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Vaniver comments on Edward Nelson claims proof of inconsistency in Peano Arithmetic - Less Wrong
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If he put a single equal sign in his 'proof', I would be more charitable. As it is, it's not clear to me that he actually did any calculations or showed any contradictions.
I'm not sure he did, but I have done the calculations and it seems to check out (although I may have made a mistake). The only laws I used were F=ma and F=Gm_1m_2/(r^2), of which the third law should emerge as an immediate consequence rather than needing to be added in on top.
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.