Categories: Mathematics, Physics.

Orthogonal curvilinear coordinates

In a 3D coordinate system, the isosurface of a coordinate (i.e. the surface where that coordinate is constant while the others vary) is known as a coordinate surface, and the intersection line of two coordinates surfaces is called a coordinate line.

A curvilinear coordinate system is one where at least one of the coordinate surfaces is curved: e.g. in cylindrical coordinates, the coordinate line of rr and zz is a circle. Here we limit ourselves to orthogonal systems, where the coordinate surfaces are always perpendicular. Examples of such orthogonal curvilinear systems are spherical coordinates, polar cylindrical coordinates, parabolic cylindrical coordinates, and (trivially) Cartesian coordinates.

Scale factors and basis vectors

Given such a system with coordinates (c1,c2,c3)(c_1, c_2, c_3). Their definition lets us convert all positions to classic Cartesian coordinates (x,y,z)(x, y, z) using functions xx, yy and zz:

x=x(c1,c2,c3)y=y(c1,c2,c3)z=z(c1,c2,c3)\begin{aligned} x &= x(c_1, c_2, c_3) \\ y &= y(c_1, c_2, c_3) \\ z &= z(c_1, c_2, c_3) \end{aligned}

A useful attribute of a coordinate system is its line element dℓ\dd{\vb{\ell}}, which represents the differential element of a line in any direction. Let e^x\vu{e}_x, e^y\vu{e}_y and e^z\vu{e}_z be the Cartesian basis unit vectors:

dℓ≡e^xdx+ e^ydy+ e^zdz\begin{aligned} \dd{\vb{\ell}} \equiv \vu{e}_x \dd{x} + \: \vu{e}_y \dd{y} + \: \vu{e}_z \dd{z} \end{aligned}

The Cartesian differential elements can be rewritten in (c1,c2,c3)(c_1, c_2, c_3) with the chain rule:

dℓ=e^x(∂x∂c1dc1+ ∂x∂c2dc2+ ∂x∂c3dc3 ⁣)+ e^y(∂y∂c1dc1+ ∂y∂c2dc2+ ∂y∂c3dc3 ⁣)+ e^z(∂z∂c1dc1+ ∂z∂c2dc2+ ∂z∂c3dc3 ⁣)=(∂x∂c1e^x+∂y∂c1e^y+∂z∂c1e^z)dc1+(∂x∂c2e^x+∂y∂c2e^y+∂z∂c2e^z)dc2+(∂x∂c3e^x+∂y∂c3e^y+∂z∂c3e^z)dc3\begin{aligned} \dd{\vb{\ell}} = \quad &\vu{e}_x \bigg( \pdv{x}{c_1} \dd{c_1} + \: \pdv{x}{c_2} \dd{c_2} + \: \pdv{x}{c_3} \dd{c_3} \!\bigg) \\ + \: &\vu{e}_y \bigg( \pdv{y}{c_1} \dd{c_1} + \: \pdv{y}{c_2} \dd{c_2} + \: \pdv{y}{c_3} \dd{c_3} \!\bigg) \\ + \: &\vu{e}_z \bigg( \pdv{z}{c_1} \dd{c_1} + \: \pdv{z}{c_2} \dd{c_2} + \: \pdv{z}{c_3} \dd{c_3} \!\bigg) \\ = \quad &\bigg( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \bigg) \dd{c_1} \\ + &\bigg( \pdv{x}{c_2} \vu{e}_x + \pdv{y}{c_2} \vu{e}_y + \pdv{z}{c_2} \vu{e}_z \bigg) \dd{c_2} \\ + &\bigg( \pdv{x}{c_3} \vu{e}_x + \pdv{y}{c_3} \vu{e}_y + \pdv{z}{c_3} \vu{e}_z \bigg) \dd{c_3} \end{aligned}

From this we define the scale factors h1h_1, h2h_2 and h3h_3 and local basis vectors e^1\vu{e}_1, e^2\vu{e}_2 and e^3\vu{e}_3:

h1e^1≡∂x∂c1e^x+∂y∂c1e^y+∂z∂c1e^zh2e^2≡∂x∂c2e^x+∂y∂c2e^y+∂z∂c2e^zh3e^3≡∂x∂c3e^x+∂y∂c3e^y+∂z∂c3e^z\begin{aligned} \boxed{ \begin{aligned} h_1 \vu{e}_1 &\equiv \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \\ h_2 \vu{e}_2 &\equiv \pdv{x}{c_2} \vu{e}_x + \pdv{y}{c_2} \vu{e}_y + \pdv{z}{c_2} \vu{e}_z \\ h_3 \vu{e}_3 &\equiv \pdv{x}{c_3} \vu{e}_x + \pdv{y}{c_3} \vu{e}_y + \pdv{z}{c_3} \vu{e}_z \end{aligned} } \end{aligned}

Where e^1\vu{e}_1, e^2\vu{e}_2 and e^3\vu{e}_3 are normalized, and orthogonal for any orthogonal curvilinear system. They are called local basis vectors because they generally depend on (c1,c2,c3)(c_1, c_2, c_3), i.e. their directions vary from position to position. Their definitions can also be inverted:

e^x≡∂c1∂xh1e^1+∂c2∂xh2e^2+∂c3∂xh3e^3e^y≡∂c1∂yh1e^1+∂c2∂yh2e^2+∂c3∂yh3e^3e^z≡∂c1∂zh1e^1+∂c2∂zh2e^2+∂c3∂zh3e^3\begin{aligned} \boxed{ \begin{aligned} \vu{e}_x &\equiv \pdv{c_1}{x} h_1 \vu{e}_1 + \pdv{c_2}{x} h_2 \vu{e}_2 + \pdv{c_3}{x} h_3 \vu{e}_3 \\ \vu{e}_y &\equiv \pdv{c_1}{y} h_1 \vu{e}_1 + \pdv{c_2}{y} h_2 \vu{e}_2 + \pdv{c_3}{y} h_3 \vu{e}_3 \\ \vu{e}_z &\equiv \pdv{c_1}{z} h_1 \vu{e}_1 + \pdv{c_2}{z} h_2 \vu{e}_2 + \pdv{c_3}{z} h_3 \vu{e}_3 \end{aligned} } \end{aligned}

In the following subsections, we use the scale factors h1h_1, h2h_2 and h3h_3 to derive general formulae for converting vector calculus from Cartesian coordinates to (c1,c2,c3)(c_1, c_2, c_3).

Differential elements

The point of the scale factors h1h_1, h2h_2 and h3h_3, as can be seen from their derivation, is to correct for “distortions” of the coordinates compared to the Cartesian system, such that the line element dℓ\dd{\vb{\ell}} retains its length. As was already established above:

dℓ=e^1h1dc1+ e^2h2dc2+ e^3h3dc3\begin{aligned} \boxed{ \dd{\vb{\ell}} = \vu{e}_1 h_1 \dd{c_1} + \: \vu{e}_2 h_2 \dd{c_2} + \: \vu{e}_3 h_3 \dd{c_3} } \end{aligned}

These terms are the differentials along each of the local basis vectors. Let us now introduce the following notation, e.g. for c1c_1:

d1 ⁣x≡∂x∂c1dc1=(∂x∂c1e^x+∂y∂c1e^y+∂z∂c1e^z)dc1=e^1h1dc1\begin{aligned} \dd{}_1\!\vb{x} \equiv \pdv{\vb{x}}{c_1} \dd{c_1} = \Big( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \Big) \dd{c_1} = \vu{e}_1 h_1 \dd{c_1} \end{aligned}

And likewise we define d2 ⁣x\dd{}_2\!\vb{x} and d3 ⁣x\dd{}_3\!\vb{x}. All differential elements (as found in e.g. integrals) can be expressed in terms of d1 ⁣x\dd{}_1\!\vb{x}, d2 ⁣x\dd{}_2\!\vb{x} and d3 ⁣x\dd{}_3\!\vb{x}.

The differential normal vector element dS^\dd{\vu{S}} in a surface integral is hence given by:

dS=d1 ⁣x×d2 ⁣x+d2 ⁣x×d3 ⁣x+d3 ⁣x×d1 ⁣x=(e^1×e^2) h1h2dc1dc2+ (e^2×e^3) h2h3dc2dc3+ (e^3×e^1) h1h3dc1dc3\begin{aligned} \dd{\vb{S}} &= \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} + \dd{}_2\!\vb{x} \cross \dd{}_3\!\vb{x} + \dd{}_3\!\vb{x} \cross \dd{}_1\!\vb{x} \\ &= (\vu{e}_1 \cross \vu{e}_2) \: h_1 h_2 \dd{c_1} \dd{c_2} + \: (\vu{e}_2 \cross \vu{e}_3) \: h_2 h_3 \dd{c_2} \dd{c_3} + \: (\vu{e}_3 \cross \vu{e}_1) \: h_1 h_3 \dd{c_1} \dd{c_3} \end{aligned}

In an orthonormal basis we have e^1×e^2=e^3\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3, e^2×e^3=e^1\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1 and e^3×e^1=e^2\vu{e}_3 \cross \vu{e}_1 = \vu{e}_2, so:

dS=e^1 h2h3dc2dc3+ e^2 h1h3dc1dc3+ e^3 h1h2dc1dc2\begin{aligned} \boxed{ \dd{\vb{S}} = \vu{e}_1 \: h_2 h_3 \dd{c_2} \dd{c_3} + \: \vu{e}_2 \: h_1 h_3 \dd{c_1} \dd{c_3} + \: \vu{e}_3 \: h_1 h_2 \dd{c_1} \dd{c_2} } \end{aligned}

Next, the differential volume dV\dd{V} must also be corrected by the scale factors:

dV=d1 ⁣x×d2 ⁣x⋅d3 ⁣x=(e^1×e^2⋅e^3) h1h2h3dc1dc2dc3\begin{aligned} \dd{V} = \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} \cdot \dd{}_3\!\vb{x} = (\vu{e}_1 \cross \vu{e}_2 \cdot \vu{e}_3) \: h_1 h_2 h_3 \dd{c_1} \dd{c_2} \dd{c_3} \end{aligned}

Once again e^1×e^2=e^3\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3, so the vectors disappear from the expression, leaving:

dV=h1h2h3dc1dc2dc3\begin{aligned} \boxed{ \dd{V} = h_1 h_2 h_3 \dd{c_1} \dd{c_2} \dd{c_3} } \end{aligned}

Basis vector derivatives

Orthonormality tells us that e^j⋅e^j=1\vu{e}_j \cdot \vu{e}_j = 1 for j=1,2,3j = 1,2,3. Differentiating with respect to ckc_k:

∂∂ck(e^j⋅e^j)=2∂e^j∂ck⋅e^j=∂∂ck1=0\begin{aligned} \pdv{}{c_k} (\vu{e}_j \cdot \vu{e}_j) = 2 \pdv{\vu{e}_j}{c_k} \cdot \vu{e}_j = \pdv{}{c_k} 1 = 0 \end{aligned}

This means that the ckc_k-derivative of e^j\vu{e}_j will always be orthogonal to e^j\vu{e}_j, for all jj and kk. Indeed, the general expression for the derivative of a local basis vector is:

∂e^j∂ck=1hj∂hk∂cje^k−δjk∑l1hl∂hj∂cle^l\begin{aligned} \boxed{ \pdv{\vu{e}_j}{c_k} = \frac{1}{h_j} \pdv{h_k}{c_j} \vu{e}_k - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_l } \end{aligned}

Where δjk\delta_{jk} is the Kronecker delta. For example, if j=1j = 1, writing this out gives:

∂e^1∂c1=−1h2∂h1∂c2e^2−1h3∂h1∂c3e^3∂e^1∂c2=1h1∂h2∂c1e^2∂e^1∂c3=1h1∂h3∂c1e^3\begin{aligned} \pdv{\vu{e}_1}{c_1} &= - \frac{1}{h_2} \pdv{h_1}{c_2} \vu{e}_2 - \frac{1}{h_3} \pdv{h_1}{c_3} \vu{e}_3 \\ \pdv{\vu{e}_1}{c_2} &= \frac{1}{h_1} \pdv{h_2}{c_1} \vu{e}_2 \\ \pdv{\vu{e}_1}{c_3} &= \frac{1}{h_1} \pdv{h_3}{c_1} \vu{e}_3 \end{aligned}

In this proof we set j=1j = 1 and k=2k = 2 for clarity, but the approach is valid for any j≠kj \neq k. We know the definitions of h1e^1h_1 \vu{e}_1 and h2e^2h_2 \vu{e}_2, and that differentiations can be reordered:

∂∂c2(h1e^1)=∂∂c2∂∂c1(xe^x+ye^y+ze^z)=∂∂c1(h2e^2)\begin{aligned} \pdv{}{c_2} (h_1 \vu{e}_1) &= \pdv{}{c_2} \pdv{}{c_1} \big( x \vu{e}_x + y \vu{e}_y + z \vu{e}_z \big) = \pdv{}{c_1} (h_2 \vu{e}_2) \end{aligned}

Expanding this according to the product rule of differentiation:

∂h1∂c2e^1+h1∂e^1∂c2=∂h2∂c1e^2+h2∂e^2∂c1\begin{aligned} \pdv{h_1}{c_2} \vu{e}_1 + h_1 \pdv{\vu{e}_1}{c_2} = \pdv{h_2}{c_1} \vu{e}_2 + h_2 \pdv{\vu{e}_2}{c_1} \end{aligned}

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

h1∂e^1∂c2=∂h2∂c1e^2+(h2∂e^2∂c1−∂h1∂c2e^1)h2∂e^2∂c1=∂h1∂c2e^1+(h1∂e^1∂c2−∂h2∂c1e^2)\begin{aligned} h_1 \pdv{\vu{e}_1}{c_2} &= \pdv{h_2}{c_1} \vu{e}_2 + \Big( h_2 \pdv{\vu{e}_2}{c_1} - \pdv{h_1}{c_2} \vu{e}_1 \Big) \\ h_2 \pdv{\vu{e}_2}{c_1} &= \pdv{h_1}{c_2} \vu{e}_1 + \Big( h_1 \pdv{\vu{e}_1}{c_2} - \pdv{h_2}{c_1} \vu{e}_2 \Big) \end{aligned}

Recall that all derivatives of e^j\vu{e}_j are orthogonal to e^j\vu{e}_j. Therefore, the first equation’s right-hand side must be orthogonal to e^1\vu{e}_1, and the second’s to e^2\vu{e}_2. We deduce that the parenthesized expressions are proportional to e^3\vu{e}_3, and call the proportionality factors λ123\lambda_{123} and λ213\lambda_{213}:

h1∂e^1∂c2=∂h2∂c1e^2+λ213e^3h2∂e^2∂c1=∂h1∂c2e^1+λ123e^3\begin{aligned} h_1 \pdv{\vu{e}_1}{c_2} &= \pdv{h_2}{c_1} \vu{e}_2 + \lambda_{213} \vu{e}_3 \\ h_2 \pdv{\vu{e}_2}{c_1} &= \pdv{h_1}{c_2} \vu{e}_1 + \lambda_{123} \vu{e}_3 \end{aligned}

Since these equations are identical, by comparing the definition of λ123\lambda_{123} to the other side of the equation, we see that λ123=λ213\lambda_{123} = \lambda_{213}:

λ123e^3=h1∂e^1∂c2−∂h2∂c1e^2=h2∂e^2∂c1−∂h1∂c2e^1=λ213e^3\begin{aligned} \lambda_{123} \vu{e}_3 &= h_1 \pdv{\vu{e}_1}{c_2} - \pdv{h_2}{c_1} \vu{e}_2 = h_2 \pdv{\vu{e}_2}{c_1} - \pdv{h_1}{c_2} \vu{e}_1 = \lambda_{213} \vu{e}_3 \end{aligned}

In general, λjkl=λkjl\lambda_{jkl} = \lambda_{kjl} for j≠k≠lj \neq k \neq l. Next, we dot-multiply λ123\lambda_{123}’s equation by e^3\vu{e}_3, using that e^2⋅e^3=0\vu{e}_2 \cdot \vu{e}_3 = 0 and consequently ∂(e^2⋅e^3)/∂c1=0\ipdv{(\vu{e}_2 \cdot \vu{e}_3)}{c_1} = 0:

λ123=h2∂e^2∂c1⋅e^3−∂h1∂c2e^1⋅e^3=h2∂e^2∂c1⋅e^3=−h2h3h3∂e^3∂c1⋅e^2=−h2h3λ132\begin{aligned} \lambda_{123} &= h_2 \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_3 - \pdv{h_1}{c_2} \vu{e}_1 \cdot \vu{e}_3 = h_2 \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_3 = - h_2 \frac{h_3}{h_3} \pdv{\vu{e}_3}{c_1} \cdot \vu{e}_2 = - \frac{h_2}{h_3} \lambda_{132} \end{aligned}

In general, λjkl=−hkλjlk/hl\lambda_{jkl} = - h_k \lambda_{jlk} / h_l for j≠k≠lj \neq k \neq l. Combining this fact with λjkl=λkjl\lambda_{jkl} = \lambda_{kjl} gives:

λjkl=−hkhlλjlk=−hkhlλljk=hkhlhjhkλlkj=hjhlλklj=−hjhlhlhjλkjl=−λjkl\begin{aligned} \lambda_{jkl} = - \frac{h_k}{h_l} \lambda_{jlk} = - \frac{h_k}{h_l} \lambda_{ljk} = \frac{h_k}{h_l} \frac{h_j}{h_k} \lambda_{lkj} = \frac{h_j}{h_l} \lambda_{klj} = - \frac{h_j}{h_l} \frac{h_l}{h_j} \lambda_{kjl} = - \lambda_{jkl} \end{aligned}

But λjkl=−λjkl\lambda_{jkl} = -\lambda_{jkl} is only possible if λjkl\lambda_{jkl} is zero. Thus λ123\lambda_{123}’s equation reduces to:

h2∂e^2∂c1=∂h1∂c2e^1  ⟹  ∂e^2∂c1=1h2∂h1∂c2e^1\begin{aligned} h_2 \pdv{\vu{e}_2}{c_1} &= \pdv{h_1}{c_2} \vu{e}_1 \qquad \implies \qquad \pdv{\vu{e}_2}{c_1} = \frac{1}{h_2} \pdv{h_1}{c_2} \vu{e}_1 \end{aligned}

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

0=e^2⋅e^1=∂∂c1(e^2⋅e^1)=∂e^2∂c1⋅e^1+e^2⋅∂e^1∂c1\begin{aligned} 0 = \vu{e}_2 \cdot \vu{e}_1 = \pdv{}{c_1} (\vu{e}_2 \cdot \vu{e}_1) = \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_1 + \vu{e}_2 \cdot \pdv{\vu{e}_1}{c_1} \end{aligned}

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

e^2⋅∂e^1∂c1=−∂e^2∂c1⋅e^1=−1h2∂h1∂c2\begin{aligned} \vu{e}_2 \cdot \pdv{\vu{e}_1}{c_1} = - \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_1 = - \frac{1}{h_2} \pdv{h_1}{c_2} \end{aligned}

Now we have the e^2\vu{e}_2-component of ∂e^1/∂c1\ipdv{\vu{e}_1}{c_1}, and can find the e^3\vu{e}_3-component in the same way:

e^3⋅∂e^1∂c1=−∂e^3∂c1⋅e^1=−1h3∂h1∂c3\begin{aligned} \vu{e}_3 \cdot \pdv{\vu{e}_1}{c_1} = - \pdv{\vu{e}_3}{c_1} \cdot \vu{e}_1 = - \frac{1}{h_3} \pdv{h_1}{c_3} \end{aligned}

Adding up the e^2\vu{e}_2- and e^3\vu{e}_3-components gives the desired formula. There is no e^1\vu{e}_1-component because ∂e^1/∂c1\ipdv{\vu{e}_1}{c_1} must be orthogonal to e^1\vu{e}_1.

Gradient of a scalar

The gradient ∇f\nabla f of a scalar field ff has the following components in (c1,c2,c3)(c_1, c_2, c_3):

(∇f)j=1hj∂f∂cj\begin{aligned} \boxed{ (\nabla f)_j = \frac{1}{h_j} \pdv{f}{c_j} } \end{aligned}

When this index notation is written out in full, the gradient ∇f\nabla f becomes:

∇f=1h1∂f∂c1e^1+1h2∂f∂c2e^2+1h3∂f∂c3e^3\begin{aligned} \nabla f = \frac{1}{h_1} \pdv{f}{c_1} \vu{e}_1 + \frac{1}{h_2} \pdv{f}{c_2} \vu{e}_2 + \frac{1}{h_3} \pdv{f}{c_3} \vu{e}_3 \end{aligned}

For any unit vector u^\vu{u}, we can project ∇f\nabla f onto it to get the component of ∇f\nabla f along u^\vu{u}. Let us choose u^=e^1\vu{u} = \vu{e}_1, then such a projection gives:

∇f⋅e^1=(∂f∂xe^x+∂f∂ye^y+∂f∂ze^z)⋅1h1(∂x∂c1e^x+∂y∂c1e^y+∂z∂c1e^z)=1h1(∂f∂x∂x∂c1+∂f∂y∂y∂c1+∂f∂z∂z∂c1)=1h1∂f∂c1\begin{aligned} \nabla f \cdot \vu{e}_1 &= \bigg( \pdv{f}{x} \vu{e}_x + \pdv{f}{y} \vu{e}_y + \pdv{f}{z} \vu{e}_z \bigg) \cdot \frac{1}{h_1} \bigg( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \bigg) \\ &= \frac{1}{h_1} \bigg( \pdv{f}{x} \pdv{x}{c_1} + \pdv{f}{y} \pdv{y}{c_1} + \pdv{f}{z} \pdv{z}{c_1} \bigg) \\ &= \frac{1}{h_1} \pdv{f}{c_1} \end{aligned}

And we can do the same for e^2\vu{e}_2 and e^3\vu{e}_3, yielding analogous results:

∇f⋅e^2=1h2∂f∂c2∇f⋅e^3=1h3∂f∂c3\begin{aligned} \nabla f \cdot \vu{e}_2 = \frac{1}{h_2} \pdv{f}{c_2} \qquad \qquad \nabla f \cdot \vu{e}_3 = \frac{1}{h_3} \pdv{f}{c_3} \end{aligned}

Finally, to express ∇f\nabla f in the new coordinate system (c1,c2,c3)(c_1, c_2, c_3), we simply combine these projections for all the basis vectors:

∇f=(∇f⋅e^1)e^1+(∇f⋅e^2)e^2+(∇f⋅e^3)e^3\begin{aligned} \nabla f = (\nabla f \cdot \vu{e}_1) \vu{e}_1 + (\nabla f \cdot \vu{e}_2) \vu{e}_2 + (\nabla f \cdot \vu{e}_3) \vu{e}_3 \end{aligned}

Divergence of a vector

The divergence of a vector field V=V1e^1+V2e^2+V3e^3\vb{V} = V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3 is given in (c1,c2,c3)(c_1, c_2, c_3) by:

∇⋅V=∑j1H∂∂cj(HVjhj)\begin{aligned} \boxed{ \nabla \cdot \vb{V} = \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j} \bigg) } \end{aligned}

Where H≡h1h2h3H \equiv h_1 h_2 h_3. When this index notation is written out in full, it becomes:

∇⋅V=1h1h2h3(∂∂c1(h2h3V1)+∂∂c2(h1h3V2)+∂∂c3(h1h2V3))\begin{aligned} \nabla \cdot \vb{V} = \frac{1}{h_1 h_2 h_3} \bigg( \pdv{}{c_1} (h_2 h_3 V_1) + \pdv{}{c_2} (h_1 h_3 V_2) + \pdv{}{c_3} (h_1 h_2 V_3) \bigg) \end{aligned}

From our earlier calculation of ∇f\nabla f, we know how to express the del ∇\nabla in (c1,c2,c3)(c_1, c_2, c_3). Now we simply take the dot product of ∇\nabla and V\vb{V}:

∇⋅V=(e^11h1∂∂c1+e^21h2∂∂c2+e^31h3∂∂c3)⋅(V1e^1+V2e^2+V3e^3)=(∑je^j1hj∂∂cj)⋅(∑kVke^k)=∑jke^j⋅1hj∂∂cj(Vke^k)=∑jk(e^j⋅e^k)1hj∂Vk∂cj+∑jk(e^j⋅∂e^k∂cj)Vkhj\begin{aligned} \nabla \cdot \vb{V} &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg) \cdot \bigg( V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3 \bigg) \\ &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \cdot \bigg( \sum_{k} V_k \vu{e}_k \bigg) \\ &= \sum_{jk} \vu{e}_j \cdot \frac{1}{h_j} \pdv{}{c_j} (V_k \vu{e}_k) \\ &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j} + \sum_{jk} \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{V_k}{h_j} \end{aligned}

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

∇⋅V=∑jk(e^j⋅e^k)1hj∂Vk∂cj+∑jke^j⋅(1hk∂hj∂cke^j−δjk∑l1hl∂hk∂cle^l)Vkhj=∑jk(e^j⋅e^k)1hj∂Vk∂cj+∑jk(e^j⋅e^j)Vkhjhk∂hj∂ck−∑jl(e^j⋅e^l)Vjhjhl∂hj∂cl=∑j1hj∂Vj∂cj+∑jkVkhjhk∂hj∂ck−∑jVjhjhj∂hj∂cj=∑j1hj∂Vj∂cj+∑j∑k≠jVkhjhk∂hj∂ck\begin{aligned} \nabla \cdot \vb{V} &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j} + \sum_{jk} \vu{e}_j \cdot \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \bigg) \frac{V_k}{h_j} \\ &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j} + \sum_{jk} (\vu{e}_j \cdot \vu{e}_j) \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} - \sum_{jl} (\vu{e}_j \cdot \vu{e}_l) \frac{V_j}{h_j h_l} \pdv{h_j}{c_l} \\ &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{jk} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} - \sum_{j} \frac{V_j}{h_j h_j} \pdv{h_j}{c_j} \\ &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} \\ \end{aligned}

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

∇⋅V=∑j1hj∂Vj∂cj+∑j∑k≠jVkhjhk∂hj∂ck=1h1∂V1∂c1+V1h1h2∂h2∂c1+V1h1h3∂h3∂c1 +V2h1h2∂h1∂c2+1h2∂V2∂c2+V2h2h3∂h3∂c2 +V3h1h3∂h1∂c3+V3h2h3∂h2∂c3+1h3∂V3∂c3=1h1h2h3(h2h3∂V1∂c1+h3V1∂h2∂c1+h2V1∂h3∂c1+h3V2∂h1∂c2+h1h3∂V2∂c2+h1V2∂h3∂c2+h2V3∂h1∂c3+h1V3∂h2∂c3+h1h2∂V3∂c3)\begin{aligned} \nabla \cdot \vb{V} &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} \\ &= \quad \frac{1}{h_1} \pdv{V_1}{c_1} + \frac{V_1}{h_1 h_2} \pdv{h_2}{c_1} + \frac{V_1}{h_1 h_3} \pdv{h_3}{c_1} \\ &\quad\: + \frac{V_2}{h_1 h_2} \pdv{h_1}{c_2} + \frac{1}{h_2} \pdv{V_2}{c_2} + \frac{V_2}{h_2 h_3} \pdv{h_3}{c_2} \\ &\quad\: + \frac{V_3}{h_1 h_3} \pdv{h_1}{c_3} + \frac{V_3}{h_2 h_3} \pdv{h_2}{c_3} + \frac{1}{h_3} \pdv{V_3}{c_3} \\ &= \frac{1}{h_1 h_2 h_3} \bigg( h_2 h_3 \pdv{V_1}{c_1} + h_3 V_1 \pdv{h_2}{c_1} + h_2 V_1 \pdv{h_3}{c_1} \\ &\qquad\qquad + h_3 V_2 \pdv{h_1}{c_2} + h_1 h_3 \pdv{V_2}{c_2} + h_1 V_2 \pdv{h_3}{c_2} \\ &\qquad\qquad + h_2 V_3 \pdv{h_1}{c_3} + h_1 V_3 \pdv{h_2}{c_3} + h_1 h_2 \pdv{V_3}{c_3} \bigg) \end{aligned}

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

Boas gives an alternative proof, which is shorter but more specialized:

We take the divergence of the c1c_1-component of V\vb{V} and expand it:

∇⋅(V1e^1)=∇⋅((h2h3V1)(e^1h2h3))=∇(h2h3V1)⋅(e^1h2h3)+(h2h3V1)(∇⋅(e^1h2h3))\begin{aligned} \nabla \cdot (V_1 \vu{e}_1) &= \nabla \cdot \bigg( \Big( h_2 h_3 V_1 \Big) \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) \bigg) \\ &= \nabla (h_2 h_3 V_1) \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) + (h_2 h_3 V_1) \bigg( \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) \bigg) \end{aligned}

The latter term is zero, because in any orthogonal basis e^2×e^3=e^1\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1, and according to our gradient formula we have ∇c2=e^2/h2\nabla c_2 = \vu{e}_2 / h_2 etc., so:

∇⋅(e^1h2h3)=∇⋅(e^2h2×e^3h3)=∇⋅(∇c2×∇c3)=∇c3⋅(∇×∇c2)−∇c2⋅(∇×∇c3)=0\begin{aligned} \nabla \cdot \bigg( \frac{\vu{e}_1}{h_2 h_3} \bigg) &= \nabla \cdot \bigg( \frac{\vu{e}_2}{h_2} \cross \frac{\vu{e}_3}{h_3} \bigg) \\ &= \nabla \cdot \big( \nabla c_2 \cross \nabla c_3 \big) \\ &= \nabla c_3 \cdot (\nabla \cross \nabla c_2) - \nabla c_2 \cdot (\nabla \cross \nabla c_3) \\ &= 0 \end{aligned}

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 e^1\vu{e}_1-term survives due to the dot product and orthogonality:

∇⋅(V1e^1)=∇(h2h3V1)⋅e^1h2h3=1h1h2h3∂∂c1(h2h3V1)\begin{aligned} \nabla \cdot (V_1 \vu{e}_1) &= \nabla (h_2 h_3 V_1) \cdot \frac{\vu{e}_1}{h_2 h_3} \\ &= \frac{1}{h_1 h_2 h_3} \pdv{}{c_1} (h_2 h_3 V_1) \end{aligned}