elI可按它展开 ex=∑∑ aImak((,,) l =0 me-I =∑∑an(kr)Ymn(,中 如取k方向在z方向(即为z轴),则 ik.r =e ikrcose e ∑ani(kr)Yn(,.)
可按它展开 如取 方向在 z 方向（即为 z 轴），则 ik r e ⋅ l ik r lm klm l 0m l e a u (r, , ) ∞ ⋅ = =− = ∑ ∑ ′ θ φ l lm l lm l 0m l a j (kr)Y ( , ) ∞ = =− = ∑ ∑ θ φ k ik r ikr cos l0 l l0 l 0 e e a j (kr)Y ( , ) ∞ ⋅ θ = = = θφ ∑
∑cj(kr)B1(cos6) 1=0 A.对kr求导,得 i∑c1j(kr) cos OP(cos0)=∑cj(kr)B1(cosθ) 1=0 cosP(C0s6)4IB-1(c0s9)+(1+1)P+1(cos9)
A. 对 kr 求导，得 c j (kr)P (cos ) l l l 0 ∑ l θ ∞ = = i c j (kr)cos P (cos ) c j (kr)P (cos ) l l l 0 l l l l 0 l θ θ θ ′ ∑ = ∑ ∞ = ∞ = [lP (cos ) (l 1)P (cos )] 2l 1 1 cos P (cos ) θ l θ l−1 θ + + l+1 θ + =
(kx)=,,[1j1-1(kr)-(1+1)j1+1(kr)] 21+1 于是有 ∑c1,,[IP-1(cosθ)+(1+1)P1+1(cos6) 021+ ∑c [j-1(kr)-(+D)j+(kr)]P(cos a) 1=021+1
于是有 [lj (kr) (l 1)j (kr)] 2l 1 1 j (kr) l l−1 − + l+1 + ′ = [lj (kr) (l 1)j (kr)]P (cos ) 2l 1 1 c l 1 l 1 l l 0 l − + θ ∞= − + + = ∑ [lP (cos ) (l 1)P (cos )] 2l 1 i c j l 1 l 1 l 0 l l − θ + θ ∞= + + + ∑
(1+1) ∑[c1+1 i(1+1) 1=0 21+3 JI+1 Pi(cos 0)+Cl21 +1 J 1+(cos 8) 1+1 2[cu+1,j (kr)+(cos 0)-CI j1+(kr)P(cos a) 1=0 21+3 21+1 2l+3 2I+1 2|-1≈221+121-1 2I+1 2I+1 2l-121-3 2l-3
2 2 l l1 l2 l2 2l 1 2l 1 2l 1 2l 1 ci c i c i c 2l 1 2l 1 2l 3 2l 3 − − − + +− + == = − −− − ∑ ∞ = + + + + + + + + l 0 l 1 l 1 l l l l 1 j P (cos )] 2 l 1 i ( l 1 ) j P (cos ) c 2 l 3 i ( l 1 ) [ c θ θ ∑ ∞ = + + + + + − + + = l 0 l 1 l l 1 l l 1 l j (kr ) P (cos )] 2 l 1 ( l 1 ) j (kr ) P (cos ) c 2 l 3 l 1 [ c θ θ l1 l 2l 3 ci c 2l 1 + + = +
c1=i(2l+1)c0 B.当kr=0时 j1(0)=80Pc0s)=1 (0
B. 当 时 l l 0 c i (2l 1)c = + kr = 0 l l0 j (0) = δ P (cos ) 1 0 θ = c0 =1