CURVILINEAR COORDINATES

8. CURVILINEAR COORDINATES

Before we discuss non-Cartesian tensors we need to talk about some properties of curvilinear coordinate systems such as spherical or cylindrical coordinates. To make the discussion concrete, we shall illustrate the ideas involved by using two familiar coordinate systems—rectangular coordinates (x, y, z) and cylindrical coordinates (r, θ, z). The elements of arc length in these two systems are given by

ds 2 = dx 2 + dy 2 + dz 2 (rectangular coordinates) (8.1) ds 2 = dr 2 +r 2 dθ 2 + dz 2 (cylindrical coordinates)

These expressions for ds are called line elements; they have much greater signifi- cance than just their use in computing arc lengths. First consider how we can find

ds 2 for a given coordinate system. In the case of a well-known coordinate system, the answer may be obvious from the geometry. For example in polar coordinates in the plane we have (from Figure 8.1 and the Pythagorean theorem)

ds 2 = dr 2 +r 2 dθ 2 .

For an unfamiliar or complicated change of variables, however, we need a systematic method of finding ds; we illustrate the method by finding the value of ds 2 for cylindrical coordinates as given in (8.1). From the equations

Figure 8.1 we get

Squaring each equation in (8.4) and adding the results, we find (8.5)

ds 2 = dx 2 + dy 2 + dz 2 = dr 2 +r 2 dθ 2 + dz 2 . Notice particularly here that all the cross products (dr dθ, etc.) canceled out.

This will not always happen, but it often does; when it does we call the coor- dinate system orthogonal. Such coordinate systems have some particularly simple and useful properties. Geometrically, an orthogonal system means that the coordinate surfaces are mutually perpendicular. For the cylindrical system (Figure 8.2), the coordinate sur- faces are r = const. (set of concentric cylinders), θ = const. (set of half-planes), and z = const. (set of planes). The three coordinate surfaces through a given point intersect at right angles. The three curves of intersec- tion of the coordinate surfaces in pairs intersect at right

Figure 8.2

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angles; these curves are called the coordinate “lines” or directions. We draw unit basis vectors tangent to the coordinate directions; for the cylindrical system (Figure

8.2) we might call them e r ,e θ ,e z (e z is identical to k). These unit vectors form an orthogonal triad like i, j, k. We refer to such coordinate systems as curvilinear coordinate systems when the coordinate surfaces (or some of them) are not planes and the coordinate lines (or some of them) are curves rather than straight lines. We shall be principally interested in orthogonal curvilinear coordinate systems.

Scale Factors and Basis Vectors In the rectangular system, if x, y, z, are the coordinates of a particle, and x changes by dx with y and z constant, then the distance the particle moves is ds = dx. However, in the cylindrical system, if θ changes by dθ with r and z constant, the distance the particle moves is not dθ, but ds = r dθ. Factors like the r in r dθ which must multiply the differentials of the coordinates to get distances are known as scale factors and are very important as

we shall see. A straightforward way to get them is to calculate ds 2 as we did in (8.5); if the transformation is orthogonal, then the scale factors can be read off from ds 2 . (Note that the coefficients in ds 2 are the squares of the scale factors.) From (8.5), we see that the scale factors for cylindrical coordinates are 1, r, 1. It is also useful to consider a vector ds which (in cylindrical coordinates) has components dr, r dθ, dz in the coordinate directions e r ,e θ ,e z :

ds = e r dr + e θ r dθ + e z dz.

Then ds 2 = ds · ds which gives (8.1), since the e vectors are orthonormal. We can write the unit basis vectors of a curvilinear coordinate system (e r ,e θ ,

e z in cylindrical coordinates) in terms of i, j, k. This is useful when we want to differentiate a vector which is expressed in terms of the curvilinear coordinate basis vectors. The unit vectors i, j, k are constant in magnitude and direction, but e r and e θ are not fixed in direction, so their derivatives are not zero. We illustrate an algebraic method of finding the relation between two sets of basis vectors by finding them for the cylindrical system. (Compare the geometrical method shown in Chapter 6, Section 4.)

Example 1. Using (8.4) and collecting coefficients of dr, dθ, and dz, we find ds = i dx + j dy + k dz

(8.7) = i(cos θ dr − r sin θ dθ) + j (sin θ dr + r cos θ dθ) + k dz = (i cos θ + j sin θ) dr + (−ir sin θ + j r cos θ) dθ + k dz.

Comparing (8.7) with (8.6), we have

Notice that e r is already a unit vector since sin 2 θ + cos 2 θ = 1, but re θ must be divided by the scale factor r to get the unit vector e θ . It is often convenient to use basis vectors which we shall call a r and a θ (which are not necessarily of unit

Section 8 Curvilinear Coordinates 523

length), given by the coefficients of dr and dθ in (8.7). Then we just have to divide each a vector by its magnitude to get the corresponding e vector. Thus from (8.7)

a r =e r is already a unit vector, (8.9)

a θ = −ir sin θ + j r cos θ has magnitude r, so

e θ = a θ = −i sin θ + j cos θ r

We can use these formulas to find the velocity and acceleration of a particle in cylindrical coordinates, and similar formulas for any coordinate system. The displacement of a particle from the origin at time t is, in cylindrical coordinates (Figure 8.3),

s = re r + ze z . Then

so Figure 8.3

ds (8.10)

= ˙re r + r ˙θe θ + ˙ze z . dt

By differentiating again with respect to t and using (8.8) to find (d/dt)(e θ ), we can find the acceleration d 2 s/dt 2 in cylindrical coordinates (Problem 2).

General Curvilinear Coordinates In general, let x 1 ,x 2 ,x 3 be the set of vari- ables or coordinates we are considering (for example, in cylindrical coordinates, x 1 = r, x 2 = θ, x 3 = z). Then the three sets of coordinate surfaces are x 1 = const., x 2 = const., x 3 = const. The three coordinate surfaces through a given point intersect in three coordinate lines.

Example 2. Given x, y, z as functions of x 1 ,x 2 ,x 3 , we can find ds and the a vectors as we did for cylindrical coordinates in (8.7) and (8.9).

=a 1 dx 1 +a 2 dx 2 +a 3 dx 3 =a n dx n , where

Now defining g =a · a , we can write ds ij 2 i j = ds · ds in matrix form as follows:

g 11 g 12 g 13 dx 1 (8.13)

ds 2 = dx 1 dx 2 dx 3  g 21 g 22 g 23   dx 2  ,

g 31 g 32 g 33 dx 3

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Note that g ij is symmetric since the dot product of two vectors is the same in either order. In simpler form using summation convention (8.13) becomes

ds 2 =g ij dx i dx j .

We will see later (Section 10) that the g ij are the components of a tensor known as the metric tensor.

If the coordinate system is orthogonal, that is, if the basis vectors (e or a) form an orthogonal triad, then ds and ds 2 can be written in terms of the scale factors as follows:

(8.15) ds = e 1 h 1 dx 1 +e 2 h 2 dx 2 +e 3 h 3 dx 3 ,  2  

h 1 0 0 dx 1 (8.16)

ds 2 = dx 1 dx 2 dx 3  0 h 2 2 0   dx 2  .

0 0 h 2 3 dx 3

Also note that the volume element in an orthogonal system is h 1 h 2 h 3 dx 1 dx 2 dx 3 (volume of a small rectangular parallelepiped with edges h 1 dx 1 ,h 2 dx 2 ,h 3 dx 3 ). For example, in cylindrical coordinates, the volume element is dr · r dθ · dz = r dr dθ dz.

PROBLEMS, SECTION 8

1. Find ds 2 in spherical coordinates by the method used to obtain (8.5) for cylindrical coordinates. Use your result to find for spherical coordinates, the scale factors, the vector ds, the volume element, the basis vectors a r ,a θ ,a φ and the corresponding unit basis vectors e r ,e θ ,e φ . Write the g ij matrix.

2. Observe that a simpler way to find the velocity ds/dt in (8.10) is to divide the vector ds in (8.6) by dt. Complete the problem to find the acceleration in cylindrical coordinates.

3. Use the results of Problem 1 to find the velocity and acceleration components in spherical coordinates. Find the velocity in two ways: starting with ds and starting with s = re r .

4. In the text and problems so far, we have found the e vectors for various coordinate systems in terms of i and j (or i, j, k in three dimensions). We can solve these equations to find i and j in terms of the e vectors, and so express a vector given in rectangular form in terms of the basis vectors of another coordinate system. Carry out this process to express in cylindrical coordinates the vector V = yi − xj + k. Hint: Use matrices (as in Chapter 3) to solve the set of equations for i and j.

5. Using the results of Problem 1, express the vector in Problem 4 in spherical coordi- nates.

As in Problem 1, find ds 2 , the scale factors, the vector ds, the volume (or area) element, the a vectors, and the e vectors for each of the following coordinate systems.

6. Parabolic cylinder coordinates u, v, z: 7. Elliptic cylinder coordinates u, v, z:

x= (u 2 −v 2 ),

x = a cosh u cos v, 2 y = a sinh u sin v,

y = uv, z = z. z = z.

Section 9 Vector Operators in Orthogonal Curvilinear Coordinates 525

8. Parabolic coordinates u, v, φ: 9. Bipolar coordinates u, v: x = uv cos φ,

a sinh u x=

, y = uv sin φ,

cosh u + cos v a sin v

z= (u 2 2 y=

−v ).

cosh u + cos v

10. Sketch or computer plot the coordinate surfaces in Problems 6 to 9. Using the expression you have found for ds, and for the e vectors, find the velocity and

acceleration components in the coordinate systems indicated. 11. Parabolic cylinder

12. Elliptic cylinder 13. Parabolic

14. Bipolar

15. Let x = u + v, y = v. Find ds, the a vectors, and ds 2 for the u, v coordinate system and show that it is not an orthogonal system. Hint: Show that the a vectors are not orthogonal, and that ds 2 contains du dv terms. Write the g ij matrix and observe that it is symmetric but not diagonal. Sketch the lines u = const. and v = const. and observe that they are not perpendicular to each other.