
 
Using the general solution for a Liouville dynamical system below, one obtains 
 
:<math>
\frac{ma^{2}}{2} \left( \cosh^{2} \xi - \cos^{2} \eta \right)^{2} \dot{\xi}^{2} = E \cosh^{2} \xi + \left( \frac{\mu_{1} + \mu_{2}}{a} \right) \cosh \xi - \gamma
</math>
 
:<math>
\frac{ma^{2}}{2} \left( \cosh^{2} \xi - \cos^{2} \eta \right)^{2} \dot{\eta}^{2} = -E \cos^{2} \eta + \left( \frac{\mu_{1} - \mu_{2}}{a} \right) \cos \eta + \gamma
</math>
 
Introducing a parameter ''u'' by the formula
 
:<math>
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}},
</math>
 
gives the [[parametric solution]]
 
:<math>
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}}.
</math>
 
Since these are [[elliptic integral]]s, 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,
 
:<math>
r_{1}^{2} r_{2}^{2} \left( \frac{d\theta_{1}}{dt} \right) \left( \frac{d\theta_{2}}{dt} \right) -
2c \left[ \mu_{1} \cos \theta_{1} + \mu_{2} \cos \theta_{2} \right],
</math>
 
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
 
:<math>
\varphi_{r} = \int dq_{r} \sqrt{v_{r}(q_{r})},
</math>
 
giving the relation
 
:<math>
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{\varphi}_{1}^{2} + \dot{\varphi}_{2}^{2} + \cdots + \dot{\varphi}_{s}^{2} = F,
</math>
 
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'',
 
:<math>
Y = \chi_{1}(\varphi_{1}) + \chi_{2}(\varphi_{2}) + \cdots + \chi_{s}(\varphi_{s}),
</math>
 
the kinetic energy can be written as
 
:<math>
T = \frac{1}{2} Y F.
</math>
 
Similarly, denoting the sum of the ω functions by ''W''
 
:<math>
W = \omega_{1}(\varphi_{1}) + \omega_{2}(\varphi_{2}) + \cdots + \omega_{s}(\varphi_{s}),
</math>
 
the potential energy ''V'' can be written as
 
:<math>
V = \frac{W}{Y}.
</math>
 
===Lagrange equation===
 
The Lagrange equation for the ''r''<sup>th</sup> variable <math>\varphi_{r}</math> is
 
:<math>
\frac{d}{dt} \left( \frac{\partial T}{\partial \dot{\varphi}_{r}} \right) =
\frac{d}{dt} \left( Y \dot{\varphi}_{r} \right) = \frac{1}{2} F \frac{\partial Y}{\partial \varphi_{r}}
-\frac{\partial V}{\partial \varphi_{r}}.
</math>
 
Multiplying both sides by <math>2 Y \dot{\varphi}_{r}</math>, re-arranging, and exploiting the relation 2''T = YF'' yields the equation
 
:<math>
2 Y \dot{\varphi}_{r} \frac{d}{dt} \left(Y \dot{\varphi}_{r}\right) =
2T\dot{\varphi}_{r} \frac{\partial Y}{\partial \varphi_{r}} - 2 Y \dot{\varphi}_{r} \frac{\partial V}{\partial \varphi_{r}} =
2 \dot{\varphi}_{r} \frac{\partial}{\partial \varphi_{r}} \left[ (E-V) Y \right],
</math>
 
which may be written as
 
:<math>
\frac{d}{dt} \left(Y^{2} \dot{\varphi}_{r}^{2} \right) =
2 E \dot{\varphi}_{r} \frac{\partial Y}{\partial \varphi_{r}} - 2 \dot{\varphi}_{r} \frac{\partial W}{\partial \varphi_{r}} =
2E \dot{\varphi}_{r} \frac{d\chi_{r} }{d\varphi_{r}} - 2 \dot{\varphi}_{r} \frac{d\omega_{r}}{d\varphi_{r}},
</math>
 
where ''E = T + V'' is the (conserved) total energy.  It follows that
 
:<math>
\frac{d}{dt} \left(Y^{2} \dot{\varphi}_{r}^{2} \right) =
2\frac{d}{dt} \left( E \chi_{r} - \omega_{r} \right),
</math>
 
which may be integrated once to yield
 
:<math>
\frac{1}{2} Y^{2} \dot{\varphi}_{r}^{2} = E \chi_{r} - \omega_{r}  + \gamma_{r},
</math>
 
where the <math>\gamma_{r}</math> are constants of integration subject to the energy conservation
 
:<math>
\sum_{r=1}^{s} \gamma_{r} = 0.
</math>
 
Inverting, taking the square root and separating the variables yields a set of separably integrable equations:
 
:<math>
\frac{\sqrt{2}}{Y} dt = \frac{d\varphi_{1}}{\sqrt{E \chi_{1} - \omega_{1} + \gamma_{1}}} =
\frac{d\varphi_{2}}{\sqrt{E \chi_{2} - \omega_{2} + \gamma_{2}}} = \cdots =
\frac{d\varphi_{s}}{\sqrt{E \chi_{s} - \omega_{s} + \gamma_{s}}}.
</math>
 
==References==
 
{{reflist|1}}
 
==Further reading==
 
* {{cite book | last = Whittaker | first = ET | year = 1937 | title = A Treatise on the Analytical Dynamics of Particles and Rigid Bodies, with an Introduction to the Problem of Three Bodies | edition = 4th | publisher = Dover Publications | location = New York | id = ISBN }}
 
[[Category:Classical mechanics]]

Latest revision as of 20:43, 25 November 2014

I would like to introduce myself to you, I am Jayson Simcox but I don't like when people use my complete name. Distributing manufacturing has been his occupation for some time. Mississippi is where his home is. To perform lacross is one of the issues she loves most.

Review my blog: telephone psychic (just click the up coming internet page)