In the foundation chapters of Electrodynamics I was introduced to concept of curl of a vector field. It was defined as follows $$ \nabla \times \mathbf A = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix} $$

Well, all right this gives us the mathematical description of curl but I wished for the physical meaning, so I did some search and found that

The curl of a vector field measures the tendency for the vector field to swirl around .

(the video of Grant Sanderson also gives the almost same physical meaning to the curl)

But let's have a look at the magnetic field created by a long staright wire on the $x$ axis and the current is flowing in the direction of positive $x$ axis. We know that field will be circular and concetric to the wire,

enter image description here by the Maxwell's equations we have for the above case $$ \nabla \times \mathbf B = \mu_0 \mathbf J$$ but my problem is for point $A$ the current density is zero, hence by the equation the curl will also be zero at point A, i.e. $$ \nabla \times \mathbf B(A) = 0$$ but we can see very clearly that there is a rotation at the point $A$ and it does have a tendency to swirl around.

Now, let's have a look at the field of a dipole, enter image description here

at point $A$ we can see very well that there is a twist but the Maxwell's Laws say $$ \nabla \times \mathbf E = 0$$ for all points.

I need an explanation of how the physical definition of curl is in agreement with the two of many scenarios that I have described above.

Can we deduce something about the field if the components of curl are known? For example, if we have $$ \left ( \nabla \times \mathbf A \right)_x = C$$ $$ \left ( \nabla \times \mathbf A\right)_y = 0 $$ $$ \left ( \nabla \times \mathbf A \right)_z = 0$$ can we deduce from the physical meaning of curl that $\mathbf A$ will swirl only in $x$ direction and will be straight with respect to $y ~\textrm{and}~z$ direction? Because if the curl gives us the amount of rotation, then it seems plausible to conclude that $\mathbf A$ will have zero rotation in $y$ and $z$ direction but it' also meaningless to have a rotation in just $x$ direction. I need an explanation of how this other way round thing (means given the curl and deducing the field) is in agreement of it's physical definition?

All these doubts are arising only because we have assigned a physical meaning to curl.

UPDATE: In this link Which @AjayMohan has given, it is stated at that “it is hard to think of rotation about a single point” and “fields don’t rotate like a solid body” but the link doesn’t seem to clarify these issues. I’m finding it very hard to think of rotation by that paddle wheel example, and how only $x$ component of curl implies (other two components are zero) that wheel will rotate along the $x$ axis.

  • 4
    $\begingroup$ I think you need to rewatch the video you linked. Curl is the amount the field swirls around about a point. You evaluate it by looking at the neighborhood of a single point only. In particular, the "twist" (as you say) of field lines has nothing to do with curl. $\endgroup$
    – knzhou
    Mar 1, 2020 at 6:50
  • $\begingroup$ @knzhou I have rewatched the video, he said that if we put something like a twig at a point then the curl is the measure of the amount of spin that would be caused. $\endgroup$
    – user240696
    Mar 1, 2020 at 7:24
  • $\begingroup$ Now, I’m confused how would the twig would rotate if the arrows push it in same direction? $\endgroup$
    – user240696
    Mar 1, 2020 at 7:26
  • $\begingroup$ @Knight If the strength of the push is different at different points on the twig, then the twig can rotate. $\endgroup$
    – Ajay Mohan
    Mar 1, 2020 at 8:13
  • $\begingroup$ Related question, or possibly a duplicate: Conceptual Understanding of Zero Curl in Ampere's Law $\endgroup$ Mar 2, 2020 at 4:01

2 Answers 2

  • In the link that OP has cited, the author writes the following.

    The curl of a vector field [at a given point] measures the tendency for the vector field to swirl around [the given point].

    Swirling is different from a mere curving of the vector field. If the sentence is misinterpreted, it would seem to imply that if a vector field merely curves at some point, then it definitely has a non-zero curl at that point.
    This misinterpretation is not true: e.g. this Math.SE answer (and the examples that OP has mentioned).
    The characteristic of swirling is difficult to ascertain from a simple visual inspection of the field.

    Instead, a good intuitive interpretation is to imagine the vector field in question to be the flow velocity of a fluid and hold a sufficiently small paddle wheel at the point of interest: If it rotates, then it has a non-zero curl at that point. This is the interpretation that 3Blue1Brown gives in the video.

  • If the $x$ component of the curl of a vector field is the only non-zero component at point $P$, then, using the paddle wheel interpretation, it means that if I hold my paddle wheel at point $P$ and orient it along the $x$ axis, it will rotate. Likewise, if I instead orient my paddle wheel along the $y$ or $z$ axis, it will not rotate.

  • $\begingroup$ But the problem is what does the swirling means? You have stated in your answer that is different from curving but you have not described what it is. Also, in your MSE link, tavien has said that it is problematic to think of rotation about a point but he also doesn’t solve it. $\endgroup$
    – user240696
    Mar 1, 2020 at 8:34

As I understood, you would like to know how "force lines" of vector field $A$ are related to $\nabla \times A$. By definition "force line" is the line which consists of infinitesimal parts $\vec{dl}=[dx,dy,dz]^T$ which are colinear to components of $\vec{A}$: $$\frac{dx}{A_x(x,y,z)} = \frac{dy}{A_y(x,y,z)} = \frac{dz}{A_z(x,y,z)}.$$

There is interesting relation, indeed. Look at curl expression: $$(\nabla\times A)_z=\frac{\partial}{\partial x} A_y - \frac{\partial}{\partial y} A_x,$$ which looks very similar to shuffled variables expression for vector direction along "force line".

When you integrate it according to Gauss-Ostrogradsky-Stokes theorem $$\int \nabla\times A\, dS = \oint_{L} A\cdot dl = \oint_{L,dl\perp A} A\cdot dl + \oint_{L,dl||A} A\cdot dl,\label{roteq}$$ you can choose path of integration consisting of two parts: split it colinear to vector and perpendicular and the first term will be always zero and second term won't have any $\cos\alpha$ type of terms.

PS. Even more, when your field has meaning of force, then $\int A\,dl$ will have meaning of work. Then having $\oint A\,dl = 0$ (which equals to $\nabla\times A=0$) will have meaning that circular work equals zero allowing you to associate potential energy with your force field $A$. That's why fields which have $\nabla\times A=0$ are also called potential fields. You can introduce scalar potential $\phi$. And after that your vector lines will coincide with $\nabla\phi$.

PS2. What you call "twist" means that force lines are having their curvature. This is totally normal because force lines are parallel to each other only at infinitesimal location for specific conditions like "no dot charges, no dot currents" (far from so-called singular points in theory of differential equations).

Physical meaning of "twist" which you observe is connected to higher order of differentials.

PS3. That equation for curl \eqref{roteq} now becomes $$ \nabla\times A = \frac{1}{dS} \Big(\int_{1\rightarrow2}^{left} A\,dl + \int_{2\rightarrow1}^{right} A\,dl \Big)=\frac{1}{dS} \int_{1\rightarrow2} (A_{left}-A_{right})\,dl, $$ where you integrate difference between left and right value along right and left force lines (along a thin strip) and then divide by square area of strip.

  • $\begingroup$ Can you please explain the definition of “force line” ? I found it quite new. By the way your first paragraph really describes my problem. Thank you for understanding me. $\endgroup$
    – user240696
    Mar 1, 2020 at 10:01
  • $\begingroup$ Yep, I clarified it a little more. $\endgroup$
    – sanaris
    Mar 1, 2020 at 10:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy