# Gauss's law: divergence and mass density

I'm watching Leonard Susskind's lectures on relativity (http://www.youtube.com/watch?v=s8UrYIZhm60&feature=youtu.be&t=31m08s), and he just introduced Gauss's law on gravity, that is:

$$\vec\nabla \cdot \vec A = -4\pi G \rho$$

To my understanding, this says that divergence at point $(x,y,z)$ is non-zero only if there is mass at that point, i.e. $\rho(x,y,z) \ne 0$.

However, this seems in conflict with the idea of divergence that I got from the definition. If $\vec A(x,y,z)$ is non-zero, and the vector field around $(x,y,z)$ is not constant, then the sum of the partial derivatives will be non-zero, regardless of whether there is any mass at $(x,y,z)$.

Is something wrong in my reasoning or am I not getting Gauss's law right?

• If the derivatives are non-zero, and there is a non-zero divergence, then there is a density at that point. The geometric understanding of divergence is related to density via Gauss' law. Dec 9, 2016 at 14:15
• Why? If I have a single mass in space at $(0,0,0)$, the acceleration vectors at $(x,y,z)$ and $(x,y,z+dz)$ will point in slightly different directions, thus the derivatives will be non-zero, but there's no density there.
– rand
Dec 9, 2016 at 14:31

The flaw with your reasoning is that the vector field being not constant around $(x,y,z)$ does not necessarily imply that the divergence is not $0$ at that point, although what you said does hold the other way around. (Think back on the definition of divergence.) For example, $(y^2,z^2,x^2)$ is not a constant vector field, but it's divergence is indeed $0$. The equation $∇⃗ ⋅A =−4πGρ$ actually arises from the integral identity $$\int_V\nabla\cdot\vec{F} dV=\oint_S\vec{F}\cdot d\vec{a}$$ which is the integral form of Gauss law in mathematics. Putting in the gravity field $\vec{A}$ due to mass distribution $\rho$ into the formula, and using the fact that it should hold anywhere, you get $∇⃗ ⋅A =−4πGρ$