Gradient

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.

Figure 16:

Consider a two-dimensional scalar field $h(x, y)$, which is (say) the height of a hill. Let $d{\bf l}=(dx, dy)$ be an element of horizontal distance. Consider $dh/dl$, where $dh$ is the change in height after moving an infinitesimal distance $d{\bf l}$. This quantity is somewhat like the one-dimensional gradient, except that $dh$ depends on the direction of $d{\bf l}$, as well as its magnitude. In the immediate vicinity of some point $P$, the slope reduces to an inclined plane (see Fig. 16). The largest value of $dh/dl$ is straight up the slope. For any other direction

$$\frac{dh}{dl}= \left(\frac{dh}{dl}\right)_{\rm max} \cos\theta.$$ (93)

Let us define a two-dimensional vector, ${\bf grad} h$, called the gradient of $h$, whose magnitude is $(dh/dl)_{\rm max}$, and whose direction is the direction up the steepest slope. Because of the $\cos\theta$ property, the component of ${\bf grad} h$ in any direction equals $dh/dl$ for that direction. [The argument, here, is analogous to that used for vector areas in Sect. 2.3. See, in particular, Eq. (13).]

The component of $dh/dl$ 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

$${\bf grad} h = \left(\frac{\partial h}{\partial x}, \frac{\partial h}{\partial y} \right).$$ (94)

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

$$h_T(x, y)= h(x_0, y_0) + \alpha (x-x_0)+\beta (y-y_0).$$ (95)

This has the same local gradients as $h(x, y)$, so

$$\alpha = \frac{\partial h}{\partial x},     \beta= \frac{\partial h}{\partial y},$$ (96)

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. We have

$$dh = \frac{\partial h}{\partial x} dx +\frac{\partial h} {\partial y} dy,$$ (97)

but ${\bf grad} h = (\partial h/\partial x, \partial h/\partial y)$ and $d{\bf l}=(dx, dy)$, so

$$dh = {\bf grad} h \cdot d{\bf l}.$$ (98)

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 $d{\bf l}$ and ${\bf grad} h$ are both proper vectors ($d{\bf l}$ 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 $(dT/dl)_{\rm max}$, and whose direction is the direction of the maximum gradient. This vector is written in component form

$${\bf grad} T = \left(\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}\right).$$ (99)

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 $d{\bf l}= (dx, dy, dz)$ is

$$dT = \frac{\partial T}{\partial x} dx +\frac{\partial T}{\partial y} dy+ \frac{\partial T}{\partial z} dz.$$ (100)

In vector form, this becomes

$$dT = {\bf grad} T \cdot d{\bf l}.$$ (101)

Suppose that $dT=0$ for some $d{\bf l}$. It follows that

$$dT = {\bf grad} T \cdot d{\bf l} = 0.$$ (102)

So, $d{\bf l}$ 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 Fig. 17).

Figure 17:

It is, of course, possible to integrate $dT$. The line integral from point $P$ to point $Q$ is written

$$\int_P^Q dT = \int_P^Q {\bf grad} T\cdot d{\bf l} = T(Q)-T(P).$$ (103)

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

In general, $\int_P^Q {\bf A}\cdot d{\bf l}$ depends on path, 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 ${\bf A} = {\bf grad} \phi$ for some scalar field $\phi$. The proof of this is straightforward. Keeping $P$ fixed we have

$$\int_P^Q {\bf A}\cdot d{\bf l} = V(Q),$$ (104)

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

$$V(Q+dx) = V(Q) + \int_Q^{Q+dx} {\bf A}\cdot d{\bf l} = V(Q) + A_x dx.$$ (105)

Hence,

$$\frac{\partial V}{\partial x} = A_x,$$ (106)

with analogous relations for the other components of ${\bf A}$. It follows that

$${\bf A} = {\bf grad} V.$$ (107)

In physics, the force due to gravity is a good example of a conservative field. If ${\bf A}$ is a force, then $\int {\bf A}\cdot d{\bf l}$ is the work done in traversing some path. If ${\bf A}$ is conservative then

$$\oint {\bf A}\cdot d{\bf l} = 0,$$ (108)

where $\oint$ corresponds to the line integral around some closed loop. The fact that zero net work is done in going around a closed loop is equivalent to the conservation of energy (this 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 $\oint {\bf A}\cdot d{\bf l} \neq 0$.

It is useful to define the vector operator

$$\nabla \equiv \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial }{\partial z}\right),$$ (109)

which is usually called the grad or del operator. This operator acts on everything to its right in a expression, until the end of the expression or a closing bracket is reached. For instance,

$${\bf grad} f = \nabla f = \left(\frac{\partial f}{\partial ... ...partial f}{\partial y}, \frac{\partial f}{\partial z}\right).$$ (110)

For two scalar fields $\phi$ and $\psi$,

$${\bf grad} (\phi \psi) = \phi {\bf grad} \psi +\psi {\bf grad} \phi$$ (111)

can be written more succinctly as

$$\nabla(\phi \psi) = \phi \nabla\psi + \psi \nabla \phi.$$ (112)

Suppose that we rotate the basis about the $z$-axis by $\theta$ degrees. By analogy with Eqs. (7)-(9), the old coordinates ($x$, $y$, $z$) are related to the new ones ($x'$, $y'$, $z'$) via

$$x = x' \cos\theta - y' \sin\theta,$$ (113)

$$y = x '\sin\theta +y' \cos\theta,$$ (114)

$$z = z'.$$ (115)

Now,

$$\frac{\partial}{\partial x'} = \left(\frac{\partial x}{\part... ...l z}{\partial x'} \right)_{y',z'} \frac{\partial}{\partial z},$$ (116)

giving

$$\frac{\partial}{\partial x'} = \cos\theta \frac{\partial}{\partial x} + \sin\theta \frac{\partial}{\partial y},$$ (117)

and

$$\nabla_{x'} = \cos\theta \nabla_x + \sin\theta \nabla_y.$$ (118)

It can be seen that the differential operator $\nabla$ transforms like a proper vector, according to Eqs. (10)-(12). This is another proof that $\nabla f$ is a good vector.