A paradox inside Newtonian world

[Tomaz Kristan]

Yes, but again, you have a point of infinite volume mass density.
I'm starting to think that that is the problem.

Every standard Newtonian mass point has an infinite density. Here, in the ball case we have an infinite density out of the structure. Nowhere inside the complex.

 

[Tomaz Kristan]

I don't get it... Wouldn't the balls move towards Jupiter as well?

It would, if there were a finite number of balls. This way, every ball has a left companions, which easily overweights the Jupiter's gravity. They are much smaller, but very much closer, too.

 

[ObsessiveMathsFreak]

Every standard Newtonian mass point has an infinite density. Here, in the ball case we have an infinite density out of the structure. Nowhere inside the complex.

Well, the density is becoming infinite as you progress to the left. The density is M/V which is 2^-n/10^-n = 5^n, so things are becoming infinitely dense. You could reduced the ball mass, but what would that do to the gravity calculation?

 

[vanesch]

Every standard Newtonian mass point has an infinite density. Here, in the ball case we have an infinite density out of the structure. Nowhere inside the complex.

This is correct, but in standard Newtonian physics (with gravity forces only), you can replace every point mass by a sphere with finite density, as long as no trajectory comes closer than the radius of the sphere. So with a point mass distribution which has a finite lower boundary on inter-point distances, you could in principle replace them by spheres with radius = half of the lower boundary. Note that this doesn't even violate Poincare's recurrence theorem, because you could even introduce *shrinking* spheres of which the constant radius, at a certain time, is always half of the "distance of closest encounter". As such, the dynamics is not perturbed, and at any finite time, the volume density is finite with a global upper boundary.

In your construction, you don't have that, because arbitrary high densities are present from the start.

 

[Tomaz Kristan]

Well, the density is becoming infinite as you progress to the left. The density is M/V which is 2^-n/10^-n = 5^n, so things are becoming infinitely dense. You could reduced the ball mass, but what would that do to the gravity calculation?

I don't see, in which section of space, we have an infinite density here. It goes over every fixed number, but still remains finite everywhere, except in the point "the rightest point of the biggest ball minus 10/9", outside the structure.

Another peculiarity.

 

[Tomaz Kristan]

In your construction, you don't have that, because arbitrary high densities are present from the start.

No, no. I have! A few post higher I gave exactly this case.

 

[ObsessiveMathsFreak]

I don't see, in which section of space, we have an infinite density here. It goes over every fixed number, but still remains finite everywhere, except in the point "the rightest point of the biggest ball minus 10/9", outside the structure.

Yes, every ball does have a finite density. But, which density is the largest?

 

[Tomaz Kristan]

Yes, every ball does have a finite density. But, which density is the largest?

No ball has the largest density. No ball is the leftmost. No natural is the biggest.

That's the "beauty" of the infinty.

 

[Gelsamel Epsilon]

It would, if there were a finite number of balls. This way, every ball has a left companions, which easily overweights the Jupiter's gravity. They are much smaller, but very much closer, too.

Despite there being infinite balls inbetween 1mm and 10/9mm there is a FIRST at 1mm. The first ball would be feeling no gravity from his left.

Unless I'm still misunderstanding?

Assuming you have an infinite line of balls forever extending so that one always does have a bigger one to the left then my reply would be that we don't know how to properly define infinity. You can prove all sorts of things if infinity was a tangible value. But it's not. You can't have 'infinity' balls.

 

[vanesch]

No, no. I have! A few post higher I gave exactly this case.

I don't think so. There is, at a given time, no upper bound to the volume density in all of space. For each given value of volume density, you will find a ball which has a higher density.
This means that I can find a sequence of points in a compact space, x_n, such that rho(x_n) > N for any N ; in other words, I have a sampling of the function rho which diverges and hence rho itself diverges.
In my description, at each given t, there is an upper density in all of space (that is, the entire density function in all of space is bounded). That bound can increase as a function of t, but for a given t-value, there can be an upper bound.
For each given t-value, you cannot find a divergent sampling of rho.
rho(x) < Max(t).

 

[vanesch]

No ball has the largest density.

That's exactly what it means, for the density to diverge.

 

[Tomaz Kristan]

Yes. Density goes over every rho, just as natural numbers go over every N.

Still, every natural is finite, as every density in a finite volume is finite in this structure. You can't show me a finite volume with an infinite density inside the structure.

Nothing illegal here, at all! At least not for the Newtonian world.

In real physics, that is a different matter, of course. You can't get arbitrary high density in the real world.

 

[ObsessiveMathsFreak]

Yes. Density goes over every rho, just as natural numbers go over every N.

Still, every natural is finite, as every density in a finite volume is finite in this structure. You can't show me a finite volume with an infinite density inside the structure.

The density is diverging, it is becoming infinite, as are all the forces on every particle. There are an infinite number of particles so to find the rate of change of the center of mass, we need to sum the rates of change of every mass. But all the numbers involved here are divergent so we no longer have the ability to find the value of the infinite sum.

In order to say that the center of mass do not move, or moves to the left, you must in fact compute this movement, or the accelleration of it instead. It's not enough to make an argument without any mathematics at all, or with only half the mathematics present. You must show your calculations if you wish to make a conclusion about the accelleration. If you do, you will find that you obtain a divergent sums, something that isn't mathematically well defined.

 

[Tomaz Kristan]

I do not care for divergent densities. Why should I?

Ever smaller mass, ever closer distance, ever greater forces - but so what?

It is by no way something illegal.

The calculations? What calculations? They are simple, almost trivial, when we talk about this "touching balls" example.

Case pretty much closed, I reckon.

 

[Galileo]

If it is in disagreement with experiment (that means, things happen in the toy world of the theory which happen observably differently in the real world, in the lab), then it is a theory which has been empirically falsified. it could have been correct, but it just isn't our world.

You're right. So I`ll just wait for experimental confirmation of the prediction that the center of mass in this construction will go to the left. Then the theory can be adjusted.

Even an axiomatic physical theory is not pure mathematics. It is not as strict. As I said: A certain intelligence has to be used when applying the theory to problems.

 

[ObsessiveMathsFreak]

The calculations? What calculations? They are simple, almost trivial, when we talk about this "touching balls" example.

There's quite a few of them though. The calculations are as you say, relatively straightforward. There may be a slight issue with the answers though.

Case pretty much closed, I reckon.

Before you go, if you could just throw down those equations, for the sake of completeness if nothing else. I'd just like to say I saw them. You'll forgive an old man his idosyncracies.

 

[vanesch]

Even an axiomatic physical theory is not pure mathematics. It is not as strict. As I said: A certain intelligence has to be used when applying the theory to problems.

Ah ? To me an axiomatic physical theory is a mathematical theory + a rule to link the mathematical objects in it to observations (which, in the case of Newtonian physics, is rather trivial).

 

[Tomaz Kristan]

ObsessiveMathsFreak,

Very well.

1 - do you agree that the force from the RIGHT side is always finite, for every ball?

2 - do you agree that the force from the LEFT side is always SMALLER than TWICE the immediate left neighbor's gravity force? [As the immediate left neighbor contains half of the mass on the left side, for every sphere.]

What means, that we have a finite force everywhere?

Do you agree then, that those forces are all balanced by the surface reaction force of every ball?

(Now, is it normal, that the tex preview doesn't work? So I can't write down this as a bunch of formulae!)

 

[Tomaz Kristan]

To me an axiomatic physical theory is a mathematical theory + a rule to link the mathematical objects in it to observations

That's the idea, I am also certain. It's no doubt, in fact.

 

[Gelsamel Epsilon]

I think the problem lies in that you cannot have 'infinite balls'

 

[Tomaz Kristan]

I think the problem lies in that you cannot have 'infinite balls'

Of course. Only that Newtonism forgets to tell you that.

Everybody knows, that you can't have the infinite number of (even ever smaller) balls in the real life.

Here, we are talking about an abstract theory, which is apparently inconsistent. What is still officially unknown and unheard of.

 

[radou]

Of course. Only that Newtonism forgets to tell you that.

So what?

P.S. I don't mean to sound rude, I'm just asking a simple question.

 

[ObsessiveMathsFreak]

1 - do you agree that the force from the RIGHT side is always finite, for every ball?

2 - do you agree that the force from the LEFT side is always SMALLER than TWICE the immediate left neighbor's gravity force? [As the immediate left neighbor contains half of the mass on the left side, for every sphere.]

I'm finding it difficult to follow in words.

What means, that we have a finite force everywhere?

But there are an infinite number of forces.

Do you agree then, that those forces are all balanced by the surface reaction force of every ball?

Are they? You haven't proven that yet. The reaction forces may or may not be cancelling everything out.

(Now, is it normal, that the tex preview doesn't work? So I can't write down this as a bunch of formulae!)

Please, do try. Otherwise I'm going to have to write them out instead.

 

[Gelsamel Epsilon]

Hmm, well I've been told numerous times that infinity amounts of things screws everything up.

Not only that but you cannot get "infinity" amount of things so the problem is irrelevant.

 

[Tomaz Kristan]

So what?

P.S. I don't mean to sound rude, I'm just asking a simple question.

It's still possible, that I am wrong. I don't see how, not do anybody else, but it could be, after all.

The other possibility is, that we will be forced to abandon those infinities, and everything will be under control, again.

I would like, that the infinity concept was illogical. Most people hate this idea, though.

Unwanted side effect could also be, that people will become more agnostic, scientifically. They will say: For more than 300 years, you had an error just before your noses, and you haven't seen it! How one can believe science?

That would be a bad thing to happen. In fact, science only harbored the magic (of infinity) for too long. Once we clean it, the science will be better than ever before.

 

[Tomaz Kristan]

Not only that but you cannot get "infinity" amount of things so the problem is irrelevant.

I repeat. Not in reality, that is quite common view lately. But in well established axiomatic systems, like Newtonian mechanics, the infinity is included as a vital part.

 

[Tomaz Kristan]

Left side forces:

2*F(x,x+1)>F(x,x+1)+F(x,x+2)+F(x,x+3)+...+F(x,x+y)+...

Is that true, ObsessiveMathsFreak?

 

[Tomaz Kristan]

F(x,x+1)=G*m(x)*m(x+1)/d(x,x+1)^2

D(x,x+1)=(10^-x+10^-(x+1))/2

 

[Tomaz Kristan]

m(x)=2^-x

So, is that true, that the force on every ball is finite, ObsessiveMathsFreak?

 

[ObsessiveMathsFreak]

What about the reactive forces? You're not solving for them.

 

[Tomaz Kristan]

What about the reactive forces? You're not solving for them.

The reactive force of the surface of any ball, is exactly opposite to the (finite) net gravity force. Fg=-Fr.

That is by the definition. It is not forbidden anywhere. So, why not?

That holds for the complex. Everything perfectly balanced and resting.

Jupiter however, far right away, is forced to drifts closer. No ball in the complex can go there to meet Jupiter, since it has a far more dominant left ball. Going there, would mean not behaving as F=M*a says.

So, we have a pathetic mass of 2 tones (even 2 grams, or 0.6 micrograms ...), which is quite anchored to the left point, and the incoming Jupiter on the right side.

The complex is a bizarre star dragger, with lots of balls. Strange as hell, isn't it?

 

[ObsessiveMathsFreak]

The reactive force of the surface of any ball, is exactly opposite to the (finite) net gravity force. Fg=-Fr.

That is by the definition. It is not forbidden anywhere. So, why not?

That holds for the complex. Everything perfectly balanced and resting.

No it isn't, or at least, you cannot simply say it is. You have assummed that the reactive forces are cancelling each net gravity force but that may or may not be the case. Essentially here you have assummed that the accelleration of all masses is zero to begin with. You don't yet know that. You must calculate. Talk will not do.

The complex is a bizarre star dragger, with lots of balls. Strange as hell, isn't it?

Things will become at lot less strange once you do those caluclations.

 

[Tomaz Kristan]

Don't expect too much from those surface reaction forces! They will do nothing.

Will they?

 

[ObsessiveMathsFreak]

Don't expect too much from those surface reaction forces! They will do nothing.

Will they?

If they fail to balance out the gravitational forces and one another, then the massess will begin accellerating either to the left or to the right. You'll have to calculate the answer to find out.

 

[Tomaz Kristan]

How could they do that? If they are solid enough, the reaction force of the surface is equal, but opposite to the gravity force.

What else?