The question I will ask makes sense in much more generality, but I will leave the translation to the experts, since I'm only looking for a special case (and it would not surprise me if the answer does not generalize). I will give some background, and then ask my question as a conjecture, set apart from the main text.

Let $\mathbb R^n$ have its usual metric, and pick a differential one-form (= vector field) $B$ (the "magnetic potential") and a differential zero-form (= function) $C$ (the "electric potential"). Then consider the following second-order ODE for parameterized paths $\gamma: [0,T] \to \mathbb R^n$: $$ 0 = \ddot \gamma + dB \cdot \dot\gamma + dC \quad\quad \text{(EOM)} $$ I'll let you pick the signs for how the two-form $dB$ eats the vector $\dot\gamma$; just be consistent.

Then (EOM) is nondegenerate, and so a solution is determined by its initial conditions $(\dot\gamma(0),\gamma(0))$. For each $T \in \mathbb R$, let $\phi_T: \mathbb R^{2n} \to \mathbb R^n$ be the "flow by time $T$" (actually, it is defined only on an open subset of $\mathbb R^{2n}$, given by $\phi_T(v,q) = \gamma(T)$ for the solution $\gamma\\,$ to (EOM) with initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$. Then $\phi_T$ is smooth; in fact, it is smooth in the $T$ variable as well. This follows from some standard fundamental result in ODEs, for which I don't have a good reference.

A path $\gamma: [0,T] \to \mathbb R^n$ is classical if it satisfies (EOM); its duration is the number $T$. We can also consider paths with negative duration by flowing backwards, although we will not need to do so.

Definition: A point $(v,q) \in \mathbb R^{2n}$ is focal for duration $T$ iff ($\phi_T(v,q)$ is defined and) $\det(\partial \phi_T(v,q)/\partial v) = 0$; i.e. fix the $q$, think of $\phi_T(-,q)$ as a function of $v$ only, and ask that its differential is degenerate. By identifying $(v,q)$ with its classical path, we will talk about "focal (classical) paths" for given durations.

It is a standard results (see e.g. Milnor's Morse Theory) that for a given point $(v,q) \in \mathbb R^{2n}$, the durations $T\in \mathbb R$ for which it is focal are discretely separated. Note that every $(v,q)$ is focal for duration $T=0$.

Proposition: Let $\gamma$ be a classical path of duration $T$. Then it is non-focal if and only if it extends to a family of classical paths smoothly parametrized by the boundary positions $(\gamma(0),\gamma(T))$.

Sketch of Proof: Being focal for duration $T$ is a closed condition on $\mathbb R^{2n}$, so we can vary $\gamma(0) = q$ while remaining non-focal. But for non-focal paths we can vary $\gamma(T)$ via the inverse function theorem.

Anyway, pick $q \in \mathbb R^n$, and $v = B(q)$ (or $-B(q)$ depending on your sign convention: for experts, I want the momentum to vanish). Then for some $\epsilon>0$, for all $T\in (0,\epsilon)$, $(v,q)$ is non-focal for duration $T$. Thus, for each $T \in (0,\epsilon)$, I can find an open neighborhood $q \in \mathcal O_0 \subseteq \mathbb R^n$ and another open neighborhood $\mathcal O_1 \subseteq \mathbb R^n$ so that for $(q_0,q_1) \in \mathcal O_0 \times \mathcal O_1$, there is a non-focal classical path $\gamma$ of duration $T$ with $\gamma(0) = q_0$, $\gamma(T) = q_1$, depending smoothly on the boundary conditions, and such that the classical path of duration $T$ and initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$ is contained within this family.

Note that as $T \to 0$, the classical path with initial conditions $(\dot\gamma(0),\gamma(0)) = (v,q)$ ends at a point very close to $q$. I don't know if I can take $\mathcal O_1$ to actually contain $q$.

I would like to reverse the direction of choices: I'd like to pick $\mathcal O_0,\mathcal O_1$ first.

Question/Conjecture: Let $q \in \mathbb R^n$. Then there exist open neighborhood $\mathcal O_0,\mathcal O_1 \subseteq \mathbb R^n$, with $q \in \mathcal O_0,\mathcal O_1$, and $\epsilon>0$ such that:

  1. There exists a family of classical paths $\gamma$ with boundary values varying in $\mathcal O_0,\mathcal O_1$ and with duration varying in $(0,\epsilon)$. I.e. let $\Delta = \{ (T,t) \in \mathbb R^2 : T \in (0,\epsilon), t\in [0,T] \}$; then there is a smooth function $\gamma: \mathcal O_0 \times \mathcal O_1 \times \Delta \to \mathbb R^n$ with: (a) $\gamma(q_0,q_1,T,-)$ is classical for each $(q_0,q_1,T) \in \mathcal O_0 \times \mathcal O_1 \times (0,\epsilon)$, and (b) $\gamma(q_0,q_1,T,0) = q_0$ and $\gamma(q_0,q_1,T,T) = q_1$.
  2. For each $T \in (0,\epsilon)$, the classical path of duration $T$ with initial conditions $(B(q),q)$ appears as some $\gamma(q,q_1,T,-)$.

For comparison, the corresponding theorem about geodesics on a Riemannian manifold is standard: around any point you can find a small neighborhood such that any two points in the neighborhood can be connected by a unique geodesic that does not leave the neighborhood. In fact, it follows from the proposition and the observation that changing the duration of a geodesic for fixed boundary conditions amounts just to a linear reparameterization.


3 Answers 3


For a sufficiently large particle energy, the original problem can be transformed to a problem of geodesic motion as follows:

The motion of a classical particle in an external magnetic field in n-dimensions can be seen as a symplectic reduction of a geodesic motion in n+1 dimensions (Rn * S1) through the Kaluza-Klein construction, given for example in section 7.6 of Marsden's book. The remaining problem is a geodesic motion in an electric potential field. Now suppose that there exists a region in the vicinity of the origin where the electric potential is bounded from above, and the particle's energy (The value of the Kaluza-Klein Hamiltonian which is a constant of motion) is larger than the maximum potential. In this case, the trajectories in this region are equivalent up to a reparametrization to a free geodesic motion in the Jacobi metric (see section 7.7). Thus, in this case, the original problem is equivalent to a Riemannian problem.

  • $\begingroup$ Thanks! I'll certainly look in the Marsden book. So you're saying that there's some metric on $\mathbb R^n \times S^1$ so that geodesic flow there is equivalent to the original mechanics on $\mathbb R^n$? Then, a question: should I worry that the solution for geodesics in general wants only a small patch in $S^1$? $\endgroup$ Commented Feb 7, 2010 at 19:18
  • $\begingroup$ I don't think that one should worry. Consider for example an isotropic harmonic oscillator on the x-y plabe together with a constant magnetic field in the z-direction. If we look on this problem in the polar plane coordinates, we see that the angular momentum on the plane and also the angular momentum in the Kaluza-Klein S1 are conserved quantities (the hamiltonian depends only on the momenta) and there is no essential difference between these two angular coordinates. $\endgroup$ Commented Feb 8, 2010 at 15:36
  • $\begingroup$ So the Jacobi metric exactly does the trick when there is no magnetic field. But I'm still dubious when there is a magnetic field. Originally I was looking for a nbhd in which I can uniquely solve the BVP with pure Dirichlet boundaries. But from the (n+1)-dimensional perspective, now I want a Dirichlet boundary at one end, and some mixed thing (Dirichlet in the original n coordinates, Neumann in the new one, because I want to fix the charge) at the other. I can solve this if I can solve the Dirichlet problem in an infinite cylinder, but I don't know how to do that. $\endgroup$ Commented Feb 10, 2010 at 5:13
  • $\begingroup$ No, this is all correct. The trick is that we only need the convex neighborhood in the (n+1)-dimensional space to include (images of, under the flow map) hyperplanes with non-zero charge, which should project diffeomorphically onto the n-dimensional base. This can always be done in small enough neighborhoods. $\endgroup$ Commented Feb 10, 2010 at 16:00
  • $\begingroup$ Sorry for the late comment, a remark on your previous comment (about the boundary conditions). If we consider a full Dirichlet BVP by selecting two arbitrary values theta_i and theta_f on the Kaluza Klein S1, for a given time T we will have a unique geodesic. but the constant of motion representing the charge will not in general have the required value. Using the time reparameterization invariance of the geodesic equation, scale the time by a constant factor inducing a similar scaling on the charge value because it is proportional to time derivatives. Thus we can get the required charge. $\endgroup$ Commented Feb 11, 2010 at 4:18

In addition to DBM's (totally correct) answer above, I realized that there's probably a much simpler answer. If I'm wrong, hopefully someone will set me right.

Let $\mathcal O$ be an open neighborhood in $\mathbb R^n$ with compact closure. Consider the family of differential equations: $$ 0 = \ddot\gamma + \epsilon \, db \cdot \dot \gamma + \epsilon^2 \,dc \quad\quad ({\rm EOM}_\epsilon)$$ The solutions to $\rm (EOM_0)$ are just straight lines. For each $\epsilon$, consider the flow $\phi_\epsilon: {\rm T}\mathcal O \to \mathbb R^{2n}$, which sends $(v,q)$ to $\bigl(\varphi_\epsilon(v,q),q\bigr)$, where $\varphi_\epsilon(v,q) \in \mathbb R^n = \gamma_\epsilon(1)$, where $\gamma_\epsilon$ solves $\rm (EOM_\epsilon)$ with initial conditions $\bigl(\dot\gamma(0),\gamma(0)\bigr) = (v,q)$. By the standard results from ODEs, $\phi_\epsilon$ is smooth when it's defined, and depends smoothly on $\epsilon$.

But the closure of $\mathcal O$ is compact, so $\phi_\epsilon$ is defined for sufficiently small $\epsilon$ depending only on $\mathcal O$. Moreover, $\phi_0(v,q) = (q+v,q)$ is one-to-one, and $\phi_\epsilon$ is too for $\epsilon$ sufficiently small depending on $\mathcal O$. But the flow by time $1$ for $\rm (EOM_\epsilon)$ is, up to a linear reparameterization, equivalent to the flow by time $\epsilon$ for $\rm (EOM_1) = (EOM)$.

Thus we have a solution to part 1. of the conjecture/question. And part 2. is essentially obvious, because this family contains all "low energy" paths that have both endpoints in $\mathcal O$. So this doesn't quite do 2. as stated, but replacing $\mathcal O$ by $\mathcal O_0 \subseteq \overline{\mathcal O_0} \subseteq \mathcal O_1 \subseteq \overline{\mathcal O_1}$ compact does the trick.


{\bf Counterexample.} (But look at my comment above please.) Take your $B$ dead zero: no magnetic field, or friction (however you are thinking of it). Your force field is now pure potential. Your inital velocities $\nu$ are all zero. Take $C = (1/2)|q|^2$ -- a harmonic oscillator --so your "EOM" is $\ddot q = - q$. Take your "base" $q \ne 0$ from your question/conjecture. Conservation of energy asserts that any $q_1 = \gamma(t)$ connected to such a $q$ by the classical path $\gamma(t)$ having initial condition $(q, \nu) = (q,0)$ satisfies $|q_1|^2 \le |q|^2$. Indeed $|\gamma(t)|^2 \le |q|^2$ with equality if and only if $t$ is an integral multiple of $\pi$. In particular for all sufficiently small time we have $|\gamma(t)| < |q|$. (You have to wait a time $2 \pi$ to get back to $q$. ) Thus if you really want your time intervals small of type $(0, \epsilon)$ with $\epsilon$ small you are screwed! You cannot have both $\mathcal O_0$ of $\mathcal O_1$ containing $q$, since $\mathcal O_0 \times \mathcal O_1$ cannot contain any point along the diagonal.

I may have gotten $q$ and $q_1$ reversed relative to your labellings of your question/conjecture, but the same trick still works. The guts of the matter is that a ball of initial conditions of the form $(q, \nu) = (q,0)$ shrinks in $q$-space under the oscillator flow: the force is attractive, after all!

The same trick is bound to work for non-zero $B$.

You might be able to `save' your conjecture by rephrasing, eg. not insisting that $\mathcal O_0 \times \mathcal O_1$ intersect the diagonal, but keep the oscillator in mind, and perhaps say more clearly where you are really headed in posting this question/conjecture.

  • $\begingroup$ I corrected some minor formatting. Annoyingly, Markdown interprets back ticks differently from the way TeX does. $\endgroup$ Commented Feb 8, 2010 at 18:44
  • $\begingroup$ Anyway, I've thought a lot about the harmonic oscillator --- it's the only one I can solve explicitly --- and maybe I misphrased my question. But the point is that for a small open neighborhood O = O_0 = O_1, for any q_0,q_1 \in O, for any \epsilon &lt; 2\pi, there is a unique path connecting q_0 to q_1. Fixing O and letting \epsilon \to 0, the path of duration \epsilon starting stationary at q\in O certainly ends in O, and so is one of the paths in my family. So this is not a counterexample. I believe that any counterexample must include a potential that grows \gg q^2. $\endgroup$ Commented Feb 8, 2010 at 18:46
  • $\begingroup$ Theo: I am up at MSRI till June: schedule to come by via email and we can talk about what you are doing if you like $\endgroup$ Commented Feb 9, 2010 at 2:48
  • $\begingroup$ I'll definitely come up and we can talk. What days are good? $\endgroup$ Commented Feb 9, 2010 at 4:34

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