# Indeterminism in Newtonian mechanics?

1. May 22, 2010

### Blue_Jaunte

This paper, written by a University of Pittsburgh professor, John Norton, describes a simple situation in which Newtonian mechanics allows for purportedly non-deterministic equations of motion:
http://www.pitt.edu/~jdnorton/papers/DomePSA2006.pdf

In the first part of the paper, he describes the problem and finds the equations of motion. In the problem, a ball rests atop a dome in a constant, downward, gravitational field. The dome is given by the equation
h = 2/(3g)r^(3/2)
(fudging the units a bit for simplicity).
Where
h = distance traveled in the vertical direction
r = distance traveled along the surface of the dome
The net force on the ball (frictionless surface) is
F = a = g*sinθ = g(dh/dr) = r^(1/2)
(assuming unit mass)
Where
θ = angle between direction of motion and the horizontal

The resulting differential equation,
r''(t) = r^(1/2)
has the expected solution, r(t)=0, as well as a class of solutions
r(t) =(1/144)[t - T]^4, for t ≥ T
for some arbitrary time T.

In the later parts of the paper, he addresses possible objections to his conclusions.

I was just wondering what other people thought of this.

-Mike

Last edited: May 22, 2010
2. May 22, 2010

### Q_Goest

Hi Mike. I read through parts of the paper and skimmed others so I'm not claiming to know everything Norton wrote. If you've read it over in detail, and assuming I miss Norton's point somewhere, please point out where in the paper I've overlooked it.

I disagree with Norton's conclusion that spontaneous motion at t=T is allowed by Newtonian mechanics, regardless of whether we are talking about a "dome" case as suggested by the paper or not. In fact, I think that's the key here. Consider the case where r approaches infinity. In other words, consider the case in which the "dome" becomes flatter and flatter, until the radius of curvature approaches infinity. Any force causing motion of the ball down the dome becomes vanishingly small for any t>T. For r = infinity, the dome is flat and the force completely disappears. Certainly, we wouldn't say that on a flat surface, there is any propensity for the ball to suddenly begin rolling. Obviously, this would require a force, and there is no force available on a flat surface, regardless of where the ball is.

Now let's compare the ball on a flat surface to a ball on a dome. Is the contact between the ball on the top of the dome the same as the contact between the ball on a flat surface? I think it is. As long as the ball is indeed at the location on the dome where the surface is perpendicular to the gravitational force, then in theory there's no difference between the two. So if a ball can't be allowed to move on a flat surface because there is no force available to move it, then how can the same ball move on a domed surface if the contact is identical?

3. May 22, 2010

### gabbagabbahey

I haven't looked at it too in-depth, but at first glance it looks as though the extraneous solutions (with spontaneous acceleration) constrain the particle to move along the dome. I'm not sure that the normal force and gravity alone would account for such motion, given the shape of the dome. Certainly, at the top of the dome, any acceleration or initial velocity would cause the particle to lose contact with the dome, and I haven't checked whether the same is true at other points on the dome, but I suspect it is.

4. Aug 14, 2010

### Petr Mugver

I think It's a question about unstable equilibrium. The point r = 0 on the dome has horizontal tangent plane, but the second derivatives of h with respect to x and y form e negatively defined matrix. I think the same problems arise in all examples of surfaces with this property, not only the dome. I'm not able to prove this assertion right now, so I go "heuristically". If the initial velocity is not zero but \epsilon, no matter how small, then the solution is unique. Letting \epsilon go to zero you find other solutions, which in my opinion are unphysical. In the real world we cannot rule out every small perturbation to the motion. In this case of unstable equilibrium, a small perturbation can cause big changes in the motion, as is the case, for example, of an egg in "equilibrium" with the longer axis in vertical position. The fact that we cannot predict the direction the ball will fall is due to the fact that we don't know the exact initial conditions at t = T.

5. Aug 14, 2010

### Petr Mugver

I thought it a bit better. Let's stay in 2D (in that article there was no need to be in 3D). So let's suppose the particle is linked to a curve y = f(x). Suppose f has a point of unstable equilibrium. By shifting the origin we can suppose:

f(0) = 0
f'(0) = 0 (equilibrium point)
f''(0) < 0 (unstable equilibrium point)

The lagrangian is

$$L=\frac{1}{2}(\dot x^2+\dot y^2)-gf=\frac{1}{2}(1+f'^2)\dot x^2-gf$$

The hamiltonian is

$$H=p\dot x-L=\frac{1}{2}(1+f'^2)\dot x^2+gf$$

This is independent of time, so it's conserved. At t = 0 we have H = 0, so

$$\dot x=\left(\frac{-2gf}{1+f'^2}\right)^{1/2}$$

This equation has (at least) two solutions: x(t) = 0 and

$$\int_{0}^{x(t)}\left(\frac{-2gf}{1+f'^2}\right)^{-1/2}dx=t>0$$

Note that this last solution exists only if f < 0 in a neighbourhood of x = 0, i.e. only if the origin is an unstable equilibrium point. Note also that the solution is of the form x(t) = x(t)u(t), where u is the step function. So x(t) and x(t - T) with T > 0 have the same initial condition, thus we get a whole family of solutions. You can add a +\dot z^2 term if you really want 3D, all you get is an extra angular momentum term, which is zero for our initial conditions, so the conclusions don't change.

Now my question is (unless I misunderstood everything, which is EXTREMELY possible): why the dome? I don't really get the point...

Last edited: Aug 14, 2010
6. Jan 24, 2011

### LuisVela

Hi there, I read the paper. I think he is jumping to deeper and more complicated explanations too soon.

Here is some curious result, that helps understand my point of view:

If the ball on the top of the dome has already been set in motion, i.e. is already falling to a given direction, we can calculate its velocity at any given time via using:
$$0=\frac{1}{2}mv^2-mgh\Rightarrow v(h)=\sqrt{2gh}$$

Now, lets imagine the opposite situation. What if we throw the ball upwards with a given speed from the position r(h) for a given h?
Well, if the speed is less than v(h), the ball wont make it to the top, and if its bigger, then it will go to the other side, and start falling there.
Whats interesting, and not too many people realise it, is this: Given that we can set the ball in motion, upwards, with the exact speed v(h). What is the time it will take the ball to reach the top?

Well, it is infinite.

I guess form here you can derive the rest, and light a bit the problem, before jumping to make any deeper assertions.