Families of orbits compatible with galactic potentials

Abstract

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We nd galactic potentials described by polynomial perturbations of harmonic oscillators, which are compatible with families of orbits \(x^{p} y=const(p\neq0)\). To this aim we apply the techniques of the planar inverse problem of dynamics.

Authors

Mira-Cristiana Anisiu
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, Cluj-Napoca, Romania,

Keywords

galactic potentials; inverse problem of dynamics.

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Anisiu Mira-Cristiana, Families of orbits compatible with galactic potentials, AIP Conference Proceedings, vol. 895, 2007, pag. 308-312

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references

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2008AnisiuBozisSpatial

Spatial Families of Orbits in 2D Conservative Fields

Mira-Cristiana Anisiu* and G. Bozis ^(†){ }^{\dagger} ^(**){ }^{*} T Popoviciu Institute of Numerical Analysis, Romanian Academy, Cluj-Napoca, Romania , e-mail: mira@math.ubbcluj.ro ^(†){ }^{\dagger} Department of Physics, Aristotle University of Thessaloniki, GR-54006, Greece, e-mail: gbozis@auth.gr

Abstract

In the framework of the 3D inverse problem of dynamics, we establish the conditions which must be fulfilled by a spatial family of curves to possibly be described by a unit mass particle under the action of a 2D potential V = v ( y , z ) V = v ( y , z ) V=v(y,z)V=v(y, z)V=v(y,z), and give a method to find the potential.

Keywords: inverse problem of Dynamics
PACS: 95.10.Ce

INTRODUCTION

In the 3D inverse problem of Dynamics, assuming that a two-parametric set of orbits
(1) f ( x , y , z ) = c 1 , g ( x , y , z ) = c 2 (1) f ( x , y , z ) = c 1 , g ( x , y , z ) = c 2 {:(1)f(x","y","z)=c_(1)","quad g(x","y","z)=c_(2):}\begin{equation*} f(x, y, z)=c_{1}, \quad g(x, y, z)=c_{2} \tag{1} \end{equation*}(1)f(x,y,z)=c1,g(x,y,z)=c2
in the O x y z O x y z OxyzO x y zOxyz space can be traced by a unit mass particle in the presence of an unknown potential V = V ( x , y , z ) V = V ( x , y , z ) V=V(x,y,z)V=V(x, y, z)V=V(x,y,z), one aims to finding the potential (see [1] for a review of the results up to 1995).
Bozis and Kotoulas [2] and also Anisiu [3] produced a system of two linear in V ( x , y , z ) V ( x , y , z ) V(x,y,z)V(x, y, z)V(x,y,z) PDEs, one of the first and one of the second order, which will be used in what follows.
We shall treat the problem for the special case of potentials V = v ( y , z ) V = v ( y , z ) V=v(y,z)V=v(y, z)V=v(y,z).

THE EQUATIONS OF THE PROBLEM

We deal with two-parametric families of orbits written in the form (1), which are in an one-to-one correspondence with a pair ( α , β α , β alpha,beta\alpha, \betaα,β ) of 'slope functions' defined by
(2) α = f z g x f x g z f y g z f z g y , β = f x g y f y g x f y g z f z g y (2) α = f z g x f x g z f y g z f z g y , β = f x g y f y g x f y g z f z g y {:(2)alpha=(f_(z)g_(x)-f_(x)g_(z))/(f_(y)g_(z)-f_(z)g_(y))","quad beta=(f_(x)g_(y)-f_(y)g_(x))/(f_(y)g_(z)-f_(z)g_(y)):}\begin{equation*} \alpha=\frac{f_{z} g_{x}-f_{x} g_{z}}{f_{y} g_{z}-f_{z} g_{y}}, \quad \beta=\frac{f_{x} g_{y}-f_{y} g_{x}}{f_{y} g_{z}-f_{z} g_{y}} \tag{2} \end{equation*}(2)α=fzgxfxgzfygzfzgy,β=fxgyfygxfygzfzgy
The indices denote partial derivatives.
Let us assume that α 0 0 α 0 0 alpha_(0)!=0\alpha_{0} \neq 0α00 and adopt the notation
(3) ε ¯ = ( 1 , α , β ) , α 0 = ε ¯ grad α , β 0 = ε ¯ grad β , Θ = 1 + α 2 + β 2 , n = Θ α 0 , n 0 = ε ¯ grad n . (3) ε ¯ = ( 1 , α , β ) , α 0 = ε ¯ grad α , β 0 = ε ¯ grad β , Θ = 1 + α 2 + β 2 , n = Θ α 0 , n 0 = ε ¯ grad n . {:(3) bar(epsi)=(1","alpha","beta)","alpha_(0)= bar(epsi)grad alpha","beta_(0)= bar(epsi)grad beta","Theta=1+alpha^(2)+beta^(2)","n=(Theta)/(alpha_(0))","n_(0)= bar(epsi)grad n.:}\begin{equation*} \bar{\varepsilon}=(1, \alpha, \beta), \alpha_{0}=\bar{\varepsilon} \operatorname{grad} \alpha, \beta_{0}=\bar{\varepsilon} \operatorname{grad} \beta, \Theta=1+\alpha^{2}+\beta^{2}, n=\frac{\Theta}{\alpha_{0}}, n_{0}=\bar{\varepsilon} \operatorname{grad} n . \tag{3} \end{equation*}(3)ε¯=(1,α,β),α0=ε¯gradα,β0=ε¯gradβ,Θ=1+α2+β2,n=Θα0,n0=ε¯gradn.
We shall consider exclusively potentials of the form V = v ( y , z ) V = v ( y , z ) V=v(y,z)V=v(y, z)V=v(y,z) and families (1) with α 0 0 α 0 0 alpha_(0)!=0\alpha_{0} \neq 0α00. In this case the system of the two PDEs mentioned in the introduction becomes
(4) v z = G v y , Θ ( α + β G ) v y y + Ψ ψ y = 0 (4) v z = G v y , Θ ( α + β G ) v y y + Ψ ψ y = 0 {:(4)v_(z)=Gv_(y)","Theta(alpha+beta G)v_(yy)+Psipsi_(y)=0:}\begin{equation*} v_{z}=G v_{y}, \Theta(\alpha+\beta G) v_{y y}+\Psi \psi_{y}=0 \tag{4} \end{equation*}(4)vz=Gvy,Θ(α+βG)vyy+Ψψy=0
where G = β 0 / α 0 , Ψ = β Θ G y + α 0 ( n 0 2 ( α + β G ) ) G = β 0 / α 0 , Ψ = β Θ G y + α 0 n 0 2 ( α + β G ) G=beta_(0)//alpha_(0),Psi=beta ThetaG_(y)+alpha_(0)(n_(0)-2(alpha+beta G))G=\beta_{0} / \alpha_{0}, \Psi=\beta \Theta G_{y}+\alpha_{0}\left(n_{0}-2(\alpha+\beta G)\right)G=β0/α0,Ψ=βΘGy+α0(n02(α+βG)).
For any compatible pair of potential v ( y , z ) v ( y , z ) v(y,z)v(y, z)v(y,z) and orbit ( α , β α , β alpha,beta\alpha, \betaα,β ) real motion is allowed in the region v y / α 0 0 v y / α 0 0 -v_(y)//alpha_(0) >= 0-v_{y} / \alpha_{0} \geq 0vy/α00 ([2], [3]).

COMPATIBLE POTENTIALS V = v ( y , z ) V = v ( y , z ) V=v(y,z)V=v(y, z)V=v(y,z) AND FAMILIES (1)

It is seen from (4a) that the function G G GGG must be independent of x x xxx, i. e.
(5) G x = 0 (5) G x = 0 {:(5)G_(x)=0:}\begin{equation*} G_{x}=0 \tag{5} \end{equation*}(5)Gx=0
We assume that α + β G 0 α + β G 0 alpha+beta G!=0\alpha+\beta G \neq 0α+βG0 and we put H = Ψ / Θ ( α + β G ) H = Ψ / Θ ( α + β G ) H=Psi//Theta(alpha+beta G)H=\Psi / \Theta(\alpha+\beta G)H=Ψ/Θ(α+βG). For the PDE ( 4 b ) PDE ( 4 b ) PDE(4b)\operatorname{PDE}(4 \mathrm{~b})PDE(4 b), now written as v y y + H v y = 0 v y y + H v y = 0 v_(yy)+Hv_(y)=0v_{y y}+H v_{y}=0vyy+Hvy=0, to have a solution of the form v ( y , z ) v ( y , z ) v(y,z)v(y, z)v(y,z) it must be
(6) H x = 0 (6) H x = 0 {:(6)H_(x)=0:}\begin{equation*} H_{x}=0 \tag{6} \end{equation*}(6)Hx=0
From (4b) we get
(7) v y = D ( z ) exp ( y H ( u , z ) d u ) (7) v y = D ( z ) exp y H ( u , z ) d u {:(7)v_(y)=D(z)exp(-int^(y)H(u,z)du):}\begin{equation*} v_{y}=D(z) \exp \left(-\int^{y} H(u, z) \mathrm{d} u\right) \tag{7} \end{equation*}(7)vy=D(z)exp(yH(u,z)du)
and from (4a) we obtain v z v z v_(z)v_{z}vz. From the compatibility condition ( v y z = v z y ) v y z = v z y (v_(yz)=v_(zy))\left(v_{y z}=v_{z y}\right)(vyz=vzy) for v y v y v_(y)v_{y}vy and v z v z v_(z)v_{z}vz, as these are given by (7) and (4a) respectively, there follows the homogeneous linear first order ODE D ( z ) = J D ( z ) D ( z ) = J D ( z ) D^(')(z)=JD(z)D^{\prime}(z)=J D(z)D(z)=JD(z), where J J JJJ depends merely on z z zzz if and only if
(8) G y y G y H G H y + H z = 0 . (8) G y y G y H G H y + H z = 0 . {:(8)G_(yy)-G_(y)H-GH_(y)+H_(z)=0.:}\begin{equation*} G_{y y}-G_{y} H-G H_{y}+H_{z}=0 . \tag{8} \end{equation*}(8)GyyGyHGHy+Hz=0.
Proposition For α 0 0 , α + β G 0 α 0 0 , α + β G 0 alpha_(0)!=0,alpha+beta G!=0\alpha_{0} \neq 0, \alpha+\beta G \neq 0α00,α+βG0 and for any family ( α , β α , β alpha,beta\alpha, \betaα,β ) satisfying the conditions (5), (6) and (8), there exists a two-dimension compatible potential v = v ( y , z ) v = v ( y , z ) v=v(y,z)v=v(y, z)v=v(y,z). The potential is given by the (compatible) equations (7) and (4a). Real motion is allowed in the region defined by the inequality D ( z ) / α 0 0 D ( z ) / α 0 0 D(z)//alpha_(0) <= 0D(z) / \alpha_{0} \leq 0D(z)/α00.
If α 0 0 , α + β G = 0 α 0 0 , α + β G = 0 alpha_(0)!=0,alpha+beta G=0\alpha_{0} \neq 0, \alpha+\beta G=0α00,α+βG=0 and Ψ 0 Ψ 0 Psi!=0\Psi \neq 0Ψ0 it follows that v = v = v=v=v= const. If α 0 0 , α + β G = 0 α 0 0 , α + β G = 0 alpha_(0)!=0,alpha+beta G=0\alpha_{0} \neq 0, \alpha+\beta G=0α00,α+βG=0 and Ψ = 0 Ψ = 0 Psi=0\Psi=0Ψ=0, equation ( 4 b ) is satisfied identically and the potential is found from ( 4 a ), provided that the condition (5) holds, and it will not be uniquely determined.
Example For the family f ( x , y , z ) = x 4 y z 3 , g ( x , y , z ) = x 2 y z f ( x , y , z ) = x 4 y z 3 , g ( x , y , z ) = x 2 y z f(x,y,z)=x^(4)yz^(3),g(x,y,z)=x^(2)yzf(x, y, z)=x^{4} y z^{3}, g(x, y, z)=x^{2} y zf(x,y,z)=x4yz3,g(x,y,z)=x2yz, we get α 0 0 , α + β G α 0 0 , α + β G alpha_(0)!=0,alpha+beta G!=\alpha_{0} \neq 0, \alpha+\beta G \neqα00,α+βG 0 and obtain the compatible potential V ( x , y , z ) = ( y 2 + z 2 ) 2 V ( x , y , z ) = y 2 + z 2 2 V(x,y,z)=-(y^(2)+z^(2))^(2)V(x, y, z)=-\left(y^{2}+z^{2}\right)^{2}V(x,y,z)=(y2+z2)2.
The case α 0 = 0 α 0 = 0 alpha_(0)=0\alpha_{0}=0α0=0 and that of potentials depending on ( x , y x , y x,yx, yx,y ) or ( x , z x , z x,zx, zx,z ) will be treated elsewhere.
Acknowledgment The work of the first author was financially supported by the scientific program 2CEEX0611-96 of the Romanian Ministry of Education and Research.

REFERENCES

  1. G. Bozis, Inverse Problems 11, 687-708 (1995).
  2. G. Bozis, and T. A. Kotoulas, Inverse Problems 21, 343-356 (2005).
  3. M.-C. Anisiu, Inverse Probl. Sci. Eng. 13, 545-558 (2005).
2007

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