# What is the inverse of the *divergence* operator?

The inverse of derivation is integral.

But what is the inverse of the divergence operator ? Doest it exist ?

• I think you're looking for the word "inverse" not "reciprocity". Mar 14, 2020 at 18:35
• @saulspatz : yes you are right. I correct the question Mar 14, 2020 at 21:39

The answer by Keith is close, except note that the divergence operator is not invertible, just like the derivative. It's "inverse" would also have some degrees of freedom.

In particular, when inverting the derivative $$F'=f$$, we have $$F(y)=\int_{x=0}^{y} f(x) dx +C$$.

If instead, we want to solve $$\nabla \cdot \boldsymbol{F}=f$$, we have $$\boldsymbol{F}(r)=\boldsymbol{F}_0(f)(r)+\boldsymbol{C}(r)$$ where $$\boldsymbol{F}_0$$ is an operator that takes $$f$$ and applies the Coulomb's integral.

$$\boldsymbol{C}$$ can be any function with a constant divergence of $$0$$. Written in unconstrained form $$\boldsymbol{C}=\nabla \times \boldsymbol{J}$$ for any vector field $$\boldsymbol{J}(r)$$.

• What is Coulomb’s integral? I can’t find very much information on it. Jul 24, 2020 at 1:27

For example, the charge density is proportional to the divergence of the electric field (Gauss's Law): $$\frac{\rho}{\varepsilon_0} = \nabla \cdot \mathbf{E}$$

The inverse is Coulomb's Law (for continuous charge distributions): $$\boldsymbol{E}(\boldsymbol{r}) = {1\over 4\pi\varepsilon_0}\int \rho(\boldsymbol{r'}) {\boldsymbol{r} - \boldsymbol{r'} \over |\boldsymbol{r} - \boldsymbol{r'}|^3} d^3r'$$ Wikipedia

• thank you. You almost answer to the question. "Divergence" is an operator, but what is the operator for the integral in 3 dimentions : there should be an "operator" in the same spirit as the "Divergence operator" ? Mar 14, 2020 at 21:41