Divergence of a Separation Vector
Formula
$$ \begin{align*} \nabla \cdot \left( \dfrac{1}{r^2}\hat{ \mathbf{r} } \right) =&\ 4\pi \delta^3(\mathbf{r}) \\[1em] \nabla \cdot \left( \dfrac{1}{\cR^{2}} \crH \right) =&\ 4\pi \delta^3(\bcR) \\[1em] \nabla^2 \left(\dfrac{1}{\cR} \right) =&\ -4\pi \delta^3 ( \bcR ) \end{align*} $$
Here, $\mathbf{r}$ is the position vector, and $\bcR$ is the separation vector.
Explanation
Let’s assume there is a vector function $\mathbf{v} = \dfrac{1}{r^2}\hat{\mathbf{r}}$. Its magnitude is inversely proportional to the square of the distance, and its direction is the direction of the radius. Now, let’s calculate the divergence of this function. Using the formula for the gradient in spherical coordinates,
$$ \nabla \cdot \mathbf{v} = \dfrac{1}{r^2}\dfrac{\partial}{\partial r}\left( r^2\dfrac{1}{r^2} \right) = \dfrac{1}{r^2}\dfrac{\partial}{\partial r}(1)=0 $$
However, if we apply the divergence theorem and calculate,
$$ \int_\mathcal{V} \nabla \cdot \mathbf{v} d\tau = \oint _{S} \mathbf{v} \cdot d \mathbf{a} $$
Assume we integrate over a sphere with the center at the origin and the radius $R$.
$$ \begin{align*} \int_\mathcal{V} \nabla \cdot \mathbf{v} d\tau &= \oint _{S} \mathbf{v} \cdot d \mathbf{a} \\ &= \int \left(\dfrac{1}{R^2}\hat{\mathbf{r}} \right) \cdot \left( R^2 \sin \theta d\theta d \phi \hat{\mathbf{r}} \right) \\ &= \int \sin \theta d\theta d\phi \\ &= \left( \int _{0} ^\pi \sin \theta d\theta \right) \left( \int _{0} ^2\pi d\phi \right) \\ &= 4\pi \end{align*} $$
According to the calculation above, since $\nabla \cdot \mathbf{v}=0$, integrating should also result in $0$. However, the result calculated following the divergence theorem is $4\pi$. There is clearly a problem somewhere. The problematic area is precisely where $r=0$ is. In $r=0$, the value of $\mathbf{v}=\dfrac{1}{r^2}\hat{\mathbf{r}}$ goes to infinity. Initially, $\dfrac{1}{r^2}$ doesn’t exist at $r=0$. The fact that $\nabla \cdot \mathbf{v}=0$ means the value is $0$ everywhere that is $r\ne 0$. But since the result of integrating, including the origin, is $4\pi$, this integral result comes solely from the point where $r=0$. To solve this problem, the Dirac Delta function was introduced. ${}\\ {}$
- In three dimensions, and
- $0$ where the value is not $0$ at all other locations, and
- To make the integral value over the whole area including the origin $4\pi$,
$$ \nabla \cdot \left( \dfrac{1}{r^2}\hat{ \mathbf{r} } \right) = 4\pi \delta^3(\mathbf{r}) $$
Generally, when represented for the separation vector,
$$ \nabla \cdot \left( \dfrac{1}{\cR^2}\crH \right) = 4\pi \delta^3(\bcR) \tag{1} $$
Since the gradient of the separation vector](../142) is $\nabla \left( \dfrac{1}{\cR} \right) = -\dfrac{1}{\cR^2}\crH$, substituting it into $(1)$ results in
$$ \begin{align*} && \nabla \cdot \left[ -\nabla \left(\dfrac{1}{\cR} \right) \right] &= 4\pi \delta^3 ( \bcR ) \\ \implies && \nabla \cdot \left[ \nabla \left(\dfrac{1}{\cR} \right) \right] &= -4\pi \delta^3 ( \bcR ) \\ \implies && \nabla^2 \left(\dfrac{1}{\cR} \right) &= -4\pi \delta^3 ( \bcR ) \end{align*} $$