Normalization and reduction of the Stark Hamiltonian

1.   Introduction

In the study of perturbations of the three degree of freedom Kepler Hamiltonian pulling back, the regularized Hamiltonian by the Kustaanheimo-Stiefel (KS) map, gives a perturbation of the four degree of freedom harmonic oscillator Hamiltonian, when restricted to the zero level set of the KS symmetry. We use the formulation of the KS transformation in ^[6] which allows us to reduce the KS symmetry using invariant theory for the first time. As an illustration, we apply this procedure to the regularized Stark Hamiltonian, which is normalized after applying the KS transformation. We do not expect this Hamiltonian to be completely integrable (see Lagrange ^[4] and also ^[5]). Our treatment follows that of ^[3] and gives the full details of obtaining the second order normal form ^[1]. We use the notation of ^[2] and note that our procedure of regularization, pull back by the KS map, normalization and reduction may be used to study three degree of freedom perturbed Keplerian systems.

2.   The basic set up

On $T_0{\mathbb{R }}^3 = ({\mathbb{R }}^3\setminus \{ 0 \}) \times {\mathbb{R }}^3$ with coordinates $(x, y)$ and standard symplectic form ${\omega }_3 = \sum^3_{i = 1}\mathrm{d} x_i \wedge \mathrm{d} y_i$ consider the Stark Hamiltonian

Here, $\langle \, \, , \, \, \rangle$ is the Euclidean inner product on ${\mathbb{R} }^3$ with associated norm $|\, \, |$. On the negative energy level $-\tfrac{ 1}{ 2} k^2$ with $k > 0$ rescaling time $\mathrm{d} t \mapsto \frac{|x|}{k}\mathrm{d} s$, we obtain

In other words, $(x, y)$ lies in the $\tfrac{1}{k}$ level set of

We assume that $f$ is small, namely, $f = \varepsilon \beta$. After the symplectic coordinate change $(x, y) \mapsto (\frac{1}{k}x, ky)$ the Hamiltonian $\widehat{K}$ becomes the preregularized Hamiltonian

on the level set ${\mathcal{K}}^{-1}(1)$.

Let $T_0{\mathbb{R} }^4 = ({\mathbb{R} }^4\setminus \{ 0 \}) \times {\mathbb{R} }^4$ have coordinates $(q, p)$ and a symplectic form ${\omega }_4 = \sum^4_{i = 1}\mathrm{d} q_i \wedge \mathrm{d} p_i$. Pull back $\mathcal{K}$ by the Kustaanheimo-Stiefel mapping

where

and

to get the regularized Stark Hamiltonian

on ${\Xi}^{-1}(0)$, since $|x| = \langle q, q \rangle = H_2+V_1$. Here

generate the algebra of polynomials invariant under the $S^1$ action ${\varphi }^{\Xi}_s$ given by the flow of $X_{\Xi}$ on $(T{\mathbb{R} }^4 = {\mathbb{R} }^8, {\omega }_4)$. The Hamiltonian $\mathcal{H}$ (2.5) is invariant under this $S^1$ action and thus is a smooth function on the orbit space ${\Xi}^{-1}(0)/S^1 \subseteq {\mathbb{R} }^{16}$ with coordinates $(K, L, H, {\Xi}; U, V)$.

3.   The first order normal form on ${\Xi}^{-1}(0)/S^1$

The harmonic oscillator vector field $X_{H_2}$ on $(T{\mathbb{R} }^4, {\omega }_4)$ induces the vector field $Y_{H_2} = \sum^4_{i = 1}\big(2V_i \frac{\partial }{\partial U_i} - 2U_i\frac{\partial }{\partial V_i }\big)$ on the orbit space ${\mathbb{R} }^8/S^1 \subseteq {\mathbb{R} }^{16}$, which leaves ${\Xi}^{-1}(0)/S^1$ invariant.

We now compute the first order normal form of the Hamiltonian $\mathcal{H}$ (2.5) on the reduced space ${\Xi }^{-1}(0)/S^1 \subseteq {\mathbb{R} }^8/S^1$.

The average of $H_2U_4-K_3V_1$ over the flow

of $Y_{H_2}$ is

The second equality above follows because $L_{X_{H_2}}K_3 = 0$ and the third because $\overline{\cos 2t} = \overline{\sin 2t } = 0$. The average of $U_4V_1$ over the flow of $Y_{H_2}$ on ${\Xi }^{-1}(0)/S^1$ is

since $\overline{{\cos }^2 \, 2t} = \overline{{\sin }^2 \, 2t} = \tfrac{ 1}{ 2}$ and $\overline{\sin 4t} = 0$. The last equality above follows from the explicit description of the orbit space ${\mathbb{R} }^8/S^1$ as the semialgebraic variety in ${\mathbb{R} }^{16}$ with coodinates $(K, L, U, V;H_2, \Xi)$ given by

So the average of $U_4V_1+ H_2U_4 - K_3V_1 - H_2K_3$ over the flow of $Y_{H_2}$ is $-\tfrac{ 3}{ 2}H_2K_3$ on ${\Xi }^{-1}(0)/S^1$. Thus the first order normal form of the regularized Stark Hamiltonian $\mathcal{H}$ (2.5) on ${\Xi }^{-1}(0)/S^1$ is

4.   The second order normal form on ${\Xi }^{-1}(0)/S^1$

In order to compute the second order normal form of the Hamiltonian $\mathcal{H}$ on ${\Xi }^{-1}(0)/S^1$, we need to find a function $F$ on ${\mathbb{R} }^{16}$ such that changing coordinates by the time $\varepsilon$ value of the flow of the Hamiltonian vector field $Y_F$ brings the regularized Hamiltonian $\mathcal{H}$ (2.5) into first order normal form. Choose $F$ so that

The following calculation shows that this choice does the job.

To determine the function $F$, we solve equation (4.1). Write $F = F_1+F_2$, where $L_{Y_{H_2}}F_1 = \beta (U_4V_1+ \tfrac{ 1}{ 2} H_2K_3)$ and $L_{Y_{H_2}}F_2 = \beta (H_2U_4-K_3V_1)$. Then

Since $L_{Y_{H_2}}V_4 = -2 U_4$ and $L_{Y_{H_2}}U_1 = 2V_1$, it follows that

Now

see ^[1], and

since $\tfrac{1}{\pi } \int^{\pi}_0 t \, \sin 4t \mathrm{d} t = -\tfrac{1}{4}$ and $\tfrac{1}{\pi } \int^{\pi}_0 t \, {\sin }^2\, 2 t \mathrm{d} t = \tfrac{1}{\pi } \int^{\pi}_0 t \, {\cos }^2\, 2 t \mathrm{d} t = \tfrac{\pi }{4}$. Thus

on ${\Xi}^{-1}(0)/S^1$, see (3.1). Hence on ${\Xi}^{-1}(0)/S^1$

We now calculate the average over the flow of $Y_{H_2}$ of

which is the ${\varepsilon }^2$ term in the transformed Hamiltonian $({\varphi }^{Y_F}_{\varepsilon})^{\ast }\mathcal{H}$, see (4.2). This determines the second order normal form of $\mathcal{H}$ on ${\Xi }^{-1}(0)/S^1$. We begin with the term

The average of

vanishes on ${\Xi}^{-1}(0)/S^1$. The term

Next we calculate $\frac{ 3}{ 2}\beta \overline{H_2(L_{Y_{K_3}}F)}$. Since $\overline{H_2V_1} = 0 = \overline{K_3U_4}$ we need only calculate the average of $U^2_1$, $U^2_4$, $V^2_1$ and $V^2_4$. We get $\overline{U^2_1} = \tfrac{ 1}{ 2} (U^2_1+V^2_1) = \overline{V^2_1}$ and $\overline{U^2_4} = \tfrac{ 1}{ 2} (U^2_4+V^2_4) = \overline{V^2_4}$. Thus $\tfrac{ 3}{ 2}\beta \overline{H_2(L_{Y_{K_3}}F)} = 0$. So the average $-\tfrac{ 3}{ 2}\beta \overline{L_{Y_F}(H_2K_3)}$ of the first term in expression (4.6) vanishes on ${\Xi}^{-1}(0)/S^1$.

Next we calculate the average of the term $L^2_{Y_F}H_2$ in expression (4.6) on ${\Xi }^{-1}(0)/S^1$. We have

We begin by finding

So the average of term I on ${\Xi}^{-1}(0)/S^1$ is

since the average of $V_1V^2_4$, $K^2_3V_1$, $H^2_2V_1$, and $U^2_1V_1$ are each $0$, $\overline{U_4V_1} = -\tfrac{ 1}{ 2} H_2K_3$ and $\overline{V^2_1} = \tfrac{ 1}{ 2} (U^2_1+V^2_1)$.

Term II is

So

For term III, we have already shown that

and for term IV we have already shown that

Term V is

So

since the average of $U^2_4V_1$ and $U_1U_4V_1$ vanish; while $\overline{U^2_4} = \tfrac{ 1}{ 2} (U^2_4+V^2_4)$ and $\overline{U_4V_1} = \tfrac{ 1}{ 2} (U_4V_1 - U_1V_4)$.

Term VI is

So

since $\overline{K_2U_4} = 0 = \overline{H_2V_1}$.

Term VII is

So

Term VIII is

So

since $\overline{U_1V_4} = \tfrac{ 1}{ 2} H_2K_3$. Collecting together the results of all the above term calculations gives

using $U_4V_1-U_1V_4 = -K_3H_2$. Thus the second order normal form of the regularized Stark Hamiltonian $\mathcal{H}$ on ${\Xi }^{-1}(0)/S^1$ is

5.   The first order normal form of ${\mathcal{H}}^{(2)}_{\mathrm{nf}}$ on $T_hS^3_1$

Since $L_{X_{H_2}}{\mathcal{H}}^{(2)}_{\mathrm{nf}} = 0$ by construction, the second order normal form ${\mathcal{H}}^{(2)}_{\mathrm{nf}}$ (4.8) is a smooth function on $(H^{-1}_2(h) \cap {\Xi}^{-1}(0))/S^1 = T_hS^3_1$, the tangent $h$-sphere bundle of the unit $3$-sphere $S^3_1$, given by

We now show that the Hamiltonian $\widetilde{\mathcal{H}}$ (5.1) on $T_hS^3_1$ can be normalized again. On $(T{\mathbb{R} }^4, {\omega }_4)$ the Hamiltonian

gives rise to the Hamiltonian vector field $X_{K_3}$, whose flow ${\varphi }^{X_{K_3}}_t(q, p)$ is

which is periodic of period $2\pi$.

The vector field $X_{K_3}$ on $T{\mathbb{R} }^4$ induces the vector field

on ${\Xi}^{-1}(0)/S^1 \subseteq {\mathbb{R} }^{16}$ with coordinates $(K, L, H_2, \Xi; U, V)$, whose flow

is periodic of period $\pi$. Since $L_{Y_{K_3}}$ maps the ideal of smooth functions which vanish identically on ${\Xi}^{-1}(0)/S^1$ into itself, $Y_{K_3}$ is a vector field on ${\Xi}^{-1}(0)/S^1$. Since $L_{X_{K_3}}H_2 = 0$, it follows that $Y_{K_3}$ induces a vector field on $T_hS^3_1$ with periodic flow. So we can normalize again.

To compute the normal form of the Hamiltonian $\widetilde{\mathcal{H}}$ (5.1) on $T_hS^3_1$ we need only calculate the average of the term

over the flow ${\varphi }^{Y_{K_3}}_s$. Since $L_{Y_{K_3}}K_3 = 0$, we need only calculate $\overline{U^2_1}$, $\overline{U^2_4}$, $\overline{V^2_1}$, and $\overline{V^2_4}$. Now

Similarly, $\overline{V^2_1} = \tfrac{ 1}{ 2} (V^2_1+V^2_4)$, $\overline{U^2_4} = \tfrac{ 1}{ 2} (U^2_1+U^2_4)$, and $\overline{V^2_4} = \tfrac{ 1}{ 2} (V^2_1+V^2_4)$. Thus

which is no surprise since $L_{Y_{K_3}}T = 0$. So the first order normal form of $\widetilde{\mathcal{H}}$ (5.1) on $T_hS^3_1$ is

6.   The reduced Hamiltonian ${\widetilde{\mathcal{H}}}^{(1)}_{\mathrm{nf}}$ on $S^2_h\times S^2_h$

The polynomial $U^2_1+V^2_1 +U^2_4+V^2_4$ is invariant under the flows ${\varphi }^{X_{H_2}}_t$, ${\varphi }^{X_{\Xi }}_u$, and thus is a polynomial on the orbit space $T_hS^3_1/S^1 = S^2_h \times S^2_h$, defined by

We now find this polynomial. From the explicit description of ${\mathbb{R} }^8/S^1$ in (3.1) it follows that on $\big(H^{-1}_2(h) \cap {\Xi}^{-1}(0) \big) /S^1$

So on $S^2_h \times S^2_h$

since $\langle U, U \rangle = h^2$, $\langle V, V \rangle = h^2$, and $\langle U, V \rangle = 0$. Thus $U^2_1+V^2_1 = K^2_1+K^2_2+K^2_3$. Again from the explicit description of $\big(H^{-1}_2(h) \cap {\Xi }^{-1}(0)\big) /S^1$ we have

So on $S^2_h \times S^2_h$

Thus $U^2_4+V^2_4 = L^2_1+L^2_2+K^2_3$. Consequently

on $S^2_h \times S^2_h$. Hence on $S^2_h \times S^2_h$

7.   The Hamiltonian system $(\widehat{\mathcal{H}}, S^2_h \times S^2_h, \{ \, \, , \, \, \})$

Using the coordinates $(\xi, \eta) = \big((K+L)/2, (K-L)/2 \big)$ on ${\mathbb{R} }^3 \times {\mathbb{R} }^3$, the space of smooth functions on the reduced space $S^2_h \times S^2_h$, defined by

has a Poisson structure with bracket relations

Since $\{ K_3, L_3 \} = 0$, it follows that $\{ K_3, \widehat{\mathcal{H}} \} = 0$. Thus the flow ${\varphi }^{Z_{K_3}}_r$ of the Hamiltonian vector field $Z_{K_3}$ on $(S^2_h \times S^2_h, \{ \, \, , \, \, \})$ generates an $S^1$ symmetry of the Hamiltonian system $(\widehat{\mathcal{H}}, S^2_h \times S^2_h, \{ \, \, , \, \, \})$. So this system is completely integrable.

We reduce this $S^1$ symmetry as follows. Consider the vector field $Z_{K_3}$ on ${\mathbb{R} }^3 \times {\mathbb{R} }^3$ corresponding to the Hamiltonian $K_3 = \tfrac{ 1}{ 2} ({\xi }_3 + {\eta }_3)$. Its integral curves satisfy

Thus the flow of $Z_{K_3}$ on ${\mathbb{R} }^3 \times {\mathbb{R} }^3$ is

which preserves $S^2_h \times S^2_h$ and is periodic of period $4\pi$.

We now determine the space $(S^2_h\times S^2_h)/S^1$ of orbits of the vector field $Z_{K_3}$. We use invariant theory. The algebra of polynomials on ${\mathbb{R} }^3 \times {\mathbb{R} }^3$, which are invariant under the $S^1$ action given by the flow ${\varphi }^{Z_{K_3}}_t$, is generated by

which are subject to the relation

In terms of invariants the defining equations of $S^2_h \times S^2_h$ become

Eliminating ${\sigma }_1$ and ${\sigma }_2$ from (7.1a) using (7.1b) and (7.1c) gives

which defines $(S^2_h \times S^2_h)/S^1$ as a semialgebraic variety in ${\mathbb{R} }^4$ with coordinates $({\sigma}_3, {\sigma }_4, {\sigma }_5, {\sigma }_6)$. Thus the reduced space $\big(K^{-1}_3(2 k) \cap (S^2_h\times S^2_h) \big)/S^1$ is defined by (7.2a) and

Consequently, $\big(K^{-1}_3(2k) \cap (S^2_h \times S^2_h) \big) /S^1$ is the semialgebraic variety

in ${\mathbb{R} }^3$ with coordinates $({\sigma }_3, {\sigma }_4, {\sigma }_6)$. When $0 < |k | < h$ the reduced space (7.3) is a smooth $2$-sphere. When $|k| = h$ it is a point. When $k = 0$ it is a topological $2$-sphere with conical singular points at $(0, 0, \pm h)$. These singular points correspond to the fixed points $h(0, 0, \pm 1, 0, 0, \mp 1)$ of the $S^1$ action on $S^2_h \times S^2_h$ generated by the flow of the vector field $Z_{K_3}$.

By (6.1) the reduced Hamiltonian on $\big(K^{-1}_3(2k) \cap (S^2_h \times S^2_h) \big)/S^1$ is

using $L_3 = {\xi }_3-{\eta }_3 = 2{\sigma }_6$, having dropped the constant $h - \tfrac{13}{8} {\varepsilon}^2 {\beta }^2 h^3 - \tfrac{3}{2} \varepsilon \beta h k - \tfrac{51}{16}{\varepsilon }^2{\beta }^2 h k^2$.

Use of AI tools declaration

The author declares that he has not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The author would like to thank the referees for their careful reading and comments on the manuscript. Especially he thanks the one who pointed out errors in his calculations.

The author receied no funding for the research in this article.

Conflict of interest

The author declares there is no conflict of interest.