The solution to the balancing ball paradox

[zerodish]

TL;DR

ASYMPTOTIC MOTION

Assume you have a perfect hill a perfect ball and a perfect putter. Then you putt the ball so it comes to rest on top of the hill. If you accept this is possible then you have to accept a ball balanced on the hill can eventually fall down the hill. That is an effect without a cause. While this is theoretically possible it would take an infinite time for the ball to come to rest. That is it approaches the top of the hill asymptotically. Also any force capable of placing the ball on the top of the hill would be working outside of the laws of physic of this universe.

 

[PeroK]

Assume you have a perfect hill a perfect ball and a perfect putter... working outside of the laws of physic of this universe.

 

[Baluncore]

Assume you have a perfect hill a perfect ball and a perfect putter.

Anything is possible, if you can assume the impossible.
Perfection is the enemy of progress.

 

[jbriggs444]

TL;DR Summary: ASYMPTOTIC MOTION

Assume you have a perfect hill a perfect ball and a perfect putter. Then you put the ball so it comes to rest on top of the hill. If you accept this is possible then you have to accept a ball balanced on the hill can eventually fall down the hill. That is an effect without a cause. While this is theoretically possible it would take an infinite time for the ball to come to rest. That is it approaches the top of the hill asymptotically. Also any force capable of placing the ball on the top of the hill would be working outside of the laws of physic of this universe.

Whether infinite time is required depends on the shape of the hill.

For a roughly hemispherical hill...

The instantaneous deceleration rate is proportional to the distance from the center. Meanwhile, the potential energy deficit and the remaining kinetic energy of the ball are proportional to the square of the distance.

If it takes time Δt to make it halfway to the top, the ball's velocity will have halved and the deceleration rate will also have halved. So we are back to essentially the same situation we started with. It will take another Δt to make it halfway again.

So yes, it takes infinite time in the hemispherical case.

But there are potential hills that do not require infinite time. If memory serves, one hill profile is $h=-e^{-\frac{1}{x^2}}$ for $x \ne 0$ and $h = 0$ for $x = 0$

The above function is not analytic. Its Taylor expansion has a zero radius of convergence at $x=0$. The derivatives of the potential function at $x=0$ are zero. To all orders. So the Taylor expansion is a constant function. Yet the actual function is not constant.

If one solves the differential equation of motion on this hill with boundary conditions for $v = 0$ and $x = 0$ at $t=0$, one will find that multiple solutions exist. Newtonian physics is not predictive for this case.

I was exposed to this situation in my differential equations class many many years ago. @john baez wrote a series of Insights articles here that exposes at least one other singularity in classical physics.

 

[zerodish]

A lot of people will come across that problem and waste a lot of time thinking about it. I came across it on youtube. I have lost my ability to save youtube videos so I don't waste my time viewing their site or saving the url of a video. I didn't waste any time on the problem it solved itself. I would make the ball zero size. Then I would try a normal probability distribution or perhaps a 2 dimensional cone. I think if you do find a solution that brings the ball to rest in a finite time the equation will not be time reversable.

 

[jbriggs444]

I think if you do find a solution that brings the ball to rest in a finite time the equation will not be time reversable.

It depends on what you mean by "time reversible".

The same solution played out in reverse obeys the laws of Newtonian mechanics. In that sense, the solution is time reversible. As it must be. The laws of Newtonian mechanics are invariant under time reversal.

However, when played out in reverse from a starting state at rest atop the hill, the time reversed solution is not the only allowed possibility. In that sense, it is not time reversible.

 

[martinbn]

https://sites.pitt.edu/~jdnorton/Goodies/Dome/index.html

 

[sophiecentaur]

TL;DR Summary: ASYMPTOTIC MOTION

then you have to accept a ball balanced on the hill can eventually fall down the hill

Imo this is just an example of unstable equilibrium, mentioned but not philosophically discussed in School Physics. it's a pencil, balanced on its point or a bell in a bell tower, waiting to be rung and kept there by a wooden bar to avoid the ringer to relax. It gives a small dimple in the potential field.

 

[jbriggs444]

Imo this is just an example of unstable equilibrium, mentioned but not philosophically discussed in School Physics. it's a pencil, balanced on its point or a bell in a bell tower, waiting to be rung and kept there by a wooden bar to avoid the ringer to relax. It gives a small dimple in the potential field.

You seem to have missed the interesting feature of the situation.

An ideal pencil, perfectly balanced on its point needs a starting impulse to begin its fall. It cannot begin falling spontaneously. A puck resting atop an appropriately shaped frictionless hill needs no starting impulse. It can begin the fall spontaneously at any time and in any direction. An effect without a cause.

 

[sophiecentaur]

You seem to have missed the interesting feature of the situation.

I think I have. What is the difference with the path of the CM of the pencil and the CM of the puck? Are they not both longitudinal lines on a sphere? Doesn't the PE vary in the same way?

 

[DaveC426913]

This is, at its core, the same problem as "can a star capture a rogue planet?" This might be a simpler case, since it eliminates a lot of the practical and intuitive factors that might come from friction and uneven surfaces and such.

One asks if a moon can enter an empty star system at just the perfect speed to get captured into a stable orbit.

Seems reasonable - until you reverse the proess, and you realize that it would also mean that a stable planet, in an otherwise empty star system, could just spontaneously spin off into outer space. This is much more intuitively false.

 

[jbriggs444]

I think I have. What is the difference with the path of the CM of the pencil and the CM of the puck? Are they not both longitudinal lines on a sphere? Doesn't the PE vary in the same way?

The shape of the potential hill is different. This means that the differential equation of motion is different. Naturally, the solution(s) to the differential equation are different. Not just quantitatively different. Qualitatively different.

This has been explained up-thread. However, I will explain again.

In the case of an approximately hemispherical hill we have a horizontal force law that is directly proportional to the distance from the center of the hilltop. Without explicitly solving the associated differential equation (second order, linear, homogenous, easy to solve) the primary solutions are two exponentials, one decaying as time increases toward $+\infty$ and the other decaying as time decreases toward $-\infty$. Any linear combination of these solutions will also be a solution.

As is usual in the case of differential equations, we use the boundary conditions for position and velocity at $t=0$ to determine which linear combination fits our scenario.

The resulting motion can be characterized as the ball or puck approaching the hill and rebounding away, approaching the hill and passing over the top, approaching the top of the hill forever, receding from the top forever or starting at the top and staying there forever.

Those are the only solutions. There is no solution where the ball or puck starts at rest at the top and then comes down.

That is for an approximately hemispherical hill.

If the hill is shaped differently, we can encounter a different differential equation where the boundary conditions (e.g. $x=0$ and $v=0$ at $t=0$) are insufficient to uniquely pick out a solution. The laws of Newtonian mechanics are not predictive in this case.

The idea that Newtons laws fail to make a prediction can be a jarring realization.

 

[willyengland]

I still don't understand what the difference between a pen on its tip and a ball on a hilltop is?

 

[martinbn]

I still don't understand what the difference between a pen on its tip and a ball on a hilltop is?

Is there a solution, where the pen is not vertical, it is given push and in finite time it goes vertical and stays there?

 

[sophiecentaur]

Not just quantitatively different. Qualitatively different.

I was trying to think this through and I may have come to terms with it. The forces acting are g and 1. the normal force from the surface or 2. The thrust along the pencil. The two accelerating forces will act in the same direction (tangential) for a given angle of displacement. I wondered about moments of inertia but. for a pencil of zero mass with a point mass at the top, the MI is zero. So the above would indicate no difference but:

Or is the difference due to the shape of the point contact of the puck which would have a finite flat bottom surface? It will have a finite MI and the contact point will be in a non-central position on the base, depending on the latitude angle rotation, angular velocity and MI? That is a lag in the net force on the puck.

Is there a solution, where the pen is not vertical, it is given push and in finite time it goes vertical and stays there?

I think that example has a discontinuity in PE at the top.

This is, at its core, the same problem as "can a star capture a rogue planet?"

I think that capture will depend on the other masses already in orbit. 'Gravity assist' by Jupiter can either slow or speed up a visiting body to bring it into the solar system permanently or kick it out faster. I would imagine that the situation type that the thread is considering would only apply for an isolated star with no planetary disc (never likely to form). The interaction that produces stable orbits will only occur with several other bodies involved, and any such orbit is not long term stable because of resonance.
Then there can be energy loss due to small collisions or even tidal losses.

 

[DaveC426913]

I think that capture will depend on the other masses already in orbit. '

"..in an otherwise empty system..." i.e. no other bodies to complicate the problem.

 

[jbriggs444]

I was trying to think this through and I may have come to terms with it. The forces acting are g and 1. the normal force from the surface or 2. The thrust along the pencil. The two accelerating forces will act in the same direction (tangential) for a given angle of displacement. I wondered about moments of inertia but. for a pencil of zero mass with a point mass at the top, the MI is zero.

None of these physical circumstances are very important. One ends up with a differential equation which is (in the region near the top anyway) negligibly different from $\frac{d^2x} {dt^2} = kx$ ($k$ being strictly positive).

[Compare to the differential equation for simple harmonic motion where $k$ is negative]

The net force is approximately proportional to displacement from the center. Any non-zero moment of inertia simply behaves like additional mass.

So the above would indicate no difference but:

Or is the difference due to the shape of the point contact of the puck which would have a finite flat bottom surface?

Do not get bogged down in physical irrelevancies. We can imagine a perfect pencil with an ideally sharp point which does not slip on the surface beneath.

It will have a finite MI and the contact point will be in a non-central position on the base, depending on the latitude angle rotation, angular velocity and MI? That is a lag in the net force on the puck.

More physical irrelevancies. Please do not try to dodge the central issue in such a way.

We should not let the fact that there is no such thing as a spherical hill prevent us from solving the equations for a spherical hill.

I think that example has a discontinuity in PE at the top.

None of the potential functions discussed in this thread are discontinuous at the top.

Note that I hate the way preview mangles $\LaTeX$ these days.

 

[sophiecentaur]

negligibly different from d2/dt2=kx (k being strictly positive).

How is that not the equation for a pencil on its point? For a pendulum the force is proportional to the displacement so for the pencil, how does the geometry not give the same equation? I'm hoping for you to switch that lightbulb on for me at any moment.

 

[sophiecentaur]

None of the potential functions discussed in this thread are discontinuous at the top.

Ouch! Thanks.

 

[jbriggs444]

How is that not the equation for a pencil on its point? For a pendulum the force is proportional to the displacement so for the pencil, how does the geometry not give the same equation? I'm hoping for you to switch that lightbulb on for me at any moment.

The characteristic equation for the differential equation with $k$ positive delivers two real roots. The characteristic equation for the differential equation with $k$ negative delivers two imaginary roots.

That is the difference between exponential solutions and sinusoidal solutions. The two situations are related by Euler's formula: $e^{ix} = \cos x + i \sin x$

Or, looking at it a different way, the sine function has a second derivative that is some negative multiple of itself. The exponential function has a second derivative that is some positive multiple of itself.

Simple harmonic motion (normal pendulum) has sinusoidal solutions.
The inverted pendulum has exponential solutions.

 

[DaveC426913]

Simple harmonic motion (normal pendulum) has sinusoidal solutions.
The inverted pendulum has exponential solutions.

Is that essentially the difference between a negative feedback loop and positive feedback loop?
(A ball rolling in a trough, versus a ball rolling along a ridge top.)

 

[sophiecentaur]

That is the difference between exponential solutions and sinusoidal solutions.

Thanks. That dug me out of the problem perfectly. That sign difference gives a 'i', which was not uppermost in my mind after all that time. My Maths muscles are very creaky these days and I often need the 'obvious' stated to me with 'hard sums'.

in an otherwise empty system..." i.e. no other bodies to complicate the problem.

OK; a very unusual situation. But there's the other thing. Gravity is an attractive force with a deep well shaped potential graph. The hyperbolic path with your two lone bodies will always send the visitor back out (symmetrically) unless there's an actual collision. So the only way for a foreigner to be captured would be with the help of other solar system objects; my earlier description. Problem with that event would be the massive amount of KE that visitor would have. Inter stellar visitors are haven't been seen many times and, to be captured, they'd need to arrive near the ecliptic plane. Comets are mostly from the Oort loud afaik. But that's a very practical requirement and doesn't follow the (idealised) topic of this thread.