1. Nov 13, 2015

### Incand

1. The problem statement, all variables and given/known data
Show that $\nabla u_i \cdot \frac{\partial \vec r}{\partial u_i} = \delta_{ij}$.

($u_i$ is assumed to be a generalized coordinate.)
2. Relevant equations
$\nabla \phi = \sum_{i=1}^3 \vec e_i \frac{1}{h_i} \frac{\partial \phi}{\partial u_i}$

3. The attempt at a solution
So what I need to show is that the gradient is orthogonal to the (other two) partial derivatives (but it doesn't have to be parallel to the third since the system doesn't have to be orthogonal!).
The expression can then be written as
$\sum_{k=1}^3 \left( \vec e_k \frac{1}{h_k} \frac{\partial u_i}{\partial u_k} \right) \cdot \frac{\partial \vec r}{\partial u_j} = \sum_{k=1}^3 \frac{1}{h_k}\frac{\partial u_i}{\partial u_k}\frac{\partial r_k}{\partial u_j}$.
I'm thinking I could use the chain rule here but I don't seem to get anywhere. Another thing of note is that $\frac{1}{h_k}\frac{\partial \vec r}{\partial u_k} = \vec e_k$ which may be of use somewhere.

2. Nov 13, 2015

### fzero

You can substitue $\partial \vec{r}/\partial u_j = h_j \vec{e}_j$ in your expression.

3. Nov 14, 2015

### Incand

Thanks!
$\sum_{k=1}^3 \left( \vec e_k \frac{1}{h_k}\frac{\partial u_i}{\partial u_k} \right) \cdot \frac{h_j}{h_j} \frac{\partial \vec r}{\partial u_j} = \sum_{k=1}^3 \left( \vec e_k \frac{1}{h_k}\frac{\partial u_i}{\partial u_k} \right) \cdot h_j \vec e_j = \frac{\partial u_i}{\partial u_j}$.
Which is obviously $1$ for $i=j$ but how do I know that it's zero when $i\ne j$? For me this isn't obvious. In a cylindrical coordinate system the unit vector $\hat \rho$ is dependent on $\hat \phi$ and the other way around so I don't see why the coordinates them self can't be dependent.

4. Nov 14, 2015

### fzero

The partial derivative $\partial/\partial u_j$ is defined so that $u_{i\neq j}$ are treated as constants for the purpose of taking the derivative.

5. Nov 14, 2015

### Incand

Right, I wasn't thinking clearly, Thanks!