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May 25

Property of Legendre transformed thermodynamic potentials

Dear volunteers,

as I understand from Legendre transformation#An equivalent definition in the differentiable case, it holds that for a differentiable, convex function $f$ and its Legendre transform $g$

Df=\left(Dg\right)^{-1}~.

An application of this transformation is, e.g., the derivation of the Helmholtz free energy $A$ from the internal energy $U$ . However, I do not see the above equation apply to this case. Since the internal energy is a function of the entropy, $U=U(S)$ , and the free energy is a function of the (entropy conjugate) temperature, $A=A(T)$ , I would expect

{\frac {\partial U}{\partial S}}=\left({\frac {\partial A}{\partial T}}\right)^{-1}

as linear Operators. As we are in the scalar case, the above derivatives are scalars $\in \mathbb {R}$ , namely

{\frac {\partial U}{\partial S}}=T

and

{\frac {\partial A}{\partial T}}=-S~.

Obviously, the composition of the linear Operators results in an Operator equivalent to multiplication by $-TS\neq 1$ .

Any hints on the location of the error(s) in the above would be highly appreciated.

Thanks, --Das O2 (talk) 20:07, 25 May 2015 (UTC)[reply]

The derivatives are not inverse linear operators of each other, they are inverse functions of each other! The derivative of a function

f:V\to \mathbb {R}

on a vector space V is a function

f':V\to V^{*}

from V (the space of x's) into the dual space

V^{*}

(the space of ps). The Legendre transform

g:V^{*}\to \mathbb {R}

is such that

g':V^{*}\to V

is the inverse function of

f':V\to V^{*}

. Perhaps a clearer way to write this: the derivative of

f(x)

is just

f'(x)=p

(in the conventions of the article Legendre transform), so the derivative of the Legendre transform

g(p)

is the inverse function

g'(p)=x

. (Thermodynamics has the signs all opposite to those used in the article, so really it is the negative inverse function in that case.) Sławomir Biały (talk) 13:21, 26 May 2015 (UTC)[reply]

Thank You! I embarrassingly mixed up derivative and differential. --Das O2 (talk) 17:55, 26 May 2015 (UTC)[reply]

Circle packing in a circle

I ran across a cute article Circle packing in a circle, which further references a site www.packomania.com , and it's interesting. There seems to be some kind of pure, fundamental meaning in which patterns of circles are regular and which are not.

To take the simplest thing for me to look at: suppose I define a "regular-edged" packing as one where every circle touching the outer rim is equally spaced, touching two circles to either side of it. Then the numbers 3, 4, 5, 6, 7, 8, 9, 10, 11?, 13*, 18, 19, 37, 56 61, 91 represent "regular-edged" numbers, and the ones between them, not. (?: the solution given for 11 differs between Packomania and the Wikipedia article - someone should sort that out *: the Wikipedia article says there are *two* equivalent solutions for #13, one of which is regular-edged and one of which isn't!)

You could likewise sort out solutions by uniqueness (whether a loose circle is present), symmetry group, etc. But are any of these things known to mean anything about anything other than how circles go in a circle? Wnt (talk) 21:20, 25 May 2015 (UTC)[reply]

The WP solution for 11 circles and the one given given at packomania have the same diameter of outer circle so they are both optimal. In fact they differ only by the placement of one circle; there are two possible "holes" where this last circle can fit. The packomania summary table has some of the things you're asking about: symmetry, #loose circles, #circles on the edge. If there are p circles packed around the outer rim then the radius of the outer circle is

1+{\frac {1}{\sin({\frac {\pi }{p}})}}

which is asymptotically a constant times p. But you can argue that the radius of the outer circle is asymptotically a constant times the √n. So it seems that the radius of the outer circle increases more and more slowly as n increases while the gaps between the radii for regular edges stay about the same size. Hence you would expect larger and larger gaps between values of n which have a regular edged solution. Whether there is any deeper meaning to this type of problem is an interesting question. On the one hand, though packomania hints of industrial applications, the problem seems to be more on the recreational side. On the other hand, it's an example of a type of multivariate optimization problem which is very difficult to solve exactly and there are sophisticated techniques to find good approximate answers. Note that the table lists the best known solutions; apparently no solution is known to be best for n>19. Also, this problem is closely related to the sphere packing problem which leads to some very serious mathematics, e.g. the Leech lattice. --RDBury (talk) 00:24, 26 May 2015 (UTC)[reply]

It seems to become sparse faster than the square root, because there are values of p that are 'unpackable', like p=11, which is not seen in the solution for any number of circles. There's some sort of trade-off that favors symmetry, i.e. the solution for fifteen circles with p=10 and five in the middle. I feel a suspicion that this is related to 11 being prime, whereas 'popular' numbers like 6 (which solves 6 and 7) or 12 (which solves 18 and 19) are highly divisible. (24 doesn't appear to turn up twice, but at 61 is one of the few 'regular-edged' solutions, followed only by p=30 as a sixfold symmetric solution for 91 circles, at least in the packomania approximations) Wnt (talk) 14:23, 27 May 2015 (UTC)[reply]