Theoretical Physics

By Georg Joos and Ira M. Freeman

Among the finest, most comprehensive treatments of theoretical physics ever written, this classic volume comprises a superb introduction to the main branches of the discipline and offers solid grounding for further research in a variety of fields. Students will find no better one-volume coverage of so many essential topics; moreover, since its first publication, the book has been substantially revised and updated with additional material on Bessel functions, spherical harmonics, superconductivity, elastomers, and other subjects.
The first four chapters review mathematical topics needed by theoretical and experimental physicists (vector analysis, mathematical representation of periodic phenomena, theory of vibrations and waves, theory of functions of a complex variable, the calculus of variations, and more). This material is followed by exhaustive coverage of mechanics (including elasticity and fluid mechanics, as well as relativistic mechanics), a highly detailed treatment of electromagnetic theory, and thorough discussions of thermodynamics, kinetic theory and statistical mechanics, quantum mechanics and nuclear physics.
Now available for the first time in paperback, this wide-ranging overview also contains an extensive 40-page appendix which provides detailed solutions to the numerous exercises included throughout the text. Although first published over 50 years ago, the book remains a solid, comprehensive survey, so well written and carefully planned that undergraduates as well as graduate students of theoretical and experimental physics will find it an indispensable reference they will turn to again and again.

• Language: English
• Publisher: Dover Publications
• Release date: Apr 22, 2013
• ISBN: 9780486318530

Book preview

shorthand.

CHAPTER I

VECTOR ANALYSIS

1. The Concept of a Vector.

Besides such magnitudes as temperature, mass, and so on, which are characterized by the assignment of a single number, and which are called scalars, there also occur in physics quantities which are not completely defined by the specification of a single number. The most important group of these is characterized by the fact that, besides the magnitude, the direction of the quantity must also be given. If, for example, we prescribe that an object, located by means of its centre of gravity P, be given a displacement of 2 cm., then the point P′ occupied by the object after displacement may be anywhere on the surface of a sphere of radius 2 cm. with centre at P. The new position of the centre of gravity will not be determined uniquely unless we specify also the direction of the displacement in some manner. Two displacements of a point may be combined into a single equivalent displacement by a simple method (fig. 1). Instead of moving the point P to P′ in a certain direction, and then displacing it in a new direction to P″, it might have been moved directly along the third side of the triangle PP′P″ to P″. Since this process may be repeated, we can use it to characterize a displacement in still another way. We place P at the origin of a rectangular co-ordinate system, and move the point first along the x-axis to P1, a distance a, then parallel to the y-axis a distance b, to P2, and finally from here parallel to the z-axis to P′, a distance c (fig. 2). Instead of this, we might have moved P at once in the direction PP′ a distance

Fig. 1

Fig. 2

We see, then, that the displacement of P is uniquely determined by specifying the three numbers a, b and c, which are called components of the displacement.

A large number of physical quantities follow the same law of combination as the displacement of a point, and may, therefore, also be defined by three numbers. Such quantities are called vectors. We may represent a vector diagrammatically by means of a directed line segment (arrow), whose length, on any convenient scale, is numerically equal to the magnitude of the physical quantity, and whose direction is that of the physical quantity. This geometric representation gives vector quantities an important advantage over other non-scalar quantities arising in physics, e.g. tensors, for which no such simple model exists.

Two vectors are said to be equal if they are identical with respect to both direction and magnitude. Since a pure displacement, i.e. a displacement without rotation, does not alter the direction, two vectors may still be equal, even when they lie in different (but parallel) lines. Physical quantities which are altered by pure displacement are therefore not immediately representable by vectors. To this class belongs, for example, the force acting on a rigid body capable of rotating about an axis: parallel displacement alters the lever arm of the force, and thus changes its effect. It is to be noted, further, that not every physical quantity which can be represented uniquely by a directed line segment may be treated as a vector; it is also necessary to investigate whether this quantity follows the Law of Composition obeyed by the displacement of a point. For instance, the rotation of a rigid body about an axis may be represented by an arrow whose direction is that of the axis of rotation, and whose length in centimetres is numerically equal to the rotation in degrees. Moreover, the direction of the arrow may be specified so that the rotation is clockwise when sighting in this direction. But we do not obtain a third rotation equivalent to the joint effect of two rotations by combining, as above, their representative arrows; the vectorial composition may be shown to be valid only for infinitesimal rotations.

In this book, vectors will be designated by heavy letters (bold-face or Clarendon type).

2. Addition and Subtraction of Vectors; Multiplication of a Vector by a Scalar.

The composition of two displacements of a point, described in § 1 (p. 7), has, as will be shown at once, all the characteristics of a summation. In general, we define, therefore, as the sum of two vectors A and B, a vector C obtained by laying off the vector B from the terminal point of A and then drawing a vector from the initial point of A to the end point of B (fig. 3). This combination is written as a vector equation.

A vector equation is equivalent to three scalar equations, since, as was shown in § 1 (p. 8), a vector is determined by specifying its three components, and two vectors are equal only if the three components are respectively equal. For if any vector is constructed as the combination of three vectors respectively equal to its components, two equal vectors can be obtained only if their corresponding components are identical.

Fig. 3

This geometric sum is commutative, like an ordinary sum; i.e. the value is independent of the order in which the summation is performed. As is seen from fig. 3, we obtain the same vector, apart from the immaterial parallel displacement, by laying off A from the end point of B and drawing a line from the initial point of B to the terminus of A. Further, a vector sum is associative; i.e. in the case of a sum of several vectors, the individual terms may be grouped in any arbitrary manner:

Fig. 4

The proof of this is evident from fig. 4.

The sum of two vectors having the same direction and sense is obtained by addition of their lengths, the direction remaining the same. From this follows at once the definition of the product of a vector and a pure number. According to the meaning of multiplication, mA means the sum of m terms, each one being A, i.e. a vector having the same direction, but m times as long. This leads to an important representation for a vector: if the direction is specified by a vector of length 1, any vector having this direction may be represented by multiplying this unit vector e by the magnitude of A, which may be designated by the corresponding Roman letter, or by | A |:

If we think of a vector as being compounded from three vectors having the directions of the axes of a rectangular co-ordinate system, and if we denote the lengths of these vectors (which we shall call the rectangular components of A), by Ax, Ay, Az and if i, j, k are the unit vectors in the directions of the axes, then the original vector is given, in terms of its components, as

The difference of the vectors C and A is a vector B which when added to A gives the vector C. As is seen from fig. 3 (p. 9), we obtain B by laying off A, with its direction reversed, from the terminus of C, and then connecting the initial point of C with the end point of A. From this it follows at once that the meaning of the multiplication of a vector by (−1) is merely reversal of direction, for, retaining the formal laws of ordinary arithmetic, we may look upon the subtraction of the vector A as the addition of a vector −A.

Ex. 1. Express, by means of an equation, the fact that three vectors A, B, and C form a closed triangle, all three vectors describing the perimeter in the same direction.

Ex. 2. Express each of the following by an equation:

(a) Two vectors A and B are parallel.

(b) Three vectors A, B and C are coplanar.

3. The Scalar Product of two Vectors.

In physics there arise certain combinations of vectors which possess the more important properties of products. We define as the scalar product of two vectors, a number (scalar) equal to the product of the magnitudes of the two vectors multiplied by the cosine of the angle included between them. The scalar product is positive if the included angle is acute, and negative if the angle is obtuse. Then, if we agree to indicate the scalar product by writing the two vectors alongside each other * the definition of the scalar product reads

where (AB) is the angle included between the two vectors. Since A cos(AB) is equal to the projection of the vector A upon the direction of B, the scalar product may also be defined as the product of the magnitude of one vector by the projection of the other upon it. A simple expression follows for this projection of a vector A upon a direction given by the unit vector e: it is merely the scalar product Ae.

The scalar product has the property of commutativity:

for, in the defining equation (5) (p. 10), the order of the factors is immaterial.

The scalar product possesses, further, the most important property of an ordinary number product—that of being distributive with respect to addition. This is expressed by the equation

It may be seen from fig. 5, that the projection, on the direction of A, of the sum of the three vectors is the same as the sum of the projections of the vectors, whence, on account of the way the scalar product was defined, equation (7) is verified.

However, the scalar product of two vectors differs from the product of two numbers in one respect. The latter vanishes only if one or both of the factors are zero; but the scalar product vanishes also when cos(AB) is zero, i.e. if the two vectors are perpendicular. Thus, for the three unit vectors of a rectangular co-ordinate system, denoted as before by i, j, k, we have

Fig. 5

The scalar product of a given vector by itself is termed the square of the vector. Since cos(AB) = 1 in this case, the result is simply the square of the magnitude of the vector:

and, reciprocally, the magnitude may be written

Thus, for the unit vectors,

If the rectangular components of two vectors are given,

we obtain, on account of the distributive property, by multiplying out in the usual manner,

which, by equations (8) and (11), reduces to

Thus the scalar product of two vectors is equal to the sum of the products of the corresponding components.

In rectangular co-ordinates, the components are identical with the projections upon the axes, so that

It is immediately evident from this result that if C is the vector sum of A and B, then each component of C is equal to the sum of the corresponding components; for if equation (1) be multiplied by i there results immediately

Equation (13) is no longer true for oblique co-ordinates, as may be seen from fig. 6, for the two-dimensional case. If we regard as components of A the partial vectors in the directions of the axes (from which A may be obtained by vector addition), then it is evident that these partial vectors are not equal to the projections upon the axes. But the terminus of A may be determined uniquely by specifying the projections upon the axes, as well as by giving the partial vectors. The former are, therefore, in a sense, also components of A. Because of their different behaviour with respect to linear transformations, the components of the first kind, which yield A by vector addition, are called the contravariant components, and the projections are called the covariant components of the vector. This distinction is important in cases where the use of oblique coordinates is essential, as in the Generalized Theory of Relativity.

Fig. 6

Ex. 3. Give the geometric significance of

Ex. 4. What is the meaning of (A + B) (A − B) for the case where A² = B² ?

Ex. 5. Calculate the angle between the two vectors

4. The Vector Product of two Vectors. The Directed Plane Area as a Vector.

Besides the scalar product, there is another equally important product-like combination of two vectors, which is itself a vector. The vector product of two vectors A and B is defined as a vector P which is perpendicular to the plane determined by A and B, and whose magnitude is equal to the area of the parallelogram formed by A and B, i.e. equal to AB sin(AB). The sign (sense) of P is so determined that, sighting along P, the shortest rotation from A (the first factor), toward B (the second factor), is clockwise (fig. 7).This stipulation denies the vector product the property of commutativity possessed by the scalar product; for if we reverse the order of the factors, then the turning of B toward A must be clockwise when looking in the direction of P, which means that P is now reversed. The vector product is indicated by writing the two vectors alongside each other, and enclosing them in square brackets.* Thus

Fig. 7

The magnitude of the product vector is

The most important property of the vector product, without which the use of the term product would not be justified, is that of distributivity with respect to addition. That is,

The proof of this formula is somewhat cumbersome, but not difficult, and will not be given here.

Just as the scalar product of two non-vanishing vectors becomes zero when the two vectors are perpendicular, so the vector product vanishes if the factors are parallel. The equation [AB] = 0 expresses the fact that A and B are parallel, provided that A and B differ from zero. In the same way, [AA] = 0. For the unit vectors along the axes,

Using these relationships, the application of the distributive property gives the following representation of the vector product in terms of the components:

This may be written as a determinant:

The change in sign of the vector product with reversal of the order of the factors manifests itself here as the well-known law that the interchange of two rows changes the sign of the determinant.

The specification of the product vector P gives the following information concerning the parallelogram formed by the two vectors A and B: (1) its position in space (the plane perpendicular to P); (2) the relative position of the sides A and B with respect to one another; (3) the area, numerically equal to the length of P. The form of the parallelogram is not given by P. In looking upon the form as being unimportant, we are recognizing the fact that every portion of a plane may be represented by a vector whose direction and magnitude are uniquely determined, according to the above rules, by the orientation, direction of description, and area of the figure.

For the vectors Si representing the faces of a closed polyhedron, the following theorem holds: If the representative vector Si is assigned to each face in such manner that it is directed towards the outside of the solid in every case, then the sum of these vectors vanishes:

This theorem is easily verified for a tetrahedron. If we denote the three vectors emanating from the vertex (fig. 8) by A, B and C, then the three faces passing through this vertex are [AB], [BC] and [CA]. The base is represented by [(C — A) (B — A)]. If this is expanded, and the resulting vector products added to the above three, the result is zero. Now any polyhedron may be subdivided into a number of tetrahedra. The above theorem holds for each of these component tetrahedra, and every surface inserted by the subdivision in the interior of the solid enters into the calculation twice, with oppositely directed normals, so that in forming the sum, the extra surfaces formed by cutting the original solid into tetrahedra cancel out, leaving only the sum over the external surfaces. We can go a step further: if we have an arbitrary closed curved surface, we can approximate to it by a polyhedron whose faces are portions of the tangent planes. In the limit, when the faces dS become vanishingly small, the polyhedron and the actual surface coincide, and the sum passes into an integral

Fig. 8

indicates integration over a closed region. A closed surface of this kind is often termed a shell ; this type of integral, which may be termed a shell integral , will be met with often in vector analysis.

In spite of the fact that we can assign the product vector P uniquely to the parallelogram formed by two vectors, there is a certain difference between the parallelogram and its representative vector. If, namely, the parallelogram be reflected in its own plane, the direction of one side with respect to the other remains unaltered, but the sign of the product vector changes. This fundamental difference between vectors like displacements (polar vectors) and those which represent a direction of rotation (axial vectors) comes out clearly when we change from a right-handed to a left-handed system of co-ordinates. But since such a change is generally unnecessary and superfluous, this difference is of no consequence in actual calculation. On the other hand, in the four-dimensional vector analysis of the Theory of Relativity, the distinction is of essential importance; in that case a polar vector has four components, while an axial vector has six.

Ex. 6. Verify (16), using an expression containing only two terms.

Ex. 7. What is the value of [AB]² + (AB)² ?

and specify directions by unit vectors, instead of using Cartesian co-ordinates. In this manner give:

(aand whose distance from the origin is p,

(b0 from this plane,

(c3.

5. Multiple Products.

(a) Product of a vector with the scalar product of two other vectors

Since a scalar product is an ordinary number, the product of a vector A with the scalar product BC means the multiplication of A with this number, according to the rule given in § 2 (p. 9), i.e. a vector having the direction of A. This is written A(BC) or (BC) A. It is to be noticed that (AB)C has an entirely different meaning from A(BC), viz. the former is a vector in the direction of C.

(b) Scalar product of a vector with the vector product of two other vectors

If P = [AB], then the product CP = C[AB] is the volume of the parallelepiped having the three vectors A, B, and C as contiguous edges, for the magnitude of P is equal to the area of the parallelogram formed by A and B (fig. 9). Thus

is the volume of the parallelepiped, since C cos (CP) is equal to its altitude. One might equally well consider the base of the parallelepiped to be formed by the vectors C and A, and then form the scalar product of this quantity with B. This means that the volume can also be represented by the product [CA]B. Because of this property, the square brackets are unnecessary, and from now on such a triple product will be denoted simply by writing the three factors alongside each other. However, the sign of the vector product changes each time the cyclic order of the factors is changed. The effect of interchanging two factors is seen from fig. 9. [BA] is a vector directed downward, and hence cos (CP) becomes negative. We may set up the following rule for the sign of the triple scalar product: The product ABC is positive if the three vectors are relatively arranged like the axes of a right-handed co-ordinate system. Thus

Fig. 9

If the components of the three vectors are given, then

This may be written as a determinant:

Equations (22) thus express the well-known property of determinants according to which the interchange of two rows causes a change in algebraic sign.

Fig. 10

(c) Vector product of a vector with the vector product of two other vectors

If P = [BC], then the vector product

signifies a vector lying in the plane of B and C, for P = [BC] is perpendicular to this plane, but R = [AP] is in turn perpendicular to P, and so falls in the plane of B and C (fig. 10).

Since any vector V lying in the plane of B and C may be written in the form

by suitably choosing the scalar multipliers α and β, R must have this form, and it remains to determine the numbers α and β. To this end, we introduce a co-ordinate system in such manner that the x-axis is in the direction of C and the y-axis is in the plane of B and C. This may be done without restricting the generality of the discussion. We then have

The vector product [BC] then assumes the simple form

If the vector AxBxCx i be added and subtracted, there results

This is equivalent to the important formula

The coefficients α and β are thus the scalar products AC and — AB.

With the aid of this formula, complex products may be converted to simpler forms. Take, for example, the scalar product of two vector products:

We put [AB] = E and obtain, by (22) (p. 16)

and then, using (24),

Ex. 9. Transform [[AB][CD]].

6. Differentiation of a Vector with respect to a Scalar; Application to the Theory of Space Curves.

be a continuous function of a continuous scalar variable u:

If the variable u is increased by Δu, the vector will change by an amount

as the limit

Derivatives of a higher order are defined similarly, e.g.:

of the points on a space curve be given as a function of the length of arc sis identical with Δs/Δs is a vector of length 1, having the direction of the tangent to the curve (fig. 11). This unit tangent vector is denoted by t, and we have

Fig. 11

The rules for the differentiation of products also correspond to those for ordinary scalar functions, as may be seen readily by writing the derivatives as limits. In the case of the vector product, however, the order of the factors is important. We thus have

but

= const.,

Since neither dv/du nor v is to vanish, this means that the derivative of a vector of constant length is perpendicular to the vector. This is also evident geometrically, for if the length is constant, the end point of the vector is restricted to move on a sphere. If the increment is infinitesimal, it is tangent to the sphere, and hence is perpendicular to the vector itself.

This result may be applied to the unit tangent t. Since this vector is always of unit length, its derivative must always be perpendicular to t, and so must be a vector in the normal plane to the curve. But this derivative, being the vector difference of two consecutive tangent vectors, must lie in the osculating plane formed by the latter, and so its direction is that of the principal normal, which direction we designate by the unit vector n. In order to calculate the magnitude of the vector dt/ds, we note that the curve, in the neighbourhood of two consecutive tangents, corresponding to three neighbouring points, may be replaced by the circle of curvature, whose centre M is determined by the intersection of the perpendiculars to the two consecutive tangents (fig. 12). The angle dϕ between these tangents is the same as that between the two radii of the circle of curvature. If ρ is the radius of this circle then, in the limit,

Fig. 12

On the other hand, | dt | = dϕ, whence

Thus is obtained the important equation

The value of the curvature is obtained from this result by squaring (scalar product) and extracting the root:

The unit vector b perpendicular to the osculating plane, gives the direction of the binormal. If we stipulate that the three vectors t, and b, in this order, form a right-handed triad, then

For a plane curve, the osculating plane (and therefore the binormal) have constant direction, so that db/ds vanishes. For a non-plane (skew) curve, this derivative may be looked upon as a measure of the second curvature, and its magnitude is known as the torsion τ. The direction of db/ds is found in the following manner: From the properties of the unit vector, db/ds must be perpendicular to b, i.e. parallel to the osculating plane. Further, since b is perpendicular to t, bt = 0, and so

Since, also, b is = ρ(dt/ds), there results

i.e. the vector db/ds is perpendicular to t. . To fix the sign, we specify that the torsion is to be reckoned positive if the rotation of b is clockwise when looking in the direction of t. This gives the vector equation

These examples show how much the use of vector analysis simplifies and clarifies the material of differential geometry. In what follows, we shall require only the principles presented above, and the reader is referred to modern textbooks on differential geometry for further applications.

?

Ex. 11. An arbitrary surface A has a plane boundary curve. Form ∫dA= s and show that [bs] = 0.

7. Space Derivatives of a Scalar Quantity.

If a scalar quantity—the temperature, for example—is given as a function of a point in a region, we speak of a scalar field. Let the scalar point function have the value u at a certain point. The change in the value of this function, corresponding to a displacement of amount ds, will depend on the direction of the displacement, i.e. the derivative du/ds may have any one of an infinity of values, depending on the direction of the displacement ds. It is easily shown, however, that we can specify the change in u in any direction, provided we know a certain vector obtainable by means of differential operations. In rectangular co-ordinates we have

Evidently, then, du may be regarded as the scalar product of ds with a vector

which is called the gradient of u, and is written grad u:

that is,

The meaning of the gradient may be seen in this way: Leaving discontinuous functions out of consideration, and dealing only with continuous point-functions, we can connect all points in the region having the same value of u by means of surfaces—so-called level surfaces (fig. 13). Each of these surfaces is characterized by the fact that displacements wholly within the surface do not alter the value of u. If, then, we let ds0 lie in one of the level surfaces, we have

Fig. 13

But since neither ds0 nor grad u is to vanish, this means that the vector grad u is perpendicular to the level surface.

Further, from the definition of the scalar product, it follows that the increment du has its greatest value—the magnitude of ds remaining constant—when ds and grad u are in the same direction, or, in other words, the vector grad u gives the direction of greatest change ( slope ) of the point-function u — hence the name gradient . According to (37), its magnitude is equal to du/ds if d s is perpendicular to the level surface. The change of u in any other direction is obtained, according to (37), by multiplying by the cosine of the angle ϕ between the gradient and the direction of displacement. This suggests a simple graphical construction for the several values of du, for a ds of fixed magnitude (fig. 14). Two lines of length equal to the magnitude of grad u are drawn through P0 and normal to the level surface, one above and one below, and the spheres having these two lines as diameters are described. Then a secant drawn in any direction gives the magnitude of the change in that direction, in relation to the change du in the direction of the gradient. It must be remembered that ds is held constant here.

Fig. 14

The derivative of u in the direction of s is denoted by ∂u/∂s, and is thus obtained by projection of grad u on the direction S.

Numerous examples of the physical occurrence of the gradient concept will be met with in many sections of this work. One example may be given now. A weather map has marked upon it the lines of equal air pressure (isobars). The direction of the wind is then given, apart from the effect of the earth’s rotation, by the direction of greatest pressure drop, which is perpendicular to the isobars; and the strength of the wind is determined by the magnitude of the pressure drop.

By introducing the gradient we have obtained a vector field, that of grad u, from a scalar field, for grad u is a vector magnitude which alters in value from place to place in the region. However, not every vector field may be obtained from a scalar field in this manner. If a vector field has this property, a result of great importance follows: Imagine a curve C drawn in the vector field (fig. 15). Let this curve consist of a succession of line-elements ds. The line integral along the curve C is defined as the limit, for ds ds, the summation to begin at P0 and end at Pis the gradient of a scalar u, this line integral between P0 and P1 is independent of the form of the curve C, for the individual magnitudes

Fig. 15

represent, according to the definition of the gradient, the change in the function u in moving along a distance ds. The sum of these changes is the difference between the values of u at P1 and at P0, and this difference is independent of the path by which one passes from P0 to P1.

If there are two paths, C and C′, then reversing the direction of motion along C′ gives with C a closed curve which is described in one direction. But since the reversal of C′ causes all its line-elements ds to reverse their sign, and since the value of the line integral is the same on both portions of the curve, the integral along the path vanishes. We thus have the important theorem: If a vector field is representable as a field of the gradient of a scalar point function, the value of the line integral of the vector, taken between two points in the region, is independent of the path; and the line integral over a closed path vanishes. Thus

and

From the definition (36) of the gradient, a useful rule follows: If u , then, since

Ex. 12. What is the meaning of grad r ? (r = distance from origin.)

Ex. 13. Let a scalar point function depend only on the distance r from the origin. Compute grad f(r).

8. The Concept of Divergence and Gauss’s Theorem.

If an arbitrary vector field is given, we may, as in the analogous case of accompanying a displacement ds. As may readily be seen, the calculation of this change necessitates a knowledge of three gradients, viz. those of the rectangular components, which give the amounts of the changes of the components. We thus arrive at a set of nine scalars (three vectors). Quantities of this nature will not be met with until later, and so we put this question aside, and take up another physically significant differential operation, which leads from a vector field to a scalar field.

Fig. 16

(fig. 16). The amount of fluid passing through the element of area ΔSΔS; for the quantity passing through is given by υs, the component of the flow perpendicular to the surface element. Let us now enclose a space τ in the vector field by means of an arbitrary closed surface. Call the elements of this surface dS and let them be so oriented that the normals are directed outward. The amount of fluid passing outward through an element dSdSdS"), has been introduced. Divergence is, then, equivalent to source-strength. Thus

are continuous point functions having continuous derivatives, this limit is finite, and independent of the form of the volume element. This latter fact is somewhat troublesome to prove, but it is easy to show that the limit is finite. Thus, let the volume element selected be a small sphere of radius ρ at whose centre the field vector has the value ν0. The departures from this value are then, in any case, of the order of magnitude of the radius ρ. Now the surface element on the sphere is ρ²dΩ, where dis a variable finite vector, whose further properties need not be given. Since by dSvanishes, and there remains a contribution from the second part, of the order of ρ³. But the volume of the sphere is 4πρ³/3, so that the result of division by this quantity must be finite.

Fig. 17

Let us now consider a finite volume which is bounded by a closed surface (fig. 17). Let this region be subdivided into volume elements. The definition of the divergence holds for each of these:

, on account of its continuity, has the same magnitude both times. There remain only the contributions of elements of the outer surface. This yields Gauss’s Theorem:

That is, the surface integral of the vector , taken over a closed surface, is equal to the volume integral of the divergence of taken throughout the enclosed volume. This is sometimes called the Divergence Theorem.

If the hydrodynamical case is taken as a basis for demonstrating this theorem, the result appears almost self-evident. The theorem, however, is true for any vector field satisfying the conditions given above.

Fig. 18

To obtain an expression for the divergence of a vector in terms of its rectangular components, we choose as volume element a rectangular parallelepiped with edges dx, dy and dz. At x, an amount of fluid dydz(υx)x passes through the face dydz contributes to the fluid passing through the surface. This fluid enters the volume, since the normal has the direction of the negative x-axis.

At x+dx, an amount dydz(υx)x+dx streams out through the opposite face. The normal is here directed along the positive x-axis. Expanding (υx)x+dx by Taylor’s Theorem, we have for the contribution of these two surface elements

Analogous contributions are furnished by the two other pairs of opposite faces. Thus we have

and, after division by the volume dxdydz,

Ex. 14. Calculate div r from first principles, and also by the formula.

Ex. 15. How may the gold balance of a country be calculated from the amount of imports and exports ? To what extent does this represent an application of Gauss’s Theorem? ( Gold = equivalent of merchandise.)

9. The Curl of a Vector, and Stokes’s Theorem.

Another equally important differential operation leads from a given vector field not to a scalar field, as in forming the divergence, but to another vector field.

We refer back to the relation is perpendicular to dS was taken to be different at the latter edges. The value of the limit

Fig. 19

is thus a function of the direction of the normal to the surface element. We now show that we can give the value of this limit for any direction, if we know the value for any three non-coplanar directions. For convenience we take these three directions to be the co-ordinate axes. As surface element we choose a small triangle ΔS = ABC, whose vertices lie in the axes (fig. 20). Instead of passing around the triangle ABC, we could describe, in turn, the triangles OBC = ΔSx, OCA = ΔSy and OAB = ΔSz. We can calculate the limit

Fig. 20

for each. In adding the line integrals along the partial triangles, each axis will be covered twice, in opposite directions. There remains only the part contributed by the sides AB, BC and CA, so that the relationship

is satisfied. On the other hand, from § 4 (p. 14),

The unit normal vector to ΔS , and is directed toward the outside of the tetrahedron, while i, j, k point inward. If cos α, cos β, cos γ , this equation is, in terms of components,

Putting these values in (44) gives

with a vector whose components are Lx, Ly, Lz ).*

We have therefore

It follows that

Just as we obtained Gauss’s Theorem, when considering the divergence, by passing from infinitesimal volume elements to finite volumes, so we obtain here a new and important relationship by considering a portion of a surface bounded by a curve. This curve need not be plane, but is to be traversed in a given sense (fig. 21). Let the surface be subdivided into infinitely small elements. The boundary of each element is to be traversed in the same sense as the outer curve. Then, for each element,

Fig. 21

If we add these equations for the separate elements, the contributions from the inner dividing lines cancel out on the left side of the resulting equation, since each segment of these lines is traversed twice, in opposite directions. The left member is then merely the integral along the bounding curve, and we obtain the important formula called Stokes’s Theorem:

The line integral of the vector , taken over a closed curve C, is equal to the surface integral of the curl of , taken over any surface having C as a boundary.

, then Gauss’s Theorem may be applied to a closed surface, and

Let this closed surface be divided into two parts by a closed curve (fig. 22). Then

Fig. 22

If, now, we apply Stokes’s Theorem to each of the two parts, we must remember that in forming the integral over the entire surface, all the normals are directed outward; but in the application of this theorem, the direction of the normals is determined by the direction in which the curve C is traversed. If we agree to take dS always toward the outside, then the sign must be reversed in transforming the second integral. We thus have

Since this equation holds for any volume, it follows that

Further, the vanishing of the line integral of the gradient of a scalar point function, taken over a closed curve [(38′) (p. 22)], may be expressed in the following way by applying Stokes’s Theorem:

Fig. 23

. From (47) (p. 27) and (43′) (p. 26), we obtain the xby traversing a surface element which lies in the yz-plane. Let this element have the form of a square of side 2h (cf. fig. 23), with sides parallel to the axes. Only the ycontributes to the integral along the horizontal sides, only the z-component contributes on the vertical sides. If we let the subscript 0 refer to the values at the centre of the square, then υy along AB has the value

Along CD

The contribution of the side AB thus is

In the same way, the side CD yields

Together, these two sides give

Similarly, the two vertical sides give

Since the area of the square is 4h², division by this quantity yields *

and, in like manner,

Hence

On the understanding that the product of the differentiating operator ∂/∂x with a quantity u means ∂u/∂x, then the last equation may be written as a determinant:

.

10. The Operator ∇.

If we compare the three formulæ

we recognize that we can express all of them symbolically in terms of a vector whose components are ∂/∂x, ∂/∂y, ∂/∂z, on the principle just explained (cf. the last sentence of the preceding section). This symbolical vector is denoted by ∇. It was introduced by Sir William Hamilton, and is called nabla , after an ancient Assyrian harp whose form ∇ resembles:*

If we multiply a scalar quantity u with this vector operator, we obtain

, we obtain, according to the definition of scalar product, the sum of the products of corresponding components:

Finally, :

From the formal correspondence of the operations grad, div, and curl with multiplication, it follows at once that these operations are distributive with respect to addition. That is,

By starting with the definition of divergence, and generalizing, we obtain a definition of the nabla operator which is independent of the co-ordinate system:

We shall illustrate the method of proving these formulæ by taking the case of the gradient. Since the form of the volume element is entirely at our disposal, we take a small cylinder, the generating lines of which are in the direction of grad u and are therefore normal to the level surface (fig. 24). Apart from terms of higher order, the value of u over the base of the cylinder is u0 — h | grad u |; at the top of the cylinder, the value is u0 + h | grad u |. It is to be remembered that, in forming the surface integral, the elements dS must always point outward from the elementary volume. The lateral surface of the cylinder does not contribute to the integral, since u has a constant value in each transverse section; and to each surface element there corresponds another which is equal and opposite. Also, we obtain the contributions of the ends of the cylinder by multiplying the area ΔS by the difference in the values of u, and by expressing the direction of dS by means of grad u:

Fig. 24

Dividing by the volume, and passing to the limit, we obtain the third equation of (58).

These formulæ yield important generalizations of Gauss’s Theorem, the proof of which is analogous to that given above:

Theorems (59) and (60) are also easily proved by writing them out in their scalar forms (which are often useful) and using (41). For (41) can be written

where A, B, C are scalar functions, and l, m, n are the direction cosines of the outward normal to S. Since A, B, C are independent, it follows that

These formulæ are closely connected with Green’s Theorem (p. 270). They are equivalent to (59), and also lead at once to (60), which involves three equations of the type

11. Calculation of the Gradient in a Vector Field; Fundamental Principles of Tensor Analysis.

In § 7 (p. 20) the question " How does a scalar point function change with a displacement ds?" change with a displacement ds?" If we start with the components of the vector, this question may be referred back to the case of a scalar point function. If a vector field be given, then at each point in space we have a field of three scalars, viz. the three components. But, according to § 7 (p. 21), the change in the components may be written down at once:

and so

Merely as an abbreviation we write

which is read: " ds-vector ."

In order to be able to calculate d , three vectors (or nine scalars) must be known. We shall see that such forms, consisting of nine numbers, have also a physical meaning. From (61) we see that the components of the vector d are linear functions of the components of the vector ds, the coefficients of these functions being the partial derivatives ∂υx/∂x, ∂υx/∂y, &c.

Let us now investigate such linear vector functions in detail. As in :

. The scheme of coefficients aik has thus an independent meaning if this correspondence is such that the passage from to is independent of the particular co-ordinate system in which the vectors are resolved into components. In this case, the coefficients aik are said to be the components of a Tensor*

A vector is a physically determined directed quantity in space, but the values of the components depend on the choice of axes. In the i, j, k system,

For a system i′, j′, k′, turned with respect to this system,

where, since i′, j′, k′ again form a rectangular system, the βik represent the direction cosines of the new axes referred to the old; e.g. βik are the direction cosines of the new axes with respect to the old. Then, as may be verified by forming the scalar products i′ j′ = 0, i′i′ = 1, &c.,

Thus the old axes are expressed in terms of the new:

become, in the new co-ordinate system,

i.e. the vector components transform in the same way as the axes. We thus have also

How must the tensor components aik ? The following equations must hold:

and

xyz xyzis determined by both triads. If we multiply the first of the unaccented equations by β11, the second by β12, the third by β13, and add, we obtain, by (67),

If we express the υx, υy, υz here in terms of the new components, according to x′ as a function of υx′, υy′, υzx′. must be identical with that defined by (70). Comparing the coefficients of υx′, υy′, υz′, we have

Thus the tensor components transform, under rotation of axes, in such manner that the new components become linear functions of all the old ones, the coefficients being the squares and products of the βik.

To find the coefficients for the tensor components aik′ set υx′ = υ1′, υx = υ1 &c., and use (67) to form the product υi′υk′ of the vector components, remembering that in expanding the brackets the order of the factors υi and υk must not be changed. Replacing υi′ υk′ by aik′ the coefficients of υiυk yield those of aik.

in with the three vectors a1, a2, a3. This amounts to regarding the three coefficients of each row as vector components:

But, as may be seen in the example of forming a gradient, the three vectors depend upon the choice of co-ordinate system, for the three vectors grad υx, grad υy and grad υz are dependent upon the co-ordinate system.

, according to (63) (p. 33), this functional correspondence Φ satisfies the relationship

with the tensor Φ. This tensor may be represented, according to (73), by three pairs of vectors (dyads):

The first factors in each term are called the antecedents, the second factors, the consequents. Thus

The two vectors are, in general, different. The tensor resulting from interchanging the antecedents and consequents is called the conjugate tensor Φ0.

Since it is immaterial whether the scalar a1 is written before or after the vector i, comparison with (77) shows that

In the component representation of a tensor, changing to the conjugate tensor means the interchange of rows and columns in the array of coefficients.

The dyad, indicated by writing two vectors side by side with a dot separating them,* also has the leading property of multiplication, viz. distributivity with respect to addition:

, in which case

since the scalar products are distributive; the result therefore follows. We may therefore look upon the union of two vectors to form a dyad

as a new kind of non-commutative product, called the direct or dyadic product. This leads to a new interpretation of the expression ds grad · , there results

If this tensor is now multiplied by the displacement ds, we obtain exactly (62), viz.

By dyadic multiplication of the nabla operator with the field vector we obtain a tensor called the vector gradient of . Multiplication of the displacement ds with this tensor as post-factor gives the change of the vector due to the displacement ds.

Let us again consider tensors in general. A tensor is said to be symmetric if

and is antisymmetric (skew-symmetric) if

By ′ be designated by p. Then

Because of the proviso that Θ is to be a symmetric tensor, we have

from which there follows

′ By ′ has the direction of the normal to the surface p = const., at the end point P of . If we draw the surface p = 1, then, according to (84), the magnitude on the direction of the normal at P (cf. fig. 25).

Fig. 25

The surface p = 1 is called the tensor ellipsoid, although without further limitation on the values of the coefficients, it might be any second degree surface. The representation by means of a second degree surface is seen to be possible, since both a surface of this kind and a symmetric tensor have six independent coefficients. Further, for every surface of the second degree there is one specially appropriate co-ordinate system, viz. that of the principal axes. In this system of axes the surface has the equation

and the tensor, referred to these new i′, j′, k′ axes, becomes simply

are thus given by

Ex. 17. Write the Frenet formulæ [equns. (32), p. 19, and (35), p. 20], using the vector gradient.

in these directions.

12. Calculation of more complicated Vector Differential Expressions with the help of the Nabla Operator.

The operations grad, div, curl and grad may be performed readily, in the case of products of scalar or vector point functions, with the help of the ∇-operator. It is only necessary to remember that ∇is a sign of differentiation as well as a vector. Everything standing to the right of such a sign is subject to the process of differentiation. In ordinary differentiation, the quantities to be held constant may always be placed to the left of the differentiating symbol. But since the vector product, and especially the dyadic product, of two vectors are not commutative—though of course in a vector product we can always change the order by changing the sign—we shall indicate by means of the subscript c the constancy of a factor on the right of the sign of differentiation. This will permit us to leave constants on the right side of the sign ∇. By following a limiting process entirely analogous to that for the ordinary differentiation of products, it may be shown that, in general, a space derivative (grad, div, curl, grad) of a product is equal to the sum of the corresponding expressions for products in which only one factor is variable. Thus we obtain the following practical rule for the space differentiation of products: Using the ∇-symbol, write the differential expression as the sum of derivatives in which only one factor is subject to