Dynamics in the Kepler problem on the Heisenberg group

Abstract

We study the nonholonomic motion of a point particle on the Heisenberg group around the fixed “sun” placed at the origin whose potential is given by the fundamental solution of the sub-Laplacian. In contrast with several recent papers that approach this problem as a variational one (hence a control problem) we study the equations of dynamical motion which are non-variational in nonholonomic mechanics. We find three independent first integrals of the system and show that its bounded trajectories are wound up around certain surfaces of the fourth order. We also describe some particular cases of trajectories.

Derivation of quadratic first integrals

By definition any first integral F of (6) must satisfy the following relation:

$$\begin{aligned} \frac{dF}{dt}= & {} \frac{\partial F}{\partial r} p_R + \frac{\partial F}{\partial \theta } \frac{p_S}{r^2} + \frac{\partial F}{\partial z} \frac{p_S}{2} + \frac{\partial F}{\partial p_R} \left( \frac{p_S^2}{r^3}-\frac{2kr^3}{(r^4+16z^2)^{3/2}} \right) \nonumber \\{} & {} \quad - \frac{\partial F}{\partial p_S} \frac{8kr^2z}{(r^4+16z^2)^{3/2}} = 0. \end{aligned}$$

(A1)

It is quite natural to search for the first integrals of (6) having the form of non-homogeneous polynomials in momenta.

We shall search for the quadratic integral of (6) in the form:

$$\begin{aligned} F = a \, p_R^2+d \, p_Rp_S+b \, p_S^2 + f \, p_R + g \, p_S+h, \end{aligned}$$

where all the coefficients are unknown functions which depend on r, $\theta$, z.

Lemma 10

If F is the first integral of (6), then functions f,  g must vanish identically.

Proof

Writing down the condition (A1) for such an integral F, we obtain the system of PDEs which splits into two parts. The first part contains relations between the unknown functions a, b, d, h only (see equations (A2)–(A7) below), the second one is between f and g and has the following form:

$$\begin{aligned} f_r=0, \qquad 2f_\theta +r^2(f_z+2g_r)=0, \qquad 2f+r^3g_z+2rg_\theta =0, \qquad rf+4zg=0. \end{aligned}$$

One can verify that the only solution to this system is $f \equiv g \equiv 0.$ We skip these calculations. $\square$

So we start our analysis with an integral of the form

$$\begin{aligned} F = a(r,\theta ,z) p_R^2 + d(r,\theta ,z) p_R p_S + b(r,\theta ,z) p_S^2 + h(r,\theta ,z). \end{aligned}$$

The condition (A1) implies:

$$\begin{aligned} a_r=0, \end{aligned}$$

(A2)

$$\begin{aligned} 2a_\theta +r^2(a_z+2d_r)=0, \end{aligned}$$

(A3)

$$\begin{aligned} 4a+r^3d_z+2r(d_\theta +r^2b_r)=0, \end{aligned}$$

(A4)

$$\begin{aligned} 2d+r^3b_z+2rb_\theta =0, \end{aligned}$$

(A5)

$$\begin{aligned} 2r^3(r^4+16z^2)^{3/2}h_r-8kr^6a-16kr^5zd=0, \end{aligned}$$

(A6)

$$\begin{aligned} r(r^4+16z^2)^{3/2}(r^2h_z+2h_\theta ) -4kr^6d-32kr^5zb=0. \end{aligned}$$

(A7)

Integrating the Eqs. (A2)–(A4) successively, we obtain

$$\begin{aligned} a(r,\theta ,z) = \alpha (\theta ,z), \qquad d(r,\theta ,z)= & {} \gamma (\theta ,z) - \frac{1}{2} r \alpha _z + \frac{\alpha _\theta }{r}, b(r,\theta ,z) = \frac{\alpha }{r^2} + \frac{\alpha _{\theta \theta }}{2r^2}\\{} & {} \quad + \frac{r^2 \alpha _{zz}}{8} + \frac{\gamma _\theta }{r} - \frac{r\gamma _z}{2} + \omega (\theta ,z), \end{aligned}$$

where $\alpha (\theta ,z)$, $\gamma (\theta ,z)$, $\omega (\theta ,z)$ are arbitrary functions. Then (A5) takes the form

$$\begin{aligned}{} & {} \alpha _{zzz}r^6 - 4\gamma _{zz}r^5 + (2\alpha _{\theta zz} + 8\omega _z)r^4 + 4(\alpha _{\theta \theta z} + 4\omega _\theta )r^2 + 16(\gamma _{\theta \theta } + \gamma )r \\{} & {} \quad + 8(\alpha _{\theta \theta \theta } + \alpha _\theta ) = 0. \end{aligned}$$

This is a polynomial in r with coefficients depending on $\theta$, z only. Since this polynomial must vanish, all its coefficients must vanish as well. This allows one to find $\alpha (\theta ,z)$, $\gamma (\theta ,z)$, $\omega (\theta ,z)$ and, consequently, the coefficients $a(r,\theta ,z)$, $b(r,\theta ,z)$, $d(r,\theta ,z)$ explicitly. We omit these long but simple calculations and skip the final form of these coefficients since they are quite cumbersome.

After that we are left with two Eqs. (A6), (A7) on the unknown function $h(r,\theta ,z)$ which take the form:

$$\begin{aligned}{} & {} (r^4+16z^2)^{3/2}h_r+ 2kr(c_8 r^2 - z(4 c_9 + z(4 c_3+c_5 r^2+4c_6z)))\cos (2\theta )\nonumber \\{} & {} \quad -2kr(c_9r^2+z(4c_8+z(4c_2-c_6r^2+4c_5z)))\sin (2\theta )- 4kr^3(c_7-c_4z^2)\nonumber \\{} & {} \quad -8kr^2z ((s_2+s_4z)\cos \theta +(s_3+s_5z)\sin \theta ) = 0, \end{aligned}$$

(A8)

$$\begin{aligned}{} & {} (r^4+16z^2)^{3/2}(r^2h_z+2h_\theta )+2c_1kr^2(r^4-16z^2) \nonumber \\{} & {} \quad -kr^2(4c_9r^2+c_2(r^4+16z^2)-2z(-8c_8+2c_3r^2+c_5r^4+6c_6r^2z-8c_5z^2))\cos (2\theta ) \nonumber \\{} & {} \quad +kr^2(-4c_8r^2+c_3(r^4+16z^2)+2z(8c_9+2c_2r^2-c_6r^4+6c_5r^2z+8c_6z^2))\sin (2\theta ) \nonumber \\{} & {} \quad -4kr^3(s_2r^2+z(8s_3-3s_4r^2+8s_5z))\cos \theta +4kr^3(-s_3r^2+z(8s_2+3s_5r^2+8s_4z))\sin \theta \nonumber \\{} & {} \quad -4kr^2z(8c_7+8s_1r^2+c_4(r^4+8z^2))=0. \end{aligned}$$

(A9)

Here $c_k$, $s_k$ are arbitrary constants. The Eq. (A8) can be integrated. However, the general solution $h(r,\theta ,z)$ to (A8) is expressed in terms of elliptic integrals. We consider the simplest case

$$\begin{aligned} s_2=s_3=s_4=s_5=0. \end{aligned}$$

In this case h(x, y, z) can be found from (A8) in terms of elementary functions as follows:

$$\begin{aligned} h(r,\theta ,z) = \psi (\theta ,z) + \frac{k}{4z \sqrt{r^4+16z^2}}\phi (r,\theta ,z), \end{aligned}$$

where $\psi (\theta ,z)$ is an arbitrary function and

$$\begin{aligned} \phi (r,\theta ,z){} & {} = (c_9r^2+z(4c_8+c_3r^2+c_6r^2z-4c_5z^2))\cos (2\theta ) + (c_8r^2\\{} & {} \quad +z(-4c_9+c_2r^2+c_5r^2z+4c_6z^2))\sin (2\theta ) \\{} & {} \quad + 8z (c_4z^2-c_7). \end{aligned}$$

The unknown function $\psi (\theta ,z)$ should be chosen such that the relation (A9) holds identically. It seems that the only possible way to satisfy this requirement is to put

$$\begin{aligned} \psi (\theta ,z) \equiv 0, \qquad c_1=c_5=c_6=c_8=c_9=s_1=0. \end{aligned}$$

In this case (A9) is satisfied. Thus we found all the coefficients of F. Notice that $\widetilde{F}=4F-8c_7H$ is also the first integral of (6) having the simpler form:

$$\begin{aligned} \widetilde{F} = \widetilde{a}\,p_R^2 + \widetilde{d}\,p_Rp_S + \widetilde{b}\,p_S^2 + \widetilde{h}, \end{aligned}$$

where

$$\begin{aligned} \widetilde{a}&= 2z(2c_4z-c_2\cos (2\theta ) +c_3\sin (2\theta )), \\ \widetilde{d}&=\Big ( c_2r+\frac{4c_3z}{r} \Big ) \cos (2\theta ) - \Big ( c_3x-\frac{4c_2z}{r} \Big ) \sin (2\theta ) -4c_4rz, \\ \widetilde{b}&= \frac{1}{r^2} \big ( (-c_3r^2+2c_2z) \cos (2\theta ) -(c_2r^2+2c_3z) \sin (2\theta ) +c_4 (r^4+4z^2) \big ), \\ \widetilde{h}&= \frac{k}{\sqrt{r^4+16z^2}} \big ( r^2(c_3\cos (2\theta )+c_2 \sin (2\theta ))+8c_4z^2 \big ). \end{aligned}$$

Here $c_2$, $c_3$, $c_4$ are arbitrary constants. It is left to notice that $\widetilde{F}$ is linear in these constants, i. e. it has the form $\widetilde{F}=c_2F_1+c_3F_2+c_4F_3$. This implies that the functions

$$\begin{aligned} F_1 =&\Big ( p_Rp_Sr-2p_R^2z+\frac{2p_S^2z}{r^2} \Big ) \cos (2\theta )+ \Big ( -p_S^2+\frac{4 p_Rp_Sz}{r}+\frac{kr^2}{\sqrt{r^4+16z^2}} \Big ) \sin (2\theta ), \\ F_2 =&- \Big ( p_Rp_Sr-2p_R^2z+\frac{2p_S^2z}{r^2} \Big ) \sin (2\theta )+ \Big ( -p_S^2+\frac{4 p_Rp_Sz}{r}+\frac{kr^2}{\sqrt{r^4+16z^2}} \Big ) \cos (2\theta ), \\ F_3&= 4z^2 p_R^2 - 4rz p_Rp_S + \frac{r^4+4z^2}{r^2}p_S^2 + \frac{8k z^2}{\sqrt{r^4+16z^2}}. \end{aligned}$$

are also integrals of (6).