aa r X i v : . [ m a t h . S G ] O c t A compactness theorem for frozen planets
Urs FrauenfelderOctober 30, 2020
Abstract
We study the moduli space of frozen planet orbits in the Helium atomfor an interpolation between instantaneous and mean interactions andshow that this moduli space is compact.
Since the beginnings of quantum mechanics the dynamics of the Helium atomis intriguing people. Different from the Hydrogen atom the dynamics of theHelium atom is not completely integrable. In particular, its phase space is notfoliated by invariant tori and the EKB method cannot be applied to it. On theother hand the Hamiltonian of Helium is invariant under simultanuous rotationof the two electrons or particle interchange so that due to these symmetriesperiod orbits are usually not isolated and Gutzwiller’s trace formula [7] cannotbe directly applied neither.An interesting periodic orbit for the Helium problem was discovered numeri-cally by Wintgen, Richter and Tanner see [9, 8], which plays an important rolein the semiclassical treatment of the Helium atom. For this orbit both electronslay on a ray on the same side of the nucleus. The inner electron collides withthe nucleus and bounces back, while the outer electron (the frozen planet) staysalmost stationary simultanuously attracted by the nucleus and repelled by theinner electron. An interesting aspect of this periodic orbit is that to the authorsknowledge it does not fit into a family of periodic orbits starting from a com-pletely integrable system. In fact if one ignores the interaction between the twoelectrons one obtains a completely integrable system. But ignoring the interac-tion between the electrons both electrons are just attracted by the nucleus andtherefore have to fall in it and there is nothing like a frozen planet orbit.In order to get a more tractable system the author replaced in [5] the instanta-neous interaction of the two electrons by a mean interaction. It turns out thatfor mean interaction between the electrons the outer electron is really frozen,i.e., it becomes stationary. One obtains a delay equation for the inner electronand it is shown analytically in [5] that there exists a unique nondegenerate so-lution. 1n this paper we interpolate linearly between the instantaneous interaction andthe mean interaction between the two electrons. It is interesting to note thatthere is a special type of frozen planet orbit. At time t = 0 both the inner andthe outer electron have vanishing velocity. The inner electron is than acceler-ated in direction of the nucleus in which it falls after some time. Suppose nowthat at the instant where the inner electron collides with the nucleus the outerelectron has vanishing velocity again. Then one can let the movie run back-wards. The inner electron jumps out of the nucleus and goes back to its initialposition. Meanwhile the outer electron goes back as well to its initial conditionso that at the end both electron are back at their initial position again with zerovelocity. One has a periodic orbit. We refer to such periodic orbits as symmetricfrozen planets . Here is the description of this moduli problem in formulas. Fora homotopy parameter r ∈ [0 ,
1] one looks at solutions q ∈ C ∞ (cid:0) [0 , , (0 , ∞ ) (cid:1) , q ∈ C (cid:0) [0 , , [0 , ∞ ) (cid:1) ∩ C ∞ (cid:0) [0 , , (0 , ∞ ) (cid:1) of the following moduli problem ¨ q ( t ) = − q ( t ) + r ( q − q ) + − r ( q ( t ) − q ( t )) , t ∈ [0 , q ( t ) = − q ( t ) − r ( q − q ) − − r ( q ( t ) − q ( t )) , t ∈ [0 , q ( t ) < q ( t ) , t ∈ [0 , q (0) = ˙ q (1) = ˙ q (0) = q (1) = 0 . (1)The variable q ( t ) describes the inner electron on the ray (0 , ∞ ) at time t andthe variable q ( t ) the outer one. The nucleus lies at the origin with whom theinner electron collides at time t = 1. The positive numbers q and q denotethe mean value of q respectively q , defined by q = Z q ( t ) dt, q = Z q ( t ) dt. For r = 0 the first two equations become the second order ODE ( ¨ q ( t ) = − q ( t ) + q − q ) , t ∈ [0 , q ( t ) = − q ( t ) − q − q ) , t ∈ [0 , . The first term on the righthand side describes the attraction by the nucleuswhose charge is two, since the nucleus of the Helium atom consists of two pro-tons. The second term describes the repulsion of the two electrons. For r = 1the first two equations become ( ¨ q ( t ) = − q ( t ) + q − q ) , t ∈ [0 , q ( t ) = − q ( t ) − q − q ) , t ∈ [0 , . In this case the instantaneous interaction between the two electrons is replacedby the interaction of their mean positions. This is not an ODE anymore, but2n equation involving delay and prolay.The main result of this paper tells us that the moduli space of solutions of(1) is compact. There are two horror scenarios which have to be ruled out. Thefirst horror scenario is ionization meaning that the outer electron escapes to in-finity. The other horror scenario is that the two electrons come arbitrary closetogether. There are two ways how the later scenario could occur. Namely bothelectrons fall simultaneously in the nucleus which leads to a triple collision, orthe energy which is not bounded a priori explodes. The following is the mainresult of this paper which rules out these horror scenarios.
Theorem 1.1
There exists a constant κ such that for every solution ( q , q , r ) of the problem (1) one has max t ∈ [0 , (cid:26) q ( t ) , q ( t ) − q ( t ) (cid:27) ≤ κ. In view of the theorem, given a sequence of solutions ( q ν , q ν , r ν ) of problem(1) one can find a convergent subsequence which then by a usual bootstrap-ping argument is again a solution of problem (1). The author does not knowif nonsymmetric frozen planet orbits actually exist. If they exist it is an openquestion if the compactness result can be extended to the nonsymmetric case.In a joint work with K. Cieliebak and E. Volkov the author is currently studyinga variational approach to frozen planet orbits [4]. Theorem 1.1 together with thevariational approach then leads to a well-defined Euler characteristic at least forsymmetric frozen planet orbits. A more difficult question is, if a homology canbe defined whose chain complex is generated by symmetric frozen planet orbits.To define such a homology one needs to generalize the compactness result togradient flow lines.In joint work with P. Albers, F. Schlenk, and J. Weber the author started togeneralize Floer homology to Hamiltonian delay equations [1, 2, 3, 6]. The au-thor believes that the frozen planet problem can trigger a lot of research in thisdirection and can become an important testing ground how far Floer homologycan be further developed. Acknowledgements:
The author acknowledges partial support by DFG grantFR 2637/2-2.
We assume that ( q , q ) is a solution of problem (1) for some r ∈ [0 , q is always negative. Therefore in view of its initial condition we have˙ q ( t ) < t ∈ (0 , q is strictly monoton decreasing. Thefollowing lemma tells us that in contrast to q the variable q is monotoneincreasing. 3 emma 2.1 If r < , then for every t ∈ (0 , we have ˙ q ( t ) > . In particular, q is strictly monotone increasing. If r = 1 , then q is constant. Proof:
That q is constant in the case r = 1 is proved in [5, Lemma 3.1]. Itsuffices therefore to consider the case r <
1. We prove it in two steps.
Step 1:
For every t ∈ [0 ,
1) such that ˙ q ( t ) = 0 we have ¨ q ( t ) > q − ¨ q = 2 q − q + 2 r ( q − q ) + 2(1 − r )( q − q ) = 2( q − q ) q q + 2 r ( q − q ) + 2(1 − r )( q − q ) > . In view of the initial condition this implies˙ q ( t ) − ˙ q ( t ) > , t > . (2)The jerk of q is given by... q = 4 ˙ q q − − r )( ˙ q − ˙ q )( q − q ) (3)In view of the initial conditions we have... q (0) = 0and in view of (2) and r < q ( t ) < q ( t ) q , t > . (4)We assume by contradiction that there exists t ∈ (0 , q ( t ) = 0 , ¨ q ( t ) ≤ . (5)We define t := inf (cid:8) t ∈ ( t ,
1] : ˙ q ( t ) ≥ (cid:9) . Here we use the convention, that if there is no t ∈ ( t ,
1] satisfying ˙ q ( t ) ≥ t = 1. In view of (3) it follows that... q ( t ) = ˙ q (0)( q ( t ) − q ( t )) < . t > t . In view of the definition of t and (4) we have... q ( t ) < , t ∈ ( t , t ) . In view of (5) this implies that¨ q ( t ) < , t ∈ ( t , t ) . Combining this once more with (5) we conclude that˙ q ( t ) < . By definition of t this implies that t = 1and therefore ˙ q (1) < . This contradicts the boundary condition in (1) and Step 1 is proved.
Step 2:
We prove the lemma.By Step 1 we conclude that q cannot have a local maximum in [0 ,
1) but allits critical points are strict local minima. By the boundary condition in (1)the function q has a critical point at time 0 which therefore has to be a strictlocal minimum. Since there are no local maxima, there cannot be any additioncritical points of q in (0 , q cannot change sign.Since q has a strict local minimum at time 0 the sign of ˙ q ( t ) is positive for t close to 0 and consequently is positive always. This finishes the proof of thelemma. (cid:3) Lemma 2.2
For the starting point of q we have the following lower bound q (0) ≥ . (6) Proof:
By (1) we have for every t ∈ [0 ,
1) the estimate¨ q ( t ) ≤ − q ( t ) ≤ − q (0) where the second inequality holds since q is monotone decreasing. Using that˙ q (0) = 0 we obtain from this the estimate q ( t ) ≤ − t q (0) + q (0) . t going to 1 we get0 ≤ − q (0) + q (0)implying 1 ≤ q (0) and therefore 1 ≤ q (0) . This finishes the proof of the lemma. (cid:3)
We abbreviate ∆ := q (0) − q (0)the distance between q and q at time t = 0. Since the variable q is decreasingand by Lemma 2 the variable q is increasing we have for every t ∈ [0 , q ( t ) − q ( t ) ≥ ∆ . (7)The following lemma tells us that q and q cannot come to close to each other. Lemma 2.3
There exists a constant c > such that ∆ ≥ (1 − r ) c . (8) Proof:
We suppose that ∆ ≤ c . We define t ∈ (0 ,
1) by the requirement q ( t ) = q (0) − ∆ . Note that ¨ q + ¨ q = − q − q ≥ − q so that for t ∈ [0 , t ] we have the estimate¨ q ( t ) + ¨ q ( t ) ≥ − q (0) − ∆) ≥ − − ∆) ≥ − . Since the velocity of both variables q and q at time t = 0 vanish we get fromthat the estimate ˙ q ( t ) + ˙ q ( t ) ≥ − t ≥ − − ˙ q ( t ) ≤
16 + ˙ q ( t ) . q is negativ but by Lemma 2.1 we have that ˙ q is positive we obtain fromthat the inequality ˙ q ( t ) ≤
512 + 2 ˙ q ( t ) . (9)Since ¨ q + ¨ q < q ( t ) − q (0) ≤ q (0) − q ( t ) = ∆so that we obtain q ( t ) − q ( t ) = (cid:0) q ( t ) − q (0) (cid:1) + (cid:0) q (0) − q (0) (cid:1) + (cid:0) q (0) − q ( t ) (cid:1) ≤ ≤ . (10)We have the following preserved quantity E = 12 (cid:16) ˙ q ( t ) + ˙ q ( t ) (cid:17) − q ( t ) − q ( t ) + r (cid:0) q ( t ) − q ( t ) (cid:1) ( q − q ) + 1 − rq ( t ) − q ( t ) . At time t = 0 this computes to be E = − q (0) + ∆ − q (0) + r ∆( q − q ) + 1 − r ∆which we estimate using (6) E ≥ − − r ∆ . (11)Note that there exists a constant 0 < ε < q ≤ (1 − ε ) q (0) . By the computations in Appendix A the constant ε can be chosen around .Hence using Lemma 2 and (6) we estimate1 q − q ≥ q (0) − q ≥ q (0) − q ≥ εq (0) ≥ ε . Evaluating the preserved quantity E at time t = t we obtain from that as wellas (9) and (10) E = 12 (cid:16) ˙ q ( t ) + ˙ q ( t ) (cid:17) − q ( t ) − q ( t ) + r (cid:0) q ( t ) − q ( t ) (cid:1) ( q − q ) (12)+ 1 − rq ( t ) − q ( t ) ≤
32 ˙ q ( t ) + 256 + 32 ε + 1 − rq (0) − q (0) + ∆= 32 ˙ q ( t ) + 256 + 32 ε + 1 − r − r ≤
32 ˙ q ( t ) + 260 + 32 ε . We abbreviate c := 260 + 32 ε . Since ˙ q ( t ) is positive by Lemma 2 we obtain the estimate˙ q ( t ) ≥ √ − r √ − c . (13)Using Lemma 2 and (6) again the acceleration of q is estimated from below forevery t ∈ [0 ,
1] by¨ q ( t ) ≥ − q ( t ) ≥ − q (0) ≥ − q (0) ≥ − . (14)Since the velocity of q at time t = 1 vanishes we obtain from this combinedwith (13) 0 = ˙ q (1)= ˙ q ( t ) + Z t ¨ q ( t ) dt ≥ √ − r √ − c − − t ) ≥ √ − r √ − c − ≥ c + 2) (1 − r )so that (8) follows with c := 13( c + 2) . This finishes the proof of the lemma. (cid:3)
We would like to replace the estimate for ∆ in Lemma 2.3 by a uniform one notdepending on r . For that purpose we need the following result on the averagepositions of q and q . Lemma 2.4
The mean values of q and q satisfy the following inequality q ≤ (cid:18) − r r r (cid:19) q . (15)8 roof: From Lemma 2 it follows that q ≤ q (1)and q attains at time t = 1 a local maximum so that¨ q (1) ≤ . Combining these inequalities with (1) we estimate0 ≥ ¨ q (1)= − q ( t ) + r ( q − q ) + 1 − r ( q ( t ) − q ( t )) = − q ( t ) + r ( q − q ) + 1 − rq ( t ) = − rq ( t ) + r ( q − q ) ≥ − rq + r ( q − q ) which we rewrite 1 + rq ≥ r ( q − q ) or equivalently q r ≤ ( q − q ) r . Taking square roots we obtain q ≤ r rr ( q − q )so that r rr q ≤ (cid:18)r rr − (cid:19) q from which (15) follows. (cid:3) Now we are in position to improve Lemma 2.5 with a uniform estimate frombelow for ∆ not depending on r anymore. Proposition 2.5
There exists a constant c > such that ∆ ≥ c . Proof:
Using (14) and ˙ q (1) = 0 we obtain the estimate˙ q (1 − t ) ≤ t q = Z q ( t ) dt = Z q (1 − t ) dt = q (0) + Z ˙ q (1 − t ) tdt ≤ q (0) + Z t dt = q (0) + 23so that q (0) ≥ q − . (16)Since ¨ q ( t ) ≤ t ∈ [0 ,
1) and ˙ q (0) = 0 we have q (0) ≤ q (17)so that combined with (6) we get the estimate q ≥ . (18)From (15) we obtain for every r ∈ [0 ,
1] the inequality q ≥ − q r r q = √ r √ r − √ r q . (19)Using (16), (17), and (19) we estimate∆ = q (0) − q (0) (20) ≥ q − − q ≥ (cid:18) √ r √ r − √ r − (cid:19) q − . Note that the function f : [0 , → (0 , ∞ ) , r
7→ √ r √ r − √ r = 11 − q r r is monotone increasing, since r r r is monotone increasing, and f (cid:0) (cid:1) = 11 − q = 11 − = 210o that it follows from (18) and (20) that∆ ≥ (cid:18) √ r √ r − √ r − (cid:19) −
23 = √ r (cid:0) √ r − √ r (cid:1) − , r ≥ . (21)Note that f (cid:0) (cid:1) = 11 − q = 11 − = so that since f is monotone increasing f ( r ) ≥ , r ≥ . Therefore it follows from (21) that∆ ≥ −
53 = 51 − , r ≥ . Combining this estimate with Lemma 2.3 proves the proposition. (cid:3)
Our next goal is to derive an upper bound for the outer electron. We firststart with an upper bound for the inner electron.
Lemma 2.6
There exists a constant c such that q (0) ≤ c . Proof:
We recall (7) which tells us that q ( t ) − q ( t ) ≥ ∆ , t ∈ [0 , , which is an immediate consequence of Lemma 2.1. In particular we have q − q ≥ ∆ . We infer from (1) that¨ q ( t ) ≥ − q ( t ) − , t ∈ [0 , . With Proposition 2.5 it follows that¨ q ( t ) ≥ − q ( t ) − c , t ∈ [0 , . Let t ∈ [0 ,
1) be the time such that q ( t ) = 1 . Because q is motonote decreasing we have¨ q ( t ) ≥ − − c , t ∈ [ t , . q (0) = 0 we deduce from that q (0) ≤ q ( t ) + t (cid:18) c (cid:19) ≤ (cid:18) c (cid:19) = 2 + 12 c . Setting c := 2 + 12 c the lemma follows. (cid:3) Now we are in position to obtain an upper bound for the outer electron.
Proposition 2.7
There exists a constant c such that q (1) ≤ c . Proof:
Since q is monotone decreasing we obtain from Lemma 2.6 that q ( t ) ≤ c , t ∈ [0 , q ≤ c . Recall the inequality (14) telling us that¨ q ( t ) ≥ − , t ∈ [0 , . Combining this with the fact that q is monotone increasing by Lemma 2.1 weobtain the estimate q (1) ≥ q ( t ) ≥ q (1) − , t ∈ [0 , q ≥ q (1) − . Together with (1) these estimates imply¨ q ( t ) ≤ − q (1) + r ( q (1) − − c ) + 1 − r ( q (1) − − c ) = − q (1) + 1( q (1) − − c ) , t ∈ [0 , . Since ˙ q (0) = ˙ q (1) = 0 there exists t satisfying ¨ q ( t ) = 0 so that we obtainthe inequality 0 ≤ − q (1) + 1( q (1) − − c ) implying q (1) ≥ (cid:0) q (1) − − c (cid:1) . Taking square roots on both sides we obtain the inequality q (1) ≥ √ (cid:0) q (1) − − c (cid:1) c + 1 ≥ (cid:0) √ − (cid:1) q (1)so that q (1) ≤ c + 1 √ − . Hence setting c := c + 1 √ − (cid:3) Proof of Theorem 1.1:
By Lemma 2.1 we know that q ( t ) is monotoneincreasing so that combined with Proposition 2.7 we have the estimate q ( t ) ≤ c , t ∈ [0 , . (22)From (7) and Proposition 2.5 we infer that q ( t ) − q ( t ) ≥ c , t ∈ [0 , q ( t ) − q ( t ) ≤ c . (23)Setting κ := max (cid:26) c , c (cid:27) the theorem follows from inequalities (22) and (23). (cid:3) A The average mean fall
The interior electron of the frozen planet problem is in the free fall. In thisappendix we discuss mean values of some free falls. This is not really neededfor the proof of the main result. On the other hand a careful analysis of themean free fall could be used to determine some constants occuring in the proofmore precisely and might be of use when trying to establish a homology theorywhich also involves compactness results for gradient flow lines. Moreover, themean free fall has its own mathematical beauty.If the acceleration is constant g > q > q (0) = q , ˙ q = 0 , ¨ q ( t ) = − g whose explicit solution is given by q ( t ) = q − g t . τ is the time of the free fall implicitly defined by q ( τ ) = 0then from the above formula one obtains explicitly τ = r q g . The average position is then given by q = 1 τ Z τ q ( t ) dt = 1 τ Z τ (cid:18) q − g t (cid:19) dt = 1 τ (cid:18) q τ − g τ (cid:19) = q − g · q g = 23 q , so that the ratio of the average position and the initial position is given by κ := qq = 23 . More generally, if the acceleration is given by the derivative f ′ of a potential f ,the free fall is a solution of the initial value problem q (0) = q , ˙ q = 0 , ¨ q ( t ) = f ′ ( q ( t )) . One has the preserved quantity12 ˙ q ( t ) + f ( q ( t )) = f ( q ) . We assume that f ′ < q is strictly decreasing. Hence the velocity attime t is given by ˙ q ( t ) = − q (cid:0) f ( q ) − f ( q ( t ) (cid:1) . Using this formula the time of the free fall can be computed as τ = Z τ dt = Z q p f ( q ) − f ( q )) dq, and the average position q = 1 τ Z τ q ( t ) dt = R q q √ f ( q ) − f ( q )) dq R q √ f ( q ) − f ( q )) dq = R q q √ f ( q ) − f ( q ) dq R q √ f ( q ) − f ( q ) dq . Note that this expression is invariant under scaling the potential f to µf for µ >
0. We now want to compute this for the homogeneous potentials f α ( q ) = − q α α >
0. We abbreviate by κ ( α ) := qq the ratio between average position and initial position for the free fall withrespect to the potential α . We have the following proposition Proposition A.1
The ratio κ ( α ) is given by κ ( α ) := Γ (cid:0) α α (cid:1) Γ (cid:0) αα (cid:1) Γ (cid:0) αα (cid:1) Γ (cid:0) α α (cid:1) . Let us look at some special values. Using Γ( x + 1) = x Γ( x ) and Γ( n + 1) = n !we obtain for the Newtonian potential − q , i.e., α = 1, κ (1) = Γ (cid:0) (cid:1) Γ(2)Γ(3)Γ (cid:0) (cid:1) = Γ (cid:0) (cid:1) (cid:0) (cid:1) = 34 . This is a bit bigger than one obtains for the free fall with constant acceler-ation. For the potential − q , i.e., α = 2, one obtains using Γ (cid:0) (cid:1) = √ π thetranszendental number κ (2) = Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ(2)Γ(1) = Γ (cid:0) (cid:1) π . Proof of Proposition A.1:
Since the average position does not change ifwe scale the potential by a positive factor we work with the potential − q α toavoid annoying factors √ q = − s q α − q α = − p q α − q α ( q q ) α so that we obtain for the time of the free fall τ = q α Z q q α p q α − q α dq Changing variables q = q (cos θ ) α , dq = − q α (cos θ ) − αα sin θ this becomes τ = 2 q α +10 αq α Z π (cos θ ) α sin θ √ − cos θ dθ = 2 q α +22 α Z π (cos θ ) α dθ = q α +22 α B (cid:18) α α , (cid:19) = q α +22 α Γ (cid:0) α α (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) αα (cid:1) B is the Betafunction. Similarly we have Z τ qdt = q α Z q q α +22 p q α − q α dq = 2 q α +42 α Z π (cos θ ) α dθ = q α +42 α B (cid:18) α α , (cid:19) = q α +42 α Γ (cid:0) α α (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) αα (cid:1) implying that q = 1 τ Z τ qdt = Γ (cid:0) α α (cid:1) Γ (cid:0) αα (cid:1) Γ (cid:0) αα (cid:1) Γ (cid:0) α α (cid:1) q This proves the proposition. (cid:3)
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