Orthogonal curvilinear coordinates

Proof. In this proof we set $j = 1$ and $k = 2$ for clarity, but the approach is valid for any $j \neq k$. We know the definitions of $h_1 \vu{e}_1$ and $h_2 \vu{e}_2$, and that differentiations can be reordered:

Expanding this according to the product rule of differentiation:

We rearrange this in two different ways. Indeed, these two equations are identical:

Recall that all derivatives of $\vu{e}_j$ are orthogonal to $\vu{e}_j$. Therefore, the first equation’s right-hand side must be orthogonal to $\vu{e}_1$, and the second’s to $\vu{e}_2$. We deduce that the parenthesized expressions are proportional to $\vu{e}_3$, and call the proportionality factors $\lambda_{123}$ and $\lambda_{213}$:

Since these equations are identical, by comparing the definition of $\lambda_{123}$ to the other side of the equation, we see that $\lambda_{123} = \lambda_{213}$:

In general, $\lambda_{jkl} = \lambda_{kjl}$ for $j \neq k \neq l$. Next, we dot-multiply $\lambda_{123}$’s equation by $\vu{e}_3$, using that $\vu{e}_2 \cdot \vu{e}_3 = 0$ and consequently $\ipdv{(\vu{e}_2 \cdot \vu{e}_3)}{c_1} = 0$:

In general, $\lambda_{jkl} = - h_k \lambda_{jlk} / h_l$ for $j \neq k \neq l$. Combining this fact with $\lambda_{jkl} = \lambda_{kjl}$ gives:

But $\lambda_{jkl} = -\lambda_{jkl}$ is only possible if $\lambda_{jkl}$ is zero. Thus $\lambda_{123}$’s equation reduces to:

This gives us the general expression for $\ipdv{\vu{e}_j}{c_k}$ when $j \neq k$, but what about $j = k$? Well, from orthogonality we know:

We just calculated one of those terms, so this equation gives us the other:

Now we have the $\vu{e}_2$-component of $\ipdv{\vu{e}_1}{c_1}$, and can find the $\vu{e}_3$-component in the same way:

Adding up the $\vu{e}_2$- and $\vu{e}_3$-components gives the desired formula. There is no $\vu{e}_1$-component because $\ipdv{\vu{e}_1}{c_1}$ must be orthogonal to $\vu{e}_1$.

Proof. For any unit vector $\vu{u}$, we can project $\nabla f$ onto it to get the component of $\nabla f$ along $\vu{u}$. Let us choose $\vu{u} = \vu{e}_1$, then such a projection gives:

And we can do the same for $\vu{e}_2$ and $\vu{e}_3$, yielding analogous results:

Finally, to express $\nabla f$ in the new coordinate system $(c_1, c_2, c_3)$, we simply combine these projections for all the basis vectors:

Proof 1. From our earlier calculation of $\nabla f$, we know how to express the del $\nabla$ in $(c_1, c_2, c_3)$. Now we simply take the dot product of $\nabla$ and $\vb{V}$:

Substituting our expression for the derivatives of the local basis vectors, we find:

Where we noticed that the latter two terms cancel out if $k = j$. Now, to proceed, it is easiest to just write out the index notation:

Which can clearly be rewritten with the product rule, leading to the desired formula.

Proof 2. We take the divergence of the $c_1$-component of $\vb{V}$ and expand it:

The latter term is zero, because in any orthogonal basis $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$, and according to our gradient formula we have $\nabla c_2 = \vu{e}_2 / h_2$ etc., so:

Where we used a vector identity and the fact that the curl of a gradient must vanish. We are thus left with the former term, to which we apply our gradient formula again, where only the $\vu{e}_1$-term survives due to the dot product and orthogonality:

We then repeat this procedure for $\vb{V}$’s other components, and simply add up the results to get the desired formula.

Proof 1. From our earlier calculation of $\nabla f$, we know how to express the del $\nabla$ in $(c_1, c_2, c_3)$. Now we simply take the cross product of $\nabla$ and $\vb{V}$:

Substituting our expression for the derivatives of the local basis vectors, we find:

Because the cross product of a vector with itself is always zero. Now, in an orthonormal basis we have $\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3$, $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$ and $\vu{e}_3 \cross \vu{e}_1 = \vu{e}_2$. This is written in index notation by summing over $l$ and multiplying by the Levi-Civita symbol $\varepsilon_{jkl}$:

Where we have used that $\varepsilon_{jkl} = -\varepsilon_{kjl}$, thereby arriving at the desired formula.

Proof 2. We take the curl of the $c_1$-component of $\vb{V}$ and apply the product rule:

The latter term disappears, because $\nabla c_1 = \vu{e}_1 / h_1$ and the curl of a gradient is always zero. Applying our gradient formula to the remaining term, we find: