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Next: Grad Operator Up: Vector Algebra and Vector Previous: Volume Integrals


A one-dimensional function $f(x)$ has a gradient $df/dx$ which is defined as the slope of the tangent to the curve at $x$. We wish to extend this idea to cover scalar fields in two and three dimensions.

Consider a two-dimensional scalar field $h(x,\,y)$, which is (say) height above sea-level in a hilly region. Let $d{\bf r}\equiv(dx,\,dy)$ be an element of horizontal distance. Consider $dh/dr$, where $dh$ is the change in height after moving an infinitesimal distance $d{\bf r}$. This quantity is somewhat like the one-dimensional gradient, except that $dh$ depends on the direction of $d{\bf r}$, as well as its magnitude. In the immediate vicinity of some point $P$, the slope reduces to an inclined plane--see Figure A.111. The largest value of $dh/dr$ is straight up the slope. It is easily shown that for any other direction

\frac{dh}{dr}= \left(\frac{dh}{dr}\right)_{\rm max}\, \cos\theta,
\end{displaymath} (1346)

where $\theta $ is the angle shown in Figure A.111. Let us define a two-dimensional vector, ${\bf grad}\,h$, called the gradient of $h$, whose magnitude is $(dh/dr)_{\rm max}$, and whose direction is the direction of steepest ascent. The $\cos\theta$ variation exhibited in the above expression ensures that the component of ${\bf grad}\,h$ in any direction is equal to $dh/dr$ for that direction.

Figure A.111: A two-dimensional gradient.
\epsfysize =2.25in

The component of $dh/dr$ in the $x$-direction can be obtained by plotting out the profile of $h$ at constant $y$, and then finding the slope of the tangent to the curve at given $x$. This quantity is known as the partial derivative of $h$ with respect to $x$ at constant $y$, and is denoted $(\partial h/\partial x)_y$. Likewise, the gradient of the profile at constant $x$ is written $(\partial h/\partial y)_x$. Note that the subscripts denoting constant-$x$ and constant-$y$ are usually omitted, unless there is any ambiguity. If follows that in component form

{\bf grad}\,h \equiv \left(\frac{\partial h}{\partial x},\, \frac{\partial h}{\partial y}
\end{displaymath} (1347)

Now, the equation of the tangent plane at $P=(x_0,\, y_0)$ is

h_T(x,\,y)= h(x_0,\,y_0) + \alpha\,(x-x_0)+\beta\,(y-y_0).
\end{displaymath} (1348)

This has the same local gradients as $h(x,\,y)$, so
\alpha = \frac{\partial h}{\partial x},~~~~~\beta= \frac{\partial h}{\partial y},
\end{displaymath} (1349)

by differentiation of the above. For small $dx=x-x_0$ and $dy=y-y_0$, the function $h$ is coincident with the tangent plane, so
dh = \frac{\partial h}{\partial x}\, dx +\frac{\partial h}
{\partial y}\, dy.
\end{displaymath} (1350)

But, ${\bf grad}\,h \equiv (\partial h/\partial x, \,\partial h/\partial y)$ and $d{\bf r}\equiv(dx,\,dy)$, so
dh = {\bf grad}\,h \cdot d{\bf r}.
\end{displaymath} (1351)

Incidentally, the above equation demonstrates that ${\bf grad}\,h$ is a proper vector, since the left-hand side is a scalar, and, according to the properties of the dot product, the right-hand side is also a scalar provided that $d{\bf r}$ and ${\bf grad}\,h$ are both proper vectors ($d{\bf r}$ is an obvious vector, because it is directly derived from displacements).

Consider, now, a three-dimensional temperature distribution $T(x,\,y,\,z)$ in (say) a reaction vessel. Let us define ${\bf grad}\,T$, as before, as a vector whose magnitude is $(dT/dr)_{\rm max}$, and whose direction is the direction of the maximum gradient. This vector is written in component form

{\bf grad}\,T \equiv \left(\frac{\partial T}{\partial x}, \,...
...artial T}{\partial y},\,
\frac{\partial T}{\partial z}\right).
\end{displaymath} (1352)

Here, $\partial T/\partial x\equiv (\partial T/\partial x)_{y, z}$ is the gradient of the one-dimensional temperature profile at constant $y$ and $z$. The change in $T$ in going from point $P$ to a neighbouring point offset by $d{\bf r} \equiv (dx,\,dy,\,dz)$ is
dT = \frac{\partial T}{\partial x}\,dx +\frac{\partial T}{\partial y}\,dy+
\frac{\partial T}{\partial z}\,dz.
\end{displaymath} (1353)

In vector form, this becomes
dT = {\bf grad}\,T \cdot d{\bf r}.
\end{displaymath} (1354)

Suppose that $dT=0$ for some $d{\bf r}$. It follows that
dT = {\bf grad}\,T \cdot d{\bf r} = 0.
\end{displaymath} (1355)

So, $d{\bf r}$ is perpendicular to ${\bf grad}\,T$. Since $dT=0$ along so-called ``isotherms'' (i.e., contours of the temperature), we conclude that the isotherms (contours) are everywhere perpendicular to ${\bf grad}\,T$--see Figure A.112.

Figure A.112: Isotherms.

It is, of course, possible to integrate $dT$. For instance, the line integral of $dT$ between points $P$ and $Q$ is written

\int_P^Q dT = \int_P^Q {\bf grad}\,T\cdot d{\bf r} = T(Q)-T(P).
\end{displaymath} (1356)

This integral is clearly independent of the path taken between $P$ and $Q$, so $\int_P^Q {\bf grad }\,T\cdot d{\bf r}$ must be path independent.

Consider a vector field ${\bf A}({\bf r})$. In general, the line integral $\int_P^Q {\bf A}\cdot d{\bf r}$ depends on the path taken between the end points, but for some special vector fields the integral is path independent. Such fields are called conservative fields. It can be shown that if ${\bf A}$ is a conservative field then ${\bf A} = {\bf grad}\,V$ for some scalar field $V$. The proof of this is straightforward. Keeping $P$ fixed, we have

\int_P^Q {\bf A}\cdot d{\bf r} = V(Q),
\end{displaymath} (1357)

where $V(Q)$ is a well-defined function, due to the path independent nature of the line integral. Consider moving the position of the end point by an infinitesimal amount $dx$ in the $x$-direction. We have
V(Q+dx) = V(Q) + \int_Q^{Q+dx} {\bf A}\cdot d{\bf r} = V(Q) + A_x\,dx.
\end{displaymath} (1358)

\frac{\partial V}{\partial x} = A_x,
\end{displaymath} (1359)

with analogous relations for the other components of ${\bf A}$. It follows that
{\bf A} = {\bf grad} \,V.
\end{displaymath} (1360)

In Newtonian dynamics, the force due to gravity is a good example of a conservative field. Now, if ${\bf A}({\bf r})$ is a force-field then $\int {\bf A}\cdot d{\bf r}$ is the work done in traversing some path. If ${\bf A}$ is conservative then

\oint {\bf A}\cdot d{\bf r} = 0,
\end{displaymath} (1361)

where $\oint$ corresponds to the line integral around a closed loop. The fact that zero net work is done in going around a closed loop is equivalent to the conservation of energy (which is why conservative fields are called ``conservative''). A good example of a non-conservative field is the force due to friction. Clearly, a frictional system loses energy in going around a closed cycle, so $\oint {\bf A}\cdot d{\bf r} \neq 0$.

next up previous
Next: Grad Operator Up: Vector Algebra and Vector Previous: Volume Integrals
Richard Fitzpatrick 2011-03-31