A Question about Unit Vectors of Cylindrical Coordinates

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SUMMARY

The discussion centers on defining the direction of the unit vector ##\hat{φ}## in cylindrical coordinates ##(ρ,φ,z)##. It confirms that ##\hat{φ}## can be derived using the right-hand rule, specifically by evaluating the cross-product of ##\hat{ρ}## and ##\hat{z}##. The user also references the equations for the Nabla, divergence, curl, and Laplacian operators in these coordinates. The definition of ##\phi## as anti-clockwise when viewed from above the z-axis is emphasized as a standard convention.

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  • Understanding of cylindrical coordinates (ρ, φ, z)
  • Familiarity with vector calculus concepts such as divergence and curl
  • Knowledge of unit vectors and their properties
  • Basic grasp of the right-hand rule in vector mathematics
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  • Study the derivation of the Nabla operator in cylindrical coordinates
  • Learn about the applications of divergence and curl in fluid dynamics
  • Explore the mathematical implications of the cross-product in three-dimensional space
  • Review the geometric interpretation of unit vectors in various coordinate systems
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sams
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I wrote the equations of the Nabla, the divergence, the curl, and the Laplacian operators in cylindrical coordinates ##(ρ,φ,z)##. I was wondering how to define the direction of the unit vector ##\hat{φ}##. Can we obtain ##\hat{φ}## by evaluating the cross-product of ##\hat{ρ}## and ##\hat{z}## or by using the right-hand rule?

Thank you so much for your help...
 
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sams said:
I wrote the equations of the Nabla, the divergence, the curl, and the Laplacian operators in cylindrical coordinates ##(ρ,φ,z)##. I was wondering how to define the direction of the unit vector ##\hat{φ}##. Can we obtain ##\hat{φ}## by evaluating the cross-product of ##\hat{ρ}## and ##\hat{z}## or by using the right-hand rule?

Thank you so much for your help...

It's all here:

http://mathworld.wolfram.com/CylindricalCoordinates.html

##\phi## (or ##\theta## to mathematicians) is usually defined to be anti-clockwise looking down the z-axis.
 
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Thank you @PeroK for your help
 

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