Relation between Vector Norms in Cylindrical and Cartesian Coordinates

Click For Summary
SUMMARY

The discussion focuses on the mathematical relationships between vector norms in cylindrical and Cartesian coordinate systems. It establishes that the transformation matrix between these systems is orthogonal, ensuring that the norms of vectors in both systems remain equivalent. The norms can be defined as the square root of the sum of the squares of their components, specifically using the formulas √(ρ² + z²) for cylindrical coordinates and r for spherical coordinates. The conversation emphasizes the importance of understanding these transformations for accurate vector analysis.

PREREQUISITES
  • Understanding of vector mathematics and norms
  • Familiarity with cylindrical and Cartesian coordinate systems
  • Knowledge of orthogonal matrices and their properties
  • Basic concepts of inner products and angles between vectors
NEXT STEPS
  • Study the properties of orthogonal matrices in linear algebra
  • Learn about vector transformations between coordinate systems
  • Explore the concept of inner products in different coordinate systems
  • Investigate applications of vector norms in physics and engineering
USEFUL FOR

Mathematicians, physicists, engineers, and students studying vector calculus or coordinate transformations will benefit from this discussion.

LagrangeEuler
Messages
711
Reaction score
22
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
<br /> \begin{bmatrix}<br /> \vec{e}_{\rho} \\[0.3em]<br /> \vec{e}_{\varphi} \\[0.3em]<br /> \vec{e}_z \\[0.3em]<br /> <br /> <br /> \end{bmatrix}<br /> =\begin{bmatrix}<br /> \cos \varphi &amp; \sin \varphi &amp; 0 \\[0.3em]<br /> -\sin \varphi &amp; \cos \varphi &amp; 0 \\[0.3em]<br /> 0 &amp; 0 &amp; 1 \\[0.3em]<br /> <br /> <br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> \vec{e}_x \\[0.3em]<br /> \vec{e}_y \\[0.3em]<br /> \vec{e}_z \\[0.3em]<br /> <br /> <br /> \end{bmatrix}<br />
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}##?
 
Physics news on Phys.org
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
 
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K