10Gauss' Law and Poisson's equation
IA Vector Calculus
10.1 Laws of gravitation
Consider a distribution of mass producing a gravitational force
F
on a point
mass
m
at
r
. The total force is a sum of contributions from each part of the
mass distribution, and is proportional to m. Write
F = mg(r),
Definition
(Gravitational field)
. g
(
r
) is the gravitational field, acceleration due
to gravity, or force per unit mass.
The gravitational field is conservative, ie
I
C
g · dr = 0.
This means that if you walk around the place and return to the same position,
the total work done is 0 and you did not gain energy, i.e. gravitational potential
energy is conserved.
Gauss’ law tells us what this gravitational field looks like:
Law
(Gauss’ law for gravitation)
.
Given any volume
V
bounded by closed
surface S,
Z
S
g · dS = −4πGM,
where
G
is Newton’s gravitational constant, and
M
is the total mass contained
in V .
These equations determine g(r) from a mass distribution.
Example.
We can obtain Newton’s law of gravitation from Gauss’ law together
with an assumption about symmetry.
Consider a total mass
M
distributed with a spherical symmetry about the
origin
O
, with all the mass contained within some radius
r
=
a
. By spherical
symmetry, we have g(r) = g(r)
ˆ
r.
Consider Gauss’ law with
S
being a sphere of radius
r
=
R > a
. Then
ˆ
n
=
ˆ
r
.
So
Z
S
g · dS =
Z
S
g(R)
ˆ
r ·
ˆ
r dS =
Z
g(R)dS = 4πR
2
g(R).
By Gauss’ law, we obtain
4πR
2
g(R) = −4πGM.
So
g(R) = −
GM
R
2
for R > a.
Therefore the gravitational force on a mass m at r is
F(r) = −
GMm
r
2
ˆ
r.
If we take the limit as
a →
0, we get a point mass
M
at the origin. Then we
recover Newton’s law of gravitation for point masses.
The condition
R
C
g ·
d
r
= 0 for any closed
C
can be re-written by Stoke’s
theorem as
Z
S
∇ ×g · dS = 0,
where S is bounded by the closed curve C. This is true for arbitrary S. So
∇ × g = 0.
In our example above,
∇×g
= 0 due to spherical symmetry. But here we showed
that it is true for all cases.
Note that we exploited symmetry to solve Gauss’ law. However, if the mass
distribution is not sufficiently symmetrical, Gauss’ law in integral form can be
difficult to use. But we can rewrite it in differential form. Suppose
M =
Z
V
ρ(r) dV,
where ρ is the mass density. Then by Gauss’ theorem
Z
S
g · dS = −4πGM ⇒
Z
V
∇ · g dV =
Z
V
−4πGρ dV.
Since this is true for all V , we must have
Law (Gauss’ Law for gravitation in differential form).
∇ · g = −4πGρ.
Since
∇ × g
= 0, we can introduce a gravitational potential
ϕ
(
r
) with
g = −∇ϕ. Then Gauss’ Law becomes
∇
2
ϕ = 4πGρ.
In the example with spherical symmetry, we can solve that
ϕ(r) = −
GM
r
for r > a.