---
title: "Orthogonal curvilinear coordinates"
sort_title: "Orthogonal curvilinear coordinates"
date: 2023-05-29 # Originally 2021-03-03, major rewrite
categories:
- Mathematics
- Physics
layout: "concept"
---

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 $$r$$ and $$z$$ 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](/know/concept/spherical-coordinates/),
[polar cylindrical coordinates](/know/concept/polar-cylindrical-coordinates/),
[parabolic cylindrical coordinates](/know/concept/parabolic-cylindrical-coordinates/),
and (trivially) [Cartesian coordinates](/know/concept/cartesian-coordinates/).



## Scale factors and basis vectors

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

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

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

$$\begin{aligned}
    \dd{\vb{\ell}}
    \equiv \vu{e}_x \dd{x} + \: \vu{e}_y \dd{y} + \: \vu{e}_z \dd{z}
\end{aligned}$$

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

$$\begin{aligned}
    \dd{\vb{\ell}}
    = \quad &\vu{e}_x \bigg( \pdv{x}{c_1} \dd{c_1} + \: \pdv{x}{c_2} \dd{c_2} + \: \pdv{x}{c_3} \dd{c_3} \!\bigg)
    \\
    + \: &\vu{e}_y \bigg( \pdv{y}{c_1} \dd{c_1} + \: \pdv{y}{c_2} \dd{c_2} + \: \pdv{y}{c_3} \dd{c_3} \!\bigg)
    \\
    + \: &\vu{e}_z \bigg( \pdv{z}{c_1} \dd{c_1} + \: \pdv{z}{c_2} \dd{c_2} + \: \pdv{z}{c_3} \dd{c_3} \!\bigg)
    \\
    = \quad &\bigg( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \bigg) \dd{c_1}
    \\
    + &\bigg( \pdv{x}{c_2} \vu{e}_x + \pdv{y}{c_2} \vu{e}_y + \pdv{z}{c_2} \vu{e}_z \bigg) \dd{c_2}
    \\
    + &\bigg( \pdv{x}{c_3} \vu{e}_x + \pdv{y}{c_3} \vu{e}_y + \pdv{z}{c_3} \vu{e}_z \bigg) \dd{c_3}
\end{aligned}$$

From this we define the **scale factors** $$h_1$$, $$h_2$$ and $$h_3$$
and **local basis vectors** $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            h_1 \vu{e}_1
            &\equiv \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z
            \\
            h_2 \vu{e}_2
            &\equiv \pdv{x}{c_2} \vu{e}_x + \pdv{y}{c_2} \vu{e}_y + \pdv{z}{c_2} \vu{e}_z
            \\
            h_3 \vu{e}_3
            &\equiv \pdv{x}{c_3} \vu{e}_x + \pdv{y}{c_3} \vu{e}_y + \pdv{z}{c_3} \vu{e}_z
        \end{aligned}
    }
\end{aligned}$$

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

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \vu{e}_x
            &\equiv \pdv{c_1}{x} h_1 \vu{e}_1 + \pdv{c_2}{x} h_2 \vu{e}_2 + \pdv{c_3}{x} h_3 \vu{e}_3
            \\
            \vu{e}_y
            &\equiv \pdv{c_1}{y} h_1 \vu{e}_1 + \pdv{c_2}{y} h_2 \vu{e}_2 + \pdv{c_3}{y} h_3 \vu{e}_3
            \\
            \vu{e}_z
            &\equiv \pdv{c_1}{z} h_1 \vu{e}_1 + \pdv{c_2}{z} h_2 \vu{e}_2 + \pdv{c_3}{z} h_3 \vu{e}_3
        \end{aligned}
    }
\end{aligned}$$

In the following subsections, we use the scale factors $$h_1$$, $$h_2$$ and $$h_3$$
to derive general formulae for converting vector calculus
from Cartesian coordinates to $$(c_1, c_2, c_3)$$.



## Differential elements

The point of the scale factors $$h_1$$, $$h_2$$ and $$h_3$$,
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 $$\dd{\vb{\ell}}$$ retains its length.
As was already established above:

$$\begin{aligned}
    \boxed{
        \dd{\vb{\ell}}
        = \vu{e}_1 h_1 \dd{c_1} + \: \vu{e}_2 h_2 \dd{c_2} + \: \vu{e}_3 h_3 \dd{c_3}
    }
\end{aligned}$$

These terms are the differentials along each of the local basis vectors.
Let us now introduce the following notation, e.g. for $$c_1$$:

$$\begin{aligned}
    \dd{}_1\!\vb{x}
    \equiv \pdv{\vb{x}}{c_1} \dd{c_1}
    = \Big( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \Big) \dd{c_1}
    = \vu{e}_1 h_1 \dd{c_1}
\end{aligned}$$

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

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

$$\begin{aligned}
    \dd{\vb{S}}
    &= \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} + \dd{}_2\!\vb{x} \cross \dd{}_3\!\vb{x} + \dd{}_3\!\vb{x} \cross \dd{}_1\!\vb{x}
    \\
    &= (\vu{e}_1 \cross \vu{e}_2) \: h_1 h_2 \dd{c_1} \dd{c_2}
    + \: (\vu{e}_2 \cross \vu{e}_3) \: h_2 h_3 \dd{c_2} \dd{c_3}
    + \: (\vu{e}_3 \cross \vu{e}_1) \: h_1 h_3 \dd{c_1} \dd{c_3}
\end{aligned}$$

In an orthonormal basis we have
$$\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3$$,
$$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$ and
$$\vu{e}_3 \cross \vu{e}_1 = \vu{e}_2$$, so:

$$\begin{aligned}
    \boxed{
        \dd{\vb{S}}
        = \vu{e}_1 \: h_2 h_3 \dd{c_2} \dd{c_3} + \: \vu{e}_2 \: h_1 h_3 \dd{c_1} \dd{c_3} + \: \vu{e}_3 \: h_1 h_2 \dd{c_1} \dd{c_2}
    }
\end{aligned}$$

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

$$\begin{aligned}
    \dd{V}
    = \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} \cdot \dd{}_3\!\vb{x}
    = (\vu{e}_1 \cross \vu{e}_2 \cdot \vu{e}_3) \: h_1 h_2 h_3 \dd{c_1} \dd{c_2} \dd{c_3}
\end{aligned}$$

Once again $$\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3$$,
so the vectors disappear from the expression, leaving:

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



## Basis vector derivatives

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

$$\begin{aligned}
    \pdv{}{c_k} (\vu{e}_j \cdot \vu{e}_j)
    = 2 \pdv{\vu{e}_j}{c_k} \cdot \vu{e}_j
    = \pdv{}{c_k} 1
    = 0
\end{aligned}$$

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

$$\begin{aligned}
    \boxed{
        \pdv{\vu{e}_j}{c_k}
        = \frac{1}{h_j} \pdv{h_k}{c_j} \vu{e}_k - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_l
    }
\end{aligned}$$

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

$$\begin{aligned}
    \pdv{\vu{e}_1}{c_1}
    &= - \frac{1}{h_2} \pdv{h_1}{c_2} \vu{e}_2 - \frac{1}{h_3} \pdv{h_1}{c_3} \vu{e}_3
    \\
    \pdv{\vu{e}_1}{c_2}
    &= \frac{1}{h_1} \pdv{h_2}{c_1} \vu{e}_2
    \\
    \pdv{\vu{e}_1}{c_3}
    &= \frac{1}{h_1} \pdv{h_3}{c_1} \vu{e}_3
\end{aligned}$$


{% include proof/start.html id="proof-deriv-basis" -%}
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:

$$\begin{aligned}
    \pdv{}{c_2} (h_1 \vu{e}_1)
    &= \pdv{}{c_2} \pdv{}{c_1} \big( x \vu{e}_x + y \vu{e}_y + z \vu{e}_z \big)
    = \pdv{}{c_1} (h_2 \vu{e}_2)
\end{aligned}$$

Expanding this according to the product rule of differentiation:

$$\begin{aligned}
    \pdv{h_1}{c_2} \vu{e}_1 + h_1 \pdv{\vu{e}_1}{c_2}
    = \pdv{h_2}{c_1} \vu{e}_2 + h_2 \pdv{\vu{e}_2}{c_1}
\end{aligned}$$

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

$$\begin{aligned}
    h_1 \pdv{\vu{e}_1}{c_2}
    &= \pdv{h_2}{c_1} \vu{e}_2 + \Big( h_2 \pdv{\vu{e}_2}{c_1} - \pdv{h_1}{c_2} \vu{e}_1 \Big)
    \\
    h_2 \pdv{\vu{e}_2}{c_1}
    &= \pdv{h_1}{c_2} \vu{e}_1 + \Big( h_1 \pdv{\vu{e}_1}{c_2} - \pdv{h_2}{c_1} \vu{e}_2 \Big)
\end{aligned}$$

Recall that all derivatives of $$\vu{e}_j$$ 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 $$\lambda_{213}$$:

$$\begin{aligned}
    h_1 \pdv{\vu{e}_1}{c_2}
    &= \pdv{h_2}{c_1} \vu{e}_2 + \lambda_{213} \vu{e}_3
    \\
    h_2 \pdv{\vu{e}_2}{c_1}
    &= \pdv{h_1}{c_2} \vu{e}_1 + \lambda_{123} \vu{e}_3
\end{aligned}$$

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

$$\begin{aligned}
    \lambda_{123} \vu{e}_3
    &= h_1 \pdv{\vu{e}_1}{c_2} - \pdv{h_2}{c_1} \vu{e}_2
    = h_2 \pdv{\vu{e}_2}{c_1} - \pdv{h_1}{c_2} \vu{e}_1
    = \lambda_{213} \vu{e}_3
\end{aligned}$$

In general, $$\lambda_{jkl} = \lambda_{kjl}$$ 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 $$\ipdv{(\vu{e}_2 \cdot \vu{e}_3)}{c_1} = 0$$:

$$\begin{aligned}
    \lambda_{123}
    &= h_2 \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_3 - \pdv{h_1}{c_2} \vu{e}_1 \cdot \vu{e}_3
    = h_2 \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_3
    = - h_2 \frac{h_3}{h_3} \pdv{\vu{e}_3}{c_1} \cdot \vu{e}_2
    = - \frac{h_2}{h_3} \lambda_{132}
\end{aligned}$$

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

$$\begin{aligned}
    \lambda_{jkl}
    = - \frac{h_k}{h_l} \lambda_{jlk}
    = - \frac{h_k}{h_l} \lambda_{ljk}
    = \frac{h_k}{h_l} \frac{h_j}{h_k} \lambda_{lkj}
    = \frac{h_j}{h_l} \lambda_{klj}
    = - \frac{h_j}{h_l} \frac{h_l}{h_j} \lambda_{kjl}
    = - \lambda_{jkl}
\end{aligned}$$

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

$$\begin{aligned}
    h_2 \pdv{\vu{e}_2}{c_1}
    &= \pdv{h_1}{c_2} \vu{e}_1
    \qquad \implies \qquad
    \pdv{\vu{e}_2}{c_1}
    = \frac{1}{h_2} \pdv{h_1}{c_2} \vu{e}_1
\end{aligned}$$

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

$$\begin{aligned}
    0
    = \vu{e}_2 \cdot \vu{e}_1
    = \pdv{}{c_1} (\vu{e}_2 \cdot \vu{e}_1)
    = \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_1 + \vu{e}_2 \cdot \pdv{\vu{e}_1}{c_1}
\end{aligned}$$

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

$$\begin{aligned}
    \vu{e}_2 \cdot \pdv{\vu{e}_1}{c_1}
    = - \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_1
    = -  \frac{1}{h_2} \pdv{h_1}{c_2}
\end{aligned}$$

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

$$\begin{aligned}
    \vu{e}_3 \cdot \pdv{\vu{e}_1}{c_1}
    = - \pdv{\vu{e}_3}{c_1} \cdot \vu{e}_1
    = -  \frac{1}{h_3} \pdv{h_1}{c_3}
\end{aligned}$$

Adding up the $$\vu{e}_2$$- 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 $$\vu{e}_1$$.
{% include proof/end.html id="proof-deriv-basis" -%}



## Gradient of a scalar

The gradient $$\nabla f$$ of a scalar field $$f$$
has the following components in $$(c_1, c_2, c_3)$$:

$$\begin{aligned}
    \boxed{
        (\nabla f)_j
        = \frac{1}{h_j} \pdv{f}{c_j}
    }
\end{aligned}$$

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

$$\begin{aligned}
    \nabla f
    = \frac{1}{h_1} \pdv{f}{c_1} \vu{e}_1
    + \frac{1}{h_2} \pdv{f}{c_2} \vu{e}_2
    + \frac{1}{h_3} \pdv{f}{c_3} \vu{e}_3
\end{aligned}$$


{% include proof/start.html id="proof-grad-scalar" -%}
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:

$$\begin{aligned}
    \nabla f \cdot \vu{e}_1
    &= \bigg( \pdv{f}{x} \vu{e}_x + \pdv{f}{y} \vu{e}_y + \pdv{f}{z} \vu{e}_z \bigg)
    \cdot \frac{1}{h_1} \bigg( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \bigg)
    \\
    &= \frac{1}{h_1} \bigg( \pdv{f}{x} \pdv{x}{c_1} + \pdv{f}{y} \pdv{y}{c_1} + \pdv{f}{z} \pdv{z}{c_1} \bigg)
    \\
    &= \frac{1}{h_1} \pdv{f}{c_1}
\end{aligned}$$

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

$$\begin{aligned}
    \nabla f \cdot \vu{e}_2
    = \frac{1}{h_2} \pdv{f}{c_2}
    \qquad \qquad
    \nabla f \cdot \vu{e}_3
    = \frac{1}{h_3} \pdv{f}{c_3}
\end{aligned}$$

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

$$\begin{aligned}
    \nabla f
    = (\nabla f \cdot \vu{e}_1) \vu{e}_1
    + (\nabla f \cdot \vu{e}_2) \vu{e}_2
    + (\nabla f \cdot \vu{e}_3) \vu{e}_3
\end{aligned}$$
{% include proof/end.html id="proof-grad-scalar" %}



## Divergence of a vector

The divergence of a vector field $$\vb{V} = V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3$$
is given in $$(c_1, c_2, c_3)$$ by:

$$\begin{aligned}
    \boxed{
        \nabla \cdot \vb{V}
        = \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j} \bigg)
    }
\end{aligned}$$

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

$$\begin{aligned}
    \nabla \cdot \vb{V}
    = \frac{1}{h_1 h_2 h_3}
    \bigg( \pdv{}{c_1} (h_2 h_3 V_1) + \pdv{}{c_2} (h_1 h_3 V_2) + \pdv{}{c_3} (h_1 h_2 V_3) \bigg)
\end{aligned}$$


{% include proof/start.html id="proof-div-vector-1" label="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 $$\vb{V}$$:

$$\begin{aligned}
    \nabla \cdot \vb{V}
    &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg)
    \cdot \bigg( V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3 \bigg)
    \\
    &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \cdot \bigg( \sum_{k} V_k \vu{e}_k \bigg)
    \\
    &= \sum_{jk} \vu{e}_j \cdot \frac{1}{h_j} \pdv{}{c_j} (V_k \vu{e}_k)
    \\
    &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j}
    + \sum_{jk} \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{V_k}{h_j}
\end{aligned}$$

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

$$\begin{aligned}
    \nabla \cdot \vb{V}
    &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j}
    + \sum_{jk} \vu{e}_j
    \cdot \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \bigg) \frac{V_k}{h_j}
    \\
    &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j}
    + \sum_{jk} (\vu{e}_j \cdot \vu{e}_j) \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
    - \sum_{jl} (\vu{e}_j \cdot \vu{e}_l) \frac{V_j}{h_j h_l} \pdv{h_j}{c_l}
    \\
    &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j}
    + \sum_{jk} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
    - \sum_{j} \frac{V_j}{h_j h_j} \pdv{h_j}{c_j}
    \\
    &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j}
    + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
    \\
\end{aligned}$$

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

$$\begin{aligned}
    \nabla \cdot \vb{V}
    &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
    \\
    &= \quad \frac{1}{h_1} \pdv{V_1}{c_1} + \frac{V_1}{h_1 h_2} \pdv{h_2}{c_1} + \frac{V_1}{h_1 h_3} \pdv{h_3}{c_1}
    \\
    &\quad\: + \frac{V_2}{h_1 h_2} \pdv{h_1}{c_2} + \frac{1}{h_2} \pdv{V_2}{c_2} + \frac{V_2}{h_2 h_3} \pdv{h_3}{c_2}
    \\
    &\quad\: + \frac{V_3}{h_1 h_3} \pdv{h_1}{c_3} + \frac{V_3}{h_2 h_3} \pdv{h_2}{c_3} + \frac{1}{h_3} \pdv{V_3}{c_3}
    \\
    &= \frac{1}{h_1 h_2 h_3} \bigg( h_2 h_3 \pdv{V_1}{c_1} + h_3 V_1 \pdv{h_2}{c_1} + h_2 V_1 \pdv{h_3}{c_1}
    \\
    &\qquad\qquad + h_3 V_2 \pdv{h_1}{c_2} + h_1 h_3 \pdv{V_2}{c_2} + h_1 V_2 \pdv{h_3}{c_2}
    \\
    &\qquad\qquad + h_2 V_3 \pdv{h_1}{c_3} + h_1 V_3 \pdv{h_2}{c_3} + h_1 h_2 \pdv{V_3}{c_3} \bigg)
\end{aligned}$$

Which can clearly be rewritten with the product rule,
leading to the desired formula.
{% include proof/end.html id="proof-div-vector-1" label="Proof 1" %}


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


{% include proof/start.html id="proof-div-vector-2" label="Proof 2" -%}
We take the divergence of the $$c_1$$-component of $$\vb{V}$$ and expand it:

$$\begin{aligned}
    \nabla \cdot (V_1 \vu{e}_1)
    &= \nabla \cdot \bigg( \Big( h_2 h_3 V_1 \Big) \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) \bigg)
    \\
    &= \nabla (h_2 h_3 V_1) \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} \Big)
    + (h_2 h_3 V_1) \bigg( \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) \bigg)
\end{aligned}$$

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

$$\begin{aligned}
    \nabla \cdot \bigg( \frac{\vu{e}_1}{h_2 h_3} \bigg)
    &= \nabla \cdot \bigg( \frac{\vu{e}_2}{h_2} \cross \frac{\vu{e}_3}{h_3} \bigg)
    \\
    &= \nabla \cdot \big( \nabla c_2 \cross \nabla c_3 \big)
    \\
    &= \nabla c_3 \cdot (\nabla \cross \nabla c_2) - \nabla c_2 \cdot (\nabla \cross \nabla c_3)
    \\
    &= 0
\end{aligned}$$

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

$$\begin{aligned}
    \nabla \cdot (V_1 \vu{e}_1)
    &= \nabla (h_2 h_3 V_1) \cdot \frac{\vu{e}_1}{h_2 h_3}
    \\
    &= \frac{1}{h_1 h_2 h_3} \pdv{}{c_1} (h_2 h_3 V_1)
\end{aligned}$$

We then repeat this procedure for $$\vb{V}$$'s other components,
and simply add up the results to get the desired formula.
{% include proof/end.html id="proof-div-vector-2" label="Proof 2" %}



## Curl of a vector

The curl of a vector field $$\vb{V}$$ has the following components in $$(c_1, c_2, c_3)$$,
where $$\varepsilon_{j k l}$$ is the *Levi-Civita symbol* in 3D:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            (\nabla \cross \vb{V})_j
            = \sum_{k l} \frac{\varepsilon_{j k l}}{h_k h_l} \pdv{}{c_k} (h_l V_l)
        \end{aligned}
    }
\end{aligned}$$

When this index notation is written out in full,
the curl $$\nabla \cross \vb{V}$$ becomes:

$$\begin{aligned}
    \nabla \times \vb{V}
    = \quad\: &\frac{1}{h_2 h_3} \bigg( \pdv{(h_3 V_3)}{c_2} - \pdv{(h_2 V_2)}{c_3} \bigg) \vu{e}_1
    \\
    + \: &\frac{1}{h_1 h_3} \bigg( \pdv{(h_1 V_1)}{c_3} - \pdv{(h_3 V_3)}{c_1} \bigg) \vu{e}_2
    \\
    + \: &\frac{1}{h_1 h_2} \bigg( \pdv{(h_2 V_2)}{c_1} - \pdv{(h_1 V_1)}{c_2} \bigg) \vu{e}_3
\end{aligned}$$


{% include proof/start.html id="proof-curl-vector-1" label="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 $$\vb{V}$$:

$$\begin{aligned}
    \nabla \cross \vb{V}
    &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg)
    \cross \bigg( V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3 \bigg)
    \\
    &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \cross \bigg( \sum_{k} V_k \vu{e}_k \bigg)
    \\
    &= \sum_{jk} \vu{e}_j \cross \frac{1}{h_j} \pdv{}{c_j} (V_k \vu{e}_k)
    \\
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} (\vu{e}_j \cross \vu{e}_k)
    + \sum_{jk} \frac{V_k}{h_j} \Big( \vu{e}_j \cross \pdv{\vu{e}_k}{c_j} \Big)
\end{aligned}$$

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

$$\begin{aligned}
    \nabla \cross \vb{V}
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} (\vu{e}_j \cross \vu{e}_k)
    + \sum_{jk} \frac{V_k}{h_j} \vu{e}_j
    \cross \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \bigg)
    \\
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} (\vu{e}_j \cross \vu{e}_k)
    + \sum_{jk} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} (\vu{e}_j \cross \vu{e}_j)
    - \sum_{jl} \frac{V_j}{h_j h_l} \pdv{h_j}{c_l} (\vu{e}_j \cross \vu{e}_l)
    \\
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} (\vu{e}_j \cross \vu{e}_k)
    - \sum_{jl} \frac{V_j}{h_j h_l} \pdv{h_j}{c_l} (\vu{e}_j \cross \vu{e}_l)
\end{aligned}$$

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 $$\varepsilon_{jkl}$$:

$$\begin{aligned}
    \nabla \cross \vb{V}
    &= \sum_{jk} \bigg( \frac{1}{h_j} \pdv{V_k}{c_j} - \frac{V_j}{h_j h_k} \pdv{h_j}{c_k} \bigg) (\vu{e}_j \cross \vu{e}_k)
    \\
    &= \sum_{jkl} \varepsilon_{jkl} \bigg( \frac{1}{h_j} \pdv{V_k}{c_j} - \frac{V_j}{h_j h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_l
    \\
    &= \sum_{jkl} \varepsilon_{jkl} \bigg( \frac{1}{h_j} \frac{h_k}{h_k} \pdv{V_k}{c_j} + \frac{V_k}{h_j h_k} \pdv{h_k}{c_j} \bigg) \vu{e}_l
    \\
    &= \sum_{jkl} \varepsilon_{jkl} \bigg( \frac{1}{h_j h_k} \pdv{}{c_j} (h_k V_k) \bigg) \vu{e}_l
\end{aligned}$$

Where we have used that $$\varepsilon_{jkl} = -\varepsilon_{kjl}$$,
thereby arriving at the desired formula.
{% include proof/end.html id="proof-curl-vector-1" label="Proof 1" %}


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


{% include proof/start.html id="proof-curl-vector-2" label="Proof 2" -%}
We take the curl of the $$c_1$$-component of $$\vb{V}$$ and apply the product rule:

$$\begin{aligned}
    \nabla \cross (V_1 \vu{e}_1)
    &= \nabla \cross \bigg( \Big( h_1 V_1 \Big) \Big( \frac{\vu{e}_1}{h_1} \Big) \bigg)
    \\
    &= \nabla (h_1 V_1) \cross \Big( \frac{\vu{e}_1}{h_1} \Big) + (h_1 V_1) \Big( \nabla \cross \frac{\vu{e}_1}{h_1} \Big)
\end{aligned}$$

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:

$$\begin{aligned}
    \nabla \cross (V_1 \vu{e}_1)
    &= \nabla (h_1 V_1) \cross \Big( \frac{\vu{e}_1}{h_1} \Big)
    \\
    &= \bigg( \frac{1}{h_1} \pdv{(h_1 V_1)}{c_1} \vu{e}_1
    + \frac{1}{h_2} \pdv{(h_1 V_1)}{c_2} \vu{e}_2
    + \frac{1}{h_3} \pdv{(h_1 V_1)}{c_3} \vu{e}_3 \bigg)
    \cross \Big( \frac{\vu{e}_1}{h_1} \Big)
    \\
    &= 0 - \frac{1}{h_1 h_2} \pdv{(h_1 V_1)}{c_2} \vu{e}_3 + \frac{1}{h_1 h_3} \pdv{(h_1 V_1)}{c_3} \vu{e}_2
\end{aligned}$$

Where we have used the fact that $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$
are related to each other by cross products thanks to orthonormality,
e.g. $$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$.
We then repeat this procedure for $$\vb{V}$$'s other components,
and simply add up the results to get the desired formula.
{% include proof/end.html id="proof-curl-vector-2" label="Proof 2" %}



## Laplacian of a scalar

The Laplacian $$\nabla^2 f$$ of a scalar field $$f$$
is calculated as follows in $$(c_1, c_2, c_3)$$:

$$\begin{aligned}
    \boxed{
        \nabla^2 f
        = \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{f}{c_j} \bigg)
    }
\end{aligned}$$

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

$$\begin{aligned}
    \nabla^2 f
    = \frac{1}{h_1 h_2 h_3} \bigg( \pdv{}{c_1} \Big( \frac{h_2 h_3}{h_1} \pdv{f}{c_1} \Big)
    + \pdv{}{c_2} \Big( \frac{h_1 h_3}{h_2} \pdv{f}{c_2} \Big)
    + \pdv{}{c_3} \Big( \frac{h_1 h_2}{h_3} \pdv{f}{c_3} \Big) \bigg)
\end{aligned}$$

This is trivial to prove: $$\nabla^2 f = \nabla \cdot (\nabla f)$$,
so combining our previous formulae is enough.



## Gradient of a divergence

The gradient of a divergence $$\nabla (\nabla \cdot \vb{V})$$
has the following components in $$(c_1, c_2, c_3)$$:

$$\begin{aligned}
    \boxed{
        \big( \nabla (\nabla \cdot \vb{V}) \big)_j
        = \frac{1}{h_j} \pdv{}{c_j} \bigg( \sum_{k} \frac{1}{H} \pdv{}{c_k} \Big( \frac{H V_k}{h_k} \Big) \bigg)
    }
\end{aligned}$$

Where $$H \equiv h_1 h_2 h_3$$.
This is trivial to prove: $$\nabla \cdot \vb{V}$$ is a scalar,
which we insert into our gradient formula.
We no longer write out the index notation,
as the formulae become quite long.



## Gradient of a vector

It also possible to take the gradient of a vector
$$\vb{V} = V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3$$,
yielding a 2nd-order tensor with the following components
in $$(c_1, c_2, c_3)$$, for $$j \neq k$$:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            (\nabla \vb{V})_{jj}
            &= \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
            \\
            (\nabla \vb{V})_{jk}
            &= \frac{1}{h_j} \pdv{V_k}{c_j} - \frac{V_j}{h_j h_k} \pdv{h_j}{c_k}
        \end{aligned}
    }
\end{aligned}$$

{% comment %}
When this index notation is written out in full,
the gradient $$\nabla \vb{V}$$ becomes:

$$\begin{aligned}
    \nabla \vb{V}
    = \quad
    \: &\bigg( \frac{1}{h_1} \pdv{V_1}{c_1} + \frac{V_2}{h_1 h_2} \pdv{h_1}{c_2} + \frac{V_3}{h_1 h_3} \pdv{h_1}{c_3} \bigg) \vu{e}_1 \vu{e}_1
    \\
    + \: &\bigg( \frac{1}{h_1} \pdv{V_2}{c_1} - \frac{V_1}{h_1 h_2} \pdv{h_1}{c_2} \bigg) \vu{e}_1 \vu{e}_2
    + \bigg( \frac{1}{h_1} \pdv{V_3}{c_1} - \frac{V_1}{h_1 h_3} \pdv{h_1}{c_3} \bigg) \vu{e}_1 \vu{e}_3
    \\
    + \: &\bigg( \frac{1}{h_2} \pdv{V_2}{c_2} + \frac{V_1}{h_1 h_2} \pdv{h_2}{c_1} + \frac{V_3}{h_2 h_3} \pdv{h_2}{c_3} \bigg) \vu{e}_2 \vu{e}_2
    \\
    + \: &\bigg( \frac{1}{h_2} \pdv{V_1}{c_2} - \frac{V_2}{h_1 h_2} \pdv{h_2}{c_1} \bigg) \vu{e}_2 \vu{e}_1
    + \bigg( \frac{1}{h_2} \pdv{V_3}{c_2} - \frac{V_2}{h_2 h_3} \pdv{h_2}{c_3} \bigg) \vu{e}_2 \vu{e}_3
    \\
    + \: &\bigg( \frac{1}{h_3} \pdv{V_3}{c_3} + \frac{V_1}{h_1 h_3} \pdv{h_3}{c_1} + \frac{V_2}{h_2 h_3} \pdv{h_3}{c_2} \bigg) \vu{e}_3 \vu{e}_3
    \\
    + \: &\bigg( \frac{1}{h_3} \pdv{V_1}{c_3} - \frac{V_3}{h_1 h_3} \pdv{h_3}{c_1} \bigg) \vu{e}_3 \vu{e}_1
    + \bigg( \frac{1}{h_3} \pdv{V_2}{c_3} - \frac{V_3}{h_2 h_3} \pdv{h_3}{c_2} \bigg) \vu{e}_3 \vu{e}_2
\end{aligned}$$
{% endcomment %}


{% include proof/start.html id="proof-grad-vector" -%}
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 dyadic product of $$\nabla$$ and $$\vb{V}$$:

$$\begin{aligned}
    \nabla \vb{V}
    &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg)
    \bigg( V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3 \bigg)
    \\
    &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \bigg( \sum_{k} V_k \vu{e}_k \bigg)
    \\
    &= \sum_{jk} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} (V_k \vu{e}_k)
    \\
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} \vu{e}_j \vu{e}_k + \sum_{jk} \frac{V_k}{h_j} \vu{e}_j \pdv{\vu{e}_k}{c_j}
\end{aligned}$$

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

$$\begin{aligned}
    \nabla \vb{V}
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} \vu{e}_j \vu{e}_k
    + \sum_{jk} \frac{V_k}{h_j} \vu{e}_j \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \bigg)
    \\
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} \vu{e}_j \vu{e}_k
    + \sum_{jk} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_j \vu{e}_j
    - \sum_{jl} \frac{V_j}{h_j h_l} \pdv{h_j}{c_l} \vu{e}_j \vu{e}_l
    \\
    &= \sum_{jk} \bigg( \frac{1}{h_j} \pdv{V_k}{c_j} \vu{e}_j \vu{e}_k
    + \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_j \vu{e}_j
    - \frac{V_j}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_j \vu{e}_k \bigg)
\end{aligned}$$

This is a 2nd-order tensor, whose diagonal components $$(\nabla \vb{V})_{ll}$$ are given by:

$$\begin{aligned}
    (\nabla \vb{V})_{ll}
    &= \vu{e}_l \cdot (\nabla \vb{V}) \cdot \vu{e}_l
    \\
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} \delta_{jl} \delta_{kl}
    + \sum_{jk} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} \delta_{jl} \delta_{jl}
    - \sum_{jk} \frac{V_j}{h_j h_k} \pdv{h_j}{c_k} \delta_{jl} \delta_{kl}
    \\
    &= \frac{1}{h_l} \pdv{V_l}{c_l} + \sum_{k} \frac{V_k}{h_l h_k} \pdv{h_l}{c_k} - \frac{V_l}{h_l h_l} \pdv{h_l}{c_l}
    \\
    &= \frac{1}{h_l} \pdv{V_l}{c_l} + \sum_{k \neq l} \frac{V_k}{h_l h_k} \pdv{h_l}{c_k}
\end{aligned}$$

Meanwhile, the off-diagonal components $$(\nabla \vb{V})_{lm}$$ are as follows, with $$l \neq m$$:

$$\begin{aligned}
    (\nabla \vb{V})_{lm}
    &= \vu{e}_l \cdot (\nabla \vb{V}) \cdot \vu{e}_m
    \\
    &= \sum_{jk} \frac{1}{h_j} \pdv{V_k}{c_j} \delta_{jl} \delta_{km}
    + \sum_{jk} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} \delta_{jl} \delta_{jm}
    - \sum_{jk} \frac{V_j}{h_j h_k} \pdv{h_j}{c_k} \delta_{jl} \delta_{km}
    \\
    &= \frac{1}{h_l} \pdv{V_m}{c_l} + \sum_{k} \frac{V_k}{h_l h_k} \pdv{h_l}{c_k} \delta_{lm} - \frac{V_l}{h_l h_m} \pdv{h_l}{c_m}
    \\
    &= \frac{1}{h_l} \pdv{V_m}{c_l} - \frac{V_l}{h_l h_m} \pdv{h_l}{c_m}
\end{aligned}$$
{% include proof/end.html id="proof-grad-vector" %}



## Advection of a vector

In physics, a common quantity is the *advection* $$(\vb{U} \cdot \nabla) \vb{V}$$
of a vector $$\vb{V}$$ according to a velocity field $$\vb{U}$$,
as found in e.g. a [material derivative](/know/concept/material-derivative/).
In $$(c_1, c_2, c_3)$$ its $$c_j$$-component is:

$$\begin{aligned}
    \boxed{
        \big( (\vb{U} \cdot \nabla) \vb{V} \big)_j
        = \sum_{k} \frac{U_k}{h_k} \pdv{V_j}{c_k}
        + \sum_{k \neq j} \frac{V_k}{h_j h_k} \bigg( U_j \pdv{h_j}{c_k} - U_k \pdv{h_k}{c_j} \bigg)
    }
\end{aligned}$$


{% include proof/start.html id="proof-adv-vector" -%}
From our earlier calculation of $$\nabla f$$,
we know how to express the del $$\nabla$$ in $$(c_1, c_2, c_3)$$.
Thanks to orthogonality,
$$\vb{U} \cdot \nabla$$ is therefore simply:

$$\begin{aligned}
    \vb{U} \cdot \nabla
    &= \bigg( U_1 \vu{e}_1 + U_2 \vu{e}_2 + U_3 \vu{e}_3 \bigg)
    \cdot \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg)
    \\
    &= \bigg( \sum_{j} U_j \vu{e}_j \bigg) \cdot \bigg( \sum_{k} \vu{e}_k \frac{1}{h_k} \pdv{}{c_k} \bigg)
    \\
    &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{U_j}{h_k} \pdv{}{c_k}
    \\
    &= \sum_{j} \frac{U_j}{h_j} \pdv{}{c_j}
\end{aligned}$$

We apply this to $$\vb{V}$$ and use the product rule of differentiation:

$$\begin{aligned}
    (\vb{U} \cdot \nabla) \vb{V}
    &= \bigg( \sum_{j} \frac{U_j}{h_j} \pdv{}{c_j} \bigg) \bigg( \sum_{k} V_k \vu{e}_k \bigg)
    \\
    &= \sum_{jk} \frac{U_j}{h_j} \pdv{}{c_j} (V_k \vu{e}_k)
    \\
    &= \sum_{jk} \frac{U_j}{h_j} \bigg( \pdv{V_k}{c_j} \vu{e}_k + V_k \pdv{\vu{e}_k}{c_j} \bigg)
\end{aligned}$$

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

$$\begin{aligned}
    (\vb{U} \cdot \nabla) \vb{V}
    &= \sum_{jk} \frac{U_j}{h_j} \bigg( \pdv{V_k}{c_j} \vu{e}_k
    + \frac{V_k}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{V_k}{h_l} \pdv{h_k}{c_l} \vu{e}_l \bigg)
    \\
    &= \sum_{jk} \frac{U_j}{h_j} \pdv{V_k}{c_j} \vu{e}_k
    + \sum_{jk} \frac{U_j V_k}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_j
    - \sum_{jl} \frac{U_j V_j}{h_j h_l} \pdv{h_j}{c_l} \vu{e}_l
\end{aligned}$$

We rename the indices such that each term contains $$\vu{e}_j$$.
Note that when $$k = j$$, the latter two terms cancel out,
so we only need to sum for $$k \neq j$$:

$$\begin{aligned}
    (\vb{U} \cdot \nabla) \vb{V}
    &= \sum_{jk} \frac{U_k}{h_k} \pdv{V_j}{c_k} \vu{e}_j
    + \sum_{jk} \frac{U_j V_k}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_j
    - \sum_{jk} \frac{U_k V_k}{h_j h_k} \pdv{h_k}{c_j} \vu{e}_j
    \\
    &= \sum_{jk} \frac{U_k}{h_k} \pdv{V_j}{c_k} \vu{e}_j
    + \sum_{j} \sum_{k \neq j} \bigg( \frac{U_j V_k}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_j
    - \frac{U_k V_k}{h_j h_k} \pdv{h_k}{c_j} \vu{e}_j \bigg)
\end{aligned}$$

Dot-multiplying by $$\vu{e}_j$$ isolates the $$c_j$$-component and gives the desired formula.
{% include proof/end.html id="proof-adv-vector" %}



## Laplacian of a vector

The Laplacian $$\nabla^2 \vb{V}$$ of a vector $$\vb{V}$$
has the following components in $$(c_1, c_2, c_3)$$:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            (\nabla^2 \vb{V})_j
            &= \sum_{k} \frac{1}{H} \pdv{}{c_k} \bigg( \frac{H}{h_k^2} \pdv{V_j}{c_k} \bigg)
            \\
            &\quad\: + \sum_{k \neq j} \frac{1}{H} \bigg( \pdv{}{c_j} \Big( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \Big)
            - \pdv{}{c_k} \Big( \frac{H V_k}{h_j h_k^2} \pdv{h_k}{c_j} \Big) \bigg)
            \\
            &\quad\: + \sum_{k \neq j} \frac{1}{h_j h_k} \bigg( \frac{1}{h_j} \pdv{V_k}{c_j} \pdv{h_j}{c_k}
            - \frac{1}{h_k} \pdv{V_k}{c_k} \pdv{h_k}{c_j} \bigg)
            \\
            &\quad\:- \sum_{k \neq j} \frac{1}{h_j h_k} \bigg( \frac{V_j}{h_j h_k} \Big( \pdv{h_j}{c_k} \Big)^2
            + \sum_{l \neq k} \frac{V_l}{h_k h_l} \pdv{h_k}{c_l} \pdv{h_k}{c_j} \bigg)
        \end{aligned}
    }
\end{aligned}$$


{% include proof/start.html id="proof-lap-vector" -%}
We already know how to calculate the Laplacian $$\nabla^2 f$$ of a scalar.
From that, we read out the $$\nabla^2$$-operator
and apply it to a vector $$\vb{V}$$ instead:

$$\begin{aligned}
    \nabla^2 \vb{V}
    &= \bigg( \sum_{j} \frac{1}{H} \pdv{}{c_j} \Big( \frac{H}{h_j^2} \pdv{}{c_j} \Big) \bigg)
    \bigg( \sum_{k} V_k \vu{e}_k \bigg)
    \\
    &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{}{c_j} (V_k \vu{e}_k) \bigg)
    \\
    &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \vu{e}_k + \frac{H V_k}{h_j^2} \pdv{\vu{e}_k}{c_j} \bigg)
    \\
    &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \vu{e}_k
    + \frac{H V_k}{h_j^2} \Big( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \Big) \bigg)
    \\
    &= \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \sum_{k} \frac{H}{h_j^2} \pdv{V_k}{c_j} \vu{e}_k
    + \sum_{k} \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \vu{e}_j
    - \sum_{l} \frac{H V_j}{h_j^2 h_l} \pdv{h_j}{c_l} \vu{e}_l \bigg)
    \\
    &= \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \sum_{k} \frac{H}{h_j^2} \pdv{V_k}{c_j} \vu{e}_k
    + \sum_{k \neq j} \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \vu{e}_j
    - \sum_{k \neq j} \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \vu{e}_k \bigg)
\end{aligned}$$

Where we have noticed that the latter two terms cancel out if $$j = k$$.
We expand according to the product rule of differentiation:

$$\begin{aligned}
    \nabla^2 \vb{V}
    &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) \vu{e}_k
    + \sum_{jk} \frac{1}{H} \frac{H}{h_j^2} \pdv{V_k}{c_j} \pdv{\vu{e}_k}{c_j}
    \\
    &\quad\: + \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_j
    + \sum_{j} \sum_{k \neq j} \frac{1}{H} \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \pdv{\vu{e}_j}{c_j}
    \\
    &\quad\: - \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_k
    - \sum_{j} \sum_{k \neq j} \frac{1}{H} \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \pdv{\vu{e}_k}{c_j}
\end{aligned}$$

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

$$\begin{aligned}
    \nabla^2 \vb{V}
    &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) \vu{e}_k
    + \sum_{jk} \frac{1}{h_j^2} \pdv{V_k}{c_j}
    \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)
    \\
    &\quad\: + \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_j
    - \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg( \sum_{l \neq j} \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_l \bigg)
    \\
    &\quad\: - \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_k
    - \sum_{j} \sum_{k \neq j} \frac{V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j \bigg)
    \\
    &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) \vu{e}_k
    + \sum_{jk} \frac{1}{h_j^2 h_k} \pdv{V_k}{c_j} \pdv{h_j}{c_k} \vu{e}_j
    - \sum_{jl} \frac{1}{h_j^2 h_l} \pdv{V_j}{c_j} \pdv{h_j}{c_l} \vu{e}_l
    \\
    &\quad\: + \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_j
    - \sum_{j} \sum_{k \neq j} \sum_{l \neq j} \frac{V_k}{h_j^2 h_k h_l} \pdv{h_j}{c_k} \pdv{h_j}{c_l} \vu{e}_l
    \\
    &\quad\: - \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_k
    - \sum_{j} \sum_{k \neq j} \frac{V_j}{h_j^2 h_k^2} \bigg( \pdv{h_j}{c_k} \bigg)^2 \vu{e}_j
    \\
    &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) \vu{e}_k
    + \sum_{j} \sum_{k \neq j} \frac{1}{h_j^2 h_k} \pdv{V_k}{c_j} \pdv{h_j}{c_k} \vu{e}_j
    - \sum_{j} \sum_{k \neq j} \frac{1}{h_j^2 h_k} \pdv{V_j}{c_j} \pdv{h_j}{c_k} \vu{e}_k
    \\
    &\quad\: + \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_j
    - \sum_{j} \sum_{k \neq j} \sum_{l \neq j} \frac{V_k}{h_j^2 h_k h_l} \pdv{h_j}{c_k} \pdv{h_j}{c_l} \vu{e}_l
    \\
    &\quad\: - \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_k
    - \sum_{j} \sum_{k \neq j} \frac{V_j}{h_j^2 h_k^2} \bigg( \pdv{h_j}{c_k} \bigg)^2 \vu{e}_j
\end{aligned}$$

Where we have once again noticed that terms #2 and #3 cancel out if $$j = k$$.
Next, we isolate the $$c_m$$-component by dot-multiplying with $$\vu{e}_m$$:

$$\begin{aligned}
    (\nabla^2 \vb{V})_m
    &= (\nabla^2 \vb{V}) \cdot \vu{e}_m
    \\
    &= \sum_{jk} \frac{\delta_{km}}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg)
    + \sum_{j} \sum_{k \neq j} \frac{\delta_{jm}}{h_j^2 h_k} \pdv{V_k}{c_j} \pdv{h_j}{c_k}
    - \sum_{j} \sum_{k \neq j} \frac{\delta_{km}}{h_j^2 h_k} \pdv{V_j}{c_j} \pdv{h_j}{c_k}
    \\
    &\quad\: + \sum_{j} \sum_{k \neq j} \delta_{jm} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg)
    - \sum_{j} \sum_{k \neq j} \sum_{l \neq j} \delta_{lm} \frac{V_k}{h_j^2 h_k h_l} \pdv{h_j}{c_k} \pdv{h_j}{c_l}
    \\
    &\quad\: - \sum_{j} \sum_{k \neq j} \delta_{km} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg)
    - \sum_{j} \sum_{k \neq j} \delta_{jm} \frac{V_j}{h_j^2 h_k^2} \bigg( \pdv{h_j}{c_k} \bigg)^2
    \\
    &= \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_m}{c_j} \bigg)
    + \sum_{k \neq m} \frac{1}{h_k h_m^2} \pdv{V_k}{c_m} \pdv{h_m}{c_k}
    - \sum_{j \neq m} \frac{1}{h_j^2 h_m} \pdv{V_j}{c_j} \pdv{h_j}{c_m}
    \\
    &\quad\: + \sum_{k \neq m} \frac{1}{H} \pdv{}{c_m} \bigg( \frac{H V_k}{h_m^2 h_k} \pdv{h_m}{c_k} \bigg)
    - \sum_{j \neq m} \sum_{k \neq j} \frac{V_k}{h_j^2 h_k h_m} \pdv{h_j}{c_k} \pdv{h_j}{c_m}
    \\
    &\quad\: - \sum_{j \neq m} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_m} \pdv{h_j}{c_m} \bigg)
    - \sum_{k \neq m} \frac{V_m}{h_m^2 h_k^2} \bigg( \pdv{h_m}{c_k} \bigg)^2
\end{aligned}$$

Which gives the desired formula after some simple index renaming and rearranging.
{% include proof/end.html id="proof-lap-vector" %}



## Divergence of a tensor

It also possible to take the divergence of a 2nd-order tensor $$\overline{\overline{\mathbf{T}}}$$,
yielding a vector with these components in $$(c_1, c_2, c_3)$$:

$$\begin{aligned}
    \boxed{
        (\nabla \cdot \overline{\overline{\mathbf{T}}})_j
        = \sum_{k} \frac{1}{h_k} \pdv{T_{kj}}{c_k}
        + \sum_{k \neq j} \frac{T_{jk}}{h_j h_k} \pdv{h_j}{c_k}
        - \sum_{k \neq j} \frac{T_{kk}}{h_j h_k} \pdv{h_k}{c_j}
        + \sum_{k} \sum_{l \neq k} \frac{T_{lj}}{h_k h_l} \pdv{h_k}{c_l}
    }
\end{aligned}$$

{% include proof/start.html id="proof-div-tensor" -%}
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 and evaluate:

$$\begin{aligned}
    \nabla \cdot \overline{\overline{\mathbf{T}}}
    &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg)
    \\
    &\quad\:\:\: \cdot \Big( T_{11} \vu{e}_1 \vu{e}_1 + T_{12} \vu{e}_1 \vu{e}_2 + T_{13} \vu{e}_1 \vu{e}_3
    \\
    &\qquad + T_{21} \vu{e}_2 \vu{e}_1 + T_{22} \vu{e}_2 \vu{e}_2 + T_{23} \vu{e}_2 \vu{e}_3
    \\
    &\qquad + T_{31} \vu{e}_3 \vu{e}_1 + T_{32} \vu{e}_3 \vu{e}_2 + T_{33} \vu{e}_3 \vu{e}_3 \Big)
    \\
    &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \cdot \bigg( \sum_{kl} T_{kl} \vu{e}_k \vu{e}_l \bigg)
    \\
    &= \sum_{jkl} \vu{e}_j \cdot \frac{1}{h_j} \pdv{}{c_j} (T_{kl} \vu{e}_k \vu{e}_l)
\end{aligned}$$

We apply the product rule of differentiation
and use that $$\vb{c} \cdot (\vb{a} \vb{b}) = (\vb{c} \cdot \vb{a}) \vb{b}$$:

$$\begin{aligned}
    \nabla \cdot \overline{\overline{\mathbf{T}}}
    &= \sum_{jkl} \bigg( (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{T_{kl}}{c_j} \vu{e}_l
    + (\vu{e}_j \cdot \vu{e}_k) \frac{T_{kl}}{h_j} \pdv{\vu{e}_l}{c_j}
    + \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg)
    \\
    &= \sum_{jkl} \bigg( \delta_{jk} \frac{1}{h_j} \pdv{T_{kl}}{c_j} \vu{e}_l + \delta_{jk} \frac{T_{kl}}{h_j} \pdv{\vu{e}_l}{c_j}
    + \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg)
    \\
    &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j}
    + \sum_{k} \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg)
\end{aligned}$$

Inserting our expressions for the derivatives of the basis vectors
in the last term, we find:

$$\begin{aligned}
    \nabla \cdot \overline{\overline{\mathbf{T}}}
    &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j}
    + \sum_{k} \vu{e}_j \cdot
    \Big( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{m} \frac{1}{h_m} \pdv{h_k}{c_m} \vu{e}_m \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg)
    \\
    &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j}
    + \sum_{k} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l
    - \sum_{m} (\vu{e}_j \cdot \vu{e}_m) \frac{T_{jl}}{h_j h_m} \pdv{h_j}{c_m} \vu{e}_l \bigg)
    \\
    &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j}
    + \sum_{k} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l - \frac{T_{jl}}{h_j h_j} \pdv{h_j}{c_j} \vu{e}_l \bigg)
    \\
    &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j}
    + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg)
\end{aligned}$$

Where we noticed that the latter two terms cancel out if $$k = j$$.
Next, rewriting $$\ipdv{\vu{e}_l}{c_j}$$:

$$\begin{aligned}
    \nabla \cdot \overline{\overline{\mathbf{T}}}
    &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l
    + \frac{T_{jl}}{h_j} \Big( \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_j - \delta_{jl} \sum_{m} \frac{1}{h_m} \pdv{h_l}{c_m} \vu{e}_m \Big)
    + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg)
    \\
    &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l
    + \frac{T_{jl}}{h_j h_l} \pdv{h_j}{c_l} \vu{e}_j - \delta_{jl} \sum_{m} \frac{T_{jl}}{h_j h_m} \pdv{h_l}{c_m} \vu{e}_m
    + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg)
    \\
    &= \sum_{jl} \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l
    + \sum_{jl} \frac{T_{jl}}{h_j h_l} \pdv{h_j}{c_l} \vu{e}_j
    - \sum_{jm} \frac{T_{jj}}{h_j h_m} \pdv{h_j}{c_m} \vu{e}_m
    + \sum_{jl} \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l
\end{aligned}$$

Renaming the indices such that each term contains $$\vu{e}_l$$,
we arrive at the full result:

$$\begin{aligned}
    \nabla \cdot \overline{\overline{\mathbf{T}}}
    &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j}
    + \frac{T_{lj}}{h_j h_l} \pdv{h_l}{c_j}
    - \frac{T_{jj}}{h_j h_l} \pdv{h_j}{c_l}
    + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_l
\end{aligned}$$

To isolate the $$c_m$$-component, we dot-multiply by $$\vu{e}_m$$
and resolve the Kronecker delta $$\delta_{lm}$$:

$$\begin{aligned}
    (\nabla \cdot \overline{\overline{\mathbf{T}}})_m
    &= (\nabla \cdot \overline{\overline{\mathbf{T}}}) \cdot \vu{e}_m
    \\
    &= \sum_{jl} \delta_{lm} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j}
    + \frac{T_{lj}}{h_j h_l} \pdv{h_l}{c_j}
    - \frac{T_{jj}}{h_j h_l} \pdv{h_j}{c_l}
    + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \bigg)
    \\
    &= \sum_{j} \frac{1}{h_j} \pdv{T_{jm}}{c_j}
    + \sum_{j} \frac{T_{mj}}{h_j h_m} \pdv{h_m}{c_j}
    - \sum_{j} \frac{T_{jj}}{h_j h_m} \pdv{h_j}{c_m}
    + \sum_{j} \sum_{k \neq j} \frac{T_{km}}{h_j h_k} \pdv{h_j}{c_k}
\end{aligned}$$

The second and third terms cancel out for $$j = m$$,
so we can sum over $$j \neq m$$ instead.
{% include proof/end.html id="proof-div-tensor" %}



## References
1.  M.L. Boas,
    *Mathematical methods in the physical sciences*, 2nd edition,
    Wiley.
2.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.
3.  B. Lautrup,
    [Orthogonal curvilinear coordinates](https://library.wolfram.com/infocenter/MathSource/5545/vectoranalysis.pdf),
    2004, unpublished.