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1.1 Free Space Maxwell's equations describe the behavior of electromagnetic fields and their interaction with mat- ter. They are one of the major building blocks of classical physics. The equations of Gauss, Ampere, and Faraday describe much of the physics of electric fields, magnetic fields, charges, and currents. Maxwell's insightful addition of the displacement current to Ampere's law connected the set and led to the prediction of electromagnetic waves. An single semester undergraduate course in electromagnetics typically will finish with a derivation of plane waves, which are the one dimensional solution to the wave equation. Here we will extend this by examining what happens when a plane wave interacts with a magnetized plasma. The de- velopment requires simultaneous solution of the plasma fluid equations and Maxwell's equations. To start let's review the derivation of the free space wave equation. Maxwell's equations in a vacuum are: Gauss's Law Faraday's Law Gauss's Law for B Ampere's Law V . E= 0 V x E = aB at V . B = 0 V X B = Hoco at Taking the curl of Faraday's equation and the time derivative of Ampere's law yields: aB aB V XV X E = -VX- (1) V X at = Hoco at2 (2) The vector calculus identity for the curl of the curl is: V X V X E = V (V . E) - V2E The divergence of E is charge density over permittivity as given by Gauss's law, and since we are examining free space there isn't any charge, so the first term is zero. With that, substituting equation 2 into equation 1 gives a wave equation for the electric field: 02 E V2E - HOCO at2 = 0 (3) If we specialize to the case where E has only an x-component that varies in the z-direction thisbecomes: ale. ale. a;2 '\"eo at2 =0 Which has the general solution: F. = Eo cos [wt kzz} it, (4) where E0 is a complex amplitude that may incorporate a phase angle as determined by the ge- ometry and initial conditions. The solution term is that ot a propagating wave. which can be seen by choosing the point on the sinusoid at time t=0 and position z=0 and following that point as time increases. To remain at the same point on the sinusoid. 2 must increase by an amount to maintain the argument equal to zero. That is: w wt kg: = 0 l- 2 = it Which shows that the phase velocity. the speed that a point on the sinusoid moves in the propa- gation direction, is given by: w v = P k: Plugging equation 4 back into the one dimensional wave equation gives: 2 _ v _ 1 k2. p _ unso' which shows that the phase velocity is determined by the properties of the medium. e and p. Plugging in the tree space values will yield 12,, is 3 > Now to find the magnetic flux density plug equation 4 into Faraday's law and assume B will have the same time dependence as E. which will give: B = lEn cos [wt kzz] 9 \"1: The magnitude of B is the magnitude of E divided by the phase velocity. the two vectors are at right angles. and the cross product of E with B gives the propagation direction [it x y = 2). 1.2 Unmagnetlzed Plasmas When a plasma is present the charged particles react to the applied fields of the wave. The force on the particles due to the wave magnetic field (qsv, x B) is smaller than that due to the wave electric field (qu) by the ratio of the particle speed to the phase speed. Which means that unless the particle speed is comparable to the phase speed (relativistic). the force due to the wave magnetic field can be ignored. In the case without a background magnetic field ions are accelerated in the direction of the electric field and electrons are accelerated in the opposite direction. Consider the idealized picture in Figure 1. Displacement oi the charges from their equilibrium positions creates a polarization field that opposes the applied electric field. Each ion and electron 12 - 1.0 - [LB - 0.0 '- n; - 0.2 - 00- 02 Figure 1: With and applied electric field, E, ions (red dots) and electrons (blue dots} are shifted from their equilibrium positions by a distance 5. resulting in a polarization field, P, that opposes the motion. pair has some separation. 6 = 515\" which creates a small dipole moment dened by the product of their charges multiplied by the separation, p = q5 [an electric dipole is defined as pair of charge particles separated by some distance}. Note that 6 points from the negative charge toward the positive charge. is. in the direction of the applied E. The polarization, P, is defined as the sum of the average dipole moments over a unit volume. For a time varying applied electric field. the motion of the ions and electrons is given by the momentum equation. For a single particle, if we consider only the force due to E: 6v; \"ha - '183 or in terms of the displacement 55: alas ms 61:2 = qu Since the displacement will be proportional to the applied field it will have the same time depen- dence. For a time harmonic field, the second derivative can be replaced by w2. Using this we can solve for the displacement: '15 5 = E s 111.5(4)2 Since the displacement is inversely proportional to mass the electrons will move significantly far- ther than the ions, which means we can pretty much ignore the ion motion and 5 = 52. s = '1\" 2E mew __ q _ mewZE where q without subscript is simply the magnitude of the charge without sign (q = q.,). To get the polarization, multiply by charge and sum over a unit volume: P = Z q5 'U 2 nq 2E, mew where n. is the number density of the plasma. In our discussion of the uid plasma equations (lecture 6) we saw that a plasma's natural frequency of oscillation when perturbed. the plasma frequency, was given by: wp = t/an/some. With that we can rewrite the polarization as: In your courses in electromagnetics, the polarization of a dielectric is defined as P = sng. where the constant of proportionality X is called the susceptibility. With that definition we see that the susceptibility of a plasma is: x = (5) It's worth noting that the plasma behaves like a dielectric. but that the susceptibility is negative and it depends on the frequency of the applied wave. Now we have the tools to examine the wave equation in the plasma. Ampere's equation allowing for the presence of currents is: 3E V x B = pa] + llama Mirroring the development of the free space equations, take the time derivative to get: an a] sin VXE=HOE+HOQ (6) The current density can be written as the sum of the charges multiplied by their velocities, so it's time derivative is: a] _ a a a Z qsvs V 3 355 _ gq' at 32 = E Z 'iss V Where we recognize that the summation is just the polarization. This can used in equation 6 to get: 33 621' 62E anmwwew (7' w2 325 = (1' :3) Wow '8' Faraday's law is not changed by the presence of the plasma, however evaluation of 1 had a term with the divergence of E, which is proportional to charge density. It the wavelength of the incident wave is larger than the Debye length, there won't be any significant charge density. Currents flow, but there is no net charge. With that, we can substitute equation 8 into equation 1 and manipulate to get the wave equation for an unmagnetized plasma: 2 2 2 \"JP aE_ We see that this expression is nearly identical to equation 3 except for the constant multiplying the time derivative include the factor relating to the polarization. Since the form of the ditferential equation is the same the term of the solution will be the same. For the plane wave: E = E, cos [wt kzpz] 2, Plugging this expression into the wave equation we get: 2 w -kPE-w2ugg( -)E=O > ki=vp= or as an index of refraction: 2 w n2 = (1 _ a?) (9) This is a significant result that illustrates a fundamental property of plasmas. The phase velocity of a wave propagating through a plasma depends on the frequency of the wave and on the plasma frequency. Hence, two waves of differing frequencies propagating through plasma will have differ- ent phase speeds. Alternatively, because the plasma frequency depends on the plasma density, different regions of a single plane wave propagating through an irregular plasma will also have different velocities. When a medium has this property, we say it is dispersive. The function relating the wave number to the frequency is called the dispersion relation, which here is given by: 2 \"JP kz'p = LIJ ( _ ) \"-060 (10) It's also worth noting that the phase velocity in the plasma is greater than the free-space propaga- tion velocity. On the surface this seems to violate the laws of physics since one of the postulates of relativity is that the vacuum speed of light is the fastest possible speed that something can travel. Further analysis would show that nothing material actually moves with the phase velocity. Proper analysis would show that the energy of the wave propagates with the group velocity which can be shown to be: v _ 6w 0.1%, 1 w2 l-Lueo Which is clearly lower than the vacuum speed. 1.3 Magnetlzed Plasmas Adding a background magnetic field adds an additional complication to developing a description cl waves propagating through a plasma. While it was possible to ignore the inuence oi the magnetic field of the wave in analyzing motion of the ions and electrons, the influence of the background field can be as large or larger than that of the wave electric field and hence must be considered. We will restrict our analysis to the electrons and will ignore motion of the ions. which means that the analysis will be valid for frequencies that are large compared to the ion plasma frequency. We will consider the case where the wave vector is aligned with the z-axis, and choose the coor- dinate system so the magnetic field lies in the x-z plane, and makes an angle a with the z-axis: Bu =Bx+Bz2 =BosinGR+Bgcose The wave will be described by: E\"; t] = Eoewtkz] where k = k2 is a vector the in propagation direction but with the equivalent properties to the wave number in the 1-d equation, and Ea is the vector magnitude. It should be understood that while the field is represented mathematically by a complex exponential, the field itself is real. Also. as it was in the case of the unmagnetized plasma, the electric field is perpendicular to the propagation direction, i.e., k . E = 0. With the background magnetic field the Lorentz force must be included in the momentum equation: ave _ qe at _ E Since the only time dependence on the right-hand side is in the electric field, the time dependence of the velocity will be the same. With that, the time derivative can be replaced with jw. To simplify the development change the cross product notation to a matrix-vector product (you can confirm this by writing out the two expressions): (E+vg x3) 12,, Ex 0 B2 0 v1 Solving for v yields: Vx jw ge BZ 0 Ex me Vu me He B z me jw ge Bx me Ey 0 me jw 0 Dividing out the j, recognizing the gyrofrequency yields: VX W joe cos 0 0 Ex Vy me -jQe cos 0 w joe sin 0 Vz 0 -joe sin 0 w 0 Taking the inverse gives: Vx w2 -02 sin20 -jwhecose -23 cos 0 sin 0 Ex qe Vy me w(w2 - 02 ) -jwhe cos 0 w2 joe sin 0 Ey -2 cos 0 sin 0 -jwhe sine Multiplying the velocity by charge and number density gives the current, which can be used in equation 6 to obtain. x w2 -2 sin20 -jwhe cose -02 cos 0 sin 0 Ex O nede at Jy = -W- me w (w2 - Q2 ) -jwhe cos 0 w2 joe sin 0 Ey -0 cos 0 sin 0 -jwhe sine The fraction nede/me can be replaced with the square of the plasma frequency cow?. With this we can now plug into the curl of Faraday's law (equation 1) to get: w2 - 12 sin20 -jwhe cose -02 cos 0 sin 0 Ex Ex V 2 E = W- HOED (2 ( w2 - ! '? ) -jwhe cos 0 w 2 joe sin 0 Ey - HOEOw2 Ey -02 cos 0 sin 0 -jwhe sine w 2 0 If we call the ugly looking matrix xp and use V2 - -k2 the equation becomes: K2E - HOcow? [1 -Xp] E = 0 where I is the identity matrix. Divide through by -Hocow2 and rearrange to get: K2 HOEDCUZ ) 2 1 - XP - E =0The constant in the parentheses multiplying the identity matrix is the ratio of c2 = 1/|.l.gg to vf, = Lei/k2, which we identify as the square of the index of refraction, n2. When the product of a matrix and a vector is equal to zero. there are two possible solutions. The trivial solution is that the vector is equal to zero. The non-trivial solution is that the determinant of the matrix is equal to zero: DET[(1n3)Ix,,] =0 The algebra of evaluating this expression is prohibitive so we'll skip it. Fortunately for us, it has been done many times and the results are available to us. This expression is known as the Appleton-flames equation, which is usually written in a form like: X{1X] (11) 2 1x IYZsinzd: (itemize) +[1X}2Y2c0526 where X = Lug/LUZ and Y = n/w. Note that the denominator of equation 11 has two possible values depending on which sign is chosen. This property results is two different wave modes that can propagate. The properties of the two modes depend on the angle 3 between the wave vector and the magnetic field. Because the magnitude of n will depend on the sign, one mode will propagate faster than the other, so sometimes the two modes are called the fast mode and the slow mode. The index of refraction is the ratio between the vacuum speed of light and the wave phase speed. It is related to the wave number k by: LU k = 'n. Cu Hence, if 11.2 is negative, the wave number will be imaginary. In regions where this is the case the wave will be attenuated with distance rather than propagating. Waves can't propagate into such regions, which is why waves at frequencies below the peak plasma frequency of the ionosphere refract toward their source rather than propagate through. Careful analysis will show that when the wave vector is parallel to B, (El = 0), the two modes represent two oppositely rotating circularly polarized waves. One of the waves will rotate in the same direction as the electrons, while the other will rotate in the direction of the ions. The two different velocities arise because of this and the large difference in mass between the ions and electrons. 2 Problems Parameter Value Plasma Frequency cup = (211]4 x 10'5 Rod 51 Electron Gyrcfrequency (13 = [2701.4 x 105 Red 51 Angle between k and Bu 9 = 10 Table 1: Constants tor use in plotting 1. (20 pt) Plot the square of the index of retraction for a magnetized plasma from the Appleton- Hartree equation (equation 11) for both modes on one set of axes over the frequency range of 0 MHz to 10 MHz using the parameters trom Table 1. Plot frequency on the horizontal axis and n2 on the vertical axis. Make two plots. first choose the range 01 the vertical axis to be from -100 to 100. then for the second plot choose the range to be from -2 to 2. Make sure you have a horizontal line on your plot for the n1 = D axis. Annotate the plot to indicate which mode corresponds to which plot trace. (you will have two plots with two traces on each plot) 2. (1 Opt) From the plot in step 1, determine the frequency ranges where 11.2 is positive tor the two modes. 3. (10 pt) On the plot of n2 with the plot range 2 ..<_ n2 g overplot vertical lines at the plasma frequency mhz and point e how do these relate to curves of on plot with range s another line for rt2 an unmagnetized comment its relationship other two lines. pt a index retraction modes as tunction angle over b in plots horizontal axis should be noose nsin a. make second same way but use>