A paradox inside Newtonian world

[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.

 

[Tomaz Kristan]

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

No. Every ball is just one. With a finite density and mass. Countably many of them.

The sequence of particles MUST END at x=0.

No ball is there at x=0 in my example.

 

[ObsessiveMathsFreak]

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.

And what is the force on the rightmost ball? Give us an equation for it.

 

[ssd]

No ball is there at x=0 in my example.

WRONG:
LOL,
if you distribute a mass on x-axis at distances 1/(10^N), N=0,1,2,...ad inf (of course the term you use is not 'ad inf' but mathematically it is so as per your description) then where do you end ...
Alternatives:
1/ At some negative value of x
2/ At some positive value of x
3/ At x=0

In other way of saying this is : what is the total length on x-axis over which mass particles are distributed?
1/ length>1
2/ length <1
3/ length=1

Another way of saying this:
Take a rod of length 1mt. Just as you place your mass particles, cut the rod at 1/(10^N) meters, N= 1,2,3,...ad inf.
That is, cut a piece of length 1/10mt . From this piece make a cut of 1/100mt... from this cut 1/1000mt and so on. Consider N as of the identical nature of your described N (i.e. countable, finite, the maximum not known etc as you have described or thought of).

Now if you add the length of all the pieces what shall you get :
1/ Something less than 1 mt
2/ Something more than 1 mt
3/ 1mt.

BTW this is probably my penultimate post in this thread since nothing attrective is left in your problem and I shall say goodbye to it after possibly looking at your answers to this post.

 

[Tomaz Kristan]

OMF,

0.992/0.81 units. The unit is the force between two 1 kg balls 1 meter apart. From center to center.

ssd,

You are not very strong at real numbers, do you?

 

[ssd]

ssd,

You are not very strong at real numbers, do you?

No ball is there at x=0 in my example.

Certainly not as strong as you : I cannot find a real number between '1' and '0.99999...recurring' . You are stronger enough to find such a number (who knows, possibly many) as the total length (starting from x=1 and going towards left) of particle distribution = sum of 9/(10^N) over N=1,2,,...ad inf ends before x=0 according to your opinion .

Unwanted side effect could also be, that people will become more agnostic, scientifically. They will say: For more than 300 years, you had an error just before your noses, and you haven't seen it! How one can believe science?

That would be a bad thing to happen. In fact, science only harbored the magic (of infinity) for too long. Once we clean it, the science will be better than ever before.

I am glad to know that a new Giant has emerged in my life time who proved all Giants from 300 or more years wrong (may I say, they were ignorant of an error) at least in one case...(consequently,may be, in many retrospective cases).

THE END from my side.
With best wishes,
Thanks and bye .

 

[ObsessiveMathsFreak]

OMF,

0.992/0.81 units. The unit is the force between two 1 kg balls 1 meter apart. From center to center.

And just to make things completely clear, how is this force on the rightmost ball calculated again?

 

[Tomaz Kristan]

OMF,

We have this situation:

Force = SUM(N=1...) 2^-N/(1-10^-N)^2

Every next ball is a little (1/10) more distant and a half as massive.

 

[Tomaz Kristan]

ssd,

The point at x=0, is NOT inside the structure described. Does not matter, if we have points or (ever smaller) balls at 10^-N, for every natural N. The 0 is outside.

 

[ObsessiveMathsFreak]

Force = SUM(N=1...) 2^-N/(1-10^-N)^2

Every next ball is a little (1/10) more distant and a half as massive.

But that's not the total force on each ball. That's only the gravitational force on each ball. You haven't taken into account the surface normal reaction force on each ball. One on the left side of the rightmost ball, and one on both sides for every other ball.

You must solve for these forces before you can calculate the force on the center of mass.

 

[Tomaz Kristan]

There is no normal reaction force at pdf example in the post #1. Balls are degenerated to points. They could be also just smaller balls, so that they don't touch.

 

[masudr]

By the way, has anyone calculated the total internal energy of the system?

 

[ObsessiveMathsFreak]

There is no normal reaction force at pdf example in the post #1. Balls are degenerated to points. They could be also just smaller balls, so that they don't touch.

They could be, but in that case the case becomes identical to the point ball case in which the force on the center of mass is a divergent sum as I explained in post 27. So the force is divergent and the center of mass cannot be said to move at all.

 

[Tomaz Kristan]

By the way, has anyone calculated the total internal energy of the system?

Yes. It's not finite. As is not the total internal energy of any two mass points system.

 

[Tomaz Kristan]

So the force is divergent and the center of mass cannot be said to move at all.

No, the force is convergent. Wana bet?

 

[masudr]

Yes. It's not finite. As is not the total internal energy of any two mass points system.

Surely the internal energy of two uncharged particles is simply

U= \frac{1}{2}m_1\dot{x}^2_1 + \frac{1}{2}m_2\dot{x}^2_2 -\frac{Gm_1 m_2}{r}

Was the "not" a typo?

 

[Tomaz Kristan]

Well, the r can become as small as you want, and the energy so defined as big as you want. For every two mass particle system.

The escape velocity arbitrary great. That's what would I call "the total internal energy".

 

[masudr]

Well, the r can become as small as you want, and the energy so defined as big as you want. For every two mass particle system.

The escape velocity arbitrary great. That's what would I call "the total internal energy".

So technically it's impossible to bring two uncharged particles together. That matches experimental evidence. Do you agree with this?

If your system has an internal energy that diverges, even when there aren't two particles in contact, then it means that it is impossible to bring particles into that configuration from particles at infinity. Do you agree with this?

 

[ObsessiveMathsFreak]

No, the force is convergent. Wana bet?

It's divergent. We've been over this before. But considering you're so adamant that it is converging, you'll be so good as to grace us all with a proof then?

 

[Tomaz Kristan]

It's divergent.

What's divergent? Where have anybody showed that?

Show me, where is shown, please.

 

[ObsessiveMathsFreak]

What's divergent? Where have anybody showed that?

Show me, where is shown, please.

Good, very good. Shifting the burden of proof. Clever.

But I'm afraid I'll have to decline your most generous invitation and insist that you, as the primary proponent of the finite sum leading to a paradox, must be the one to provide this, the final step in your argument, namely, the proof that the force on the center of mass is a finite sum.

 

[Tomaz Kristan]

All you need is in the post #1, to understand the whole problem. That all forces are finite and pointed to the left is almost trivial to see.

I am not forcing you to make any new calculations. You are free to think what you want.

 

[masudr]

Yes, but given that the energy diverges, it is impossible to bring particles into this configuration from particles at rest at infinity.

 

[Eli Botkin]

That "all forces" are "pointed to the left" is neither trivial nor true. For every force pointed to the left there is an equal and opposite force (pointed to the right). That is what is meant by the 3rd law, action and reaction.

 

[ObsessiveMathsFreak]

I am not forcing you to make any new calculations. You are free to think what you want.

Well, I think I need some solid calculations. So does this thread.

 

[Tomaz Kristan]

Yes, but given that the energy diverges, it is impossible to bring particles into this configuration from particles at rest at infinity.

Of course it's possible in the abstract Newtonian world. Not in the real life, sure.

 

[Tomaz Kristan]

That "all forces" are "pointed to the left" is neither trivial nor true. For every force pointed to the left there is an equal and opposite force (pointed to the right). That is what is meant by the 3rd law, action and reaction.

I know. The third law does not permit this. But the second law demands it.

That's why we have a paradox here.

 

[masudr]

Of course it's possible in the abstract Newtonian world. Not in the real life, sure.

It's not even possible there. It is not a point in the phase of space of possible states.

 

[Tomaz Kristan]

masdur,

You can have two mass points of 1 kg each, without spliting one 2 kg mass point into two. What would require an infinite amount of energy.

In the same way, one can just set my construction #1 as the initial situation.

 

[Eli Botkin]

Tomaz:
As you know, this is a long thread and I entered very late. Very possibly you’ve already replied to my following questions, in which case I ask your indulgence.

You say “But the second law demands it. That’s why we have a paradox here.”

But the second law states only that F = ma, so it seems to me that you are basing your argument on an assumption that for this infinite sequence of masses ALL accelerations are to the left, and consequently ALL forces must be to the left as well.

Suppose we were to truncate the sequence to the first googolplex of masses, would you still hold the same view? Maybe not. And if not, isn't it possible (and maybe likely) that an ‘infinite’ sequence of masses brings about a result that is beyond our ken if it isn't correct to extrapolate its behavior from an ever increasing ‘finite’ set?

So maybe the “paradox”, based on such an assumption, is of your own making and not nature’s.

 

[Tomaz Kristan]

Suppose we were to truncate the sequence to the first googolplex of masses, would you still hold the same view?

Not at all. With any finite number of masses, everything is just fine. No paradox there.

it isn't correct to extrapolate its behavior from an ever increasing ‘finite’ set?

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.

For the reals, maximum may be there for the infinite set of real numbers. Sometimes. Sometimes not.

So maybe the “paradox”, based on such an assumption, is of your own making and not nature’s.

Nature does not have this problem. Newton's abstract world, our theory has it. That's my point.