Calculating d\rr in Curvilinear Coordinates

Section 8.5 Calculating \(d\rr\) in Curvilinear Coordinates

In Section 8.2, you discovered how to write \(d\rr\) in rectangular coordinates. However, this coordinate system would be a poor choice to describe a path on a cylindrically or spherically shaped surface. We will now find appropriate expressions in these cases.

Activity 8.5. The Vector Differential in Cylindrical Coordinates.

Figure 8.6. An infinitesimal box in cylindrical coordinates

You will now use geometry to determine the general form for \(d\rr\) in cylindrical coordinates by determining \(d\rr\) along the specific paths below. See Figure 8.6.

Note that an infinitesimal element of length in the \(\shat\) direction is simply \(ds\text{,}\) just as an infinitesimal element of length in the \(\xhat\) direction is \(dx\text{.}\) But, an infinitesimal element of length in the \(\phat\) direction is not just \(d\phi\text{,}\) since this would be an angle and does not have the units of length.

Geometrically determine the length of the three paths leading from \(a\) to \(b\) and write these lengths in the corresponding boxes on the diagram.

Now, remembering that \(d\rr\) has both magnitude and direction, write the infinitesimal displacement vector \(d\rr\) along the three bold paths from the point \((s, \phi, z)\) to \(b\text{.}\) Notice that, along any of these three paths, only one coordinate \(s\text{,}\) \(\phi\text{,}\) or \(z\) is changing at a time (i.e. along path 1, \(dz\ne0\text{,}\) but \(d\phi=0\) and \(ds=0\)).

Path 1: \(d\rr=\)

Path 2: \(d\rr=\)

Path 3: \(d\rr=\)

If all three coordinates are allowed to change simultaneously, by an infinitesimal amount, we could write this \(d\rr\) for any path as:

\(d\rr\)=

This is the general line element in cylindrical coordinates.

Hint.

Make sure you think about what the correct lengths are in each coordinate direction. Angles are not lengths!

Solution.

Upon completing this activity, you should have obtained the expressions

\begin{align*} d\rr \amp = dx\,\xhat + dy\,\yhat + dz\,\zhat\\ \amp = ds\,\shat + s\,d\phi\,\phat + dz\,\zhat \end{align*}

in rectangular and cylindrical coordinates, respectively.

Activity 8.6. The Vector Differential in Spherical Coordinates.

Figure 8.7. An infinitesimal box in spherical coordinates

You will now use geometry to determine the general form for \(d\rr\) in spherical coordinates by determining \(d\rr\) along the specific paths below. See Figure 8.7. As in the cylindrical case, note that an infinitesimal element of length in the \(\that\) or \(\phat\) direction is not just \(d\theta\) or \(d\phi\text{.}\) You will need to be more careful. Geometrically determine the length of the three bold paths leading from \((r, \theta, \phi)\) and write these lengths below. Now, remembering that \(d\rr\) has both magnitude and direction, write down below the infinitesimal displacement vector \(d\rr\) along the three paths. Notice that, along any of these three paths, only one coordinate \(r\text{,}\) \(\theta\text{,}\) or \(\phi\) is changing at a time (i.e. along path 1, \(d\theta\ne0\text{,}\) but \(dr=0\) and \(d\phi=0\)).

Path 1: \(d\rr=\)

Path 2: \(d\rr=\qquad\) (Be careful, this is the tricky one.)

Path 3: \(d\rr=\)

If all three coordinates are allowed to change simultaneously, by an infinitesimal amount, we could write this \(d\rr\) for any path as:

\(d\rr\)=

This is the general line element in spherical coordinates.

Hint.

As before, make sure you think about what the correct lengths are in each coordinate direction, and recall that angles are not lengths. Also make sure you know the radius of a circle of constant latitude.

Solution.

Upon completing the worksheet, you should have obtained the expressions

\begin{align*} d\rr \amp = dx\,\xhat + dy\,\yhat + dz\,\zhat\\ \amp = ds\,\shat + s\,d\phi\,\phat + dz\,\zhat\\ \amp = dr\,\rhat + r\,d\theta\,\that + r\,\sin\theta\,d\phi\,\phat \end{align*}

in rectangular, cylindrical, and spherical coordinates, respectively.