## Electromagnetic Theory: Volume 3

Four Cases of Elementary Waves. Representation of a Row of Impulses by Taylors Theorem. On Operational Solutions and their Interpretation. Oct 9 General Case of an Intermediate Impressed.

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First Genera Cam s Z Z. Equal Positive ami Negative Terminal Inductances. Impiwed Force in the Gas Z Zx. Terminal Coil and Condenser in Sequence. Closed Cable and Leak Another Way. Same as last without Initial Splitting. Closed Cable with Discontinuous Potential and Current. Theory of a Leak Normal Systems. Theory of a Leak Operational Solution. Evaluation of Energy in Normal Systems. Two Cable witi different Constants in Sequence with an In eertion. A Fourier and a Beasel Cable in Sequence. Operational Solution for Sources in the General Case. Special Case of Zeroth Beasel Solution.

Construction of General Solution by the Convergent Formulas. April 24 me Rationality in p of Operational Solutions. Impulsive Impressed Voltages and the Impulsive Waves. Wave due to Va varying as fP log t. Application to Terminal Resistances Full Solutions with. This implies that A is a complex number, and tiiat E, B differ in their arguments as well as moduli.

J both amplitudes were real — or had the same complex arguments. If a wave containing more than one frequency enters the conductor its profile and energy distribution alter as it travels. However, the dominant effect of conductivity is that of attenuation, or absorption, of the wave in the conductor. Because of this a wave never penetrates very far into a conducting medium, the higher the frequency, or the greater the conductivity, the less the penetra- tion.

For a perfect conductor a — oo the penetration is zero and there is no field at all inside a perfect conductor. The extent of the penetration is measured by the skin-depth 5. Therefore These are the cases when the effect is important. Because of this fact radio waves cannot penetrate the layer, but arc reflected back to the surface of the earth. We assume a plane vrave incident on the surface together with a reflected and a refracted wave and the boundary conditions, corresponding to But for the refracted ray complex values must be used for k", cos 0", sin 0" though the forms of the results remain the same.

This is yet a further complication in the effect of the metallic surface on the wave. When we look at the Fresnel formulae in This latter result implies that all the energy in the incident ray is reflected by the surface. The detailed discus- sion of the relations between the amplitudes of the various rays is given in advanced books on electromagnetic theory; the calculations are tedious but mostly elementary.

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Exercises 11,6 1. Determine the complete electromagnetic field, and find the ratio of the amplitudes of the rcflecicd and incident electric fields. Determine k and the magnetic field. A Itne. A uniform isotropic material has conductivity a, permeability ft and permittivity e. An infinite slab of this materia! Determine the reflected and transmitted waves. Prove that in a medium of conductivity cr.

In this. Waveguides used in practice approximate to this, but have imperfectly con- ducting walls, may have bends or junctions in the tube, and may even be partly occupied by dielectric media. We do not discuss the effects of these latter modifications, but merely establish the basic types of wave that can travel in a straight waveguide. This is the chief difference between our analysis and our previous discussion of plane waves. Second, we see what restrictions the presence of boundary walls of perfect conductors place on the solutions so obtained.

We use complex exponentials to represent the oscillating fields and assume no charge or currents inside the tube.

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There will, of course, be charges and currents on the walls of the tube to correspond with the fields. We consider, when necessary, complex values for y which will correspond to attenuation of the wave. But, so long as y is real and positive the solution corresponds to propagation without attenuation. These are called TEM-waves. The general solution of the form We continue the discussion further for TM-waves, and give, without derivation, the corresponding results for TE-waves.

Hence, to determine thcTM-waves given by cqns. From the derivation of Also, if attenuation of the wave is elementary electromagnetic theory, VOL. Hence frequencies below cvj 2jt cannot be transmitted down the tube in the mode corresponding to there is a cut-off frequency. Note that different frequencies corresponding to a given mode v. This effect is familiar in acoustics where the sound of a voice after it has travelled along a pipe is very different from the original.

The situation with TE-waves is closely similar.

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When we apply the bound- ary conditions to eqns. Therefore the possible TE-waves are obtained from the solution of eqn. The other considerations apply as for TM-waves. The solution for TEM-waves is trivial unless the region in the. The discussion of this case is given in the example below p.

The determination of the TM- or TE-waves depends first of all on the solution of eqns. Provided that or ip, obtained from the solution of In the TM-modc, the energy in unit length of the tube is therefore the real part of where the integral is taken over the area of cross-section of the lube. This is From Green's theorem and cqn. Note that coy 2rV y c-. This dependence on frequency occurs here. The discussion for TEM-waves. These waves correspond to case 3 on p. The remaining equations from n. This also implies that eqns. We can satisfy 4 by writing i.

Now the imposition of the boundary conditions on the solution of 10 , viz. Hence there can be no waves of type TEM in a single hollow pipe. Such waves can be propagated in a coaxial cable, where one conductor encloses the other. We cannot pursue the discussion of propagation along coaxial cables here; for this the reader should consult special treatises. If the field is established in the space outikk a conducting tube, the area is doubly conneaed and the TEM mode c. Rectangular waveguides We develop the solution for the special case of a tube with rectangular- cros.

We will consider only TM-waves. The solution is found by the method of separation of variables- which gives where tlic eigenvalues arc given by ,. ITe various components of tlic field are given by i. For a square section of side 5 cm the maximum wavelength transmitted is cm. The parallel plate waveguide Tills is 11 waveguide with wails consisting of the planes. We include this because the solution enables us to understand how the more complicated phenomena uiih rectangular, and other shaped, waveguides arise physically. In this solution a TM-wave B is always transverse, while E has a component in the z-direction; the corresponding TE-wave has — E, — 0 and B has a z-component, the solution being given by reflection as before.

We can now understand how the reflection from another pair of faces in the rectangular waveguide, or from the curved surface of some other shaped tube, builds up the more complicated modes in the general investigation. Find the condition that such a solution should represent a field between two perfectly conducting planes. A cos a. Tlic boundao' conditions to be satisfied on.

The first of these is identically satisfied, and the second is satisfied for. For propap. Determme the critical wavelength and the wavelength m the guide correspondmg to a watelength! This case was discussed in the text, with a shght difference of notation, viz. With these changes the field vector components are given in eqns Hence the frequency of the oscillation must exceed the cut-off frequency. Example 4. Show that, for a given n, there is a critical value wp which to must exceed if waves of this- iv'pc are to be transmitted in the cylinder.

We include this example as an indication of how non-reciangular wave guides may be disaisscd, and to show how' there is a corresponding pattern of possible modes of oscillation. We return to cqns. IHcrc fl replaces y of the previous discussion. Example 5. If the field docs not vanish identically, show thot the least allowed value of a is.

The verification of Maxwell's equations is a matter of manipulation of the vector diffe- rential operations. B Hence div B -- 0, identically. Show that the electric field component along the axis of the guide is of the form Determine the other field components, and obtain an expression for the cncrgj'-flow along the guide.

Shmv that at all points in the field, the electric and magnetic fields are perpendicular and that there is a phase difference of. Whereas the waveguide is a hollow conductor wth the field propagated down the inside, a transmission line consists, usually, of two parallel con- ductors and the field is established between them.

One kind of line is exempli- fied in telegraph or telephone unres, or power cables, where the conductors are outside one another. Another kind is the coaxial cable where one con- ductor completely encloses the other and the field is established in the space between them. We shall discuss chiefly the simplest case of two straight conductors, having infinite electrical conductivity, which are embedded in a uniform medium.

This medium may have a finite electrical conductivity correspond- ing, for example, to a submarine cable. In practice the medium surrounding the conductors is not uniform, e. Since the transmission line field is established outside tlie conductors these reflections from the outer walls do not take place. Hence, correspond- ing to this field there arc distributions of current, cliarge and voltage on the conductors of the transmission line.

In most practical applications these currents and voltages — the signals transmitted — are of more importance than the field strengths. Also the quantities we consider all vary harmonically with time and so, effectively, the conductors cany alternating currents and voltages. So it is natural to describe the behaviour of the line in terms of impedances, which relate the currents and voltages, as in alternating current theory. The arrangement we consider is made up of one or more, usually two, long, straight conductors with uniform cross-section perpendicular to the r-dircciion.

We look for fields which correspond to the propagation of wave. The analysis is identical as far as cqn. Wc cannot give a thorough discussion of these here, and content ourselves ttitli the following account which draws an analogy with the conditions for uniqueness in electrostatics. In piiysical terms this means that the potential and field correspond to the ticlcl of a line charge through the origin; this is the. The conditions are sufficient for uniqueness; this does not imply that they are also necessary. Equation The consequence of applying We illustrate the above by considering the case of a perfectly conducting cylinder of radius a.

In cylindrical polars this becomes, because cj is independent of 0, d-d 1 di. It is clear that the first of conditions A similar argument applies to solutions which depend on the polar angle 0. Again there arc no TE- or TM- solutions. Since only TEM-waves arc possible the field vectors satisfy eqns.

Tlic solution of eqns. At infinity 0 must satisfy the radiation conditions Since the first of conditions 1 1. If a charge, for example, is situated on part of the line a disturbance is propagated, both ways, along the line with velocity v, but this disturbance cannot be represented by a wave of the kind considered here with a single frequency. If the line is not a perfect conductor, then a wave can be propagated along it, but this wave is not of the TEM-type. We now investigate what charges, currents and voltages on the conductors of the line correspond to these fields.

We consider two curves and Sz in the xy-plane Fig. The static field 0 corresponds to charges 0. Hence we see that 9i ——32 and the current flowing along one conductor returns along the other. If we denote From cqn. We consider now what modifications become necessary if the medium surrounding the conductors is a conducting medium with a conductivity small compared with that of the line itself.

Tltcn ' ff? The second consequence is the addition of an internal inductance L to L. The further modifications necessary in eqns. The equations for the network, when the time variation is given by the f. We can make the resemblance complete if we regard the element. In the hitter two identifications we arc using the rules for connecting rc. Then eqns. After division, by these become identical with. This model of a network of impedances is usually used in discussions of transmission lines, and because the final equations are identical with 1 1.

In this model the conductor. This differs from our original specification of two conductors. The discrepancy can be avoided if we regard the two conductors as separated by an infinite conducting plane at zero potential. The second line is then the image of the line of the first conductor in this plane.

This plane is taken to be the earth conductor in the model. We now complete our discussion of the transmission line starting from theequations The mathematical discussion consists of finding a solution of eqns. Because of the resistance in the conductors the signal suffers attenuation as it is transmitted down the line; the important practical case is that in which attenuation occurs without distortion. If the attenuation depends on the frequency, a given signal, which contains many different frequencies, will have these different components attenuated to different degrees when it reaches the end of the line, and so suffers distortion.

Similarly distortion occurs if the different frequencies are transmitted with velocities which depend on the frequency. We shall see that by adjusting the values of the parameters i? Such a line can have amplifiers inserted into it at suitable points along its length to over- come the attenuation, and a clear signal can then be transmitted over long distances. If the quantities a, arc independent of to, then all frequencies arc pro- p.

In this case the line is distortionless. If we eliminate a from cqns. T- GR. The general solution of the eqns. Thus far we have not considered the effect of the ends of the line. There are three particular results: 1. In general, the solution of problems concerning transmission lines requires ideas and techniques very closely similar to those required for w'aves on strings, both finite and infinite in length. In a irnnsmission line each of the elements AiA-, A complex alternating current.

Vo is fed in at the terminal Ai,, and. Sliow that if the terminal yl,. A, Jj liy VLliy L. By this means Kirchholfs fitM l. Hence this transmission line will transmit only signals with this frequency without diminishing the amplitude, i. A uniform cable has constant resistance R, capacity C and inductancei per unit length. Show that it induces a periodic potential at distance. At t 0 the end a- 0 is earthed. Show that at subsequent time t the potential at.

State Maxwell's equations for an isotropic conducting medium of conductivity tr, pcrmc. Show that It, li and I! A plane electromagnetic wave, whose wavelength is! Obtain from. Maxwell's equations for a homogeneous medium of cUclectric constant. If tlic medium is highly conducting, show tliai the E waves and 11 waves are out of phase bv approximately. State the other conditions which must hold. Determine the current in the sheet necessary to maintain the oscillation.

A general dielectric medium is divided into two regions, denoted 1 and 2, by a metal sheet 5 which carries a current of surface density i. The sheet S is in the shape of an infinite cylinder of radius a, whose axis lies along the z-axis; the space inside and outside S is empty. Kinp, r nnd 6. Determine the airrent in the sheet ncccssarv- to maintain this c'cilbiion. Prmc that in an cleciromapnctic field independent of the coordinate y, where the axes Ox, Oy, Or are rectangular cartesian, and i.

J, k arc unit vectors along them, the vectors E es. Show that Maxwell's equations for anisotropic homogeneous non-conducting charge- free medium can be satisfied by t. Find the reflected wave. It is now time to take up the general question of invariance. A scientific theory is, generally speaking, invariant under some group of transforma- tions. Tltc reader may be reminded that by a group is meant a set of quan- tities Ixitwecn which a binary operation is defined.

This binarj' operation, often regarded as a product, for transformations consists of applying two traii'-formaiions in succession. The binary operation is associative and for the existence of a group there must be an identity element and a reciprocal of every element.

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In the ease of transformations the identity element is tlie identity transformation and the reciprocal element is the inverse transfor- mation. These ideas arc already familiar in the case of the orthogonal group in three dimensions i. Indeed the whole invariance under this group is automatically built into a theory' as soon as it is c,xprcssed in vectorial form. Accordingly the theory' of electromagnetism must be invariant under some group which has the orthogonal group in three dimensions as a subgroup. Exactly what group this is wc shall return to shortly.

Before that, let us consider exactly what it means to. For example, in the case of electromagnetic theory the numbers might be the six components of the electric field and the magnetic flux density, E and B. On the other band, if they are merely numbers represent- ing in part the properties of the coordinate system, it must make a difference whether one proceeds from the first to the third system directly or via some intermediate stage, and in this ease the conditions ul — v will not be ful- filled, Tlic condition just expressed on the transformations of the sets of numbers is known technically as requiring the transformations to be a representation of the original group and the argument which we have just given which is the general form of the principle of relativity can be put by saying that any physically significant set of numbers must transform under a representation of the group of transformations of the theory'.

It is instructive to look first at the example of the orthogonal group in three dimensions. Here we know of one example of a representation in the components of an ordin- ary' vector. However, not all linear transfor- mations of this kind arc permitted,. In this case, then, we see that the components of the vectors transform in the same way as the unit vectors themselves, but this is a coincidence.

What has basically been done here is to derive a representation of the group by regarding a vector as a displacement; in other words, one chooses something which, from its geometrical or physical interpretation, is known to be independent of the coordinate system and so to transform under a representation of the group, and uses its representation to define a whole class of such objects.

In this way one could construct a series of more com- plicated representations. These more complicated transformations are called the tensor representations of the group. The particular transformation above is of a tensor of rank 2. By taking products of more than two vectors we get the higher order tensor representations. These are by no means all the possible representations, although they are important ones, as can be seen from the following example. The quanlitie. In this ease ue have a rather obvious choice to begin with, when we seek for sets of quantities obsiously transforming under a representation of the group, for we can choose i, r as a prototype.

The next step is to inquire about analogues of the vector products in ordinary vector analysis. It is, of course, obvious that under the orthogonal group, for example, we could consider, instead of the vector product, the array of nine quantities transforming under the second-order tensor repre- sentation but it is not convenient to do so here because electromagnetic theoiy is not concerned, in general, with such quantities.

They enter, for instance, in rigid mechanics, in the definition of moments and products of inertia, and in certain other branches of applied mathematics, such as elasti- city. Accordingly we shall consider products of jp. By the very way in which these are written down they must be automatically invariant under the orthogonal group in three dimensions but we are con- cerned also in transforming between coordinate systems in uniform relative S TI 1 C only remaining dilTiculty is that the transformation involves, as well as the particular combination of vectors with which wc started, the components of the vector product of the two vectors.

But the transformation of these, A;,Bs-A;B'. Bs-A:iB2, It is interesting to notice that for transformations in which the velocity V is small compared with that of light, eqns. Example I. The remaining results follow directly from the transformations. Then A'. The calculations of the transfomiation of the remaining components are carried out in similar manner. Example 3. Tlicn Example 1 follows, by using the result of Example 2, with 6, A taken as?! We know already that it has the orthogonal group in three dimen- sions as a subgroup.

In order to establish this we have to show that the product of any two elements of the group again belongs to the group. As far as the product of two rotations is concerned this is well known and amounts to the theorem that any member of the rotation group is a rotation in the cle- ntemary sense about a certain axis Euler's theorem. If we are concerned with the product of a Lorentz transformation and a rotation, it is dear from the way in which weareabic to write our Lorentz transformation in s'cetorial form th.

Indeed we tacitly take account of this u hcncM-r rvc choose the. It only remains therefore to consider the product of two Lorentz transformations. The reader may instantly verify this by using the definitions. If the two velocities are not in the same direction, we consider first the case in which they are perpendicular, and we choose these directions as the X- and y-axes. In order to show that the transformation t, x, y to t", x", y" is a transformation of the group we?

Wc now eliminate t between cqns. Pv Pu The chief importance of this result is that by a suitable choice of u, v the angle a, i.

This result, combined with that for two transfor- mations in the same direction, suffices to prove that we are dealing with a complete group of transformations. Tlic first result is that the charge, being simply the number of electrons present, is unchanged by the transformation. Let us adopt two coordinate. We arc concerned witii an element of volume in he new primed coordinate system, which is d. We have already considered the fonnul. Accordingly if one defines the charge and current four-vector as g, J — 2 pb the expression will have as its scalar time component the density m any frame.

Accordingly we take the charge-current four-vector as the basis needed to estimate the transformation properties of the whole set of equations. In discussing the invariance of the equations it is necessary to take up again the method of derivation which was discussed in the last volume. At this stage the main reason for this assumption is that it is logically consistent to assume these three equations, whereas the remaining one cannot hold for non-steady currents.

Kow it follows from Example 1 of p. By a similar argument, 5, —E is also a six-vector. But the difference appears, of course, in the constitutive relations. In the.

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The second pair, eqns. An immediate consequence of these transformations is the following. Suppose now that a transformation of coordinates of the usual form is made. Another reduction, which is alway. From the. Another interesting application of the transformations is to the e. This dielectric is between the plates of a moving condenser and the plates of this condenser are short-circuited by means of brushes and a wire which passes through a ballistic galvanometer. The condenser moves with a uniform speed in the direction of the x-axis, the x-f plane being chosen parallel to the plates of the condenser.

A magnetic field is then applied in the y-direction see Fig. We have to apply the constitutive relations in the dielec- tric, and accordingly we must transform to a frame of reference in which the dielectric is at rest. The mag- niludc of this charge is in good agreement with the results predicted by the theory. This identification is confirmed by observing that the invariant expression i is the one expected from the usual Lorentz condition [eqn. X and sirrularly for those of O. Obtain expressions for D and B in terms of E, H, e, p when the medium is moving with uniform velocity o.

Choose 0 in the direction of the ar-axis. Example S. A linearly polarized plane electromagnetic wave is propagated in free space from a transmitter fixed in an inenial frame S. The fields of the transmitted wave observed in. Hence the reflected frequency is as stated. A particle of rest mass and charge e moves from rest in a uniform electric field Ej and a uniform magnetic field Bk, where E cB. Tront the last tsso equations i. Find the possible states of motion of the particle for which the acceleration will be zero. By considering the four-vector property of Ap, deduce that a, must transform as a four-vector under the Lorentz transformation and that kpXp must be a scalar, so that kp also transforms as a four- vector.

S, Describe bricfls a method for transforming tltc components of the electromagnetic field in free space from one Lorentz frame to another. Accordingly the ordinary current in the Maxwell equations is negli- gible compared with the displacement current and we may take the equations oo as so that, since curlJ?

It is dear from the equations, since they arc wave equations, that tlic field will be transmitted with a finite vdocity c. One can. Since the field propagates with a finite speed, the field linc. In other words, as frequency increases, the field lines tend to separate. Those which are near enough to the dipole will move away from it and back towards it, but there will be some critical surface which separates thc.

These more distant ones correspond to the radiation field in which we arc principally interested here. In the ease of the static or slowly changing field the most con- M-menl further restriction i-. However, it will prove to be more satisfactory to use a different restriction here. It would be very convenient at this point if the vector A also satisfied the wave equation. Let us choose, then. Now the. Substituting for the poten- tials the field strengths t.

This relates the Hertz potential to the distribution of electric dipoles. The first kind of symmetry which springs to mind is that of spherical symmetry. This suggests that. It is therefore necessary to go up to the ne. We shall suppose that the charges present, which produce the field, are located very near to the origin and are fluctuating in some way so as to produce the radiation conditions nccc-ssary.

In fact the solution which we shall find is mainly applicable to a dipole whose strength varies with the lime, but we shall also consider the general case. It is natural to use polar coordinates for the more detailed calculations; because the axis of sym- metry is the z-axis, the angle between this and the radius vector r may be taken as 0. Tlic Hertz vector, JJ, in order to prcsers'c the axial symmetry can only have components radially outwards and parallel to the z-axis. Both of these components can only depend upon the distance from the origin.

We shall I'C particularly interested in the. The distinction between these tw'o parts corresponds to that between the near zone and the distant zone mentioned in our earlier intuitive argument. Taking again the particular ca. The part of the electric field which is important nearest to the origin falls off as the cube of the distance and the part which is important a great way off again falls off inversely as the distance. Approximately, one can say that the values of the fields near to the oscillating system are M cos 6 iTtEo! These conditions do not depend upon the particular Lorentz frame of reference.

The field of a dipole of moment m directed along tlie polar axi? Tfor' 4rreori' so that the electric field corresponds to such a dipole with varying moment m. Now such a varying moment may be considered as due to a current element. This is in complete agreement with the electrical picture. Thus both the fields imply that the sohnion we have found corresponds to an oscillating dipole at the origin directed up the r-axis.

It is to be noticed that from the equations for the field in the near zone the squares of the field strength which enter into the expression for the energy arc inversely proportional to the sixth power of r. Moreover, the electric and magnetic field strength dilTer in phase by T-t. There is therefore an energy-flow outwards. As a result the energy decreases only as the inverse square of the distance.

Thus apart from the two zones dificring by the size of the fields and the manner in which these fields fall off with distance there is also another important distinction in the behaviour of the energy in them. In the near zone the energy flows out and back, so that in so far as the transmission of radiation is concerned the whole behaviour is really a pretence. The region in which the pretence takes place will be larger when the near zone is larger, that is, the lower the frequency or the longer the wavelength.

For a wavelength of about 60 cm the near zone has a radius of about 5 cm, whereas for a wavelength of 6 km the near zone has a radius of about m. The large amount of energy consumed by the long-wave transmitter is mostly used in setting up the pretence of transmission in the near zone which is in fact only a pumping in and out of energy. Certain plausible assumptions nbout the scalar and vector potentials lead to a solution which can then, by looking at the near field, be seen to correspond to an oscillating dipole.

Two different directions suggest themselves for proceeding from this point. In the present section we look at one example of a solution found in this way. Alternatively we miglit seek to find the radiation field from a.