# What condition is fulfilled by the path of a mass sliding down a lumpy hill?

Suppose I have a hill that goes up and down unevenly. It's frictionless, and I want to slide a point mass down the hill. I am interested in the path it takes. (By "path" I mean the trail it leaves behind, not the time-dependent trajectory.)

I can find the equations of motion of the point mass and, from there, find the path by finding the entire time-dependent motion. Is there a way to find the path directly, given the initial velocity of the particle? For example, does the correct path extremize something?

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I think what Raskolnikov says is incorrect (I'd post this as a comment but can't because of low reputation). What he says would be correct in the absence of gravity, so it is just about finding geodesics on some manifold given by your surface.

Your problem is harder, though, because what you are interested in are not geodesics, but trajectories that correspond to non-zero force acting on the particle. I am not sure what more could I write about this, because your question is too general: it is equivalent to solving equations of motion on (almost) arbitrary two-dimensional manifold with (almost) arbitrary potential.

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Thanks, Marek. I do not completely understand your second paragraph. It seems to me the problem is well defined. We could go find some real hills and slide heavy things down them and look at where the grass was flattened. Everything is obeying the basic laws of physics. Why is it hard to say something intelligible about the paths left behind in the grass? – Mark Eichenlaub Nov 13 '10 at 19:33
I think what Marek wants to say is that if you want an analytical solution, you'll have to specify the landscape (and the starting point), for instance as a function $z=f(x,y)$ or $g(x,y,z)=0$. – Raskolnikov Nov 13 '10 at 19:37
Sure it is well-defined. But it doesn't mean it's easy to solve it in general, or even say anything about the solution. What can you say about arbitrary function on the reals? Well, nothing. But if you specialize to polynomials, then you can find some interesting properties. In the same vein, you can't say anything about general surface: there's just too many of them. If you give us something concrete, a reasonable family of surfaces (e.g. a parabolic surface, or surfaces of constant curvature) then we might be able to say something about the solutions. – Marek Nov 13 '10 at 19:40
@Raskolnikov Suppose I do that. Then what procedure can I follow, aside from solving the entire dynamics problem, to go from $z = f(x,y)$ to a path $g(x,y)=0$? – Mark Eichenlaub Nov 13 '10 at 19:41
Well, if you solve it using the procedure, you'll obtain a path parametrized by the time $t$. So $(x(t),y(t),z(t))$. Now, depending on how complicated your problem is, you can try to eliminate $t$. Think of it as two equations $x=x(t)$ and $y=y(t)$. But the parametrized equations are just as good in my opinion. If you can get them. – Raskolnikov Nov 13 '10 at 19:52

I think you're simply looking for the shortest path between the starting point and the arrival point under the constraint that the path should be on the surface.

So as Mark pointed out, this can't be right. But what is still correct is the principle of least action:

$$S=\int \; \left[ \frac{1}{2}m v^2 - mgz - \lambda (z-f(x,y)) \right] \; dt$$

I chose a function of the form $z=f(x,y)$ to describe the landscape, but more general forms are possible.

Extremizing this action leads to the following equations of motion:

\begin{align} m\frac{d^2 x}{dt^2} & = \lambda \frac{\partial f}{\partial x} \\\\ m\frac{d^2 y}{dt^2} & = \lambda \frac{\partial f}{\partial y} \\\\ m\frac{d^2 z}{dt^2} & = -mg -\lambda \\\\ z=f(x,y) \end{align}

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The point does not follow the shortest path. It is easy to construct counterexamples. For example, consider the interior of an ellipsoid that has high eccentricity. Make gravity point down the long axis, and start the point at the "equator" line. The point slides all the way down to the bottom of the ellipsoid and back up the other side, but it is shorter to go around the equator. – Mark Eichenlaub Nov 13 '10 at 19:20
You're right. I have been sloppy. In that case, I think the only thing that is being extremized must be the classical action for the problem. – Raskolnikov Nov 13 '10 at 19:26
I think I get your point, but I'm not sure about the equations of motion. It looks to me like the ones you have consider $v^2 = dx/dt^2 + dy/dt^2$, and ignore motion in the $z$ direction. – Mark Eichenlaub Nov 13 '10 at 20:17
I just noticed the third equation should have $z$ instead of $x$. Corrected it. So $v^2 = (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2$. Hope I didn't make any sign mistake. I just took the variation off the top of my head. :p – Raskolnikov Nov 13 '10 at 21:14
But $z=f(x,y)$. It shouldn't have its own equation of motion. It's dependent. – Mark Eichenlaub Nov 13 '10 at 21:34

If you know the path $\vec{r}(t)$ then you know the acceleration $\vec{a}=\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$ and from that you get the force needed. $\vec{F}=m\cdot\vec{a}$ .This is also known as a forward dynamics problem.

Given the hill shape $\vec{r}(s)$ where $s$ is a parameter and the derivatives $\vec{r}'=\frac{\mathrm{d}\vec{r}}{\mathrm{d}s}$, $\vec{r}''=\frac{\mathrm{d}^{2}\vec{r}}{\mathrm{d}s^{2}}$ fully describe the hill properties. To follow the hill, the particle velocity has to be tangent to the hill $\vec{v}=\upsilon\vec{r}'$ and the acceleration $\vec{a}=\dot{\upsilon}\vec{r}'+\upsilon^{2}\vec{r}''$.
The equations of motion are $\sum\vec{F}=m\,\vec{a}$ and the reaction forces have to provide no work (or power) making them normal to the path. Why? $\vec{F}_{\mathrm{reaction}}\cdot\vec{v}=\vec{F}_{\mathrm{reaction}}\cdot\upsilon\vec{r}'=0$ only if $\vec{F}_{\mathrm{reaction}}$ is perpendicular to $\vec{r}'$.

If you can define the instanteneous tangential and normal unit vectors at each point $s$ then the equations of motion are decomposed nicely along those directions allowing to solve for the tangential acceleration $\dot{\upsilon}$ and the reaction force $F_{\mathrm{reaction}}$.

I am not fully understanding your question of, given a shape (path) what is the path without the equations of motion? You either know the path and get the forces, or you know the forces and get the path.

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The question is: given a hill, find a path down that hill. Your answer would then be to derive the forces from the shape of the hill. I was curious whether there was some particularly simple way to find the path without dynamics i.e. whether the path had some geometric quality. If you read the above responses I think this will become clear. – Mark Eichenlaub Nov 13 '10 at 20:56