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Monday, May 19, 2008

Spherical packing problem

I concluded in the mid-nineties that I had a proof of the spherical packing problem with a relatively short time of pondering, and held that view.

So I concluded within about a year or two of working on spherical packing that I'd solved the problem.

I did write a paper and presented it to the Proceedings of the AMS, and the editor who reviewed it while not claiming an error that I can remember did reject it, of course, as that was back I think in 1996 (or 1995), but I think part of his objecting to the paper was the idea that the answer was too simple. He couldn't accept if I remember correctly that a couple of pages and some relatively simple calculations solved the problem, so it was rejected as not complicated enough.

Which brings up the question of what was the argument, except I've since lost the paper--I'm not good with keeping up with old research as I tend to just lose the paperwork over time which is why modern technology is good as now I just keep things on my blogs or much of what is on it would be lost as well--and haven't been motivated to go into detail explaining it much since because it was so simple, so it's just not exciting enough.

But now as I think back it's worth putting out what I think was the approach I used, and you can judge for yourself, if you think it is a route to the solution to the problem.

Ok, so I would do a lot of drawing of 10 of the 13 spheres that closely pack in a hexagonal close-packing structure, which I knew was the closest packing.

I only drew 10 because the arrangement is symmetric to a plane that slices through so that you see a circular close packing of 6 circles surrounding one central circle.

I then connected the dots of the centers and drew lines on the top half connecting all the top half spheres together and then I noted that I now had an inner structure made up of 4 triangular pyramids and three rectangular pyramids.

I then calculated the density of each type of pyramid, and noted that the triangular pyramid has the maximum possible density as it is the closest packing possible of 4 spheres, while the rectangular pyramid has a much lesser density.

And then I'd guess that I proved that any deformation of the triangular pyramid, as that would decrease its density, could not be made up for by the deformation of the rectangular pyramids, so that any deformation would lead to a smaller density, and that was it.

(Oh, now I think I remember what I did. It was a linear argument as I think I had one line that was like 0.77x + 0.72y equals something or other where x=4 and y=3, for the hexagonal close, and then noted that the line would be something else for the deformation and did some kind of slope thing to prove. Really damn simple I think it was. Like I said, the reviewer objected because it was too simple. Nothing else.)

Even as a thought experiment you can note that if you smoothly deform the rectangular pyramids in order to increase their density, as you move the spheres in tighter you pull them out further with the triangular pyramids decreasing their density, until at the limit you have turned your rectangular pyramid into a triangular pyramid, and deformed out your triangular pyramid into a rectangular one, while on the top you have a triangular pyramid at all times but the lengths along its base deform outward until they pull back in to the close packed structure.

So as you imagine going from hexagonal close-packing and move the spheres, if you had a density increase then you'd get a density curve that has to increase at some point and then drop back down, as at the end of the deformation you just have what you started with--4 triangular pyramids and 3 rectangular pyramids in a hexagonal close-packing.

Then you can simply prove that deformation in any direction leads to a drop in density, or more rigorously you can look for maxima or minima along any gradient, and note that you only get minima and no maxima showing the density decreases steadily with deformation until it increases back to the hexagonal close-packing density.

(Yeah, but I think that as I mentioned above, as I've come back to edit this post after puzzling over what my past self did, I did something incredibly simple using lines and slope!!! Wow. So it was like incredibly simple. But still kind of boring as I moved on and pondered Fermat's Last Theorem for years later.)

Not sure what I did in that paper back then but I concluded the solution was easy enough that I didn't worry much about it afterwards, and moved on to concentrating on Fermat's Last Theorem exclusively.

(Ok, so no need to be coy: if d is the packing density then approximately you have

(0.77x + 0.72y)/7 = d

where x = 4 and y = 3, for hexagonal close packing, but with any smooth deformation you get another linear equation, with alternate densities:

(4d1 + 3d2)/7 = d*

but d1 is less than 0.77 for one set while d2 is slightly more than 0.72 for the other but the 4 versus 3 makes the difference. So if you look at the slope of the line for any intervening values it cannot cross the previous line, so then that solves the spherical packing problem. Wow. I figured that out over ten years ago, if that's how I did it, and it seems familiar so I think it is.)

Oh, so how does that apply to packings in general if I'm just using 13 spheres?

Well, I noted that if you expand out to more spheres you can see them as being built up by these units that have 8 triangular pyramids and 6 rectangular pyramids--remember I was looking at half the structure before--which could be assembled to build a larger structure of arbitrary size, and since each individual element was as closely packed as possible and locked in perfectly with the larger structure, the hexagonal close-packing provided the maximum packing.

The entire proof if I remember correctly flowed easily from noting that 4 spheres could most closely pack only one way, which was in a triangular pyramid.

There really isn't any way to objectively attack the approach, but it's also more of a straight-forward no-nonsense way of solving the problem which is how I do things, and in my opinion, modern mathematicians, like people in computer tech fields, don't like simple and straightforward for the "hard" problems, so my solution is not something that they'd care to know, even though there is no way mathematically to invalidate the approach.


James Harris
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