Synthetic Vector Analysis

DISCLAIMER: This is just a "work in progress" -- very much so -- some of the assertions may be false! Nothing has been checked, as yet. :-)

Synthetic -- versus analytic -- vector analysis has gone out of style half a century ago. Quaternions -- versus vectors -- have gone out of style a century ago, shortly after Willard Gibbs invented vectors in c 1905.. And yet, they have great elegance. Here, we attempt a revival.
In the mid 1950's, this author had employed quaternions to design an antenna, completely mystifying his colleagues. Since both a scalar and a vector potential was involved, a quaternion was more concise than an ordered pair of a scalar and a vector.
Since we make extensive use of the cross-product, which is peculiar to three-dimensional space, much of what we do here does not generalize easily to higher dimensional space.

Synthetic versus analytic. The analytic -- with its emphasis upon a specific coordinate system -- is more popular and admittedly is a little easier at the start, e.g., the proof of the triple cross-product. The synthetic inherently is independent of any coordinate system; thus, does not require any proofs to that effect. It is more concise.

Directional derivative. For any given function of position f=f(R), the product u . del(f) is the directional derivative of f in the direction of the unit vector u. Indeed, this property characterizes the differential vector operator del and, thus, may be taken as its definition. Hence, in the Cartesian coordinate system, it is obvious that the operator del is

del = i d/dx + j d/dy + k d/dz

The operator del, when it is operating -- by scalar multiplication -- upon a scalar is called a gradient, when operating -- by the inner (dot) product multiplication -- upon a vector is called the divergence, when operating -- by the cross-product multiplication -- upon a vector is called the curl. The square of del, that is, the inner-product of del by itself del.del is called the LaPlacian operator.

In the Cartesian coordinate system, it is obvious that the operator del operating upon the radius vector R yields: The divergence del . R = 3, the curl del x R = 0, and del R = ii + jj + kk = I = the identity diadic. The gradient of the square of the radius vector is del R^2 = 2 R. In summary, we have

gradient of R = del R = identity diadic
divergence of R = del . R = 3
curl of R = del x R = 0
LaPlacian = del^2 = del . del

Since none of these four equations contains any reference to the specific coordinate system, each must be true in any coordinate system. Hence, the easiest way of obtaining del in a specific coordinate system is to solve the first equation for del. Then, if desired, check it in second or third equation. Finally, evaluate the fourth equation.

Since the magnitude r of R is defined as r = sqrt(R^2), we have the gradient of r^2, likewise, is del r^2 = 2 R. Then, the gradient of r, itself, is del r = del sqrt(r^2) = R / r; that is, the unit radius vector.

r^2 = R^2
del r^2 = 2 R
del r = R / r

In the following two theorems, the scalar function of position phi = phi(R) is called the scalar potential and the vector function of position A = A(R) is called the vector potential.
A vector function of position F = F(R) is said to be irrotational if del x F = 0 is an identity in R and solenoidal if del . F = 0 is an identity in R.

Theorem irrotational: Iff (= if and only if) F is the gradient of the scalar potential phi = phi(R); then, F is irrotational. That is, if

F = del phi

then

0 = del x F = del x del phi

is an identity in R and conversely.

Theorem solenoidal: Iff F is the curl of a vector potential A = A(R); then, F is solenoidal. That is, if

F = del x A

then

0 = del . F = del . del x A

is an identity in R and conversely.

Proof (for each of these two theorems): In the forward direction, the two del's cancel out. In the backward direction, it is just the Cauchy condition for the existence of the respective potential function.

In general, F is the sum of an irrotational and a solenoidal component. Is it indeed true that any F can be decomposed into just these two components????? In the quaternion notation, we write the quaternion F in terms of its quaternion potential (phi, A) as

F = (0, del) (phi, A) = (del . A, del phi + del x A).

Question: Is del . A indeed identically zero?????

Quaternion, A quaternion is an ordered-pair of a scalar and a vector, subject to the following axioms.
Addition is just the addition of the components. Scalar multiplication likewise is just the scalar multiplication of the components. The product quaternion Q of the two quaternions Q1 and Q2 is defined as

Q = Q1 Q2 = (s1 s2 + v1 . v2, v1 x v2 + s1 v2 + v1 s2).

Alternatively, we may write a quaternion as Q = a + bi + cj + dk, with the multiplication table: i . i = 1, etc.; i . j = k, j . i = - k, etc. While this multiplication table does generalize to higher dimensional space, beyond three-dimensional space, we lose the property that the product quaternion is in the null-space of its factors. Whoops! On Friday 20-th May 2005, in an e-mail to me, Doug pointed out that i . i = - 1. Gramercy, Doug, for demonstrating the notorious fallibility of my memory. :-) At least, I had it correct on another page, where I go further to show that Quaternions constitute a division ring.

The triple cross-product a x (b x c) = (a . c) b - (a . b) c is proven here.

Finally, for the inverse-square central-force, we obtain the gradient of its potential phi = (- 1 / r) as F = del phi = del (- 1 / r) = R / r^3. The field F is irrotational. Of course, any expression with r in the denominator is not defined at the origin; that is, at r = 0.
Consider the vector potential A = - b / r, where b is a constant vector. Then, the curl of A is F = del x A = R x b / r^3. The field F is solenoidal. The proof is left as an exercise for the reader.

Green's theorem:

Stoke's theorem: For a closed curve C, the directional line-integral -- the integral of F . dR -- is equal to the surface integral of the normal component of the curl of F -- the integral of n . del x F. The former integral is performed in the positive direction around the normal to the surface S, while the latter is performed over the surface S bounded by the curve C. Hence, an irrotational F has a null directional line-integral over a closed curve. However, nothing is said about a curve which is not closed.

Flux-divergence theorem: For a closed surface S, the flux -- the integral of n . F, where n is the outward pointing normal to the surface -- is equal to the volume integral of the divergence of F. The former integral is performed over the surface S, while the latter is performed over the volume bounded by the surface S.. Hence, a solenoidal F has a null flux over a closed surface. and conversely only if it additionally is known that the divergence does not change sign. However, nothing is said about a surface which is not closed.

Proof (for each of these three theorems) is by writing the right-hand side as a double- integral; then, observing that the inner-integral is the inverse of the respective del operator. The corollaries are obvious.

Here we make use of a portion of Maxwell's equations. Given that the field F has a null divergence (0 = del . F) throughout the space, except in a to-be-specified bounded region about the origin. The charge q possesses an inverse-square scalar potential.
Lemma: The flux through a sphere of radius r about the origin of a point charge q at the origin is (4 pi q), where q is a constant scalar.
Proof: The outward-pointing unit-normal to the sphere is (R / r). The scalar potential phi of the charge q is phi = - q / r^2. The corresponding field F is its gradient F = del phi = R q / r^3. The area of the sphere is area = 4 pi r^2. Multiply these together, to obtain flux = (R / r) . (R q / r^3) (4 pi r^2) = 4 pi q. QED.
We immediately have the Gauss theorem: The flux through any closed surface enclosing any shape region, with any distribution of the total charge q, is 4 pi q.
A pair of point charges q at b / 2 and -q at -b / 2 constitute a dipole of moment b q, where b is a constant vector. Corollary: The flux through any closed surface enclosing a dipole is zero. Proof follows from the observation that the total charge of a dipole is zero. The field F of a dipole approaches asymptotically F = b q / r^3. Thus, the far-field of a dipole is inverse-third power.

Jacobian. The Jacobian is defined as the determinant of the matrix, whose rows are the derivatives of the components of the new basis-vectors wrt (= with respect to) those of the old. It can be shown that the product of the Jacobian of the forward and inverse transformations is one. A transformation is said to be proper iff (= if and only if) its Jacobian is greater that zero. Any point at which the Jacobian is zero is a singular point of the transformation..

Various coordinate systems. We already have said that del R is the identity diadic. It has to be the same identity diadic in any coordinate system. We may use this property to find the del operator is some specific coordinate systems. In the following, each derivative is the *partial* derivative; but, since we do not have the appropriate glyph to so indicate, we employ the d, instead. The unit vectors in the Cartesian coordinate system are the usual i, j, and k. In each of the other coordinate systems, we employ m, n, and p. Unfortunately, we have no distinguishing type-font for them. Also, please read rho for each occurrence of r.

Chirality. While the two chiralities are distinguishable, there is no absolute way of specifying a particular one. The best one can do, is to state that the usual chirality is that of the dominant hand of the majority of Homo sapiens sapiens. However, once we have adopted a reference item with an arbitrary chirality, we usually require that all other items posses the same chirality; that is, that they are a proper transformation.

Speed and arc-length. The speed v is the magnitude of the velocity V. It also is the derivative of the arc-length s. As an equation, we have

ds/dt = v = |V| = sqrt(v^2).

The arc-length then is the line-integral of v, along a specified path. Often, the resulting quadrature is not elementary and has to be evaluated numerically.

Cartesian coordinates.

i = i, j = j, k = k
i . i = j . j = k . k = 1; i . j = j . k = k . i = 0; i x j = k, j x k = i, k x i = j
displacement = R = i x + j y + k z = i x + j y + k z
d i / dx = d j / dx = d k / dx = d i / dy = d j / dy = d k / dy = d i / dz = d j / dz = d k / dz = 0
del = i d/dx + j d/dy + k d/dz
LaPlacian = del . del = d^2 / dx^2 + d^2 / dy^2 + d^2 / dz^2
Jacobian = J((x, y, z) / (x, y z)) = 1
velocity = V = dR/dt = i dx/dt + j dy/dt + k dz/dt
speed = v = sqrt(V^2) = sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)
acceleration = A = d^2 R / dt^2 = i d^2 x / dt^2 + j d^2 y / dt^2 + k d^2 z / dt^2
on the interval [0, 2 pi], the orthonormal basis is {1/sqrt(2 pi), cos(n x)/sqrt(pi), sin(n x)/sqrt(pi), | n a natural number}

Cylindrical coordinates

m = i cos(theta) + j sin(theta), n = - i sin(theta) + j cos(theta), k = k
m . m = n . n = k . k = 1; m . n = n . k = k . m = 0; m x n = k, n x k = m, k x m = n
displacement = R = r (i cos(theta) + j sin(theta)) + k z = m r + k z
d m / d r = d n / d r = d k / d r = 0; d m / d theta = n, d n / d theta = - m, d k / d theta = 0; d m / d z = d n / d z = d k / d z = 0
del = m d / d r + n (1 / r) d / d theta + k d / d z
LaPlacian = del . del = d^2 / dr^2 + (1 / r) d / dr + (1 / r^2) d^2 / d theta^2 + d^2 / d z^2
Jacobian = J((x, y, z) / (r, theta, z)) = (r dr d-theta dz) / (dx dy dz)
velocity = V = dR/dt = m dr/dt + n r d-theta / dt + k dz/dt
speed = v = sqrt(V^2) = sqrt((dr/dt)^2 + (r d-theta / dt)^2 + (dz/dt)^2)
acceleration = A = d^2 R / dt^2 = m d^2 r / dt^2 + 2 n dr/dt d-theta / dt + n r d^2 theta / dt^2 - m r (d-theta / dt)^2 + k d^2 z / dt^2 =
= m (d^2 r / dt^2 - r (d-theta / dt)^2) + n (r d^2 theta / dt^2 + 2 dr/dt d-theta / dt) + k d^2 z / dt^2
the orthonormal basis are the Bessel functions ????

Special cases: (1) Polar coordinates -- The third coordinate z is identically zero.
(2) Constant angular speed in polar coordinates -- The radius r is a constant. The angle theta = omega t, where the angular-speed omega is a constant.

velocity = V = n r omega; speed = v = |V| = r omega
acceleration = A = - m r omega^2 = - m v^2 / r; magnitude of acceleration = a = |A| = r omega^2 = v^2 / r
frequency = n; omega = 2 pi n; period = T = 1 / n [The n on this line is NOT the unit vector. It is a scalar constant.]

Spherical coordinates

m = (i cos(theta) + j sin(theta)) cos(phi) + k sin(phi), n = - i sin(theta) + j cos(theta)), p = - (i cos(theta) + j sin(theta)) sin(phi) + k cos(phi)
m . m = n . n = p . p = 1; m . n = n . p = p . m = 0; m x n = p, n x p = m, p x m = n
displacement = R = r ((i cos(theta) + j sin(theta)) cos(phi) + k sin(phi)) = m r
i cos(theta) + j sin(theta) = m cos(phi) - p sin(phi), k = m sin(phi) + p cos(phi)
d m / d r = dn / dr = d p / d r = 0;
d m / d theta = n cos(phi), d n / d theta = - m cos(phi) + p sin(phi), d p / d theta = - n sin(phi)
d m / d phi = p, d n / d phi = 0, d p / d phi = - m
del = m d/dr + (n / (r cos(phi))) d/d theta + (p / r) d/d phi
LaPlacian = del . del = d^2 / dr^2 + (1 / (r cos(phi))^2) d^2 / d theta^2 + (p / r)^2 d^2 / d phi^2
Jacobian = J((x, y, z) / (r, theta, phi)) = (r^2 cos(phi) dr d-theta d-phi) / (dx dy dz)
velocity = V = dR/dt = m dr/dt + r (n cos(phi) d theta / dt + p d phi / dt)
speed = v = sqrt(V^2) = sqrt((dr/dt)^2 + (r^2 ((cos(phi) d-theta / dt)^2 + (d-phi / dt)^2))
acceleration = A = d^2 R / dt^2 =
= m d^2 r / dt^2 + r (n cos(phi) d^2 theta / dt^2 + p d^2 phi / dt^2) + (n cos(phi) d theta / dt + p d phi / dt) dr/dt - r n sin(phi) d phi / dt d theta / dt +
+ (n cos(phi) d theta / dt + p d phi / dt) dr/dt +
+ r ((- m cos(phi) d theta / dt) cos(phi) d theta / dt - (n sin(phi) d theta / dt + m d phi / dt) d phi / dt) ???? =
= m (d^2 r / dt^2 - r cos^2(phi) (d theta / dt)^2 + (d phi / dt)^2) +
+ n (r cos(phi) d^2 theta / dr^2 + cos(phi) d theta / dt dr/dt - 2 r sin(phi) d theta / dt d phi / dt + cos(phi) d theta / dt dr/dt) +
+ p (d^2 phi / dt^2 + 2 d phi / dt dr/dt) ?????
the orthonormal basis are the Legender (sp?) polynomials ????

Spheroidal coordinates

This coordinate system deserves a detailed derivation. The general spheroidal-coordinates are defined as proportional to the triple Q

Q = Q(a cos(theta) cos(phi), b sin(theta) cos(phi), c sin(phi))

where each of the constants {a, b, c} is in the open-interval (0, oo). There is no restriction as to their relative magnitudes. The angle theta is the spheroidal-longitude and the angle phi is the spheroidal-latitude. There are three special cases:

a = b > c oblate-spheroidal
a = b = c spherical
a = b < c prolate-spheroidal

Observe that

(x / a)^2 + (y / b)^2 + (z / c)^2 = (cos(theta) cos(phi))^2 + (sin(theta) cos(phi))^2 + (sin(phi))^2 = 1.

Thus justifying the name "spheroidal". The most general ortho-normal three-dimensional matrix is given by the matrix

(cos(theta) cos(phi), sin(theta) cos(phi), sin(phi); - sin(theta), cos(theta), 0; -cos(theta) sin(phi), -sin(theta) sin(phi), cos(phi))

Let the vectors {M, N, P} be defined by

M = (c (i b cos(theta) + j a sin(theta)) cos(phi) + k a b sin(phi)) cos(phi)
N = - i a sin(theta) + j b cos(theta) cos(phi)
P = - (i a cos(theta) + j b sin(theta)) sin(phi) + k c cos(phi)

We have constructed M as M = N x P. where N is the partial-derivative of the aforementioned triple wrt (= with respect to) theta and P is that wrt phi. Eventually, we will need the square of each of these vectors {M, N, P} and of the triple Q.

M^2 = N^2 P^2
N^2 = ((a sin(theta))^2 + (b cos(theta))^2) (cos(phi))^2
P^2 = ((a cos(theta))^2 + (b sin(theta))^2) (sin(phi))^2 + (c cos(phi))^2
Q^2 = (a cos(theta))^2 + (b sin(theta))^2 (cos(phi))^2 + (c sin(phi))^2

It is obvious that these vectors {M, N, P} are orthogonal

M . N = N . P = P . M = 0

Let the corresponding unit vectors be

m = M / |M|
n = N / |N|
p = P / |P|
q = Q / |Q|

Then, they are normal

m . m = n . n = p . p = q . q = 1

Hence, they are ortho-normal

m . n = n . p = p . m = 0

Since for any cyclically-permuted ortho-normal set of three vectors, we have either

m x n = p
n x p = m
p x m = n

xor their negatives, depending upon the chirallity of the set; all that is required is to verify the sign of any one of these cross-products. By the original construction of M as M = N x P, we have the second of the preceding set of equations. Hence, the signs must be correct for each of the three equations. See how clever we are? Without actually performing any of these cross-product multiplications, we have established all three! From its value in Cartesian-coordinates, the displacement vector R is

R = i x + j y + k z = m r = (M / |M|) r,

where (since m is a unit-vector) r is the magnitude of R

r = |R|

Relationship to spherical-coordinates. We may compute the corresponding spherical-coordinates by means of two steps: First, find the Cartesian-coordinates. Second, convert to spherical-coordinates. The definition of Q (the point on the spheroid) and that of R (the local displacement-vector) already provides the Cartesian-coordinates. Then, the spherical-coordinates are

For Q
- spherical-radius = sqrt(Q^2) = sqrt((a cos(theta))^2 + (b sin(theta))^2 (cos(phi))^2 + (c sin(phi))^2)
- spherical-longitude = Atan((b / a) tan(theta))
- spherical-latitude = Atan(c / sqrt((a cos(theta))^2 + (b sin(theta))^2) tan(phi))
For R
- spherical-radius = r [That is how we defined it.]
- spherical-longitude = Atan((a / b) tan(theta))
- spherical-latitude = Atan((a b / (c sqrt((b cos(theta))^2 + (a sin(theta))^2))) tan(phi))

Each of the four inverse-tangents is to be taken in the same quadrant as the corresponding argument of the tangent function therein. Please observe that the internal multiplier in the expression for the longitudes are the reciprocals of each other and that for the latitudes are nearly so.

For any vector U, we have

|U|^2 = U^2 = U . U

Since half its derivative is

|U| d|U|/dt = U . dU/dt

we obtain

d|U|/dt = U . dU/dt / |U| = u . dU/dt

where u is the corresponding unit-vector

u = U / |U|

By the quotient-rule, its derivative is

du/dt = (|U| dU/dt - U d|U|/dt) / |U|^2 = (dU/dt - u d|U|/dt) / |U| = (U^2 dU/dt - U (|U| d|U|/dt)) / |U|^3

If U is a function of {theta, phi}; then, we have

dU/dt = dU/d-theta d-theta/dt + dU/d-phi d-phi/dt.

Substitution yields

d|U|/dt = u . (dU/d-theta d-theta/dt + dU/d-phi d-phi/dt)

Further simplification is interesting; but, not practical:

du/dt = ((dU/d-theta d-theta/dt + dU/d-phi d-phi/dt) - u u . (dU/d-theta d-theta/dt + dU/d-phi d-phi/dt)) / |U| =
= (dU/d-theta d-theta/dt + dU/d-phi d-phi/dt) / |U| - u u . ((dU/d-theta d-theta/dt + dU/d-phi d-phi/dt) / |U|) =
= w - u u . w = (U^2 (w |U|) - U U . (w |U|)) / |U|^3

where, for brevity, we have defined the vector w as

w = (dU/d-theta d-theta/dt + dU/d-phi d-phi/dt) / |U|

For the vector M, we have

dM/dr = dm/dr = 0
dM/d-theta = (- i a sin(theta) + j b cos(theta)) (cos(phi))^2 = N cos(phi)
dM/d-phi = - (i a cos(theta) + j b sin(theta)) sin(phi) + k c cos(phi) = P
dM/dt = N cos(phi)) d-theta/dt + P d-phi/dt
|M| d|M|/dt = M . dM/dt = M . (0 dr/dt + N d-theta/dt + P d-phi/dt) = 0

Hence, the derivative of m wrt t is

dm/dt = dM/dt / |M| = (N d-theta/dt + P d-phi/dt) / sqrt(N^2 P^2)

The velocity V and speed v are

velocity = V = dR/dt = d(m r)/dt = m dr/dt + r dm/dt = m dr/dt + r ((N d-theta/dt + P d-phi/dt) / sqrt(N^2 P^2))
speed = v = sqrt(V^2) = sqrt((m dr/dt)^2 + r^2 (N^2 (d-theta/dt)^2 + P^2 (d-phi/dt)^2) / (N^2 P^2))) = sqrt((m dr/dt)^2 + r^2 ((d-theta/dt)^2 / P^2 + (d-phi/dt)^2 / N^2))

For the vector N, we have

dN/dr = dn/dr = 0
dN/d-theta = - (i a cos(theta) + j b sin(theta)) cos(phi)
dN/d-phi = (- i a sin(theta) + j b cos(theta)) sin(phi)
dN/dt = - (i a cos(theta) + j b sin(theta)) cos(phi) d-theta/dt + (- i a sin(theta) + j b cos(theta)) sin(phi) d-phi/dt
|N| d|N|/dt = N . dN/dt = (- i a sin(theta) + j b cos(theta) cos(phi)) . (- (i a cos(theta) + j b sin(theta)) cos(phi) d-theta/dt + (- i a sin(theta) + j b cos(theta)) sin(phi) d-phi/dt) =
= (a^2 + b^2) sin(theta) cos(theta) (cos(phi))^2 d-theta/dt + ((a sin(theta))^2 + (b cos(theta))^2) sin(phi) cos(phi) d-phi/dt

Hence, the derivative of n wrt t is

dn/dt = (N^2 (dN/dt) - N (|N| d|N|/dt)) / |N|^3 =
= ((((a sin(theta))^2 + (b cos(theta))^2) (cos(phi))^2) () -
- () ((a^2 + b^2) sin(theta) cos(theta) (cos(phi))^2 d-theta/dt + ((a sin(theta))^2 + (b cos(theta))^2) sin(phi) cos(phi) d-phi/dt)) /
/ (((a sin(theta))^2 + (b cos(theta))^2) (cos(phi))^2)^3/2
=

For the vector P, we have

dP/dr = dp/dr = 0
dP/d-theta =
dP/d-phi =
dP/dt =
|P| d|P|/dt = P . dP/dt =
=

Hence, the derivative of p wrt t is

dp/dt = (P^2 (dP/dt) - P (|P| d|P|/dt)) / |P|^3 =
=

For any vector U, we may resolve it in terms of our ortho-normal basis {m, n, p} as

U = m m . U + n n . U + p p . U = M M . U / M^2 + N N . U / N^2 + P P . U / P^2

We begin by resolving each of the unit-vectors {i, j, k}.

i = M M . i / M^2 + N N . i / N^2 + P P . i / P^2 = M b c cos(theta) (cos(phi))^2 / (N^2 P^2) - N a sin(theta) cos(phi) / N^2 - P a cos(theta) sin(phi) / P^2
j = M M . j / M^2 + N N . j / N^2 + P P . j / P^2 = M a c sin(theta) (cos(phi))^2 / (N^2 P^2) + N b cos(theta) cos(phi) / N^2 - P b sin(theta) sin(phi) / P^2
k = M M . k / M^2 + N N . k / N^2 + P P . k / P^2 = M a b sin(phi) cos(phi) / (N^2 P^2) + c cos(phi) / P^2

Second, we compute certain sums

i a cos(theta) + j b sin(theta) = M a b c (cos(phi))^2 / (N^2 P^2) - N (a^2 - b^2) sin(theta) cos(theta) cos(phi) / N^2 - P ((a cos(theta))^2 + (b sin(theta))^2) sin(phi) / P^2
- i a sin(theta) + j b cos(theta) = 2 M a b c sin(theta) cos(theta) (cos(phi))^2 / (N^2 P^2) - N ((a sin(theta))^2 - (b cos(theta))^2) cos(phi) / N^2 - P (a^2 - b^2) sin(theta) cos(theta) sin(phi) / P^2

This is as far as I consider it to be feasible to go by hand. In the Autumn, I expect to have access to the Maple symbolic-mathematics program. I will verify the foregoing derivations and carry the computations further.

Central force
To make it easier to write the expressions, for the purpose of this discussion, we take the origin at the center. However, in general, they would not coincide.
Definition of central force: A force F is said to be central iff (= if and only if) it is radial and dependent only upon an integrable function f() of the distance from the center of force, whose domain need not include this center. That is, F = f(r) R / r.
Definition of a line integral: A line integral is an integral along a directed curve. This line certainly need not be straight. The glyph for a line integral is a letter 'c' superimposed upon the glyph for an integral. If the curve is closed, we employ the glyph of a letter 'o' superimposed upon the glyph for an integral.
Definition of a work integral: A work integral is a line integral of the inner-product of a force and the differential of the radius vector, namely, the line-integral of F . dR.
Definition of a conservative force: A force is said to be conservative iff the work-integral around any closed curve in its domain is zero.
Theorem: A central force is conservative. Proof: Step one. Employ spherical coordinates. Consider a circle of non-zero radius, centered upon the origin. It is obvious that the work-integral around this circle is zero. Step two. It is obvious that the work-integral around any closed curve, which does not surround the origin is zero. Combining these two results, we have that the work-integral around any closed curve is zero. QED.
Corollary: The inverse-square force F = k R / r^3 is conservative. Proof: This is a special case of the theorem.
Theorem: The superimposition of any finite amount of central forces -- each with its own center -- is a conservative force. Proof follows immediately from the fact that the Lebesque-integral is linearly-additive.