In classical mechanics , a Liouville dynamical system is an exactly solvable 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:[1]
T = 1 2 { u 1 ( q 1 ) + u 2 ( q 2 ) + ⋯ + u s ( q s ) } { v 1 ( q 1 ) q ˙ 1 2 + v 2 ( q 2 ) q ˙ 2 2 + ⋯ + v s ( q s ) q ˙ s 2 } {\displaystyle 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\}} V = w 1 ( q 1 ) + w 2 ( q 2 ) + ⋯ + w s ( q s ) u 1 ( q 1 ) + u 2 ( q 2 ) + ⋯ + u s ( q s ) {\displaystyle 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
2 Y d t = d φ 1 E χ 1 − ω 1 + γ 1 = d φ 2 E χ 2 − ω 2 + γ 2 = ⋯ = d φ s E χ s − ω s + γ s {\displaystyle {\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}}}}} where E = T + V is the conserved energy and the γ s {\displaystyle \gamma _{s}} qs to φs , and the functions us and ws substituted by their counterparts χs and ωs . 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
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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
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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 ) = − μ 1 ( x − a ) 2 + y 2 − μ 2 ( x + a ) 2 + y 2 . {\displaystyle 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}}}}.} 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 ξ cos η , {\displaystyle x=a\cosh \xi \cos \eta ,} y = a sinh ξ sin η , {\displaystyle y=a\sinh \xi \sin \eta ,} the potential energy can be written as
V ( ξ , η ) = − μ 1 a ( cosh ξ − cos η ) − μ 2 a ( cosh ξ + cos η ) = − μ 1 ( cosh ξ + cos η ) − μ 2 ( cosh ξ − cos η ) a ( cosh 2 ξ − cos 2 η ) , {\displaystyle 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)}},} and the kinetic energy as
T = m a 2 2 ( cosh 2 ξ − cos 2 η ) ( ξ ˙ 2 + η ˙ 2 ) . {\displaystyle T={\frac {ma^{2}}{2}}\left(\cosh ^{2}\xi -\cos ^{2}\eta \right)\left({\dot {\xi }}^{2}+{\dot {\eta }}^{2}\right).} 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 ξ − cos 2 η {\displaystyle Y=\cosh ^{2}\xi -\cos ^{2}\eta } and the function W equals
W = − μ 1 ( cosh ξ + cos η ) − μ 2 ( cosh ξ − cos η ) {\displaystyle W=-\mu _{1}\left(\cosh \xi +\cos \eta \right)-\mu _{2}\left(\cosh \xi -\cos \eta \right)} Using the general solution for a Liouville dynamical system below, one obtains
m a 2 2 ( cosh 2 ξ − cos 2 η ) 2 ξ ˙ 2 = E cosh 2 ξ + ( μ 1 + μ 2 a ) cosh ξ − γ {\displaystyle {\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 } m a 2 2 ( cosh 2 ξ − cos 2 η ) 2 η ˙ 2 = − E cos 2 η + ( μ 1 − μ 2 a ) cos η + γ {\displaystyle {\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 } Introducing a parameter u by the formula
d u = d ξ E cosh 2 ξ + ( μ 1 + μ 2 a ) cosh ξ − γ = d η − E cos 2 η + ( μ 1 − μ 2 a ) cos η + γ , {\displaystyle 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 = ∫ d ξ E cosh 2 ξ + ( μ 1 + μ 2 a ) cosh ξ − γ = ∫ d η − E cos 2 η + ( μ 1 − μ 2 a ) cos η + γ . {\displaystyle 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
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The bicentric problem has a constant of motion, namely,
r 1 2 r 2 2 d θ 1 d t d θ 2 d t + 2 c ( μ 1 cos θ 1 − μ 2 cos θ 2 ) , {\displaystyle r_{1}^{2}\,r_{2}^{2}{\frac {d\theta _{1}}{dt}}{\frac {d\theta _{2}}{dt}}+2\,c\left(\mu _{1}\cos \theta _{1}-\mu _{2}\cos \theta _{2}\right),} from which the problem can be solved using the method of the last multiplier.
Derivation
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New variables
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To eliminate the v functions, the variables are changed to an equivalent set
φ r = ∫ d q r v r ( q r ) , {\displaystyle \varphi _{r}=\int dq_{r}{\sqrt {v_{r}(q_{r})}},} giving the relation
v 1 ( q 1 ) q ˙ 1 2 + v 2 ( q 2 ) q ˙ 2 2 + ⋯ + v s ( q s ) q ˙ s 2 = φ ˙ 1 2 + φ ˙ 2 2 + ⋯ + φ ˙ s 2 = F , {\displaystyle 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,} 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 = χ 1 ( φ 1 ) + χ 2 ( φ 2 ) + ⋯ + χ s ( φ s ) , {\displaystyle Y=\chi _{1}(\varphi _{1})+\chi _{2}(\varphi _{2})+\cdots +\chi _{s}(\varphi _{s}),} the kinetic energy can be written as
T = 1 2 Y F . {\displaystyle T={\frac {1}{2}}YF.} Similarly, denoting the sum of the ω functions by W
W = ω 1 ( φ 1 ) + ω 2 ( φ 2 ) + ⋯ + ω s ( φ s ) , {\displaystyle W=\omega _{1}(\varphi _{1})+\omega _{2}(\varphi _{2})+\cdots +\omega _{s}(\varphi _{s}),} the potential energy V can be written as
V = W Y . {\displaystyle V={\frac {W}{Y}}.} Lagrange equation
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The Lagrange equation for the r th variable φ r {\displaystyle \varphi _{r}}
d d t ( ∂ T ∂ φ ˙ r ) = d d t ( Y φ ˙ r ) = 1 2 F ∂ Y ∂ φ r − ∂ V ∂ φ r . {\displaystyle {\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}}}.} Multiplying both sides by 2 Y φ ˙ r {\displaystyle 2Y{\dot {\varphi }}_{r}} T = YF yields the equation
2 Y φ ˙ r d d t ( Y φ ˙ r ) = 2 T φ ˙ r ∂ Y ∂ φ r − 2 Y φ ˙ r ∂ V ∂ φ r = 2 φ ˙ r ∂ ∂ φ r [ ( E − V ) Y ] , {\displaystyle 2Y{\dot {\varphi }}_{r}{\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)=2T{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2Y{\dot {\varphi }}_{r}{\frac {\partial V}{\partial \varphi _{r}}}=2{\dot {\varphi }}_{r}{\frac {\partial }{\partial \varphi _{r}}}\left[(E-V)Y\right],} which may be written as
d d t ( Y 2 φ ˙ r 2 ) = 2 E φ ˙ r ∂ Y ∂ φ r − 2 φ ˙ r ∂ W ∂ φ r = 2 E φ ˙ r d χ r d φ r − 2 φ ˙ r d ω r d φ r , {\displaystyle {\frac {d}{dt}}\left(Y^{2}{\dot {\varphi }}_{r}^{2}\right)=2E{\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}}},} where E = T + V is the (conserved) total energy. It follows that
d d t ( Y 2 φ ˙ r 2 ) = 2 d d t ( E χ r − ω r ) , {\displaystyle {\frac {d}{dt}}\left(Y^{2}{\dot {\varphi }}_{r}^{2}\right)=2{\frac {d}{dt}}\left(E\chi _{r}-\omega _{r}\right),} which may be integrated once to yield
1 2 Y 2 φ ˙ r 2 = E χ r − ω r + γ r , {\displaystyle {\frac {1}{2}}Y^{2}{\dot {\varphi }}_{r}^{2}=E\chi _{r}-\omega _{r}+\gamma _{r},} where the γ r {\displaystyle \gamma _{r}}
∑ r = 1 s γ r = 0. {\displaystyle \sum _{r=1}^{s}\gamma _{r}=0.} Inverting, taking the square root and separating the variables yields a set of separably integrable equations:
2 Y d t = d φ 1 E χ 1 − ω 1 + γ 1 = d φ 2 E χ 2 − ω 2 + γ 2 = ⋯ = d φ s E χ s − ω s + γ s . {\displaystyle {\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}}}}.} References
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Further reading
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![{\displaystyle 2Y{\dot {\varphi }}_{r}{\frac {d}{dt}}\left(Y{\dot {\varphi }}_{r}\right)=2T{\dot {\varphi }}_{r}{\frac {\partial Y}{\partial \varphi _{r}}}-2Y{\dot {\varphi }}_{r}{\frac {\partial V}{\partial \varphi _{r}}}=2{\dot {\varphi }}_{r}{\frac {\partial }{\partial \varphi _{r}}}\left[(E-V)Y\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1addebe0acd83dc477577658fa1c849a6171128d)
