Is there a general formula for the del operator $\nabla$ in different coordinate systems?

Question

The $\nabla$-operator is simple in cartesian coordinates, $[\partial_x,\partial_y,\partial_z]$, but in spherical coordinates, it becomes $[\partial_r, \frac{1}{r}\partial_\theta, \frac{1}{r\sin\theta}\partial_\varphi]$ and in cylindrical coordinates $[\partial_\rho, \frac{1}{\rho}\partial_\varphi, \partial_z]$; is there a general formula for converting into a different coordinate system, perhaps in terms of a Jacobian?

(Sub-question: Is there any reason why similar operators like $\hat{\nabla} = [\partial_r, \partial_\theta, \partial_\varphi]$ aren't in use?)

Answer 1

There is a large body of literature on this subject. I found a section in an old calculus book on Orthogonal Curvilinear Coordinates in the chapter on Vector Analysis. It's too complex to present here. (The book is Advanced Calculus for Applications by Hildebrand.) However, you can get started here: Curvilinear coordinates.

Answer 2

Disclosure: I'm the author of the preprint "Coordinates Last: Vector Analysis Done Fast" (Creative Commons). The following answer is developed late in that document.

Consider general, possibly non-orthogonal coordinates $u^i$ (where $i=1,2,3$), with natural (covariant) basis vectors $$\mathbf{h}_i = \partial_i \mathbf{r}$$ (where $\mathbf{r}$ is position and $\partial_i$ means $\frac{\partial}{\partial u^i}$), and dual (contravariant) basis vectors $$\mathbf{h}^i = \nabla u^i$$ where, as usual, the gradient is defined in terms of the directional derivative. In these coordinates we find that the gradient of a general scalar field $p$ is $$\nabla p = \mathbf{h}^i \partial_i p$$ with implicit summation over $i$. This suggests the operational definition $$\color{red}{\nabla = \mathbf{h}^i \partial_i}$$ (with implicit summation).

What then shall we make of the operators $$\nabla{\cdot} = \mathbf{h}^i \partial_i \cdot$$ and $$\nabla{\times} = \mathbf{h}^i \partial_i \times$$ and $$\nabla{\cdot}\nabla = \mathbf{h}^i \partial_i \cdot \nabla ~?$$ Well, if we insert an operand and formally evaluate the dot- or cross-product in these coordinates, the dot or cross is eliminated, so that the result is as if we had $$\nabla{\cdot} = \mathbf{h}^i \cdot \partial_i$$ and $$\nabla{\times} = \mathbf{h}^i \times \partial_i$$ and $$\nabla{\cdot}\nabla = \mathbf{h}^i \cdot \partial_i \nabla$$ —that is, as if (in the words of E.B. Wilson) the operator $\partial_i$ could "pass by" the dot or the cross.

Now here's the prize: Provided that we account for any non-uniformity of the basis vectors in the differentiations, the last three results are correct expressions for the divergence, curl, and Laplacian operators in general coordinates! (The first two of the three are apparently due to C.T. Tai, who was otherwise hostile to the del-dot and del-cross notations and the "pass by" argument.)

In Cartesian coordinates, the first two of the three reduce to $$\nabla{\cdot} = \mathbf{i} \cdot \partial_x + \mathbf{j} \cdot \partial_y + \mathbf{k} \cdot \partial_z$$ and $$\nabla{\times} = \mathbf{i} \times \partial_x + \mathbf{j} \times \partial_y + \mathbf{k} \times \partial_z \,.$$ These agree with the initial definitions of the divergence and curl given by J. Willard Gibbs in his Elements of Vector Analysis (first installment, 1881).

Answer 3

Suppose $x^i$ is a coordinate system and $v_i$ be the tangent vectors to the coordinates, i.e.

$$D_{v_i}f = \frac{\partial f}{\partial x^i}.$$

The definition of the gradient vector is the equation $$w\cdot\vec\nabla f = D_w f$$ for any vector $w$. Suppose that $a^i$ are the components of the gradient of $f$ with respect to the basis $v_i$: $$\vec \nabla f = \sum_i a^i v_i$$ We can solve for $a^i$ by taking the scalar product of both sides of this equation with $v_j$. Writing $g_{ij} = v_{i}\cdot v_j$ for the (symmetric, positive definite) Gram matrix of the $v_i$, you have $$ \sum_i a^ig_{ij} = v_j \cdot\vec \nabla f = \frac{\partial f}{\partial x^j}$$ so, writing $g^{ij}$ for the components of the matrix inverse of $(g_{ij})$, we have $$a^i = \sum_j g^{ij} \frac{\partial f}{\partial x^j}.$$ Thus, $$\vec\nabla f = \sum_{i,j} g^{ij} \frac{\partial f}{\partial x^j}v_i.$$

How do you actually calculate the $g_{ij}$?

You asked for a Jacobian. This happens when you start with some other coordinate system $y^i$ in which you already know the scalar products $w_i\cdot w_j$ of its tangent vectors. So if your coordinates are functions $x_i=x^i(y^1,\dots,y^n)$ then in order to find $g_{ij}=v_i\cdot v_j$ you can want to write $v_i$ in terms of $w_i$, which you do using the chain rule: $$D_{v_i} f = \frac{\partial f}{\partial y^i} = \sum_j\frac{\partial x^j}{\partial y^i}\frac{\partial f}{\partial x^j} = \sum_j\frac{\partial x^j}{\partial y^i} D_{w_i}f$$ from where it follows that $$v_i = \sum_j\frac{\partial x^j}{\partial y^i} w_i$$ and hence $$g_{ij} = v_i\cdot v_j = \sum_{k,l}\frac{\partial x^k}{\partial y^i}\frac{\partial x^l}{\partial y^j} w_k \cdot w_l.$$ For example, if $y^i$ are Cartesian coordinates, the vectors $w_i$ are the standard basis vectors, so $w_i\cdot w_j$ are all zero except they are $1$ when $i=j$. Thus, the metric coefficients are $$g_{ij} = \sum_{k}\frac{\partial x^k}{\partial y^i}\frac{\partial x^k}{\partial y^j} $$ which you can write as $J^TJ$, where $J$ is the Jacobian matrix of the change of coordinates $x^i=x^i(y^1,\dots,y^n)$.