Answer to the Question 05/02

LINE OF CHARGES

The question was:

Consider N identical "charges" placed along a straight line. A pair of charges at positions xi and xj repel each other via potential -ln|xi-xj|. Each charge is attracted to the origin of coordinates by a parabolic potential xi2/2. Show that the equilibrium positions of the charges are given by the zeroes of the Hermite polynomial HN.


(9/02) The problem was originally considered by T.J. Stieltjes, Comp. Rend. Acad. Sci., Paris, 100, 439 and 620 (1885). (It can be found the book: T. J. Stieltjes, Oevres Complés, published by Groningen, P. Noordhoff (1914-18).)

The problem has been solved (22/5/02) by Oded Farago from Materials Research Laboratory in University of California at Santa Barbara (e-mail farago@mrl.ucsb.edu), and (16/8/02) by Markus Walser (e-mail walser@web.de) (see his solution in the following postscript file), as well as (2/9/02) by Armin Rahmani (e-mail armin_rahmani@hotmail.com).

A very useful ('classical') reference on properties of orthogonal polynomials is G. Szegö Orthogonal Polynomials, American Math. Soc. Colloquium Publications, vol. XXIII (1959). Some recent developments in the area of "electrostatic analogies" of orthogonal poynomials can be found in the paper of M.E.H. Ismail, Pacific J. Math., 193, 355 (2000) (here is a copy in PDF format). It also contains many useful references.

The solution:

Hermite polynomial HN(x) of order N satisfies differential equation

HN"- 2 x HN'+2 N HN=0,

where ' and " denote the first an second derivatives with respect to x. If x1 is one the zeroes of the polynomial then we immediately see from the above equation that

HN"(x1)/[2HN'(x1)= 1/x1.

Now consider an arbitrary polynomial G(x) of order N-1, with zeroes at x2, x3,... , xN, and an additional function F(x)=(x-x1)G(x). For such polynomials it is true that

1/(x1-x2)+ 1/(x1-x3)+...+ 1/(x1-xN)=G'(x1)/G(x1)= F"(x1)/2F'(x1).

If function F happens to be HN(x), and xi are the zeroes of the polynomial, then the above relation will have the form

1/(x1-x2)+ 1/(x1-x3)+...+ 1/(x1-xN)=1/(2x1).

Obviously, x1 in the above relation can denote any zero of the polynomial. However, the above relation is exactly the equilibrium condition of a "charge" at position x1, that is attracted to the origin via potential x12, and is repelled by any other charges by potential -Sumi=2,Nln|x1-xi|. The equilibrium position is determined by equating the derivative of the potential with respect to x1 and equating the result to zero. This coincides with the equation above!

Thus we proved that the condition of equilibrium of each charge coincides with the condition satisfied by the zero of Hermite polynomial.
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