Line elements in space
REFERENCE: Hartle
We continue our look at differential geometry with spatial coordinates.
Recall that in the plane, our line element looked like this:
dS2 = dl12 + dl22.
In three dimensions we simply add a third length dimension to get our
general line element:
[Eq. 1]
dS2 = dl12 + dl22 + dl32.
Now that we have three dimensions we can define three different general
area elements
[Eq. 2]
dA = dl1dl2, dA = dl1dl3, dA = dl2dl3,
and one general volume element
dV = dl1dl2dl3.
[Eq. 3]
In the case of the Cartesian coordinate system we have
dl1 = dx,
dl2 = dy,
dl3 = dz,
so that
dS2 = dx2 + dy2 + dz2.
[Eq. 4]
[Eq. 5]
(z)
We call Eq. 5 the Cartesian line
element.
Of course, it is just an
expression of the Pythagorean theorem
in three dimensions.
We usually use
Cartesian coordinates when there are
not cylindrical or spherical symmetries
associated with a problem (see Fig. 1).
Cartesian
coordinates
(y)
dS
(x,y,z)
(x + dx,y + dy ,z + dz)
dl3 = dz
dl1 = dx
dl2 = dy
(x)
[Fig. 1]
In the case of cylindrical symmetry, we add the z-axis to the polar
plane to get a system of coordinates called polar-cylindrical (or
cylindrical, for short) shown in Fig. 2. We have
dl1 = dr,
dl2 = rd,
dl3 = dz,
[Eq. 6]
(z)
dl1 = dr
dS2 = dr2 + r2d2 + dz2. [Eq. 7]
The
coordinate
transformations
between
Cartesian
and
polarcylindrical coordinates are left as
an exercise. The transformations can
be used to show that dS is invariant
under the transformations.
In other
words, the distance dS between two
points is the same, whether you are
using
Cartesian
or
cylindrical
coordinates.
(r + dr, + d ,z + dz)
dS
(r,,z)
(x)
dl3 = dz
(y)
dl2 = rd
Polar-cylindrical
coordinates
[Fig. 2]
Finally, in the case of spherical symmetry
coordinates, shown in Fig. 3.
dl1 = dr,
dl2 = rd,
dl3 = r sin  d,
we
can
use
spherical
[Eq. 8]
dS2 = dr2 + r2d2 + r2sin2 d2.
or
[Eq. 9]
dS2 = dr2 + r2(d2 + sin2 d2).
The coordinate transformation from
spherical coordinates to Cartesian
coordinates are given here:
(z)
(r + dr, + d, + d)
dl1 = dr
x = r sin  cos ,
y = r sin  sin ,
z = r cos .
dl2 = rd
dS
[Eq. 10]

d
dl3 = r sin  d
(r,,)
(y)
d

As is the case in polar-cylindrical
Spherical
(x)
coordinates, we can show that if Eq.
coordinates
5 is true, then so is Eq. 10. To do
so, we find the differentials dx, dy
[Fig. 3]
and dz from Eq. 10, and substitute
them into Eq. 5.
dx = dr sin  cos  + r cos  cos  d - r sin  sin  d,
dy = dr sin  sin  + r cos  sin  d + r sin  cos  d,
[Eq. 10]
dx = dr cos  - r sin  d.
Squaring Eq. 10 and substitution into Eq. 5 will produce Eq. 9 showing that Eq. 9 is equivalent to Eq. 5 for finding the distance
between two points.
From Eq. 3 recall that dV = dl1dl2dl3.
Substitution of Eq. 8 yields
dV = drrdr sin  d
= r2sin  dr d d.
[Eq. 11]
We call Eq. 11 the spherical volume element. Integrating Eq. 11 for
0  r  R, 0    , and 0    2,
we obtain
2  R
[Eq. 12]
V =    r2sin  dr d d = 4R3/3.
0 0 0
Eq. 12 is the expected result for the volume of a sphere
From Eq. 2 and Eq. 8 we have (for r = R)
[Eq. 13]
dA = dl2dl3 = R2sin  d d,
so that
2 
A =   R2sin  d d = 4R2.
0 0
[Eq. 14]
Eq. 14 is the expected result for the surface area of a sphere.