Curvilinear Coordinates
\[(u_1 , u_2 , u_3 )\]
We can express the ordinary Cartesian coordinates in terms of these curvilinear coordinates.
\[x=x(u_1 , u_2 ,u_3 )\]
\[y=y(u_1 , u_2 ,u_3 )\]
\[z=z(u_1 , u_2 ,u_3 )\]
Hence
\[\mathbf{r}=\mathbf{r}(u_1 , u_2 ,u_3 )\]
A tangent vector to the
\[u_1\]
curve is
\[\frac{\partial \mathbf{r}}{\partial u_1}\]
and a unit vector tangent to the \[u_1\]
curve is \[ \mathbf{e_1} = \frac{\partial \mathbf{r} / \partial u_1}{ |\partial \mathbf{r} / \partial u_1 |}\]
Similarly, unit vectors tangent to the
\[u_2 , \; u_3\]
curves are\[\mathbf{e_2} = \frac{\partial \mathbf{r} / \partial u_2}{ |\partial \mathbf{r} / \partial u_2 |} , \: \mathbf{e_3} = \frac{\partial \mathbf{r} / \partial u_2}{ |\partial \mathbf{r} / \partial u_3 |}\]
A curvilinear system is said to be orthogonal if the coordinates axes meet at right angles. If the vectors
\[\mathbf{e_1} , \: \mathbf{e_2} , \: \mathbf{e_3} \]
are mutually perpendicular the the system is orthogonal and if the dot product of each is 1, the system is also 1, then the system is orthonormal.\[\mathbf{e_i} \cdot \mathbf{j} = \left\{ \begin{array}{cc} 0 & i \neq j \\ 1 & i=j \end{array} \right. \]
The system is orthonirmal
Also the unit tangent vectors
\[\mathbf{e_1} , \: \mathbf{e_2} , \: \mathbf{e_3} \]
are perpendicular to their respective coordinate surfaces so that the vector \[\mathbf{e_1}\]
would be perpendicular to the surface \[u_2 = constant\]
Also the system is right handed so that
\[\mathbf{e_1} \times \mathbf{e_2} = \mathbf{e_3}\]
