How to prove that this coordinate system is ortoghonal?

Question

I have a coordinate system $(u,v)$ given by:

$$ \begin{eqnarray} u&=&xy \\ v&=&\ln(y) \end{eqnarray} $$

How can i check if it is ortoghonal, find the unit vectors and find the Jacobian determinant for this C.S?.

What i have done, is search the definitions.

It will be OG if the inner product of his unit vectors is $0$:

$$<v_i,v_j>=0 \quad i \neq j$$

But i need the unit vectors, and those are:

$$e_i=\dfrac{\partial r}{\partial q_i}$$

For a point $P(q_1,q_2,...,q_n)$ where

$$\vec{r}=h_1 \textrm{d} q_1 \hat{q}_{1}+h_2\textrm{d} q_2 \hat{q}_{2}+h_3 \textrm{d}q_3 \hat{q}_{3}$$

But i dont know how to check this.

Answer 1

One way is to find line element in the new coordinate system. According to the definition:

$$ \begin{eqnarray} y &=& e^v \\ x &=& u e^{-v} \end{eqnarray} $$

so one can write:

$$ \begin{eqnarray} dy &=& e^v dv \\ dx &=& - u e^{-v} dv + e^{-v} du \end{eqnarray} $$

so the line element is: $$ \textrm{d}s^2 = \textrm{d}x ^2 + \textrm{d}y^2 = (u^2 e^{-2v}+ e^{2v}) \textrm{d}v^2 -2 u e^{-2v} \textrm{d}u \textrm{d}v + e^{-2u} \textrm{d}u^2 $$

so this is not an orthogonal system otherwise coefficient of $\textrm{d}u \textrm{d}v$ was zero.

Answer 2

Calculate partials and scalar product: $$\begin{array}{l} u = xy,v = \ln (y)\\ \left( {\begin{array}{*{20}{c}} {{\partial _x}u}\\ {{\partial _y}u} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} y\\ x \end{array}} \right),\left( {\begin{array}{*{20}{c}} {{\partial _x}v}\\ {{\partial _y}v} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\\ {\frac{1}{y}} \end{array}} \right)\\ \left( {\begin{array}{*{20}{c}} {{\partial _x}u}\\ {{\partial _y}u} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} {{\partial _x}v}\\ {{\partial _y}v} \end{array}} \right) = y \cdot 0 + x \cdot \frac{1}{y} = \frac{x}{y} \ne 0 \end{array}$$ Jacobian is regular: $$\det \left( {\begin{array}{*{20}{c}} {{\partial _x}u}&{{\partial _x}v}\\ {{\partial _y}u}&{{\partial _y}v} \end{array}} \right) = \det \left( {\begin{array}{*{20}{c}} y&0\\ x&{\frac{1}{y}} \end{array}} \right) = 1$$