Relation between Vector Norms in Cylindrical and Cartesian Coordinates

In summary, the relations between vectors in cylindrical and Cartesian coordinate systems involve a rotation by an angle phi around the z-axis. The vectors can be written in matrix form, with the matrix being orthogonal. The norms of the vectors are the same, and the norm of a vector in cylindrical coordinates is defined as the square root of rho squared plus z squared. For the inner product, the angle between the two vectors must be considered, and it may be easier to first convert to Cartesian coordinates.
  • #1
LagrangeEuler
717
22
Relations between vectors in cylindrical and
Cartesian
coordinate systems are given by
[tex]\vec{e}_{\rho}=\cos \varphi \vec{e}_x+\sin \varphi \vec{e}_y[/tex]
[tex]\vec{e}_{\varphi}=-\sin \varphi \vec{e}_x+\cos \varphi \vec{e}_y [/tex]
[tex] \vec{e}_z=\vec{e}_z [/tex]
We can write this in form
[tex]
\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}
[/tex]
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}##?
 
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  • #2
LagrangeEuler said:
Relations between vectors in cylindrical and
Cartesian
coordinate systems are given by ...
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
LagrangeEuler said:
But how to define norm of vector
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.
 
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