Gradient in Orthogonal Curvilinear Coordinates.

Let us consider the transformation of rectangular coordinates $$(x,y,z)$$ into an orthogonal coordinates system $$(u_1,u_2,u_3)$$

Let $$r = xi + yj + zk$$ be the general position vector that can be written as $$r=r(u_1,u_2,u_3)$$; Now unit tangent vector in the direction of $$u_1$$ can be written as $$e_1=\frac{\frac{\partial r}{\partial u_1}}{\lvert \frac{\partial r}{\partial u_1} \rvert}\\Or,\ \ \ e_1=\frac{\frac{\partial r}{\partial u_1}}{h_1} \ \ \ \ where \ \ \ h_1=\frac{\partial r}{\partial u_1}$$

Similarly $$e_2$$ and $$e_3$$ can be written.

Now I'm trying to follow the proof of showing that $$\frac{\partial r}{\partial u_1}, \frac{\partial r}{\partial u_2},\frac{\partial r}{\partial u_3}$$ and $$\nabla u_1,\nabla u_2,\nabla u_3,$$ form a reciprocal system of vectors. But the first step in the proof is: $$\nabla u_1 = \frac{e_1}{h_1}$$

How is this expression obtained?

My attempt:

$$\nabla u_1 = \frac{\partial u_1}{\partial x}i + \frac{\partial u_1}{\partial y}j + \frac{\partial u_1}{\partial z}k$$

This doesn't lead me to involve $$h_1$$ term. So alternately,

$$\nabla u_1 = \frac{\partial u_1}{\partial u_1} e_1 + \frac{\partial u_1}{\partial u_2}e_2 + \frac{\partial u_1}{\partial u_3}e_3 = e_1 + 0 + 0 = e_1$$

I'm missing something conceptual. Kindly clarify.

• The last formula is not the correct way to compute the gradient. But if you take your next to last formula and compute the dot product with $\mathbf e_i = (\partial x/\partial u_i, \partial y/\partial u_i, \partial z/\partial u_i)/h_i$, you'll get the expected result. – Maxim Nov 8 '18 at 14:23
• @Maxim Could you please add an answer or elaborate? If I do the dot product as you said I'm getting $\nabla u_1 . e_1 = \frac{3}{h_1}$ – yathish Nov 17 '18 at 12:32
• $\frac {\partial u_1} {\partial x} \frac {\partial x} {\partial u_1} \neq 1$. What holds is $\frac {\partial(u_1, u_2, u_3)} {\partial(x, y, z)} \frac {\partial(x, y, z)} {\partial(u_1, u_2, u_3)} = \frac {\partial(u_1, u_2, u_3)} {\partial(u_1, u_2, u_3)}$. Putting it slightly differently: write $\frac {\partial u_1} {\partial u_i}$ as the result of applying the chain rule to $u_1 = u_1(x(u_1, u_2, u_3), y(u_1, u_2, u_3), z(u_1, u_2, u_3))$. – Maxim Nov 17 '18 at 18:39
• @Maxim major clarification that I needed! So now I get $\nabla u_1 \cdot \mathbf e_1 = \frac{1}{h_1}$. This is of the form $\mathbf{\vec{a} \cdot \hat{b}}=k.$ Would this necessarily mean $\mathbf{\vec{a}}=k\mathbf{\hat{b}}?$ How do I get $\nabla u_1 = \frac{\mathbf e_1}{h_1}$? – yathish Nov 18 '18 at 4:15
• Evaluate $(\nabla u_1) \cdot \mathbf e_2$ and $(\nabla u_1) \cdot \mathbf e_3$ in the same way. Then you know the projections of $\nabla u_1$ on all three basis vectors and can write out the decomposition of $\nabla u_1$ over that basis. – Maxim Nov 18 '18 at 15:03