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530    CHAPTER 15  Special Functions and Eigenfunction Expansions

                                 15.2.5 Distribution of Charged Particles
                                 Suppose N charged beads are distributed on the x-axis between −1 and 1. The beads are free to
                                 move along the wire on this interval. There is a planar charge of +1 at the ends of the interval
                                 and each bead has a charge of +2. The like charges of the beads repel each other, and the ends
                                 repel the beads as well. If left alone, the beads will eventually reach an equilibrium position in
                                 which they are at rest, with the repelling forces in balance. We claim that the beads will end up
                                 at the zeros of P N (x).
                                    To show this, we will use the fact that two particles (beads) carrying planar charge q 1 and q 2
                                 and located r units apart exert a force of magnitude kq 1 q 2 /r on each other.
                                    Begin with the simple case that N = 2. This will help us understand the forces involved. If
                                 the beads are at x 1 and x 2 , with x 1 < x 2 , then the force q 1 at x 1 is
                                                                 1       2       1

                                                        F 1 = 2k     −       +        .
                                                               x 1 + 1  x 2 − x 1  x 1 − 1
                                 The force on q 2 at x 2 is
                                                                 1       2       1

                                                        F 2 = 2k     +       +        .
                                                               x 2 + 1  x 2 − x 1  x 2 − 1
                                 To be in equilibrium it is necessary that F 1 = 0 and F 2 = 0. Add these two equations to obtain
                                                                2x 1   2x 2
                                                                    +       = 0.
                                                                       2
                                                                2
                                                               x − 1  x − 1
                                                                1
                                                                       2
                                 Simplify this to obtain
                                                             2(x 1 x 2 − 1)(x 2 + x 2 ) = 0,
                                 hence x 1 =−x 2 .From F 1 = 0 this yields
                                                              1     2      1
                                                                 +     +      = 0,
                                                            x 1 + 1  2x 1  x 1 − 1
                                     2
                                 or 3x − 1 = 0. Then
                                     1
                                                                    1          1
                                                             x 1 =−√ and x 2 = √
                                                                    3          3
                                 and these are the zeros of P 2 (x).
                                    This gives us some confidence in proceeding to the case of N beads. Suppose they are
                                 located at x 1 < x 2 < ··· < x N . The forces acting on q k at x k are 2k/(x k + 1) due to the charge of
                                 +1at x =−1, 2k/(x k − 1) due to the charge of +1at x =+1, 4k/(x k − x i ) due to the charge of
                                 +2at x i < x k , and −4k/(x i − x k ) due to the charge of +2at x i > x k . The total force on q k is

                                                                             N
                                                             2k      2k           4k
                                                        F k =    +       +            .
                                                            x k + 1  x k − 1    x k − x i
                                                                           i=1,i =k
                                 The beads are in equilibrium if and only if the forces are zero, hence if and only if, after dividing
                                 out k, x 1 ,··· , x N satisfy
                                                                       N
                                                               2x k          2
                                                                    =
                                                              1 − x  2 k  i=1,i =k  x k − x i
                                 for k =1,2,··· , N. This is a necessary and sufficient condition for the beads to be in equilibrium.
                                 We will show that the N simple zeros of P N (x) (labeled in increasing order) also satisfy this
                                 condition.




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                                   October 14, 2010  15:20  THM/NEIL   Page-530        27410_15_ch15_p505-562
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