Appendix 7 — Derivation of the Laplace and Poisson Equations
A vector field for which the path between two points does not matter for the work required
is called a conservative field.
In such a field, every path from point A to point B requires the same amount of energy.
This implies that there exists a scalar potential \(\phi\) such that:
\begin{equation}
\begin{aligned}
\vec{F} = \vec{\nabla} \phi
\end{aligned}
\end{equation}
The nabla operator is defined as:
\begin{equation}
\begin{aligned}
\vec{\nabla}
= \frac{\partial}{\partial x}\,\hat{e}_x
+ \frac{\partial}{\partial y}\,\hat{e}_y
+ \frac{\partial}{\partial z}\,\hat{e}_z
= \left(
\frac{\partial}{\partial x},
\frac{\partial}{\partial y},
\frac{\partial}{\partial z}
\right)
\end{aligned}
\end{equation}
The gravitational field \(\vec{F}_g\) is an example of a conservative field:
\begin{equation}
\vec{F}_g = \vec{\nabla} \phi
\label{eq:R03}
\end{equation}
According to Gauss's theorem, for any closed surface:
\begin{equation}
\begin{aligned}
\oiint_{\partial A} \vec{F}_g \cdot d\vec{A}
=
\iiint_V (\vec{\nabla} \cdot \vec{F}_g)\, dV
\end{aligned}
\end{equation}
Vacuum: no mass, no source
In a vacuum, there is no mass, and therefore no source of gravity:
\begin{equation}
\vec{\nabla} \cdot \vec{F}_g = 0
\label{eq:R05}
\end{equation}
Substitute (\ref{eq:R03}) into (\ref{eq:R05}):
\begin{equation}
\begin{aligned}
\vec{\nabla} \cdot \vec{F}_g = 0
\quad\Rightarrow\quad
\vec{\nabla} \cdot (\vec{\nabla} \phi) = 0
\end{aligned}
\end{equation}
Explicitly:
\begin{equation}
\begin{aligned}
\vec{\nabla} \cdot \vec{\nabla} \phi
=
\left(
\frac{\partial}{\partial x},
\frac{\partial}{\partial y},
\frac{\partial}{\partial z}
\right)
\cdot
\left(
\frac{\partial \phi}{\partial x},
\frac{\partial \phi}{\partial y},
\frac{\partial \phi}{\partial z}
\right)
= 0
\end{aligned}
\end{equation}
Since \(x,y,z\) are orthogonal:
\begin{equation}
\begin{aligned}
\frac{\partial^2 \phi}{\partial x^2}
+
\frac{\partial^2 \phi}{\partial y^2}
+
\frac{\partial^2 \phi}{\partial z^2}
= 0
\end{aligned}
\end{equation}
This is written as:
\begin{equation}
\begin{aligned}
\nabla^2 \phi = 0
\quad\text{or}\quad
\Delta \phi = 0
\end{aligned}
\end{equation}
The operator \(\nabla^2\), the Laplacian, is:
\begin{equation}
\begin{aligned}
\nabla^2
= \frac{\partial^2}{\partial x^2}
+ \frac{\partial^2}{\partial y^2}
+ \frac{\partial^2}{\partial z^2}
\end{aligned}
\end{equation}
Thus, in vacuum the Laplace equation holds:
\begin{equation}
\begin{aligned}
\nabla^2 \phi = 0.
\end{aligned}
\end{equation}
Inside a region with mass
Inside matter there is a source of gravity.
According to Newton’s law of gravitation, the gravitational field is:
Application of Gauss's theorem to the gravitational field
The gravitational field of a point mass is:
\begin{equation}
\begin{aligned}
\vec{F}_g = \frac{Gm}{r^{2}}\,\hat{r}
\end{aligned}
\end{equation}
where \(\hat{r}\) is the unit vector in the radial direction.
Apply Gauss's theorem again:
\begin{equation}
\begin{aligned}
\iiint_V (\vec{\nabla}\cdot\vec{F}_g)\, dV
=
\oiint_{\partial A} \vec{F}_g\cdot d\vec{A}
\end{aligned}
\end{equation}
Since \(\vec{F}_g = \vec{\nabla}\phi\), it follows that:
\begin{equation}
\begin{aligned}
\iiint_V \Delta\phi\, dV
=
\oiint_{\partial A}
\frac{Gm}{r^{2}}\,\hat{r}\cdot d\vec{A}
=
= \oiint_A G\frac{m}{r^2} \, dA
\end{aligned}
\end{equation}
For a spherical surface:
\begin{equation}
\begin{aligned}
A = 4\pi r^{2}
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
V = \frac{4\pi}{3} r^{3}
\end{aligned}
\end{equation}
Since \(r\) is constant over the spherical surface:
\begin{equation}
\iiint_V \Delta\varphi \,
dV = \oiint_A G\frac{m}{r^2} \,
dA = G\frac{m}{r^2}\oiint_A dA =
G\frac{m}{r^2} \cdot 4\pi r^2 = 4\pi G m
\label{eq:R17}
\end{equation}
With the mass density:
\begin{equation}
\begin{aligned}
\rho = \frac{m}{V}
\end{aligned}
\end{equation}
equation (\ref{eq:R17}) becomes:
\begin{equation}
\begin{aligned}
\iiint_V \Delta\phi\, dV
=
4\pi G m
=
4\pi G \iiint \rho \, dV
=
\iiint_V 4\pi G \rho\, dV
\end{aligned}
\end{equation}
Since the volume is arbitrary:
\begin{aligned}
\Delta\phi = 4\pi G \rho
\label{eq:R20}
\end{aligned}
\end{equation}
This is the Poisson equation, valid in regions where mass is present.
Summary
-
In a region with mass density \(\rho\):
\begin{equation}
\begin{aligned}
\Delta\phi = 4\pi G \rho
\end{aligned}
\end{equation}
or:
\begin{equation}
\begin{aligned}
\nabla^{2}\phi = 4\pi G \rho
\quad\text{(Poisson equation)}
\end{aligned}
\end{equation}
-
In empty space (vacuum):
\begin{equation}
\begin{aligned}
\Delta\phi = 0
\end{aligned}
\end{equation}
or:
\begin{equation}
\begin{aligned}
\nabla^{2}\phi = 0
\quad\text{(Laplace equation)}
\end{aligned}
\end{equation}
Consideration
The presence of mass generates gravitational flux.
When you are inside a mass sphere and move outward, the amount of enclosed mass changes, and therefore the total flux also changes:
\begin{equation}
\begin{aligned}
\nabla^{2}\phi = 4\pi G \rho.
\end{aligned}
\end{equation}
When you are outside the mass sphere, the enclosed mass remains constant, and the total flux therefore remains constant:
\begin{equation}
\begin{aligned}
\nabla^{2}\phi = 0.
\end{aligned}
\end{equation}
Appendix 7.1 — Application of the Laplace Operator to the Gravitational Potential
In this chapter we apply the Laplace operator to the gravitational potential,
both outside and inside a static sphere.
The relevant formulas for the Newtonian potential are derived in:
- Appendix 7.1.1 — Outside a sphere
- Appendix 7.1.2 — Inside a sphere
Newtonian gravity
Gravity according to Newton:
\begin{equation}
\begin{aligned}
F = mg = \frac{GmM}{r^{2}}
\end{aligned}
\end{equation}
Gravitational field:
\begin{equation}
\begin{aligned}
g = \frac{GM}{r^{2}}
\end{aligned}
\end{equation}
Gravitational potential:
\begin{equation}
\begin{aligned}
\phi_{\text{newton}} = -\frac{GM}{r},
\qquad \text{where}\qquad
g = \frac{d\phi_{\text{newton}}}{dr}
\end{aligned}
\end{equation}
Here \(r\) is the distance from the center of the sphere, \(R\) is the radius of the sphere,
\(M\) is the mass of the sphere, and \(m\) is the mass of a test particle.
Gravitational potential in general relativity
Outside a sphere (see Chapter 2.8.7,
equation (210):
\begin{equation}
\begin{aligned}
\phi = g_{00}
= 1 - \frac{2GM}{c^{2}r}
= 1 + \frac{2\phi_{\text{newton}}}{c^{2}}
\end{aligned}
\end{equation}
Thus:
\begin{equation}
\begin{aligned}
\phi_{\text{newton, outside}}
= -\frac{GM}{r}
\end{aligned}
\end{equation}
Inside a sphere (see Appendix 7.1.4,
equation (107):
\begin{equation}
\begin{aligned}
\phi
= 1 - \frac{3GM}{c^{2}R}
+ \frac{GM}{c^{2}}\frac{r^{2}}{R^{3}}=
1+\frac{2}{c^2} \cdot \left(-\frac{3GM}{2R}
+ \frac{GM}{2}\frac{r^{2}}{R^{3}}\right)
\end{aligned}
\end{equation}
Newtonian limit:
\begin{equation}
\phi_{\text{newton, inside}}
= -\frac{3GM}{2R}
+ \frac{GM}{2}\frac{r^{2}}{R^{3}}
\label{eq:R32}
\end{equation}
Preparation: relation between \(r\) and Cartesian coordinates
\begin{equation}
\begin{aligned}
r^{2} = x^{2} + y^{2} + z^{2}
\end{aligned}
\end{equation}
Take the derivative with respect to \(x\):
\begin{equation}
\begin{aligned}
\frac{\partial r}{\partial x}
\quad\Rightarrow\quad
2r\,\frac{\partial r}{\partial x} = 2x
\end{aligned}
\end{equation}
Thus:
\begin{equation}
\begin{aligned}
\frac{\partial r}{\partial x} = \frac{x}{r}
\end{aligned}
\end{equation}
Appendix 7.1.1 — Outside a Sphere (Laplace)
The gravitational potential outside a sphere is:
\begin{equation}
\begin{aligned}
\phi_{\text{newton, buiten}} = -\frac{GM}{r}
\end{aligned}
\end{equation}
We now apply the Laplace operator:
\begin{equation}
\begin{aligned}
\nabla^{2}\phi
= \frac{\partial^{2}\phi}{\partial x^{2}}
+ \frac{\partial^{2}\phi}{\partial y^{2}}
+ \frac{\partial^{2}\phi}{\partial z^{2}}
\end{aligned}
\end{equation}
Since \(r = \sqrt{x^{2}+y^{2}+z^{2}}\), we use:
\begin{equation}
\begin{aligned}
\frac{\partial r}{\partial x} = \frac{x}{r},
\qquad
\frac{\partial r}{\partial y} = \frac{y}{r},
\qquad
\frac{\partial r}{\partial z} = \frac{z}{r}
\end{aligned}
\end{equation}
and:
\begin{equation}
\begin{aligned}
\frac{\partial \phi}{\partial r}
= \frac{d}{dr}\left(-\frac{GM}{r}\right)
= \frac{GM}{r^{2}}
\end{aligned}
\end{equation}
The full derivation follows in the next section.
Appendix 7.1.1 — Outside a Sphere (Laplace)
The Newtonian gravitational potential outside a sphere is:
\begin{equation}
\begin{aligned}
\phi_{\text{newton, buiten}} = -\frac{GM}{r}
\end{aligned}
\end{equation}
We use:
\begin{equation}
\begin{aligned}
r = \sqrt{x^{2} + y^{2} + z^{2}},
\qquad
\frac{\partial r}{\partial x} = \frac{x}{r}.
\end{aligned}
\end{equation}
First derivative with respect to x
\begin{equation}
\begin{aligned}
\frac{\partial \phi}{\partial x}
= \frac{\partial \phi}{\partial r}\,\frac{\partial r}{\partial x}
= \frac{GM}{r^{2}} \cdot \frac{x}{r}
= \frac{GM\,x}{r^{3}}
\end{aligned}
\end{equation}
Second derivative with respect to x
\begin{equation}
\begin{aligned}
\frac{\partial^{2}\phi}{\partial x^{2}}
= \frac{\partial}{\partial x}\left(\frac{GM\,x}{r^{3}}\right)
= GM\left(
-3\frac{x^{2}}{r^{5}} + \frac{1}{r^{3}}
\right)
\end{aligned}
\end{equation}
Analogously for \(y\) and \(z\):
\begin{equation}
\begin{aligned}
\frac{\partial^{2}\phi}{\partial y^{2}}
= GM\left(
-3\frac{y^{2}}{r^{5}} + \frac{1}{r^{3}}
\right),
\qquad
\frac{\partial^{2}\phi}{\partial z^{2}}
= GM\left(
-3\frac{z^{2}}{r^{5}} + \frac{1}{r^{3}}
\right)
\end{aligned}
\end{equation}
Sum over the three directions
\begin{equation}
\begin{aligned}
\Delta\phi
= \frac{\partial^{2}\phi}{\partial x^{2}}
+ \frac{\partial^{2}\phi}{\partial y^{2}}
+ \frac{\partial^{2}\phi}{\partial z^{2}}
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\Delta\phi
= GM\left[
-3\frac{x^{2}+y^{2}+z^{2}}{r^{5}}
+ 3\frac{1}{r^{3}}
\right]
\end{aligned}
\end{equation}
Since \(x^{2}+y^{2}+z^{2} = r^{2}\):
\begin{equation}
\begin{aligned}
\Delta\phi
= GM\left(
-3\frac{r^{2}}{r^{5}} + 3\frac{1}{r^{3}}
\right)
= GM\left(
-3\frac{1}{r^{3}} + 3\frac{1}{r^{3}}
\right)
= 0
\end{aligned}
\end{equation}
Thus, outside the sphere:
\begin{equation}
\begin{aligned}
\Delta\phi_{\text{newton}} = 0
\end{aligned}
\end{equation}
The gravitational potential satisfies the Laplace equation outside the mass.
Appendix 7.1.2 — Inside a Sphere (Poisson)
As derived above, for Cartesian coordinates inside a sphere:
\begin{equation}
\begin{aligned}
\frac{\partial r}{\partial x} = \frac{x}{r}.
\end{aligned}
\end{equation}
The Newtonian gravitational potential inside a homogeneous sphere is
(see equation (\ref{eq:R32}) in Appendix 7.1):
\begin{equation}
\begin{aligned}
\phi_{\text{newton, binnen}}
=
-\frac{3GM}{2R}
+
\frac{GM}{2}\frac{r^{2}}{R^{3}}.
\end{aligned}
\end{equation}
Derivatives with respect to \(x\)
First:
\begin{equation}
\begin{aligned}
\frac{\partial \phi}{\partial x}
=
\frac{\partial \phi}{\partial r}
\frac{\partial r}{\partial x}.
\end{aligned}
\end{equation}
Since:
\begin{equation}
\begin{aligned}
\frac{\partial \phi}{\partial r}
=
\frac{GM}{R^{3}}\, r,
\qquad
\frac{\partial r}{\partial x} = \frac{x}{r},
\end{aligned}
\end{equation}
it follows:
\begin{equation}
\begin{aligned}
\frac{\partial \phi}{\partial x}
=
\frac{GM}{R^{3}}\, r \cdot \frac{x}{r}
=
\frac{GM}{R^{3}}\, x.
\end{aligned}
\end{equation}
Second derivative:
\begin{equation}
\begin{aligned}
\frac{\partial^{2}\phi}{\partial x^{2}}
=
\frac{GM}{R^{3}}.
\end{aligned}
\end{equation}
The same holds for \(y\) and \(z\).
Thus:
\begin{equation}
\begin{aligned}
\Delta \phi_{\text{newton}}
=
\frac{\partial^{2}\phi}{\partial x^{2}}
+
\frac{\partial^{2}\phi}{\partial y^{2}}
+
\frac{\partial^{2}\phi}{\partial z^{2}}
=
3\,\frac{GM}{R^{3}}.
\end{aligned}
\end{equation}
Now use:
\begin{equation}
\begin{aligned}
M = \frac{4\pi}{3} R^{3}\rho
\quad\Rightarrow\quad
\frac{GM}{R^{3}} = \frac{4\pi G}{3}\rho.
\end{aligned}
\end{equation}
Thus:
\begin{equation}
\begin{aligned}
\Delta \phi_{\text{newton}}
=
3 \cdot \frac{4\pi G}{3}\rho
=
4\pi G \rho.
\end{aligned}
\end{equation}
Inside the sphere, the Newtonian potential therefore satisfies the Poisson equation:
\begin{equation}
\begin{aligned}
\boxed{
\Delta \phi_{\text{newton}} = 4\pi G \rho
}
\end{aligned}
\end{equation}
Relativistic potential
In the weak-field approximation:
\begin{equation}
\begin{aligned}
\phi = 1 + \frac{2\phi_{\text{newton}}}{c^{2}}.
\end{aligned}
\end{equation}
Therefore:
\begin{equation}
\begin{aligned}
\Delta \phi
=
\frac{2}{c^{2}}\, \Delta \phi_{\text{newton}}
=
\frac{2}{c^{2}}\, 4\pi G \rho
=
\frac{8\pi G}{c^{2}}\, \rho.
\end{aligned}
\end{equation}
Thus:
\begin{equation}
\begin{aligned}
\boxed{
\Delta \phi = \frac{8\pi G}{c^{2}}\, \rho
}
\end{aligned}
\end{equation}
This is precisely the weak-field limit of the Einstein equations.
Appendix 7.1.3 — Simplification of the Application of the Laplace/Poisson Operator
We consider a function \(f(r)\) to which the Laplace operator is applied.
The distance from the origin is:
\begin{equation}
\begin{aligned}
r^{2} = x^{2} + y^{2} + z^{2}.
\end{aligned}
\end{equation}
Gradient of \(f(r)\)
The gradient is:
\begin{equation}
\begin{aligned}
\vec{\nabla} f(r)
= \left(
\frac{\partial f}{\partial x},
\frac{\partial f}{\partial y},
\frac{\partial f}{\partial z}
\right).
\end{aligned}
\end{equation}
Since:
\begin{equation}
\begin{aligned}
\frac{\partial r}{\partial x} = \frac{x}{r},
\end{aligned}
\end{equation}
it follows:
\begin{equation}
\frac{\partial f(r)}{\partial x}
= \frac{df}{dr}\,\frac{\partial r}{\partial x}
= \frac{df}{dr}\,\frac{\vec{x}}{r}.
\label{eq:R65}
\end{equation}
Analogously for \(y\) and \(z\).
Therefore:
\begin{equation}
\begin{aligned}
\vec{\nabla} f(r)
= \frac{df}{dr}\left(
\frac{x}{r},\frac{y}{r},\frac{z}{r}
\right)=
\frac{df}{dr}\left(
\frac{\vec{x}}{r}+\frac{\vec{y}}{r}+\frac{\vec{z}}{r}
\right)
= \frac{df}{dr}\,\frac{\vec{r}}{r}
=\frac{df}{dr}\cdot\hat{r}.
\end{aligned}
\end{equation}
Second derivatives
Differentiate equation (\ref{eq:R65}) again with respect to \(x\):
\begin{equation}
\begin{aligned}
\frac{\partial^{2} f}{\partial x^{2}}
=
\frac{d^{2}f}{dr^{2}}\cdot\left(\frac{x}{r}\right)^{2}
+
\frac{df}{dr}\cdot\frac{1}{r}\cdot\left(
1 - \frac{x^{2}}{r^{2}}
\right).
\end{aligned}
\end{equation}
Analogously for \(y\) and \(z\).
Summing the three directions:
\begin{equation}
\begin{aligned}
\frac{\partial^{2} f}{\partial x^{2}}
+
\frac{\partial^{2} f}{\partial y^{2}}
+
\frac{\partial^{2} f}{\partial z^{2}}
=
\frac{d^{2}f}{dr^{2}}\cdot
\frac{x^{2}+y^{2}+z^{2}}{r^{2}}
+
\frac{df}{dr}\cdot\frac{1}{r}\cdot
\left(
3 - \frac{x^{2}+y^{2}+z^{2}}{r^{2}}
\right).
\end{aligned}
\end{equation}
Since \(x^{2}+y^{2}+z^{2} = r^{2}\), it follows:
\begin{equation}
\Delta f(r)
=
\frac{d^{2}f}{dr^{2}}
+
\frac{2}{r}\frac{df}{dr}.
\label{eq:R69}
\end{equation}
This is the well-known form of the Laplace operator in spherical symmetry.
Application to the general Newtonian potential
Take the general form:
\begin{equation}
\begin{aligned}
\phi_{\text{newton}} = L + K r^{n},
\end{aligned}
\end{equation}
where \(L\) and \(K\) are constants.
Then:
\begin{equation}
\begin{aligned}
\frac{d\phi}{dr} = n Kr^{\,n-1},
\qquad
\frac{d^{2}\phi}{dr^{2}} = n(n-1) Kr^{\,n-2}.
\end{aligned}
\end{equation}
Substituting into (\ref{eq:R69}):
\begin{equation}
\begin{aligned}
\Delta\phi
= n(n-1) Kr^{\,n-2}
+ \frac{2}{r} n Kr^{\,n-1}
&= n(n-1 + 2) Kr^{\,n-2}
\\ &\quad
= n(n+1) Kr^{\,n-2}
\label{eq:R72}
\end{aligned}
\end{equation}
This shows that:
- for \(n = -1\): \(\Delta\phi = 0\) (outside a sphere → Laplace)
- for \(n = 2\): \(\Delta\phi = 4\pi G\rho\) (inside a sphere → Poisson)
Relativistic correction
From:
\begin{equation}
\begin{aligned}
\phi = 1 + \frac{2\phi_{\text{newton}}}{c^{2}}
\end{aligned}
\end{equation}
it follows:
\begin{equation}
\begin{aligned}
\Delta\phi
= \frac{2}{c^{2}}\,\Delta\phi_{\text{newton}}
= \frac{2}{c^{2}}(4\pi G\rho)
= \frac{8\pi G\rho}{c^{2}}.
\end{aligned}
\end{equation}
Thus:
\begin{equation}
\begin{aligned}
\boxed{
\Delta\phi = \frac{8\pi G\rho}{c^{2}}
}
\end{aligned}
\end{equation}
Application to the gravitational potentials
1. Outside a sphere
\begin{equation}
\begin{aligned}
\phi_{\text{newton}} = -\frac{GM}{r}
\end{aligned}
\end{equation}
This corresponds to:
\begin{equation}
\begin{aligned}
n = -1, \qquad L = 0, \qquad K = -GM.
\end{aligned}
\end{equation}
Substituting into equation (72):
\begin{equation}
\begin{aligned}
\Delta\phi
= (-1)(-1+1)(-GM)\, r^{-3}
= 0 \cdot GM\, r^{-3}
= 0.
\end{aligned}
\end{equation}
Thus, outside the sphere:
\begin{equation}
\begin{aligned}
\Delta\phi_{\text{newton}} = 0.
\end{aligned}
\end{equation}
This is exactly the Laplace equation.
2. Inside a sphere
\begin{equation}
\begin{aligned}
\phi_{\text{newton}}
= -\frac{3GM}{2R}
+ \frac{GM}{2}\frac{r^{2}}{R^{3}}
\end{aligned}
\end{equation}
This corresponds to:
\begin{equation}
\begin{aligned}
n = +2,
\qquad
L = -\frac{3GM}{2R},
\qquad
K = \frac{GM}{2R^{3}}.
\end{aligned}
\end{equation}
Substituting into equation (72):
\begin{equation}
\begin{aligned}
\Delta\phi
= 2(2+1)\frac{GM}{2R^{3}} r^{\,2-2}
= 6\frac{GM}{2R^{3}}
= 3\frac{GM}{R^{3}}.
\end{aligned}
\end{equation}
Use:
\begin{equation}
\begin{aligned}
M = \frac{4\pi}{3}R^{3}\rho
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\Delta\phi
= 3G\frac{4\pi}{3}R^{3}\rho \frac{1}{R^{3}}
= 4\pi G\rho.
\end{aligned}
\end{equation}
Thus, inside the sphere:
\begin{equation}
\begin{aligned}
\Delta\phi_{\text{newton}} = 4\pi G\rho.
\end{aligned}
\end{equation}
This is exactly the Poisson equation.
Observation
From the general formula:
\begin{equation}
\begin{aligned}
\Delta\phi = n(n+1)K r^{\,n-2}
\end{aligned}
\end{equation}
it follows directly that:
- \(\Delta\phi = 0\) for \(n = 0\) (constant potential)
- \(\Delta\phi = 0\) for \(n = -1\) (point mass → outside the sphere)
- \(\Delta\phi \to 0\) for \(r \to \infty\) when \(n < 2\)
This is fully consistent with the physical interpretation of Laplace (no source) and Poisson (with source).
Appendix 7.1.4 — Derivation of the Gravitational Potential Inside a Static Sphere
We derive the gravitational potential inside a static, homogeneous sphere based on the
Poisson equation:
\begin{equation}
\begin{aligned}
\Delta \phi_{\text{newton}} = 4\pi G \rho.
\end{aligned}
\end{equation}
We take the general form:
\begin{equation}
\begin{aligned}
\phi_{\text{newton}} = L + K r^{n}.
\end{aligned}
\end{equation}
According to equation (72) from Appendix 7.1.3:
\begin{equation}
\begin{aligned}
\Delta \phi_{\text{newton}}
= n(n+1)K r^{\,n-2}.
\end{aligned}
\end{equation}
Substitution into the Poisson equation gives:
\begin{equation}
\begin{aligned}
4\pi G\rho = n(n+1)K r^{\,n-2}.
\end{aligned}
\end{equation}
Since the right-hand side must be independent of \(r\), it follows:
\begin{equation}
\begin{aligned}
n = 2.
\end{aligned}
\end{equation}
Thus:
\begin{equation}
\begin{aligned}
6K = 4\pi G\rho
\quad\Rightarrow\quad
K = \frac{2}{3}\pi G\rho.
\end{aligned}
\end{equation}
Using:
\begin{equation}
\begin{aligned}
\rho = \frac{3M}{4\pi R^{3}}
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
K = \frac{2}{3}\pi G \cdot \frac{3M}{4\pi R^{3}}
= \frac{1}{2}\frac{GM}{R^{3}}.
\end{aligned}
\end{equation}
The gravitational potential inside the sphere
\begin{equation}
\begin{aligned}
\phi_{\text{newton}}(r)
= L + \frac{1}{2}\frac{GM}{R^{3}} r^{2}.
\end{aligned}
\end{equation}
At the surface \(r = R\), the interior potential must match the exterior potential:
\begin{equation}
\begin{aligned}
\phi_{\text{newton}}(R)
= -\frac{GM}{R}
\end{aligned}
\end{equation}
Thus:
\begin{equation}
\begin{aligned}
-\frac{GM}{R}
= L + \frac{1}{2}\frac{GM}{R^{3}}R^{2}
= L + \frac{1}{2}\frac{GM}{R}.
\end{aligned}
\end{equation}
From this it follows:
\begin{equation}
\begin{aligned}
L = -\frac{3}{2}\frac{GM}{R}.
\end{aligned}
\end{equation}
The full potential inside the sphere then becomes:
\begin{equation}
\begin{aligned}
\phi_{\text{newton}}(r)
= -\frac{3}{2}\frac{GM}{R}
+ \frac{1}{2}\frac{GM}{R^{3}} r^{2}.
\end{aligned}
\end{equation}
Or more compactly:
\begin{equation}
\begin{aligned}
\boxed{
\phi_{\text{newton}}(r)
= -\frac{3GM}{2R}
+ \frac{GM}{2R^{3}}\, r^{2}
}
\end{aligned}
\end{equation}
Acceleration inside the sphere
From the derivative of the Newtonian potential it follows:
\begin{equation}
\begin{aligned}
g_r = \frac{d\phi_{\text{newton}}}{dr}
= \frac{GM}{R^{3}}\, r.
\end{aligned}
\end{equation}
Thus:
This is exactly the classical gravitational acceleration at the surface of a spherical mass.
Relativistic gravitational potential inside the sphere
The relation between the relativistic potential \(\phi\) and the Newtonian potential is:
\begin{equation}
\begin{aligned}
\phi = 1 + \frac{2\phi_{\text{newton}}}{c^{2}}.
\end{aligned}
\end{equation}
With:
\begin{equation}
\begin{aligned}
\phi_{\text{newton}}(r)
= -\frac{3GM}{2R}
+ \frac{GM}{2R^{3}} r^{2},
\end{aligned}
\end{equation}
it follows:
\begin{equation}
\begin{aligned}
\phi(r)
= 1
+ \frac{2}{c^{2}}
\left(
-\frac{3GM}{2R}
+ \frac{GM}{2R^{3}} r^{2}
\right).
\end{aligned}
\end{equation}
Expanding gives:
\begin{equation}
\begin{aligned}
\boxed{
\phi(r)
= 1
- \frac{3GM}{c^{2}R}
+ \frac{GM}{c^{2}R^{3}}\, r^{2}
}
\label{eq:R107}
\end{aligned}
\end{equation}
This is the relativistic time component \(g_{00}\) inside a homogeneous sphere in the weak-field limit.