A paradox inside Newtonian world

[ObsessiveMathsFreak]

No, it is not always correct just to extrapolate that way. For example, every finite set of naturals have a maximum element. But there is not so for an infinite set of them. Never.

But extrapolating from finite sets is the only way we have of evalutating convergent sums. Without this extrapolation, all infinite sums are undefined.

 

[Eli Botkin]

Tomaz:
But why insist that Newton's theory has to apply to infinite sets of masses?
Newton proposed a physical theory. Infinite sets are not within that scope.
Your "paradox" may be entertaining but I don't see it as any challenge to the theory.

 

[Tomaz Kristan]

But extrapolating from finite sets is the only way we have of evalutating convergent sums. Without this extrapolation, all infinite sums are undefined.

I respect this. Gave you several examples, related to this case.

The infinite sum of forces, all converge to a finite value.

 

[Tomaz Kristan]

But why insist that Newton's theory has to apply to infinite sets of masses?

Be cause there is no rule to forbid that. It is allowed, and it's enough.

 

[K.J.Healey]

I skipped over like 15 pages here cause it was getting tedious, but did you ever show WHY you believe the collective masses on the left (of the ball +jupiter experiment) do not move toward jupiter? I don't understand that.
What would keep them from moving as a whole? The collective forces acting on one of the masses is all of the surrounding masses plus jupiter. While the near masses may keep it from moving, each one has some effect from Jupiter that I don't see how gets canceled out. Each one would have an effect, that would press it a little more on the one to its right, which in turn would result in a normal on the next one toward the left, which would move it to the right, all the way down the chain. What am i missing?

Also we're assuming an infinite number of balls? Because each ball would have a resultant force to the left EXCEPT the left-most one. It would have the resultant force to the right, and it would be great, correct? Basically the force of all balls to its right. By making the system infinite its almost like looking at the whole picture, minus the leftmost ball. Right?

 

[Tomaz Kristan]

What would keep them from moving as a whole?

It's the same thing. You can look every subsystem as a whole, or only the "whole system", it's motion would be the same.

For example, you may cut the Moon into 777 pieces in your calculation and still the orbit must be the same.

Also we're assuming an infinite number of balls?

Yes, we do. Here in this construction. (Maybe I should call it Kristan's complex ).

I don't think - no, I am quite sure in fact! - that nothing like this lives in the real world. It's just a problem inside the abstract Newtonian world.

It would have the resultant force to the right, and it would be great, correct?

Correct. In the finite case, it's all okay.

 

[Tomaz Kristan]

WHY you believe the collective masses on the left (of the ball +jupiter experiment) do not move toward jupiter?

They (each and every ball) have the only liberty to move in the accordance to the resulting force. And the resulting force is a finite force to the left, for any ball you choose.

Still, I do not necessarily insist in a moving thing. The fact, that at the moment t=0, when all the velocities are still zero, when the third Newton's law is already not obeyed -- is enough.

 

[xnick]

I skipped over like 15 pages here cause it was getting tedious

Same here

I smell a problem in setting up the original configuration...
(i.e. related to the unbounded density of the progressive balls and... oh yeah! we need infinity balls! If they start packed together, how could we unpack them? If they start separated, how do we move them to their final positions?)

 

[K.J.Healey]

They (each and every ball) have the only liberty to move in the accordance to the resulting force. And the resulting force is a finite force to the left, for any ball you choose.

As I've read in previous pages, Newtonian physics allows for infinite particle masses, correct? Or even non-particle. Does it allow for infinite spatial dimensions?

What if instead of having them go farther to the left, you say, there are an infinite number of them with the same pattern as stated, from some distance X=-D to X=0. So the infinite particles are not the endpoints but rather the inbetweens. Does the same problem occour? I would imagine not and the system of balls would move together toward Jupiter in the expected manner. The problem is lying not that its just an infinite series or the number of items, which seems trivial, but that the dimensional distance is infinite as well. At least that's what it looks like. But I am no PhD mathematician.

 

[Tomaz Kristan]

If they start packed together, how could we unpack them? If they start separated, how do we move them to their final positions?)

Doesn't matter. You can't even "separate" two mass points. But you can always set the original position every way you want.

 

[ObsessiveMathsFreak]

And the resulting force is a finite force to the left, for any ball you choose.

But the force on the center of mass is infinite.

 

[Tomaz Kristan]

But the force on the center of mass is infinite.

Not true. The mass center is at 10/19 and the force there is of course finite.

 

[heusdens]

Three weeks ago I have constructed this apparent paradox inside the Newtonian world.



http://critticall.com/alog/Antinomy_inside_mechanics.pdf"

Am I wrong or not? What's your say?

- Thomas

Yeah.

For every mass particle you can take the combined force of all the particles left from it, acting as a force to the left relative to the center of mass of all particles on the left side, and all the particles right from it, acting as a combined force to the right relative to the particles to the right. This holds true for every mass particle, except the right most one.

But at the same time, the center of mass of all the masses on the left side, will move to the right, and the center of mass of all the masses on the right side, move to the left.

This then however makes clear that, while at first we say that for every particle it moves to the left, it also belongs to a constellation of particles which center of mass moves to the right.

So, your initial hypotheses that ALL masses move to the left, is not true.

This could be shown also by in advance calculating the center of mass of all particles, and calculate the nett force from every particle as acted upon by that center of mass. So, most part of the particles move to the right.
It's center of mass doesn't move.

 

[Tomaz Kristan]

Nothing moves at t=0. What happens later, I don't even care that much. Situation is illegal already.

At t=0, all net forces, to every ball, point left. And that's bad enough.

 

[ObsessiveMathsFreak]

Not true. The mass center is at 10/19 and the force there is of course finite.

What, the force field? That's not the force on the center of mass. The force on the center of mass is a divergent sum of the forces on every particle.

 

[Tomaz Kristan]

You know what? I don't even care for that. All I need is the situation at t=0, and the finite (all negative!) forces to every ball.

What may going on later, is irrelevant. The paradox has already happened at zero time.

 

[quantumdude]

I've just noticed this thread, and I haven't read the whole thing. So apologies if this has already been brought up. But it seems to me that there is a very basic problem with this. Tomaz, you describe your scenario as "a mass of 2^{-N} kg 10^{-N} for every N". So each mass point can be indexed by a natural number.

You then go on to talk about "the rightmost mass point". But what would its index be? It can only be the largest natural number. But there is no largest natural number. So it seems to me that you can't say anything about the rightmost mass point. It simply doesn't exist.

 

[arildno]

The whole thread is silly.
OP simply doesn't understand that using conditionally convergent distributions is unphysical to start with; he insists that his own unphysicality is Newton's fault.

I haven't bothered with this nonsense for a long time.

 

[Tomaz Kristan]

Tom,

I describe it as a mas of 2^-N kg at position 10^-N meter for every natural N, 0 included.

So the rightmost ball is at 1 meter, has 1 kg. The whole system has of course 2 kg.

And the leftmost ball does not exists.

That's how I describe the damn thing.

 

[ObsessiveMathsFreak]

You know what? I don't even care for that. All I need is the situation at t=0, and the finite (all negative!) forces to every ball.

You don't even care about the force on the center of mass? Isn't that the whole point of this entire thread? Everything I have said applys exclusively at t=0.

What may going on later, is irrelevant. The paradox has already happened at zero time.

You still haven't proven that mathematically.

 

[Tomaz Kristan]

OMF ...

Do you agree, that a finite force is to every mass point?

Do you agree, that every one of those forces is directed toward the left?

(At t=0, of course!)

It would be enough, thank you.

 

[Tomaz Kristan]

Resume?

So, here we are. This simplified version, at t=0, with no moving (yet), is sky clear and nobody can raise an objection?

Good.

 

[K.J.Healey]

Im sorry, have you gotten around to doing any of the math so I could see what you're stating?

 

[Tomaz Kristan]

The left pointing component of the force is equal to 0.992*G*(m/d)^2 for every ball. Where d is the distance to the nearest left ball.

The right pointing component is always smaller. It's also always a finite sum of a finite number of finite right pointed forces.

The resulting force is always negative, for each ball.

Need to be even more specific?

 

[ObsessiveMathsFreak]

The right pointing component is always smaller. It's also always a finite sum of a finite number of finite right pointed forces.

The resulting force is always negative, for each ball.

Yes and the resulting sum of all such forces is undefined.

Consider the force on each ball simply from the one to the left of it.

The mass of the right ball is 2^{-N}. The mass of the ball to its left is 2^{-N-1}. The distance between the two balls is 10^{-N}-10^{-N-1}.

For the left pulling (negative) force on the Nth ball due to only the ball to its immediate left is

F_N = \frac{G 2^{-N} 2^{-N-1}}{\left(10^{-N}-10^{-N-1}\right)^2}= \frac{G 2^{-2N-1} }{10^{-2N}\left(1-10^{-1}\right)^2}
F_N = G\frac{5^{2N}}{2 \left(\frac{9}{10}\right)^2}=\frac{100 G}{162} 5^{2N}

So the sum of all these leftmost forces(a sum less than the total leftmost forces is:

S = \sum_{N=0}^{\infty} F_N = \sum_{N=0}^{\infty} \frac{100 G}{162} 5^{2N}
S = \frac{100 G}{162} \sum_{N=0}^{\infty} 5^{2N}

5^{2N} is obviously, no immediately obviously a divergent sum. In case you don't believe me, here's the first terms 1, 25 , 625 , 15625, 390625, 9765625, 244140625. This isn't going to diverge.

What more do you want? The force on the center of mass is not defined. It doesn't have a mathematically defined motion.

 

[Tomaz Kristan]

Your calculation is wrong, OMF.

 

[ObsessiveMathsFreak]

Your calculation is wrong, OMF.

Words fail me.

 

[Tomaz Kristan]

I know that, already. But do try!

Do you agree, that the sum of masses on the left side of every ball is equal to the mass of that ball?

1/2+1/4+... =1.

Yes?

 

[Hurkyl]

If it's wrong, then you could certainly point out a specific error now, couldn't you?


Incidentally, if I may guess where you're going next, you should remember that the gravitational field exerted by a system of particles is usually not equal to the gravitational field exerted by a hypothetical point particle located at their center of mass whose mass is the sum of the individual masses.

 

[Tomaz Kristan]

Shall we bet around this?

I am all for, are you also?

 

[Tomaz Kristan]

The honesty is the only tool required, as that lady said.

And a little bit of knowledge about limits also. To see that the left side force is always finite here.

 

[Tomaz Kristan]

The force, to every ball, from its left side is smaller, than if it was, if all the masses on the left were concentrated at the point, where only the nearest left ball is.

This way the mass of the nearest left ball would be only doubled and the left force would be double as much, as it is now, from this nearest ball alone.

The way it is described at the post #1, it's only 992/1000 as much.

It's just trivial to see this.

 

[ObsessiveMathsFreak]

Wakalixes.

Would you mind please explaining what in my last calculation was wrong exactly?

 

[Tomaz Kristan]

Where is your error?

The every next ball in the row is not ever closer. Their distancies go as 1.0d, 1.1d, 1.11d, 1.111d ... and so on.

I can't imagine, you didn't know this.

 

[CPL.Luke]

there is also another way of approximating this, that should become more accurate as n goes to infinity.

if you merely define a desnity function that represents the distribution continuously (would the op be opposed to this?)

you can write the gravitational field as being equal to -G int[1,0] den(x) dx/x^2

which is found by finding the contribution to the total field by any infinitesimal segment of mass.

I tried doing this with the density function as e^-x (because the calculations involving this exponential function are easier) and quickly found that the integral is divergent, as you end up with ln(0) in a number of the terms, as the divergent natural log appeared due to the division by x and not the exponential I think that its safe to infer that any exponential distribution will result in a divergent gravitational field.

thus the situation is unphysical and mathematically undefinable, and so is not worth talking about.

but things like this happen quite often when working in Newtonian gravity, just try and calculate the amount of energy that a particle gains when it falls into a point mass.

 

[Mentz114]

This is ridiculous. Almost "how many angels" stuff.

Tomaz, if you start with a physically imposible situation, it doesn't matter what physical model you use to evolve it, it will always remain a physically impossible situation.

My reaction to your 'paradox' - very amusing, so what ? It means nothing.

Stop wasting your time with these fantasies and study the real thing - GR cosmology.

 

[Tomaz Kristan]

Luke,

I don't give any guaranties to my example changed in any way. A modification still may be no good (what's the whole intent), but I stand only behind my construction.

Here, it's not the infinite amount of energy, what is the problem. It's the rebellion against the Third Law of Newton.

The bastard between Cantor's Infinity hotel and the Newtonian mechanics is the ugly one. The paradox. No matter of good advices I get.

 

[ObsessiveMathsFreak]

The every next ball in the row is not ever closer. Their distancies go as 1.0d, 1.1d, 1.11d, 1.111d ... and so on.

I can't imagine, you didn't know this.

My calculation considered the left force on every ball due mearly to the ball to its immediate left. If this sum is divergent, then the overall left force sum, which you yourself "proved" is "greater", is also a divergent sum. PLease read the argument again.

 

[Tomaz Kristan]

My calculation considered the left force on every ball due mearly to the ball to its immediate left.

This is one reason, why your calculation is wrong.

If this sum is divergent

Means nothing. The sum of all forces between various parts of a rigid body can easily be divergent. So what?

 

[Hurkyl]

Means nothing. The sum of all forces between various parts of a rigid body can easily be divergent. So what?

So you've been claiming for dozens of pages that the sum of all of the forces converges.

 

[Tomaz Kristan]

The sum of all the forces, to every ball is finite, yes. Negative, but finite.

 

[ObsessiveMathsFreak]

The sum of all the forces, to every ball is finite, yes. Negative, but finite.

And the sum of all the forces on all the balls is infinite. This has been proven twice now. Frankly, you're trolling at this point.

 

[Tomaz Kristan]

> And the sum of all the forces on all the balls is infinite.

The sum of all forces on all balls doesn't count. You can tile a cube to have the same "effect" of the "infinite sum of all forces".

It is not I who trolls here, OMF.

What only matters is:

The sum of all the forces, to every ball is finite, yes. Negative, but finite.

Now, if that's not true, is the only question.

 

[ObsessiveMathsFreak]

The sum of all the forces, to every ball is finite, yes. Negative, but finite.

To every individual ball. Look this would go a lot faster if you bothered to do some calculations.

 

[Tomaz Kristan]

Calculation

The right side force to mass point m is:

(G*m*m/2)/(d*d)+(G*m*m/4)/(d*d*1.1*1.1)+...+... = (0.992*G*m*m)/(d*d)

What else do you need?

 

[K.J.Healey]

The right side force to mass point m is:

(G*m*m/2)/(d*d)+(G*m*m/4)/(d*d*1.1*1.1)+...+... = (0.992*G*m*m)/(d*d)

What else do you need?

What you have written there is what, this?:

\sum_{i=0}^{\infty} \frac{Gm^2}{2^{i+1} d^{2} \sum_{j=0}^{i} 10^{-j} }= \frac{0.922 Gm^2}{d^2}

Im just turning your ...'s into sums.

 

[Tomaz Kristan]

Looks quite good to me.

 

[K.J.Healey]

but its not right. Left side approaches zero.

 

[Tomaz Kristan]

The left side is quite weak, but I don't care. At least not more than for the OMF's calculation somewhere above.

But the right side is correct and that's enough.

 

[K.J.Healey]

how is it correct? i was just putting what you said into calculable terms. How did you get that number, please show the math.