Verifying that the divergence of a gravitational field is zero

Question

This is an example from the book Calculus: a complete course, of Adams and Essex. There're several reasonings presented in the solution I don't really understand.

EXAMPLE 3. Ch 17.1

Verify that the vector field $\mathbf{F} = \frac{m \mathbf{r}}{|\mathbf{r}|^3}$, due to a source of strength $m$ at $(0, 0, 0)$, has zero divergence at all points in $\mathbb{R}^3$ except the origin.

$\textbf{Solution}$
Since $ \mathbf{F}(x, y, z) = \frac{m}{r^3} \left( x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \right), $ where $ r^2 = x^2 + y^2 + z^2, $
and since $\partial r/\partial x = x/r$, we have
$ \frac{\partial F_1}{\partial x} = m \frac{\partial}{\partial x} \left(\frac{x}{r^3} \right) = m \frac{r^3 - 3xr^2 \left(\frac{x}{r}\right)}{r^6} = m \frac{r^2 - 3x^2}{r^5}. $
Similarly, $ \frac{\partial F_2}{\partial y} = m \frac{r^2 - 3y^2}{r^5} $ and $ \frac{\partial F_3}{\partial z} = m \frac{r^2 - 3z^2}{r^5}. $
Adding these up, we get $ \nabla \cdot \mathbf{F}(x, y, z) = 0 $ if $ r > 0 $.

$\textbf{My questions}$

1. Why is the absolute sign $|\mathbf{r}|^3$ in the denominator removed in the solution?
2. Where does $ r^2 = x^2 + y^2 + z^2$ come from?
3. Why is $\partial r/\partial x = x/r$?
4. Can this vector field $\mathbf{F} = \frac{m \mathbf{r}}{|\mathbf{r}|^3}$ expressed in terms of $\mathbf{F}=F_1\vec{i}+F_2\vec{j}+F_3\vec{k}$, so that $\frac{\partial F_1}{\partial x}$ is more easily calculated?

Answer 1

Before answering your questions in turn, it's important to know the following.

In spherical coordinates, we have a scalar $r$ and a vector $\mathbf{r}$. The vector is defined

$$\mathbf{r}=x \,\vec{i}+y\,\vec{j} + z\,\vec{k}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\,.$$

I will use a bold face for a vector instead of the over-arrows from now on.

The scalar $r$ is defined as $r=|\mathbf{r}|$ and the absolute value sign is defined as $|\mathbf{r}|=\sqrt{x^2+y^2+z^2}$.

Now, the four questions.

1. In the solution it looks like it is written $\frac{1}{r^3}$. Notice that this is the scalar $r$, rather than the vector $\mathbf{r}$. Recalling what I write at the top, we see that $r=|\mathbf{r}|$. So you can either write $\frac{1}{r^3}$ or $\frac{1}{|\mathbf{r}|^3}$ but never $\frac{1}{\mathbf{r}^3}$.

2. $r^2=x^2+y^2+z^2$ is Pythagoras' theorem! If we consider a general vector $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$, it's absolute value squared is defined as $|\mathbf{a}|^2=(a_1)^2+(a_2)^2+(a_3)^2$. Then, remembering $\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ and substituting $a_1=x$, $a_2=y$, and $a_3=z$, we arrive at $r^2=|\mathbf{r}|^2=x^2+y^2+z^2$.

3. The origin of $\frac{\partial r}{\partial x}$. We start with $$r=\sqrt{x^2+y^2+z^2}=(x^2+y^2+z^2)^{1/2}.$$ To calculate the partial derivative with respect to $x$, we need to use the chain rule and the power rule. The power rule that we need says $\frac{\mathrm{d}}{\mathrm{d}x}x^{1/2}=\frac{1}{2}x^{-1/2}$. Then, using the chain rule, we have $$\frac{\partial r}{\partial x}=\frac{\partial}{\partial x}(x^2+y^2+z^2)^{1/2}=\left(\frac{\partial}{\partial x}\left(x^2+y^2+z^2\right)\right)\frac{1}{2}(x^2+y^2+z^2)^{-1/2}.$$ The term in brackets comes from the chain rule and performing this derivative, we see this factor equals $2x$ (the $2$ and the $1/2$ cancel). Putting this together, we see $$\frac{\partial r}{\partial x}=x(x^2+y^2+z^2)^{-1/2}=\frac{x}{\sqrt{x^2+y^2+z^2}}.$$ We recall now that $r=\sqrt{x^2+y^2+z^2}$, which is the factor in the denominator! This results in $$\frac{\partial r}{\partial x}=\frac{x}{r}.$$

4. You can always express a vector $\mathbf{F}$ in terms of the basis vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ as $\mathbf{F}=F_1\mathbf{i}+F_2\mathbf{j}+F_3\mathbf{k}$. As we see in this case, $F_1=\frac{m}{r^3}x$, $F_2=\frac{m}{r^3}y$, and $F_3=\frac{m}{r^3}z$. When calculating a quantity such as $\frac{\partial F_1}{\partial x}$, you will need to use the definition $r=\sqrt{x^2+y^2+z^2}$ as $r$ also depends on $x$!