Dot products in spherical polars? In the book classical mechanics, it said that since the three unit vectors $\hat r$, $\hat \theta$ and $\hat \phi$ are mutually prependicular, we can evaluate dot products in spherical polars in just the same way as in Cartesians.
If
$$a=a_r\hat r+a_\theta \hat \theta + a_\phi \hat \phi$$
and  $$b=b_r\hat r+b_\theta \hat \theta + b_\phi \hat \phi,$$
then
$$a\cdot b=a_rb_r+a_\theta b_\theta+a_\phi b_\phi$$
I can't understand why the mutually perpendicular can deduce it?
 A: The scalar product is bilinear. It means that if you take a vector $\overrightarrow{a} = A_1 \overrightarrow{a_1} + A_2 \overrightarrow{a_2}$ and $\overrightarrow{b} = B_1 \overrightarrow{b_1} + B_2 \overrightarrow{b_2}$, the scalar product will write:
\begin{align}
\overrightarrow{a} \cdot \overrightarrow{b} &= (A_1 \overrightarrow{a_1} + A_2 \overrightarrow{a_2}) \cdot (B_1 \overrightarrow{b_1} + B_2 \overrightarrow{b_2}) \\
&= A_1 B_1  \overrightarrow{a_1} \cdot \overrightarrow{b_1} +  A_1 B_2  \overrightarrow{a_1} \cdot \overrightarrow{b_2} + A_2 B_1  \overrightarrow{a_2} \cdot \overrightarrow{b_1} + A_2 B_2  \overrightarrow{a_2} \cdot \overrightarrow{b_2}.
\end{align}
Now what happens in your case is that $\overrightarrow{a}$ and $\overrightarrow{b}$ are decomposed over the same orthonormal basis (vectors have a unit length and are "mutually perpendicular"). Two vectors being "perpendicular" means nothing more than the fact that their dot product is $0$. So if you develop your expression to let appear all the terms just as I did in my example, you will find the usual $a_r b_r + a_{\theta} b_{\theta} + a_{\phi} b_{\phi}$ (just like in Cartesian coordinates), plus all the terms proportional to $\overrightarrow{e_r} \cdot \overrightarrow{e_{\theta}}$,  $\overrightarrow{e_r} \cdot \overrightarrow{e_{\phi}}$ and $\overrightarrow{e_{\theta}} \cdot \overrightarrow{e_{\phi}}$, that will be equal to $0$ because $\overrightarrow{e_r}, ~ \overrightarrow{e_{\theta}}$ and $\overrightarrow{e_{\phi}}$ are mutually perpendicular.
A: The result stems from the fact that the spherical coordinates are orthogonal (i.e., mutually perpendicular), which makes the unit vectors orthonormal, so we should have that
$$\mathrm{e}_i\cdot\mathrm{e}_j=\begin{cases}1&\text{for }i=j \\ 0&\text{otherwise}\end{cases}\tag{1}$$
where $\mathrm e_i$ is the unit vector.
Another way of seeing this is through another common definition of the dot product as the magnitude of the two vectors times the angle between them:
$$a\cdot b=\Vert a\Vert \,\Vert b\Vert \cos\delta
$$
Perpendicular (orthogonal) axes have $\delta=90^\circ$ while aligned axes have $\delta=0^\circ$, which means that,
$$
\mathrm{e}_i\cdot\mathrm{e}_j=\begin{cases}\cos0^\circ\equiv1&\text{for }i=j \\ \cos90^\circ\equiv0&\text{otherwise}\end{cases}
$$
which is the same as (1) above.
