Gradient

Consider a two-dimensional scalar field that represents (say) height above sea-level in a hilly region. Let be an element of horizontal distance. Consider , where is the change in height after moving an infinitesimal distance . This quantity is somewhat like the one-dimensional gradient, except that depends on the direction of , as well as its magnitude. In the immediate vicinity of some point , the slope reduces to an inclined plane. (See Figure A.19.) The largest value of is straight up the slope. It is easily shown that for any other direction

(A.103) |

where is the angle shown in Figure A.19. Let us define a two-dimensional vector, , called the

The component of
in the
-direction can be obtained by plotting out the
profile of
at constant
, and then finding the slope of the tangent to the
curve at given
. This quantity is known as the *partial derivative* of
with respect to
at constant
, and is denoted
.
Likewise, the gradient of the profile at constant
is written
. Note that the subscripts denoting constant
and
constant
are usually omitted, unless there is any ambiguity. It follows that
in component form

(A.104) |

Now, the equation of the tangent plane at is

(A.105) |

This has the same local gradients as , so

(A.106) |

For small and , the function is coincident with the tangent plane. It follows that

(A.107) |

But, and , so

(A.108) |

Incidentally, the previous equation demonstrates that is a proper vector, because 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 and are both proper vectors ( is an obvious vector, because it is directly derived from displacements).

Consider, now, a three-dimensional temperature distribution in (say) a reaction vessel. Let us define , as before, as a vector whose magnitude is , and whose direction is the direction of the maximum gradient. This vector is written in component form

(A.109) |

Here, is the gradient of the one-dimensional temperature profile at constant and . The change in in going from point to a neighboring point offset by is

(A.110) |

In vector form, this becomes

(A.111) |

Suppose that for some . It follows that

(A.112) |

So, is perpendicular to . Because along so-called ``isotherms'' (i.e., contours of the temperature), we conclude that the isotherms (contours) are everywhere perpendicular to . (See Figure A.20.)

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

(A.113) |

This integral is clearly independent of the path taken between and , so must be path independent.

Consider a vector field
. In general, the line integral
depends on the path
taken between the end points. However, for some special vector fields the integral is path independent. Such fields
are called *conservative* fields. It can be shown that if
is a
conservative field then
for some scalar field
.
The proof of this is straightforward. Keeping
fixed, we have

(A.114) |

where 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 in the -direction. We have

(A.115) |

Hence,

(A.116) |

with analogous relations for the other components of . It follows that

(A.117) |

The force field due to gravity is a good example of a conservative field. Now, if is a force-field then is the work done in traversing some path. If is conservative then

(A.118) |

where 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 field due to friction. Clearly, a frictional system loses energy in going around a closed cycle, so .