Mathematical Methods for Physics and Engineering - Matematica.NET

21.3 SEPARATION OF VARIABLES IN POLAR COORDINATESsurprising that the form of the expression for u changes there. Let us therefore take twoseparate regions.In the region r>a(i) we must have u → 0asr →∞, implying that all A =0,and(ii) the system is axially symmetric and so only m = 0 terms appear.With these restrictions we can write as a trial form∞∑u(r, θ, φ) = B l r −(l+1) Pl 0 (cos θ). (21.59)l=0The constants B l are still to be determined; this we do by calculating directly the potentialwhere this can be done simply – in this case, on the polar axis.Considering a point P on the polar axis at a distance z (> a) from the plane of the ring(taken as θ = π/2), all parts of the ring are at a distance (z 2 + a 2 ) 1/2 from it. The potentialat P is thus straightforwardlyGMu(z,0,φ)=− , (21.60)(z 2 + a 2 )1/2where G is the gravitational constant. This must be the same as (21.59) for the particularvalues r = z, θ =0,andφ undefined. Since Pl 0(cos θ) =P l(cos θ) withP l (1) = 1, puttingr = z in (21.59) gives∞∑ B lu(z,0,φ)= . (21.61)zl+1 l=0However, expanding (21.60) for z>a(as it applies to this region of space) we obtainu(z,0,φ)=− GM [1 − 1 ( a) 2 3( a) 4+ − ···],z 2 z 8 zwhich on comparison with (21.61) gives §B 0 = −GM,B 2l = − GMa2l (−1) l (2l − 1)!!for l ≥ 1, (21.62)2 l l!B 2l+1 =0.We now conclude the argument by saying that if a solution for a general point (r, θ, φ)exists at all, which of course we very much expect on physical grounds, then it must be(21.59) with the B l given by (21.62). This is so because thus defined it is a function withno arbitrary constants and which satisfies all the boundary conditions, and the uniquenesstheorem states that there is only one such function. The expression for the potential in theregion r>ais therefore[]u(r, θ, φ) =− GM ∞∑ (−1) l (2l − 1)!!( a) 2l1+P2l (cos θ) .r2 l l! rl=1The expression for r