next up previous
Next: Useful Vector Identities Up: Vectors and Vector Fields Previous: Laplacian Operator


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

For a small loop, we expect $ \oint {\bf A}\cdot d{\bf r}$ to be proportional to the area of the loop. Moreover, for a fixed-area loop, we expect $ \oint {\bf A}\cdot d{\bf r}$ to depend on the orientation of the loop. One particular orientation will give the maximum value: $ \oint {\bf A}\cdot d{\bf r}
= 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

$\displaystyle \vert{\bf curl}\,{\bf A}\vert = \lim_{dS\rightarrow 0}\frac{\oint {\bf A}\cdot d{\bf r}}{dS}$ (A.150)

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 a right-hand circulation rule.

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 r}$ around a small rectangle in the $ y$ -$ z$ plane, as shown in Figure A.25. The contribution from sides 1 and 3 is

$\displaystyle A_z(y+dy)\,dz - A_z(y)\,dz = \frac{\partial A_z}{\partial y} \,dy\,dz.$ (A.151)

The contribution from sides 2 and 4 is

$\displaystyle -A_y(z+dz)\,dy + A_y(z)\,dy = -\frac{\partial A_y}{\partial y}\,dy\,dz.$ (A.152)

So, the total of all contributions gives

$\displaystyle \oint {\bf A}\cdot d{\bf r} = \left(\frac{\partial A_z}{\partial y}- \frac{\partial A_y}{\partial z}\right)\,dS,$ (A.153)

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

Figure: A vector line integral around a small rectangular loop in the $ y$ -$ z$ plane.
\epsfysize =2in

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 r}$ 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

$\displaystyle \oint {\bf A} \cdot d{\bf r} = \left(\frac{\partial A_z}{\partial y}- \frac{\partial A_y}{\partial z}\right) dS_x$ (A.154)

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

$\displaystyle \oint {\bf A} \cdot d{\bf r} = \left(\frac{\partial A_x}{\partial z}- \frac{\partial A_z}{\partial x}\right) dS_y.$ (A.155)

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

$\displaystyle \oint {\bf A} \cdot d{\bf r} = \left(\frac{\partial A_y}{\partial x}- \frac{\partial A_x}{\partial y}\right) dS_z.$ (A.156)

Imagine an arbitrary loop of vector area $ d{\bf S} = (dS_x, \,dS_y, \,dS_z)$ . We can construct this out of three vector areas, $ 1$ , $ 2$ , and $ 3$ , directed in the $ x$ -, $ y$ -, and $ z$ -directions, respectively, as indicated in Figure A.26. 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,

$\displaystyle \oint {\bf A} \cdot d{\bf r} = \oint {\bf A}\cdot d{\bf r}_1 + \oint {\bf A}\cdot d{\bf r}_2+\oint {\bf A}\cdot d{\bf r}_3,$ (A.157)


$\displaystyle \oint {\bf A}\cdot d{\bf r} = {\bf curl}\,{\bf A} \cdot d{\bf S} = \vert{\bf curl}\,{\bf A}\vert\,\vert d {\bf S}\vert\,\cos\theta,$ (A.158)


$\displaystyle {\bf curl}\,{\bf A} = \left(\frac{\partial A_z}{\partial y}- \fra...
...x},\, \frac{\partial A_y}{\partial x}- \frac{\partial A_x} {\partial y}\right),$ (A.159)

and $ \theta $ is the angle subtended between the directions of $ {\bf curl}\,{\bf A}$ and $ d{\bf S}$ . Note that

$\displaystyle {\bf curl}\,{\bf A} = \nabla\times {\bf A}.$ (A.160)

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

Figure A.26: Decomposition of a vector area into its Cartesian components.
\epsfysize =2.5in

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

$\displaystyle {\bf v} =$   $\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \times {\bf r}.$ (A.161)

[See Equation (A.52).] Let us evaluate $ \nabla\times{\bf v}$ on the axis of rotation. The $ x$ -component is proportional to the integral $ \oint {\bf v}\cdot d{\bf r}$ 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 r}/ dS$ around some loop in the $ x$ -$ y$ plane. Consider a circular loop. We have $ \oint {\bf v}\cdot d{\bf r} = 2\pi \,r \,\omega \,r$ with $ dS = \pi \,r^2$ . Here, $ r$ is the perpendicular distance from the rotation axis. It follows that $ (\nabla\times {\bf v})_z = 2\,\omega$ , which is independent of $ r$ . So, on the axis, $ \nabla\times{\bf v} = (0\,,0\,,2\,\omega)$ . Off the axis, at position $ {\bf r}_0$ , we can write

$\displaystyle {\bf v} =$   $\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \times ({\bf r}-{\bf r}_0) +$   $\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \times {\bf r}_0.$ (A.162)

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

Another important result of vector field theory is the curl theorem:

$\displaystyle \oint_C {\bf A} \cdot d{\bf r} = \int_S \nabla\times{\bf A}\cdot d{\bf S},$ (A.163)

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 Equation (A.158) for each of the small loops, we can see that the contribution from all of the loops is also equal to the integral of $ \nabla\times{\bf A}
\cdot d{\bf S}$ across the whole surface. This proves the theorem.

One immediate consequence of the curl theorem is that $ \nabla\times {\bf A}$ is ``incompressible.'' Consider any two surfaces, $ S_1$ and $ S_2$ , that share the same rim. (See Figure A.23.) It is clear from the curl theorem that $ \int \nabla\times {\bf A}\cdot d{\bf S}$ is the same for both surfaces. Thus, it follows that $ \oint\nabla\times {\bf A}\cdot d{\bf S} = 0$ for any closed surface. However, we have from the divergence theorem that $ \oint\nabla\times{\bf A}\cdot d{\bf S} = \int \nabla\cdot(\nabla\times{\bf A}) \,dV
=0$ for any volume. Hence,

$\displaystyle \nabla\cdot\,(\nabla\times{\bf A}) \equiv 0.$ (A.164)

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

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

$\displaystyle \nabla\times(\nabla\phi)\equiv {\bf0}.$ (A.165)

Thus, a conservative field is also an irrotational one.

next up previous
Next: Useful Vector Identities Up: Vectors and Vector Fields Previous: Laplacian Operator
Richard Fitzpatrick 2016-03-31