The operation mathematically means $$(\nabla \times \vec A)\cdot\hat n = \lim_{\Delta S\to\ 0} \frac{\oint\vec A\cdot\ d\vec l }{\left | \Delta S \right |}$$ and the proof of this is quite logical.
In textbooks, I have found that they relate curl to rotation.
But how does the above mathematical treatment show this relation?

Any mathematical help is welcome.

  • $\begingroup$ @user36790 I have posted the question on MathSE here. $\endgroup$ Oct 10, 2015 at 18:52
  • $\begingroup$ The curl should be a vector, not a scalar. $\endgroup$
    – littleO
    Oct 10, 2015 at 19:11
  • $\begingroup$ @littleO Can you please edit the equation suitably? I found this equation in a textbook of mine. $\endgroup$ Oct 10, 2015 at 19:25
  • $\begingroup$ You could state it as $(\nabla \times \vec A)\cdot n = \lim_{\Delta S \to 0} \frac{\oint \vec A \cdot d \vec \ell}{| \Delta S |}$, where $\Delta S$ is a planar oriented surface and with unit normal vector $n$, and $| \Delta S|$ is the area of $\Delta S$. Here $\oint \vec A \cdot d \vec \ell$ is the line integral of $\vec A$ around the boundary of $\Delta S$. Note that to avoid difficulties in defining this limit, a rigorous approach usually defines $\nabla \times A$ using the standard formula for the curl. $\endgroup$
    – littleO
    Oct 10, 2015 at 19:39
  • $\begingroup$ It might be helpful to imagine what happens if $\vec A $ describes a fluid that's flowing in circles, in planes orthogonal to $\vec n $. This kind of flow makes the quantity on the right very large. $\endgroup$
    – littleO
    Oct 10, 2015 at 20:52

2 Answers 2


Let $\bf{C}$ be any vector-field. Let there be a closed-curve $\Gamma$. Take the line-integral of $\bf C$ around this complete loop that is, take the tangential-component of the field at any point on the curve & take dot-product of $\mathbf{C}_\text{tangential}\cdot d\mathbf{s}$ & integrate over the whole loop. This is circulation : $$\text{Circulation}_{\Gamma}=\int_{\Gamma} \mathbf{ C} \cdot d\mathbf{s}.$$


Now make partitions on $C$ with an intermediary-bridge, say $B$; this then will make two loops viz; $\Gamma_1 \;\&\; \Gamma_2.$ Now circulation over the whole loop is just line-integral around each sub-loop that is, $$\text{Circulation}_{\Gamma} \\= \text{Circulation}_{\Gamma_1} + \text{Circulation}_{\Gamma_2}\\= (\text{Circulation}_{\Gamma_a} + \text{Circulation}_{\Gamma_{ab}}) + (\text{Circulation}_{\Gamma_b} -\text{Circulation}_{\Gamma_{ab}})\\ =\text{Circulation}_{\Gamma_{a}} + \text{Circulation}_{\Gamma_{b}}.$$

division of loop

Break into still further smaller loops as $\Gamma_1,\Gamma_2,\Gamma_3 \ldots \Gamma_N.$

further division of the loop

Circulation over the whole-loop $$\int_{\Gamma} \mathbf{C}\cdot d\mathbf{s}= \sum_{i=1}^N\int_{\Gamma_i} \mathbf{C}\cdot d\mathbf{s}_i.$$

Now we want to get the infinitesimal-circulation characteristic to a certain coordinate. We then make $N\to \infty;$ however each integral that is $\int_{\Gamma_i} \mathbf{C}\cdot d\mathbf{s}_i \to 0$ as $N\to \infty.$ So, in order to get the finite characteristic which is associated with circulation locally to a point, we divide each integral by the area enclosed by each sub-loop. That is, take the ratio of circulation to loop area$$\frac{\text{Circulation}_{\Gamma_i}}{a_i}.$$ This is our local-property that is the infinitesimal circulation of the vector-field at a certain point (around it) is given by $$\lim_{a_i\to 0} \frac{\int_{\Gamma_i} \mathbf{C}\cdot d\mathbf{s}_i}{a_i}.$$

  • $\begingroup$ Your answer has nothing to do with rotations, as was asked by the OP. $\endgroup$
    – Alex M.
    Oct 11, 2015 at 16:07
  • $\begingroup$ @Alex M.: He wanted to know how circulation is related to the formula; I deduced it using argument that relates how circulation is related to the formula; lf you don't like then there is an answer-box, I think.... $\endgroup$
    – user142971
    Oct 11, 2015 at 16:10
  • $\begingroup$ Dear Aniket, it is my duty to ask whether my answer helped you or not. Just comment on what are your problems that still persist so that I may help. $\endgroup$
    – user142971
    Oct 11, 2015 at 16:21
  • $\begingroup$ @user36790 Your answer is quite good and self-explanatory. I have understood it. Only one thing: In the fist paragraph, you first take dot product of $F_{tangential}⋅ds$ but then you write $\tau = F⋅ds$. So is it $F_{tangential}$ or just $F$? $\endgroup$ Oct 11, 2015 at 16:37
  • 1
    $\begingroup$ @user36790 yup now the pics are crystal clear. $\endgroup$ Oct 16, 2015 at 5:35

First a small disclaimer as the rotation may not be rotation in the same sense as the question-maker intends, but anyways. Please tell me if I misunderstand the question and I can remove it.

A rotation $\phi$ radians can be described as $\left[\begin{array}{rr} a&-b\\b&a\end{array}\right], {a = \cos(\phi), b = \sin(\phi)}$, and an infinitesimal rotation in vector analysis is $$\nabla \times {\bf v} = \frac{\partial {\bf v}_x}{\partial y} - \frac{\partial {\bf v}_y}{\partial x}$$ So how do these relate to each other, or more interestingly how does $\phi$ relate to the cross product above.

An interesting thing is that the matrix operator above is in fact how the vector changes, not how it moves i.e. the additive part. To get the additive part, we need to take I minus the rotated vector, i.e. the change vector.

We need some discrete filters to measure the rotation, and here are some basic traditional Sobel filters. These are linear filters which are applied on a discretization of a function, usually on a regular cartesian grid. In mathematical terms we perform a discrete convolution with the filters

$$D_x = \left[\begin{array}{rr} 1&0&-1\\2&0&-2\\1&0&-1\end{array}\right] \hspace{1cm} D_y = \left[\begin{array}{rrr} 1&2&1\\0&0&0\\-1&-2&-1\end{array}\right]$$

This convolution is defined as $$F(i,j) = \sum_{k,l \in \{0,1,2,3\}} I(i+k,j+l)D_x(3-k,3-l)$$ and produces the filter response $F$. You can read more about linear filtering (discrete convolution) and Sobel filters in particular.

enter image description here Local displacement vector field. Each vector points out from the middle with the difference in coordinates.

enter image description here

After applying the rotation described above for $\phi = \frac{\pi}{2}$. Each vector contains it's own displacement compared to before rotation.

enter image description here

Applying the curl filters according to curl formula and fitting to a $sin$ curve shows that we can do curl on a proper rotation field and estimate phi., the scale 16 (sin maximum) can be adjusted by normalizing the discrete filters accordingly.

  • $\begingroup$ Please complete the solution as soon as possible. $\endgroup$ Oct 11, 2015 at 13:15
  • $\begingroup$ Oh yes sorry. I forgot about it, working on some illustrations now. $\endgroup$ Oct 11, 2015 at 13:34
  • $\begingroup$ Fixed some embarrasing bugs in my code. This should make more sense. Largest "rotation" for $\pm$ $\pi / 4$ and there is a sign change for half a revolution, as we know should be the case for curl. Negative or positive $\hat {\bf z}$ deciding counter clockwise or clockwise rotation. $\endgroup$ Oct 11, 2015 at 15:44
  • $\begingroup$ Any explanation for the downvote? $\endgroup$ Oct 11, 2015 at 16:22
  • $\begingroup$ Me in a dilemma too. Someone has downvoted my question and both the answers. I don't why. $\endgroup$ Oct 11, 2015 at 16:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .