21.3 SEPARATION OF VARIABLES IN POLAR COORDINATES
surprising that the form of the expression foruchanges there. Let us therefore take two
separate regions.
In the regionr>a
(i) we must haveu→0asr→∞, implying that allA=0,and
(ii) the system is axially symmetric and so onlym= 0 terms appear.
With these restrictions we can write as a trial form
u(r, θ, φ)=∑∞
=0Br−(+1)P^0 (cosθ). (21.59)The constantsBare still to be determined; this we do by calculatingdirectlythe potential
where this can be done simply – in this case, on the polar axis.
Considering a pointPon the polar axis at a distancez(>a) from the plane of the ring
(taken asθ=π/2), all parts of the ring are at a distance (z^2 +a^2 )^1 /^2 from it. The potential
atPis thus straightforwardly
u(z, 0 ,φ)=−GM
(z^2 +a^2 )^1 /^2, (21.60)
whereGis the gravitational constant. This must be the same as (21.59) for the particular
valuesr=z,θ=0,andφundefined. SinceP^0 (cosθ)=P(cosθ)withP(1) = 1, putting
r=zin (21.59) gives
u(z, 0 ,φ)=∑∞
=0B
z+1. (21.61)
However, expanding (21.60) forz>a(as it applies to this region of space) we obtain
u(z, 0 ,φ)=−GM
z[
1 −
1
2
(az) 2
+
3
8
(az) 4
−···
]
,
which on comparison with (21.61) gives§
B 0 =−GM,B 2 =−GMa^2 (−1)(2−1)!!
2 !for≥ 1 , (21.62)B 2 +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 theBgiven by (21.62). This is so because thus defined it is a function with
no arbitrary constants and which satisfies all the boundary conditions, and the uniqueness
theorem states that there is only one such function. The expression for the potential in the
regionr>ais therefore
u(r, θ, φ)=−GM
r[
1+
∑∞
=1(−1)(2−1)!!
2 !
(ar) 2
P 2 (cosθ)]
.
The expression forr<acan be found in a similar way. The finiteness ofuatr=0and
the axial symmetry give
u(r, θ, φ)=∑∞
=0ArP^0 (cosθ).§(2−1)!! = 1× 3 ×···×(2−1).