Mathematical Tools for Physics

9—Vector Calculus 1 269

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) (41)

There are now four arbitrary constants to examine. Start withC′. It’s the coefficient of 1 /rin the domain where
r < 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 ( 40 ) 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 the the equations ( 37 ) and ( 39 ) 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 (42)

Where do these continuity conditions come from? Assume for a moment that the first one is false, thatV 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

V dV/dr d^2 V/dr^2

• Watch out for the similar looking equations that appear in electromagnetism. Only the first of these
  equations holds there; the second must be modified.