A paradox inside Newtonian world

[Hurkyl]

V = - G*SUM(over n){ (1/2)^n/|r - (1/10)^n|}.

It's a fair bet to claim that this series does not converge. So, we get a different answer, non-answer in fact, than above.

Diverges at r=0, and is undefined whenever r is the coordinate of one of the mass points. Converges everywhere else.

 

[Tomaz Kristan]

Converges everywhere else.

And above all, converges for the system as stated in post #1.

 

[Hurkyl]

No, it diverges at r=0 and is undefined when r is the coordinate of the mass points.

 

[reilly]

Diverges at r=0, and is undefined whenever r is the coordinate of one of the mass points. Converges everywhere else.

Absolutely correct. But that's sufficient to make my point.
Regards,
Reilly Atkinson

 

[Tomaz Kristan]

The sum of forces to every ball is finite and very well defined.

And you know that, I guess.

 

[Tomaz Kristan]

Let take the rightmost ball! It had been all the left balls condensed into into the second rightmost ball, the force between those two would be greater than it is now. And quite finite, still. Sure.

See? See the implication to every other ball?

 

[ObsessiveMathsFreak]

The force on the center of mass is zero for every configuration of N masses. In the limit as N-> infinity, the force is still zero. With an infinite number of particles, the force is undefined. We've been over this before.

You keep claiming that teh center of gravity moves to the left, but you haven't proved that. If it does move the the left, how fast does it go? What is the value, in Newtons, of the force on the center of gravity. Please don't say something like:

Every ball has a finite force to the left. There are an infinite number of balls. No leftmost ball.

Then the force is positive to the left.

Do you agree?

Just try and come up with an actual figure. Some number that represents how much force is on the center of mass. You claim it will accellerate. If so, by how much? If you can't give this answer, then the problem is not well posed.

 

[Tomaz Kristan]

I don't care what happens later. All the forces in one direction is already an illegal situation. Remember the third law of sir Isaac Newton?

 

[my_wan]

If this -theoretical paradox- holds it would allow a propulsion device in a device similar to the one about to be described. It wasn't thought of by myself and I,ve never seen an answer. I haven't ever seriously considered it because of the source I have now forgotten. Originally I thought the containment parameters were impossible but as I am now considering it apparently it is not.

Containment surface (hole)
Cut a hole in the center of a flat sheet of metal with the following shape.
Start at position x=0, y=1 as the center point to draw a partial circular line through the lower two quadrants beginning at x=-5 through the -y axis and ending at x=5. Define L as the distance from x=-5 to x=5. Now for each point on the partial circular line using the straight line formula of length L define a corresponding point in the upper two quadrants where the straight line always passes through point x=0, y=0.

The beginning parameters x=1, center of partial circle, as well as x=-5 and x=5 or length can be changed to taste. More efficient shapes are definable that reduces torque or acceleration when the rod passes the x axis.

Device
Now cut a rod of length L that can spin inside the container. Mount a motor with the shaft at position x=0, y=0. Cut a slot in the rod such that the motor will spin the rod but allow the rod to slid back and forth, and mount it within the container. Add a control devise to the motor such that the RPM remains constant under changing torque.

If the paradox holds and you spin this device clockwise an excess of MV will always exist in the direction of the +y axis offset in the +x direction due to the forced containment accelerations. Using the device in pairs with opposite spin would be necessary to balance rotational torque. Proof of concept could be done with one driving a toy car up an incline.

Theoretical implications
It has occurred to me that even if it works to whatever degree it doesn't have to violate the second law.
Law - No motion or system of motions may change the total momentum of an (enclosed) system.
The question then is; 'Is this an enclosed system?'
Just as a pure thought experiment the answer is no. First we are providing energy to maintain a constant RPM. Secondly this system is by definition radiating gravitational radiation. Could that be why it doesn't work?
This is worth figuring out even for purely theoretical reasons.

 

[ObsessiveMathsFreak]

I don't care what happens later.

Considering that what happens later determines whether you are right or wrong, I would have thought otherwise.

All the forces in one direction is already an illegal situation. Remember the third law of sir Isaac Newton?

Unless of course, all the forces happen to sum to zero. Or infinity. If you could just check that, that would be great.

 

[Tomaz Kristan]

The sums of all those forces are finite - at every ball center. It is simple to see that.

 

[Hurkyl]

I don't care what happens later. All the forces in one direction is already an illegal situation. Remember the third law of sir Isaac Newton?

Yes:

To every action (force applied) there is an equal and opposite reaction (equal force in the opposite direction).

Newton's third law does not say that the sum of all internal forces is zero. It doesn't even say that the sum of all internal forces is well-defined.

 

[Amr Morsi]

My_Wan,

I agree with you on this. And this can be easily deduced by a fundamental situation and then to be safely and securely generalized. The rule is concerned with locally-fixed-particles body, not any system. And this is done in many of the mechanics-specialized books.

But, I just want to wonder about gravitational waves. Away from GR, if there is no gravitational waves, then is a particle (affecting another particle) was moving, the new Force at any space-point after a time will depend upon that instantaneous distance from the particle at that time. So as to say, effect will be transmitted at ZERO-TIME! Or... Remote Effect. And this was totally rejected by Newton, although not having spoken about Gravitational Waves. It was a paradox for him, especially in the last of his age.
What do you think buddy?

I think that all the calulations of the problem may be done in the below:

Meq * Rcm> = Sum{ mi * Int[ Int[ G * Meq (ri'>)/(ri^2) ]dt ]dt }.

Where: 1. > denotes a vector, 2. '> denotes a unit vector, 3. mi is the mass of the particle, 4. ri> is the displacement vector of the particle from the Rcm>, 5. Rcm> is the position vector of the centre of mass 6. Meq is the summation of all masses, 7. Sum: summation and 8. Int: integration.

Differentiating both sides, we get the acceleration of the CM. Of course, R.H.S is also in both sides of this derived ODE, but, we can test the R.H.S. to check this special system. It may help.
By the way, we can reach:

Meq * Acm> = Sum(over i){ Sum(over j<>i) { mi * mj * G * [ (ri'>)/(ri^2)+(rj'>)/(rj^2) ] } }.
where Acm> is the acceleration vector of the CM.

Please forgive me not to go into these details. As I already got determined that the Rule was somewhat commonly generalized (and understood) in error.
It is related to rigid bodies.


Yours,
Amr Morsi.

 

[ObsessiveMathsFreak]

The sums of all those forces are finite - at every ball center.

It is simple to see that.

I suppose it is, but we aren't talking about the forces at every ball center. We're talking about the overall force on the center of mass. Could you just give that, in Newtons please.

Now remember, it's not enough to say it points to the left. You have to say by how much it points to the left. So off you go.

 

[Tomaz Kristan]

You are right!

F(N,N-1)+F(N,N-2)+F(N,N-3)+ ... + ... = 1.984*F(N,N-1)

 

[ObsessiveMathsFreak]

You are right!

F(N,N-1)+F(N,N-2)+F(N,N-3)+ ... + ... = 1.984*F(N,N-1)

I'm afraid that's the force on only one ball. Your entire argument has been about the center of mass. If you could just calculate the force on the center of mass please.

 

[Tomaz Kristan]

I'm afraid that's the force on only one ball.

I am afraid it's not. It's only to the left side part. Always finite as you see.

But you have to add the right side force to each. Also a finite number, so the sum is a finite one, for every ball.

Do we agree so far?

 

[ObsessiveMathsFreak]

I am afraid it's not. It's only to the left side part. Always finite as you see.

But you have to add the right side force to each. Also a finite number, so the sum is a finite one, for every ball.

Do we agree so far?

Yes. We are all agreeded that the sum of forces is finite on every ball. But there are an infinite number of balls, so what is the force on the center of gravity?

 

[Tomaz Kristan]

Yes. We are all agreeded that the sum of forces is finite on every ball. But there are an infinite number of balls, so what is the force on the center of gravity?

Also finite. At any moment.

 

[ObsessiveMathsFreak]

Also finite. At any moment.

And so, what is this finite value?

 

[Tomaz Kristan]

And so, what is this finite value?

It's almost twice the acceleration of the rightmost ball. At time t=0.

 

[ObsessiveMathsFreak]

It's almost twice the acceleration of the rightmost ball. At time t=0.

Great. Now all we need to know is the accelleration of the rightmost ball. In Newtons. So what is that please?

 

[ssd]

If I understood the OP's assertions correctly then the particle at x=0 (let us name it as P1) has mass=0 and has infinitely high density. Also that, the particles in the line are countable ( that is, one can index them, say as, P1, P2,P3, P4,...etc from the left). Consider a situation when we have just two balls P1 and P2 (or, for that matter P1 and Pn,for any arbitrary n>1). What will be the gravitational force between them? (I mean, how to write the expression of gravitational force as per Newton's law of gravitation?) I am saying this because of the fact that as per OP's assertion, the balls towards the left of a ball is supposed to anchor that.

 

[Tomaz Kristan]

If I understood the OP's assertions correctly then the particle at x=0 (let us name it as P1) has mass=0

There is NO particle at x=0. Only at 10^N for every finite N. Nothing at x=0.

 

[ssd]

There is NO particle at x=0. Only at 10^N for every finite N. Nothing at x=0.

If N is finite the paradox vanishes and along with that the sum of all the mass particles is not equal to 2 kg. Since 2= sum of 2^(-N) over N=0,1,2,...ad inf

If I am not wrong, you automatically asserted N-->infinity by asserting the total mass to be 2 kg.

The limit of the sum of distances between two consecutive particles from the 1kg particle (which is at 1mt from origin) = sum of (9/10^N ) over N=1,2,3...ad inf,
which is =1. So starting from 1mt on x-axis and moving towards left (as stated by you), we end up (in limit) at x=0. Therefore at x=0 one shall have a particle with mass=0 in limit (which is not the same as having "NO PARTICLE") and infinitely high density.

I do not argue to make you agree with this. But if this basic thing is ignored in judgement, then the paradox is just assumed, not proved just as in the way Achilles Paradox stands.

 

[Tomaz Kristan]

If N is finite the paradox vanishes

Wrong. N is not the maximal number, that doesn't exist at all. It's an infinite number of finite numbers.

Every ball is finite, but no one is maximal. Okay?

 

[ObsessiveMathsFreak]

Every ball is finite, but no one is maximal. Okay?

We knew that to begin with. What about the center of mass? What is the force on that?

 

[Tomaz Kristan]

I've told you. Almost twice as big, as the force to the rightmost ball. Which is almost twice as big, as between the two rightmost balls.

 

[masudr]

I've told you. Almost twice as big, as the force to the rightmost ball. Which is almost twice as big, as between the two rightmost balls.

I think he wants an answer that doesn't involve any unknowns.

 

[ssd]

Every ball is finite, but no one is maximal. Okay?

"Every ball is finite" ... what does that mathematically mean? Do you mean "countable"?

Whatever the term may be, my question is not answered. As per your setup, for any N, we can discretely observe the ball or its position on x-axis can be discretely located.

The sequence of particles MUST END at x=0. How do we write the gravitational force of the particle at x=0 with the particle at x=1, say.

While you are stating the problem, you are doing it philosophically with 2 balls at right, then the next ..and then "similarly" ...and there after you are reclutent to describe any thing mathematically. You are unable to suggest any solution to questions like "rearrangement of terms of conditionally convergent series" or "impossibility to solve infinite equations of force on a ball"... Note that all these happen as N--> infinity. That is why I posed my question in a way that you have to take the particle at x=0 into account.

In the same way, being as philosophical as you one can say that,
the particle at x=0 has no force towards its left ... it will be accelerated to the next ball and it shall move more way than the next(towards each other) because it will have higher acceleration..."similarly" the combined mass of these two will move towards right ... until the CG is reached...
Note that the fastest car on Earth cannot cross a moving tortoise if the car is initally 1mt behind at the start of the race... this happens iff we look with a sole philosophical angle upon the problem without going into mathematics or iff do not care answering mathematical quaries.