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Mathematical Methods for Physicists and Engineers
By Royal Eugene Collins Dover Publications, Inc.
Copyright © 1999 Royal Eugene Collins
All rights reserved.
ISBN: 978-0-486-15012-3
CHAPTER 1
Elementary vector calculus; the vector field
Here we present a brief but essentially complete presentation of those elements of vector calculus essential for the physicist or engineer. A knowledge of the elementary algebra of vectors is assumed.
THE POSITION VECTOR AND THE LINE ELEMENT; COORDINATE TRANSFORMATIONS
We consider a point in space defined with reference to a set of rectangular cartesian coordinates by the three numbers, x1, x2, x3, which are the coordinates of the point, or the orthogonal projections of the point, on the three axes. The directed line from the origin to the point is the position vector of the point. We define the base set of unit vectors on these axes as 11, 12, 13 where
(1.1) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Then we write,
(1.2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
as the representation of the position vector on these axes.
Now the differential displacement vector is just given by,
(1.3) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
since these unit vectors, 1i, i = 1, 2, 3, have the same magnitude and direction at all points of space, i.e., dr is constructed as,
(1.4) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
or
(1.5) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
but, since for any two points we have 1'i = 1i, the unit vectors factor out and Eq. (1.3) results.
The magnitude of dr is the line element, ds. This can be constructed as
(1.6) ds = |dr| = (dr · dr)1/2
or
(1.7) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
by virtue of Eq. (1.1).
Now we do not always use rectangular orthogonal axes. Thus we now examine what happens to dr and ds under some arbitrary change of variables,
(1.8) xi = xi(q1, q2, q3), [??] = 1, 2, 3
Thus,
(1.9) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and these can be inserted into Eq. (1.3) for the dx 's and the order of summation interchanged to give,
(1.10) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
In these equations the dq's may have any dimensions, or units, but the x's must all have the same dimensions in order to maintain dimensional homogeniety of our equations. We also note that the quantities in parenthesis in Eq. (1.10) are vectors in the rectangular space, but having components, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], which are also not dimensionally homogeneous. Thus now multiply and divide each term here by a factor, hj, so that we have,
(1.11) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and require that the hj be chosen so that the factors [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] have the same dimensions as the xi.
Furthermore we now note that the vectors forming each parenthesis above, denoted now as,
(1.12) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
are dimensionless and will be unit vectors, if we require,
(1.13) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
for each j. This condition then determines the hj in terms of the coordinate transformation by the relations,
(1.14) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
so the hj are fixed except for sign by these equations.
We now have the expression for dr in the form,
(1.15) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and here let us again look at the square of the line element, ds2. We have,
(1.16) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
This is distinctly different from our Eq. (1.7) for the rectangular system but it reduces to a very similar form if we require,
(1.17) 1'j · 1'k = 0, j ≠ k
This states that these new unit vectors are mutually orthogonal, or perpendicular. Hence for any orthogonal coordinate system (i.e., Eq. (1.17) satisfied) we have,
(1.18) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and it is just such systems of coordinates that we will nearly always employ.
Note that the criterion of orthogonality,Eq. (1.17), has the form,
(1.19) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
if we make use of Eqs. (1.12) which define the 1'j vectors.
SCALE FACTORS; AREAS AND VOLUMES
The elements hj, j = 1, 2, 3, introduced above are called the scale factors of the "curvilinear" coordinate system, qj. Here we note how area elements, and volume elements, are expressed in terms of these hj and the dqj.
Consider Eq. (1.15) for dr and choose dq2 = dq3 = 0, then we have a vector displacement we will call dr1 given by
(1.20) dr1 = 1'1h1 dq1
This is in the direction of 1'1, for dq1 > 0, and is a line element of length h1dq1, from the definition of dr. Similarly if at the same point in space we let dq1 = dq3 = 0 we form another line element,
(1.21) dr2 = 1'2h2dq2
in the direction of 1'2, dq2 > 0, and of length, h2dq2.
The vector product, or cross product, is,
(1.22) dr1 × dr2 = (1'1 × 1'2)h1 h2 dq1 dq2 = dA3
which in magnitude is the area of the parallelogram indicated in Fig. 1-1, with
(1.23) |1'1 × 1'2| = sin θ
Obviously if the coordinate system is orthogonal, sin θ = 1. Also we note that the direction of this vector element of area is in the direction of 1'3, i.e., from our definitions above we have the cyclic rule
(1.24) 1'i × 1'j = 1'k (i, j, k in cyclic order)
just as for the original unit vectors.
If we now form the scalar, or dot, product of a line element,
(1.25) dr3 = 1'3h3dq3
with dA3 we obtain,
(1.26) dτ = (1'1 × 1'2 · 1'3) h1h2h3 dq1dq2dq3
as the volume of the elemental parallelopiped. Obviously, for orthogonal systems this is just
(1.27) dτ = h1h2h3 dq1dq2dq3
and we will restrict our consideration to such systems. Here h1h2h3 is the Jacobian of the transformation from the xi to the qj.
INVERSE TRANSFORMATION
We call attention to the fact that we generally require that it be possible to carry out the inverse transformation,
(1.28) qj = qj(x1, x2, x3), j = 1, 2, 3
so that,
(1.29) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Substitution of these expressions for the dqj in Eq. (1.9) shows that the necessary conditions for this inverse to exist are,
(1.30) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Also, in order that we should be able to solve Eq. (1.9) for the dqj as functions of the dxi, it is required that the determinant of the coefficients should not vanish, i.e.,
(1.31) Det([partial derivative]xi/[partial derivative]qj) ≠ 0
These ideas become more concrete when we consider an example.
Example (1)
We here examine the spherical coordinate system in terms of the general ideas outlined above. The equations of transformation from the rectangular system, now denoted as x, y, z, to the spherical system, r, θ, φ, I, are:
(1.32) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
as the equivalent of Eq. (1.8).
The scale factors are
(1.33) h2r = ([partial derivative]x/[partial derivative])2 + ([partial derivative]y/[partial derivative]r)2 + ([partial derivative]z/[partial derivative]r)2
(1.34) h2θ = ([partial derivative]x/[partial derivative]θ)2 + ([partial derivative]y/[partial derivative]θ)2 + ([partial derivative]z/[partial derivative]θ)2
and
(1.35) h2φ = ([partial derivative]x/[partial derivative]φ)2 + ([partial derivative]y/[partial derivative]φ)2 + ([partial derivative]z/[partial derivative]φ)2
which reduce to,
(1.36) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
if we take the positive square roots.
Next we verify that the r, θ, I system is an orthogonal system, i.e.,
(1.37) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and similarly with (r, φ), (r, φ) replacing (r, φ) here. This does indeed prove to be true for all these combinations as required by Eq. (1.14).
Having constructed the scale factors and shown that the r, θ, φ system is an orthogonal system we know from our discussion above that we can form a unit vector 1r in the direction of increasing r at any point in space, and similar vectors, 1θ and 1I, in the directions of increasing θ and φ respectively, at this same point. These, 1r, 1θ, 1I form a local orthogonal rectangular set of axes, i.e., as depicted in Fig. 1-2.
However, as can be seen from the equations defining these unit vectors, 1r for example,
(1.38) 1r = 1x sin θ cos φ + 1y sin θ sin φ + 1z cos θ
formed as prescribed by Eq. (1.12), is not a constant vector. That is, although its length is always unity, its direction varies from point to point in space. For example we can differentiate 1r with respect to θ or φ and the result is not zero, whereas all derivatives of 1x, 1y, and 1z are zero.
From our general formulation we see that, for example,
(1.39) 1θ × 1φr2 sin θ dθ dφ = 1rhθhφ dθdφ
is an element of area oriented in space such that the vector r from the origin is orthogonal to this area, i.e., this is the magnitude of the element of area on a spherical surface of radius r at the point r, θ, φ, and directed outward from the surface.
EXECUTE PROBLEM SET ([1-1)
GRADIENT OF A SCALAR
We consider some continuous function, F(gj), of a set of orthogonal coordinates, qj, j = 1, 2, 3, having continuous derivatives; at least continuous first derivatives. The total differential of this function is given by,
(1.40) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and represents the difference in the values of F at the two points [(q1 + dq1), (q2 + dq2), (q3 + dq3)] and (q1, q2, q3).
Note that if we multiply and divide the first term by h1, the second by h2, and the third by h3 this appears as,
(1.41) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and this has the form of the scalar, or dot product, of the displacement vector, dr, having components hi dqi, with another vector,
(1.42) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
which we call the gradient of F, or just "grad F". Thus we see
(1.43) f(r + dr) - F(r) = [nabla]F · dr = dF
in any orthogonal coordinate system, with [nabla]F defined by Eq. (1.42).
Note that the magnitude of the gradient, |[nabla]F|, is
(1.44) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence we can write Eq. (1.43) above as,
(1.45) dF = |[nabla]F| ds cos γ
since ds = |dr| and we here define γ as the angle between the direction of [nabla]F and dr. This follows from the definition of the scalar product. Since |[nabla]F|, being just a function of the qj, has a fixed value at any point in space, and we can choose some fixed magnitude for the displacement ds, we ask in what direction must we make the displacement in order to achieve maximum change in F, i.e., dF = maximum. Quite obviously this occurs for γ = 0, or cos γ = 1. Thus the gradient vector is always directed in the direction of the maximum space rate of change of F.
Also observe that dF = 0 corresponds to γ = π/2, but since dF = 0 defines the surface of constant F, we see that the vector [nabla]F is always at right angles to the surfaces of constant F. This is indicated in Fig. 1-3.
EXECUTE PROBLEM SET (1-2)
SURFACE INTEGRALS IN VECTOR FIELDS
Just above we saw that we could form a vector, [nabla]F, from some scalar function, F(qj) in a certain way. The components of this vector were each a function of the coordinates. We call such a vector a vector field. A vector field assigns, at each point of space, a vector, say V(r), whose direction and magnitude will in general be different at each point, r, of this space.
(Continues...)
Excerpted from Mathematical Methods for Physicists and Engineers by Royal Eugene Collins. Copyright © 1999 Royal Eugene Collins. Excerpted by permission of Dover Publications, Inc..
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