Mathematics of Physics and Engineering

144 Functions of Several Variables

then, using (3.1.36) and dropping (dt)^2 , we find that the line element in
Q coordinates is

(ds)^2 = ^2h^2 k(qi,q2,q 3 )(dqk)^2. (3.1.39)

fc=i

EXERCISE 3.1.27.^5 (a) Verify (3.1.39).
(b) Verify that, for small a, the distance between the two nearby points
with the Q coordinates (ci,02,03) and (ci +0,02,03) is approximately h\a.
(c) Verify that in cylindrical coordinates, (ds)^2 = (dr)^2 +r^2 (d0)^2 + (dz)^2.
(d) Verify that in spherical coordinates, (ds)^2 = (dr)^2 + r^2 sin (p (d9)^2 +
r^2 (dtp)^2. Hint for (b) and (c): while direct substitution into (3.1.39) will work,
getting closer to the basics produces an easier argument. For example, in spherical
coordinates, if r and ip are fixed and 6 is changing, the corresponding coordinate
curve is a circle of radius rsimp (draw a picture). Then a change of A0 in 8
produces the change r sin tp A9 in the arc length. Similar arguments work for the
other two coordinate curves, and then it remains to use orthogonality. Note that,
with this approach, you get the values of hk without using (3.1.37).
(e) Verify that if the Q coordinate are not orthogonal, then

(ds)^2 = Y^ Ha dQi d1ji where H^ = ]P -a^-*^-

i,i=i fc=i 0Qi OQj

(3.1.40)

We already used a four-dimensional analog of this relation in the study of
general relativity; see (2.4-23) on page 107.

For future reference, let us summarize the values of the functions hk for
cylindrical and spherical coordinates:

Cylindrical

Spherical

hi

1

1

hi

r

rs'm<p

h 3

1

r

We can now derive the expression for the GRADIENT in Q coordinates.
The value of a function at a point does not depend on the coordinate
system, but the formula for computing that value does: if / = X1X2 +x% in
cartesian coordinates, the same function in spherical coordinates becomes
r^2 cos#sin#sin tp + r cosip. Given an expression /(gi,92,(73) for a function
in Q coordinates, we define the partial derivatives df/dqk, k — 1,2,3 in