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I am reading the book "Div, Grad, Curl, and All that" and I got to the section about curl. In this section, the author defines the curl to be

$$ (\nabla \times \mathbf{F})\cdot \mathbf{\hat{n}} \ \overset{\underset{\mathrm{def}}{}}{=} \lim_{S \to 0}\left( \frac{1}{|S|}\oint_C \mathbf{F} \cdot d\mathbf{r}\right) \tag{1}$$

The author gives a "physicist's rough-and-ready proof" of how this expression reduces to each of the $x,y$ and $z$ components of $$ \nabla \times \mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \boldsymbol{\hat\imath} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \boldsymbol{\hat\jmath} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \boldsymbol{\hat k} \tag{2} $$ when analyzing a closed path $C$ which is parallel to the $yz, xz$, and $xy$ planes respectively.


After this, I started wondering if there's a more rigorous way to show that, in cartesian coordinates, equation $(2)$ satisfies the definition in equation $(1)$. I'm specifically interested in a proof where you assume any arbitrary closed path $C$.

If someone could tell me how this is can be done, or could point me in a direction where this proof is already given I would greatly appreciate it. Thank you!

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    $\begingroup$ It's depressing that very few vector calculus textbooks prove this... $\endgroup$ Jul 13, 2020 at 0:11
  • $\begingroup$ Are you trying to go from $(1)$ to $(2)$? Or from $(2)$ to $(1)$? In any case, a rigorous development would first require you to formulate precisely what is meant by the limit in $(1)$, so are you comfortable with $\epsilon, \delta$ type definitions of limits (and some basic topology like open sets etc)? $\endgroup$
    – peek-a-boo
    Jul 13, 2020 at 3:46
  • $\begingroup$ @peek-a-boo I want to get from $(2)$ to $(1)$ since (I think) it would be easier to show this direction given that $(2)$ is an explicit definition, contrary to $(1)$. As for the second part, I am also comfortable with the subjects you mention. If there's a proof in an undergrad multivariable calculus book I think I should be able to understand it. $\endgroup$
    – Robert Lee
    Jul 13, 2020 at 4:01
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    $\begingroup$ I'm not sure about any rigorous references for vector calculus (because I directly studied the General Stokes theorem on manifolds, for which I used Spivak's Calculus on Manifolds). In going from (2) to (1) rigorously, I can't see any approach besides invoking Stokes theorem (also, I feel that the first definition is slightly problematic without the second, because how do you even know a-priori that the limit on the RHS of $(1)$ exists)? Of course, I'm not saying it's not possible... just that I can't see an approach without already invoking Stokes theorem. $\endgroup$
    – peek-a-boo
    Jul 13, 2020 at 4:10
  • $\begingroup$ To me, once you make the temporary assumption that the limit in $(1)$ exists, it is pretty easy (compared to the other direction) to prove that $(1) \implies (2)$, because here, you just have to choose a small enough planar surface with normal vectors along $e_1, e_2,e_3$ respectively (and use the mean-value theorem for derivatives and integrals few times). The hard thing though is showing that the limit actually exists, and the only way I see how to prove this is to invoke Stokes theorem. $\endgroup$
    – peek-a-boo
    Jul 13, 2020 at 4:20

2 Answers 2

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Let's formulate the definition of curl slightly more precisely in the form of a definition/theorem. I'll also not use boldface objects, simply for ease of typing

Definition/Theorem.

Let $A\subset \Bbb{R}^3$ be open, $F: A \to \Bbb{R}^3$ be $C^1$. Then, there is a unique continuous function $H:A \to \Bbb{R}^3$, such that for every $p\in A$, for every $\epsilon > 0$, for every "nice" surface $S\subset A$, there is an open neighbourhood $U$ of $p$ in $S$ such that for every "nice" oriented surface $M$ with $p\in M\subset U$, with outward normal vector field $n(\cdot)$ (which is simply the restriction of the outward normal of $S$ to $M$), boundary $\partial M$ and tangent vector field $\tau(\cdot)$ on $\partial M$, we have: \begin{align} \left|\dfrac{1}{|M|}\int_{\partial M}\langle F, \tau\rangle\, dl - \langle H(p), n(p)\rangle\right| < \epsilon\tag{1} \end{align} (this is the more precise meaning of the limit you're talking about) In this case, because $H$ is unique, we can give it the name $\text{curl}(F)$. In fact, we can show that \begin{align} H = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \boldsymbol{\hat\imath} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \boldsymbol{\hat\jmath} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \boldsymbol{\hat k} \tag{2} \end{align}

In the above paragraph, "nice oriented surface" means nice enough so that Stoke's theorem can be applied; for example a smooth two-dimensional, oriented manifold-with-boundary (or however much you wanna weaken the hypothesis... because each book presents it with varying levels of generality... just as long as Stoke's theorem can be applied).

Note that for the above definition of curl to make sense, we have to first show the existence and uniqueness of such a vector field $H$. We shall start by showing the uniqueness of $H$. So, we assume such a $H$ exists, and then prove its components are determined according to the formula $(2)$; this will complete the proof of uniqueness.


Proof of Uniqueness

I'll carry out the computation in detail for proving $H_x = \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z}$, and leave the other two to you (it's simply a matter of renaming $x,y,z$). We prove this equality pointwise of course. So, fix a point $p \in A$; then for any $\delta > 0$ such that the closed cube $C_{p,\delta} = p + [-\delta,\delta]^3$ (which is the closed cube centered at $p$ of sidelength $2\delta$) lies entirely inside $A$ (note that since $A$ is open, there are infinitely many such $\delta>0$), we define $M^{\delta} := \{p_1\}\times [p_2-\delta, p_2 + \delta]\times [p_3-\delta, p_3 + \delta]$. This is a piece of a plane which we shall orient so that it has a constant outward normal vector field $n = e_1 \equiv \boldsymbol{i}$. Now, we calculate: note that $\partial M^{\delta}$ has $4$-pieces, and the unit tangent vector along these boundary paths is constant, so (if you're very careful with signs... which I hope I didn't make any sign mistakes), we get \begin{align} \dfrac{1}{|M^{\delta}|} \int_{\partial M^{\delta}}\langle F, \tau \rangle\, dl &= \dfrac{1}{4\delta^2} \bigg[\int_{-\delta}^{\delta} F_2(p_1, p_2 + y, p_3-\delta) - F_2(p_1, p_2 + y, p_3+\delta) \, dy\bigg]\\ &+\dfrac{1}{4\delta^2}\bigg[ \int_{-\delta}^{\delta} F_3(p_1, p_2+\delta, p_3+z) - F_3(p_1, p_2-\delta, p_3+z)\, dz\bigg] \end{align} For each term, we apply the mean-value theorem for integrals (which we can use since everything is continuous), to obtain some $\eta \in [p_2-\delta, p_2+\delta]$ and some $\zeta\in [p_3-\delta, p_3+\delta]$ such that

\begin{align} \dfrac{1}{|M^{\delta}|} \int_{\partial M^{\delta}}\langle F, \tau \rangle\, dl &= \dfrac{1}{4\delta^2}\bigg[2\delta \cdot F_2(p_1, \eta, p_3-\delta) - 2\delta \cdot F_2(p_1, \eta, p_3+\delta)\bigg] \\ &+ \dfrac{1}{4\delta^2}\bigg[2\delta \cdot F_3(p_1, p_2+\delta, \zeta) - 2\delta \cdot F_3(p_1, p_2-\delta, \zeta)\bigg] \\\\ &=\dfrac{F_3(p_1, p_2+\delta, \zeta) - F_3(p_1, p_2-\delta, \zeta)}{2\delta} -\dfrac{F_2(p_1, \eta, p_3+\delta) - F_2(p_1, \eta, p_3-\delta)}{2\delta} \\ &= \dfrac{\partial F_3}{\partial y}(p_1, \alpha, \zeta) - \dfrac{\partial F_2}{\partial z}(p_1, \eta, \beta), \end{align} for some $\alpha\in [p_2-\delta, p_2+\delta], \beta\in [p_3-\delta, p_3+\delta]$, using the mean-value theorem for derivatives (which can certainly be applied since we assumed $F$ is $C^1$).

Quick summary: what we showed so far is that for every $p\in A$ and every $\delta>0$ such that the cube $C_{p,\delta}$ lies inside $A$, if we define $M^{\delta}$ as above to be the plane centered at $p$ with normal pointing in $e_1$ direction, then, there exist points $a_{p,\delta},b_{p,\delta} \in M^{\delta} \subset C_{p,\delta}$ inside the surface (in particular inside the cube) such that

\begin{align} \dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle \, dl &= \dfrac{\partial F_3}{\partial y}(a_{p,\delta}) - \dfrac{\partial F_2}{\partial z}(b_{p,\delta}) \end{align} From here, it is a simple matter of using the continuity of partial derivatives. Here's the full $\epsilon,\delta$ argument to finish it off: let $p\in A$ and $\epsilon> 0$ be arbitrary. By our hypothesis of $(1)$, there is an open $U$ such that blablabla. Now, for this given $\epsilon > 0$, let's choose $\delta > 0$ small enough so that

  • the cube $C_{p,\delta}$ lies inside $U$
  • the $\delta$ "works" for the continuity of $\dfrac{\partial F_3}{\partial y}$ and $\dfrac{\partial F_2}{\partial z}$ at the point $p$

(so really we have to take a minimum of several $\delta$'s). Then, we choose the oriented plane $M^{\delta}$ as defined above (this plane lies inside $U$ by construction, because of how small $\delta$ is). \begin{align} \left|\left(\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_2}{\partial z}(p)\right) - H_1(p)\right| &= \left|\left(\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_2}{\partial z}(p)\right) - \langle H(p), n(p)\rangle\right| \\\\ &\leq \left|\dfrac{\partial F_3}{\partial y}(p) - \dfrac{\partial F_3}{\partial y}(a_{p,\delta})\right| + \left|-\dfrac{\partial F_2}{\partial z}(p) + \dfrac{\partial F_2}{\partial z}(b_{p,\delta})\right|\\ &+ \left|\left(\dfrac{\partial F_3}{\partial y}(a_{p,\delta}) - \dfrac{\partial F_2}{\partial z}(b_{p,\delta})\right) - \dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle\, dl \right| \\ &+ \left|\dfrac{1}{|M^{\delta}|}\int_{\partial M^{\delta}}\langle F, \tau\rangle\, dl - \langle H(p), n(p) \rangle \right| \\\\ &\leq 4\epsilon \end{align} (each absolute value is $\leq \epsilon$ based on everything I've said above, and by the choice of $\delta$). Since the point $p$ and $\epsilon > 0$ are arbitrary, the inequality above shows that \begin{align} H_1 &= \dfrac{\partial F_3}{\partial y} - \dfrac{\partial F_2}{\partial z} \end{align}

As a recap of the proof idea: choose a small plane $M^{\delta}$ with outward normal pointing along $e_1$; it is the flatness of the plane (which is inherently adapted to cartesian coordinates), along with the ease of boundary parametrization which makes the resulting line integral easy to calculate. Then, simply calculate everything, and use the mean-value theorems for derivatives and integrals (this is one way to fill in the gaps for the arugments you typically see in physics texts, which say "let's keep things up to first order only" and where they use $\approx$ everywhere); finally we complete it off with a standard $\epsilon,\delta$ continuity argument.

Some remarks is that for this argument to work I've had to assume $F$ is $C^1$, so that I can apply the mean-value theorems twice, and finally finish it off with a continuity argument. I'm not sure if this proof can be strengthened so that we only have to assume $F$ is differentiable (rather than $C^1$).


Proof of Converse

Now we show the existence of such a vector field $H$; for this we'll show that $\text{curl}F$, defined by equation $(2)$ satisfies the conditions of $(1)$. Like I mentioned in the comments, I'm not sure how to do this without already appealing to Stokes theorem. With Stokes' theorem, this becomes quite simple.

Let $p\in A$, $\epsilon > 0$ and let $S\subset A$ be any "nice surface". Since $\langle\text{curl}(F), n\rangle$ is a continuous function on $S$, there is an open neighbourhood $U$ around $p$ in $S$ such that for all $q\in U$, \begin{align} \left|\langle \text{curl}F(q), n(q)\rangle - \langle \text{curl}F(p), n(p)\rangle\right| & \leq \epsilon \end{align} Now, for any "nice surface" $M\subset U$ (with unit normal being the restriction of the one already on $S$), we have by Stokes theorem: \begin{align} \left|\dfrac{1}{|M|}\int_{\partial M}\langle F,\tau\rangle \, dl - \langle \text{curl} F(p), n(p) \rangle\right| &= \dfrac{1}{|M|}\left|\int_M \langle \text{curl}F, n\rangle \, dA - \int_M \langle \text{curl} F(p), n(p) \rangle\, dA\right| \\ &\leq \dfrac{1}{|M|}\int_M \left|\langle\text{curl }F, n\rangle - \langle\text{curl }F(p), n(p)\rangle \right| \, dA \\ & \leq \dfrac{1}{|M|} \epsilon |M| \\ &= \epsilon. \end{align} This completes the proof of existence.

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  • $\begingroup$ Thank you for your answer! It's now much clearer to me what you meant in your comments. I really liked the use of the mean-value theorems to fill the gaps in the argument and make it rigorous. It got me wondering, could I use a similar construction of a cube around an arbitrary point $p \in V$ to show that $\left|\dfrac{1}{|V|} \iint_{\partial V}\langle \mathbf{F} , \mathbf{\hat{n}} \rangle\, dA - \left( \frac{\partial F_x}{\partial x}(p)+\frac{\partial F_y}{\partial y}(p)+\frac{\partial F_z}{\partial z}(p)\right)\right| < \epsilon$? Which would justify the divergence definition as well. $\endgroup$
    – Robert Lee
    Jul 18, 2020 at 4:58
  • $\begingroup$ @RobertLee yes a very similar argument holds for divergence as well. $\endgroup$
    – peek-a-boo
    Jul 18, 2020 at 5:14
  • $\begingroup$ I have a question about the $\epsilon,\delta$ argument at the end of the uniqueness proof. I understand that the first $2$ absolute values are less than $\epsilon$ since $|\alpha - p| \le \delta_1$ and $|\beta- p| \le \delta_2$ because of the MVT, and the last absolute value is less than $\epsilon$ because of the hypothesis in $(1)$, but did we not show that the third absolute value is $0$? If I understood correctly, we defined $\alpha_{p,\delta}$ and $\beta_{p,\delta}$ using the MVT's such that the partial derivatives at those points would equal the integral. Am I misunderstanding something? $\endgroup$
    – Robert Lee
    Jul 19, 2020 at 4:01
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    $\begingroup$ @RobertLee ah yes, of course that's right, the third one is $0$. What I actually meant to do was add $0$ in the form of $\left(\dfrac{\partial F_3}{\partial y}(a_{p,\delta}) - \dfrac{\partial F_2}{\partial z}(b_{p,\delta})\right) - \dfrac{1}{|M^{\delta}|}\displaystyle\int_{\partial M^{\delta}}\langle F,\tau\rangle\, dl$, and then apply the triangle inequality. But anyway, as I've written things, it's still true (because $0 \leq \epsilon$), just perhaps a little confusing... sorry about that $\endgroup$
    – peek-a-boo
    Jul 19, 2020 at 4:28
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The basic geometric object here is that of a (piecewise) smooth closed curve $c$ bounding a smooth 2D surface $S$ in ${\Bbb R}^3$. Let the curve be parametrized as $r(t)=(x(t),y(t),z(t))$, $0\leq t\leq 1$, with $r(0)=r(1)$ since closed.

Projecting $S$ to e.g. the $xy$-plane yields a 2D domain which has an area denoted $S_z$. Choosing a sign convention we may calculate the area as $$ S_z = \oint_c x \,dy = \int_0^1 x(t) \frac{dy}{dt} dt .$$ It also equals $-\oint y dx$ since $d(xy)=x dy + y dx$ is a differential of a smooth function, whence integrates to zero along $c$. The area-vector associated with $S$ (or $c$) is then the vector consisting of the areas (with fixed sign conventions) of the 3 natural projections. It is given by: $$ \vec{S}(c) = \left( \oint_c y \, dz,\ \oint_c z\, dx,\ \oint_c x\,dz \right) = \left( -\oint_c z \, dy,\ -\oint_c x\, dz,\ -\oint_c z\,dx \right).$$ We also write is as $\vec{S}(c) = \vec{n} |S|$ with $\vec{n}$ a unit vector. One may think of $\vec{n}$ as a normal to the surface $S$ (certainly true if $c$ and $S$ lies in a 2D hyperplane). Now, let $\vec{F}=(F_x,F_y,F_z)$ be a smooth vector field that we wish to integrate along $c$. Thus, we want to calculate $\oint_c \vec{F}\cdot d\vec{r}$ and relate the result to the above area-vector $\vec{S}(c)$. For this we need some kind of approximation. So assume first that $\vec{F}$ is a linear function of the form $\vec{F}=(0,0,F_z)$ with $F_z = {\it const} + ax +by + cz$. Then $$ \oint_c \vec{F}\cdot d\vec{r} = \oint_c (ax\, dz + by\, dz + cz\,dz) = a (-S_y) + b S_x + 0= -\frac{\partial F_z}{\partial x} S_y + \frac{\partial F_z}{\partial y} S_x.$$ We have obtained $ \oint_c \vec{F}\cdot d\vec{r}= (\nabla \times \vec{F}) \cdot \vec{n} |S|$, which remains valid when adding contributions also from $F_x$ and $F_y$ (writing up all details is cumbersome, however). This formula is exact for $F$ linear. As $c$ shrinks towards a point $O$ you would like to show that the non-linear contribution disappears in the limit so you have to be careful about how the shrinking takes place. It suffices e.g. if the ratio of $|S|$ to the square-length of $c$ stays uniformly bounded from below under the shrinking process. I'll leave this part aside.

The natural framework is really differential forms and stokes formula (in any dimension). I personally like the book of V.I.Arnold, Mathematical methods of Classical Mechanics, but there is a vast literature on the subject.

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