I just read about this amazing result and I have a very short question. I've read that five pieces is enough to get from one sphere to two spheres: is the underlying process sort of the same to get from one sphere to a sphere of any size (is five enough?)? Sorry if this has been asked before, I couldn't find anything.
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Five is enough for doubling the volume (i.e. two spheres), if you want more then you need more parts. – Asaf Karagila Dec 18 '13 at 14:37
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To begin with, see here.
I do not know of a written account where details are worked out more precisely, but the basic idea is the following: Lebesgue outer measure on $\mathbb R^n$, $\mu^*$, is defined for all subsets of $\mathbb R^n$, whether measurable or not, extends Lebesgue measure $\mu$, and satisfies monotonicity: $\mu^*(A)\le\mu^*(B)$ for $A\subseteq B$, and subadditivity: $\mu^*(\bigcup_{n\in\mathbb N}A_n)\le\sum_n\mu^*(A_n)$.
Suppose a sphere of volume $V$ is split into $k$ pieces $A_i$. Then $\mu^*(A_i)\le V$ for all $i$, so $\sum_i\mu^*(A_i)\le kV$, that is, no more than a sphere of volume at most $kV$ (or $k$ spheres of the same radius as the original) can be obtained if only $k$ pieces are used. In the link given above, Blackadar mentions that, in his estimation, about $10^{30}$ pieces are needed to turn a pea-sized ball into one of the size of the sun.
Barry Cipra pointed out in the comments that the outer measure argument is an overkill for the pea-sun analysis: If a sphere $S$ has diameter larger than $kd$, where $d$ is the diameter of the original sphere $s$, then along any diameter it has at least $k+1$ points at distance larger than $d$ from each other so, if $S$ is obtained by splitting and rearranging $s$, at least $k+1$ pieces are needed.
We can say a bit more: For $n>1$, let $f(n)$ be the smallest $k$ such that a solid sphere can be split into $k$ pieces that can be rearranged to form $n$ solid spheres of the same radius as the original. Each sphere must use at least $2$ pieces, so $f(n)\ge 2n$, and we know that $f(2)=5$. It would be tempting to say that $f(n)\le 2n+1$: For example, start with a sphere, split into $5$ pieces, and rearrange, so we have pieces $A,B,C,D,E$, with $A,B$ forming a sphere and $C,D,E$ another. (It would be natural to expect that) we can split $E$ into three pieces, isomorphic to $A,B,E$, and further rearrange; but this is a bit careless: Even if $E$ is chosen of size continuum, the pieces $C,D$ must be moved in this process, and end up being split as well. The best I can see at the moment is $f(n)\le 5n-2$ for $n>2$.
Andrés E. Caicedo
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2@Igor: That's easy. Just use a particle accelerator. Destroy the lead and create gold! No need to use the axiom of choice there! – Asaf Karagila Dec 18 '13 at 15:55
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A cruder estimate, not using any measure theory, is that if a sphere of radius $d$ is split into $k$ pieces that reassemble into a sphere of radius $D$, then $D\le kd$. – Barry Cipra Dec 18 '13 at 21:42
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@BarryCipra, how do you argue about this without measure theoretic ideas? – Andrés E. Caicedo Dec 18 '13 at 21:45
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1@AndresCaicedo, I was mainly referring to your $\mu^*$'s. If you have a sphere of diameter greater than $kd$, then you can pick $k+1$ points (along a diameter) each of which is at a distance greater than $d$ from the others, so no two of which can come from the same piece of the sphere of diameter $d$. – Barry Cipra Dec 18 '13 at 22:17
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@bof We cannot, as far as I can see. I was sketching what seemed like a natural approach, that does not appear to work even if we pick $E$ of size continuum. – Andrés E. Caicedo Dec 18 '13 at 22:50
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@bof I edited the paragraph to make explicit that it is not a proof (even if we are careful about picking $E$ of the right cardinality.) – Andrés E. Caicedo Dec 18 '13 at 22:54
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Now I feel dumber than usual because I neither see what's tempting about that false argument, nor do I understand your critique of it. First, I don't see why anyone would think $E$ can be split into three pieces isomorphic (I guess that means congruent) to $A,B,E$. Second, if you really could split $E$ into sets $A',B',E'$ congruent to $A,B,E$, I don't see why you would have to mess with $C,D$. Why couldn't you just remove $E$ from the second ball, split it into $A',B',E'$, put $E'$ in the position formerly occupied by $E$, and use $A',B'$ to make the third ball? – bof Dec 19 '13 at 05:22
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@bof Hehe. Maybe I should think about it more and see if there is anything there. But surely the right way of improving the bound on $f(n)$ is to go directly through the usual argument and adapt it to produce $n$ copies of the sphere rather than merely two. – Andrés E. Caicedo Dec 19 '13 at 05:55