Orthogonal curvilinear coordinates (derivatives of unit vectors)

Suppose that $\{u_i\}_{1\le i\le 3}$ is a set of orthogonal curvilinear coordinates with unit vectors $\{\mathbf{\hat{e}_i}\}_{1\le i\le 3}$. I proved that $$\frac{\partial \mathbf{\hat{e}_i}}{\partial u_j} = \frac{\mathbf{\hat{e}_j}}{h_i}\frac{\partial h_j}{\partial u_i},\tag{1}$$ where $h_i$ is a scale factor such that for a position vector $\mathbf r$ we have $\dfrac{\partial \mathbf r}{\partial u_i}= h_i \mathbf{\hat{e}_i}$. Eq. $(1)$ is valid for $i\neq j$. I'm trying to prove that

$$\frac{\partial \mathbf{\hat{e}_i}}{\partial u_i} = -\sum\limits_{k\neq i}\frac{\mathbf{\hat{e}_k}}{h_k}\frac{\partial h_i}{\partial u_k}.$$

First of all I find it strange that the variation of $\mathbf{\hat{e}_i}$ with respect to the $i$-th coordinate is not pointing towards $\mathbf{\hat{e}_i}$. For instance in Eq. (1) we have that $\dfrac{\partial \mathbf{\hat{e}_i}}{\partial u_j}\parallel \mathbf{\hat{e}_j}$, which makes sense. And secondly this is what I tried to prove the equality (I followed some steps that were suggested, I will mark these steps with $\overset{*}{=}$):

\begin{align} \mathbf{\hat{e}_i} = \mathbf{\hat{e}_j}\times \mathbf{\hat{e}_k} = \sum\limits_{i=1}^3\epsilon_{ijk}\mathbf{\hat{e}_i} \overset{*}{=}\frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\mathbf{\hat{e}_j}\times \mathbf{\hat{e}_k}, \end{align}

where $\epsilon_{ijk}$ is the Levi-Civita tensor. \begin{align} \frac{\partial \mathbf{\hat{e}_i}}{\partial u_i} &= \frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\frac{\partial}{\partial u_i}(\mathbf{\hat{e}_j}\times \mathbf{\hat{e}_k}) = \frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\left(\frac{\partial\mathbf{\hat{e}_j}}{\partial u_i}\times \mathbf{\hat{e}_k} + \frac{\partial\mathbf{\hat{e}_k}}{\partial u_i}\times \mathbf{\hat{e}_j}\right)\\ &\overset{*}{=} \frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\left(\frac{\mathbf{\hat{e}_i}}{h_j}\frac{\partial h_i}{\partial u_j}\times \mathbf{\hat{e}_k} + \frac{\mathbf{\hat{e}_i}}{h_k}\frac{\partial h_i}{\partial u_k}\times \mathbf{\hat{e}_j}\right)=\frac12\sum\limits_{j,k=1}^3\epsilon_{ijk}\left(-\frac{\mathbf{\hat{e}_j}}{h_j}\frac{\partial h_i}{\partial u_j} + \frac{\mathbf{\hat{e}_k}}{h_k}\frac{\partial h_i}{\partial u_k}\right). \end{align}

I feel so close...

I have the hint that one should get to this result:

$$\frac{\partial \mathbf{\hat{e}_i}}{\partial u_i} = \sum_{jkl}\epsilon_{ijk}\epsilon_{ilj}\frac{\mathbf{\hat{e}_l}}{h_k}\frac{\partial h_i}{\partial u_k}.$$

But I don't know how to get to that. Any help is very much appreciated. Thanks.

• Are you sure $\frac{\partial \mathbf r}{\partial u_i}= h_i \mathbf{\hat{e}_i}$ is correct? I think the basis vector there should be of the curvilinear system, not of the Cartesian system. See here for details. – wltrup Aug 17 '15 at 0:48
• @wltrup $\{\bf{\hat e}_i\}$ is the basis of the curvilinear system, – Vladimir Vargas Aug 17 '15 at 0:54
• Oh, right. I misread or missed the first sentence of your question the first time I read it. It's late (here in London)... maybe I should go to bed now. – wltrup Aug 17 '15 at 0:54

Not familar with Levi-Civita, but here is an alternative proof: \begin{equation} \mathbf{\hat e_i} \cdot \mathbf{\hat e_i}=1 \\ \frac{\partial\mathbf{\hat e_i}}{\partial u_i}\cdot \mathbf{\hat e_i} = 0 \end{equation} Thus $\frac{\partial \mathbf{\hat e_i}}{\partial u_i}$ has no $i$ component. For $i\neq j$ \begin{equation} \mathbf{\hat e_i} \cdot \mathbf{\hat e_j}=0 \\ \frac{\partial \mathbf{\hat e_i}}{\partial u_i}\cdot \mathbf{\hat e_j} = -\frac{\partial \mathbf{\hat e_j}}{\partial u_i}\cdot \mathbf{\hat e_i} \end{equation} So the $j$ component of $\frac{\partial\mathbf{\hat e_i}}{\partial u_i}$ is $- \mathbf{\hat e_i}\cdot \frac{\partial\mathbf{\hat e_j}}{\partial u_i}$.
\begin{equation} - \mathbf{\hat e_i}\cdot \frac{\partial\mathbf{\hat e_j}}{\partial u_i}\\ =-\frac{\partial \mathbf{r}}{h_i\partial u_i}\cdot\frac{\partial}{\partial u_i}(\frac{\partial \mathbf{r}}{h_j\partial u_j})\\ =-\frac{\partial \mathbf{r}}{h_i\partial u_i}\cdot(\frac{\partial}{\partial u_i}(\frac{1}{h_j})\frac{\partial \mathbf{r}}{\partial u_j} + \frac{1}{h_j}\frac{\partial^2 \mathbf{r}}{\partial u_i \partial u_j}) \end{equation} The first item is in the direction of $j$, thus \begin{equation} - \mathbf{\hat e_i}\cdot \frac{\partial\mathbf{\hat e_j}}{\partial u_i}\\ =-\frac{1}{h_i h_j}\frac{\partial \mathbf{r}}{\partial u_i}\cdot\frac{\partial^2 \mathbf{r}}{\partial u_i \partial u_j}\\ =-\frac{1}{2h_i h_j}(\frac{\partial}{\partial u_j}(\frac{\partial \mathbf{r}}{\partial u_i}\cdot \frac{\partial \mathbf{r}}{\partial u_i}) )\\ =-\frac{1}{2h_i h_j}\frac{\partial}{\partial u_j}h_i^2\\ =-\frac{1}{ h_j}\frac{\partial}{\partial u_j}h_i \end{equation} Thus, \begin{equation} \frac{\partial \mathbf{\hat e_i}}{\partial u_i}\cdot \mathbf{\hat e_j}=-\frac{1}{ h_j}\frac{\partial}{\partial u_j}h_i \end{equation} Similarly, \begin{equation} \frac{\partial \mathbf{\hat e_i}}{\partial u_i}\cdot \mathbf{\hat e_k}=-\frac{1}{ h_k}\frac{\partial}{\partial u_k}h_i \end{equation}