XiXiDu comments on Edward Nelson claims proof of inconsistency in Peano Arithmetic

Comment author: RichardKennaway 27 September 2011 07:38:26PM * 4 points [-]

Vg vf pbeerpg gung ng g=0, rnpu cnegvpyr unf n yrsgjneqf irybpvgl. Ohg vg vf abg pbeerpg gung gur prager bs tenivgl ng g=0 unf n yrsgjneqf irybpvgl. Guvf vf orpnhfr gur flfgrz orunirf qvfpbagvahbhfyl ng g=0, naq fb gur irybpvgl bs gur prager bs tenivgl pnaabg or qrgrezvarq ol gnxvat n jrvtugrq nirentr bs gur vavgvny irybpvgvrf bs gur pbzcbaragf.

Va zber qrgnvy:

Yrg P(g) or gur cbfvgvba bs gur prager bs tenivgl ng gvzr g.

Gur irybpvgl bs gur prager bs tenivgl ng g=0 vf gur yvzvg nf g -> 0 bs (P(g)-P(0))/g.

Ng nal gvzr g, P(g) vf gur fhz bire a bs Z(a)C(a,g), jurer Z(a) vf gur znff bs gur agu cbvag naq C(a,g) vgf cbfvgvba.

Gur nethzrag eryvrf ba pbzchgvat qP/qg nf gur fhz bire a bs Z(a)qC(a,g)/qg. Guvf fjncf gur beqre bs gjb yvzvgvat cebprffrf, fhzzngvba naq qvssreragvngvba, juvpu vf inyvq sbe jryy-orunirq fhzf, ohg abg sbe guvf bar.

For intuition, imagine that there were just the first three particles in the sequence, and that the number 10 is replaced by something much larger. The two leftmost particles will rapidly oscillate about one another, while drifting slowly towards the rightmost. Now imagine the first four particles are present. The two leftmost will oscillate madly about each other, while that system as a whole oscillates rapidly about the third from the left (I'm conveniently ignoring what happens at the instants when particles meet and pass through each other), and the ensemble of the three leftmost drifts towards the rightmost. Repeat.

Comment author: AlexMennen 27 September 2011 08:32:13PM 1 point [-]

It's acceleration that matters, not velocity (the initial velocity of all points is zero, or at least that's how I thought of it). However, your argument does generalize nicely to acceleration, and could possibly be the correct resolution.

Comment author: RichardKennaway 27 September 2011 09:47:03PM * 0 points [-]

Yes, stupid mistake of mine. Replace velocity by acceleration and position by velocity and the same argument works. The fallacy is in the vagrepunatr bs yvzvgf.

Comment author: Alejandro1 29 September 2011 04:37:30PM 2 points [-]

Similar, for those who enjoyed discussing this problem: Did you know that Newtonian mechanics is indeterministic?

Comment author: Sniffnoy 01 October 2011 07:04:41AM * 1 point [-]

These equations don't make sense dimensionally. Are there supposed to be constants of proportionality that aren't being mentioned? Are they using the convention c=1? Well, I doubt it's relevant (scaling things shouldn't change the result), but...

Edit: Also, perhaps I just don't know enough differential equations, but it's not obvious to me that a curve such as he describes exists. I expect it does; it's easy enough to write down a differential equation for the height, which will give you a curve that makes sense when r>0, but it's not obvious to me that everything still works when we allow x=0.

Comment author: Alejandro1 01 October 2011 03:00:40PM 0 points [-]

These equations don't make sense dimensionally. Are there supposed to be constants of proportionality that aren't being mentioned?

That is my guess. The simplest way IMO would be replace the g in eq. 1 by a constant c with units of distance^(-1/2). The differential equation becomes r'' = g c r^1/2, which works dimensionally. The nontrivial solution (eq. 4) is correct with an added (c g)^2 in front.

it's not obvious to me that a curve such as he describes exists.

I'm not sure what you mean here. What could be wrong in principle with a curve h = c r^3/2 describing the shape of a dome, even at r = 0?

Comment author: Sniffnoy 01 October 2011 08:29:45PM 1 point [-]

Notice "r" here does not mean the usual radial distance but rather the radial distance along the curve itself. I don't see any obvious barrier to such a thing and I expect it exists, but that it actually does isn't obvious to me; the resulting differential equation seems to have a problem at h=0, and not knowing much about differential equations I have no idea if it's removable or not (since getting an explicit solution is not so easy).

Comment author: Alejandro1 02 October 2011 01:52:21AM 1 point [-]

You piqued my curiosity, so I sat to play a little with the equation. If R is the usual radial coordinate, I got:

dh/dR = c y^(1/3) / [(1 - c^2 y^(2/3))^(1/2)]

with y = 3h/(2c), using the definition of c in my previous comment. (I got this by switching from dh/dr to dh/dR with the relation between sin and tan, and replaciong r by r(h). Feel free to check my math, I might have made a mistake.) This was easily integrated by Mathematica, giving a result that is too long to write here, but has no particular problem at h = 0, other than dR/Dh being infinite there. That is expected, it just means dh/dR = 0 so the peak of the dome is a smooth maximum.

Comment author: Vaniver comment score below threshold [+] (28 children)

Comment author: Vaniver 27 September 2011 08:03:25PM 1 point [-]

I am curious why this received downvotes. Is it because people expected me to rot13 my answer?

Comment author: benelliott 27 September 2011 08:17:06PM 5 points [-]

I didn't downvote, but if I had it would have been because you seem to be missing the point a bit.

It may well be the case that we live in a discrete universe without infinities (I assume that's what you meant by integer) but this is not the case for Newtonian Mechanics, which asserts that both space and time remain continuous at every scale. Thomas never claimed to have a paradox in actual physics, only in Newtonian mechanics, so what you are saying is irrelevant to his claim, you might was well bring up GR or QM as a solution.

Appealing to the third law also doesn't help, his whole point is that if you do the calculations one way you get one answer and if you do them another way you get another answer, hence the use of the word paradox.

Comment author: Vaniver 27 September 2011 09:46:40PM -2 points [-]

Appealing to the third law also doesn't help, his whole point is that if you do the calculations one way you get one answer and if you do them another way you get another answer, hence the use of the word paradox.

His point is if you only do some of the calculations, you can come up with the wrong answer. I fail to see how that implies a paradox rather than just sloppiness.

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.

continue this thread »

Comment author: ArisKatsaris 27 September 2011 10:15:01PM * 3 points [-]

I didn't downvote you, however both your objections were rather besides the point. One doesn't deal with the issue of Newtonian mechanics, and whether they were indeed as obviously flawed from the beginning -- it feels a bit like explaining the paradox of Achilles reaching the turtle by saying we're living in an integer universe. That may be true, but it feels a rather weak dodging-the-issue explanation.

As for the second answer, it doesn't really state where the calculations in the presentation of the paradox are wrong or are missing something. It just says that they are, so it's a non-solution.

Comment author: Vaniver 27 September 2011 11:06:07PM 0 points [-]

As for the second answer, it doesn't really state where the calculations in the presentation of the paradox are wrong or are missing something. It just says that they are, so it's a non-solution.

Maybe this is because I was a physics student, but to me the missing pieces implied by the second answer are so obvious as to make it not worth my time to type them and yours to read them. Apparently I was mistaken, so here they are.

By the principle of superposition, we can break down a N-body problem into (N choose 2) 2-body problems. Consider the mass M and the mass m, a distance R away from each other (with m lighter and at the leftward position x). Their CoM is at x+RM/(M+m). The force FMm, leading to the acceleration am, is GMm/R^2, leading to GM/R^2. The force FmM, leading to the acceleration aM, is -GMm/R^2, leading to -Gm/R^2 (these are negative because it is being pulled leftward). To determine the acceleration of the center of mass, we calculate m*am+M*aM=GMm/R^2-GMm/R^2=0. The CoM of that pair will not move due to forces exerted by that pair. This is independent of M, m, and R. When we add a third mass, that consists of adding two new pair systems- each of which has a CoM acceleration of 0. This can be continued up to arbitrarily high N.

That may be true, but it feels a rather weak dodging-the-issue explanation.

Whenever there's a paradox that mentions infinity, delete the infinity and see if the paradox still exists. Odds are very high it won't. A lot of diseased mathematical thinking is the result of not being clear with how you take limits, and so when I find a "paradox" that disappears when you get rid of infinity, that's enough for me to drop the problem.

Comment author: JoshuaZ 28 September 2011 02:55:37AM 3 points [-]

Whenever there's a paradox that mentions infinity, delete the infinity and see if the paradox still exists. Odds are very high it won't. A lot of diseased mathematical thinking is the result of not being clear with how you take limits, and so when I find a "paradox" that disappears when you get rid of infinity, that's enough for me to drop the problem.

This seems like an unproductive attitude. If everyone took this attitude they would never have hammered out the problems in calculus in the 19th century. And physicists would probably not have every discovered renormalization in the 20th century.

A better approach is to try to define rigorously what is happening with the infinities. When you try that, either it becomes impossible (that is there's something that seems intuitively definable that isn't definable) or one approach turns out to be correct, or you discover a hidden ambiguity in the problem. In any of those cases one learns a lot more than simply saying that there's an infinity so one can ignore the problem.

Comment author: Vaniver 28 September 2011 03:57:26AM -1 points [-]

This seems like an unproductive attitude. If everyone took this attitude they would never have hammered out the problems in calculus in the 19th century. And physicists would probably not have every discovered renormalization in the 20th century.

I will make sure to not spread that opinion in the event that I travel back in time.

A better approach is to try to define rigorously what is happening with the infinities. When you try that, either it becomes impossible (that is there's something that seems intuitively definable that isn't definable) or one approach turns out to be correct, or you discover a hidden ambiguity in the problem. In any of those cases one learns a lot more than simply saying that there's an infinity so one can ignore the problem.

I agree with you that this is a better approach. However, the problem in question is "find the error" not "how conservation of momentum works," and so as soon as you realize "hm, they're not treating this infinity as a limit" then the error is found, the problem is solved, and your curiosity should have annihilated itself.

Comment author: lessdazed 27 September 2011 08:51:43PM 0 points [-]

Vf vg gung gur vavgvny irybpvgl bs gur znggre vf mreb, gur vavgvny naq riraghny nppryrengvba bs gur pragre bs tenivgl vf mreb, ohg gur pragre bs tenivgl vgfrys ortvaf jvgu n yrsgjneq irybpvgl?

Comment author: ArisKatsaris 27 September 2011 07:16:49PM * 0 points [-]

This is a good one, but I won't try doing the math to figure out the specific error with it (other than the obvious problem of (rot13) gur arrq gb pnyphyngr gur rssrpg bs nyy gur znffrf, abg whfg gur vzzrqvngryl ba gur evtug naq gur vzzrqvngryl ba gur yrsg.)

Here's what my intuition is telling me: Gung jung jvyy npghnyyl unccra vf gung gur znffrf jvyy pbairetr fbzrjurer va gur pragre, jvgu fbzr bs gur znffrf nsgre gur zvqqyr cbvag zbivat evtugjneqf vafgrnq. V vzntvar gung V'q unir gb hfr vagrtengvba bire gur ahzore bs vasvavgr znffrf ba gur yrsg, gb frr gung gur zngu sbepr gur jubyr guvat gb onynape.

Comment author: Oscar_Cunningham 27 September 2011 06:31:22PM 0 points [-]

Gur prager bs znff vf ng gur bevtva naq fgnlf gurer?

Comment author: [deleted] 27 September 2011 06:29:37PM 0 points [-]

Clever! It took me a couple of minutes to get it: Gur yrsgzbfg cbvag (gur 1 xt znff) jvyy zbir gb gur evtug, xrrcvat gur pragre bs tenivgl jurer vg vf. Fb vg'f abg gehr gung "jr pna cebir guvf sbe rirel cbvag.

Comment author: ArisKatsaris 27 September 2011 07:11:25PM 0 points [-]

You are misreading which one's the rightmost mass. The rightmost mass is the one with 1kg.

Comment author: [deleted] 27 September 2011 07:18:10PM 0 points [-]

Whoops, you're right. I assumed +1/10 meant +1/10th from +1. I had the whole thing backwards.

Comment author: AlexMennen 27 September 2011 07:11:30PM 0 points [-]

Gurer vf ab reebe. Gur cevapvcyr gung gur pragre bs znff pnaabg nppryrengr pbzrf sebz gur snpg gung, sbe nal cbvag znff, gur sbepr vg rkregf ba bgure znffrf vf rknpgyl bccbfvgr gb gur sbepr bgure znffrf rkreg ba vg. Guvf qbrf abg nccyl gb vasvavgr pbyyrpgvbaf bs znffrf sbe gur fbzr ernfba gung na vasvavgr frevrf bs barf qbrf abg nqq gb mreb, rira gubhtu vg vf gur qvssrerapr orgjrra gjb vqragvpny frevrf, rnpu gur fhz bs nyy cbfvgvir vagrtref.

Comment author: Douglas_Knight 27 September 2011 10:01:11PM 0 points [-]

Vs lbh guvax gurer vf n fcrpvsvp qviretrag frevrf, anzr vg.

Comment author: AlexMennen 28 September 2011 12:49:12AM * 0 points [-]

Gur fhz bs gur gbgny sbeprf ba nyy cnegvpyrf.

sum(n=0 to inf)
{
[sum(m=0 to n-1) G*(1/2)^n*(1/2)^m/((1/10)^m-(1/10)^n)^2]
- [sum(m=n-1 to inf) G*(1/2)^n*(1/2)^m/((1/10)^n-(1/10)^m)^2]
}

vf yrff guna 0 sbe rirel grez va gur bhgre fhz, rira gubhtu lbh pna pnapry bhg nyy gur grezf ol ernpuvat vagb gur vaare fhzf.

Comment author: wedrifid 28 September 2011 02:26:26AM * 1 point [-]

To put code blocks in mark down put four spaces before the lines you want to be code. I assume that is what you intended?

Comment author: AlexMennen 28 September 2011 02:45:17AM 0 points [-]

Edited, although I'm not sure it's all that much more readable now.

Comment author: wedrifid 28 September 2011 02:52:22AM 0 points [-]

I was thinking of including a line break or two. ;)

Comment author: AlexMennen 28 September 2011 02:58:41AM * 0 points [-]

Good idea.

Comment author: Douglas_Knight 28 September 2011 06:43:43PM 0 points [-]

bx, gung vf n ceboyrzngvp erneenatrzrag. Ohg juvpu fvqr vf pbeerpg? Lbh znqr n pynvz nobhg guvf va lbhe bevtvany pbzzrag, ohg V guvax lbh tbg vg onpxjneqf.

Comment author: AlexMennen 29 September 2011 02:26:02AM * 0 points [-]

My previous attempt at a reply didn't make much sense. And this is deep enough into the thread that I won't bother with rot13.

Let's try again: The description given by Thomas accurately models how each individual particle accelerates (towards the origin). It thus also accurately models how the center of mass accelerates in the non-limiting case.

Comment author: Douglas_Knight 29 September 2011 02:54:04AM 0 points [-]

I think Richard Kennaway's answer is correct and it seems to me to be opposite to your original answer. The non-limiting case is not interesting. The question is whether there is a coherent extension of Newtonian mechanics to infinitely many particles. It does cover this case and the center of mass still works.

Comment author: AlexMennen 29 September 2011 03:18:49AM * 0 points [-]

The question is whether there is a coherent extension of Newtonian mechanics to infinitely many particles.

I think this paradox shows that the answer is no.

It does cover this case and the center of mass still works.

I can't figure out what this sentence was intended to mean.

Comment author: Douglas_Knight 29 September 2011 03:23:39AM 1 point [-]

Richard Kennaway's answer

Comment author: AlexMennen 28 September 2011 11:30:24PM 0 points [-]

V guvax gung va gur aba-yvzvgvat pnfr, gur neenatrzrag fubjvat rirelguvat qevsgvat gbjneqf gur bevtva vf pbeerpg naq gur neenatrzrag fubjvat gur pragre bs znff fgnlvat fgvyy vf va reebe. Bs pbhefr, va gur yvzvg nf gur ahzore bs cbvag znffrf vapyhqrq tbrf gb vasvavgl, vg'f gur bgure jnl nebhaq.

Comment author: rwallace 27 September 2011 07:07:13PM 0 points [-]

An elegant puzzle.

Gur fbyhgvba vf rnfvrfg gb frr vs jr pbafvqre n svavgr ahzore bs vgrengvbaf bs gur cebprqher bs fcyvggvat bss n fznyyre znff. Va rnpu vgrengvba, gur yrsgzbfg znff onynaprf gur obbxf ol haqretbvat irel encvq nppryrengvba gb gur evtug. Va gur yvzvg nf gur ahzore bs vgrengvbaf tbrf gb vasvavgl, jr unir na vasvavgrfvzny znff onynapvat gur obbxf jvgu vasvavgryl ynetr nppryrengvba.

Comment author: [deleted] 28 September 2011 12:44:20AM 1 point [-]

I think the solution I came up with is, in spirit, the same as this one.

Pbafvqre gur nzbhag bs nppryrengvba rnpu cnegvpyr haqretbrf. Gur snegure lbh tb gb gur yrsg, gur terngre gur nppryrengvbaf naq gur fznyyre gur qvfgnaprf, naq, guhf, gur yrff gvzr vg gnxrf orsber n pbyyvfvba unccraf. (V pbhyq or zvfgnxra gurer.) Sbe rirel cbfvgvir nzbhag bs gvzr, pbyyvfvbaf unccra orsber gung nzbhag bs gvzr. Gurersber, gur orunivbe bs gur flfgrz nsgre nal nzbhag bs gvzr vf haqrsvarq.

Comment author: Tyrrell_McAllister 28 September 2011 03:33:47PM 1 point [-]

Jung vs jr fhccbfr gung gur obqvrf pna cnff guebhtu rnpu bgure? Ubj qbrf gur flfgrz orunir nf gvzr cebterffrf?

Comment author: pengvado 28 September 2011 08:49:37PM 0 points [-]

Vg'f rnfl rabhtu gb rkgraq arjgbavna zrpunavpf gb unaqyr gjb-cnegvpyr pbyyvfvbaf. Whfg cvpx bar bs "rynfgvp pbyyvfvba" be "cnff guebhtu rnpu bgure", naq nccyl gur pbafreingvba ynjf naq flzzrgevrf. Ohg V'z abg njner bs nal cebcbfnyf gb unaqyr guerr-be-zber-cnegvpyr pbyyvfvbaf: va gung pnfr, pbafreingvba bs raretl naq zbzraghz vfa'g rabhtu gb qrgrezvar gur bhgchg fgngr, lbh'q arrq fbzr npghny qlanzvpf gung ner qrsvarq ba gur vafgnag bs pbyyvfvba.

Comment author: Douglas_Knight 29 September 2011 03:17:20AM * 0 points [-]

Do you have a reference for how to extend Newtonian mechanics to collisions or passing-through of point particles subject to gravity? how about people complaining about 3 body collisions?

Bringing in energy and momentum doesn't sound helpful to me because they are both infinite at the point of collision.

My understanding is that both of the choices has a unique analytic extension to the complex plane away from the point of collision. Most limiting approaches will agree with this one. In particular, Richard Kennaway's approach (to non-collision) is perturb the particles into another dimension, so that the particles don't collide. The limit of small perturbation is the passing-through model.

For elastic collisions, I would take the limit of small radius collisions. I think this is fine for two body collisions. In dimension one, I can see a couple ways to do three body collisions, including this one. The other, once you can do two body collisions, is to perturb one of bodies, to get a bunch of two body collisions; in the limit of small perturbation, you get a three body collision. But if you extend this to higher dimension, it results in the third body passing through the collision (which is a bad sign for my claim that most limiting approaches agree). When I started writing, I thought the small radius approach had the same problem, but I'm not sure anymore.

Comment author: rwallace 28 September 2011 04:27:58AM 0 points [-]

Yes, I think so too.

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

Article Navigation

Comments (115)

Nearest Meetups

Virtual Study Room

Recent Comments

Recent Posts

Recent Rationality Quotes

Recent Wiki Edits

Top Contributors, 30 Days

Recent Karma Awards

Comments (115)

Top Contributors, 30 Days

You'll need to login or register to do that

(Don't worry, it only takes a few seconds)

Create

Login