17—Densities and Distributions 423There are two terms; start with the first one.r > 0 , soeiqr→ 0 asq→+i∞. The contour is along
the real axis, so push it towardi∞and pick up the residue at+ik.
−
1
8 π^2 ir
∫
=−
1
8 π^2 ir
2 πiRes
q=ikq
(q−ik)(q+ik)
eiqr=−
1
8 π^2 ir
2 πi
ik
2 ik
e−kr (17.41)
For the second term, see problem17.15. Combine it with the preceding result to get
G=−
1
4 πr
e−kr (17.42)
This is the solution to Eq. (17.40), and if I now letkgo to zero, it isG=− 1 / 4 πr.
Just as in one dimension, once you have the Green’s function, you can write down the solution
to the original equation.
∇^2 V=−ρ/ 0 =⇒ V=−
1
0
∫
d^3 r′G(~r,~r′)ρ(~r′) =
1
4 π 0
∫
d^3 r′
ρ(~r′)
|~r−~r′|
(17.43)
State this equation in English, and it says that the potential of a single point charge isq/ 4 π 0 r
and that the total potential is the sum over the contributions of all the charges. Of course this
development also provides the Green’s function for the more complicated equation (17.42).
Applications to Potentials
Let a charge density beqδ(~r). This satisfies
∫
d^3 rρ=q. The potential it generates is that of a point
charge at the origin. This just reproduces the Green’s function.
φ=
1
4 π 0
∫
d^3 r′qδ(~r′)
1
|~r−~r′|
=
q
4 π 0 r
(17.44)
What ifρ(~r) =−p∂δ(~r)/∂z? [Why−p? Patience.] At least you can say that the dimensions
of the constantpare charge times length. Now use the Green’s function to compute the potential it
generates.
φ=
1
4 π 0
∫
d^3 r′(−p)
∂δ(~r′)
∂z′
1
|~r−~r′|
=
p
4 π 0
∫
d^3 r′δ(~r′)
∂
∂z′
1
|~r−~r′|
=
p
4 π 0
∂
∂z′
1
|~r−~r′|
∣
∣∣
∣
~r′=0(17.45)
This is awkward, so I’ll use a little trick. Make a change of variables in (17.45)~u=~r−~r′, then
∂
∂z′
→ −
∂
∂uz
,
p
4 π 0
∂
∂z′
1
|~r−~r′|
∣∣
∣
∣
~r′=0=−
p
4 π 0
∂
∂uz
1
u
(17.46)
(See also problem17.19.) Cutting through the notation, this last expression is just