Appendix 7 — Derivation of the Laplace and Poisson Equations

A vector field for which the work required does not depend on the path between two points 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 the existence of a scalar potential \(\phi\) such that:

\[ \vec{F} = \vec{\nabla} \phi \tag{1} \]

The nabla operator is defined as:

\[ \vec{\nabla} = \frac{\partial}{\partial x}\,\hat{e}_x + \frac{\partial}{\partial y}\,\hat{e}_y + \frac{\partial}{\partial z}\,\hat{e}_z = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \tag{2} \]

The gravitational field \(\vec{F}_g\) is an example of a conservative field:

\[ \vec{F}_g = \vec{\nabla} \phi \tag{3} \]

According to Gauss’s theorem, for any closed surface:

\[ \oint_{\partial A} \vec{F}_g \cdot d\vec{A} = \iiint_V (\vec{\nabla} \cdot \vec{F}_g)\, dV \tag{4} \]

Vacuum: no mass, no source

In vacuum there is no mass, and therefore no source of gravitation:

\[ \vec{\nabla} \cdot \vec{F}_g = 0 \tag{5} \]

Substitute (3) into (5):

\[ \vec{\nabla} \cdot \vec{F}_g = 0 \quad\Rightarrow\quad \vec{\nabla} \cdot (\vec{\nabla} \phi) = 0 \tag{6} \]

Explicitly:

\[ \vec{\nabla} \cdot \vec{\nabla} \phi = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \cdot \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right) = 0 \tag{7} \]

Since \(x,y,z\) are orthogonal:

\[ \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} = 0 \]

This is written as:

\[ \nabla^2 \phi = 0 \quad\text{or}\quad \Delta \phi = 0 \tag{8} \]

The operator \(\nabla^2\), the Laplacian, is:

\[ \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \]

Thus, in vacuum the Laplace equation holds:

\[ \nabla^2 \phi = 0. \]

Inside a volume containing mass

Inside a mass distribution, there is a gravitational source. 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:

\[ \vec{F}_g = \frac{Gm}{r^{2}}\,\hat{r} \tag{9} \]

where \(\hat{r}\) is the radial unit vector.

Apply Gauss’s theorem again:

\[ \iiint_V (\vec{\nabla}\cdot\vec{F}_g)\, dV = \oint_{\partial A} \vec{F}_g\cdot d\vec{A} \tag{10} \]

Since \(\vec{F}_g = \vec{\nabla}\phi\), it follows:

\[ \iiint_V \Delta\phi\, dV = \oint_{\partial A} \frac{Gm}{r^{2}}\,\hat{r}\cdot d\vec{A} \]

For a spherical surface:

\[ A = 4\pi r^{2} \tag{11} \] \[ V = \frac{4\pi}{3} r^{3} \tag{12} \]

Since \(r\) is constant over the spherical surface:

\[ \oint_{\partial A} \frac{Gm}{r^{2}}\, dA = \frac{Gm}{r^{2}} \cdot 4\pi r^{2} = 4\pi G m \tag{13} \]

With the mass density:

\[ \rho = \frac{m}{V} \tag{14} \]

equation (13) becomes:

\[ \iiint_V \Delta\phi\, dV = 4\pi G m = \iiint_V 4\pi G \rho\, dV \]

Since the volume is arbitrary:

\[ \Delta\phi = 4\pi G \rho \tag{15} \]

This is the Poisson equation, valid in regions where mass is present.

Summary

• In a region with mass density \(\rho\):
  \[ \Delta\phi = 4\pi G \rho \tag{16} \] or: \[ \nabla^{2}\phi = 4\pi G \rho \quad\text{(Poisson equation)} \]
• In empty space (vacuum):
  \[ \Delta\phi = 0 \tag{17} \] or: \[ \nabla^{2}\phi = 0 \quad\text{(Laplace equation)} \]

Consideration

The presence of mass generates gravitational flux. When moving outward inside a massive sphere, the enclosed mass changes, and therefore the total flux changes:

\[ \nabla^{2}\phi = 4\pi G \rho. \]

When outside the mass sphere, the enclosed mass remains constant and so does the total flux:

\[ \nabla^{2}\phi = 0. \]

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 gravitation

Gravity according to Newton:

\[ F = mg = \frac{GmM}{r^{2}} \]

Gravitational field:

\[ g = \frac{GM}{r^{2}} \]

Gravitational potential:

\[ \phi_{\text{newton}} = -\frac{GM}{r}, \qquad \text{where}\qquad g = \frac{d\phi_{\text{newton}}}{dr} \]

Here \(r\) is the distance to the center of the sphere, \(R\) the radius of the sphere, \(M\) the mass of the sphere, and \(m\) the mass of a test particle.

Gravitational potential in general relativity

Outside a sphere (see Chapter 2.8, Equation 5):

\[ \phi = g_{00} = 1 - \frac{2GM}{c^{2}r} = 1 + \frac{2\phi_{\text{newton}}}{c^{2}} \]

Thus:

\[ \phi_{\text{newton, outside}} = -\frac{GM}{r} \tag{1} \]

Inside a sphere (see Appendix 7.1.4, Equation 3):

\[ \phi = 1 - \frac{3GM}{c^{2}R} + \frac{GM}{c^{2}}\frac{r^{2}}{R^{3}}= 1+\frac{2}{c^2} \cdot \left(-\frac{3GM}{2R} + \frac{GM}{2}\frac{r^{2}}{R^{3}}\right) \]

Newtonian limit:

\[ \phi_{\text{newton, inside}} = -\frac{3GM}{2R} + \frac{GM}{2}\frac{r^{2}}{R^{3}} \tag{2} \]

Preparation: relation between \(r\) and Cartesian coordinates

\[ r^{2} = x^{2} + y^{2} + z^{2} \]

Taking the derivative with respect to \(x\):

\[ \frac{\partial r}{\partial x} \quad\Rightarrow\quad 2r\,\frac{\partial r}{\partial x} = 2x \]

Thus:

\[ \frac{\partial r}{\partial x} = \frac{x}{r} \]

Appendix 7.1.1 — Outside a Sphere (Laplace)

The gravitational potential outside a sphere is:

\[ \phi_{\text{newton, outside}} = -\frac{GM}{r} \]

We now apply the Laplace operator:

\[ \nabla^{2}\phi = \frac{\partial^{2}\phi}{\partial x^{2}} + \frac{\partial^{2}\phi}{\partial y^{2}} + \frac{\partial^{2}\phi}{\partial z^{2}} \]

Since \(r = \sqrt{x^{2}+y^{2}+z^{2}}\), we use:

\[ \frac{\partial r}{\partial x} = \frac{x}{r}, \qquad \frac{\partial r}{\partial y} = \frac{y}{r}, \qquad \frac{\partial r}{\partial z} = \frac{z}{r} \]

and:

\[ \frac{\partial \phi}{\partial r} = \frac{d}{dr}\left(-\frac{GM}{r}\right) = \frac{GM}{r^{2}} \]

The full calculation is given in the next section.

Appendix 7.1.1 — Outside a Sphere (Laplace)

The Newtonian gravitational potential outside a sphere is:

\[ \phi_{\text{newton, outside}} = -\frac{GM}{r} \]

We use: \[ r = \sqrt{x^{2} + y^{2} + z^{2}}, \qquad \frac{\partial r}{\partial x} = \frac{x}{r}. \]

First derivative with respect to \(x\)

\[ \frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial r}\,\frac{\partial r}{\partial x} = \frac{GM}{r^{2}} \cdot \frac{x}{r} = \frac{GM\,x}{r^{3}} \]

Second derivative with respect to \(x\)

\[ \frac{\partial^{2}\phi}{\partial x^{2}} = \frac{\partial}{\partial x}\left(\frac{GM\,x}{r^{3}}\right) = GM\left( -3\frac{x^{2}}{r^{5}} + \frac{1}{r^{3}} \right) \]

Analogously for \(y\) and \(z\):

\[ \frac{\partial^{2}\phi}{\partial y^{2}} = GM\left( -3\frac{y^{2}}{r^{5}} + \frac{1}{r^{3}} \right), \qquad \frac{\partial^{2}\phi}{\partial z^{2}} = GM\left( -3\frac{z^{2}}{r^{5}} + \frac{1}{r^{3}} \right) \]

Sum over the three directions

\[ \Delta\phi = \frac{\partial^{2}\phi}{\partial x^{2}} + \frac{\partial^{2}\phi}{\partial y^{2}} + \frac{\partial^{2}\phi}{\partial z^{2}} \]

\[ \Delta\phi = GM\left[ -3\frac{x^{2}+y^{2}+z^{2}}{r^{5}} + 3\frac{1}{r^{3}} \right] \]

Since \(x^{2}+y^{2}+z^{2} = r^{2}\):

\[ \Delta\phi = GM\left( -3\frac{r^{2}}{r^{5}} + 3\frac{1}{r^{3}} \right) = GM\left( -3\frac{1}{r^{3}} + 3\frac{1}{r^{3}} \right) = 0 \]

Thus, outside the sphere:

\[ \Delta\phi_{\text{newton}} = 0 \]

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:

The Newtonian gravitational potential inside a homogeneous spherical mass is (see Equation 2 in Appendix 7.1):

Derivatives with respect to \(x\)

First:

Since:

it follows:

Second derivative:

The same holds for \(y\) and \(z\). Thus:

Now use:

Therefore:

Inside the sphere, the Newtonian potential therefore satisfies the Poisson equation:

Relativistic potential

In the weak-field approximation:

Therefore:

Thus:

This is exactly 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 to the origin is:

\[ r^{2} = x^{2} + y^{2} + z^{2}. \]

Gradient of \(f(r)\)

The gradient is:

\[ \vec{\nabla} f(r) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right). \]

Since: \[ \frac{\partial r}{\partial x} = \frac{x}{r}, \] it follows:

\[ \frac{\partial f(r)}{\partial x} = \frac{df}{dr}\,\frac{\partial r}{\partial x} = \frac{df}{dr}\,\frac{\vec{x}}{r}. \tag{1} \]

Analogously for \(y\) and \(z\). Therefore:

\[ \vec{\nabla} f(r) = \frac{df}{dr}\left( \frac{x}{r},\frac{y}{r},\frac{z}{r} \right)= \frac{df}{dr}\,\frac{\vec{r}}{r} =\frac{df}{dr}\cdot\hat{r}. \]

Second derivatives

Differentiate Equation (1) again with respect to \(x\):

\[ \frac{\partial^{2} f}{\partial x^{2}} = \frac{d^{2}f}{dr^{2}}\cdot\left(\frac{x}{r}\right)^{2} + \frac{df}{dr}\cdot\frac{1}{r}\cdot\left( 1 - \frac{x^{2}}{r^{2}} \right). \]

Analogously for \(y\) and \(z\). Summing over the three directions:

\[ \frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}} = \frac{d^{2}f}{dr^{2}}\cdot \frac{x^{2}+y^{2}+z^{2}}{r^{2}} + \frac{df}{dr}\cdot\frac{1}{r}\cdot \left( 3 - \frac{x^{2}+y^{2}+z^{2}}{r^{2}} \right). \]

Since \(x^{2}+y^{2}+z^{2} = r^{2}\), it follows:

\[ \Delta f(r) = \frac{d^{2}f}{dr^{2}} + \frac{2}{r}\frac{df}{dr}. \tag{2} \]

This is the well-known form of the Laplace operator in spherical symmetry.

Application to the general Newtonian potential

Take the general form:

\[ \phi_{\text{newton}} = L + K r^{n}, \tag{3} \]

where \(L\) and \(K\) are constants.

Then:

\[ \frac{d\phi}{dr} = n Kr^{\,n-1}, \qquad \frac{d^{2}\phi}{dr^{2}} = n(n-1) Kr^{\,n-2}. \]

Substitute into (2):

\[ \Delta\phi = n(n-1) Kr^{\,n-2} + \frac{2}{r} n Kr^{\,n-1} = n(n+1) Kr^{\,n-2} \tag{4} \]

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: \[ \phi = 1 + \frac{2\phi_{\text{newton}}}{c^{2}} \] it follows:

\[ \Delta\phi = \frac{2}{c^{2}}\,\Delta\phi_{\text{newton}} = \frac{8\pi G\rho}{c^{2}}. \]

Thus:

\[ \boxed{ \Delta\phi = \frac{8\pi G\rho}{c^{2}} } \]

Application to the gravitational potentials

1. Outside a sphere

\[ \phi_{\text{newton}} = -\frac{GM}{r} \]

This corresponds to:

\[ n = -1, \qquad L = 0, \qquad K = -GM. \]

Substitution into (4):

\[ \Delta\phi = 0. \]

Thus, outside the sphere:

\[ \Delta\phi_{\text{newton}} = 0. \]

This is exactly the Laplace equation.

2. Inside a sphere

\[ \phi_{\text{newton}} = -\frac{3GM}{2R} + \frac{GM}{2}\frac{r^{2}}{R^{3}} \]

This corresponds to:

\[ n = +2, \qquad L = -\frac{3GM}{2R}, \qquad K = \frac{GM}{2R^{3}}. \]

Substitution into (4):

\[ \Delta\phi = 4\pi G\rho. \]

Thus, inside the sphere:

\[ \Delta\phi_{\text{newton}} = 4\pi G\rho. \]

This is exactly the Poisson equation.

Observation

From the general formula:

\[ \Delta\phi = n(n+1)K r^{\,n-2} \]

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 perfectly matches 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:

\[ \Delta \phi_{\text{newton}} = 4\pi G \rho. \]

We assume the general form:

\[ \phi_{\text{newton}} = L + K r^{n}. \]

According to Equation (4) from Appendix 7.1.3, we have:

\[ \Delta \phi_{\text{newton}} = n(n+1)K r^{\,n-2}. \tag{2} \]

Substitution into the Poisson equation yields:

\[ 4\pi G\rho = n(n+1)K r^{\,n-2}. \]

Since the right-hand side must be independent of \(r\), it follows that:

\[ n = 2. \]

Thus:

\[ 6K = 4\pi G\rho \quad\Rightarrow\quad K = \frac{2}{3}\pi G\rho. \]

Using: \[ \rho = \frac{3M}{4\pi R^{3}} \]

\[ K = \frac{2}{3}\pi G \cdot \frac{3M}{4\pi R^{3}} = \frac{1}{2}\frac{GM}{R^{3}}. \]

The gravitational potential inside the sphere

\[ \phi_{\text{newton}}(r) = L + \frac{1}{2}\frac{GM}{R^{3}} r^{2}. \]

At the surface \(r = R\), the interior potential must match the exterior potential:

\[ \phi_{\text{newton}}(R) = -\frac{GM}{R} \tag{1} \]

Thus:

\[ -\frac{GM}{R} = L + \frac{1}{2}\frac{GM}{R^{3}}R^{2} = L + \frac{1}{2}\frac{GM}{R}. \]

This implies:

\[ L = -\frac{3}{2}\frac{GM}{R}. \]

The complete potential inside the sphere then becomes:

\[ \phi_{\text{newton}}(r) = -\frac{3}{2}\frac{GM}{R} + \frac{1}{2}\frac{GM}{R^{3}} r^{2}. \]

Or, more compactly:

\[ \boxed{ \phi_{\text{newton}}(r) = -\frac{3GM}{2R} + \frac{GM}{2R^{3}}\, r^{2} } \tag{2} \]

Acceleration inside the sphere

From the derivative of the Newtonian potential it follows that:

\[ g_r = \frac{d\phi_{\text{newton}}}{dr} = \frac{GM}{R^{3}}\, r. \]

Thus:

• At \(r = 0\): \[ g_r = 0. \]
• At \(r = R\): \[ g_r = \frac{GM}{R^{2}}. \]

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:

\[ \phi = 1 + \frac{2\phi_{\text{newton}}}{c^{2}}. \]

With: \[ \phi_{\text{newton}}(r) = -\frac{3GM}{2R} + \frac{GM}{2R^{3}} r^{2}, \] it follows that:

\[ \phi(r) = 1 + \frac{2}{c^{2}} \left( -\frac{3GM}{2R} + \frac{GM}{2R^{3}} r^{2} \right). \]

Evaluating this expression gives:

\[ \boxed{ \phi(r) = 1 - \frac{3GM}{c^{2}R} + \frac{GM}{c^{2}R^{3}}\, r^{2} } \tag{3} \]

This is the relativistic time component \(g_{00}\) inside a homogeneous sphere in the weak-field limit.