A paradox inside Newtonian world

[greedangerfoolishego]

So what you are saying, is that after all these masses fall in on each other, assuming collision is inelastic and that it forms one big mass eventually (ideally anyway), is that there is a net force still acting on the system and the resultant mass will continue to accelarate to the left?

Have you considerred your frame of reference for this problem? If you take x=o at the smallest mass (albeit infinite mass density), then this means that your frame of reference will be accelerating once that particle at x=0 is acted upon by gravitational force from the other masses. Maybe that's the real problem? If I remember correctly, Newton's Laws don't hold for non-inertial frames of reference?

Or is it the case that there is no mass at x=0, therefore won't be affected by gravitational forces, albeit infinite mass density?

 

[Tomaz Kristan]

So what you are saying, is that after all these masses fall in on each other

At this point, I have long ago lost my interest. I am gazing them only until Jupiter is still far away. The conservation of momentum has been broken, I don't care anymore, what allegedly happened after that.

Even more. Don't want to speculate about an impossible case further. It would be like to ask, if there are integers inside the set of all sets, which do not contain themselves. Game already over.

assuming collision is inelastic and that it forms one big mass eventually (ideally anyway), is that there is a net force still acting on the system and the resultant mass will continue to accelarate to the left?

Yes, as a matter of fact, it does seems so.

Have you considerred your frame of reference for this problem?

Well, it is not. I am safely parked at T(10,10,10) watching the show.

If you take x=o at the smallest mass (albeit infinite mass density)

No mass at all there at zero. But the infinite density, yes, and only at this point.

Or is it the case that there is no mass at x=0, therefore won't be affected by gravitational forces, albeit infinite mass density?

This one! Nothing at 0, but something at every point right of the zero to the rightmost point of the 1 tone ball. [Until Jupiter comes, at least.]

 

[ObsessiveMathsFreak]

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

If the mass was "solid", i.e. unmoving. But if you assume that then you've assumed what you were trying to prove. You haven't proven what you set out to prove.

You still haven't proved that the collective mass doesn't move anywhere.

 

[Tomaz Kristan]

If the mass was "solid", i.e. unmoving. But if you assume that then you've assumed what you were trying to prove. You haven't proven what you set out to prove.

You still haven't proved that the collective mass doesn't move anywhere.

I proved for every ball on the left side. Therefore the complex also can't move.

 

[greedangerfoolishego]

At this point, I have long ago lost my interest. I am gazing them only until Jupiter is still far away. The conservation of momentum has been broken, I don't care anymore, what allegedly happened after that.

I thought the whole point of conservation of momentum was that total momentum beforehand = total momentum afterwards. How can you ignore further events if there is that time factor involved?

Why are you saying that there is a violation of conservation laws here? Are you saying that because all of the forces in your system don't cancel out and give zero, there is a brakedown of momentum conservation, and thus Newtonian mechanics?

I am stuggling with what exactly the paradox is here. What SHOULD happen?

 

[ObsessiveMathsFreak]

I proved for every ball on the left side. Therefore the complex also can't move.

No you didn't. You assummed that the reactive forces were exactly cancelling the gravity + previous reactive forces. That assumption forced the accelleration of each ball to be zero, but that's not a valid proof.

 

[Tomaz Kristan]

I am stuggling with what exactly the paradox is here. What SHOULD happen?

The mass center should not move without an external force. But it does. Jupiter comes more to the left, but complex stays put, instead of traveling toward Jupiter.

So the gravity center moves, and no external force acted.

 

[Tomaz Kristan]

No you didn't. You assummed that the reactive forces were exactly cancelling the gravity + previous reactive forces. That assumption forced the accelleration of each ball to be zero, but that's not a valid proof.

How do you expect, a surface reaction can move something? It could just be a 'passive' response to other force acting there.

The desk reactive force will not throw you pencil up. Not by itself. It can transfer some other force from bellow.

Do we have some here?

 

[greedangerfoolishego]

Having read the question again I think this is just incorrect application of Newtons laws, with terminology being misunderstood / things being mislabelled in the supposed paradox.

A collective group of body's is not itself one body.

The centre of mass reffered to in Newtons laws therefore isn't the centre of mass for the whole system, it is the centre of mass for one body i.e. one mass, not the net mass. So we have many centre of masses in this system.

Subsequently, gravitational force isn't an internal force at all, it is an external force acting on each of the body's.

Correct me if I am wrong!

 

[Tomaz Kristan]

A collective group of body's is not itself one body.

Nobody talks about that, at all.

The centre of mass reffered to in Newtons laws therefore isn't the centre of mass for the whole system

Wrong.

it is the centre of mass for one body i.e. one mass, not the net mass. So we have many centre of masses in this system.

Wrong.

Subsequently, gravitational force isn't an internal force at all, it is an external force acting on each of the body's.

Wrong.

Correct me if I am wrong!

Done.

 

[greedangerfoolishego]

Nobody talks about that, at all.
Wrong.
Wrong.
Wrong.
Done.

Fair enough!

In case I have been a bit hasty in thinking that Newton wasn't refferring to the centre of mass of a system of particles but rather the centre of mass of a rigid body,...

I think your mistake is setting x = 0 as a fixed reference point in your system.
This point interacts with the positions of the other particles in a way that gives rise to a non-inertial frame of reference, in which Newtons Laws don't hold.

Instead, how about using the centre of mass as a frame of reference?

http://en.wikipedia.org/wiki/Center_of_mass_frame

 

[ObsessiveMathsFreak]

How do you expect, a surface reaction can move something? It could just be a 'passive' response to other force acting there.

If you don't believe me, ride an elevator! Reactive forces can contribute to overall motion in their direction.

The desk reactive force will not throw you pencil up. Not by itself. It can transfer some other force from bellow.

Do we have some here?

We might. There's an infinite number of particles involved here after all. Who's to say.

Anyway, you need to actually write down some mathematical equations and solve for the accelleration of each particle. Otherwise this is all just philosophy.

P.S.

 

[Tomaz Kristan]

If you don't believe me, ride an elevator! Reactive forces can contribute to overall motion in their direction..

Well, there is a motor which runs the elevator somewhere, isn't it? And the force is translated from there.

Anyway, you need to actually write down some mathematical equations and solve for the accelleration of each particle. Otherwise this is all just philosophy.

What equations do you need? Fg=-Fr?

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

None.

 

[greedangerfoolishego]

'A non inertial frame of reference is one in which a body violates Newton's Laws of Motion, mainly the First Law. In such a frame, despite no real force acting on a body at rest, it might move; or one that was already moving come at rest or change it's direction of motion.

Newton's first and second laws of motion do not hold in non-inertial reference frames. Specifically, masses in non-inertial reference frames appear to feel fictitious forces (such as the Coriolis force or the centrifugal force) that derive from the acceleration of the reference frame itself. Fictitious forces cause apparent accelerations in objects without any physical force causing the acceleration. Fictitious forces are proportional to the mass upon which they act; if such forces are observed, scientists will recognize that they are in a non-inertial reference frame. For example, the rotation of the Earth can be observed from the Coriolis force acting on a Foucault pendulum.'

I'm pretty sure by that by not using the centre of mass frame, you are in fact using a non inertial frame of reference. Incidentally, an elevator is a non-inertial frame of reference too, you feel imaginary forces as it comes to rest / starts moving because it is accelerating.

 

[Tomaz Kristan]

I'm pretty sure by that by not using the centre of mass frame, you are in fact using a non inertial frame of reference. Incidentally, an elevator is a non-inertial frame of reference too, you feel imaginary forces as it comes to rest / starts moving because it is accelerating.

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

 

[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 :)