How to calculate divergence of the given function?

Question

The vector function is:

$$\mathbf{v}=\frac{1}{r^2}\hat{\mathbf{r}}$$

$r$ is the magnitude of position vector and

$\hat{\mathbf{r}}$ is the unit vector along the position vector

Now divergence will be

$$\nabla \cdot \mathbf{v}={\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)}\cdot \mathbf{v}$$

How is this evaluated?

Answer 1

Without switching coordinate systems, this is my favorite method, since it breaks down the identity into small pieces.

Let $\mathbf{r} = x\mathbf{i} + y \mathbf{j} + z \mathbf{k}$, and $r = \sqrt{x^2 + y^2 + z^2}$. Notice that \begin{align*} \mathbf{v} &= \frac{\mathbf{r}}{r^3}\\ \mathbf{r}\cdot\mathbf{r} &= r^2 \\ \nabla r &= \frac{\mathbf{r}}{r} \\ \nabla \cdot \mathbf{r} &= 3 \\ \end{align*} We can use the product rule for the divergence, and the power rule for the gradient: \begin{align*} \nabla \cdot \mathbf{v} &= \nabla\cdot(r^{-3} \mathbf{r}) \\ &= (\nabla r^{-3}) \cdot \mathbf{r} + r^{-3} \nabla \cdot \mathbf{r} \\ &= (-3)r^{-4}\nabla r \cdot \mathbf{r} + 3 r^{-3} \\ &= (-3)r^{-4}(r^{-1}\mathbf{r})\cdot \mathbf{r} + 3 r^{-3} \\ &= (-3)r^{-5}\mathbf{r}\cdot\mathbf{r} + 3r^{-3} \\ &= (-3)r^{-5}r^2 + 3r^{-3} \\ &= (-3)r^{-3} + 3r^{-3} = 0 \end{align*}

Answer 2

Note, in Cartesian coordinates $r = \sqrt{x^2+y^2+z^2}$ and $\mathbf{r} = x\hat{\mathbf{i}}+y\hat{\mathbf{j}}+z\hat{\mathbf{k}}$.

So, we have $\mathbf{v} = \left( \frac{1}{r^2} \mathbf{\hat{r}} \right)$. It might be instructve to write the unit vector $\mathbf{\hat{r}} = \frac{\mathbf{\mathbf{r}}}{r}$. This is the definition of the unit vector.

Putting that together we have

$$\mathbf{v} = \left( \frac{\mathbf{r}}{r^3} \right), $$

which is equivalent to

$$\mathbf{v} = \frac{x}{(x^2+y^2+z^2)^{3/2}}\hat{\mathbf{i}}+ \frac{y}{(x^2+y^2+z^2)^{3/2}}\hat{\mathbf{j}}+\frac{z}{(x^2+y^2+z^2)^{3/2}}\hat{\mathbf{k}}.$$

The divergence is then given by

$$ \nabla \cdot \mathbf{v} = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z} \right) \cdot \left( \frac{x}{(x^2+y^2+z^2)^{3/2}}, \frac{y}{(x^2+y^2+z^2)^{3/2}}, \frac{z}{(x^2+y^2+z^2)^{3/2}} \right).$$

Can you take it from here?

As some of the comments mention, using spherical coordinates is far easier in this example.