9—Vector Calculus 1 228This is a scalar equation instead of a vector equation, so it will often be easier to handle. Apply it to
the same example as above, the uniform spherical mass.
The Laplacian,∇^2 is the divergence of the gradient, so to express it in spherical coordinates,
combine Eqs. (9.24) and (9.31).
∇^2 V=
1
r^2
∂
∂r
(
r^2
∂V
∂r
)
+
1
r^2 sinθ
∂
∂θ
(
sinθ
∂V
∂θ
)
+
1
r^2 sin^2 θ
∂^2 V
∂φ^2
(9.43)
Because the mass is spherical it doesn’t change no matter how you rotate it so the same thingholds for the solution,V(r). Use this spherical coordinate representation of∇^2 and for this case theθ
andφderivatives vanish.
1
r^2
d
dr
(
r^2
dV
dr
)
= 4πGρ(r) (9.44)
I changed from∂todbecause there’s now one independent variable, not several. Just as with Eq. (9.38)
I’ll divide this into two cases, inside and outside.
Outside:1
r^2
d
dr
(
r^2
dV
dr
)
= 0, so r^2
dV
dr
=C
Continue solving this and you have
dV
dr
=
C
r^2
−→V(r) =−
C
r
+D (r > R) (9.45)
Inside:1
r^2
d
dr
(
r^2
dV
dr
)
= 4πGρ 0 so r^2
dV
dr
= 4πGρ 0
r^3
3
+C′
Continue, dividing byr^2 and integrating,
V(r) = 4πGρ 0
r^2
6
−
C′
r
+D′ (r < R) (9.46)
There are now four arbitrary constants to examine. Start withC′. It’s the coefficient of 1 /rin the
domain wherer < R. That means that it blows up asr→ 0 , but there’s nothing at the origin to
cause this.C′= 0. Notice that the same argument doesnoteliminateCbecause (9.45) applies only
forr > R.
Boundary Conditions
Now for the boundary conditions atr =R. There are a couple of ways to determine this. I find
the simplest and the most general approach is to recognize that the equations (9.42) and (9.44) must
be satisfiedeverywhere. That means not just outside, not just inside, butatthe surface too. The
consequence of this statement is the result*
V is continuous atr=R dV/dris continuous atr=R (9.47)
Where do these continuity conditions come from? Assume for a moment that the first one is false, that
V is discontinuous atr=R, and look at the proposition graphically. IfV changes value in a very
small interval the graphs ofV, ofdV/dr, and ofd^2 V/dr^2 look like
- Watch out for the similar looking equations that appear in electrostatics. Only the first of these
equations holds there; the second must be modified by a dielectric constant.