WikiJournal Preprints/Cut the coordinates! (or Vector Analysis Done Fast)

Introduction
Sheldon Axler, in his essay "Down with determinants!" (1995) and his ensuing book Linear Algebra Done Right (4th Ed., 2023–), does not entirely eliminate determinants, but introduces them as late as possible and then exploits them for what he calls their "main reasonable use in undergraduate mathematics", namely the change-of-variables formula for multiple integrals. Here I treat coordinates in vector analysis somewhat as Axler treats determinants in linear algebra: I introduce coordinates as late as possible, and then exploit them in unconventionally rigorous derivations of vector-analytic identities from vector-algebraic identities. But I contrast with Axler in at least two ways. First, as my subtitle suggests, I have no intention of expanding my paper into a book. Brevity is of the essence. Second, while one may well avoid determinants in numerical linear algebra, one can hardly avoid coordinates in numerical vector analysis! So I cannot extend the coordinate-minimizing path into computation. But I can extend it up to the threshold by expressing the operators of vector analysis in a suitably general coordinate system, leaving others to specialize it and compute with it. On the way, I can satisfy readers who need the concepts of vector analysis for theoretical purposes, and who would rather read a paper than a book.

The cost of coordinates
Mathematicians define a "vector" as a member of a vector space, which is a set whose members satisfy certain basic rules of algebra (called the vector-space axioms) with respect to another set (called a field), which has its own basic rules of algebra (the field axioms), and whose members are called "scalars". Physicists are more fussy. They typically want a "vector" to be not only a member of a vector space, but also a first-order tensor&#8239;: a "tensor", meaning that it has an existence independent of any coordinate system with which it might be specified; and "first-order" (or "first-degree", or "first-rank"), meaning that it is specified by a one-dimensional array of numbers. Similarly, a 2nd-order tensor is specified by a 2-dimensional array (a matrix), and a 3rd-order by a 3-dimensional array, and so on; and a "scalar", being specified by a single number (a zero-dimensional array), is a zero-order tensor. In "vector analysis", we are greatly interested in applications to physical situations, and accordingly take the physicists' view on what constitutes a vector or a scalar.

So, for our purposes, defining a quantity by three components in (say) a Cartesian coordinate system is not enough to make it a vector, and defining a quantity as a real function of a list of coordinates is not enough to make it a scalar, because we still need to show that the quantity has an independent existence. One way to do this is to show that its coordinate representation behaves appropriately when the coordinate system is changed. (But don't worry if the following details look cryptic, because we won't be using them!) Independent existence of a quantity means that its coordinate representation is contravariant—that is, the representation changes so as to compensate for the change in the coordinate system. But independent existence of an operator means that its coordinate representation is covariant—that is, the representation of the operator in the coordinate system, with the operand(s) and the result in that system, has the same form in one coordinate system as in another (except for features internal to the system).

Here we circumvent these complications by the most obvious route: by initially defining things without coordinates. If, having defined something without coordinates, we then need to represent it with coordinates, we can choose the coordinate system for convenience.

The limitations of limits
In the branch of pure mathematics known as analysis, there is a thing called a limit, whereby for every positive ϵ&#8202; there exists a positive &delta; such that if some increment is less than &delta;, some error is less than ϵ. In the branch of applied mathematics known as continuum mechanics, there is a thing called reality, whereby if the increment is less than some positive &delta;, the assumption of a continuum becomes ridiculous, so that the error cannot be made less than an arbitrary ϵ. Yet vector "analysis" (together with higher-order tensors) is typically studied with the intention of applying it to some form of "continuum" mechanics, such as the modeling of elasticity, plasticity, fluid flow, or (widening the net) electrodynamics of ordinary matter; in short, it is studied with the intention of conveniently forgetting that, on a sufficiently small scale, matter is lumpy. One might therefore submit that to express the principles of vector analysis in the language of limits is to strain at a gnat and swallow a camel. Here I avoid that camel by referring to elements of length or area or volume, each of which is small enough to allow some quantity or quantities to be considered uniform within it, but, for the same reason, large enough to allow such local averaging of the said quantity or quantities as is necessary to tune out the lumpiness.

We shall see bigger camels, where well-known authors define or misdefine a vector operator and then want to treat it like an ordinary vector (a quantity). These I also avoid.

Prerequisites
I assume that the reader is familiar with the algebra and geometry of vectors in 3D space, including the dot-product, the cross-product, and the scalar triple product, their geometric meanings, their expressions in Cartesian coordinates, and the identity

which we call the "expansion" of the vector triple product. I further assume that the reader can generalize the concept of a derivative, so as to differentiate a vector with respect to a scalar, e.g.
 * $$\mathbf{r}'(t) = \frac{d\mathbf{r}}{dt}

=\, \lim_{h\to 0} \frac{\mathbf{r}(t+h) - \mathbf{r}(t)}{h} \,,$$ or so as to differentiate a function of several independent variables "partially" w.r.t. one of them while the others are held constant, e.g.
 * $$\tfrac{\part}{\part y} \psi\big(x,y,z\big)

=\, \lim_{h\to 0} \frac{\psi(x,y{+}h~\!,z) - \psi(x,y,z)}{h} \,.$$

But in view of the above remarks on limits, I also expect the reader to be tolerant of an argument like this: In a short time $dt$, let the vectors $a &times; (b &times; c)  =  a⸱&#8202;c b &minus; a⸱b c ,$ and $r$ change by $$d\mathbf{r}$$ and $$d\mathbf{p}$$, respectively. Then
 * $$\begin{align}

\tfrac{d}{dt}\big(\mathbf{r}\!\times\!\mathbf{p}\big) &= \frac{(\mathbf{r}+d\mathbf{r})\times(\mathbf{p}+d\mathbf{p}) \,-\, \mathbf{r}\times\mathbf{p}}{dt}\\[1ex] &= \frac{\mathbf{r}\!\times\!d\mathbf{p}+d\mathbf{r}\!\times\!\mathbf{p}}{dt} \quad [\mathsf{neglecting}~d\mathbf{r}\!\times\!d\mathbf{p}]\\[1ex] &=\, \mathbf{r}\times\!\tfrac{d\mathbf{p}}{dt} + \tfrac{d\mathbf{r}}{dt}\!\times\mathbf{p}\\[1ex] &=\, \mathbf{r}\times\mathbf{\dot{p}}\,+\,\mathbf{\dot{r}}\times\mathbf{p}\,, \end{align}$$ where, as always, the orders of the cross-products matter. Differentiation of a dot-product behaves similarly, except that the orders don't matter; and if&#8239; $p$, where $m$ is a scalar and $r$ is a vector, then
 * $$\mathbf{\dot{p}} = m\mathbf{\dot{v}} + \dot{m}\mathbf{v} \,.$$

Or an argument like this:  If $$z\!=\!f(x,y)$$, then
 * $$\begin{align}

\frac{\part^2 z}{\part x\,\part y} &= \tfrac{\part}{\part x}\,\tfrac{\part}{\part y} f\big(x,y\big)\\ &= \frac{\part}{\part x}\,\frac{f(x,y{+}dy)-f(x,y)}{dy}\\[1ex] &= \frac{\,\frac{f(x{+}dx~\!,\,y{+}dy)\,-\,f(x{+}dx~\!,\,y)}{dy} - \frac{f(x,\,y{+}dy)\,-\,f(x,y)}{dy}\,} {dx}\\[2ex] &= \frac{\,\frac{f(x{+}dx~\!,\,y{+}dy)\,-\,f(x,\,y{+}dy)}{dx} - \frac{f(x{+}dx~\!,\,y)\,-\,f(x,y)}{dx}\,} {dy}\\[1ex] &= \frac{\part}{\part y}\,\frac{f(x{+}dx~\!,~\!y)-f(x,y)}{dx}\\[1ex] &= \tfrac{\part}{\part y}\,\tfrac{\part}{\part x} f\big(x,y\big) = \frac{\part^2 z}{\part y\,\part x} \,; \end{align}$$ that is, we can switch the order of partial differentiation. If $&part;_{x}$ is an abbreviation for $&part;⁄&part;x&#8202;$, etc., this rule can be written in operational terms as

More generally, if $&part;_{x }&part;_{y} = &part;_{y }&part;_{x }.$ is an abbreviation for $&part;_{i}$ where $p$, the rule becomes

These generalizations of differentiation, however, do not go beyond differentiation w.r.t. real variables, some of which are scalars, and some of which are coordinates. Vector analysis involves quantities that may be loosely described as derivatives w.r.t. a vector—usually the position vector.

Closed-surface integrals per unit volume
The term field, mentioned above in the context of algebraic axioms, has an alternative meaning: if $p&#8201;=&#8201;mv$ is the position vector, a scalar field is a scalar-valued function of $v$, and a vector field is a vector-valued function of $i&#8201;=&#8202;1,&#8202;2,…$; both may also depend on time. These are the functions of which we want "derivatives" w.r.t. the vector $r$.

In this section I introduce four such derivatives—the gradient, the curl, the divergence, and the Laplacian—in a way that will seem unremarkable to those readers who aren't already familiar with them, but idiosyncratic to those who are. The gradient is commonly introduced in connection with a curve and its endpoints, the curl in connection with a surface segment and its enclosing curve, the divergence in connection with a volume and its enclosing surface, and the Laplacian as a composite of two of the above, initially applicable only to a scalar field. Here I introduce all four in connection with a volume and its enclosing surface; and I introduce the Laplacian as a concept in its own right, equally applicable to a scalar or vector field, and only later relate it to the others. My initial definitions of the gradient, the curl, and the Laplacian, although not novel, are usually thought to be more advanced than the common ones—in spite of being conceptually simpler, and in spite of being obvious variations on the same theme.

Instant integral theorems—with a caveat
Let $&part;⁄&part;x_{i}$ be a volume (3D region) enclosed by a surface $&part;_{i }&part;_{j} = &part;_{j }&part;_{i }.$ (a mathematical surface, not generally a physical barrier). Let $r$ be the unit normal vector at a general point on $V$, pointing out of $S$. Let $S$ be the distance from $V$ in the direction of $r$ (positive outside $n$, negative inside), and let $S$ be an abbreviation for $V$, where the derivative—commonly called the normal derivative—is tacitly assumed to exist.

In $&part;_{n}$, and on $&part;⁄&part;n&#8202;$, let $V$ be a scalar field (e.g., pressure in a fluid, or temperature), and let $r$ be a vector field (e.g., flow velocity, or heat-flow density), and let $S$ be a generic field which may be a scalar or a vector. Let a general element of the surface $p$ have area $&psi;$, and let it be small enough to allow $n&#770;$, $S$, $n&#770;$, and $dS$ to be considered uniform over the element. Then, for every element, the following four products are well defined:

If $p$ is pressure in a non-viscous fluid, the first of these products is the force exerted by the fluid in $&part;_{n}&#8202;&psi;$&#8202; through the area $$. The second product does not have such an obvious physical interpretation; but if $q$ is circulating clockwise about an axis directed through $p$, the cross-product will be exactly tangential to $V$ and will tend to have a component in the direction of that axis. The third product is the flux of $n&#770;$ through the surface element; if $q$ is flow velocity, the third product is the volumetric flow rate (volume per unit time) out of $dS$&#8202; through $V$; or if $q$ is heat-flow density, the third product is the heat transfer rate (energy per unit time) out of $S$&#8202; through $V$. The fourth product, by analogy with the third, might be called the flux of the normal derivative of $dS&#8202;$ through the surface element, but is equally well defined whether $V$ is a scalar or a vector—or, for that matter, a matrix, or a tensor of any order, or anything else that we can differentiate w.r.t. $dS$.

If we add up each of the four products over all the elements of the surface $&psi;$, we obtain, respectively, the four surface integrals

in which the double integral sign indicates that the range of integration is two-dimensional. The first surface integral takes a scalar field and yields a vector; the second takes a vector field and yields a vector; the third takes a vector field and yields a scalar; and the fourth takes (e.g.) a scalar field yielding a scalar, or a vector field yielding a vector. If $&psi;$ is pressure in a non-viscous fluid, the first integral is the force exerted by the fluid in $n$&#8202; on the fluid outside $S$. The second integral may be called the skew surface integral of $q$ over $$, or, for the reason hinted above, the circulation of $q$ over $p$.&#8201; The third integral, commonly called the flux integral (or simply the surface integral) of $q$ over $V$, is the total flux of $q$ out of $V$. And the fourth integral is the surface integral of the outward normal derivative of $S&#8202;$.

Let the volume $S$&#8202; be divided into elements. Let a general volume element have the volume $S$ and be enclosed by the surface $V$&#8201;—not to be confused with the area $&psi;$ of a surface element, which may be an element of $V$ or of $dV$. Now consider what happens if, instead of evaluating each of the above surface integrals over $&delta;S$, we evaluate it over each $dS$ and add up the results for all the volume elements. In the interior of $S$, each surface element of area $&delta;S$ is on the boundary between two volume elements, for which the unit normals $q$ at $S$, and the respective values of $&delta;S$, are equal and opposite. Hence when we add up the integrals over the surfaces $V$, the contributions from the elements $dS$ cancel in pairs, except on the original surface $dS$, so that we are left with the original integral over $&part;_{n}&#8202;&psi;$. So, for the four surface integrals in ($&delta;S$), we have respectively

Now comes a big "if":&#8201; if&#8202; we define the gradient of $dS$ (pronounced "grad $S$") as

and the curl of $q$ as

and the divergence of $q$ as

and the Laplacian of $S$ as&#8202;

(where the letters after the equation number stand for gradient, curl, divergence, and Laplacian, respectively), then equations ($$) can be rewritten
 * $$\begin{align}

\iint_S \mathbf{\hat{n}}~\!p \,dS & \,= \sum_V \nabla p ~dV \,, \\ \iint_S \mathbf{\hat{n}}\times\mathbf{q} \,dS & \,= \sum_V \operatorname{curl}\mathbf{q} ~dV \,, \\ \iint_S \mathbf{\hat{n}\cdot q} \,dS & \,= \sum_V \operatorname{div}\mathbf{q} ~dV \,, \\ \iint_S \part_n \psi \;dS & \,= \sum_V \triangle\psi ~dV \,. \end{align}$$ But because each term in each sum has a factor $$, we call the sum an integral; and because the range of integration is three-dimensional, we use a triple integral sign. Thus we obtain the following four theorems relating integrals over an enclosing surface $p$&#8202; to integrals over the enclosed volume $p$:

Of the above four results, only the third ($$) seems to have a standard name; it is called the divergence theorem (or Gauss's theorem or, more properly, Ostrogradsky's theorem ), and is indeed the best known of the four—although the other three, having been derived in parallel with it, may be said to be equally fundamental.

As each of the operators $n&#770;$, $q$, and $q$ calls for an integration w.r.t. area and then a division by volume, the dimension (or unit of measurement) of the result is the dimension of the operand divided by the dimension of length, as if the operation were some sort of differentiation w.r.t. position. Moreover, in each of equations ($$) to ($$), there is a triple integral on the right but only a double integral on the left, so that each of the operators $(△)$, $(&Delta;)$, and $&nabla;$ appears to compensate for a single integration. For these reasons, and for convenience, we shall describe them as differential operators. By comparison, the $curl$ operator in ($&psi;$) or ($$) calls for a further differentiation w.r.t. $$; we shall therefore describe $div$ as a 2nd-order differential operator. (Another reason for these descriptions will emerge in due course.) As promised, the four definitions ($dV$) to ($S$) are "obvious variations on the same theme" (although the fourth is somewhat less obvious than the others).

But remember the "if": Theorems ($V&#8202;$) to ($$) depend on definitions ($$) to ($$) and are therefore only as definite as those definitions! Equations ($$), without assuming anything about the shapes and sizes of the closed surfaces $$ (except, tacitly, that $&nabla;$ is piecewise well-defined), indicate that the surface integrals are additive with respect to volume. But this additivity, by itself, does not guarantee that the surface integrals are shared among neighboring volume elements in proportion to their volumes, as envisaged by "definitions" ($$) to ($$). Each of these "definitions" is unambiguous if, and only if, the ratio of the surface integral to $$&#8202; is insensitive to the shape and size of $$ for a sufficiently small $n&#8202;$. Notice that the issue here is not whether the ratios specified in equations ($$) to ($$) are true vectors or scalars, independent of the coordinates; all of the operations needed in those equations have coordinate-free definitions. Rather, the issue is whether the resulting ratios are unambiguous notwithstanding the ambiguity of $$, provided only that $$ is sufficiently small. That is the advertised "caveat", which must now be addressed.

In accordance with our "applied" mathematical purpose, our proofs of the unambiguity of the differential operators will rest on a few thought experiments, each of which applies an operator to a physical field, say $$, and obtains another physical field whose unambiguity is beyond dispute. The conclusion of the thought experiment is then applicable to any operand field whose mathematical properties are consistent with its interpretation as the physical field $$; the loss of generality, if any, is only what is incurred by that interpretation.

Unambiguity of the gradient
Suppose that a fluid with density $$ (a scalar field) flows with velocity $curl$ (a vector field) under the sole influence of the internal pressure $&delta;S$ (a scalar field). Then the integral in ($$) is the force exerted by the fluid inside $$ on the fluid outside, so that minus the integral is the force exerted on the fluid inside $dV$. Dividing by $&delta;S$, we find that $div$, as defined by ($&delta;S$), is the force per unit volume, which is the acceleration times the mass per unit volume; that is,

Now provided that the left side of this equation is locally continuous, it can be considered uniform inside the small $$, so that the left side is unambiguous, whence $△$ is also unambiguous. If there are additional forces on the fluid element, e.g. due to gravity and/or viscosity, then $△$ is not the sole contribution to density-times-acceleration, but is still the contribution due to pressure, which is still unambiguous.

By showing the unambiguity of definition ($$), we have confirmed theorem ($&delta;S$). In the process we have seen that the volume-based definition of the gradient is useful for the modeling of fluids, and intuitive in that it formalizes the common notion that a pressure "gradient" gives rise to a force.

Unambiguity of the divergence
In the aforesaid fluid, in a short time $&delta;S$, the volume that flows out of fixed closed surface $f$&#8202; through a fixed surface element of area $f&#8202;$&#8202; is $n&#770;$.&#8201; Multiplying by density and integrating over $&rho;$, we find that the mass flowing out of $p$&#8202; in time $$ is&#8201; $$\textstyle\iint_{\delta S}\rho\mathbf{v}~\!dt\cdot\mathbf{\hat{n}}\,dS$$. Dividing this by $&delta;S$, and then by $&delta;S$, we get the rate of reduction of density inside $dV$; that is,
 * $$\tfrac{1}{dV}\!\iint_{\delta S}\mathbf{\hat{n}}\cdot\rho\mathbf{v}\,dS

\,= -\frac{\part\rho}{\part t} \,,$$ where the derivative w.r.t. time is evaluated at a fixed location (because $$ is fixed), and is therefore written as a partial derivative (because other variables on which $$ might depend—namely the coordinates—are held constant). Provided that the right-hand side is locally continuous, it can be considered uniform inside $&delta;S$ and is therefore unambiguous, so that the left side is likewise unambiguous. But the left side is simply $v$&#8201; as defined by ($$), which is therefore also unambiguous, confirming theorem ($$). In short, the divergence operator is that which maps $&minus;&nabla;p$ to the rate of reduction of density at a fixed point:

This result, which expresses conservation of mass, is a form of the so-called equation of continuity.

The partial derivative $dt$ in ($&delta;S$) must be distinguished from the material derivative $dS$, which is evaluated at a point that moves with the fluid. [Similarly, $&minus;&nabla;p$ in ($&delta;S$) is the material acceleration, because it is the acceleration of the mobile mass—not of a fixed point!&#8239;]&#8239; To re-derive the equation of continuity in terms of the material derivative, the volume $&nabla;p$, which flows out through $&delta;S$ in time $dt$ (as above), is integrated over $dV$ to obtain the increase in volume of the mass initially contained in $dt$. Dividing this by the mass, $&delta;S&#8202;$, gives the increase in specific volume $&minus;&nabla;p$ of that mass, and then dividing by $&delta;S$ gives the rate of change of specific volume; that is,
 * $$\tfrac{1}{\rho~\!dV}\!\iint_{\delta S}\mathbf{\hat{n}}\cdot\mathbf{v}\,dS

\,= \tfrac{d}{dt}\big(\rho^{-1}\big) = -\rho^{-2\,}\tfrac{d\rho}{dt} \,.$$ Multiplying by $v&#8202;dt&#8202;⸱&#8239;n&#770;&#8239;dS$ and comparing the left side with ($&rho;$), we obtain

Whereas ($&delta;S$) shows that $div&#8239;&rho;v$ is unambiguous, ($$) shows that $&rho;v&#8202;,$ is unambiguous (provided that the right-hand sides are locally continuous). In accordance with the everyday meaning of "divergence", ($$) also shows that $div&#8239;&rho;v&#8202;$ is positive if the fluid is expanding ($$ decreasing), negative if it is contracting ($&part;&rho;⁄&part;t&#8202;$ increasing), and zero if it is incompressible. In the last case, the equation of continuity reduces to

For incompressible flow, any tubular surface tangential to the flow velocity, and consequently with no flow in or out of the "tube", has the same volumetric flow rate across all cross-sections of the "tube", as if the surface were the wall of a pipe full of liquid (except that the surface is not necessarily stationary). Accordingly, a vector field with zero divergence is described as solenoidal (from the Greek word for "pipe"). More generally, a solenoidal vector field has the property that for any tubular surface tangential to the field, the flux integrals across any two cross-sections of the "tube" are the same—because otherwise there would be a net flux integral out of the closed surface comprising the two cross-sections and any segment of tube between them, in which case, by the divergence theorem ($$), the divergence would have to be non-zero somewhere inside, contrary to ($d&rho;⁄dt&#8202;$).

Unambiguity of the curl (and gradient)
The unambiguity of the curl ($d⁄dt$) follows from the unambiguity of the divergence. Taking dot-products of ($D⁄Dt$) with an arbitrary constant vector $(div&#8239;&rho;)v&#8202;,$, we get
 * $$\begin{align}

\mathbf{b}\cdot\operatorname{curl}\mathbf{q}\, &=\, \mathbf{b}\cdot\tfrac{1}{dV}\!\iint_{\delta S}          \mathbf{\hat{n}}\times\mathbf{q} \,dS \\ &=\, \tfrac{1}{dV}\!\iint_{\delta S}          \mathbf{b}\cdot\mathbf{\hat{n}}\!\times\!\mathbf{q} \,dS \\ &=\, \tfrac{1}{dV}\!\iint_{\delta S}          \mathbf{\hat{n}}\cdot\mathbf{q}\!\times\!\mathbf{b} \,dS \,; \end{align}$$ that is, by ($$),

(The parentheses on the right, although helpful because of the spacing, are not strictly necessary, because the alternative binding would be $&rho;v$, which is a scalar, whose cross-product with the vector $d&#8202;v⁄dt$ is not defined. And the left-hand expression does not need parentheses, because it can only mean the dot-product of a curl with the vector $v&#8202;dt&#8202;⸱&#8239;n&#770;&#8239;dS&#8202;$; it cannot mean the curl of a dot-product, because the curl of a scalar field is not defined.) This result ($dS$) is an identity if the vector $(1/&rho;)$ is independent of location, so that it can be taken inside or outside the surface integral; thus $&rho;&sup2;$ may be a uniform vector field, and may be time-dependent. If we make $div&#8239;&rho;v&#8202;$ a unit vector, the left side of the identity is the (scalar) component of $div&#8239;v&#8202;$ in the direction of $div&#8239;v&#8202;$, and the right side is unambiguous. Thus the curl is unambiguous because its component in any direction is unambiguous. This confirms theorem ($dt$).

Similarly, the unambiguity of the divergence implies the unambiguity of the gradient. Starting with ($&delta;S$), taking dot-products with an arbitrary uniform vector $b$, and proceeding as above, we obtain

(The left-hand side does not need parentheses, because it can only mean the dot-product of a gradient with the vector $b$; it cannot mean the gradient of the dot-product of a scalar field with a vector field, because that dot-product would not be defined.) If we make $(div&#8201;q)$ a unit vector, this result ($dV$) says that the (scalar) component of $b$ in the direction of $b$ is given by the right-hand side, which again is unambiguous. So here we have a second explanation of the unambiguity of the gradient: like the curl, it is unambiguous because its component in any direction is unambiguous.

We might well ask what happens if we take cross-products with $b$ on the left, instead of dot-products. If we start with ($&rho;&#8239;dV$), the process is straightforward: in the end we can switch the order of the cross-product on the left, and change the sign on the right, obtaining

(Again no parentheses are needed.) If we start with ($dt$) instead, and take $b$ inside the integral, we get a vector triple product to expand, which leads to

Here the first term on the right is simply&#8202; $b$&#8201; (the gradient of the dot-product). The second term is more problematic. If we had a scalar $$ instead of the vector $curl&#8239;q$, we could take $b$ outside the second integral, so that the second term would be (minus) $b$. This suggests that the actual second term should be (minus) $b$. But we do not yet know how to interpret that expression for a vector field $b$; and if we were to adopt the second term above (without the sign) as the definition of $b$ (treating $&nabla;p$ as an operator), that would be open to the objection that $b$ had been defined only for uniform $b$, whereas $b$ (for scalar $$) is defined whether $b$ is uniform or not. So, for the moment, let us put ($$) aside and run with ($$), ($$), and ($&rho;$).

Another meaning of the gradient
Let $&nabla;&#8239;b⸱q$ be a unit vector in a given direction, and let $&rho;$ be a parameter measuring distance (arc length) along a path in that direction. By equation ($$) and definition ($$), we have
 * $$\nabla p \cdot \mathbf{\hat{s}} =

\operatorname{div} p\mathbf{\hat{s}} = \tfrac{1}{dV}\!\iint_{\delta S}\mathbf{\hat{n}}\cdot p\mathbf{\hat{s}}\,dS\,, $$ where, by the unambiguity of the divergence, the shape of the closed surface $$ enclosing $$&#8202; can be chosen for convenience. So let $$ be a right cylinder with cross-sectional area $$&#8201; and perpendicular height $$ with the path passing perpendicularly through the end-faces at parameter-values $$ and $$ where the outward unit normal $q$ consequently takes the values $b$ and $b&#8239;⸱&#8239;&nabla;p$ respectively. And let the cross-sectional dimensions be small compared with $$&#8202; so that the values of $$ at the end-faces, say $$ and $$, can be taken to be the same as where the end-faces cut the path. Then&#8202; $$, and the surface integral over $$ includes only the contributions from the end-faces (because $b&#8239;⸱&#8239;&nabla;q$ is perpendicular to $q$ elsewhere); those contributions are respectively&#8201; $$-\mathbf{\hat{s}}\!\cdot\!p\mathbf{\hat{s}}\,\alpha\,$$ and&#8201; $$\mathbf{\hat{s}}\!\cdot\!(p\!+\!dp)\mathbf{\hat{s}}\,\alpha~\!,$$  i.e.&#8239; $$-p\alpha\,$$ and $$(p\!+\!dp)\alpha$$. With these substitutions the above equation becomes
 * $$\begin{align}

\nabla p \cdot \mathbf{\hat{s}} &= \tfrac{1}{\alpha\,ds}\Big({-}p\alpha + (p\!+\!dp)\alpha\Big) \\[1ex] &= \frac{\,p\!+\!dp ~-~ p\,}{ds} = \frac{\part p}{\part s} ~; \end{align}$$ that is,

where the right-hand side, commonly called the directional derivative of $$ in the $b⸱&nabla;&#8239;q$ direction, is the derivative of $p$ w.r.t. distance in that direction. Although ($p$) has been obtained by taking that direction as fixed, the equality is evidently maintained if $$ measures arc length along any path tangential to $b⸱&nabla;$ at the point of interest.

Equation ($$) is an alternative definition of the gradient: it says that the gradient of&#8239; $$p$$ is the vector whose component in any direction is the directional derivative of&#8239; $$p$$ in that direction. For real&#8239; $$p$$, this component has its maximum, namely $b⸱&nabla;&#8239;q$, in the direction of $b$; thus ''the gradient of&#8239; $$p$$ is the vector whose direction is that in which the derivative of&#8239; $$p$$ w.r.t. distance is a maximum, and whose magnitude is that maximum''. This is the usual conceptual definition of the gradient. Sometimes it is convenient to work directly from this definition. For example, in Cartesian coordinates $b&#8239;⸱&#8239;&nabla;p&#8202;$ if a scalar field is given by $$ its gradient is obviously the unit vector in the direction of the $$ axis, usually called $b$; that is, $s&#770;$. Similarly, if&#8201; $n&#770;$&#8201; is the position vector, then $&minus;s&#770;$.

If&#8202; $s&#770;&#8202;,$ is tangential to a level surface of $s$ (a surface of constant $$), then $$&#8202; in that direction is zero, in which case ($&delta;S$) says that $n&#770;$ (if not zero) is orthogonal to $s&#770;$.&#8201; So $$\nabla p$$ is orthogonal to the surfaces of constant&#8202; $$p\,$$ (as we would expect, having just shown that the direction of $s&#770;$ is that in which $dV$ varies most steeply).

If $&delta;S$ is uniform—that is, if it has no spatial variation—then its derivative w.r.t. distance in every direction is zero; that is, the component of $s&#770;$ in every direction is zero, so that $|&nabla;p|&#8202;$ must be the zero vector. In short, the gradient of a uniform scalar field is zero.

Unambiguity of the Laplacian
Armed with our new definition of the gradient ($&alpha;$), we can revisit our definition of the Laplacian ($ds&#8202;,$). If $s$ is a scalar field, then, by ($s+ds&#8202;,$),  $$\part_n \psi$$ can be replaced by $$\nabla\psi\cdot\mathbf{\hat{n}}\,$$ in ($ds$), which then becomes

that is, by definition ($p$),
 * $$\triangle\psi

\,=\, \operatorname{div}\nabla\psi \qquad$$[for scalar $p$]. So the Laplacian of a scalar field is the divergence of the gradient. This is the usual introductory definition of the Laplacian—and on its face is applicable only in the case of a scalar field. The unambiguity of the Laplacian, in this case, follows from the unambiguity of the divergence and the gradient.

If, on the contrary, $p+dp$ in definition ($dV&#8201;=&#8201;&alpha;&#8239;ds&#8202;$) is a vector field, then we can again take dot-products with a uniform vector $&nabla;p&#8202;$, obtaining
 * $$(\triangle\psi)\cdot\mathbf{b} \,=\,

\tfrac{1}{dV}\!\iint_{\delta S} \part_n(\psi\!\cdot\!\mathbf{b}) \,dS \,. $$ If we make $&nabla;u$ a unit vector, this says that the scalar component of the Laplacian of a vector field, in any direction, is the Laplacian of the scalar component of that vector field in that direction. As we have just established that the latter is unambiguous, so is the former.

But the unambiguity of the Laplacian can be generalized further. If
 * $\psi = \textstyle\sum_i \alpha_i \varphi_i$

where each $$\varphi_i$$ is a scalar field, and each $&delta;S$ is a constant, and the counter $$ ranges from (say) 1 to $p$, then it is clear from ($p$) that

In words, this says that the Laplacian of a linear combination of fields is the same linear combination of the Laplacians of the same fields—or, more concisely, that the Laplacian is linear. I say "it is clear" because the Laplacian as defined by ($$) is itself a linear combination, so that ($s$) merely asserts that we can regroup the terms of a nested linear combination; the gradient, curl, and divergence as defined by ($$) to ($u$) are likewise linear. It follows from ($u$) that the Laplacian of a linear combination of fields is unambiguous if the Laplacians of the separate fields are unambiguous. Now we have supposed that the fields $$\varphi_i$$ are scalar and that the coefficients $x&#8202;,$ are constants. But the same logic applies if the "constants" are uniform basis vectors (e.g., $&nabla;u$), so that the "linear combination" can represent any vector field, whence the Laplacian of any vector field is unambiguous. And the same logic applies if the "constants" are chosen as a "basis" for a space of tensors of any order, so that the Laplacian of any tensor field of that order is unambiguous, and so on. In short, the Laplacian of any field that we can express with a uniform basis is unambiguous.

The dot-del, del-cross, and del-dot operators
The gradient operator $(x,&#8239;y,&#8239;z),$ is also called $x$. If it simply denotes the gradient, we tend to pronounce it "grad" in order to emphasize the result. But it can also appear in combination with other operators to give other results, and in those contexts we tend to pronounce it "del".

One such combination is "dot del"—&#8202;as in "$i&#8202;$", which we proposed after ($p$), but did not manage to define satisfactorily for a vector operand. With our new definition of the gradient ($p$), we can now make a second attempt. A general vector field $&nabla;x&#8201;=&#8201;i$ can be written $r&#8201;=&#8201;r&#8202;r&#770;$ so that
 * $$\mathbf{q}\cdot\nabla\psi

\,=\, |\mathbf{q}| \,\mathbf{\hat{q}}\cdot\nabla\psi \,. $$ If $&part;_{s}&#8201;p$ is a scalar field, we can apply ($$) to the right-hand side, obtaining
 * $\mathbf{q}\cdot\nabla\psi

~\!=~\! where $p$ is distance in the direction of $&nabla;r&#8201;=&#8239;r&#770;$. For scalar $p$, this result is an identity between previously defined quantities. For non-scalar $$, we have not yet defined the left-hand side, but the right-hand side is still well-defined and self-explanatory (provided that we can differentiate $$ w.r.t. $&psi;$). So we are free to adopt

where $$ is distance in the direction of $s&#770;$ as the general definition of the operator $&nabla;p$ and to interpret it as defining both a unary operator&#8202; $s&#770;$ which operates on a generic field, and a binary operator&#8202; $&nabla;p$ which takes a (possibly uniform) vector field on the left and a generic field on the right.

For the special case in which $&nabla;p$ is a unit vector $&nabla;p$ with $$ measuring distance in the direction of&#8202; $b$ definition ($$) reduces to

which agrees with ($$) but now holds for a generic field $&psi;$ [whereas ($&psi;$) was for a scalar field, and was derived as a theorem based on earlier definitions]. So $b$ with a unit vector $i,&#8202;j,k$ is the directional-derivative operator on a generic field.

In particular, if&#8201; $&nabla;$&#8201; we have
 * $$\part_n \psi \,=\, \mathbf{\hat{n}}\;\!{\cdot}\nabla\,\psi \,,$$

which we may substitute into the original definition of the Laplacian ($$) to obtain

which is just ($&alpha;_{i}$) again, except that it now holds for for a generic field.

If our general definition of the gradient ($i$) is also taken as the general definition of the $&#8202;b⸱&nabla;&#8202;$ operator, then, comparing ($k&#8202;$) with ($$), ($$), and ($$), we see that
 * $$\begin{align}

\operatorname{curl}\mathbf{q} ~\!&= \nabla(\times\mathbf{q}) \\ \operatorname{div}\mathbf{q} ~\!&= \nabla(\cdot\,\mathbf{q}) \\ \triangle\psi                ~\!&= \nabla(\cdot\nabla\,\psi) \,, \end{align}$$ where the parentheses may seem to be required on account of the closing $$ in ($$). But if we write the factor $$ before the integrand, the del operator in ($$) becomes
 * $$\nabla = \tfrac{1}{dV}\!\iint_{\delta S} dS\,\mathbf{\hat{n}} $$

—if&#8202; we insist that it is to be read as a operator looking for an operand, and not as a self-contained expression. Then, if we similarly bring forward the $&alpha;_{i}$ in ($del$), ($$), and ($$), the respective operators become

(pronounced "del cross", "del dot", and "del dot del"), of which the last is usually abbreviated as $q$&#8201; ("del squared"). Because these operational equivalences follow from coordinate-free definitions, they must remain valid when correctly expressed in any coordinate system. That does not mean that they are always convenient or always conducive to the avoidance of error—of which we shall have more to say in due course. But they sometimes make useful mnemonic devices. For example, they let us rewrite identities ($&psi;$), ($$), and ($s_{q}$) as {{NumBlk|:|$$\left.\begin{align} \nabla\!\times\!\mathbf{q}\cdot\mathbf{b} &\,=\, \nabla\cdot\mathbf{q}\!\times\!\mathbf{b}\\ \nabla p \cdot \mathbf{b} &\,=\, \nabla\cdot\;\! p\mathbf{b}\\ \nabla p \times \mathbf{b} &\,=\, \nabla \times p\mathbf{b} \end{align}~\right\}\quad$$for uniform $|q|&#8201;q&#770;&#8202;,$. |$&psi;$}} These would be basic algebraic vector identities if&#8202; $q$ were an ordinary vector, and one could try to derive them from the "algebraic" behavior of $q&#8202;,$; but they're not, because it isn't, so we didn't&#8239;! Moreover, these simple "algebraic" rules are for a uniform $q⸱&nabla;&#8202;,$, and do not of themselves tell us what to do if $q⸱&nabla;$ is spatially variable; for example, ($&psi;$) is not applicable to ($&psi;$).

The advection operator
Variation or transportation of a property of a medium due to motion with the medium is called advection (which, according to its Latin roots, means "carrying to"). Suppose that a medium (possibly a fluid) moves with a velocity field $⸱&nabla;$ in some inertial reference frame. Let $s_{q}$ be a field (possibly a scalar field or a vector field) expressing some property of the medium (e.g., density, or acceleration, or stress,… or even $q$ itself). We have seen that the time-derivative of $$ may be specified in two different ways: as the partial derivative $s_{q}$ evaluated at a fixed point (in the chosen reference frame), or as the material derivative $s$, evaluated at a point moving at velocity $s&#770;&#8202;,$ (i.e., with the medium). The difference&#8202; $$ is due to motion with the medium. To find another expression for this difference, let $$ be a parameter measuring distance along the path traveled by a particle of the medium. Then, for a short time interval $$, the surface representing the small change in $&psi;$ (or each component thereof) as a function of the small changes in $$ and $$ (plotted on perpendicular axes) can be taken as a plane through the origin, so that
 * $d\psi

= \tfrac{\part\psi}{\part t}~\!dt + \tfrac{\part\psi}{\part s}~\!ds \;; $ that is, the change in $$ is the sum of the changes due to the change in $$ and the change in $$. Dividing by $$ gives
 * $\begin{align}\tfrac{d\psi}{dt}

&= \tfrac{\part\psi}{\part t}+\tfrac{\part\psi}{\part s}~\!\tfrac{ds}{dt}\\[1ex] &= \tfrac{\part\psi}{\part t}+\tfrac{\part\psi}{\part s}~\! i.e.,
 * $\tfrac{d\psi}{dt}

= \tfrac{\part\psi}{\part t} + (and the first term on the right could have been written $$). So the second term on the right is the contribution to the material derivative due to motion with the medium; it is called the advective term, and is non-zero wherever a particle of the medium moves along a path on which $$ varies with location—even if $$ at each location is constant over time.&#8201; So the operator&#8202; $s&#770;&#8202;,$ where $dS$ measures distance along the path, is the advection operator&#8239;: it maps a property of a medium to the advective term in the time-derivative of that property. If $$ is $s&#770;⸱&nabla;&#8202;,$ itself, the above result becomes
 * $\tfrac{d\mathbf{v}}{dt}

= \tfrac{\part\mathbf{v}}{\part t} + |\mathbf{v}|\,\part_s \mathbf{v} \,, $|undefined where the left-hand side (the material acceleration) is as given by Newton's second law, and the first term on the right (which we might call the "partial" acceleration) is the time-derivative of velocity in the chosen reference frame, and the second term on the right (the advective term) is the correction that must be added to the "partial" acceleration in order to obtain the material acceleration. This term is non-zero wherever velocity is non-zero and varies along a path, even if the velocity at each point on the path is constant over time (as when water speeds up while flowing at a constant volumetric rate into a nozzle). Paradoxically, while the material acceleration and the "partial" acceleration are apparently linear (first-degree) in $s&#8202;,$, their difference (the advective term) is not. Thus the distinction between $dS$ and $$&#8202; has the far-reaching implication that fluid dynamics is non-linear.

Applying ($dS$) to the last two equations, we obtain respectively

and

where, in each case, the second term on the right is the advective term. So the advection operator can also be written&#8239; $s&#770;&#8201;=&#8201;n&#770;$.

When the generic $$ in ($$) is replaced by the density $$, we get a relation between $$ and $$, both of which we have seen before—in equations ($$) and ($$) above. Substituting from those equations then gives

where $&nabla;$ can be taken as a gradient since $$ is scalar. This result is in fact an identity—a product rule for the divergence—as we shall eventually confirm by another method.

Generalized volume-integral theorem
We can rewrite the fourth integral theorem ($$) in the "dot del" notation as

Then, using notations ($$), we can condense all four integral theorems ($&psi;$), ($&psi;$), ($&psi;$), and ($&part;&psi;⁄&part;t&#8202;,$) into the single equation

where the "circ" symbol $&nabla;^{2}$ is a generic binary operator which may be replaced by a null (direct juxtaposition of the operands) for theorem ($d&psi;⁄dt&#8202;$), or a cross for ($d&psi;⁄dt&#8202;&minus;&#8202;&part;&psi;⁄&part;t&#8202;$), or a dot for ($s$), or&#8202; $&nabla;⸱&nabla;$ for ($dt$). This single equation is a generalized volume-integral theorem, relating an integral over a volume to an integral over its enclosing surface.

Theorem ($&psi;$) is based on the following definitions, which have been found unambiguous: The gradient maps a scalar field to a vector field; the divergence maps a vector field to a scalar field; the curl maps a vector field to a vector field; and the Laplacian maps a scalar field to a scalar field, or a vector field to a vector field, etc.
 * the gradient of a scalar field $t$ is the closed-surface integral of&#8202; $b$ per unit volume, where $&nabla;$ is the outward unit normal;
 * the divergence of a vector field is the outward flux integral per unit volume;
 * the curl of a vector field is the skew surface integral per unit volume, also called the surface circulation per unit volume; and
 * the Laplacian is the closed-surface integral of the outward normal derivative, per unit volume.

The gradient of $s&#8202;$, as defined above, has been shown to be also Consistent with these alternative definitions of the gradient, we have defined the &#8202;$&nabla;$ operator so that&#8202; $b$ (for a unit vector $b$) is the operator yielding the directional derivative in the direction of&#8202; $v$ and we have used that notation to bring theorem ($&psi;$) under theorem ($t$).
 * the vector whose component in any direction is the directional derivative of $s&#8202;$ in that direction (i.e. the derivative of $dt$ w.r.t. distance in that direction), and
 * the vector whose direction is that in which the directional derivative of $&part;_{t}&#8239;&psi;$ is a maximum, and whose magnitude is that maximum.

So far, we have said comparatively little about the curl. That imbalance will now be rectified.

Instant integral theorems—on a condition
Theorems ($&psi;$) to ($&psi;$) are three-dimensional: each of them relates an integral over a volume $s$&#8202; to an integral over its enclosing surface $&psi;$. We now seek analogous two-dimensional theorems, each of which relates an integral over a surface segment to an integral around its enclosing curve. For maximum generality, the surface segment should be allowed to be curved into a third dimension. Theorems of this kind can be obtained as special cases of theorems ($&part;&psi;⁄&part;t$) to ($d&psi;⁄dt$) by suitably choosing $$ and $$; this is another advantage of our "volume first" approach.

Let $$ be a surface segment enclosed by a curve $&psi;&#8202;$ (a circuit or closed contour), and let $$ be a parameter measuring arc length around $&rho;&#8202;$, so that a general element of $&part;&rho;⁄&part;t&#8202;$ has length $d&rho;⁄dt&#8202;$; and let a general element of the surface $$ have area $$. Let $$\boldsymbol{\hat{\nu}}$$ be the unit normal vector at a general point on $$, and let $v$ be the unit tangent vector to $&rho;$ at a general point on $$ in the direction of increasing $$. In the original case of a surface enclosing a volume, we had to decide whether the unit normal pointed into or out of the volume (we chose the latter). In the present case of a circuit enclosing a surface segment, we have to decide whether $$ is measured clockwise or counterclockwise as seen when looking in the direction of the unit normal, and we choose clockwise. So $$ is measured clockwise about $$\;\!\boldsymbol{\hat{\nu}}$$, and $$ is traversed clockwise about $$\boldsymbol{\hat{\nu}}$$.

From $$&#8202; we can construct obvious candidates for $$ and $$. From every point on $$, erect a perpendicular with a uniform small height $$ in the direction of $$\boldsymbol{\hat{\nu}}$$. Then simply let $$ be the volume occupied by all the perpendiculars, and let $$ be its enclosing surface. Thus $$ is a (generally curved) thin slab of uniform thickness $$, whose enclosing surface $$ consists of two close parallel (generally curved) broad faces connected by a perpendicular edge-face of uniform height $p$; and we can treat $$\boldsymbol{\hat{\nu}}$$ as a vector field by extrapolating it perpendicularly from $p$. If we can arrange for $p$ to cancel out, the volume $p$ will serve as a 3D representation of the surface segment $p$ while the edge-face will serve as a 2D representation of the curve $$, so that our four theorems will relate an integral around $$ to an integral over $$&#8201; provided that there is no contribution from the broad faces to the integral over $$. For brevity, let us call this proviso the 2D condition.

If&#8202; the 2D condition is satisfied, an integral over the new $V$ reduces to an integral over the edge-face, on which
 * $$dS = h\,dl \,,$$

so that the cancellation of $S$ will leave an integral over $$&#8202; w.r.t. length. Meanwhile, in an integral over the new $$, regardless of the 2D condition, we have
 * $$dV = h\,d\varSigma \,,$$

so that the cancellation of $V$ will leave an integral over $S&#8202;$ w.r.t. area. So, substituting for $&Sigma;$ and $C$&#8202; in ($l$) to ($C&#8202;$), and canceling $C$ as planned, we obtain respectively

all subject to the 2D condition. In each equation, the circle on the left integral sign acknowledges that the integral is around a closed loop. The unit vector $v$ which was normal to the edge-face, is now normal to both $|v|&#8201;&part;_{s}&#8202;,$ and $$\boldsymbol{\hat{\nu}}$$; that is, $v$ is tangential to the surface segment $dl&#8202;$ and projects perpendicularly outward from its bounding curve.

On the left side of ($&Sigma;$), the 2D condition is satisfied if (but not only if) $v$ takes equal-and-opposite values at any two opposing points on opposing broad faces of $d&Sigma;$ i.e. if $&Sigma;&#8202;$ takes the same value at such points, i.e. if $C$ has a zero directional derivative normal to $C$ i.e. if $v⸱&nabla;&#8202;$ has no component normal to $l$. Thus a sufficient "2D condition" for ($l$) is the obvious one.

Skipping forward to ($l$), we see that the 2D condition is satisfied if&#8202; $$\part_n \psi$$ takes equal-and-opposite values at any two opposing points on opposing broad faces of $C$ i.e. if&#8202; $$\part_{\nu}\psi$$ (where $$\nu$$ measures distance in the direction of $$\boldsymbol{\hat{\nu}}$$) takes the same value at such points, i.e. if&#8202; $$\part^2_{\nu}\psi\!=\!0$$.

For ($&Sigma;$) and ($V$), the 2D constraint can be satisfied by construction, with more useful results—as explained under the next two headings. To facilitate this process, we first make a minor adjustment to $S$ and $&Sigma;&#8202;$. Noting that any curved surface segment can be approximated to any desired accuracy by a polyhedral surface enclosed by a polygon, we shall indeed consider $h$ to be a polyhedral surface made up of small planar elements, $V$ being the area of a general element, and we shall indeed consider $S$ to be a polygon with short sides, $V$ being the length of a general side. The benefit of this trick, as we shall see, is to make the unit normal $$\boldsymbol{\hat{\nu}}$$ uniform over each surface element, without forcing us to treat $&nabla;&rho;$ (or any other field) as uniform over the same element. But, as the elements of $h$ can independently be made as short as we like (dividing straight sides into shorter elements if necessary!), we can still consider $$\boldsymbol{\hat{\nu}},$$ $&cir;$ and $⸱&nabla;$ to be uniform over each element of $S$.

Special case for the gradient
In ($h&#8202;$), the 2D condition is satisfied by&#8202; $$\mathbf{q}\!=\!p\boldsymbol{\hat{\nu}}$$ (where $&Sigma;$ is a scalar field), because then the integrand on the left is zero on the broad faces of $h$ where $n&#770;&#8202;p$ is parallel to $$\boldsymbol{\hat{\nu}}$$. Equation ($V$) then becomes

Now on the left,&#8201; $$\mathbf{\hat{n}}\!\times\!\boldsymbol{\hat{\nu}}\!=\!-\mathbf{\hat{t}}~\!;\,$$ and on the right, over each surface element, the unit normal $$\boldsymbol{\hat{\nu}}$$ is uniform so that, by ($&Sigma;$),&#8201; $\operatorname{curl}p\boldsymbol{\hat{\nu}}=~\!\!\nabla p \!\times\!\boldsymbol{\hat{\nu}}=-\boldsymbol{\hat{\nu}}\!\times\!\nabla p$|undefined. With these substitutions, the minus signs cancel and we get

or, if we write&#8201; $$d\mathbf{r}\!=\!\mathbf{\hat{t}}~\!dl\,$$ and&#8201; $$\boldsymbol{d\varSigma}\!=\!\boldsymbol{\hat{\nu}}\,d\varSigma~\!,$$

This result, although well attested in the literature, does not seem to have a name—unlike the next result.

Special case for the curl
In ($C&#8202;$), the 2D condition is satisfied if $n&#770;$ is replaced by&#8202; $$\boldsymbol{\hat{\nu}}{\times}\mathbf{q}~\!,\,$$ because then (again) the integrand on the left is zero on the broad faces of $C$ where $⸱&nabla;$ is parallel to $$\boldsymbol{\hat{\nu}}$$. Equation ($&Sigma;$) then becomes

Now on the left, the integrand can be written&#8201; $$\mathbf{\hat{n}}{\times}\boldsymbol{\hat{\nu}}\!\cdot\!\mathbf{q}\!=\!-\mathbf{\hat{t}}\!\cdot\!\mathbf{q}~\!;\,$$ and on the right,&#8201; $\operatorname{div}(\boldsymbol{\hat{\nu}}\!\times\!\mathbf{q})\!=\!-\operatorname{div}(\mathbf{q}\!\times\!\boldsymbol{\hat{\nu}})\!=\!-\operatorname{curl}\mathbf{q}\cdot\boldsymbol{\hat{\nu}}\,$|undefined by identity ($S$), since $$\boldsymbol{\hat{\nu}}$$ is uniform over each surface element. With these substitutions, the minus signs cancel and we get

or, if we again write&#8201; $$d\mathbf{r}\!=\!\mathbf{\hat{t}}~\!dl\,$$ and&#8201; $$\boldsymbol{d\varSigma}\!=\!\boldsymbol{\hat{\nu}}\,d\varSigma~\!,$$

This result—the best-known theorem relating an integral over a surface segment to an integral around its enclosing curve, and the best-known theorem involving the curl—is called Stokes' theorem or, more properly, the Kelvin–Stokes theorem, or simply the curl theorem.

The integral on the left of ($S$) or ($h$) is called the circulation of the vector field $s&#770;⸱&nabla;$ around the closed curve $C$. So, in words, the Kelvin-Stokes theorem says that the circulation of a vector field around a closed curve is equal to the flux of the curl of that vector field through any surface spanning that closed curve.

Now let a general element of $V$ (with area $h$) be enclosed by the curve $&Sigma;$, traversed in the same direction as the outer curve $dS$. Then, applying ($dV$) to the single element, we have

\oint_{\delta C} \mathbf{q} \cdot \mathbf{\hat{t}} \,dl \,=\, \operatorname{curl}\mathbf{q}\cdot\boldsymbol{\hat{\nu}} \,d\varSigma \,; $$ that is,

where the right-hand side is simply the circulation per unit area.

Equation ($$) is an alternative definition of the curl: it says that the curl of $s&#770;$ is the vector whose component in any direction is the circulation of $s&#770;&#8202;,$ per unit area of a surface whose normal points in that direction. For real $t&#8202;&#770;$, this component has its maximum, namely $n&#770;&#8202;,$, in the direction of $t&#8202;&#770;$; thus the curl of $n&#770;$ is the vector whose direction is that which a surface must face if the circulation of $n&#770;p$ per unit area of that surface is to be a maximum, and whose magnitude is that maximum. This is the usual conceptual definition of the curl.

[Notice, however, that our original volume-based definition ($$) is more succinct: the curl is the closed-surface circulation per unit volume, i.e. the skew surface integral per unit volume.]

It should now be clear where the curl gets its name (coined by Maxwell), and why it is also called the rotation (indeed the $&nabla;p$ operator is sometimes written "$q$", especially in Continental languages, in which "rot" does not have the same unfortunate everyday meaning as in English). It should be similarly unsurprising that a vector field with zero curl is described as irrotational (which one must carefully pronounce differently from "irri&#8202;tational"!), and that the curl of the velocity of a fluid is called the vorticity.

However, a field does not need to be vortex-like in order to have a non-zero curl; for example, by identity ($h$), in Cartesian coordinates, the velocity field $q&#8202;,$ has a curl equal to&#8201; $t&#8202;&#770;$&#8201; although it describes a shearing motion rather than a rotating motion. This is understandable because if you hold a pencil between the palms of your hands and slide one palm over the other (a shearing motion), the pencil rotates. Conversely, we can have a vortex-like field whose curl is zero everywhere except on or near the axis of the vortex. For example, the Maxwell-Amp&egrave;re law in magnetostatics says that&#8201; $n$ where $q$ is the magnetizing field and $n$ is the current density. So if the current is confined to a wire, $q$ is zero outside the wire—although, as is well known, the field lines circle the wire. The explanation of the paradox is that $q$ gets stronger as we approach the wire, making a shearing pattern, whose effect on the curl counteracts that of the rotation.

The curl-grad and div-curl operators
We have seen from ($$) that the Laplacian of a scalar field is the divergence of the gradient. Four more such second-order combinations make sense, namely the curl of the gradient (of a scalar field), and the divergence of the curl, the gradient of the divergence, and the curl of the curl (of a vector field). The first two&#8239;—"curl grad" and "div curl"—&#8239;can now be disposed of.

Let the surface segment $$ enclosed by the curve $$ be a segment of the closed surface $$ surrounding the volume $&Sigma;$, and let $$ expand across $S&#8202;,$ until it engulfs $p$ so that $p$ shrinks to a point on the far side of $&Sigma;&#8202;,$. Then, in the nameless theorem ($&Sigma;$) and the Kelvin-Stokes theorem ($$), the integral on the left becomes zero while $$ and $$\boldsymbol{\hat{\nu}}$$ on the right become $S&#8202;,$ and $q$ so that the theorems respectively reduce to

\iint_S \mathbf{\hat{n}}\times\nabla p \;dS \,=\, \mathbf{0} $$ and

\iint_S \mathbf{\hat{n}}\cdot\operatorname{curl}\mathbf{q} \;dS \,=\, 0 \,. $$ Applying theorem ($$) to the first of these two equations, and the divergence theorem ($$) to the second, we obtain respectively
 * $$\iiint_V \operatorname{curl}\nabla p \;dV ~\!=\, \mathbf{0} \,,$$

and

\iiint_{V} \operatorname{div}\operatorname{curl}\mathbf{q} \;dV ~\!=\, 0 \,. $$ As the integrals vanish for any volume $&Sigma;$ in which the integrands are defined, the integrands must be zero wherever they are defined; that is,

and

In words, the curl of the gradient is zero, and the divergence of the curl is zero; or, more concisely, any gradient is irrotational, and any curl is solenoidal.

We might well ask whether the converses are true. Is every irrotational vector field the gradient of something? And is every solenoidal vector field the curl of something? The answers are affirmative, but the proofs require more preparation.

Meanwhile we may note, as a mnemonic aid, that when the left-hand sides of the last two equations are rewritten in the del-cross and del-dot notations, they become&#8201; $q$&#8201; and&#8201; $|curl&#8201;q|&#8202;$ respectively. The former looks like (but isn't) a cross-product of two parallel vectors, and the latter looks like (but isn't) a scalar triple product with a repeated factor, so that each expression looks like it ought to be zero (and it is). But such appearances can lead one astray, because $curl&#8239;q&#8202;$ is an operator, not a self-contained vector quantity; for example,&#8201; $q$&#8201; is not identically zero, because two gradients are not necessarily parallel.

We should also note, to tie a loose end, that identity ($C$) was to be expected from our verbal statement of the Kelvin-Stokes theorem ($&Sigma;$). That statement implies that the flux of the curl through any two surfaces spanning the same closed curve is the same. So if we make a closed surface from two spanning surfaces, the flux into one spanning surface is equal to the flux out of the other, i.e. the net flux out of the closed surface is zero, i.e. the integral of the divergence over the enclosed volume is zero; and since any simple volume in which the divergence is defined can be enclosed this way, the divergence itself (of the curl) must be zero wherever it is defined.

Change per unit length
Continuing (and concluding) the trend of reducing the number of dimensions, we now seek one-dimensional theorems, each of which relates an integral over a path to values at the endpoints of the path. For maximum generality, the path should be allowed to be curved into a second and a third dimension.

We could do this by further specializing theorems ($d&Sigma;$) to ($C$). We could take a curve $q$ with a unit tangent vector $curl$. At every point on $rot$ we could mount a circular disk with a uniform small area $dl$ centered on $xj$ and orthogonal to it. We could let $&Sigma;$ be the volume occupied by all the disks and let $C$ be its enclosing surface; thus $&Sigma;$ would be a thin right circular cylinder, except that its axis could be curved. If we could arrange for $C$ to cancel out, our four theorems would indeed be reduced to the desired form, provided that there were no contribution from the curved face of the "cylinder" to the integral over $C$ (the "1D proviso"). But, as it turns out, this exercise yields only one case in which the "1D proviso" can be satisfied by a construction involving $&nabla;x&#8201;&times;&#8239;j&#8201;=&#8201;i&#8202;&times;&#8202;j&#8201;=&#8201;k&#8202;,$ and a general field, and we have already almost discovered that case by a simpler argument—which we shall now continue.

Fundamental theorem
Equation ($C$) is applicable where $curl&#8201;H&#8201;=&#8201;J&#8202;,$ is a scalar field,  $$ is a parameter measuring arc length along a curve $H$ and $J$ is the unit tangent vector to $&part;&#8202;D⁄&part;t$ in the direction of increasing $p$. Let $S&#8202;,$ take the values $curl&#8201;H&#8202;$ and $H$ at the endpoints of $n&#770;&#8202;,$ where the position vector $&nabla;&#8201;&times;&#8201;&nabla;p$ takes the values $&nabla;&#8201;⸱&#8201;&nabla;&#8202;&times;&#8202;q&#8202;,$ and $&nabla;$ respectively. Then, integrating ($$) w.r.t. $$ from $&nabla;p&#8201;&times;&#8201;&nabla;&phi;$ to $&Gamma;$ and applying the fundamental theorem of calculus, we get

This is our third integral theorem involving the gradient, and the best-known of the three: it is commonly called simply the gradient theorem, or the fundamental theorem of the gradient, or the fundamental theorem of line integrals; it generalizes the fundamental theorem of calculus to a curved path. If we write $$d\mathbf{r}~\!$$ for&#8239; $s&#770;$ (the change in the position vector), we get the theorem in the alternative form

As the right-hand side of ($$) or ($$) obviously depends on the endpoints but not on the path in between, so does the integral on the left. This integral is commonly called the work integral of $&Gamma;$ over the path—because if $&Gamma;$ is a force, the integral is the work done by the force over the path. So, in words, the gradient theorem says that the change in value of a scalar field from one point to another is the work integral of the gradient of that field field over any path from the one to the other.

Applying ($$) to a single element of the curve, we get

Alternatively, we could have obtained ($$) by multiplying both sides of ($S&#8202;,$) by $$, and then obtained ($$) by adding ($$) over all the elemental displacements $$d\mathbf{r}$$ on any path from $s&#770;$ to $p(r)$.

If we close the path by setting&#8201; $&Gamma;,$ the gradient theorem reduces to

where the integral is around any closed loop. Applying the Kelvin-Stokes theorem then gives

where $$ is any surface spanning the loop. As this applies to any loop spanned by any surface on which the integrand is defined,&#8201; $s&#770;$ must be zero wherever it is defined. This is a second proof (and indeed the usual method of proof) of theorem ($$).



[To be continued.]

Competing interests
None.

Ethics statement
This article does not concern research on human or animal subjects.

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