# A paradox inside Newtonian world

1. Nov 12, 2006

### Tomaz Kristan

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

http://critticall.com/alog/Antinomy_inside_mechanics.pdf" [Broken]

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

- Thomas

Last edited by a moderator: May 2, 2017
2. Nov 12, 2006

### arildno

You're calculations are wrong. That's all there is to it.
I'll see if I am going to bother to show you your computational mistakes.

3. Nov 12, 2006

### Galileo

Hello Tomaz,

That's correct.
The one on the right is farther away than the one on the left, but the right one is also heavier. The gravitational force also increases with mass. But in this case, sure, the force the left one exerts is way higher than the one on the right.
You can in general show that the force the particle on the left exerts is 25 times higher than force from the right one. But to determine the motion of a particle you have to consider the NET force on it, which is the sum of the gravitational forces from ALL other particles, not just the ones next to it. Take these into account in your analysis and you'll see that the center of mass at 10/19 meters will stay in place.

4. Nov 12, 2006

### Tomaz Kristan

The NET force IS negative, it is pointing to the left.

We have only a finite number of masses on the right. Every one of them, more than balanced with those on the left.

For example. The mass at 1/10 is pulled to the right by the mass at 1, with 25 times less force, than to to the left, by the mass at 1/100.

Since that one at 1/100 is 4 times smaller than that at 1 - and 10 times closer.

Holds for any other also. Those big masses on the right are quickly outweighed by those smaller on the left, which are much closer.

Right?

EDIT: typo.

5. Nov 12, 2006

### arildno

Haven't you heard of Newton's 3.law?
Add all the component F=ma together, the sum of the force terms necessarily reduce to 0 due to N 3.law, hence the right-hand side also equals zero, meaning that the C.M has acceleration 0.

So, your conclusion is dead wrong; all that remains is to do a bit of tedious calculation to show exactly where the flaw in your argument is located.

6. Nov 12, 2006

### Tomaz Kristan

Of course I've heard about the Newton Third Law.

That's why, we have a paradox here. A theorem, which opposes this law.

The calculation is quite simple, however. We have LESS force to the left, as it had been only a mass twice as big in the neighboring point, as it is. As stated, it is dispersed almost to 0.

The force to the right is even much smaller, due to a greater distance. 10 times the distance, means 100 times less force, per mass unit.

Don't you see?

7. Nov 12, 2006

### Staff: Mentor

How silly. Distribute your masses any way you want. If gravity between the masses is the only force acting, they all move towards the center of gravity. So what? You don't seriously think that they all move left, do you?

8. Nov 12, 2006

### Tomaz Kristan

The center of mass is at 10/19.

Do you think, that the mass at 1/10 goes there? Ignoring the forces at its left side?

9. Nov 12, 2006

### chronon

It looks to me like you're right; there does seem to be a paradox.

Newtonian gravity does have problems with infinity, e.g. it's possible to devise a system of bodies where one goes to infinity in finite time, and of course there's always the question of how an infinite universe can remain static. However, I've not come across your paradox before.

One thing to take into account is that you're only considering the instantaneous force acting on each particle. It might be interesting to consider what happens to the system over a period of time - my feeling is that as you get closer to the origin, the time spent travelling left by each particle would get shorter and shorter before it met its neighbour and started travelling to the right. Hence it might turn out that the centre of mass doesn't actually move.

Last edited by a moderator: May 2, 2017
10. Nov 12, 2006

### Tomaz Kristan

chronon,

I have thought about that. That the half kilo mass could fall back, doe to the escaping masses on the left side and therefore prevailing gravitational pull from the biggest 1 kilo mass. But that would mean, that the mass center travels left even faster.

And also, you may initially put the solid rods with a negligible mass between every two consequent balls. Then remove only the rightmost one.

This way the shape on the left side and therefore forces are stable long enough. Only the biggest ball is slowly drifting left, eventualy passes the 10/19 mark, where the old center of gravity stood.

Smaller balls can't go right for the gravitational pull from the left, nor left for the strong solid rods between them.

We have a Newtonian version of the so called Infinity Hotel, weighting 2 kilograms. And doing essentially more troubles.

EDIT: typo

Last edited: Nov 12, 2006
11. Nov 12, 2006

### chronon

But the masses will hit each other eventually. What I am saying is that the time taken for this will get shorter and shorter as you go towards the origin, and when the masses on the left have joined together they will start moving to the right. There may be no interval of time in which the centre of mass is actually moving to the left.
But if you join the masses on the left together then the local forces between them will be cancelled out by the force from the rods, and only the force of the rightmost mass will remain. Hence the paradox will disappear.

12. Nov 12, 2006

### robphy

So, what happens to the left-most mass?
The net-force on it points to the right.

It seems to resolve this, consider a finite number (N<infinity) of masses arranged according to your scheme. Then, take the N-> infinity limit.

13. Nov 12, 2006

### Hurkyl

Staff Emeritus
Although there is ostensibly zero mass at the origin, it is being flung rightwards with infinite force. When you figure out how to handle that singularity, it will almost certainly provide the momentum contribution that you're missing.

14. Nov 12, 2006

### Staff: Mentor

Why don't you do the calculations and post the results. You're not likely to find someone here to do them for you here.

15. Nov 12, 2006

### Galileo

Alright, the net force on every mass is indeed pointing to the left. The problem is that you have infinitely many particles and the forces are tending towards infinity. In this way, the expression for the TOTAL force is the system is a sum which will diverge if counted one way (particle wise), but will be zero if the forces are counted pairwise.

I'd say the way to solve the paradox is to say the situation is unphysical and the correct way to handle it is like robphy said and calculate it for a finite number of particles, then take the limit N-> infinity.

16. Nov 12, 2006

### Hurkyl

Staff Emeritus
That doesn't work; the apparent paradox doesn't hinge on the assumption that it's a physical situation. (Of course, it does hinge on the assumption that this is an allowed mass distribution in the mathematical formalism)

17. Nov 12, 2006

### Tomaz Kristan

There is NO mass at 0. Since every mass is at 10^N. None at 0 at the beginning.

And I did the calculations. No right pointing NET force.

And the case with solid rods between every pair of neighboring masses, except between the last two, is astonishingly simple. Let understand and agree with this simpler case first!

18. Nov 12, 2006

### Hurkyl

Staff Emeritus
I'm pretty sure there is zero mass at the origin, not "NO mass".

I was talking about the gravitational field at the origin.

That's easy. The left complex exerts a leftward gravitational force of approximately 1.122 G on the right mass. The right mass exerts a rightward gravitational force of approximately 1.122 G on the left complex. (assuming the rods have zero mass)

19. Nov 12, 2006

### Tomaz Kristan

So, you say, the half kilo mass is falling toward the right one with the same acceleration as every other? No tidal force inside the left complex?

The rod between two balls just keeps the distance. Doesn't glue them together!

So how this uniform acceleration is possible?

20. Nov 12, 2006

### Tomaz Kristan

In fact, every ball is falling, as the combined force of all other balls demands.

And this is the net force, always pointing to the left, balanced by the rod, like standing on the floor. The Moon will not pull you up, Jupiter might, when close enough.

Here we have an infinite cascade of ever smaller, yet closer balls. Every sphere is tide down by all bellow.

And so to speak, Jupiter is falling down.

21. Nov 12, 2006

### Hurkyl

Staff Emeritus
There is certainly a considerable amount of tension in the rods, but you already hypothesized that they are perfectly rigid, so they can withstand the stress without expanding. Your masses are point particles, so nothing is happening to them either.

Or, did you mean that your masses are unattached to the rods? In that case, the behavior is, as far as I can tell, identical to your original problem. (unless you suppose the existence of a "point rod" between the origin and your complex... in which case you will get the uniform acceleration I mentioned)

What does Jupiter have to do with anything?

Last edited: Nov 12, 2006
22. Nov 12, 2006

### Galileo

Well, it IS a physical paradox, since it apparantly violates conservation of momentum. Mathematically there's nothing weird; the proof that the sum of internal forces is zero doesn't work here because that sum is a divergent series.

I did the calculation. The total force can be gotten from:

$$F_T=G\sum_{i=0}^{\infty}(25)^i\left(\sum_{k=1}^i \frac{2^k}{(10^k-1)^2}-\sum_{i=0}^{\infty} \frac{50^k}{(10^k-1)^2}\right)$$
the rightmost sum is about equal to 1.1226 (in appropriate units).
This sum is clearly diverging. But as long as you have a finite number of particles, say N, then the first sum goes from 1 to N and the last one from 1 to N-i and you can rearrange the terms in the sum to see they cancel pairwise.

Last edited: Nov 12, 2006
23. Nov 12, 2006

### Tomaz Kristan

You can also rearrange those balls on the following way:

First, you have a 1/256 kg ball. Very small and dense, holding in the air. You hang a bigger ball of 1/128 kg on the bottom of that one. Since both spheres are small enough, the gravity will press them together. Still holding the top one by an external force.

Now, you can put another, two times bigger, beneath the two. Until you come to the 1 kilo, holding almost two kilos in the air, supporting only the highest and smallest one.

Now you extend this complex at the top with an ever smaller and more dense ball.

You do that for every natural N and there is exactly 2 kg construction, with no top ball, which hangs in the air with no external support at all! Every ball clings under the one above by the pure gravity.

However, the planet will slowly drift up. Very, very slowly of course.

And yes, this is only possible inside an ideal Newtonian world. Not here. What is a shame, anyway.

24. Nov 13, 2006

### Tomaz Kristan

You can point this 2 kg construction of infinite number of spheres, each 10 times smaller by radius and 2 times smaller by mass to the nearest star. With wider end closer to the star and wait for that star - to be dragged over here!

Every ball of this meter or so long device will be coined to rest by those ever smaller balls with the enormous gravitational pull. When the star will come close eventually, the biggest ball will fly to that star, but you still have plenty of them in the row.

Of course, not in the real life but in the ideal world of Newtonian laws an Newtonian gravity it must work this paradoxical way.