# determinant of matrix of transformation from Cartesian to orthogonal curvilinear

Let $(x_1, x_2)$ and $(y_1, y_1)$ be two orthogonal coordinate system with unit vectos $(\hat i_1, \hat i_2)$ and $(\hat e_1, \hat e_2)$ respectively defined by the $x_1 = x_1(y_1,y_2)$ and $x_2 = x_2(y_1,y_2)$, $(x_1,x_2)$ be the Cartesian coordinate system.

The transformation of unit vectors between two system is given by the relation $$\begin{bmatrix} \hat e_1\\ \hat e_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{h_1} \frac{\partial x_1}{\partial y_1} & \frac{1}{h_1} \frac{\partial x_2}{\partial y_1}\\ \frac{1}{h_2} \frac{\partial x_1}{\partial y_2} & \frac{1}{h_2} \frac{\partial x_2}{\partial y_2} \end{bmatrix} \times \begin{bmatrix} \hat i_1\\ \hat i_2 \end{bmatrix}$$ Where $h_1 = \sqrt{ \left ( \partial x_1 \over \partial y_1 \right )^2 + \left ( \partial x_1 \over \partial y_1 \right )^2} , h_2 = \sqrt{ \left ( \partial x_1 \over \partial y_2 \right )^2 + \left ( \partial x_1 \over \partial y_2 \right )^2}$

My question is, for orthogonal coordinate system, is $$\begin{vmatrix} \frac{1}{h_1} \frac{\partial x_1}{\partial y_1} & \frac{1}{h_1} \frac{\partial x_2}{\partial y_1}\\ \frac{1}{h_2} \frac{\partial x_1}{\partial y_2} & \frac{1}{h_2} \frac{\partial x_2}{\partial y_2} \end{vmatrix} = 1$$ Thank you for your help!!

• If $U=AV$ with $U,V$ orthogonal, then $A=UV^{-1}$ is orthogonal, hence $\det A=\pm1$.
– anon
Commented Jun 26, 2012 at 17:15

Notice that the two matrices formed out of unit orthogonal vectors will be orthogonal matrices (in fact, the condition $U^TU=I$ encodes the fact they are orthogonal and normalized). The set of ortho-gonal matrices is closed under multiplication and inversion (it is a group!), so if $U=AV$ with the matrices $U,V$ orthogonal, then $A=UV^{-1}$ is orthogonal and hence $\det A=\pm1$.