Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The $s$ squares problem is to count the number $r_s (n)$ of integer solutions $(x_1,x_2,...,x_s)$ of the Diophantine equation $x_{1}^{2}+x_{2}^{2}+...+x_{s}^{2}=n$ in which changing the sign or order of the $x_i$ ’s gives distinct solutions. If looking into the statistical mechanics for classical ideal gas in 3D, we meet with the same thing with $s=3N$, $N$ is the number of particles. But now the $3N$ squares problem is to count the number of the microstates in the so-called microscope ensemble. The following asymptotic expression of $r_{3N}(n)$ is experimentally validated, so it is physically proved:

$r_{3N}(n)\approx \frac{{\pi}^{3N/2}}{\Gamma (3N/2)} {{n}^{3N/2-1}}$, in thermodynamic limit $n/N=const.$ and $n \to \infty$ .

My question is: How to give an estimate of the error, and does anyone know such a formula in mathematical literature?


S.C.Miline, New infinite families of exact sums of squares formulas, Jacobi elliptic functions and Ramanujan’s tau function, Proc. Natl. Acad. Sci. USA, 1996, 93:15004-15008, and references cited therein.

share|improve this question
What do you mean there is no such formula in mathematical literature? It seems to me that you are just giving a crude estimate based on the number of lattice points inside a sphere. –  Gjergji Zaimi Sep 11 '10 at 11:04
Is what you are claiming this: that for each positive integer $k$ $$r_{3N}(kN)\sim\frac{\pi^{3N/2}}{\Gamma(3N/2)}(kN)^{3N/2-1}$$ as $N\to\infty$? You claim this is proved; have you a reference to the proof? (Is Milne's paper such a reference?) Are you sure there is "no such formula in mathematical literature"? (There's an awful lot of mathematical literature). –  Robin Chapman Sep 11 '10 at 11:05
QHLIU: in the original question, you said the limit was "proved". Now you say it's a "problem". What exacty do you mean? –  Robin Chapman Sep 11 '10 at 11:30
QHLIU: so, when you said the statement was proved, that wasn't actually what you meant :-( Also now both $n$ and $N$ are "linearly indpendent" and $n/N$ is constant. :-( –  Robin Chapman Sep 11 '10 at 11:40
QHLIU: have you a mathematical question here? Are you still asserting that the assertion you claimed was "proved" is proved? If so please can you give a reference? –  Robin Chapman Sep 11 '10 at 11:44

1 Answer 1

The main reason is that, unlike physicists, mathematicians do not put wrong formulae in their papers and textbooks (well, I admit, there are exceptions, but it isn't the general culture, not yet at least)

Why is it wrong? Let's denote $3N$ by $N$ and the ratio $n/N$ (my $N$) by $k$ to simplify the matters. Then we are to find the $n$-th coefficient of the function $F(z)^N$ where $F(z)=1+2z+2z^4+2z^9+\dots$. It is the average of $F(z)^Nz^{-n}$ over any circle of radius less than $1$. The last expression can also be written as $G(z)^N$ where $G=F(z)z^{-k}$. Now choose the radius $r_k$ that minimizes $F(r)r^{-k}$ on $(0,1)$. Then we can use the classical Laplace formula on that circle and get $(F(r_k)r_k^{-k})^N$ as the main exponential factor. I do not know what $F(r_k)r_k^{-k}$ is, but I know it is not $\sqrt{\pi ek}$ because I can invert the Legendre transform of the latter in logarithmic coordinates. Admitted, it becomes close to it when $k$ is allowed to grow, but in the declared range (small $k$) the OP's approximation is off by an exponential factor, which makes that $-1$ in the power of $n$ totally pathetic.

The actual question is why something like that appears in phisics textbooks. My answer would be "because we teach our students Thomas' calculus instead of Fichtenholtz' analysis, thus wasting both our and their time", but I'd better not start this discussion now.

share|improve this answer
fedja: I admit that physical way of reasoning usually loses the mathematical rigor, but it is supported by the experiments. Besides, physical reasoning enriches the mathematical studies, such as the Fourier series, distribution theory, etc. All primitive physical results are then mathematically proved correct; no exception is so far identified. Why isn’t it wrong? The approximate solution to the squares problem is easily attainable by treating the lattice (hyper-)spherical surface as a smooth one, and the number of the solutions is then nothing but the area of the surface area. –  QHLIU Sep 11 '10 at 13:55
Downvoted, because this really feels like a diatribe about Why Physicists Are Bad. –  Michael Lugo Sep 11 '10 at 17:09
I'm not sure what you mean by "supported by experiments". Usually it means that the formulae and the models are tailored to what is observed so that the discrepancy is no longer obvious. Anyway, I do not mind back of envelope approximations: I do them all my life myself. What I mind is pretending they work in the ranges of parameters beyond where they can be proved. Yes, the surface area is a reasonable approximation in general, but the declared range is exactly where it fails miserably. –  fedja Sep 11 '10 at 17:23
At Michael: If you change it to "diatribe about Why Physicists Are So Poorly Trained In Mathematics", I'll agree. "Bad" is a meaningless word in such contexts. –  fedja Sep 11 '10 at 17:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.