D.1 Orthogonal coordinate derivatives

Since there is no fundamental difference between the three orthogonal coordinates, it suffices to show that

$$\frac{\partial{\hat\imath}_1}{\partial u_2} = \frac{1}{h_1} \frac{\partial h_2}{\partial u_1} {\hat\imath}_2$$

The other derivatives with $i\ne{}j$ go the same way, and the derivatives with $i=j$ can then be evaluated by writing ${\hat\imath}_i$ as a cross product of the other two unit vectors.

Now to derive the above result, the only thing we know a priori is that the derivative of a unit vector is normal to the unit vector, so:

$$\frac{\partial{\hat\imath}_1}{\partial u_2} = c_2 {\hat\imath}_2 + c_3 {\hat\imath}_3$$

But it is not obvious why $c_2$ would have to be $\partial{}h_2/h_1\partial{}u_1$ and $c_3$ would have to be zero.

Recall however that ${\hat\imath}_1$ was defined as

$$\frac{\partial {\skew0\vec r}}{\partial u_1} \equiv h_1 {\hat\imath}_1$$

and if we differentiate this with respect to $u_2$, one of the terms will involve the desired derivative of ${\hat\imath}_1$:

$$\frac{\partial^2{\skew0\vec r}}{\partial u_1\partial u_2} ... ...\imath}_1 + h_1 \frac{\partial{\hat\imath}_1}{\partial u_2}$$

Since we can change the order of differentiation without changing the derivative, we must have

$$\frac{\partial h_1}{\partial u_2} {\hat\imath}_1 + h_1 \... ...\imath}_2 + h_2 \frac{\partial{\hat\imath}_2}{\partial u_1}$$

so, writing out the derivative

$$\frac{\partial h_1}{\partial u_2} {\hat\imath}_1 + h_1 (... ...\imath}_2 + h_2 \frac{\partial{\hat\imath}_2}{\partial u_1}$$

Now compare ${\hat\imath}_2$ components in both sides, noting that the derivative of ${\hat\imath}_2$ is normal to ${\hat\imath}_2$. Then you see that the value of $c_2$ given above is indeed correct.

But it is not clear from the above why $c_3$ would have to be zero. To find that out, we dot the derivative $\partial^2{\skew0\vec r}/\partial{}u_1\partial{}u_2$ with $\partial{}{\skew0\vec r}/\partial{}u_3$, because the latter derivative is by definition $h_3{\hat\imath}_3$, so the dot product will give

$$\frac{\partial^2{\skew0\vec r}}{\partial u_1\partial u_2} ... ...t \frac{\partial{\skew0\vec r}}{\partial u_3} = h_1 c_3 h_3$$

So to show that $c_3$ is zero, we must show that the dot product above is zero. That can be done with a bit of manipulation. The only thing you can do, of course, is shuffle the derivatives around. In particular, if you pull the derivative with respect to $u_1$ to the front of the entire thing,

$$\frac{\partial^2{\skew0\vec r}}{\partial u_1\partial u_2} ... ...dot \frac{\partial{\skew0\vec r}}{\partial u_1\partial u_3}$$

where the final term corrects for the additional term generated by pulling the $u_1$ derivative out. Note now that the dot product in the parentheses above is zero since the vectors are orthogonal. Only the final term survives.

Next, ask yourself: why pull the $u_1$ derivative out? Why not $u_2$? After all, $u_1$ and $u_2$ appear completely symmetrically in the expression. That suggests that really, what we should do is pull $u_1$ out of half the term and $u_2$ out of the other half. That gives

$$- {\textstyle\frac{1}{2}} \frac{\partial{\skew0\vec r}}{... ...dot \frac{\partial{\skew0\vec r}}{\partial u_2\partial u_3}$$

That is seen to be the same as

$$- {\textstyle\frac{1}{2}} \frac{\partial}{\partial u_3} ... ...} \cdot \frac{\partial{\skew0\vec r}}{\partial u_1} \right)$$

and the dot product in parenthesis is zero. So indeed $c_3$ is zero.