2 Background .1 Cosmological .11 Mathematics .10 Book


Book 10 - Hints and Answers 2

Chapter V


5.1 a) By the limiting process (20.9), one obtains, since I 'n(la) = 0, instead of (20.19)

whence by means of Bessel's equation

Normalization to 1 yields then from I n(lr) = 0

b) By (21.11), relating yn(r) and I n+?(r), the normalizing integral becomes

The boundary condition in the task

yields then, similarly as in a), by going to the limit

and from there by Bessel's equation and further contraction

whence one finds the to 1 normalized form of yn

5.2 The proof follows from Green's theorem

by setting v = 1/r - the distance of the integration point from P. Due to the singularity of v at P, as we know, P is surrounded by a sphere Kr with radius . If one extends the integration on the left hand side over the space between Ka and Kr, the left hand side vanishes and the right hand side is equal to the sum of the surface integrals over Ka and Kr (n is both times the external normal to the surface). By the limit in the integral over Kr, one finds from (1)

However, the third term on the right hand side vanishes, because everywhere inside S one has Du = 0. Thus Equation (2) already contains the proof of the theorem regarding the arithmetic mean.

5.3 Equation (27.14) and the prescribed condition u = U on the sphere r0 = a yield

whence, by multiplication by and and integration with respect to j0 from 0 to 2p,

next, multiplication by and integration with respect to from 0 to 2p yields

which agrees, apart from the notation, with (27.13a).

Comparing (27.13) and (27.14) with regard to their dependence on r, now yields

which is valid for n = 1, 2, ???. The summation is extended over all roots k = knl of the equation Yn(ka)=0 or, what is the same, of the equation yn(ka) = 0. For the recalculation of Yn into yn, employ (21.11):

and (20.19), which is also valid for non-integer n,

in which k is a root of In+?(ka) = 0, i.e., also of yn(ka)=0. (5) and (5a) yield then

If you set now Yn = Nyn and demand that

you find that

Hence, the conversion of (4) from Yn to yn becomes with a = r0/a

By (21.11), for n = 0,


In this especially simple case, Equation (8) becomes

This yields especially for a = ?

i.e., Leibnitz's sequence (2.8). In general, using (2.9), one obtains with x = 2pa a representation of the saw profile

Confirm this result as an example for Chapter I. However, it is clear that for n > 0 there are hidden in Equation (8) much more general and deeper analytical relations.

6.1 a) (31.4) shows that P has the z-direction and, apart from z, only depends on r? = x? + y?

With a) of the task,

This vanishes for z = 0 because then R = R'. As a consequence, there also vanish the expressions for , obtained by differentiation with respect to x, y or r.

b) To start with, by (31.4),

However, since by assumption P vanishes for z = 0, also vanish for all points x, y on Earth's surface.

c) (35.1) yields now

These derivatives vanish for z = 0, since Pz itself, according to c), vanishes for z = 0.

d) Now, (35.1) yields

However, by d),

This vanishes for z = 0 when R' = R.

6.2 This exercise is not only valuable for the understanding of Zenneck-waves, but also for the general insight into magnetic rotation fields and their representation by complex operators.

(32.20) yields by (31.4) for the air space

Multiplied by the exponential time factor, these expressions represent an elliptic oscillation which is know from Optics. The principal axes of this ellipse are, due to the complex nature of the right hand side of (1), rotated with respect to the x- and z-directions. If one forms the absolute values of , they yield, taken positive and negative, the limits between which oscillate to and fro, and thereby describe a rectangle circumscribing the ellipse. The ratio of its sides is the absolute value of

This present value 1/n follows directly from the definitions (32.16a) and (32.2) of p and n. Since , the rectangle is narrow and high (Fig. 42a).

On the other hand, in Earth, by (32.20a) and (31.7),

One has again an elliptic oscillation, embedded in a rectangle, the ratio of the sides of which is given by the absolute value of

where the last value, as before, follows from (32.16a) and (32.2). Since , the rectangle is now wide and low (Fig. 42b). The present ellipse is travelled in the opposite direction to the previous one, which follows from the reciprocity of the two values n and 1/n in (4) and (2).

If one imagines that the field advances in air forward at its phase velocity to the positive x side, one arrives at the idea that is drags the field in Earth behind it against the resistance ruling there.

6.3 Let be the local electric field strength in the direction of the antenna. It performs at the antenna, short compared with the wave length, in the time dt the work . Hence, by (36.20), the time mean value of the work (t = period) is

With the numerical method of 6.36, one finds

Hence one can also write

a) Vertical antenna: (31.4) and the differential equation of P yields with

Use (32.9) for P and express eikR/R as well as eikR'/R' by (31.14). Since, for all three terms, their dependence on r is given by I0(lr), an application of the operator

results simply under the integral sign in the common factor +l?I0(lr), i.e., at r = 0 on the antenna the factor l?. Hence, (32.9) yields

Since m is real for l > k, there disappears in the first row during the formation of the real part the integration over and one can advance without problems of convergence to z = h, i.e., to the location of the antenna. Thus,

If one takes the values of these integrals from (36.13) - (36.17), substitutes (4) in (1) and converts by addition of the factor (36.22) into our system of MKSQ-units, one finds exactly the value of W from (36.23).

b) Horizontal antenna: With follows from (31.4)

Since, following (33.12) and (33.15), the dependence on x in Px is given by I0(lr), in Pz by xI1(lr)/r and for small x, y

one finds for r = 0

These factors ?(2k? - l?) and l/2 appear during the calculation of (5) under the integral sign of 33.12, (31.15), respectively, where one can employ for the first two terms on the right hand side in (33.12) Equation (31.14) and in (33.15), apart from the differentiation with respect to x, employ still that with respect to z (the factor -m under the integral sign). Thus, one finds instead of (5)

The first term in the brackets [] arises from the third term in 33.12, the second from (33.15). If one transforms the two terms to a common denominator and notes, as ahead of (33.13c), that m?-m?E=k?(n?-1), one finds after simplifications

Thus, the second row of (7), after setting still z = 0, becomes identical to the integral in (36.17a) for L. If one steps in the first row of (7) by (3) to , one can again replace the upper limit by k and execute the integrals as in (36.16a). Thus, one obtains during transition to our system of units exactly the representation of (36.23a).