What Is the Divergence of the Gravitational Field?

The divergence of a vector field measures its local net outward flow per unit volume. For the Newtonian gravitational field, the key result is that its divergence is proportional to the mass density, with a minus sign because gravity points inward toward positive mass.

This result is the differential form of Gauss’s law for gravity. It also clarifies an apparent puzzle: a point mass creates a gravitational field throughout space, but the divergence is zero at every point away from the mass itself.

The divergence equation

Let $\mathbf{g}$ denote gravitational acceleration and let $\rho$ be mass density. Newtonian gravity gives

$$\nabla\cdot\mathbf{g}=-4\pi G\rho,$$

where $G$ is the gravitational constant. The equation holds pointwise wherever the field and density are sufficiently regular; more generally, it holds in the distributional sense, including for idealized point masses. In a vacuum region, $\rho=0$, so

$$\nabla\cdot\mathbf{g}=0.$$

This does not mean the gravitational field is zero in vacuum. It means that a sufficiently small volume containing no mass has no net gravitational flux out through its boundary. Field lines can pass through the volume without beginning or ending there.

The sign follows from the direction of the field. Positive mass acts as a sink for gravitational field lines: the flux through a closed surface enclosing mass is inward, so it is negative when outward flux is taken as positive.

Deriving it from Gauss’s law

For any closed surface $S$ enclosing volume $V$, the gravitational version of Gauss’s law is

$$\oint_S \mathbf{g}\cdot d\mathbf{A}=-4\pi G M_{\text{enc}},$$

where $d\mathbf{A}$ points outward and $M_{\text{enc}}$ is the mass inside the surface. Since $M_{\text{enc}}=\int_V\rho\,dV$, the divergence theorem gives

$$\int_V \nabla\cdot\mathbf{g}\,dV =\oint_S \mathbf{g}\cdot d\mathbf{A} =-4\pi G\int_V\rho\,dV.$$

Because this holds for arbitrary volumes, the integrands must agree, giving $\nabla\cdot\mathbf{g}=-4\pi G\rho$.

The same relation follows from the gravitational potential $\Phi$. With the usual convention $\mathbf{g}=-\nabla\Phi$, the potential satisfies Poisson’s equation, $\nabla^2\Phi=4\pi G\rho$. Taking the divergence of $\mathbf{g}$ then gives the same result.

Worked example: a uniform-density sphere

Consider a sphere with constant density $\rho$. At a distance $r$ from its center, the mass enclosed is

$$M(r)=\frac{4}{3}\pi r^3\rho.$$

Inside the sphere, spherical symmetry gives the gravitational field

$$\mathbf{g}(r)=-\frac{G M(r)}{r^2}\hat{\mathbf{r}} =-\frac{4\pi G\rho}{3}r\hat{\mathbf{r}}.$$

Since $r\hat{\mathbf{r}}=\mathbf{r}$, this is a vector field of the form $\mathbf{g}=a\mathbf{r}$, where $a=-4\pi G\rho/3$. In three dimensions, $\nabla\cdot(a\mathbf{r})=3a$, so

$$\nabla\cdot\mathbf{g}=3\left(-\frac{4\pi G\rho}{3}\right)=-4\pi G\rho.$$

Outside the sphere, the field is that of a point mass at the center, $\mathbf{g}=-GM\hat{\mathbf{r}}/r^2$. Its divergence is zero at every exterior point. At the boundary, the density jumps and the classical divergence is not defined exactly on the surface. The field is continuous there, so there is no delta-function surface source; the equation holds piecewise and in the distributional sense.

What happens for a point mass?

A point mass $M$ at the origin has field

$$\mathbf{g}(\mathbf{r})=-\frac{GM}{r^2}\hat{\mathbf{r}}.$$

For a radial field $f(r)\hat{\mathbf{r}}$, the divergence at $r>0$ is $r^{-2}\,d(r^2f)/dr$. Here $r^2f=-GM$, a constant, so the divergence is zero wherever $r\ne0$.

At the origin the ordinary field formula is singular. In the distributional sense, the complete result is

$$\nabla\cdot\mathbf{g}=-4\pi GM\,\delta^{(3)}(\mathbf{r}),$$

where $\delta^{(3)}$ is the three-dimensional Dirac delta. This concentrated source accounts for the inward flux through every sphere surrounding the mass. Equivalently, the point mass has density $\rho=M\delta^{(3)}(\mathbf{r})$.

Common mistakes

• Saying the divergence is negative everywhere around a point mass. It is zero at every ordinary point with $r>0$; the source is concentrated at the mass in the distributional description.
• Treating zero divergence as zero field. A field can be nonzero and still have zero divergence locally, as the exterior field of a spherical mass demonstrates.
• Dropping the minus sign. Gauss’s law uses outward surface normals, while gravity from positive mass points inward, producing negative flux.
• Confusing the field with the potential. The potential obeys $\nabla^2\Phi=4\pi G\rho$; the acceleration field is $\mathbf{g}=-\nabla\Phi$, so its divergence has the opposite sign.