Coordinate transformation on del operator

Question

If there are two sets of orthogonal bases, $\hat{x}_1, ...,\hat{x_n}$ and $\hat{u}_1, ...,\hat{u_n}$, and a point in space can be expressed as

$$\vec{r}=x_1\hat{x}_1+...+x_n\hat{x}_n$$

and

$$\vec{r}=u_1\hat{u}_1+...+u_n\hat{u}_n$$

with the coordinate transformation

$$x_i=f_i(u_1,...,u_n), i=1,2,...,n$$

So the volume element transformation shall be

$$dx_1dx_2...dx_n=||A||du_1du_2...du_2$$

where $A$ is a $n\times n$ matrix with

$$A_{ij}=\partial{f_i}/\partial{u_j}$$

Is there a similar relationship with the del operator $\nabla$? Given all $f_i$ and the del operator for the first set of bases, and all the projections $\hat{x}_i \cdot \hat{u}_i$, how do I get the del operator for the second set?

Answer 1

To calculate $\hat{u}_j$ you can calculate $\nabla u_j$ then normalize it; $\hat{u}_j = \frac{1}{||\nabla u_j||}\nabla u_j$. In terms of Cartesian coordinates: $$ \nabla u_j = \langle \frac{\partial u_j}{\partial x_1}, \frac{\partial u_j}{\partial x_2}, \dots , \frac{\partial u_j}{\partial x_n} \rangle .$$ Now, if you have also orthogonalized the $u$-coordinates then you can define the components in terms of dot-products. For $\vec{F} = F_1 \hat{x_1}+F_2 \hat{x_2}+ \cdots F_n \hat{x_n}$ we can convert to $u$-coordinates, $$ \vec{F} = (\vec{F} \cdot \hat{u_1})\hat{u_1}+(\vec{F} \cdot \hat{u_2})\hat{u_2}+ \cdots + (\vec{F} \cdot \hat{u_n})\hat{u_n}$$ So, this gives us the conversion of a vector to orthonormalized u-coordinates. If non-orthogonal coordinates are used then we have no choice but face the inversion problem directly, as in linear algebra solve $n$-equations simulataneously. I hope your coordinates are orthonormal. Let us apply the comment above to $\nabla f$. By definition, in Cartesian coordinates, $$ \nabla f = \frac{\partial f}{\partial x_1}\hat{x_1}+\frac{\partial f}{\partial x_2}\hat{x_2}+ \cdots +\frac{\partial f}{\partial x_n}\hat{x_n}$$ We want to express $\nabla f$ in $u$-coordinates. So, we wish to find functions $c_1, \dots c_n$ for which: $$ \nabla f = c_1\hat{u_1}+c_2\hat{u_2}+\cdots + c_n\hat{u_n} $$ In view of my assumption of orthonormality of the $u$-coordinate system, I look to unravel the $j$-th component as follows: $$ c_j = \nabla f \cdot \hat{u_j} = \frac{1}{||\nabla u_j||}\nabla f \cdot \nabla u_j = \frac{1}{||\nabla u_j||}\sum_{i=1}^n \frac{\partial u_j}{\partial x_i }\frac{\partial}{\partial x_i}[f]$$ by the chain-rule. Consequently, we find: $$ \nabla = \hat{u_1}\frac{1}{||\nabla u_1||}\sum_{i=1}^n \frac{\partial u_1}{\partial x_i }\frac{\partial}{\partial x_i}+ \cdots + \hat{u_n}\frac{1}{||\nabla u_n||}\sum_{i=1}^n \frac{\partial u_n}{\partial x_i }\frac{\partial}{\partial x_i}. $$ What we want though are partial derivatives in the $u$-coordinates. The chain-rule informs us how to convert: $\frac{\partial}{\partial x_i} = \sum_{k=1}^n \frac{\partial u_k}{\partial x_i} \frac{\partial}{\partial u_k}$. So, we can use that to convert to the desired formula. I think this can be made clearer, but I must stop working on it for now, again, I would encourage you to surf more questions. In particular, there is a method based on the infinitesimal line-elements which is slick, for example, $ds^2 = dx^2+dy^2= dr^2+r^2d\theta^2$ hence $\nabla = \hat{r}\frac{\partial}{\partial r}+ \hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}$.