# A Unitary matrix

1. May 1, 2017

### LagrangeEuler

Relations between vectors in cylindrical and
Cartesian
coordinate systems are given by
$$\vec{e}_{\rho}=\cos \varphi \vec{e}_x+\sin \varphi \vec{e}_y$$
$$\vec{e}_{\varphi}=-\sin \varphi \vec{e}_x+\cos \varphi \vec{e}_y$$
$$\vec{e}_z=\vec{e}_z$$
We can write this in form
$$\begin{bmatrix} \vec{e}_{\rho} \\[0.3em] \vec{e}_{\varphi} \\[0.3em] \vec{e}_z \\[0.3em] \end{bmatrix} =\begin{bmatrix} \cos \varphi & \sin \varphi & 0 \\[0.3em] -\sin \varphi & \cos \varphi & 0 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix} \begin{bmatrix} \vec{e}_x \\[0.3em] \vec{e}_y \\[0.3em] \vec{e}_z \\[0.3em] \end{bmatrix}$$
where matrix $\begin{bmatrix} \cos \varphi & \sin \varphi & 0 \\[0.3em] -\sin \varphi & \cos \varphi & 0 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix}$ is orthogonal. Then means that norms of the vectors $\begin{bmatrix} \vec{e}_x \\[0.3em] \vec{e}_y \\[0.3em] \vec{e}_z \\[0.3em] \end{bmatrix}$ and
$\begin{bmatrix} \vec{e}_{\rho} \\[0.3em] \vec{e}_{\varphi} \\[0.3em] \vec{e}_z \\[0.3em] \end{bmatrix}$ are the same. But how to define norm of vector
$\begin{bmatrix} \vec{e}_{\rho} \\[0.3em] \vec{e}_{\varphi} \\[0.3em] \vec{e}_z \\[0.3em] \end{bmatrix}$?

2. May 4, 2017

### BvU

What is it you are saying here ? Because what you describe is the cartesian coordinates after a rotation over an angle $\phi$ around the z-axis. A vector $(a,b,c)$ in cartesian coordiates is not identical to a vector $(\rho, \phi, z)$ in cylindrical coordinates with $\rho = a\cos\phi+b\sin\phi\,,\ \$ etc

3. May 4, 2017

### BvU

Generally: norm squared of $\,\vec v\ \$ is $\ \ \vec v\cdot\vec v$ .
Since the angle between a vector and itself is zero, this is pretty easy:
In cylindrical coordinates $\sqrt{\rho^2+z^2\ }\$ and for spherical coordinates simply $r$.

For the inner product in general you need the angle between the two vectors. Probably easiest to first convert to cartesian coordinates and work it out for the various coordinate systems.