Derivation of Poisson’s Equation
by Matthew Fox
Derive Poisson’s equation
∇2 φ(~r) = 4πGρ,
where φ(~r) is the gravitational potential function and ~r is the position vector
from some mass density ρ.
1
Matthew Fox
Derivation of Poisson’s Equation
Our Solution
Poisson’s equation asserts that the Laplacian for the gravitational scalar function
−GM
,
|~r|
φ(~r) ≡
(1)
where M is the mass of some body in free-space, is given by
∇2 φ(~r) = 4πGρ,
(2)
where ρ is the mass-density in some volume of free-space. The derivation begins
with (1). We compute
∂φ ∂φ ∂φ
~
,
,
∇φ(~r) =
∂x ∂y ∂z
*
+
∂
−GM
∂
−GM
∂
−GM
p
p
p
=
,
,
∂x x2 + y 2 + z 2 ∂y x2 + y 2 + z 2 ∂z x2 + y 2 + z 2
GM x
GM y
GM z
=
,
,
(x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )3/2 (x2 + y 2 + z 2 )3/2
GM
r̂.
=
|~r|2
From Newton’s law of gravity, we know
GM m
GM
F~ = −
r̂ = m~g (~r) =⇒ ~g (~r) = − 2 r̂,
|~r|2
|~r|
(3)
from which it follows that
~ r) = −~g (~r).
∇φ(~
(4)
Hold this equation in memory for now, as it will be important shortly. We now
transition and consider a lump of mass M suspended in free space, which we
surround by a Gaussian surface of radius ~r so that our test mass m lies at this
surface. From the divergence theorem, we know that
‹
˚
~=
~ · ~g (~r)dV,
~g (~r) · dA
∇
∂Σ
(5)
Σ
where Σ is the surface over which we intend to integrate. For our Gaussian
surface, which is a sphere,
‹
‹
‹
~=
~g (~r) · dA
−g(~r)r̂ · (dA)r̂ =⇒
∂Σ
∂Σ
∂Σ
2
−g(~r)dA = −g(~r)(4π|~r|2 )
Matthew Fox
Derivation of Poisson’s Equation
~ From (3), it follows
because ~g (~r) is radially inward, and so is antiparallel to dA.
once more that
‹
~ = −4πGM,
~g (~r) · dA
(6)
∂Σ
which is Gauss’s law for the gravitational field and is synonymous with Gauss’s
law for the electric field in Maxwell’s equations. From here, we employ the divergence theorem, which, as (5) suggests, relates the total flux of the gravitational
field to the field’s divergence. Note that the mass M in (6) is not strictly the
mass of some body, but instead represents the mass enclosed in the Gaussian
surface Σ. It then follows that
˚
ρ dV,
M=
(7)
Σ
where ρ is the mass-density within the surface Σ. Substituting this into (6) and
then incorporating the divergence theorem, we have that
˚
˚
~ · ~g (~r)dV.
∇
−4πG
ρ dV =
Σ
(8)
Σ
Notice that our choice of Σ is arbitrary, and so (8) is true for every possible
choice of Σ. We are then forced to conclude that (8) must be true regardless
of Σ, which is possible only if the integrands of each volume integral are equal.
Consequently,
~ · ~g (~r) = −4πGρ.
∇
As we derived earlier,
~ r),
~g (~r) = −∇φ(~
making (9) identical to
~ · ~g (~r) = ∇
~ · (−∇φ(~
~ r)) = −4πGρ =⇒ ∇2 φ(~r) = 4πGρ,
∇
which is (2) and is Poisson’s equation.
3
(9)