Div and curl operators in a left-handed coordinate system?

  • Thread starter Aidyan
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In a right-handed cartesian coordinate system the divergence and curl operators are respectively:

[itex]\nabla \cdot A= \frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z}[/itex]

[itex]\nabla \times \mathbf{A}= \begin{vmatrix}
\widehat{x} & \widehat{y} & \widehat{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
A_{x} & A_{y} & A_{z} \\
\end{vmatrix}= (\frac{\partial A_{z}}{\partial y}-\frac{\partial A_{y}}{\partial z}) \widehat{x}-(\frac{\partial A_{z}}{\partial x}-\frac{\partial A_{x}}{\partial z}) \widehat{y}+(\frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y}) \widehat{z}[/itex]

While, for the (still right-handed) cylindrical coordinate system they are:

[itex]\nabla \cdot \mathbf{A}=\frac{A_{r}}{r} + \frac{\partial A_{r}}{\partial r} + \frac{1}{r} \frac{\partial A_{\theta}}{\partial \theta}+\frac{\partial A_{z}}{\partial z}[/itex]


[itex]\nabla \times \mathbf{A}= \left( \frac{1}{r} \frac{\partial A_{z}}{\partial \theta}- \frac{\partial A_{\theta}}{\partial z} \right) \overrightarrow{e_{r}} +
\left( \frac{\partial A_{r}}{\partial z}- \frac{\partial A_{z}}{\partial r} \right) \overrightarrow{e_{\theta}} +
\left( \frac{A_{\theta}}{r} + \frac{\partial A_{\theta}}{\partial r}- \frac{1}{r} \frac{\partial A_{r}}{\partial \theta} \right) \overrightarrow{e_{z}}[/itex]

For a left-handed cartesian and cylindrical coordinate system is it just a matter of changing some sign in the third component? Or is it not that immediate? I'm bit unsure and confused about that... can someone help?
 

fresh_42

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Orientation is a deliberate order. You see from the formulas, that the cross products depend on this order, so there is a left and right for it, i.e. a change of sign for the cross product if you switch from one to the other system.
 

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