What happens if you put a sphere (ball) on the top of a pyramid?

[DrStupid]

Aesthetics?

That's a matter of taste.

The reason is “because it usually works” (for either assumption).

If both assumption work than scientists actually don't have to use some criterion to reject some of the solutions. They can also keep them all.

 

[Dale]

They can also keep them all.

Certainly. I never said otherwise. My whole objection is to the claim that the ball “must” fall. You are allowed to keep them all, but you are also allowed to reject the acausal ones. As you say, it is a matter of taste

 

[A.T.]

Let's give some thought to some of the alternatives in a deterministic world:

Suppose the one determined outcome is that the object falls to the NNE after an hour, the same direction every time with that exact initial condition.

This would for one violate symmetry: The situation is identical in every way to the setup in a rotated coordinate system where the object now falls to the south, and the setup thus produces a sort of test for absolute coordinates. This alone is not fatal, but still implies a loss of principle of relativity.
The NNE fall is after an hour, and yet the condition after 50 minutes is in every way identical to the original condition, so by that argument, the solution is nondeterministic, or there is a hidden variable (timer) required to account for the determined behavior.If the one determined solution is being balanced forever, we lose the time symmetry property of the physics. The object can go to the top and stop there, but the reverse situation cannot happen. That's also true of course of the above scenario where the ball can roll in from the west but cannot roll back to the west, only to the NNE after an hour.

So to have deterministic roll-down you:
- lose time symmetry
- lose spatial symmetry / relativity
- have to introduce hidden variables

To have deterministic stay-on-top you:
- lose time symmetry

Seems like an easy choice to me.

 

[A.T.]

No. They do not. If two results are both consistent with Newton's Laws then Newton's laws do not choose between them.

The correct conclusion that can be drawn from Newton's laws is that if the net force on the ball is zero, the second derivative of the ball's position is zero. That property is preserved in the solution where the ball begins moving.

You may be tempted to refer to the first law and suggest that an object subject to no net force must persist in a state of rest. However, the first law is still upheld. At every instant when the ball is accelerated, there is an open interval within which it is subject to a non-zero net force.

I would say it depends on how you interpret Newton's Laws:

- If you interpret them as purely relational (stating how forces are related to acceleration), then you indeed have multiple solutions.

- If you interpret them as causal (stating how forces cause acceleration), then you cannot justify the departure from rest by pointing out, that once it starts moving there will be a net force, that caused it to start moving.

Usually the purely relational interpretation is sufficient, because adding the causal part doesn't change the quantitative prediction. This special case however is an example where only the causal part allows you to make/pick a prediction.

 

[Halc]

- If you interpret them as causal (stating how forces cause acceleration), then you cannot justify the departure from rest by pointing out, that once it starts moving there will be a net force, that caused it to start moving.

If you interpret them as causal (stating how forces cause acceleration), then you cannot justify the departure from motion by pointing out a net force that caused it to stop moving. Doesn't this make the uphill case contradict causality just as much?

 

[jbriggs444]

- If you interpret them as causal (stating how forces cause acceleration), then you cannot justify the departure from rest by pointing out, that once it starts moving there will be a net force, that caused it to start moving.

Usually the purely relational interpretation is sufficient, because adding the causal part doesn't change the quantitative prediction. This special case however is an example where only the causal part allows you to make/pick a prediction.

Can you give a definition for "causal" that is not a synonym for "deterministic"?

I agree that if the laws of physics are correct, and always make a deterministic prediction based on the current system state and if the current system state is completely characterized by the position, shape and velocities of all components then, if the ball stays on top for any time at all, the laws of physics must predict that it stays there forever.

I do not agree that Newton's laws alone make a deterministic prediction.

 

[wrobel]

Finally, an elegant example of apparent violation of determinism in classical physics has been created by John Norton (2003). As illustrated in Figure 4, imagine a ball sitting at the apex of a frictionless dome whose equation is specified as a function of radial distance from the apex point. This rest-state is our initial condition for the system; what should its future behavior be? Clearly one solution is for the ball to remain at rest at the apex indefinitely.

In this example the conditions of the existence and uniqueness theorem for ODE fail. So that there are several solutions with the same initial values. All ODE textbooks contain such examples. It is just another example showing that physics is not the same as mathematical models. But this stupid text by Mr John Norton arises again and again. Especially philosophers like such a stuff.

 

[DrStupid]

I do not agree that Newton's laws alone make a deterministic prediction.

Yes, that's what Norton's dome shows. If you just use Newon's laws you get infinite different solutions. If you want to have a single solution you need something else in addition to Newton's laws. I wonder why we want to have a single solution only. We know that there are systems that cannot be predicted in reality (at least because it is impossible to know the starting conditions with sufficient precision). In case of Norton's dome the prediction that the ball will start sliding after an unknown time is actually closer to reality than the prediction that it will stay forever. If we know that reality is not full predictable why should we implement something like that into the theory?

 

[bahamagreen]

Here is the approach in the tilting pencil, mentioned a ways back...

Tilting Pencils

 

[wrobel]

On the other hand there is a kind of nonuniqueness which is encountered in reality indeed.

Consider a match box on the table. Perhaps this match box has been laying here from yesterday or perhaps I pushed it here from another place a minute ago.

Systems of classical mechanics must meet uniqueness forwards but they are not obliged to have uniqueness backwards

We know that there are systems that cannot be predicted in reality (at least because it is impossible to know the starting conditions with sufficient precision).

yes and it is well described in terms of deterministic ODE

 

[jbriggs444]

Systems of classical mechanics must meet uniqueness forwards

Why?

 

[wrobel]

Why?

because for example that the classical object cannot occupy different places of the space simultaneously

 

[jbriggs444]

because for example that the classical object cannot occupy different places of the space simultaneously

Possibly I am not understanding what you are getting at. You are suggesting that actual physical reality (as distinct from our model thereof) must realize exactly one unique future?

I do not see a way to test such a claim.

 

[wrobel]

imagine how could it nice to have paid a dollar in a shop and still have the same dollar in your pocket

 

[jbriggs444]

imagine how could it nice to have paid a dollar in a shop and still have the same dollar in your pocket

I fail to see the relevance. An undetermined future is not the same as both having the dollar in the shop and in your pocket.

An undetermined future means that there is a future-you who has spent the doller and has the can of beans and a future-you who has the dollar and has left a can of beans on the shelf. Neither one can eat a can of beans with a dollar still in his pocket.

 

[wrobel]

Possibly I am not understanding what you are getting at. You are suggesting that actual physical reality (as distinct from our model thereof) must realize exactly one unique future?

I do not see a way to test such a claim.

I fail to see the relevance.

It is not a claim it is just what we see: we see only one reality.

 

[jbriggs444]

It is not a claim it is just what we see: we see only one reality.

Which is exactly what we would see if there is more than one reality. It is an untestable claim.

 

[wrobel]

Which is exactly what we would see if there is more than one reality. It is an untestable claim.

ok the opposite claim is also untestable
we discuss Russell's teapot do not we?

 

[jbriggs444]

ok the opposite claim is also untestable
we discuss Russell's teapot do not we?

Indeed.

Edit: I agree to abandon this digression into the philosophy of determinism.

 

[wrobel]

Indeed. You made the unfalsifiable claim that cannot be defended.

Actually not. Because

Which is exactly what we would see if there is more than one reality. It is an untestable claim

this means that the phrase "more than one reality" merely does not make sense while my phrase just fixes what we see

 

[Baluncore]

The Vickers hardness test indents a surface with a diamond pyramid.
The Brinell hardness test indents a surface with a hard sphere.

The point of the pyramid will offer an infinite pressure to the sphere and form a pyramidal socket in the sphere. Or, the sphere will crush the point of the pyramid, forming a concave platform close to the top of the pyramid. One or both of those two mechanisms will effectively lock the sphere in a stable position at the top of the pyramid.

The sphere therefore, will not fall from the apex of the pyramid.

 

[Vanadium 50]

A classical pyramid (assuming made out of a continuous material and not atoms) will have an infinitely sharp point. When the sphere is placed on it, the pyramid point will penetrate it (because the pressure is infinite) to a depth determined by Young's modulus. The sphere will not roll without a push.

If your response is "this doesn't count", I would respond that the question needs to be better defined then.

 

[DrStupid]

yes and it is well described in terms of deterministic ODE

It is described in terms of ODEs that may have different solutions for the same starting conditions. We discussed an example above. The ODEs alone don't guarantee that the system is full predictable - not even in theory.

 

[Halc]

Concerning reality instead of mathematical idealizations:

A classical pyramid (assuming made out of a continuous material and not atoms) will have an infinitely sharp point. When the sphere is placed on it, the pyramid point will penetrate it (because the pressure is infinite) to a depth determined by Young's modulus. The sphere will not roll without a push.

OK, but we've already discarded actual physics if we're talking about some fictional continuous material and not a collection of particles held in a vague shape by mutual forces. Young's modulus need not apply here since there is no reason to assume limited hardness for the purposes of this topic.

Yes, a physical (atoms this time) ball-ish shape perched perfectly on a symmetrical (ish) location like that will deform enough to form a finite surface area which can reasonably be expected to make the equilibrium stable. We're not unaware of this.

The Vickers hardness test indents a surface with a diamond pyramid.
The Brinell hardness test indents a surface with a hard sphere.

The point of the pyramid will offer an infinite pressure to the sphere and form a pyramidal socket in the sphere.

This set of comments makes a sort of implication that a mathematical sphere resting on a flat or other non-sharp surface will not offer infinite pressure. If we're talking physical and not mathematical, then the deformation (strain) will occur well before pressure becomes infinite, so this singularity is never reached. Colliding billiard balls do not generate infinite pressure.

 

[Vanadium 50]

OK, but we've already discarded actual physics

Yes we have.

Classical physics has a well-defined answer. The ball is impaled and doesn't move.
Quantum mechanics has a well-defined answer: the state cannot be prepared as specified.

What this whole argument about involves some sort of part-classical part-quantum set of rules that have never been clearly specified.

 

[samalkhaiat]

I cannot find the original article about the pencil, which computed that a perfect balanced pencil shaped object has a 50% chance of falling before 30 seconds at Earth gravity.

Actually, the pencil will fall after maximum time of 3-4 sec. Try to do the experiment.
If you solve Newton’s equation for the CM of the pencil and assume that the best initial conditions (i.e., the angle from the vertical and the associated angular velocity) are determined by the uncertainty principle, \dot{\theta}_{0} = \frac{\hbar}{mr^{2}\theta_{0}}, you find t_{max} \approx 3.6 \ \mbox{sec}. The example in the OP is no different from the pencil.

 

[Baluncore]

This set of comments makes a sort of implication that a mathematical sphere resting on a flat or other non-sharp surface will not offer infinite pressure.

A “sort of implication” is your subjective response to the pressure of conflating the mathematics of perfect geometry with the physics of real materials. If you insist on cross-coupling those different domains you will continue to trip over the idiosyncratic contradictions.

Only for perfect geometry can the area of contact be zero. The sphere would require mass, and acceleration to generate an infinite mathematical contact pressure. In the OP, the mathematical sphere has zero specified mass, so gravity is irrelevant, and nothing can fall. Indeed, a perfect mathematical sphere will have lower density than our real atmosphere, so it cannot fall but will rise to escape the perfect-Platonic-solids model of the Solar system. As is obvious, the two anti-domains must be disentangled, or all hell breaks loose.

For the real material world there is deformation at the point of contact. That real deformation may be elastic, plastic, or a melding of the materials to form a welded connection. As Hertz identified, the applied mathematics of real material contacts are tractable when the geometry can be simplified. That is why the well defined geometry of a sphere, or the point of a pyramid, is used as an indenter for material hardness testing.

 

[Halc]

Back to the pencil thing. I certainly take back my off-the-cuff claim about the ideal pencil claim when initially brought up. It seems that an ideal pencil is far more stable than a ball on Norton's dome. A pencil moving upward would never actually become vertical, taking infinite time to do so. This was illustrated by the link in post 59 here:

Here is the approach in the tilting pencil, mentioned a ways back...

Tilting Pencils

To quote the bottom line of that: "This idealized pencil would take 1 year ... to tip over (theoretically) if its initial stationary position is θ0 = 10^–117498396 degrees from vertical."
So they managed to have it take a year to fall, and only by cheating by giving it an initial displacement from vertical. Norton's ball will fall at any time spontaneously, and take a very finite (short) time to do so once it starts.

As for the physical pencil, of course it will fall quickly.

If you solve Newton’s equation for the CM of the pencil and assume that the best initial conditions (i.e., the angle from the vertical and the associated angular velocity) are determined by the uncertainty principle, \dot{\theta}_{0} = \frac{\hbar}{mr^{2}\theta_{0}}, you find t_{max} \approx 3.6 \ \mbox{sec}. The example in the OP is no different from the pencil.

Yes, you forgot that the real pencil is made of atoms and is not balanced on a point, but on a limited area, and random forces like moon tides and the bug in the next room will surely throw the thing. I'm not disputing any of that.

Actually, the pencil will fall after maximum time of 3-4 sec.

OK, I dispute this statement. There is a finite probability that the random forces and whatnot actually correct any imbalance that arise, and the pencil stays up for minutes at least (indefinitely actually). You can say there some very high probability of it falling within 3-4 seconds, but that probability for an object placed within tolerances of vertical can never be 100%.

 

[jbriggs444]

In the OP, the mathematical sphere has zero specified mass, so gravity is irrelevant, and nothing can fall.

That depends on how one models gravity acting on an entity with zero mass. If we are going to extend Newtonian gravity to such objects, I prefer a model where objects with and without mass accelerate identically.

 

[wrobel]

It is described in terms of ODEs that may have different solutions for the same starting conditions. We discussed an example above. The ODEs alone don't guarantee that the system is full predictable - not even in theory.

ODE alone does not even guarantee existence but ODE+ well-known regularity assumption do guarantee it all

We discussed an example above.

I thought you discussed physics because such examples of ODE are discussed in textbooks

 

[samalkhaiat]

As for the physical pencil, of course it will fall quickly.
you forgot that the real pencil is made of atoms and is not balanced on a point, but on a limited area,

No, I did not "forget" that. The thermal motions of atoms in the contact “area” prevent the pencil from being in stable equilibrium state for more than 4 sec. To account for this experimental result (the < 4sec.), one has to assume that the best initial conditions are restricted by the uncertainty principle which accounts for the fact that “the real pencil is made of atoms”.

OK, I dispute this statement.

Sir, I talk physics not gibberish. Unless you do the experiment and manage to balance the pencil for more than 4 seconds, you cannot dispute physical statement which is based on a very realistic mathematical model that predicts the experimental (< 4 seconds) result.

There is a finite probability that the random forces and whatnot actually correct any imbalance that arise, and the pencil stays up for minutes at least (indefinitely actually).

Again, I ask you to make a realistic mathematical model and calculate for us this “finite probability” of yours. If you cannot do that, then you should listen to the experts.

 

[jbriggs444]

Unless you do the experiment and manage to balance the pencil for more than 4 seconds, you cannot dispute physical statement which is based on a very realistic mathematical model that predicts the experimental (< 4 seconds) result.

The mathematical model in question has flaws which have been pointed out. To whit, it assumes that the only relevant uncertain inputs are present in the initial setup.

Direct experimental testing [balance a pencil and time the fall] is clearly infeasible or impractical for the sort of quibble that is being suggested. Indirect experimental evidence has been known for nearly 200 years [Brownian motion].

 

[.Scott]

I don't see any point in getting too attached to any specific real-world model for this experiment. Perfect spheres, perfect pyramids, and perfect balance are inconsistent with Physics. Spheres and pyramids are made of subatomic particles and those objects can only be positioned to within the limits of the HUP. That may seem to be a unimportant limit, but the falling pencil link by @bahamagree in post #59 shows that astronomically small offsets from "perfect" will quickly compound.

In the most vanilla worlds were "perfect" is possible, the sphere will remain in place.

 

[Baluncore]

In the most vanilla worlds were "perfect" is possible, the sphere will remain in place.

That is also consistent with Brownian motion being a martingale.
Any rotation of the pencil or sphere, will gyroscopically overcome the minor θ₀ = 10^–117498396 degrees from vertical.

 

[.Scott]

Any rotation of the pencil or sphere, will gyroscopically overcome the minor θ₀ = 10^–117498396 degrees from vertical.

Since that would result in the pencil toppling in about a year, the rotation would have to be on the order of once a year at a minimum.