MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(M,g,V)$ be a Riemannian manifold with potential $V\in\mathcal{C}^{\infty}(M)$. Let $\gamma : I\to M$ be a smooth curve, $I\subseteq \mathbb{R}$ an open interval. We say that $\gamma$ is the trajectory of a constrained unit-mass point-particle (or, simply, a constrained trajectory) if

$\nabla_{\dot{\gamma}}\dot{\gamma}=\langle X, \nu \rangle \cdot \nu $

for every $t\in I$, where $\nabla$ is the Levi-Civita connection, $X:=\mathrm{grad}_gV$ is the "force of gravity" and $\nu:=\dot{\gamma}/||\dot{\gamma}||$ is the tangent direction (whenever $\dot{\gamma}\neq 0$). I think this definition reduces to the "standard" one when $M$ is flat $\mathbb{R}^3$ and $V=-x_3$ (if it doesn't, please correct it so that it fits the physical viewpoint).

Let's say a constrained trajectory $\gamma$ is a brachistochrone between points $a, b\in M$, with $V(a)>V(b)$, if

1) $\gamma (0)=a$, $\dot{\gamma}(0)=0$, there is $T>0$ with $\gamma(T)=b$, and

2) for every other constrained trajectory $\eta$, with $\eta(0)=a$ and $\dot{\eta}(0)=0$ with $\eta(T^{*})=b$ for $T^{*}>0$, we have $T\leq T^{*}$.

Has this concept already been studied? Does it have any interesting applications?

Can anything interesting be said if $M$ is, say, the round $2$-sphere in $\mathbb{R}^3$ and $V$ is $-x_3$? Or what in the case $M$ is homogeneus (such as the hyperbolic plane) ?

Note: this question was just out of curiosity, I'm not doing research on this.

share|cite|improve this question
Do you really mean covariantly costant in the last sentence of your question? The hyperbolic plane has no covariantly constant vector fields. Indeed, the existence of a covariantly constant vector field implies that the universal cover has a flat factor. – José Figueroa-O'Farrill Aug 3 '11 at 12:29
Thank you. I was just thinking of a vector field that resembled, in the hyperbolic context, the classic case in $\mathbb{R}^3$. So maybe this just makes no sense. Edited. – Qfwfq Aug 3 '11 at 12:32
Could you please check your equation for the CONSTRAINED TRAJECTORY. For me it implies that the acceleration is proportional to the velocity which means that the trajectories are reparameterized geodesics of the metric $g$ (the choice of the function $V$ affects the "time" on the trajectory but not the trajectory considered as an unparameterized curve). – Vladimir S Matveev Aug 3 '11 at 13:00
@VSM: perhaps I indeed don't remember well how physics works! I just thought that the force by the constraint would compensate the force the body exerts on it, which I thought is the normal component of grav. force, so the resulting force would be the tang. component of grav. force. And then write " F=ma ". Feel absolutely free to edit according to the right physical equation. [I will have a look to the thing in a later moment] – Qfwfq Aug 3 '11 at 13:57
Maybe I'm wrong but I think the equation of motion should be $\nabla_\dot{\gamma} \dot{\gamma}=X$. – Thomas Richard Aug 3 '11 at 15:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.