Prerequisites

Curl

Figure 1: Computing the horizontal contribution to the circulation around a small rectangular loop.

Consider a small rectangular loop in the $yz$-plane, with sides parallel to the coordinate axes, as shown in Figure 1. What is the circulation of $\AA$ around this loop?

Consider first the horizontal edges, on each of which $d\rr=dy\,\yhat$. However, when computing the circulation of $\AA$ around this loop, we traverse these two edges in opposite directions. In particular, when traversing the loop in the counterclockwise direction, $dy<0$ on top and $dy>0$ on the bottom. We would like to compare these two edges, but we have \begin{equation} dy_{\hbox{top}} = -dy_{\hbox{bot}} \end{equation} so we conventionally set $dy=dy_{\hbox{bot}}$, leading to $dy_{\hbox{top}}=-dy$. 1) Thus, \begin{eqnarray} \label{topbot} \sum_{\rm top+bottom} \AA \cdot d\rr &=& - \AA(z+dz)\cdot\yhat\> dy + \AA(z)\cdot\yhat\> dy \\ &=& - \Bigl( A_y(z+dz) - A_y(z) \Bigr) \> dy \nonumber\\ &=& - \frac{A_y(z+dz) - A_y(z)}{dz} \> dy\,dz \nonumber\\ &=& - \Partial{A_y}{z} \>\> dy\,dz \nonumber \end{eqnarray} where we have multiplied and divided by $dz$ to obtain the surface element $dA=dy\,dz$ in the third step, and used the definition of the derivative in the final step.

Just as with the divergence, in making this argument we are assuming that $\AA$ doesn't change much in the $x$ and $y$ directions, while nonetheless caring about the change in the $z$ direction. We are again subtracting the values at two points separated in the $z$ direction, so we are canceling the zeroth order term in $z$, and therefore need the next order term. This can be made precise using a multivariable Taylor series expansion.

Repeating this argument for the remaining two sides leads to \begin{eqnarray} \sum_{\rm sides}\AA \cdot d\rr &=& \AA(y+dy)\cdot\zhat\> dz - \AA(y)\cdot\zhat\> dz \\ &=& \Bigl( A_z(y+dy) - A_z(y) \Bigr) \> dz \nonumber\\ &=& \frac{A_z(y+dy) - A_z(y)}{dy} \> dy\,dz \nonumber\\ &=& \Partial{A_z}{y} \>\> dy\,dz \nonumber \end{eqnarray} where care must be taken with the signs, which are different from those in ($\ref{topbot}$). Adding up both expressions, we obtain \begin{equation} \hbox{total $yz$-circulation} = \left( \Partial{A_z}{y} - \Partial{A_y}{z} \right) dx\,dy \end{equation} Since this is proportional to the area of the loop, it approaches zero as the loop shrinks down to a point. The interesting quantity is therefore the ratio of the circulation to area. This ratio is almost the curl, but not quite.

We could have oriented our loop any way we liked. We have a circulation for each orientation, which we can associate with the normal vector to the loop; curl is a vector quantity. Our loop has counterclockwise orientation of the loop as seen from the positive $x$-axis; we are computing the $\xhat$-component of the curl.

Putting this all together, we define the $\xhat$-component of the curl of a vector field $\AA$ to be \begin{equation} \hbox{curl}(\AA)\cdot\xhat = \frac{\hbox{$yz$-circulation}}{\hbox{unit area}} = \Partial{A_z}{y} - \Partial{A_y}{z} \end{equation} The rectangular expression for the full curl now follows by cyclic symmetry, yielding \begin{equation} \hbox{curl}(\AA) = \left( \Partial{A_z}{y} - \Partial{A_y}{z} \right) \xhat + \left( \Partial{A_x}{z} - \Partial{A_z}{x} \right) \yhat + \left( \Partial{A_y}{x} - \Partial{A_x}{y} \right) \zhat \end{equation} which is more easily remembered in the form \begin{equation} \hbox{curl}(\AA) = \grad\times\AA = \left| \matrix{\xhat& \yhat& \zhat\cr \noalign{\smallskip} \Partial{}{x}& \Partial{}{y}& \Partial{}{z}\cr \noalign{\smallskip} A_x& A_y& A_z\cr} \right| \end{equation} An analogous construction can be used in curvilinear coordinates; the results for spherical and cylindrical coordinates can be found in § {Formulas for Div, Grad, Curl}. as well as on the inside front of Griffiths' textbook, Introduction to Electrodynamics.

1) A similar argument using a finite box would require integrating around the loop, in which case the limits of integration would be equal and opposite for these two edges, leading to an equivalent minus sign when adding the integrals.

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