# Gauss' Law for Magnetism Derivative Form: With or without volume integral?

I've been reading through FLP Vol. II, and he has proven that as the flux through a closed surface is: $\ \int_{surface} \mathbf{F} \space \mathrm{d}\mathbf{a}$, according to the divergence theorem, the flux through a surface can be defined as: $\ \int_{volume} \nabla \cdot \mathbf{F} \space \mathrm{d}V$, where $\ \mathbf{F}$ is any vector field, and the volume is that which is enclosed by the surface.

Previously he had stated as a word equation that: $\ \text{Flux of } \mathbf{B} \text{ through any closed surface}=0.$ I would therefore assume that $\ \int_{volume} \nabla \cdot \mathbf{B} \space \mathrm{d}V = 0$, however Gauss' law for magnetism states that: $\ \nabla \cdot \mathbf{B} = 0$. Does that mean that $\ \nabla \cdot \mathbf{B} = 0$ and $\ \int_{volume} \nabla \cdot \mathbf{B} \space \mathrm{d}V = 0$ are equivalent statements, or am I making a fundamental error somewhere?

• They're equivalent alright. This is exactly how you go between the integral and differential formulations of Maxwell's laws. – David H Apr 6 '14 at 14:45

Intuitively, if the volume integral of a function is 0 over any arbitrary volume, the function itself must be 0 at all points in space.

More concretely, consider a function for which $\int_V \, f \, \mathrm{d}x = 0$ for any volume $V$. Then, $\int_{V+dV} \, f \, \mathrm{d}x = 0$ for any infinitesimal addition to V.

$$\int_{V+dV} \, f \, \mathrm{d}x - \int_V \, f \, \mathrm{d}x = \int_{dV} \, f \, \mathrm{d}x = f(\text{at dV}) = 0$$

In your case, $f = \nabla \cdot B$, so $\nabla \cdot B = 0$.

(Note: I was a bit lazy with my notation above, so it's not a formal proof. However, it should still provide the intuitive answer to your question.)

As Draksis said, the condition is that the integral over any volume has to be zero. If you want a formal proof, here you go:

Let's call $f(\mathbf{x}) = \nabla \cdot \mathbf{B}$, and assume it is continuous. Suppose there's some $\mathbf{x}_0 \in \mathbb{R}^3$ with $f(\mathbf{x}_0) \neq 0$, and let's say that $f(\mathbf{x}_0) > 0$ (the proof for $f < 0$ is identical). Then since $f$ is a continous function, there is some ball $B$ around $\mathbf{x}_0$ where $f$ is positive. Therefore, $\int_B f > 0$, which is a contradiction with the assumption that the integral should be zero for any volume.