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A paradox inside Newtonian world

by Tomaz Kristan
Tags: inside, newtonian, paradox, world
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ObsessiveMathsFreak
#181
Nov22-06, 11:46 AM
P: 406
Quote Quote by Tomaz Kristan View Post
There is NO leftmost ball at all.
But there is in the limiting case as the number of particles goes to infinity.
Tomaz Kristan
#182
Nov22-06, 11:56 AM
P: 218
Limit for what? For the "leftmost" ball speed after a second? For the force between the two "leftmost" balls?

Nothing like that exists.
ObsessiveMathsFreak
#183
Nov22-06, 01:02 PM
P: 406
Quote Quote by Tomaz Kristan View Post
Limit for what? For the "leftmost" ball speed after a second? For the force between the two "leftmost" balls?

Nothing like that exists.
In the limit as the number of balls goes towards infinity. As we logically increase the number of balls, in each of our cases, there is a leftmost ball.
Tomaz Kristan
#184
Nov22-06, 02:16 PM
P: 218
Quote Quote by ObsessiveMathsFreak View Post
In the limit as the number of balls goes towards infinity. As we logically increase the number of balls, in each of our cases, there is a leftmost ball.
No, it doesn't go that way and you know that.
Hurkyl
#185
Nov22-06, 04:02 PM
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There is NO leftmost ball at all.
There is, in the nonstanard model. It contains H balls, where H is a transfinite (hyper)integer. The H-th ball is the leftmost.
radou
#186
Nov22-06, 05:13 PM
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This discussion seems to be an ideal example of an infinite process.
Gelsamel Epsilon
#187
Nov22-06, 07:49 PM
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What does transfinite mean? Boundless but not infinite?
Hurkyl
#188
Nov22-06, 07:55 PM
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The standard natural numbers form an (external) subset of the hypernatural numbers. We (externally) define that a hypernatural number is transfinite if and only if it is larger than every natural number.

The word "transfinite" is used to distinguish it from the standard usage of "infinite", since, for example, a transfinite sum is something different than an infinite sum. (But their values are infinitessimally close, if the summand is well behaved)
Gelsamel Epsilon
#189
Nov22-06, 10:56 PM
P: 316
So basically transfinite numbers are numbers which are larger then any finite number but smaller then infinity?
Hurkyl
#190
Nov22-06, 11:04 PM
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Quote Quote by Gelsamel Epsilon
So basically transfinite numbers are numbers which are larger then any finite number but smaller then infinity?
That is an accurate statement.

(resisting urge to go into what is probably unnecessary detail)
Tomaz Kristan
#191
Nov23-06, 01:20 AM
P: 218
So, you say, that you are going to clean up the mess by adding some more infinite stuff?

Well, maybe, who knows, but currently those transfinite shadow balls are nowhere defined, inside the Newtonian world. It's still to be done, if it's of any use, anyway.

Now, it's no solution.
Gelsamel Epsilon
#192
Nov23-06, 03:55 AM
P: 316
Quote Quote by Hurkyl View Post
That is an accurate statement.

(resisting urge to go into what is probably unnecessary detail)
Go into more detail if you really want too, I'm always keen on learning things.
Tomaz Kristan
#193
Nov23-06, 04:19 AM
P: 218
Don't steal me my thread, please. Go elsewhere. Unless he somehow solve the paradox I gave, with those hypernaturals. Hypernumbers deserve a new topic. Here are welcome only iff something become clear using them.
ObsessiveMathsFreak
#194
Nov23-06, 06:20 AM
P: 406
Quote Quote by Tomaz Kristan View Post
No, it doesn't go that way and you know that.
It does go that way. That's what an infinite sum means.

When we write [tex]\sum_{n=0}^{\infty} a_n[/tex], what we mean is;

[tex]\lim_{k \to \infty } \sum_{n=0}^{k} a_n[/tex]

It's clear as crystal. In each limiting case, there is a leftmost ball and the center of mass remains fixed. In the limit as [tex]k \to \infty[/tex], the accelleration of the center of mass is zero. And that is all we can say without progressing to very esoteric arguments about things like hyperreal numbers etc.
Tomaz Kristan
#195
Nov23-06, 06:44 AM
P: 218
You argument is false, OMF.

If it hadn't been, you could just as well proved, that the biggest natural number exists. Bigger than any other.

Well, it's a basic mistake on your side, trust me!
vanesch
#196
Nov23-06, 07:28 AM
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Quote Quote by ObsessiveMathsFreak View Post
It does go that way. That's what an infinite sum means.

When we write [tex]\sum_{n=0}^{\infty} a_n[/tex], what we mean is;

[tex]\lim_{k \to \infty } \sum_{n=0}^{k} a_n[/tex]

It's clear as crystal. In each limiting case, there is a leftmost ball and the center of mass remains fixed. In the limit as [tex]k \to \infty[/tex], the accelleration of the center of mass is zero.
Yes, this is correct. It is because you approach the final situation as a sequence of situations in which there are each time a finite number of balls, but more and more of them. However, there's no guarantee that this "adding balls" is the one and only correct "limiting procedure".

The other approach, the one that leads to the paradox, is, by NOT setting up a sequence of situations with more and more balls, but by considering all balls at once, and calculate the total force on each individual ball. If you do that, it turns out that the sign of the force on each individual ball, by the entire set of all balls, is the same.
Adding now the forces of all balls together will then result of course in a net force with the same sign.

There are similar situations where you cannot just consider sequences of physical setups, and take the limit of a quantity in this sequence, as the value of the quantity that would occur in the limiting situation. Another example is this:

Consider an Euclidean space with a homogeneous, constant mass density. Turns out (by symmetry) that this mass density doesn't result in any (Newtonian) gravitation force on a test mass.
However, if you approach this situation by considering a sphere of radius R with homogeneous mass density, and 0 outside, then your test mass will undergo, for each value of R, a specific force towards the center of the sphere. If R > d (distance between test particle and center of sphere), then this force will not change anymore. So the force, as a function of R, grows first, and becomes a constant from the moment R > d. Taking the limit R - > infinity gives you this constant force.

Nevertheless, the physical situation with R-> infinity is a space filled with a homogeneous mass density, where the force should be 0.
ObsessiveMathsFreak
#197
Nov23-06, 08:00 AM
P: 406
Quote Quote by Tomaz Kristan View Post
You argument is false, OMF.

If it hadn't been, you could just as well proved, that the biggest natural number exists. Bigger than any other.

Well, it's a basic mistake on your side, trust me!
I assure you, an infinite number of non zero numbers sums to infinity. When we speak of "infinte" sums, we are in fact speaking about the limiting case as the number of terms in the sum increases without bound. That's what a "sum to infinity" really means. The limiting case.
Tomaz Kristan
#198
Nov23-06, 08:21 AM
P: 218
> I assure you, an infinite number of non zero numbers sums to infinity.

1/2+1/4+1/8+ ... = 1

Don't you think so?


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