James Clerk Maxwell forbidden knowledge ?? Notice The abstract, bibliography and other common items are missing in this document. The reason is not impolite. It is necessity. I beg to be excused. CHAPTER 1 – Several waves in one (1-a) What will we treat? Ambient: the vacuum Phenomenon: electromagnetic propagation Case: monodimensional plane wave, without polarization of any kind in the transverse fields. Task: analyze the electrical displacement Let us highlight a very common bad habit in the classroom. It is to insist too much on the waves of the E and H fields, without paying attention to other fields included in electromagnetic propagation, which also undulate. The D field wave (electric shift) is included, but I have no newsfrom someone who pays attention to you in the classrooms. This is rare since Maxwell specifically needed to analyze D to formulate a complete, coherent and consistent theory. They show us that D was the key that allowed Maxwell to access a new and wonderful environment, without showing us how that key is applied in each specific case, for example in the propagation of a wave. With the intention of filling this gap, I am writing this document, intended for people who have basic knowledge of electrodynamics. (1-b) What does it mean without polarization? In the following sections, mathematical developments await us that are not difficult, but are long and laborious. Then a little conceptual support will do us good, which is the best compass to guide us in the middle of the mathematical jungle. The propagation direction is the x axis. The (x, y) plane contains that axis. Also the plane (x, z). And many other planes the same (there are infinite planes that contain it). one
Consequently there are infinite pairs of mutually perpendicular planes, like those we see on the left of the image . Each pair of such planes gives rise to an electromagnetic wave, like the one we see in the (x, y) and (x, z) planes (electric field and magnetic field). The energy of propagation is distributed in all those infinite planes. The elemental wave operating on a pair of mutually perpendicular planes has an infinitesimal energy. This is implicit in the condition that interests us, propagation without polarization of any kind. Mentioning it explicitly is a conceptual necessity. The infinite planes intersect each other on a line. That line is the axis of symmetry of the propagation. From that axis to what radiusdoes the field of the wave cover? Even infinite radius or even a infinite radius whose measure we do not know? I don’t ask the question to answer it in this section of the document. I just want to motivate a reflection in the person who reads. Although we ignore the measurement of the radius, the cylindrical symmetry is evident. The wave field forms a cylinder, the radius of which should be determined by the equationsfrom Maxwell. If the Maxwellian theory is coherent, consistent, and complete, then the radius of the cylinder will be deduced from it. In each of the infinite planes the electric field of one elemental wave and the magnetic field of another coexist. This is inevitable because no plane lacks an electric field or a magnetic field. Does this coexistence imply any incompatibility? Both fields are transversal. Then the vectors of both in the same plane are parallel and give a vector product equal to zero. Poynting’s theorem indicates that the power flow is given by that vector product. This means there is no power associated with vectors E and H on the same plane. The power is given by the vectors E and H located in different planes, as we learn in the classrooms and in the bibliography. We can foresee two modes of cylindrical symmetry, by semi-planes (we will call it SPL mode) and by planes (PL mode). In SPL mode the direction changes on the axis of symmetry (the direction in one half plane is inverse to the direction in the other). In PL mode the field has the same direction throughout the plane. two
Can SPL mode exist physically? In this mode each vector in a half plane has an equal and opposite partner in the other half plane. So the vector sum is zero. ?? With no net nonzero fields perpendicular to the direction of propagation, the Poynting vector equals zero. There is no power flow, there is no energy and there is no wave. For that reason SPL mode is not physically possible. The only possibility is PL mode. What would we do to build a three-dimensional model of the PL mode? An easy way would be to populate with transversal vectors of the same direction a plane that contains the propagation axis. Then we rotate that plane half a turn around the axis. Each half plane sweeps half the volumecylindrical and with that the cylinder is completely full of field. The graph on the left showsone green and one blue plane mutually perpendicular. We populate the green plane with vectors and rotate it half a turn. Thus we generate pairs of symmetrical vectors with respect to the blue plane. For example, the pair of black vectors represented in the graph on the right. Remember that the idea of populating a plane with vectors and rotating it serves to colloquially describe how the fields are arranged in the cylindrical symmetry. It is not the description of a physical turn. It is a didactic resource. What results in this form of field filling the entire volume around the axis of propagation? The resultant of the symmetric pair is perpendicular to the direction of propagation and belongs to the blue plane. The same happens with the results of all the symmetric pairs. In resulting terms, the wave with cylindrical symmetry supports the simplest wave function that we know, with two mutually perpendicular planes, the electric field in one of them and the magnetic field in the other. It is the flat representation that we learned in the classrooms. In this document we will analyze the resulting wave function, without forgetting that it physically corresponds to cylindrical symmetry. 3
D = x D cos (? T – kx) + y D sin (? T – kx) (1-c) The impossible and the possible In the case that interests us, E and H cannot have a longitudinal component. Can D have it? It is equivalent to asking if the following is admissible. ?? ?? (1) x, y are verses of the coordinate axes Remembering the identityde Moivre ei? = cos? + i sen? (2) we note that (1) corresponds to the complex exponential solution of the wave equation, since in the geometric representation of a complex the real part and the imaginary part are mutually perpendicular, like the components longitudinal and transverse in (1). In the vacuum E and H cannot have a longitudinal component and, for that reason, they do not admit the complex exponential solution. We want to find out if D supports it. That is the task. Now we can understand the title of this document (From James Clerk Maxwell to Max Planck). In quantum theory complex exponentials abound. If we use the same type of function, let’s prepare for the same type of results. The essential thing is to test the consequences of (1) and find out if it is a valid solution for D. (1-d) Something evident in the complex solution The divergence is calculated by deriving each component from the respective coordinate axis. In (1) both components are spatiallyfunctions only of x. The derivative of the transverse component with respect to its axis is equal to zero. The derivative of the longitudinal component with respect to its axis is different from zero. So D has nonzero divergence. D divergence equals densityof electric charge in the region where the field is present. Then (1) attributes a non-zero electrical charge density to the region where the wave propagates. The complex solution describes a wave in an electrically sensitive and electrically active space. We call this sensitive and active space empty. (1-e) Does it load in a vacuum? It’s possible ? To answer we need to start from the definition of D. D = P + eo E definition of the electrical displacement oe (3) The term eo E is transversal, because E is. Geometrically, P must be longitudinal for there to be correspondence between (1) and (3). Is that physically possible? In terms of local resultants, polarization is a field with collinear symmetry that does not alter electrical neutrality. That means that, within a segment of infinite length, there are a pair of equal and opposite vectors, resulting from all local contributions. In the case we are dealing with, could the polarization be transversal? Impossible, since two transversal vectors that correspond toDifferent x values are not collinear. Only 4
D = x D cos (? T – kx) + and D sin (? T – kx) are two longitudinal vectors corresponding to different values of x. This is the property physics that agrees with the geometric condition. The polarization establishes a charge density, which corresponds to bound charge. But we are looking at the spread in a vacuum. Is Maxwellian electrodynamics ready to include vacuum polarization? (1-f) Maxwell and the vacuum It does not matter what Maxwell and other scientists of the time thought about the vacuum. The following questions are interesting. Is Maxwellian electrodynamics sufficient to include the polarization of the vacuum? In the case that interests us, does the polarization imply any movement of charged particles, such as electrons and positrons? The value? eito limit, does it require conceiving of vacuum as a medium with real dielectric properties, capable of polarizing like the known dielectrics? We will not be able to respond without taking everything into account. And to take everything into account at the same time we will need a little math. CHAPTER 2 – The bound charge of the vacuum (2-a) Peak value and rotary vector of the D field For convenience we repeat here equation (1). ?? ?? (1) Where we have (? T – kx) = (2n – 1) p 2 will we have ?? D = and D (at the zeros of P) (4) Let’s repeat (3). For (? T – kx) = (2n – 1) p 2 D = P + eo E (3) results in P = 0, then we have D = eo E (at the zeros of P) (5) In those places the m The module of E has the peak value. So, we have ?? ?? E = y E In (5) we replace E as indicated by (6) D = y eo E (6) (7) (in the zeros of P) (in the zeros of P) 5
D = x eo E cos (? T – kx) + y eo E sin (? T – kx) D = x D cos (? T – kx) + y D sin (? T – kx) Dx = D cos (? t – kx) · D = D k sin (? t – kx) ?? ?? ?? ?? ?? In (7) we replace vector D as indicated by (4) and then simplify D = eo E Too many steps to get to (8), which is trivial! It is true. I request to be excused. Let’s replace in (1) D as indicated by (8) ?? ?? Let’s extract a common factor eoE D = eo E [x cos (? T – kx) + and sin (? T – kx)] (8) (9) (10) ´ ?? The right-hand bracket in (10) is a vector sum whose resultant always has a module equal to 1. The only thing that changes is the angle formed by the resultant with the direction of propagation. A detector located at a fixed point on the x-axis simply sees a rotary vector of module e or E, whosespeedangular is equal to? . In a graphic animation, the vector would look like a clockwise hand. If a detector could travel in a vacuum with the same speed as the wave, it would see a vector that does not rotate. It would look like the hand of a stopped clock. And if the detector were to travel in the propagation direction with an intermediate speed between 0 and C, it would see the vector spin, but more slowly than when the detector was at a point on the x-axis. If someone were to use a detector like this as a clock, the measured times would depend on the speed of the clock with respect to the coordinate origin. (2-b) Load density Let us remember the deviation definition (divD = · D) · D =? Dx? X +? Dy? Y +? Dz? Z (11) The part in coordinate z does not exist . The derivative with respect to y is equal to zero. Then it remains? Dx? X (12) · D = For convenience we repeat here equation (1). ?? ?? (1) (13) (14) According to (1) do we have ?? We carry out the derivative indicated in (12) ?? We see in (14) that the charge density obeys a wave function. Does that involve movement of charge in the direction of propagation? 6
?? Equation (14) does not imply charge movement of any kind. That equation describes how polarization works. In the region where a wave propagates, the polarization of the vacuum is not static, it is dynamic. So the charge density at each point on the x-axis varies periodically. In this case sine wave. Each point in the vacuum is a place where the charge density changes continuously. Are some effects of dynamic polarization equivalent to the effects of mobile charges? Yes, some effects are equivalent and with an analogy we will understand better. (2-c) Analogy of the luminous advertisement Let’s think of a luminous advertisement of those we see in the streets of the city. It is a board that contains many, many small cells capable of lighting up. A programmable electronic device is responsible for controlling the lighting of each cell. This makes it possible to form mobile images. The images move without the cells moving. You can program a wave function in twocolors , red for the positive half cycle and blue for the negative half cycle. In this case, we will see a train of alternating red and blue semi-cycles, which move in the direction of propagation. Colors travel. The cells do not. The seen is very similar to see a train pass by. Each colorIt looks like a wagon that maintains its shape and geometric measurements while traveling. The color is very intense at the peak of the function and gradually fades towards the nodes. Something like the following figure. If the color intensity represents the charge density value and the color represents the sign, then the figure gives an idea of how polarization works in a vacuum. There is no actual movement of cargo. There is a train of virtual electric dipoles traveling in the direction of propagation. The segment that each dipole spans on the x-axis measures a wavelength. In other words, each dipole covers an entire space cycle, as the following figure suggests. Let’s notice something. The systemThe emitter starts its operation emitting the first cycle. When the second emits, from any point in the second cycle the polarization of the first gives zero and the net charge of the first is equal to zero. That is reciprocal. From any point in the first cycle the polarization of the entire second cycle results in zero and the net charge is zero. In other words, each entire cycle is physically autonomous and independent of the others. 7
(2-d) Pause to re ex ect We are starting to become awareof what the complex exponential solution of the wave equation implies in a vacuum. It involves the polarization of the vacuum with immobile bound charges. It implies a continuous distribution of charge in the region where the wave is present. Its dynamics are virtually equivalent to dipoles traveling in the direction of propagation. In this virtual trip, the geometry and measurements of each dipole are kept constant. ?? ?? ?? A dose of mathematics as small as calculating the divergence has revealed profound physical properties. The wave charge density we analyzed cannot be produced by scattered particles in the propagation region. Equation (14) expresses a continuous distribution, whatever the wavelength. For very small wavelengths, a set of scattered particles cannot give a continuous distribution. And for a wavelength less than the size of the particle, there could not be a single particle in the region spanned by a half cycle. The following is definitely imposed. The dynamic polarization of the vacuum corresponds to a physically continuous charge density. Continuity is not a statistical average. It is a physical reality. At each point in the region where the wave is present there is an infinite portion of electrical charge. Error in sight! It is known that the charge is quantized and that it can only appear as an integer number of units and equal to the charge of the electron. A physically continuous load distribution down to the inimitesimal level is absurd! We can stop this now! Right, we can. With the same vehemence we could have left him from the first word. Before leaving it, I would like to analyze what is the basis of the idea that the previous is wrong. Charge is known to be a quantized quantity. What do we omit when saying we know? Let us express fullythe available knowledge regarding the electric load. Experiments based on the ionization of matter have detected particles that are not neutral. In this type of particle, the net charge is a multiple of a minimum unit equal to e. Can the information obtained with ionized matter be extrapolated in a vacuum? That is the question based legitimately on the available knowledge. What do we know about it? Is an extrapolation like that grounded or not? Yes now. We can drop this. We can also continue it, because a legitimate question can lead to a legitimate answer. My decision was to continue. This document shows what was found on the way. 8
j = x CD k sin (? t – kx) × H = x? D sin (? T – kx) + (2-e) Verify the rotors The electromagnetic wave equation is derived from Maxwell’s equations. The deduction is based on the rotors in fields E and B. If the longitudinal component altered a rotor or both we would be in trouble . If we do not alter them, we will know that this component is consistent with Maxwell’s equations. Rotor of the magnetic field In the vacuum we have B = µo H (15) Since being µo constant, it is true × B = µo × H (16) × B does not change if × H does not change. For convenience we will analyze × H. In the propagation direction there is virtual motion of electric charge. In other words, there is a virtual current. We say virtual because we base the conceptin the polarization of the vacuum. A current sensitive instrument will detect the same effects that it detects in vulgar cases. In effective terms, the virtual current fulfills the same lawsthan all currents. By the way of arriving at the concept, we will hardly forget that the dynamic polarization produces a virtual current. So I propose to simply call it current and symbolize it i. This current is present at all points in the region where the wave propagates. We then have a current density j, equal to the charge density multiplied by the velocity of propagation vp. vp? speed of propagation o (17) i? current or ´ (dynamic polarization) (18) or ´ (dynamic polarization) · D (19) (20) j? current density j = vp · D as indicated by (14) In (20) we replace ?? ?? (21) (22) (23) j = vp D k sin (? T – kx) We apply vp = x C ?? We apply C k =? j = x? D sin (? T – kx) × H is given by thelaw of Ampere-Maxwell. ? D? T (24) × H = j + In (24) we replace j as indicated by (23) ?? ? D? T (25) 9
D = x D cos (? T – kx) + y D sin (? T – kx) = -x? D sin (? T – kx) + y? D cos (? T – kx) × H = x? D sin (? T – kx) – x? D sin (? T – kx) + y? D cos (? T – kx) × H = y? eo E cos (? t – kx) ja = x? D sin (? T – kx) jb = -x? D sin (? T – kx) D is given by equation (1), which for convenience we will repeat here. (one) ?? ?? We drift with respect to time, as indicated by (25). ? D? T ?? ?? (26) In (25) do we replace? D? T as indicated (26) ?? ?? ?? (27) In (27) there are two longitudinal current densities that have opposite directions and equal modules. Obviously the sum of both gives zero, but there is an essential physical detail, which we will discuss in another section. Now let us simplify and observe how it is × H. ?? ?? ?? × H = y? D cos (? T – kx) In (28) we replace D with e or E, as indicated by (8). ?? (28) (29) We note in (29) that the rotor of the magnetic field is identical to the result obtained assuming that the longitudinal component does not exist. In other words, it is the same rotor that is used to deduce the wave equation from Maxwell’s equations. The longitudinal component, by definition, does not alter the electric field. Consequently it does not alter × E. ?? ?? ?? (27) Both rotors have the shape that we traditionally know. This means that the complex exponential solution is consistent with Maxwell’s equations. CHAPTER 3 – Inductance of propagation (3-a) The two longitudinal current densities For convenience we repeat here equation (27). × H = x? D sin (? T – kx) – x? D sin (? T – kx) + y? D cos (? T – kx) Let’s symbolize ja to the first monomial of (27), jb to the second and jc to the third. ?? (30) (31) (32) ?? ?? jc = y? D cos (? T – kx) 10
ja and jb are two longitudinal current densities, which have opposite directions and equal absolute values. They are symmetrical. Then we can ignore them, because the vector sum gives zero! Just a moment. Ignore something that explicitly appears in Ampere-Maxwell’s law? Yes! Although we would like to do something with that symmetric pair, we cannot, because on any longitudinal line the vector sum is null! A line is a line! There is no escape ! That protest gives the key. A line is a line … In static mode, where everything is given and finalized beforehand, it is true. A line is a single line. In analytical mode, where there are algebraic and geometric terms that are brought to the infinite limit, a line is as many lines as we need. In the case that we deal with, we only need two parallel lines, one for ja and one for jb. We start with two parallel lines separated by a distance g, we make the necessary statement and then we go to the limit for g? 0, which converts the two lines into the same line (FIGURE 1). In the case of jc we can also start with two parallel transverse lines, separated by a distance equal to g. The same distance can be assumed in both cases and we save In the case of jc we can also start with two parallel transverse lines, separated by a distance equal to g. The same distance can be assumed in both cases and we save In the case of jc we can also start with two parallel transverse lines, separated by a distance equal to g. The same distance can be assumed in both cases and we savenomenclature. In the transverse direction there is a single current density and on both lines it would have the same direction. Then we will have a square with current densities on its sides (FIGURE 2). Physically we can and should think that a line is equivalent to a rod whose cross section is infinitesimal. The shape of the section is of no interest and we can imagine it as best suited, square, circular, rectangular, elliptical section, whatever, because at the infinite limit all forms end up giving a section. on in? nitesimal which obviously has an in? nitesimal area ds. An infinite current density passing through an infinitesimal area gives an infinitesimal current. On the longitudinal sides of the square we have two infinite currents of opposite directions. On the transverse sides two of the same direction (FIGURE 3). Let us represent in perspective the infinite dynamics. (B and ds belong to the same point) 11
(3-b) Physical meaning of (27) and calculation of inductance The magnetic field is perpendicular to the surface of the square represented in Figure 3. That is, the surface vector and the magnetic field are parallel. In the limit for g? 0 we have an infinite square normally crossed by the magnetic field. So the infinite magnetic flow dfB is dfB = B · ds (33) ds? infinitesimal area Since B and ds are parallel is dfB = B ds (34) B? modulus of B The set of infinite currents represented in Figure 3 is the cause of the magnetic field within the infinite square. In the limit for g? 0, the distance between the center of the square and one of the sides is infinite. At a point located at an infinitesimal distance from an infinite current, the magnetic induction due to this current is unique. Each side of the square produces, with its infinite current, an infinite induction in the center of the square. The inductions produced by the transverse currents dic are equal and opposite. That is why this pair of infinite currents does not contribute to induction in net terms. The inductions produced by dia and dib are the same and in the same direction. Then the longitudinal pair is responsible for the net induction. In the limit for g? 0, the infinite day path continues on the dib path and vice versa. That is to say, that pair of infinite currents is equivalent to a loop of infinite radius traveled by an infinite current di = dia = dib. Is this the physical meaning of equation (27), whose discussion was previously started. Let us write the definition of inductance. L = dfB di (35) In the context of the infinite square, the derivative indicated in (35) is equal to the quotient 12
