A paradox inside Newtonian world

[greedangerfoolishego]

As I've already said. I am sitting at X=10, Y=10, Z=10, watching down the show.

I don't understand.
What does that mean?
You need to be in a frame of reference that's moving along with the centre of mass so that it appears to remain still.
If it moves it means that you are in a non inertial frame of reference, where it is known that Newton's Laws don't hold.
Sitting at (10,10,10) just means your at a point in the same frame of reference, does it not?

 

[Gelsamel Epsilon]

Where is the center of gravity?

Infinitely to the left?

 

[Tomaz Kristan]

Where is the center of gravity?

Infinitely to the left?

The center of gravity is not far away from the Jupiters's center of gravity. A little left.

 

[Gelsamel Epsilon]

If the masses of the ball keep getting bigger and bigger every millimeter and the balls go on forever wouldn't that mean that the center of gravity lies infinitely to the left?

 

[Tomaz Kristan]

Sum of all (mass*distance)/(Sum of all masses)

There is the gravity enter. Somewhere left of the Jupiter's center.

 

[Gelsamel Epsilon]

And the masses approach infinity...

 

[Tomaz Kristan]

And the masses approach infinity...

No, do they not. They approach 0, if you go left. Forces approach infinity, bat that does not matter for the gravity center.

 

[Gelsamel Epsilon]

What if I go right? Don't they get infinitely big?

Unless there is a first ball, if there is a first ball then that changes everything.

Edit: To elaborate. If they get infinitely big then the center of gravity is undefined.

If there is a first ball, then that first ball feels no gravity force from the left, and hence moves right.

 

[Tomaz Kristan]

No, if you go right, there is the last ball of 1 tone, 1 mm. Then there is a lot of empty space, than you meet Jupiter on the same line.

For every natural number N, from zero up, there is a ball of 2^-N tons and 10^N mm diameter.

So, a small, 10/9 mm and 2 tons complex ... and a planet size and planet mass ball, a light year or so, to the right, on the same line.

OK?

 

[Gelsamel Epsilon]

I'm not sure I get your model totally.

How big is the space between the center of mass of each ball?

 

[Tomaz Kristan]

How big is the space between the center of mass of each ball?

In the 2 tones complex, every ball touches one on the left and one on the right side. Except the biggest one. That one has only the left side neighbor.

The one on the left is smaller by diameter 10 times, the one on the right is 10 times bigger by diameter.

The one on the left is smaller by mass 2 times, the one on the right is 2 times bigger by mass.

The Jupiter is on the far right side.

 

[Gelsamel Epsilon]

What difference do masses touching have on the center of gravity of the balls?

 

[Tomaz Kristan]

What difference do masses touching have on the center of gravity of the balls?

By their existence, the center of gravity is not at the Jupiter's center. Let say, it's about a meter more closer to the complex, at the beginning.

Then it goes on a travel with the Jupiter. Toward the complex!

The complex would go also, but every of its balls are under much greater force of the left side neighbors.

See the whole picture now?

 

[Gelsamel Epsilon]

I mean the center of gravity of the balls themself. If they are touching wouldn't that distinguishe themselves as the same mass? (Touching of course in a pure and abstract method, ie when I hold stuff I'm not touching them directly, I can't get close enough for that, but in this case they are?)

What distinguishes the gravity I get from 2 lego peices joined, not joined, and just touching?

For instance if I get 2 lego blocks like this
_____
|____|
|____|

Then the center of gravity is right at the break in the middle line.

_____
|__'__|
|____|

however if they're just apart then there is 2 centers of gravity. Both at the Xs

______
|__x__|

______
|__x__|

But if they are so close that they're touching, so so so close that they're atoms are basically touching then which describes the centers of gravity? The former or latter? Do you know?

Edit: What distinguishes us from the mountain from the crust from the core of the Earth in terms of what is generating gravity from the Earth's center of gravity? And you can take that question down to individual particles.

 

[Tomaz Kristan]

There is nothing like "Earth and people" for gravity. Just masses interacting.

Every chunk with every other.

It is no problem here.

 

[ObsessiveMathsFreak]

What equations do you need? Fg=-Fr?

What forces are present here, except those two? Gravity and surface reaction?

None.

Let the overall gravitation pull on each ball be given by $$G_n$$. These can be calculated for each individual ball. Let the surface reaction on the ball on its left be given by $$L_n$$, and the surface reation on its right be given by $$R_n$$

The total force to the left on the nth ball is given by
$$G_n + R_n - L_n$$

Let the accelleration of each ball to the left be $$a_n$$. Let the mass be $$m_n$$. From F=ma we have.

$$m_n a_n = G_n + R_n - L_n$$

Let us assume for the sake of simplicity that the accelleration of every ball is equal, otherwise our surface reactions will cause problems. $$a_n = a$$

So for all n
$$a = \frac{G_n + R_n - L_n}{m_n}$$

As the accelleration is equal for all n, we have
$$\frac{G_n + R_n - L_n}{m_n} = \frac{G_{n+1} + R_{n+1} - L_{n+1}}{m_{n+1}}$$

By Newtons third law, the left surface reaction on the nth ball is equal in magnitude(but opposite in direction) to the right surface reaction on the (n+1)th ball.
$$L_n=R_{n+1} \text{ or } R_n = L_{n-1}$$
Thus,

$$\frac{G_n + L_{n-1} - L_n}{m_n} = \frac{G_{n+1} + L_n -L_{n+1}}{m_{n+1}}$$
$$G_n + L_{n-1} - L_n = m_n \frac{G_{n+1} + L_n -L_{n+1}}{m_{n+1}}$$
$$G_n + L_{n-1} - L_n = m_n \frac{G_{n+1} + L_n -L_{n+1}}{m_{n+1}}$$
$$L_{n-1} = L_n - G_n +m_n \frac{G_{n+1} + L_n -L_{n+1}}{m_{n+1}}$$

or to make it a little easier to work with, bump up the n's by one to get
$$L_{n} = L_{n+1} - G_{n+1} +m_{n+1} \frac{G_{n+2} + L_{n+1} -L_{n+2}}{m_{n+2}}$$

Starting off with the first particle (n=1), to find the total force on it we need to find the leftmost reaction force using the formula;
$$L_{1} = L_{2} - G_{2} +m_{2} \frac{G_{3} + L_{2} -L_{3}}{m_{3}}$$
We know $$G_2 , G_3, m_2, m_3$$, but we do not know $$L_2, L_3$$. Thus in order to find them we must use;
$$L_{2} = L_{3} - G_{3} +m_{3} \frac{G_{4} + L_{3} -L_{4}}{m_{4}}$$
$$L_{3} = L_{4} - G_{4} +m_{4} \frac{G_{5} + L_{4} -L_{5}}{m_{5}}$$

but we do not know $$L_4, L_5$$, so we must find those. But again their formulae will involve $$L_6, L_7$$, whose formulae involve $$L_8, L_9$$, etc, etc , etc.

In short we cannot solve the infinite system of linear equations for every $$L_n$$. Thus we cannot solve for the force, and so cannot find the accelleration of the system. Hence we find that applying mathematics to a physical problem is often more elucidating than verbal argument or consideration.

 

[Tomaz Kristan]

We know that all the forces from the left side are SMALLER, than if only one ball was there, encompasing all mass of others, in the same distance.

That's true for every ball.

No "infinite system of linear equations required".

 

[Tomaz Kristan]

Do you expect, an infinite force somewhere, ObsessiveMathsFreak?Do you expect, an undefined force somewhere, ObsessiveMathsFreak?

 

[Tomaz Kristan]

Diving left, toward 0, increases forces, just like diving to an ever denser ocean.

Those reaction forces, deep down are not felt anywhere closer to surface.

The structure is not that strange for the Newtonian space. Rather quite common.

Only the result is strange. Very strange!

 

[Jheriko]

Wouldn't it be more reasonable to assume that Newtonian gravity is broken than Newton's laws? Even better, can't we just say that Newtonian gravity is not consistent with Newton's laws?

The way I see it the force calculated by using Newtonian gravity is causing the "paradox", not Newton's laws themselves. If one of Newton's laws could be shown to lead to consequences that contradict another of those laws then I would be more willing to say there is a problem with the laws. As it stands all I see is that infinties arising from the Newtonian gravity model applied to point particles are producing an inconsistency with Newton's third law. Imo this isn't a fatal problem for Newtonian gravity, and it is certainly not a paradox. It only becomes a paradox if you assume that the two laws are consistent.

P.S. Interesting thread, very long and lots of interesting discussions on infinity :)

 

[Tomaz Kristan]

Wouldn't it be more reasonable to assume that Newtonian gravity is broken than Newton's laws? Even better, can't we just say that Newtonian gravity is not consistent with Newton's laws?

Can be gravity or any arbitrary other force here. Removing gravity is not enough.

Probably removing infinity in EVERY form is enough. But it is not certain.

p.s.

I wonder how many readers agree with me, that something IS very wrong.

 

[ObsessiveMathsFreak]

No "infinite system of linear equations required".

How else are you going to solve for the accelleration of the system? You need to find the reaction forces if you want to find the total force experieinced by each particle.

 

[Tomaz Kristan]

How else are you going to solve for the accelleration of the system? You need to find the reaction forces if you want to find the total force experieinced by each particle.

I don't care for the accelleration of the "system". I care only for the every ball, which is static. Except Jupiter.

The rightmost ball of the complex doesn't go anywhere. Why?

Be cause it is under some finite gravity, balanced by the surface reaction force. So for every next ball.

Where could it go? Do you think, they are going somewhere? What made them do that?

 

[ObsessiveMathsFreak]

The rightmost ball of the complex doesn't go anywhere. Why?

Be cause it is under some finite gravity, balanced by the surface reaction force. So for every next ball.

It will only be balanced by the surface reaction force if the reaction force is equal and opposite to the resultant gravity force. You haven't proven that. The reaction force could be greater than the gravity force, pushing the system to the right. It could be less than the gravity force, and the system will move to the left.

You need to compute the force. You haven't done that yet. You're assumming all the time that the reaction forces do in fact balance the gravity forces, but you haven't proved it yet.

 

[greedangerfoolishego]

Can you not have the same situation with only 2 balls?
If the one on the right has more mass than the one on the left?
Can someone tell me why (if) I am on completely the wrong track with regards to the frame of regerence being the problem here?
Are you quite sure that the centre of gravity is accelerating to the left, as opposed to just drifting at constant velocity? The centre of gravity is allowed to move, just not accelarate, right? So, assuming that it is accelerating like you say I suppose using the centre of mass frame doesn't make a difference because it would be an accelerating frame of reference? Am I right?
So if you can prove that it is defnitley accelerating then I can see the problem. So far all you've got is that it's moving left, I think.
ObsessiveMaths, am I on the right track or being foolish?

 

[Tomaz Kristan]

The centre of gravity is allowed to move, just not accelarate, right?

Wrong. Since everything stands still at the beginning, the moving Jupiter and standing complex IS a HUGE problem. GC should stay put in the absence of the external force.

 

[Tomaz Kristan]

You need to compute the force. You haven't done that yet. You're assumming all the time that the reaction forces do in fact balance the gravity forces, but you haven't proved it yet.

No. I can just assume the surface reaction forces big just enough to balance gravity for every ball.

It is nothing to prove here, at all.

 

[ObsessiveMathsFreak]

No. I can just assume the surface reaction forces big just enough to balance gravity for every ball.

And what justifies that assumption? What makes it valid? You can't just talk it into validity. At some point, you need to write down an equation. I did in post 136. When you run the numbers and actually try to solve the problem, you end up with an infinite system of linear equations.

In response to my presentation of actual equations, you've responded with some vagaries about infinities and frankly what looks like complete nonsense. I don't like saying this, but I do wonder if you are presenting your arguments here in good faith.

From the get go, you've presented a model and made only verbal arguments as to why there is a paradox. At no point have you presented any substantial equations whatsoever. Considering this is a mathematical paradox more so than anything else, I would have expected at least some mathematics presentation. Your original paper doesn't even contain the formulae for the gravity on each mass.

In post 27 I demonstrated why the motion of the center of mass is not a mathematically defined value. Later in post 136, largely because you would not present any equation whatsoever, I had to actually present the relevant equations for you. The system again turned out to be "unsolvable".

If you have issues with anything in either of these posts, then please be a little more specific as to what exactly you have an issue with. As in, if you think $$a_n=0$$, please explain why you think so, other than to say it "must" be zero.

Your paradox is inherantly a mathematical one, based on a physical problem. As such it cannot be solved, answered or even understood in purely verbal terms. It must be examined in a mathematical framework. Please keep that in mind.

 

[Quaoar]

Can't we just say that conditionally convergent series, which can be manipulated to produce any answer, should not be used in physical problems? As far I know those, they AREN'T used in any physical situation.

So how about this: Any supposed physical situation that reduces to a conditionally convergent series is unphysical.

 

[Tomaz Kristan]

ObsessiveMathsFreak,

There is nothing (almost nothing) to calculate. The surface reaction forces are defined to balance any negative gravitational net force.

They are not defined to balance any positive (repulsive) gravitational net force in this case. They could also be, but they are not.

Gravity is chosen, but any other force would do, had it been set right.

The paradox arises, if the R^3 set and the Newton's axioms are combined. Who is "to blame", I don't know.

I know, it's not a real world problem, it's just a problem of an axiomatic system, we believed it was sane.