578 Real Analysis
Proof.For the sake of completeness we will prove Green’s identity. Consider the vector
field
−→
F =f∇g. Thendiv−→
F =
∂
∂x(
f∂g
∂x)
+
∂
∂y(
f∂g
∂y)
+
∂
∂z(
f∂g
∂z)
=f(
∂^2 g
∂x^2+
∂^2 g
∂y^2+
∂^2 g
∂z^2)
+
(
∂f
∂x∂g
∂x+
∂f
∂y∂g
∂y+
∂f
∂z∂g
∂z)
.
So the left-hand side is
∫∫∫
Rdiv−→
FdV. By the Gauss–Ostrogradski divergence theorem
this is equal to
∫∫S(f∇g)·−→ndS=∫∫
Sf(∇g·−→n)dS=∫∫
Sf
∂g
∂ndS.Writing Green’s first identity for the vector fieldg∇fand then subtracting it from
that of the vector fieldf∇g, we obtain Green’s second identity
∫∫∫R(f∇^2 g−g∇^2 f)dV=∫∫
S(
f
∂g
∂n−g
∂f
∂n)
dS.The fact thatfandgare constant along the lines passing through the origin means that
on the unit sphere,∂f∂n=∂g∂n=0. Hence the conclusion.
531.Because
−→
F is obtained as an integral of the point-mass contributions of the masses
distributed in space, it suffices to prove this equality for a massMconcentrated at one
point, say the origin.
Newton’s law says that the gravitational force between two massesm 1 andm 2 at
distanceris equal tom^1 rm 22 G. By Newton’s law, a massMlocated at the origin generates
the gravitational field
−→
F (x, y, z)=MG1
x^2 +y^2 +z^2·
x−→
i +y−→
j +z−→
k
√
x^2 +y^2 +z^2=−MG
x−→
i +y−→
j +z−→
k
(x^2 +y^2 +z^2 )^3 /^2.
One can easily check that the divergence of this field is zero. Consider a small sphereS 0
of radiusrcentered at the origin, and letVbe the solid lying betweenS 0 andS. By the
Gauss–Ostrogradski divergence theorem,
∫∫
S−→
F·−→ndS−∫∫
S 0−→
F ·−→ndS=∫∫∫
Vdiv−→
FdV= 0.Hence it suffices to prove the Gauss law for the sphereS 0. On this sphere the flow
−→
F·−→nis
constantly equal to−GMr 2. Integrating it over the sphere gives− 4 πMG, proving the law.