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

[Aidyan]

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

\nabla \cdot A= \frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z}

\nabla \times \mathbf{A}= \begin{vmatrix}<br /> \widehat{x} &amp; \widehat{y} &amp; \widehat{z} \\<br /> \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \\<br /> A_{x} &amp; A_{y} &amp; A_{z} \\<br /> \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}

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

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

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]

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.