Do there exist small neighborhoods in a classical mechanical system without pairs of focal points? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:39:01Z http://mathoverflow.net/feeds/question/13873 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13873/do-there-exist-small-neighborhoods-in-a-classical-mechanical-system-without-pairs Do there exist small neighborhoods in a classical mechanical system without pairs of focal points? Theo Johnson-Freyd 2010-02-02T21:38:54Z 2010-10-23T15:09:43Z <p>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.</p> <p>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.</p> <p>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.</p> <p>A path $\gamma: [0,T] \to \mathbb R^n$ is <em>classical</em> if it satisfies (EOM); its <em>duration</em> is the number $T$. We can also consider paths with negative duration by flowing backwards, although we will not need to do so.</p> <p><strong>Definition:</strong> A point $(v,q) \in \mathbb R^{2n}$ is <em>focal for duration $T$</em> 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.</p> <p>It is a standard results (see e.g. Milnor's <em>Morse Theory</em>) 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 $T=0$.</p> <p><strong>Proposition:</strong> 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 $(\gamma(0),\gamma(T))$.</p> <p><strong>Sketch of Proof:</strong> 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.</p> <p>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.</p> <p>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 $q$.</p> <p>I would like to reverse the direction of choices: I'd like to pick $\mathcal O_0,\mathcal O_1$ first.</p> <blockquote> <p><strong>Question/Conjecture:</strong> 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:</p> <ol> <li>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 $\gamma(q_0,q_1,T,T) = q_1$.</li> <li>For each $T \in (0,\epsilon)$, the classical path of duration $T$ with initial conditions $(B(q),q)$ appears as some $\gamma(q,q_1,T,-)$.</li> </ol> </blockquote> <p>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.</p> http://mathoverflow.net/questions/13873/do-there-exist-small-neighborhoods-in-a-classical-mechanical-system-without-pairs/14494#14494 Answer by David Bar Moshe for Do there exist small neighborhoods in a classical mechanical system without pairs of focal points? David Bar Moshe 2010-02-07T14:35:58Z 2010-02-07T14:35:58Z <p>For a sufficiently large particle energy, the original problem can be transformed to a problem of geodesic motion as follows:</p> <p>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 <a href="http://books.google.com/books?id=I2gH9ZIs-3AC&amp;printsec=frontcover&amp;dq=Marsden+%2B+Introduction+to+mechanics+and+symmetry&amp;source=bl&amp;ots=STIWna6Ib8&amp;sig=qrkEUDWwL1mbzkoNPA8OPiAWyEY&amp;hl=en&amp;ei=ps5uS8KQDpWCnQPN3ZnSBA&amp;sa=X&amp;oi=book%5Fresult&amp;ct=result&amp;resnum=2&amp;ved=0CBEQ6AEwAQ#v=onepage&amp;q=&amp;f=false" rel="nofollow">Marsden's book. </a> 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.</p> http://mathoverflow.net/questions/13873/do-there-exist-small-neighborhoods-in-a-classical-mechanical-system-without-pairs/14623#14623 Answer by Richard Montgomery for Do there exist small neighborhoods in a classical mechanical system without pairs of focal points? Richard Montgomery 2010-02-08T08:16:08Z 2010-02-08T18:38:51Z <p>{\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 $|\gamma(t)| &lt; |q|$. (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.</p> <p>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! </p> <p>The same trick is bound to work for non-zero $B$.</p> <p>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.</p> http://mathoverflow.net/questions/13873/do-there-exist-small-neighborhoods-in-a-classical-mechanical-system-without-pairs/14927#14927 Answer by Theo Johnson-Freyd for Do there exist small neighborhoods in a classical mechanical system without pairs of focal points? Theo Johnson-Freyd 2010-02-10T17:59:46Z 2010-02-10T17:59:46Z <p>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.</p> <p>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 $\epsilon$.</p> <p>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 $\rm (EOM_1) = (EOM)$.</p> <p>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 <em>quite</em> 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.</p>