Question

show that the curvilinear coordinate system defined by the following equation is orthogonal :
X=UVcosA
Y=UVsinA
Z=(u^2-v^2)/2

Answer 1

Answer on Question#38738 – Math - Other

We have the curvilinear coordinate system

$$x = u v \cos \alpha, \ x = u v \sin \alpha, \ z = \frac {u ^ {2} - v ^ {2}}{2}$$

In curvilinear coordinates, a point in space is specified by the coordinates, and at every such point there is bound a set of basis vectors, which generally are not constant. In orthogonal coordinates the basis vectors vary, they are always orthogonal with respect to each other. In other words,

$$\vec {e} _ {i} \cdot \vec {e} _ {j} = 0 \ \text{if} \ i \neq j,$$

These basis vectors are by definition the tangent vectors of the curves obtained by varying one coordinate, keeping the others fixed:

$$\vec {e} _ {i} = \frac {\partial \vec {r}}{\partial q _ {i}}$$

where $\vec{r}$ is some point and $q_{i}$ is the coordinate for which the basis vector is extracted.

We have:

$$\vec {r} = (x, y, z) = \left(u v \cos \alpha , u v \sin \alpha , \frac {u ^ {2} - v ^ {2}}{2}\right)$$

Let's determine the tangent vectors and check their orthogonality:

$$\vec {e} _ {1} = \frac {\partial \vec {r}}{\partial u} = (v \cos \alpha , v \sin \alpha , u),$$ $$\vec {e} _ {2} = \frac {\partial \vec {r}}{\partial v} = (u \cos \alpha , u \sin \alpha , - v),$$ $$\vec {e} _ {3} = \frac {\partial \vec {r}}{\partial \alpha} = (- u v \sin \alpha , u v \cos \alpha , 0).$$

Then

$$\vec {e} _ {1} \cdot \vec {e} _ {2} = u v \cdot (\cos \alpha \cdot \cos \alpha + \sin \alpha \cdot \sin \alpha) - u v = u v - u v = 0,$$ $$\vec {e} _ {2} \cdot \vec {e} _ {3} = u ^ {2} v \cdot (- \cos \alpha \cdot \sin \alpha + \sin \alpha \cdot \cos \alpha) = u ^ {2} v \cdot 0 = 0,$$ $$\vec {e} _ {1} \cdot \vec {e} _ {3} = u v ^ {2} \cdot (- \cos \alpha \cdot \sin \alpha + \sin \alpha \cdot \cos \alpha) = u v ^ {2} \cdot 0 = 0.$$

This shows that all pairs of tangent vectors are orthogonal and therefore that the coordinate system is an orthogonal one.