Curl

Consider a vector field ${\bf A}$, and a loop which lies in one plane. The integral of ${\bf A}$ around this loop is written $\oint {\bf A}\cdot d{\bf l}$, where $d{\bf l}$ is a line element of the loop. If ${\bf A}$ is a conservative field then ${\bf A} = {\bf grad} \phi$ and $\oint {\bf A}\cdot d{\bf l}=0$ for all loops. In general, for a non-conservative field, $\oint {\bf A}\cdot d{\bf l} \neq 0$.

For a small loop we expect $\oint {\bf A}\cdot d{\bf l}$ to be proportional to the area of the loop. Moreover, for a fixed area loop we expect $\oint {\bf A}\cdot d{\bf l}$ to depend on the orientation of the loop. One particular orientation will give the maximum value: $\oint {\bf A}\cdot d{\bf l} = I_{\rm max}$. If the loop subtends an angle $\theta$ with this optimum orientation then we expect $I= I_{\rm max}\cos\theta$. Let us introduce the vector field ${\bf curl} {\bf A}$ whose magnitude is

$$\vert{\bf curl} {\bf A}\vert = \lim_{dS\rightarrow 0}\frac{\oint {\bf A}\cdot d{\bf l}}{dS}$$ (143)

for the orientation giving $I_{\rm max}$. Here, $dS$ is the area of the loop. The direction of ${\bf curl} {\bf A}$ is perpendicular to the plane of the loop, when it is in the orientation giving $I_{\rm max}$, with the sense given by the right-hand grip rule.

Figure 22:

Let us now express ${\bf curl} {\bf A}$ in terms of the components of ${\bf A}$. First, we shall evaluate $\oint {\bf A}\cdot d{\bf l}$ around a small rectangle in the $y$-$z$ plane (see Fig. 22). The contribution from sides 1 and 3 is

$$A_z(y+dy) dz - A_z(y) dz = \frac{\partial A_z}{\partial y} dy dz.$$ (144)

The contribution from sides 2 and 4 is

$$-A_y(z+dz) dy + A_y(z) dy = -\frac{\partial A_y}{\partial y} dy dz.$$ (145)

So, the total of all contributions gives

$$\oint {\bf A}\cdot d{\bf l} = \left(\frac{\partial A_z}{\partial y}- \frac{\partial A_y}{\partial z}\right) dS,$$ (146)

where $dS=dy dz$ is the area of the loop.

Consider a non-rectangular (but still small) loop in the $y$-$z$ plane. We can divide it into rectangular elements, and form $\oint {\bf A}\cdot d{\bf l}$ over all the resultant loops. The interior contributions cancel, so we are just left with the contribution from the outer loop. Also, the area of the outer loop is the sum of all the areas of the inner loops. We conclude that

$$\oint {\bf A} \cdot d{\bf l} = \left(\frac{\partial A_z}{\partial y}- \frac{\partial A_y}{\partial z}\right) dS_x$$ (147)

is valid for a small loop $d{\bf S} = (dS_x, 0, 0)$ of any shape in the $y$-$z$ plane. Likewise, we can show that if the loop is in the $x$-$z$ plane then $d{\bf S} = (0, dS_y, 0)$ and

$$\oint {\bf A} \cdot d{\bf l} = \left(\frac{\partial A_x}{\partial z}- \frac{\partial A_z}{\partial x}\right) dS_y.$$ (148)

Finally, if the loop is in the $x$-$y$ plane then $d{\bf S} = (0, 0, dS_z)$ and

$$\oint {\bf A} \cdot d{\bf l} = \left(\frac{\partial A_y}{\partial x}- \frac{\partial A_x}{\partial y}\right) dS_z.$$ (149)

Figure 23:

Imagine an arbitrary loop of vector area $d{\bf S} = (dS_x, dS_y, dS_z)$. We can construct this out of three loops in the $x$-, $y$-, and $z$-directions, as indicated in Fig. 23. If we form the line integral around all three loops then the interior contributions cancel, and we are left with the line integral around the original loop. Thus,

$$\oint {\bf A} \cdot d{\bf l} = \oint {\bf A}\cdot d{\bf l}_1 + \oint {\bf A}\cdot d{\bf l}_2+\oint {\bf A}\cdot d{\bf l}_3,$$ (150)

giving

$$\oint {\bf A}\cdot d{\bf l} = {\bf curl} {\bf A} \cdot d{\b... ... = \vert{\bf curl} {\bf A}\vert \vert d S\vert \cos\theta,$$ (151)

where

$${\bf curl} {\bf A} = \left(\frac{\partial A_z}{\partial y}-... ...ial A_y}{\partial x}- \frac{\partial A_x} {\partial y}\right).$$ (152)

Note that

$${\bf curl} {\bf A} = \nabla\times {\bf A}.$$ (153)

This demonstrates that ${\bf curl} {\bf A}$ is a good vector field, since it is the cross product of the $\nabla$ operator (a good vector operator) and the vector field ${\bf A}$.

Consider a solid body rotating about the $z$-axis. The angular velocity is given by $\mbox{\boldmath$\omega$} = (0, 0, \omega)$, so the rotation velocity at position ${\bf r}$ is

$${\bf v} = \mbox{\boldmath$\omega$}\times {\bf r}$$ (154)

[see Eq. (43)]. Let us evaluate ${\bf curl} {\bf v}$ on the axis of rotation. The $x$-component is proportional to the integral $\oint {\bf v}\cdot d{\bf l}$ around a loop in the $y$-$z$ plane. This is plainly zero. Likewise, the $y$-component is also zero. The $z$-component is $\oint {\bf v}\cdot d{\bf l}/ dS$ around some loop in the $x$-$y$ plane. Consider a circular loop. We have $\oint {\bf v}\cdot d{\bf l} = 2\pi r \omega r$ with $dS = \pi r^2$. Here, $r$ is the radial distance from the rotation axis. It follows that $({\bf curl} {\bf v})_z = 2 \omega$, which is independent of $r$. So, on the axis, ${\bf curl} {\bf v} = (0 ,0 ,2 \omega)$. Off the axis, at position ${\bf r}_0$, we can write

$${\bf v} = \mbox{\boldmath$\omega$}\times ({\bf r}-{\bf r}_0) + \mbox{\boldmath$\omega$}\times {\bf r}_0.$$ (155)

The first part has the same curl as the velocity field on the axis, and the second part has zero curl, since it is constant. Thus, ${\bf curl} {\bf v} = (0 ,0 ,2 \omega)$ everywhere in the body. This allows us to form a physical picture of ${\bf curl} {\bf A}$. If we imagine ${\bf A}$ as the velocity field of some fluid, then ${\bf curl} {\bf A}$ at any given point is equal to twice the local angular rotation velocity: i.e., 2$\omega$. Hence, a vector field with ${\bf curl} {\bf A} ={\bf0}$ everywhere is said to be irrotational.

Another important result of vector field theory is the curl theorem or Stokes' theorem,

$$\oint_C {\bf A} \cdot d{\bf l} = \int_S {\bf curl A}\cdot d{\bf S},$$ (156)

for some (non-planar) surface $S$ bounded by a rim $C$. This theorem can easily be proved by splitting the loop up into many small rectangular loops, and forming the integral around all of the resultant loops. All of the contributions from the interior loops cancel, leaving just the contribution from the outer rim. Making use of Eq. (151) for each of the small loops, we can see that the contribution from all of the loops is also equal to the integral of ${\bf curl} {\bf A} \cdot d{\bf S}$ across the whole surface. This proves the theorem.

One immediate consequence of of Stokes' theorem is that ${\bf curl} {\bf A}$ is ``incompressible.'' Consider two surfaces, $S_1$ and $S_2$, which share the same rim. It is clear from Stokes' theorem that $\int {\bf curl} {\bf A}\cdot d{\bf S}$ is the same for both surfaces. Thus, it follows that $\oint{\bf curl} {\bf A}\cdot d{\bf S} = 0$ for any closed surface. However, we have from the divergence theorem that $\oint{\bf curl} {\bf A}\cdot d{\bf S} = \int {\mit div} ({\bf curl} {\bf A}) dV =0$ for any volume. Hence,

$${\mit div} ({\bf curl} {\bf A}) \equiv 0.$$ (157)

So, ${\bf curl A}$ is a solenoidal field.

We have seen that for a conservative field $\oint {\bf A}\cdot d{\bf l}=0$ for any loop. This is entirely equivalent to ${\bf A} = {\bf grad} \phi$. However, the magnitude of ${\bf curl} {\bf A}$ is $\lim_{ dS\rightarrow 0}\oint{\bf A}\cdot d{\bf l} /dS$ for some particular loop. It is clear then that ${\bf curl} {\bf A} ={\bf0}$ for a conservative field. In other words,

$${\bf curl} ({\bf grad} \phi)\equiv {\bf0}.$$ (158)

Thus, a conservative field is also an irrotational one.

Finally, it can be shown that

$${\bf curl} ({\bf curl} {\bf A} ) = {\bf grad} ({\mit div} {\bf A}) - \nabla^2 {\bf A},$$ (159)

where

$$\nabla^2{\bf A} = (\nabla^2 A_x, \nabla^2 A_y, \nabla^2 A_z).$$ (160)

It should be emphasized, however, that the above result is only valid in Cartesian coordinates.