JoshuaZ comments on Edward Nelson claims proof of inconsistency in Peano Arithmetic - Less Wrong

13 Post author: JoshuaZ 27 September 2011 12:46PM

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Comment author: JoshuaZ 27 September 2011 10:31:54PM 2 points [-]

If I can do a calculation and stop and then do another calculation and get a contradiction that's still a contradiction. it doesn't matter that I can do other calculations that would lead to a non-contradiction.

Comment author: Vaniver 27 September 2011 11:07:24PM 0 points [-]

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.

Comment author: benelliott 27 September 2011 11:32:26PM 1 point [-]

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.

Comment author: Vaniver 28 September 2011 12:07:03AM -2 points [-]

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.

Comment author: benelliott 28 September 2011 12:12:03AM 1 point [-]

you calculate the force caused by N+1 particles on N particles

I don't think I do this.

If you did the calculations for the system of N particles, then took the limit as N approached infinity

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.

Comment author: Vaniver 28 September 2011 03:53:49AM -1 points [-]

I don't think I do this.

Does your leftmost particle have a rightward acceleration which makes the weighted average of acceleration (i.e. CoM acceleration) 0?

If you had said that, instead of saying other irrelevant things, then I doubt anyone would have objected.

I have edited the ancestral post to say that. Hopefully, the superior articulation will cause its karma to rise into the positives.

Comment author: benelliott 28 September 2011 08:40:55AM 2 points [-]

Since I wasn't using a limit I didn't have a leftmost particle.

Comment author: Vaniver 28 September 2011 02:05:54PM -1 points [-]

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.

Comment author: benelliott 28 September 2011 03:44:54PM 1 point [-]

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.

Comment author: Vaniver 28 September 2011 03:47:44PM *  -1 points [-]

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.