Question

Show that the curvilinear coordinate system defined by the following equations is
orthogonal:
x=uv cos ∅; y= uv sin ∅; z=1/2 (u^2-v^2 )

Answer 1

Answer on Question #46381, Engineering, Other

Task: Show that the curvilinear coordinate system defined by the following equations is orthogonal:

$$x = u v \cos \varphi ;$$ $$y = u v \sin \varphi ;$$ $$z = \frac{(u^2 - v^2)}{2}$$

Solution:

$$\vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} uv \cos \varphi \\ uv \sin \varphi \\ \frac{(u^2 - v^2)}{2} \end{pmatrix}$$

Derivatives of the radius vector:

$$\vec{r}_u = \begin{pmatrix} x_u \\ y_u \\ z_u \end{pmatrix} = \begin{pmatrix} v \cos \varphi \\ v \sin \varphi \\ u \end{pmatrix}; \vec{r}_v = \begin{pmatrix} x_v \\ y_v \\ z_v \end{pmatrix} = \begin{pmatrix} u \cos \varphi \\ u \sin \varphi \\ -v \end{pmatrix}; \vec{r}_\varphi = \begin{pmatrix} x_\varphi \\ y_\varphi \\ z_\varphi \end{pmatrix} = \begin{pmatrix} -uv \sin \varphi \\ vu \cos \varphi \\ 0 \end{pmatrix};$$

Scalar products:

$$\vec{r}_u \cdot \vec{r}_v = uv \cos^2 \varphi + uv \sin^2 \varphi - uv = 0$$ $$\vec{r}_u \cdot \vec{r}_\varphi = -uv^2 \cos \varphi \sin \varphi + uv^2 \sin \varphi \cos \varphi = 0$$ $$\vec{r}_v \cdot \vec{r}_\varphi = -u^2 v \cos \varphi \sin \varphi + u^2 v \sin \varphi \cos \varphi = 0$$

It means that $\vec{r}_u, \vec{r}_v, r_\varphi$ can be chosen as a basis and these vectors are orthogonal.

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