Dot product in spherical coordinates

[FrankJ777]

TL;DR

I'm trying to understand how to take the dot product in spherical coordinates, if that is indeed what I'm supposed to be doing here?

I'm learing about antennas in a course, and we are using Jin's Electromagnetic text.
This isn't a homework problem, I'm just trying to understand what I'm supposed to do in this situation.

This part of the text discusses how to evaluate a radiation pattern.
One of the steps to evaluate the pattern is we have to find an expression for r'cosψ. (See diagram below.)
It says to find the expression, take the dot product between the vector r' and unit vector r.
Like so...r'⋅ \hat r
It gives the formula:
r' cos ψ = \textbf{r'} ⋅ \hat r = \textbf{r'} ⋅ \hat x cosθsinφ + \textbf{r'} ⋅ \hat y sinθsinφ + \textbf{r'} ⋅ \hat z cosθ

I'm confused about what I'm supposed to do, and what I'm supposed to get as an output.

I see that, that expression is the identity for the r component when converting from cartesian to spherical, I'm not sure what's going on.
Am I supposed to convert r' and r_hat to cartesian then take the dot product?

What are the components. \hat x cosθsinφ \hat y sinθsinφ \hat z cosθ
Do I find the θ and φ of r_hat?

I appreciate any help. I'm very confused.

 

[PeroK]

A unit vector $\hat n$ defined by the polar angle $\theta$ and azimuthal angle $\varphi$ is:
$$\hat n = \sin \theta\cos \varphi \ \hat x + \sin \theta\sin \varphi \ \hat y + \cos \theta \hat z$$
Note that this is the "physics" convention for $\theta, \varphi$. The "maths" convention uses these angles the other way round. See, for example:
https://en.wikipedia.org/wiki/Spherical_coordinate_system

 

[vanhees71]

Let's work with a Cartesian basis system $\vec{e}_j$. The spherical coordinate system is then usually introduced by choosing $\vec{e}_3$ as the polar axis. The position vector is parametrized by spherical coordinates $(r,\vartheta,\varphi)$ as
$$\vec{x}=\begin{pmatrix} r \sin \vartheta \cos \varphi \\ r \sin \vartheta \sin \varphi \\ r \cos \vartheta \end{pmatrix}.$$
Where here and in the following all column vectors are referring to the Cartesian coordinates.

The holonomic basis of the spherical coordinates are given by
$$\vec{b}_r = \partial_r \vec{x}=\begin{pmatrix} \sin \vartheta \cos \varphi \\ \sin \vartheta \sin \varphi \\ \cos \vartheta \end{pmatrix},$$
$$\vec{b}_{\vartheta} = \partial_{\vartheta} \vec{x} =r \begin{pmatrix} \cos \vartheta \cos \varphi \\ \cos \vartheta \sin \varphi \\ -\sin \vartheta \end{pmatrix},$$
$$\vec{b}_{\varphi}=\partial_{\varphi} \vec{x} = r \begin{pmatrix} -\sin \vartheta \sin \varphi \\ \sin \vartheta \cos \varphi \\ 0\end{pmatrix}.$$
It's easy to see by takting the scalar products that these vectors are pairwise orthogonal and thus one usually uses a normalized basis, i.e., in each point there's a Cartesian basis spanned by tangent vectors on the coordinate lines. These vectors are
$$\vec{e}_r=\vec{b}_r=\begin{pmatrix} \sin \vartheta \cos \varphi \\ \sin \vartheta \sin \varphi \\ \cos \vartheta \end{pmatrix},$$
$$\vec{e}_{\vartheta}=\vec{b}_{\vartheta}/r=\begin{pmatrix} \cos \vartheta \cos \varphi \\ \cos \vartheta \sin \varphi \\ -\sin \vartheta \end{pmatrix},$$
$$\vec{e}_{\varphi}=\vec{b}_{\varphi}/(r \sin \vartheta) = \begin{pmatrix} -\sin \varphi \\ \cos \varphi \\ 0\end{pmatrix}.$$
In this order they build a right-handed Cartesian basis, $\vec{e}_r \times \vec{e}_{\vartheta}=\vec{e}_{\varphi}$ (and cyclic).

Now you can express any vector field with components wrt. this basis
$$\vec{V}(\vec{r})=V_r(r,\vartheta,\varphi) \vec{e}_r + V_{\vartheta}(r,\vartheta,\varphi) \vec{e}_{\vartheta}+ V_{\varphi}(r,\vartheta,\varphi).$$

Position vectors are somewhat special. Here you often need the relation between the angle $\psi$ between two position vectors $\vec{x}$ and $\vec{x}'$. To evaluate this, it's more convenient to use the Cartesian components in terms of the spherical coordinates, i.e.,
$$\vec{x}=r \begin{pmatrix} \sin \vartheta \cos \varphi \\ \sin \vartheta \sin \varphi \\ \cos \vartheta \end{pmatrix}, \quad \vec{x}'=r' \begin{pmatrix} \sin \vartheta' \cos \varphi' \\ \sin \vartheta' \sin \varphi' \\ \cos \vartheta' \end{pmatrix}.$$
Then by definition of the scalar product you have after using the cosine addition theorem
$$\vec{x} \cdot \vec{x}'=r r' \cos \psi = r r' [\sin \vartheta \sin \vartheta' \cos(\varphi-\varphi') + \cos \vartheta \cos \vartheta')].$$