In Cartesian coordinates, consider a vector field $\boldsymbol{f}=\boldsymbol{f}(x,y)$ defined over some domain $\Omega$ in the $xy$ plane. Taking a closed contour $\Gamma$ that encloses surface area $A$ at any point inside the same domain ($\Omega$), then we can define the two integrals

$$I_{1}=\int_{\Gamma}\boldsymbol{f}\cdot\boldsymbol{\hat{n}}dl,$$ $$I_{2}=\int_{\Gamma}\boldsymbol{f}\cdot\boldsymbol{dl},$$

where $dl$ is the length element along the contour, $\hat{n}$ is the unit vector normal (perpendicular) to the element $dl$ at each point in the plane, and bold symbols represent vector quantities.

To make $I_{2}=0$, we can choose the contour $\Gamma$ shape to be orthogonal to the field flowlines at each point, whereas to make $I_{1}=0$, we choose them to be parallel at each point. For example, if the field is generated radially from a point source in the plane, we can choose the contour shapes to be concentric circles around the source point, so that these circles will always cross the radial field lines perpendicularly and give $I_{2}=0$. Or if the field is circular flow, we can choose the contour to also be circular, and thus get $I_{1}=0$. These examples should work regardless of how large/small the enclosed area $A$ is taken.

The question:

Is there any nontrivial example of a family of fields and contour shapes that we can choose that will give us $I_{1}=0$ or $I_{2}=0$ at every point (not necessarily concentric) and for any size of enclosed area $A$ (including when $A\rightarrow0 $) in the domain of the problem? Can you given any practical examples of scenarios that could lead to this?

And if this is impossible to satisfy at every point, is there a theorem/proof about this?


It seems like you don't know Green's Theorem (or, more generally, Stokes's Theorem and Gauss's Theorem, also called the Divergence Theorem). Assuming your vector field is continuously differentiable, you certainly can't even accomplish your goal for one point unless you have strong conditions on $\text{div}\, \boldsymbol f$ and $\text{curl}\,\boldsymbol f$. To get $I_1=0$ in a non-instantaneous fashion, you'll need $\text{div}\, \boldsymbol f=0$ on a region containing your point, and, likewise, to get $I_2=0$ you'll need $\text{curl}\,\boldsymbol f = \mathbf 0$ on a region. The latter happens, in particular, for conservative force fields, i.e., when $\boldsymbol f = \nabla\phi$ for some function $\phi$. [Geometrically, you can't get curves as you described unless the field flowlines form closed curves; usually that won't happen. Nor, in general, will the orthogonal trajectories of the field lines form closed curves. The examples you have in your mind are special cases indeed.]

Side remark: You can get vanishing of both $I_1$ and $I_2$ when $\boldsymbol f = \nabla\phi$ and $\phi$ is harmonic on your region in the plane. Then both the curl and divergence will vanish.

  • $\begingroup$ (1/2) Thanks for your answer. Yes, I know Stokes' and Divergence theorems, and I understand that $I_{1}$ and $I_{2}$ can be described in terms of curl and div. But what I was interested in knowing was whether it can be possible to make the integrals vanish by some appropriate choise of contour shapes, rather than by making the integrand itself vanish (as in curl $f$=0 or div $f$=0). Imagine a 2D domain (like a planar cavity) where the wave equation is to be solved subject to the boundary conditions at the domain edges, and it gave eigenfunctions as solutions. Now let $f$ be some linear... $\endgroup$ – user135626 Jun 24 '18 at 0:32
  • $\begingroup$ (2/2)...combination of these solutions, taken as our integrand. Then can one find families of contours over which the integrals will vanish at every point (even when the contours become very small and basically defining points)? Because then the integrand will also need to vanish at every point in the domain and this may have interesting consequences... $\endgroup$ – user135626 Jun 24 '18 at 0:32
  • $\begingroup$ OK, well, your idea was to make the line integral integrand vanish at every point. As I pointed out, that surely can't happen with closed curves, in general. So both of us need to pick shapes to make the integral of div/curl vanish, in general (if there is any symmetry or anti-symmetry at all). $\endgroup$ – Ted Shifrin Jun 24 '18 at 0:34
  • $\begingroup$ So, if I've from physical facts the following relation held true:$\int_{\sigma\Omega}f\cdot \hat{n} dl=0$ (ie. at the domain boundary, assuming simple domain, like a square or disk) And I wish to do a similar integral on interior regions of arbitrary shape,to see if it can vanish there too. How would you suggest I do it? My initial idea was to try to find contours similar in shape to the outter boundary of the domain, then start shrinking them slowly and see if the integrals will still vanish. Then I became interested in knowing if there exist fields or contours that will make it 0 everywhere. $\endgroup$ – user135626 Jun 24 '18 at 0:44
  • $\begingroup$ Let's talk about a specific $\boldsymbol f$. $\endgroup$ – Ted Shifrin Jun 24 '18 at 0:45

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.