Elliptical Coordinates

This method of starting with a complex function has the advantage of the availability of all of the tools of Complex-Variables. The disadvantage is that only solids of revolution can be generated. An alternative would be to define the three-dimensional shape of a closed-surface. Then, find the normal as the cross-product of the two orthogonal directions along the surface. Please see the spheroidal-coordinates for my exposition of the general spheroid -- each of the three semi-axes are independent.

These so-called elliptical coordinates are based upon the four trigonometric functions: the hyperbolic or circular cosine or sine. Each of which, in turn is the sum or difference of two exponential functions. From each of these trigonometric functions, we obtain an orthogonal net (pair of related families) of functions, in two-dimensional space. Each of these four nets we revolve about the y-axis, to obtain a spheroidal (usually called ellipsoidal) three-dimensional coordinate system. We begin with the

Oblate Spheroidal

Definition

The elliptical coordinates (u, v) may be defined in terms of the Cartesian coordinates (x, y) by the complex function

x + i y = z = c cosh(w) = c cosh(u + i v) = c (cosh(u) cos(v) + i sinh(u) sin(v)),

where c is a real constant and the complex variables w and z are decomposed into their respective components as shown. Compute its purely-imaginary and real components as

y = (z - z*) / (2 i) = c ((cosh(u) cos(v) + i sinh(u) sin(v)) - (cosh(u) cos(v) - i sinh(u) sin(v))) / (2 i) = c sinh(u) sin(v)
x = (z + z*) / 2 = c ((cosh(u) cos(v) + i sinh(u) sin(v)) + (cosh(u) cos(v) - i sinh(u) sin(v))) / 2 = c cosh(u) cos(v).

This pair of equations actually constitute the usual definition of the elliptical coordinates. We have employed various properties of the exponential, hyperbolic-trigonometric, and circular-trigonometric functions in the preceding transformations.

Hyperbola

The equation of a hyperbola, in its standard form, is

1 = (x / A)^2 - (y / B)^2,

where I have employed the capital A and B for the constants to distinguish them from those for the forthcoming ellipse. Let these two constants be

A = c cos(v)
B = c sin(v).

Substitute x, y, A, and B into the equation for the hyperbola, to obtain

1 =? ((c cosh(u) cos(v)) / (c cos(v)))^2 - ((c sinh(u) sin(v)) / (c sin(v)))^2 = (cosh(u))^2 - (sinh(u))^2 = 1.

It checks by the identity for the hyperbolic cosine.

Ellipse

The equation of an ellipse, in its standard form is

1 = (x / a)^2 + (y / b)^2,

where a and b are constants. Let these two constants be

a = c cosh(u)
b = c sinh(u).

Observe that

0 < b < a

for any u in (0, oo). Substitute x, y, a, and b into the equation for the ellipse, to obtain

1 =? ((c cosh(u) cos(v)) / (c cosh(u)))^2 - ((c sinh(u) sin(v)) / (c sinh(u)))^2 = (cos(v))^2 + (sin(v))^2 = 1.

It checks by the identity for the circular cosine.

Orthogonal Families

There are several ways to show that these two families -- the hyperbolas and the ellipses -- are orthogonal. Find the partial derivatives of the defining equations as

dy/du = c cosh(u) sin(v)
dy/dv = c sinh(u) cos(v)
dx/du = c sinh(u) cos(v)
dx/dv = - c cosh(u) sin(v).

Tangent Vector

The tangent vector along the hyperbola and that along the ellipse are, respectively,

(dx/du, dy/du) = c (sinh(u) cos(v), cosh(u) sin(v)) = (b cos(v), a sin(v))
(dx/dv, dy/dv) = c (- cosh(u) sin(v), sinh(u) cos(v)) = (- a sin(v), b cos(v)).

For them to be orthogonal, their inner (dot) product has to be zero:

0 =? (dx/dy, dy/du) . (dx/dv, dy/dv) = (b cos(v), a sin(v)) . (- a sin(v), b cos(v)) = 0.

Gradient

The gradient vector to the hyperbola and that to the ellipse are, respectively,

(dx/dv, dy/dv)
(dx/du, dy/du).

These are just be previous vectors, but taken in reverse order. For these two families to be orthogonal, the inner (dot) product of the gradient vectors has to be zero, as already demonstrated.

Cauchy Condition

Without the necessity of finding the complex function explicitly, we see that the Cauchy condition is satisfied:

dx/du = dy/dv
dx/dv = - dy/du.

Hence, the function is analytical. No surprise here. The hyperbolic cosine is a familiar function. We then know that any analytic function provides a conformal -- isotropic and angle-preserving -- mapping. Thus, the orthogonal (u + i v) is mapped into the orthogonal (x + i y).

Polar Coordinates.

The polar coordinates (r, phi) may be defined in terms of the Cartesian coordinates (x, y) by the set of two equations

y = r sin(phi)
x = r cos(phi).

Set these (x, y) equal to the previous ones, to obtain

r sin(phi) = b sin(v)
r cos(phi) = a cos(v).

Divide, to obtain

tan(phi) = (b / a) tan(v).

Hence, we have the polar angle phi as

phi = Atan((b / a) tan(v)),

where phi is to be taken in the same quadrant as v. Square each of the two preceding equations, and add, to obtain

r^2 = (b sin(v))^2 + (a cos(v))^2,

where we have employed the identity for the circular cosine. Substitute for a and b and employ the identity for the hyperbolic cosine, to obtain

r^2 = (c sinh(u) sin(v))^2 + (c cosh(u) cos(v))^2 = c^2 ((sinh(u))^2 + (cos(v))^2) = c^2 ((cosh(u))^2 - (sin(v))^2).

The radius r is the positive square-root, namely

r = sqrt((b sin(v))^2 + (a cos(v))^2) = |c| sqrt((sinh(u))^2 + (cos(v))^2) = |c| sqrt((cosh(u))^2 - (sin(v))^2).

Likewise, we have the elliptical angle v as

v = Atan((a / b) tan(phi),

where, again, v is to be taken in the same quadrant as phi. Thus, we have shown how to find the polar coordinates (r, phi) from v and how to find r and v from phi.

Spheroidal Coordinates.

Revolve the elliptical coordinates about the y-axis to obtain the oblate-spheroidal coordinates from the elliptical coordinates.

The angle theta through which we have revolved is called the longitude.
The angle v is the oblate-spheroidal (usually called the ellipsoidal) latitude.
The angle phi is the spherical latitude.
The r is the local (that is, at the given latitude) radius.
The a is the equatorial radius and is greater than the b, which is the polar radius.

Unit Vectors

The magnitude of each of the first two (the gradient to the ellipse and to the hyperbola) of the three vectors is

|(b cos(v), a sin(v))| = sqrt((a sin(v))^2 + (b cos(v))^2) = |(- a sin(v), b cos(v))|.

That each of these two vectors has the same magnitude confirms that the mapping is isotropic, as we said previously. Let us add an altitude h, along the gradient to the ellipse, to obtain

y = b sin(v) + h a sin(v) / sqrt(...) = (b + h a / sqrt(...)) sin(v) = r sin(phi)
x = a cos(v) + h b cos(v) / sqrt(...) = (a + h b / sqrt(...)) cos(v) = r cos(phi).

Divide, to obtain

tan(phi) = ((b + h a / sqrt(...)) / (a + h b / sqrt(...))) tan(v).

Hence, we have the circular polar angle phi as

phi = Atan((( b + h a / sqrt(...)) / ( a + h b / sqrt(...))) tan(v)),

where phi is to be taken in the same quadrant as v. Square each of the two preceding equations, and add, to obtain

r^2 = ((a cos(v))^2 + (b (sin(v))^2) + 2 h a b / sqrt(...) + h^2.

The radius r is the positive square-root, namely

r = sqrt(((a cos(v))^2 + (b (sin(v))^2) + 2 h a b / sqrt(...) + h^2).

At h=0, these two results check the previous. The three-dimensional Cartesian coordinates are

(x, y, z) = (r cos(phi) cos(theta), r cos(phi) sin(theta), r sin(phi)) =
= (w cos(theta), w sin(theta), (b + h a / sqrt(...)) sin(v))
w = (a + h b / sqrt(...)) cos(v) [for brevity]
sqrt(...) = sqrt((a sin(v))^2 + (b cos(v))^2) [for brevity]
phi = Atan(z / w). [the spherical latitude]

Any plane (except the equatorial plane) through the center of the oblate spheroid is an ellipse. Hence, the arc-length of any segment is an elliptic integral. Furthermore, these ellipses are not the geodesics. On the other hand, the straight-line Euclidean distance between a pair of points is just

dist = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2).

Actually, we just have begun the study of this coordinate system. In the synthetic vector analysis page, we provide the unit vectors, the displacement, the derivative of each of the unit vectors, the del, the LaPlacian, the Jacobian, the velocity and acceleration, and the orthonormal basis for the Cartesian, cylindrical, and spherical coordinate systems. The same should be provided for each of these four ellipsoidal coordinate systems, as well. Maybe someday? :-)

Prolate Spheroidal

Now, do it all over again; but, employing the hyperbolic sine function.

Jacobian

At the kindly suggestion by Professor Emeritus ___ ___, at SUNY, I am adding the Jacobians.

We will employ the identities to substitute for the square of any sine, in terms of the corresponding cosine. These identities are for the hyperbolic and circular.

We will suggest the domain for each variable, as we introduce it. The conventions as to the preferred domains vary and, in any case, are pre-empted by the requirements of a specific situation. Hence, pray, feel free to disregard these suggestions.

Any Jacobian that is negative, means that the transformation -- by definition -- is improper; that is, it changes the chirality of the coordinate system.

Each of the following tables provides the derivation of a Jacobian. The rows of the table are:

The name of the Jacobian.
The names of the dependent variables, one per column, beginning in the second column.
The function of each dependent variable, in terms of the independent variables.
Each subsequent row provides the partial derivative of the dependent variable wrt (=with respect to) to the indicated independent variable.

Then, the stated Jacobian matrix is displayed in the resulting bottom right-hand corner of the table. It is a square of one less than the width of the table.

ellipse / hyperbola

Let the new variables lambda and mu be defined as

lambda = cosh(u), u in the open domain (0, oo). Hence, the range of lambda is in the open interval (1, oo)
mu = cos(v), v in the semi-closed domain (- pi, pi]. Hence, the range of mu is in the closed interval [-1, 1]

Jacobian of (lambda, mu) wrt (u, v)
	lambda	mu
function	cosh(u)	cos(v)
d/du	sinh(u)	0
d/dv	0	-sin(v)

The Jacobian matrix is just the 2x2 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((lambda, mu) / (u, v)) = - sinh(u) sin(v).

Jacobian of (x, y) wrt (u, v)
	x	y
function	c cosh(u) cos(v)	c sinh(u) sin(v)
d/du	c sinh(u) cos(v)	c cosh(u) sin(v)
d/dv	- c cosh(u) sin(v)	c sinh(u) cos(v)

The domain of c is in the open interval (0, oo). The range of the ordered pair (x, y) is the whole real plane, except the origin.

The Jacobian matrix is just the 2x2 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((x, y) / (u, v)) = c^2 ((sinh(u))^2 (cos(v))^2 + (cosh(u))^2 (sin(v))^2) = c^2 ((cosh(u))^2 - (cos(v))^2).

J((x, y) / (lambda, mu)) = J((x, y) / (u, v)) / J((lambda, mu) / (u, v)) = c^2 ((cosh(u))^2 - (cos(v))^2) / ( -sinh(u) sin(v)) = -+ c^2 (lambda^2 - mu^2) / sqrt((lambda^2 - 1) (1 - mu^2)). (The - sign if the square-root is positive; otherwise, the + sign.)

Could this Jacobian have been computed directly from (x, y) to (lambda, mu)? Aye, forsooth. We could have employed

x = c lambda mu
y = c sqrt((lambda^2 - 1) (1 - mu^2)).

However, we would have had to deal with a surfeit of square-roots.

prolate spheroid / hyperboloid of revolution of two sheets

Now, let us rotate the ellipse about the x-axis, that is, the axis through the focci.

Jacobian of (lambda, mu, phi) wrt (u, v, phi)
	lambda	mu	phi
function	cosh(u)	cos(v)	phi
d/du	sinh(u)	0	0
d/dv	0	-sin(v)	0
d/d-phi	0	0	1

The Jacobian matrix is just the 3z3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((lambda, mu, phi) / (u, v, phi)) = - sinh(u) sin(v).

Jacobian of (x, y, z) wrt (u, v, phi)
	x	y	z
function	c cosh(u) cos(v)	c sinh(u) sin(v) cos(phi)	c sinh(u) sin(v) sin(phi)
d/du	c sinh(u) cos(v)	c cosh(u) sin(v) cos(phi)	c cosh(u) sin(v) sin(phi)
d/dv	- c cosh(u) sin(v)	c sinh(u) cos(v) cos(phi)	c sinh(u) cos(v) sin(phi)
d/d-phi	0	- c sinh(u) sin(v) sin(phi)	c sinh(u) sin(v) cos(phi)

Conventions vary. Either constrain phi to the closed domain [- pi / 2, pi / 2] xor constrain v to the closed sub-domain [0, pi] and then let the domain of phi be the semi-closed interval (- pi, pi].

The Jacobian matrix is just the 3x3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((x, y, z) / (u, v, phi)) = c^3 sinh(u) (((sinh(u))^2 (cos(v))^2 (cos(phi))^2 + (cosh(u))^2 (sin(v))^2 (sin(phi))^2 + (sinh(u))^2 (cos(v))^2 (sin(phi))^2 + (cosh(u))^2 (sin(v))^2 (cos(phi))^2) sin(v) = c^3 sinh(u) ((cosh(u))^2 - (cos(v))^2) sin(v).

J((x, y, z) / (lambda, mu, phi)) = J((x, y, z) / (u, v, phi)) / J((lambda, mu), phi / (u, v, phi)) = c^3 sinh(u) ((cosh(u))^2 - (cos(v))^2) sin(v). / (- sinh(u) sin(v)) = - c^3 ((cosh(u))^2 - (cos(v))^2) = - c^3 (lambda^2 - mu^2).

Pray, observe that we have removed the removable singularities at u = 0 and at each integral multiple of pi for v by cancelling the sinh(u) and sin(v). This singularity will come back to haunt us in the next section.

oblate spheroid / hyperboloid of revolution of one sheet

Now, let us rotate the ellipse about the y-axis, that is, the axis which does not go through the focci.

Jacobian of (lambda, mu, phi) wrt (u, v, phi)
	lambda	mu	phi
function	cosh(u)	cos(v)	phi
d/du	sinh(u)	0	0
d/dv	0	-sin(v)	0
d/d-phi	0	0	1

The Jacobian matrix is just the 3z3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((lambda, mu, phi) / (u, v, phi)) = - sinh(u) sin(v).

[This is a copy of the Jacobian at the beginning of the previous section. We repeat it here, so that it would be in front of us for the next step of the calculation.]

Jacobian of (x, y, z) wrt (u, v, phi)
	x	y	z
function	c cosh(u) cos(v) cos(phi)	c sinh(u) sin(v)	c cosh(u) cos(v) sin(phi)
d/du	c sinh(u) cos(v) cos(phi)	c cosh(u) sin(v)	c sinh(u) cos(v) sin(phi)
d/dv	- c cosh(u) sin(v) cos(phi)	c sinh(u) cos(v)	- c cosh(u) sin(v) sin(phi)
d/d-phi	- c cosh(u) cos(v) sin(phi)	0	c cosh(u) cos(v) cos(phi)

Conventions vary. Either constrain phi to the closed domain [- pi / 2, pi / 2] xor constrain v to the closed sub-domain [- pi / 2, pi / 2] and then let the domain of phi be the semi-closed interval (- pi, pi].

The Jacobian matrix is just the 3x3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((x, y, z) / (u, v, phi)) = c^3 cosh(u) ((sinh(u))^2 (cos(v))^2 (cos(phi))^2 + (cosh(u))^2 (sin(v))^2 (sin(phi))^2 + (sinh(u))^2 (cos(v))^2 (sin(phi))^2 + (cosh(u))^2 (sin(v))^2 (cos(phi))^2) cos(v) = c^3 cosh(u) ((cosh(u))^2 - (cos(v))^2) cos(v).

J((x, y, z) / (lambda, mu, phi)) = J((x, y, z) / (u, v, phi)) / J((lambda, mu), phi / (u, v, phi)) = c^3 cosh(u) ((cosh(u))^2 - (cos(v))^2) cos(v) / (- sinh(u) sin(v)) = -+ c^3 lambda (lambda^2 - mu^2) mu / sqrt((lambda^2 - 1) (1 - mu^2)). (The - sign if the square-root is positive; otherwise, the + sign.)

Pray, observe that this time we have been unable to circumvent the singularities at u = 0 and at each integral multiple of pi for v. Specifically, these are poles of order one-half. Hence, they should integrate-out in the typical use of the Jacobian. As will be seen in the next section, the singularity u = 0 is the closed equatorial disk, bounded by the foci. The other singularity is the remainder of the equatorial plane.

Confocal

From a comparison with the geometrical development of the conic sections, it is apparent that, in the plane case:

The foci are located at (-c, 0) and (c, 0);
- hence, the net is confocal.
- the distance between the foci is 2 c, which often is written as rho = 2 c.
lambda = (|(x, y) - (-c, 0)| + |(x, y) - (c, 0)|) / (2 c),
mu = (|(x, y) - (-c, 0)| - |(x, y) - (c, 0)|) / (2 c),

Let

g = |(x, y) - (-c, 0)| / (2 c) = sqrt((x + c)^2 + ;y^2) / (2 c). Its range is the open interval (0, oo).
h = |(x, y) - (c, 0)| / (2 c) = sqrt((x - c)^2 + y^2) / (2 c). Its range is the open interval (0, oo).

Then, we have

lambda = g + h.
mu = g - h.

We also have

Jacobian of (lambda, mu, phi) wrt (h, g, phi)
	lambda	mu	phi
function	g + h	g - h	phi
d/dh	1	- 1	0
d/dg	1	1	0
d/d-phi	0	0	1

The Jacobian matrix is just the 3z3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((lambda, mu, phi) / (h, g, phi)) = 1.

Then, the prolate-spheroidal Jacobian becomes

J((x, y, z) / (h, g, phi)) = J((x, y, z) / (lambda, mu, phi)) J((lambda, mu, phi) / (h, g, phi)) = c^3 (lambda^2 - mu^2) * 1 = c^3 ((g + h)^2 - (g - h)^2) = 4 c^3 g h.

and the oblate-spheroidal Jacobian becomes

J((x, y, z) / (h, g, phi)) = J((x, y, z) / (lambda, mu, phi)) J((lambda, mu, phi) / (h, g, phi)) = -+ c^3 lambda (lambda^2 - mu^2) mu / sqrt((lambda^2 - 1) (1 - mu^2)) * 1 = -+ c^3 (g + h)((g + h)^2 - (g - h)^2) (g - h) / sqrt(((g + h)^2 - 1) (1 - (g - h)^2)) = -+ 4 c^3 g (g^2 - h^2) h / sqrt(((g + h)^2 - 1) (1 - (g - h)^2)). (The - sign if the square-root is positive; otherwise, the + sign.)

Special cases

So far, we tacitly have assumed that c is not zero. Obviously, any attempted division by c is impossible, when c is zero. Forsooth, even multiplication by c, when c is zero, yields an unusable result. Thus, we have to deal with the values of zero and infinity as limiting special cases.

Polar

The limit as c approaches zero, from the right, is a circle. We call the resulting coordinate system, polar.

Jacobian of (x, y) wrt (r, theta)
	x	y
function	r cos(theta)	r sin(theta)
d/dr	cos(theta)	sin(theta)
d/d-theta	- r sin(theta)	r cos(theta)

The domain of theta usually is taken as the semi-closed interval (- pi, pi[.

The Jacobian matrix is just the 2x2 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((x, y) / (r, theta)) = r^2 ((cos(theta))^2 + (sin(theta))^2) = r^2.

Spherical

The corresponding three-dimensional coordinate system is the spherical.

Jacobian of (x, y, z) wrt (r, theta, phi)
	x	y	z
function	r cos(theta) cos(phi)	r sin(theta) cos(phi)	r sin(phi)
d/dr	cos(theta) cos(phi)	sin(theta) cos(phi)	sin(phi)
d/d-theta	- r sin(theta) cos(phi)	r cos(theta) cos(phi)	0
d/d-phi	- r cos(theta) sin(phi)	- r sin(theta) sin(phi)	r cos(phi)

Ordinarily, we take the domain of theta to be the semi-closed interval (- pi, pi[ and that of phi to be the closed interval [- pi / 2, pi / 2].

The Jacobian matrix is just the 3x3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((x, y, z) / (r, theta, phi)) = r^2 ((cos(theta))^2 (cos(phi))^2 + (sin(theta))^2 (sin(phi))^2 + (cos(theta))^2 (sin(phi))^2 + (sin(theta))^2 (cos(phi))^2) cos(phi) = r^2 cos(phi).

Cylindrical

The limit as c increases without bound is a rectangle, in two dimensions. The corresponding solid of revolution is a cylinder. However, the cylinder also may be considered as a circle, which ha been extruded. Thus, we just adjoin the z to the polar coordinate system, to obtain:

Jacobian of (x, y, z) wrt (r, theta, z)
	x	y	z
function	r cos(theta)	r sin(theta)	z
d/dr	cos(theta)	sin(theta)	0
d/d-theta	- r sin(theta)	r cos(theta)	0
d/dz	0	0	1

The domain of theta is the semi-closed interval (- pi, pi].

The Jacobian matrix is just the 3z3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((x, y, z) / (r, theta. z)) = r ((cos(theta))^2 + (sin(theta))^2) = r.

It also is possible to extrude an ellipse / hyperbola. Then, we obtain the several corresponding cylinders.

As a preliminary, we compute these two Jacobians.

Jacobian of (lambda, mu, z) wrt (u, v, z)
	lambda	mu	z
function	cosh(u)	cos(v)	z
d/du	sinh(u)	0	0
d/dv	0	- sin(v)	0
d/dz	0	0	1

The Jacobian matrix is just the 3z3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((lambda, mu, z) / (u, v. z)) = - sinh(u) sin(v).

Jacobian of (lambda, mu, z) wrt (h, g, z)
	lambda	mu	z
function	g + h	g - h	z
d/dh	1	- 1	0
d/dg	1	1	0
d/dz	0	0	1

The Jacobian matrix is just the 3z3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((lambda, mu, z) / (g, h,. z)) = 1.

ellipsoidal cylinder / hyperboloidal cylinder

Jacobian of (x, y, z) wrt (u, v, z)
	x	y	z
function	c cosh(u) cos(v)	c sinh(u) sin(v)	z
d/du	c sinh(u) cos(v)	c cosh(u) sin(v)	0
d/dv	- c cosh(u) sin(v)	c sinh(u) cos(v)	0
d/dz	0	0	1

The Jacobian matrix is just the 3z3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((x, y, z) / (u, v. z)) =c^2 ((sinh(u))^2 (cos(v))^2 + (cosh(u))^2 (sin(v))^2) = c^2 ((cosh(u))^2 - (cos(v))^2).

J((x, y, z) / (lambda, mu, z)) = J((x, y, z) / (u, v, z)) / J((u, v, z) / (lambda, mu, z)) = c^2 ((cosh(u))^2 - (cos(v))^2) / (- sinh(u) sin(v)) = -+ c^2 (lambda^2 - mu^2) / sqrt((lambda^2 - 1) (1 - mu^2)). (The - sign if the square-root is positive; otherwise, the + sign.)

J((x, y, z) / (h, g, z)) = J((x, y, z) / (lambda, mu, z)) J((lambda, mu, z) / (g, h,. z)) = -+ c^2 (lambda^2 - mu^2) / sqrt((lambda^2 - 1) (1 - mu^2)) * 1 = -+ c^2 ((g + h)^2 - (g - h)^2) / sqrt((g + h)^2 - 1) (1 - (g - h)^2)) = 4 c^2 g h / sqrt((g + h)^2 - 1) (1 - (g - h)^2)). (The - sign if the square-root is positive; otherwise, the + sign.)

circular cylinder

Jacobian of (x, y, z) wrt (r, theta, z)
	x	y	z
function	r cos(theta)	r sin(theta)	z
d/dr	cos(theta)	sin(theta)	0
d/d-theta	- r sin(theta)	r cos(theta)	0
d/dz	0	0	1

The Jacobian matrix is just the 3x3 array in the bottom right-hand corner of the foregoing table. Its determinant, called the Jacobian, becomes

J((x, y, z) / (r, theta. z)) = r ((cos(theta))^2 + (sin(theta))^2) = 2.

Are there any other coordinate systems in use? Aye, forsooth. The cartographers have devised a myriad of coordinate systems to map -- however imperfectly -- a sphere to a plane. Upon request, I might consider describing some.