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Vaniver comments on Edward Nelson claims proof of inconsistency in Peano Arithmetic - Less Wrong
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The reason they "check out" is because you calculate the force caused by N+1 particles on N particles. Because your calculation has an external particle, the CoM has an acceleration. This is entirely an artifact of how the limit is taken, and is thus a sign of sloppiness and incompleteness.
If you did the calculations for the system of N particles, then took the limit as N approached infinity, you would get no CoM acceleration. This really has nothing to do with Newtonian physics.
I don't think I do this.
Obviously the problem is with an infinity not taken as a limit. If you had said that, instead of saying other irrelevant things, then I doubt anyone would have objected.
Does your leftmost particle have a rightward acceleration which makes the weighted average of acceleration (i.e. CoM acceleration) 0?
I have edited the ancestral post to say that. Hopefully, the superior articulation will cause its karma to rise into the positives.
Since I wasn't using a limit I didn't have a leftmost particle.
Then you are calculating the force caused by N+1 particles on N particles. For every particle i, you look at the i-1 to the right, add up their gravitational force, and see that it is dwarfed by force from the particle to the left- particle number i+1.
If you have a finite number of particles, the mistake vanishes. If you have an infinite number of particles but you add particles all at once instead of half of i and half of i+1 at once, the mistake vanishes.
I calculated using all the particles to the left rather than just one, and so every pair got taken into account once for each member of that pair.
Then you were calculating the gravitational effect of an infinite number of external particles on a finite number of particles, which makes things worse.
[Edit]: Note that each pair has to be taken into account twice. The N+1 vs. N error results from having an extra force floating around; the N+Infinity vs. N error results from having an infinite number of extra forces floating around. If you add forces as linked pairs, the error disappears.
Once for each member of the pair = twice
No, I calculated it for every particle, and therefore I calculated it on an infinite number of particles. Obviously my calculations only considered one particle at any time, but the same could be said of calculations in problems involving finitely many particles.
Suppose you have a system of N+1 particles.
You calculate the gravitational force on the first particle. There are N forces on it; you now have N unpaired forces.
Now you calculate the gravitational force on the second particle. There are N forces on it; you now have 2N-2 unpaired forces (you added N more forces, but one of the new ones and one of the old ones are paired).
Then you calculate the gravitational force on the third particle. There are N forces on it; you now have 3N-6 unpaired forces (you added N more forces, and 3 choose 2 are paired).
The number of unpaired forces will grow until it peaks and then shrinks down to zero (as the number of forces that pair up will eventually overcome the added forces).
What happens if you take this approach with an infinite number of particles? Well, at each moment you add an infinite number of new forces, and only a finite number pair up. Thus, as soon as you start the calculations you have an infinite number of unpaired forces and that number only grows.
The right way to do this is the buddy system: never let a force enter the calculations without its reverse force. To do this, you add particles to your consideration one by one. The number of considered forces is N choose 2, and you never have unpaired forces.You let the number of considered particles grow to infinity, and it still works.
It certainly doesn't grow, it becomes infinity and stays there.
Okay, this is getting into set theory, but its actually the case that adding infinitely many things and then taking finitely many away can eventually leave you with zero.
Suppose at step n you add the following numbers, 2n-1, 2(2n-1), 4(2n-1)... (2^k)(2n-1) for all k. Then you take away the number n.
This procedure adds infinitely many at every step, and only takes one away, so by your argument it should only ever grow and end up as infinity. However, its pretty clear that every number gets added once and then taken away once, so we end up with nothing.
I think its pretty clear that a similar thing is going on here.
Again, I agree that the limits are being done wrong, I just disagree with your description of the N vs N+1 thing.