《力学》课程教学资源（参考资料）拉普拉斯-隆格-楞次矢量 Runge-Lenz-Symmetries

KARLSTAD UNIVERSITY DEPARTMENT OF ENGINEERING AND PHYSICS Analytical mechanics RUNGE-LENZ-SYMMETRIES Author: Supervisor: Rasmus Laven Juirgen Fuchs January 21,2016

KARLSTAD UNIVERSITY DEPARTMENT OF ENGINEERING AND PHYSICS Analytical mechanics Runge-Lenz-Symmetries Author: Rasmus Lav´en Supervisor: J¨urgen Fuchs January 21, 2016

Abstract In this report we study the symmetries that correspond to the conservation of the Runge-Lenz vector in the Kepler problem.In section 2 we use Noether's theorem to define a Runge-Lenz vector as a consequence of an invariance of the action integral.It's shown that such a vector exists for all central potentials. In section 3 we describe the Kepler problem in space-time.By choosing a nice parametrization we show that the equations of motion and the conservation of energy describe a harmonic oscillator with an extra derivative in four dimensions and a four dimensional sphere,respectively.From this we define a conserved tensor.The components of this tensor correspond to the Runge-Lenz vector and angular momentum. 1

Abstract In this report we study the symmetries that correspond to the conservation of the Runge-Lenz vector in the Kepler problem. In section 2 we use Noether’s theorem to define a Runge-Lenz vector as a consequence of an invariance of the action integral. It’s shown that such a vector exists for all central potentials. In section 3 we describe the Kepler problem in space-time. By choosing a nice parametrization we show that the equations of motion and the conservation of energy describe a harmonic oscillator with an extra derivative in four dimensions and a four dimensional sphere, respectively. From this we define a conserved tensor. The components of this tensor correspond to the Runge-Lenz vector and angular momentum. 1

CONTENTS Contents 1 Introduction 3 2 Runge-Lenz-vector from a symmetry of the action integral 5 2.1 General aspects of conserved quantities 5 22 runge-Lenz-vector...:·················.····· 7 3 Extra symmetries in the Kepler problem 13 2

CONTENTS Contents 1 Introduction 3 2 Runge-Lenz-vector from a symmetry of the action integral 5 2.1 General aspects of conserved quantities . . . . . . . . . . . . . . . . 5 2.2 Runge-Lenz-vector . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 3 Extra symmetries in the Kepler problem 13 2

Introduction 1 Introduction One of the most famous problems in classical mechanics is the Kepler problem. This is the problem of a point mass in a central force field of the form F(r)= 一k (1) A special thing about this problem is that there exists an extra conserved quantity besides the total energy and the angular momentum.This quantity is a vector called the Runge-Lenz vector.The Runge-Lenz vector A for a particle of mass m moving n a cra force fielddefine A=p×Lx (2) Here p is the momentum of the particle,L is the angular momentum,m the mass and r the position vector of the particle.In the Kepler-problem the angular momentum and energy are conserved.One might then think that there exist seven conserved quantities.This is not the case,because the variables are not independent of each other.From Figure 1 one sees that the Runge-Lenz-vector lies in the plane of motion and thus A.L=0.Further on by taking the dot product A.A one obtains A2 =m2k2+2mEL2.From this one can see that there are only five independent constants of motion in the Kepler-problem [1].The orbits in the Kepler-problem are conic-sections.A nice way to realize this is by using the Runge-Lenz-vector.By denoting 6 as the angle between the position vector and the Runge-Lenz-vector one has A·r=Arcost9=r·p×L-mkr=L·r×p-mkr=L2-mkr. (3) 3

Introduction 1 Introduction One of the most famous problems in classical mechanics is the Kepler problem. This is the problem of a point mass in a central force field of the form F(r) = −k r 2 er. (1) A special thing about this problem is that there exists an extra conserved quantity besides the total energy and the angular momentum. This quantity is a vector called the Runge-Lenz vector. The Runge-Lenz vector A for a particle of mass m moving in a central force field F = − k r 2 er is defined as A := p × L − mk |r| r. (2) Here p is the momentum of the particle, L is the angular momentum, m the mass and r the position vector of the particle. In the Kepler-problem the angular momentum and energy are conserved. One might then think that there exist seven conserved quantities. This is not the case, because the variables are not independent of each other. From Figure 1 one sees that the Runge-Lenz-vector lies in the plane of motion and thus A·L = 0. Further on by taking the dot product A · A one obtains A2 = m2k 2 + 2mEL2 . From this one can see that there are only five independent constants of motion in the Kepler-problem [1]. The orbits in the Kepler-problem are conic-sections. A nice way to realize this is by using the Runge-Lenz-vector. By denoting θ as the angle between the position vector and the Runge-Lenz-vector one has A · r = Arcosθ = r · p × L − mkr = L · r × p − mkr = L 2 − mkr. (3) 3

Introduction A pxL mkt pxL 1 A mkf 37 城≥ 4 Figure 1:Illustration of how the Runge-Lenz-vector is oriented in the Kepler orbits [3. Thus we can solve for r as mk (④) which is the equation of a conic section with eccentricity e=A/mk provided A is constant [3].Thus we see that the conservation of the Runge-Lenz-vector actually is the reason that the orbits of the Kepler-problem are closed.The Runge-Lenz- vector can in principle be generalized to any central potential as we shall see in Section 2.2.However these generalized Runge-Lenz-vectors are often complicated functions and usually not expressible in closed form[3].Since the conservation of the Runge-Lenz-vector implies closed orbits for the Kepler-problem one might expect that there exists some analogue of this derivation for the isotropic harmonic oscillator.This is indeed the case but we shall leave this question here [1]. 4

Introduction . Figure 1: Illustration of how the Runge-Lenz-vector is oriented in the Kepler orbits [3]. Thus we can solve for r as 1 r = mk L2  1 + A mk cosθ , (4) which is the equation of a conic section with eccentricity e = A/mk provided A is constant [3]. Thus we see that the conservation of the Runge-Lenz-vector actually is the reason that the orbits of the Kepler-problem are closed. The Runge-Lenz￾vector can in principle be generalized to any central potential as we shall see in Section 2.2. However these generalized Runge-Lenz-vectors are often complicated functions and usually not expressible in closed form[3]. Since the conservation of the Runge-Lenz-vector implies closed orbits for the Kepler-problem one might expect that there exists some analogue of this derivation for the isotropic harmonic oscillator. This is indeed the case but we shall leave this question here [1]. 4

Runge-Lenz-vector from a symmetry of the action integral 2 Runge-Lenz-vector from a symmetry of the ac- tion integral 2.1 General aspects of conserved quantities Noether's theorem relates conservation laws with symmetries.A somewhat simpli- fied way to state this theorem is to say that if the action integral is invariant under some transformation then there is some conserved quantity.Hamilton's principle says that the physical path of a system is such that the action is stationary.This simply means that the action is invariant under an infinitesimal variation of the path as g(t)(t)=qi(t)+oni.Besides the Euler-Lagrange-equations,this also implies conservations laws.A trivial example is a Lagrangian with some cyclic co- ordinate,then the canonical momentum conjugate to this coordinate is conserved. This follows directly from the Euler-Lagrange-equations and thus this conservation law follows from Hamilton's principle.Noether's theorem can somewhat simplified be stated as,an invariance of the Lagrangian corresponds to a conservation law. To see how this works we vary the path as 9→q=9+0q, (5) where the variation ogi is such that it is zero at the endpoints.The velocity then becomes 4→6=+0： (6) The Lagrangian of the new coordinates can be expanded in a power series as 60=e创+影+∑ 6+ (7)

Runge-Lenz-vector from a symmetry of the action integral 2 Runge-Lenz-vector from a symmetry of the ac￾tion integral 2.1 General aspects of conserved quantities Noether’s theorem relates conservation laws with symmetries. A somewhat simpli- fied way to state this theorem is to say that if the action integral is invariant under some transformation then there is some conserved quantity. Hamilton’s principle says that the physical path of a system is such that the action is stationary. This simply means that the action is invariant under an infinitesimal variation of the path as q(t) 7→ q 0 i (t) = qi(t) + δηi . Besides the Euler-Lagrange-equations, this also implies conservations laws. A trivial example is a Lagrangian with some cyclic co￾ordinate, then the canonical momentum conjugate to this coordinate is conserved. This follows directly from the Euler-Lagrange-equations and thus this conservation law follows from Hamilton’s principle. Noether’s theorem can somewhat simplified be stated as, an invariance of the Lagrangian corresponds to a conservation law. To see how this works we vary the path as qi 7→ q 0 i = qi + δqi , (5) where the variation δqi is such that it is zero at the endpoints. The velocity then becomes q˙i 7→ q˙ 0 i = ˙qi + δq˙i . (6) The Lagrangian of the new coordinates can be expanded in a power series as L(q 0 i , q˙ 0 i ) = L(qi , q˙i) +X i ∂L ∂qi δqi + X i ∂L ∂q˙i δq˙i + . . . (7) 5

2.1 General aspects of conserved quantities Taking only the linear terms we can write the variation of the Lagrangian as 6L= (8) We now assume that the variation of the action can be written as the integral of the variation of the Lagrangian 0=6S= a∑ aL d aLl δqi +, dt agi| (9) In the last equality we used integration by parts and that the variation is zero at the endpoints of integration.By the fundamental lemma of the calculus of variations this integral is zero only if the integrand is zero.Also we assume that the coordinates are independent,which implies that the coefficient in front of each Ogi is zero,and thus we obtain the Euler-Lagrange-equations ∂Ld∂L =0. (10) Oqi dt aqi These are one equation for each coordinate gi.By the Euler-Lagrange-equations we can write ∑- 三 d aL Zdt∂4 (11) By inserting this expression in the expression for the variation of the Lagrangian we can write =∑ d aL sqs dt aq 0q (12)

2.1 General aspects of conserved quantities Taking only the linear terms we can write the variation of the Lagrangian as δL = X i ∂L ∂qi δqi + X i ∂L ∂q˙i δq˙i . (8) We now assume that the variation of the action can be written as the integral of the variation of the Lagrangian 0 = δS = Z t2 t1 dtδL = Z t2 t1 dtX i  ∂L ∂qi δqi + ∂L ∂q˙i δq˙i  = Z t2 t1 dtX i  ∂L ∂qi − d dt ∂L ∂q˙i  δqi . (9) In the last equality we used integration by parts and that the variation is zero at the endpoints of integration. By the fundamental lemma of the calculus of variations this integral is zero only if the integrand is zero. Also we assume that the coordinates are independent, which implies that the coefficient in front of each δqi is zero, and thus we obtain the Euler-Lagrange-equations ∂L ∂qi − d dt ∂L ∂q˙i = 0. (10) These are one equation for each coordinate qi . By the Euler-Lagrange-equations we can write X i ∂L ∂qi = X i d dt ∂L ∂q˙i . (11) By inserting this expression in the expression for the variation of the Lagrangian we can write δL = X i  d dt ∂L ∂q˙i δqi  + X i ∂L ∂q˙i δq˙i = X i d dt ∂L ∂q˙ δqi  . (12) 6

2.2 Runge-Lenz-vector This expressions is a good way of defining constants of motion.As an example we consider the case of a central potential.The Lagrangian can then be written as L=罗2++)-V) (13) This Lagrangian of this system is obviously invariant under rotations,e.g rotation about the x-axis by a small angle 60 as r→r'=r+60rXex (14) From Equation (12)we can now define a constant of motion as 1 [aLs g=∑nc×eh=py-p.=L, (15) which is the r-component of angular momentum.Although this seems as a useful method it is not possible to define the Runge-Lenz-vector as a consequence of an invariance of the Lagrangian,but for the whole action integral.By this method it is only possible to define constants of motions such as linear and angular momentum which are conserved due to some cyclic coordinate. 2.2 Runge-Lenz-vector This section is based on the work done in [2].We will now see how we can define the Runge-Lenz-vector by a similar variational calculation.We study a system with n generalized coordinates g1,...gn,described by a time independent Lagrangian L(ge,)We now try a slightly more complicated variation of both the path and time and then demand that the action integral is invariant.The path and time is 7

2.2 Runge-Lenz-vector This expressions is a good way of defining constants of motion. As an example we consider the case of a central potential. The Lagrangian can then be written as L = m 2 ( ˙x 2 + ˙y 2 + ˙z 2 ) − V (r). (13) This Lagrangian of this system is obviously invariant under rotations, e.g rotation about the x-axis by a small angle δθ as r 7→ r 0 = r + δθr × ex. (14) From Equation (12) we can now define a constant of motion as 1 δθ X i  ∂L ∂q˙i δqi  = X i pi(r × ex)i = zpy − pzy = Lx, (15) which is the x-component of angular momentum. Although this seems as a useful method it is not possible to define the Runge-Lenz-vector as a consequence of an invariance of the Lagrangian, but for the whole action integral. By this method it is only possible to define constants of motions such as linear and angular momentum which are conserved due to some cyclic coordinate. 2.2 Runge-Lenz-vector This section is based on the work done in [2]. We will now see how we can define the Runge-Lenz-vector by a similar variational calculation. We study a system with n generalized coordinates q1, . . . qn , described by a time independent Lagrangian L(qk, q˙k). We now try a slightly more complicated variation of both the path and time and then demand that the action integral is invariant. The path and time is 7

2.2 Runge-Lenz-vector varied as 9→q=q+6a(qk,t) and (16) t→t=t+δB, (17) with some small parameter 6.Here the variation of the coordinates is a function of the original variables and time.The variation added to time 03 is taken to be constant.Since the variation of time is a constant and the Lagrangian is assumed to be time-independent,the transformation of time will not change the action integral.But the variation of time is still introduced for later calculations.Also here the variations are such that they vanish at the endpoints of integration.The velocities then become =:+dd: (18) We can as before expand the Lagrangian in a power series and write the variation of the Lagrangian as dL=∑w6 6a+入6oiy (19) We now demand that the action integral is invariant under this transformation i.e L(qk,q)dt= dt[L(g,g)+gg,] dt], (20) t t where we added a total time derivative which we can do since the integrand is determined only up to a total time derivative.The variation of the action integral then becomes 0=6S 广+割 (21) where f is defined to be the variation of g due to the coordinate transformation. We now simply focus on the integrand since it is the most important part.We can 8

2.2 Runge-Lenz-vector varied as qi 7→ q 0 i = qi + δαi(qk, t) and (16) t 7→ t 0 = t + δβ, (17) with some small parameter δ. Here the variation of the coordinates is a function of the original variables and time. The variation added to time δβ is taken to be constant. Since the variation of time is a constant and the Lagrangian is assumed to be time-independent, the transformation of time will not change the action integral. But the variation of time is still introduced for later calculations. Also here the variations are such that they vanish at the endpoints of integration. The velocities then become q˙ 0 i = ˙qi + δα˙ i . (18) We can as before expand the Lagrangian in a power series and write the variation of the Lagrangian as δL = X i ∂L ∂qi δα + X i ∂L ∂q˙ii δα˙ i . (19) We now demand that the action integral is invariant under this transformation i.e Z t2 t1 L(qk, q˙k)dt = Z t2 t1 dt L(q 0 k , q˙ 0 k ) + dg(q, t) dt , (20) where we added a total time derivative which we can do since the integrand is determined only up to a total time derivative. The variation of the action integral then becomes 0 = δS = Z t2 t1 dt δL + df dt  , (21) where f is defined to be the variation of g due to the coordinate transformation. We now simply focus on the integrand since it is the most important part. We can 8

2.2 Runge-Lenz-vector rewrite the integrand in the variation of the transformed action integral as 6L+ df =6L+ +d业8- dt dtdt dt d 歌-+69+小- d aL aL dt dt∂9k (6ak-9k6B) 8qk [∑0m-40+u9+小 d (22) dt In the last step we simply used the Euler-Lagrange equations.We can see that an invariance of the action integral under this more general transformation implies both the Euler-Lagrange equations and another equation which is a conservation law.As before the variation of the action integral is zero only if the integrand is zero and thus we obtain the expression d dt a亚(6ak-gi0)+L6B+f =0. (23) This is an expression involving the conserved quantities of the system.The terms in the time derivative can be identified as a linear combination of the conserved quantities of the system.By rewriting this we get 0= d dt 0Las-H6+f小 (24) where H is the Hamiltonian of the system.This expression has three terms:the Hamiltonian which comes from the variation of time,the conjugate momentum which comes from the variation of coordinates and the function f which we will see is important for understanding how the Runge-Lenz vector can be conserved. To see more conserved quantities we need to determine the functions f(g,t)and a(g,t)such that this expression holds.To proceed further we restrict ourself to 9

2.2 Runge-Lenz-vector rewrite the integrand in the variation of the transformed action integral as δL + df dt = δL + df dt + dL dt δβ − dL dt δβ = d dtX k ∂L ∂q˙k ￾ δα − q˙kδβ
+ Lδβ + f  − X k  d dt ∂L ∂q˙k − ∂L ∂qk  ￾ δαk − q˙kδβ) = d dtX k ∂L ∂q˙k ￾ δα − q˙kδβ
+ Lδβ + f  . (22) In the last step we simply used the Euler-Lagrange equations. We can see that an invariance of the action integral under this more general transformation implies both the Euler-Lagrange equations and another equation which is a conservation law. As before the variation of the action integral is zero only if the integrand is zero and thus we obtain the expression d dtX k ∂L ∂q˙k ￾ δαk − q˙kδt
+ Lδβ + f  = 0. (23) This is an expression involving the conserved quantities of the system. The terms in the time derivative can be identified as a linear combination of the conserved quantities of the system. By rewriting this we get 0 = d dtX k ∂L ∂q˙k δαk − Hδβ + f  , (24) where H is the Hamiltonian of the system. This expression has three terms: the Hamiltonian which comes from the variation of time, the conjugate momentum which comes from the variation of coordinates and the function f which we will see is important for understanding how the Runge-Lenz vector can be conserved. To see more conserved quantities we need to determine the functions f(q, t) and δα(q, t) such that this expression holds. To proceed further we restrict ourself to 9