I am given a curvilinear coordinates $a$, $b$ and $\phi$ as follows: $$a=\frac{x^2+y^2+z^2}{2z},\ b=\frac{x^2+y^2+z^2}{2\sqrt{x^2+y^2}},\ \phi=\arctan\frac{y}{x}.$$ I am asked to find covariant and contravariant basis vectors for it. But I cannot get the relation between $x$, $y$ and $z$ based on $a$, $b$ and $\phi$ to calculate these vectors and I am not given sufficient examples to solve these problems. I would appreciate any suggestions.

  • $\begingroup$ I think a starting point is $\tan \phi = y/x$. $\endgroup$ – IAmNoOne Oct 10 '18 at 0:15
  • $\begingroup$ Actually, I can convert this into spherical coordinates, but that way getting the Contravariant basis becomes more difficult than using the Cartesian coordinates. $\endgroup$ – Alex Parker Oct 10 '18 at 0:21

The tangential $(\mathbf{t}_i$) and the normal $\mathbf{n}_i$ basis are given by

$$\begin{cases} \mathbf{t}_i&=\frac{\partial \mathbf{r}}{\partial u_i} \\ \mathbf{n}_i&=\nabla u_i\end{cases}$$

I suggest you start with the normal basis (sometimes called the covariant basis) $\mathbf{n}_i$. It is rather straightforward to calculate the gradient of a, b and $\phi$

In this case, you will eventually find that

$$\nabla u_1\cdot \nabla u_2=\nabla u_1\cdot \nabla u_3= \nabla u_2\cdot \nabla u_3=0$$

Which means that your basis vectors are orthogonal.

Establish the corressponding scale factors $h_i=\frac{1}{|\nabla u_i|}$

Finally, use

$$\frac{1}{h_i}\frac{\partial \mathbf{r}}{\partial u_i}=h_i\nabla u_i$$

to find the contravariant basis $\mathbf{t}_i=\frac{\partial \mathbf{r}}{\partial u_i}$


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