The curl of a 3D vector field is a 3D vector itself and has two definitions - one in integral form and one in differential form.

  • Definition 1: $$ \operatorname{curl}\vec{F}(x,y,z) \, \cdot \, \hat{n} =\lim_{A_\hat{n} \to 0} \frac{1}{A_\hat{n}} \oint{\vec{F} \, \cdot \, \vec{ds}} $$

    Where $\hat{n}$ is an arbitrary unit normal vector (you would substitute $\hat{\imath}, \hat{\jmath}$ and $\hat{k}$ in to find the three components of the curl vector. $A_\hat{n}$ is the magnitude of the area enclosed by the loop in the plane perpendicular to $\hat{n}$.

  • Definition 2: $$ \operatorname{curl} \vec{F}(x,y,z) = \vec{\nabla} \, \times \, \vec{F} = \left({\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}}\right) \cdot \hat{\imath} \, - \, \left({\frac{\partial F_z}{\partial x} - \frac{\partial F_x}{\partial z}}\right) \cdot \hat{\jmath} \, + \left({\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}}\right) \cdot \hat{k} \, $$

I would like to know how you would show that these two equations are equal with respect to each component of the curl.

Thank you.

  • 2
    $\begingroup$ ... and similarly to your other recent thread, questions about mathematics belong on Mathematics, not here. $\endgroup$ – E.P. Oct 17 '19 at 13:06
  • $\begingroup$ Okay, but if I could reason it using the divergence answer I wouldn't have asked this question I'm afraid. Also, I would argue this absolutely is physics (mathematical physics) as it is intrinsically linked to classical electromagnetism. $\endgroup$ – Jamie Smith Oct 18 '19 at 8:02

Let's just consider the $\hat k$ component of curl.

For the first definition, take a plane through $(x,y,z)$ parallel to the $x,y$ plane. Let the loop consist of the four directed segments $(x+h,y-h)$ to $(x+h,y+h),$ $(x+h,y+h)$ to $(x-h,y+h),$ $(x-h,y+h)$ to $(x-h,y-h),$ and $(x-h,y-h)$ to $(x+h,y-h).$

Assuming that $\vec F$ is differentiable at $(x,y,z),$ the function $F_x$ is approximated by $$ F_x(x+\Delta x, y+\Delta y, z) = F_x(x,y,z) + (\Delta x)\frac{\partial F_x}{\partial x} + (\Delta y)\frac{\partial F_x}{\partial y} \tag1$$ and the function $F_y$ is approximated by $$ F_y(x+\Delta x, y+\Delta y, z) = F_y(x,y,z) + (\Delta x)\frac{\partial F_y}{\partial x} + (\Delta y)\frac{\partial F_y}{\partial y} \tag2$$ in a neighborhood of $(x,y,z).$ By making $h$ as small as needed we can make that approximation as good as we want in a neighborhood that includes the loop.

On the segment from $(x+h,y-h,z)$ to $(x+h,y+h,z),$ $\vec{ds}=\hat\jmath,$ so the $x$ and $z$ components of $\vec F\cdot\vec{ds}$ are zero and $\vec F\cdot\vec{ds} = F_y.$ On that segment the value of $\Delta x$ in Equation $(2)$ is $h,$ so the value of $(\Delta x)\frac{\partial F_y}{\partial x}$ on the segment is uniformly $h\frac{\partial F_y}{\partial x}.$ Meanwhile, the value of $\Delta y$ ranges from $-h$ to $h$ with mean value $0.$ Hence the mean value of $(\Delta y)\frac{\partial F_y}{\partial y}$ also is zero. That is, its contribution to the integral of $\vec{F} \cdot \vec{ds} = F_y$ along that segment is zero. So the mean value of $F_y$ along the segment is $F_y(x,y,z) + h\frac{\partial F_y}{\partial x},$ meaning we get the integral of $F_y$ on that segment by multiplying this quantity by the length of the segment, $2h.$

On the opposite segment (from $(x-h,y+h,z)$ to $(x-h,y-h,z)$), again the $x$ and $z$ components of $\vec F\cdot\vec{ds}$ are zero, but now $\vec F\cdot\vec{ds} = -F_y$ (since the direction of the segment is opposite from $\hat\jmath$) and the mean value of $F_y$ is $F_y(x,y,z) - h\frac{\partial F_y}{\partial x},$ so the integral is $-2h$ times this last quantity.

Similarly, along the other two segments, two of the components of $\vec F$ contribute nothing to the product $\vec F\cdot\vec{ds},$ but this time it is the $y$ and $z$ components that drop out and we need look only at $F_x.$ The mean values of $F_x$ along the segments $(x+h,y+h,z)$ to $(x-h,y+h,z)$ and $(x-h,y-h,z)$ to $(x+h,y-h,z)$ are $F_x(x,y,z) + h\frac{\partial F_x}{\partial y}$ and $F_x(x,y,z) - h\frac{\partial F_x}{\partial y}$, respectively, and the integrands are equal to $-F_x$ and $F_x$ respectively.

Integrating along the four segments of the loop in sequence, \begin{align} \oint\vec{F} \cdot \vec{ds} &\approx 2h\left(F_y(x,y,z) + h \frac{\partial F_y}{\partial x}\right) - 2h\left(F_x(x,y,z) + h \frac{\partial F_x}{\partial y}\right)\\ &\qquad - 2h\left(F_y(x,y,z) - h \frac{\partial F_y}{\partial x}\right) + 2h\left(F_x(x,y,z) - h \frac{\partial F_x}{\partial y}\right) \\ &= 4h^2 \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \end{align}

The area inside the loop is $4h^2,$ so $$ \frac{1}{A_\hat{k}} \oint\vec{F} \cdot \vec{ds} \approx \frac{1}{4h^2} \left( 4h^2 \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\right) = \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}. $$

In the limit, this approximation becomes exact. The other two components work similarly.

That's the general intuition, anyway. You might be able to get a rigorous proof out of this by keeping track of the error terms in each approximation.

I'm not going to try to rewrite this just now, but it might help to observe that $\vec F$ is approximated by a constant ($\vec F(x,y,z)$) plus a linear term based on the components of the gradient of $\vec F$, and that the integral of the constant term around the loop is zero so we can ignore it and look only at the integral of the linear term. That way one might be able to completely avoid writing $F_x(x,y,z)$ and $F_y(x,y,z)$.

  • $\begingroup$ Thanks for your answer! I'm confused by your statement that "the mean value of $F_y$ along the segment is $h\frac{\partial F_y}{\partial x}$" for very small h. Could you explain why that is please? $\endgroup$ – Jamie Smith Oct 22 '19 at 8:38
  • $\begingroup$ I made a mistake: I left out a term of the approximation. That term cancels out in the integral, so the answer came out right in the end (and I knew it would cancel, which is probably why I "canceled" it before writing it), but this made nonsense of the middle parts of the answer. I think it's fixed now. $\endgroup$ – David K Oct 22 '19 at 12:15
  • $\begingroup$ Okay, that makes more sense now. Just a final question - where does equation one explicitly come from? I'm struggling to remember how to show that it's true. Is it simply an extension of the partial sum of a function F? I.e. $dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy + \frac{\partial F}{\partial z} dz$ and therefore F(x+dx, y+dy, z+dz) = F(x,y,z) + dF? $\endgroup$ – Jamie Smith Oct 23 '19 at 9:36
  • $\begingroup$ Also, why is it just $F_y$ that you are looking at along the lines?? It's the line integral of the full 3D vector field $\vec{F}$ no? $\endgroup$ – Jamie Smith Oct 23 '19 at 9:47
  • $\begingroup$ This would probably have gone better if I had done it on paper first rather than direct-to-answer-window. It's harder to make cut-and-paste errors with a pencil. But note that in all versions of the answer, I also mention $F_x$ (in $\frac{\partial F_x}{\partial y}$). But when I inserted the constant term on those segments I put the wrong one (which again was easy to overlook because I knew they would cancel out). Hopefully no more errors. $\endgroup$ – David K Oct 23 '19 at 12:19

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