Laptag lectures on resonance cones and what you need to derive them

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1 Laptag lectures on resonance cones and what you need to derive them Walter Gekelman Summer 213 The object of this is to derive the formula for resonance cones in a plasma. There are a number of steps along the way. We have to discuss: a) What is the index of refraction of a plasma? b) Wave propagation and - What type of plasma waves are we talking about? c) What is a dielectric in free space and in a plasma? d) Coordinate systems e) Electric potential and electric fields f) Fourier analysis (This is done in detail in other LAPTAG notes on the website, download them and read them NOW!) g) Vectors and complex numbers h) Tensors and matrices i) and of course Maxwell s Equations. As they are covered elsewhere we review them briefly First of all coordinates systems. The first coordinate system you learn about is rectangular coordinates. There a vector r is given by its x,y and z coordinate positions. r = xî + yĵ + z ˆk = x, y,z ( ). Here î, ĵ, ˆk are vectors of length 1 (in any coordinate system, in mks they are 1 meter long, in the English system they are 1 inch long, etc). The vector may be drawn as:!! 1

2 We can also express the vector r in spherical coordinates. Here the vector is determined not by how long its projection is on the three axis but by its length in the ˆr direction (this is a pointer from the origin towards the tip of the vector, it also has length 1). The vector is described not by 3 lengths but by one length and two angles r = r,θ,φ ( ) shown in figure 1. To go from spherical coordinates to rectangular the transformation is: x = r cosθ sinφ y = r sinθ sinϕ z = r cosφ Problem : Find the inverse transformation. That is what is r,θ,φ if you are given x,y,z Let us Briefly Review Electric Fields and the Maxwell Equation that deals with them Electric Force. The force on one charge due to another is: Note we are in the mks system of units (1) F = q q 1 2 ˆr It is a vector quantity. 2 4πε r This is the force on Q due to the presence of charge q. The direction is on the straight line joining them. This is of the same form as gravitation (accident?) It is attractive or repulsive depending on the product of qq. What if you have 3 charges? Then we have to decide which charge we wish to calculate the force on. For example Figure 2-3 charges 2

3 The two red charges are positive and the blue one is negative. What are the forces on charge 2 on the right? The negative charges repel and F21 is the force on 2 due to charge 1. F23 as shown is attractive. To calculate them we simply use the force law: Suppose our test charge q is exposed to more than one charge. The electric field (or force it experiences) is the vector sum of the forces from all the charges. There may be discrete charges, line charges surface charges of volume filled charges. The electric field (also a vector is defined as the force on an infinitesimally small charge: (2) E = F / q = 1 4πε Q r 2 r Lim q Which is real F or E? The utility of the electric field is that all that occurs in it is the source charge Q. The test charge serves as a point indicator telling us how big the field is and which way it points. The next figure (Figure 3) is the electric field due to a single point charge: 3

4 The charge (red ) is at the center. It must be positive because the field lines point outwards. The vectors shown are unit vectors. They all have the same size to make the picture clear. Actually they get smaller by 1 as you go outwards. 2 r If there are more than one charge than For the case of a uniform charge density, ρ (charge per unit volume), dq = ρdv here small volume element contain s a small amount of charge. Then we can re-write (2) as (3) E = 1 4πε filling a volume V. ρ ˆr dv This is how to calculate the the electric field due to a charge density r 2 In the early 18 s Carl Fredrich Gauss (shown above) through experimentation discovered that : (4) Ei ˆndA = q enclosed = 1 ρ dv ε ε The term on the left is the total charge divided by ε and the formal name for the term on the left is the electric flux. Consider a closed surface such as the one in figure 4 4

5 Figure 4 closed Gaussian surface this bounds a volume, think of it as the skin stretched over the volume. da is a small patch of surface area and the black arrow is the normal vector to the surface (perpendicular to the surface). Φ i = E i ΔA i cosθ i = E i i ˆndA i. This is an element of electric flux through the surface. The definition of electric flux is (5) Φ E = Ei ˆndA There are other types of flux (magnetic flux, particle flux ). The integral tells us the total flux crossing the blue surface. Consider this rectangular patch. In figure 5 Figure 5 Electric field lines crossing a square area. The electric flux is proportional to the number of lines that cross. As an example. Suppose we chose our area as the surface of a closed cube and there is an electric field constant in space as shown below in figure 6: 5

6 Figure 6. da1 and da2 point in opposite directions but the field points only to the right and is constant therefore EdA 2 EdA 1 =. You can do this face by face and see that the integral is zero therefore there must be no charge in the cube. Now let us do the same problem of a sphere of charge. Now the patch is a patch of the surface and: 6

7 φ E = surface Ei ˆndA = q enclosed ε = Ei ˆr da = E da surface = E da φ E = E4πr 2 = q ε q E = 4πr 2 ε Note that is q was negative the answer would be negative implying the field lines point inwards. We have Gauss s law as in integral. To derive our plasma equations we need it in differential form. We do this using a famous theorem called the divergence theorem: (6) iadv = iˆnda Where A is any vector. Applying this to equation (4) vol closed area Ei ˆndA = ie dv = 1 ρ dv ε. Since both integrals are over the same volume whats inside of them (the integrand) must be equal. (7) i E = ρ ε This is the differential form of Gauss s law. This equation is derived for media where the dielectric constant ε which reflects the property of the media is just a number. This constant relates charge to electric field. In a plasma with a magnetic field this is no longer a constant. It becomes a mathematical function which depends upon the plasma density, magnetic field and so on. Because of the magnetic it has different values in different directions. It could be a vector but it is not because if you multiply two vectors you get 9 components. It must be a tensor. We will have more to say about tensors as we go on. But now we must re-write (7) as (8) i( εe ) = ρ ext Here ε ε Coulomb s Law and E = Φ Note that is the gradient operator. When operates on a scalar function it produces a vector. The gradient tells us how rapidly a vector quantity varies in space. In rectangular coordinates = î + ĵ + ˆk. x y z The gradient tell us what the slope of a curve is in every direction. 7

8 Consider a field that varies in the x direction only Φ = Φ î ΔΦ î (Lim Δx ). The gradient is the derivative. If increases with x it x Δx points in the direction opposite to that. This is illustrated below ; Φ(x) ΔΦ(x) x slop Above dφ points down where we are taking the derivative (in the middle of the red line) dx therefore the gradient points to the left. In the presence of a dielectric or a plasma the (9) i ε E ( ) = i ε κ E ( )ρ ext where we define κ = ε ε when the electric field vector is multiplied by the tensor κ we get another vector as a result. Each substance has its own κ. We will write down what this is for a plasma. It is derived under the assumptions that for this problem only the electron motion counts (that is because resonance cones are a high frequency phenomena) and that the plasma is in a magnetic field which is constant and points in the z direction. We will also assume that the plasma is cold. This means that we wont worry about the thermal motion of the electrons and only consider their motion in due to the electric field of the resonance cone waves. (1) κ = κ xx κ xy κ yx κ yy κ ω 2 pe (11) κ xx = κ yy = 1 ( ω 2 ω 2 ce ) 8

9 (12) κ xy = κ yx = iω 2 ceω pe ( ω 2 ω 2 ce ) (13) κ = 1 ω 2 pe ω 2 The quantities in the dielectric tensor are where the plasma physics comes in (14) f ce = qb m e = B (B in Gauss) f pe = ne2 m e = n (n # cm 3 ) fce is the electron cyclotron frequency and it tells us how many times a second an electron spins in the magnetic field. fpe is the electron plasma frequency and it depends on the plasma density. Before we go to the nest step we have to expand upon the concept of gradient and potential gradient We first start with the definition of work. b (15) W = ( Fidr ) ; unit of work is the Joule (in the mks system) a ( ). Consider a spring. The force to contract/stretch it is : F = x Here x is the equilibrium position. If it is compressed x<x and the force is negative (you have to exert a force in the negative x direction. The spring is anchored to a wall located at negative x. The work done to compress/stretch the spring is : b (16)W = F x dx = k xdx = 1 2 kx2 ; note it is always positive. a 2 Consider again a system where U+W=E. (W is the kinetic energy, U the potential energy, E the total energy which is constant if energy is conserved). Take the differential of equation (9) In 1D (17) ΔW = F x dx ; ΔU dx = F x In three dimensions F= - U where = î + ĵ + x y z ˆk (This is the gradient) Diagrams of potential are incredibly useful in predicting how a system will go. 9

10 We have now the key equations but before we go on to solve them we have to introduce one more tool we will need. This is Fourier analysis. 1

11 Fourier s theorem lets us express solutions to mathematical equations as an infinite series of sines and cosines. f t ( ) = a + a n cos( nωt ) + b n sin( nωt ) n=1 (To find out more about this download the pdf on Fourier Transforms Whats a Fourier Transform.pdf in the Laptag lectures website ) Fortunately in practice, we usually only need a few terms to get close to the solution. Furthermore Fourier found a prescription to find all of the a s and b s. Instead of using sines and cosines we will switch to exponential notation using the famous Euler s theorem e iθ = cosθ + isinθ. Theta can be anything so let us set it θ = k i r ωt e i ( ki r ωt ) = cos k i r ωt ( ) + isin( k i r ωt), i = 1 k r = x + y + k z z. k is the wavenumber k = k = 2π λ, where lambda is the wavelength of the wave in question. Note that kr = 2πr λ has the dimensions of an angle (pi times something 2π (radians) = 18(degrees) ). Since ω = 2π f (the angular frequency), ωt also has dimensions of an angle in radians. We will use exponential notation and the agree that when we are done with the problem drop the imaginary part of the solution (Got to be real). We are doing this to make life easier. If we stay with sines and cosines it turns out that we will be awash in trig identities such as sina+sinb, sin2a etc.., in our calculations and we are apt to make a mistake. We will now attack the problem by asking What is the plasma response to a rapidly oscillating electrical charge placed at the origin of our coordinate system? First we need a mathematical expression that describes this. Using Euler s (or De Moivre s theorm. Both E and D were contemporaries that came up with the theorem at the same time!) we write (14) ρ ext = qe iωt δ ( r ). We now have a new kid in town, the Dirac delta function. This may be ( ) δ r written as (15) δ r ( ) = r 1 r = r 11

12 This means the charge density is zero everywhere but at the origin and at the origin oscillates like a cosine. The are many useful mathematical approximations to the delta function. For fun some of them are: Try writing an IDL program for sin(nx) nx and make n larger and larger and then plot it as in the graph above. There are many alternate approximate definitions of the delta function some are: δ ( x) = 1 2L + 1 L δ ( x) = 2 L n=1 n=1 sin kx δ ( x) = = 1 πk 2π δ ( x) = k π e k2 x 2 cos nπ x L sin nπ x L cos nπς L, ς L e ikx dk ; Fourier Transform δ ( x) = k 1 π 1+ k 2 π 2 The figure below illustrates the limiting process for two different cases. 12

13 Using Fourier Analysis equations (8) and (9) become: (16) ε ik κ i Φ = qe iωt δ ( r ) or (17) ε ik κ iikφ = qe iωt δ ( r ) = ε k κ ikφ Using Fourier analysis it can be proven that : ik ; i ik i ; ik ; t iω To demonstrate E t = E e i ( ki r ωt ) = iω E e i ( ki r ωt ). Try it for the gradient and curl! t Now we have to do the algebra (18) ε k κ i kφ = ε k z κ xx κ xy κ yx κ yy κ k z Φ 13

14 Equation 18 is an exercise in matrix multiplication. A matrix is an extension of a vector into more dimensions. They are extremely useful objects (they can be used to rotate and translate objects and matrix functions are built into the hardware of all video games) and come with a set of rules on how to multiply, add and subtract them. Consider a matrix A. (19) A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 It has three rows and three columns, like the dielectric tensor any element of the matrix can be written as a ij. For matrix (19) i and j can be 1,2,3. To multiply a matrix by a number ( a scalar) then you simply multiply every term by that scalar. α A = α A ( ). To multiply two matricies, one by another, the rule is given by; (2) ( ) = αa ij A B = C only if c ij = a ik b kj where c ij = a i1 b 1 j + a i2 b 2 j + a i2 b 3 j +... k Note that when you multiply two numbers (3)(7) = (7)*(3). This is NOT rue for matricies Try proving A B B A AB Here are some examples: 2 1 A = ; B = 3 4 ( ) C = A ( B C ) A B = ; A B = ( 2) ( ) and finally A B = Lets go back to equation (18) first multiply the center matrix by the vector on the right (Note a vector is the case of a matrix with one row or one column. The general formula then becomes (21) AB = C = c i = j a ij b j a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 b 1 b 2 b 3 = a 11 b 1 + a 12 b 2 + a 13 b 3 a 21 b 1 + a 22 b 2 + a 23 b 3 a 31 b 1 + a 32 b 2 + a 33 b 3 14

15 Here another example is: = As mentioned matricies represent rotations. Consider two coordinate systems, The primed one is rotated with respect to the unprimed one as shown in the figure below: The distance from the origin to point P must be the same for both observers. What is different is the projections of point P on the axis of both. Show that the transformation is 15

16 Let is get back to equation (18). First we wish to do: κ xx κ xy κ yx κ yy κ k z κ xx κ xy = κ yx +κ yy κ k z. Using the formulas above The equation is now: (22) ε k κ i kφ = ε k z κ xx κ xy κ yx +κ yy κ k z Φ = ε k z κ xx κ xy κ yx +κ yy κ k z Φ ε k κ i kφ = ε κ xx k 2 x κ xy +κ yx +κ yy k 2 2 ( y +κ k z )Φ But since kxx=kyy (see components of tensor) let us call them both k. = y, κ = k zz k 2 = k 2 2 x + ε κ xx k 2 x +κ xy κ yx +κ yy k 2 2 ( y +κ k z )Φ = ε (k 2 +κ k 2 z )Φ 16

17 Finally our equation is κ 2 k 2 2 ( +κ k z )Φ = q e iωt δ ( r ) ε (23) Φ( k) = q ( ) e iωt δ r ε 2 k 2 +κ k z 2 ( ) Note we are in cylindrical coordinates as shown in the coordinates above. The z axis is along the magnetic field Then use on of the definitions of the 3D delta function (it is best suited to this problem) (24) δ ( r ) = 1 ( 2π ) e i2π 3 (25) Φ( k) = qe iωt ε kir d 3 k e i kir κ 2 k 2 2 ( +κ k z ) e iωt d 3 k ( 2π ) 3 The quantity we are interested in,, is a function of k. Since the intergral is over k are actually working in k space or wavenumber space. The cylindrical coordinate system above has axis labels kx,ky,kz. Note that for a wave the wavenumber is defined as k = 2π λ where l is the wavelength of the Fourier component. This is a direct result of using Fourier analysis, But k = ( k sinφ,k cosφ,k ), r = ( ρ,,z) (in cylindrical coordinates, k space). Therefore ki r = ρk sinφ + zk d 3 k = k dk dk dφ For example for an area of element in k space 17

18 ! This is a volume element in cylindrical coordinates. But it is in k space. Why. A cylinder volume is an area times a length. The area is a circle. In r space (rather than k space) V = πr 2 z. Here r is perpendicular to z. The differential is the differential of the area time the differential in z dv = d(πr 2 )dz = 2πrdrdz. In k space k is the component of the wavenumber in the plane perpendicular to the magnetic field so its just like a radius. 2π = 2π dφ. If nothing depends on angle then you get 2π but if things are angularly dependent then you have to do the integral. As you see the latter is the case. and equation 21 becomes (26) Φ( ρ,z) = qeiωt ( ) e i(ρk sinφ+zk ) ε κ 2 k 2 2 +κ k z k dk dφdz = qeiωt ( 2π ) 3 2π κ 2 k 2 2 ( +κ k z ) ( ) 2 ε e ikz k dk dk k = k = 2π e iρk sinφ We are integrating over all possible wavenumbers both parallel and perpendicular to B and all angles so the answer depend on space and time only. It turns out that the integral over phi can be done! Its complicated and leads to dφ 2π 2π e i( nβ xsinβ ) dβ = 2π J n x ( ) J is a Bessel function 2π Let n=; 2π e i( xsinβ ) dβ = J x ( ) Let x = k ρ Then (27) Φ k ( ) == qe iωt k e ikz dk dk ( 2π ) 2 ε k = k = ( ) κ 2 k 2 2 ( +κ k z ) J k o ρ What are Bessel Functions? They are in a way like sines and cosines. The differential equation and solution for sines and cosines is (28) 2 y x 2 = α 2 y ; y=acos( αx) + Bsin( αx) 18

19 Bessel functions also are solutions to (of course) Bessel s equation which is 2 y (29) x 2 x + x y 2 x + ( x2 α 2 ) y = This equation pops up quite a bit in Physics. It is the equation for the motion of waves on the head of a drum when it it struck by a drumstick. The Bessel function solution is (26) J α ( x) = m= ( 1) m 1 m!γ(m + α + 1) 2 x ( 2m+α ) and in our case (26 ) J ( x) = m= ( 1) m 1 m!γ(m + 1) 2 x ( 2m) The gamma function is defined as : 1i 2 i 3i 4 i in ( ) Lim n z( z +1) z + 2 Γ z ( )(z + 3)i i z + n Where z is any number. If m is an integer ( ) n2 Γ( n) = ( n 1)! The! symbol means factorial for example 3!=3*2*1 = 6. Also! = 1 For m=4 for example 1 Γ( m 1) = 1 3! = 1. Therefore the J s are all infinite series. But so are the sines and cosines. 6 The first fews gammas are Γ(1) = 1 ( ) = 1 ( ) = 2 ( ) = n 1 Γ 2 Γ 3 Γ n ( )! Homework calculate and plot z,γ z ( ) where z=.1,.2, 5. Gamma has many other definitions it can also be expressed as an integral. ( ) = e t t z 1 Γ z dt 19

20 Homework plot the Jn s for various values of n. The places where J goes to zero are called the zeros of the Bessel functions. J ( k ρ) =. Find the first 2 and plot verses the number of the zero The gamma functions have been studied for quite some time and one of their unusual properties is Γ( z)γ( 1 z) = π sin mπ where m is the same as in the above factorial definition. Let s plot the zero order Bessel function Note other functions can be expressed as infinite series for example: cos x = 1 x2 2! + x4 4! x6 6!... ; cos x = 1 m= ( ( x 2m ) )m ( 2m)! IDL has all the Bessel functions built in! Note another definition of the delta function is: ( ) = ρ J m ( kρ) (28) δ ρ ρ' J m ( kρ' )dk This is analogous to a similar integral for sines 2

21 2π sin mxsinnx dx = πδ mn Note the sines can used used in a Fourier series to build up any well behaved mathematical function. Since the Bessel functions behave in a similar fashion they too can be used in an infinite series to build up any function and in fact that is what we just did. They are part of the k space Fourier analysis of the potential. There is a special name for mathematical function that have the this property, they are called orthogonal functions. Aside from the sines and cosine there are an infinite amount of other orthogonal functions. Next we integrate (27) by looking it up! First of all since the integral over k is an even function of k (that means it is symmetric about k =) (29) φ( k) = qe iωt 4π 2 ε k cosk z k = k dk dk 2 k 2 +κ k z 2 ( ) ( ) J o k ρ = qeiωt 2π 2 ε cosk z k = k dk dk 2 k 2 +κ k z 2 ( ) ( ) J o k ρ We will do this for the special case of κ <. What does this mean? (3) κ = 1 1 ω 2 pe ω 2 ω 2 pe ( ω 2 ω 2 ce ) approximate (29) by κ ω 2 pe ω 2 ω 2 pe ω 2 ce ω 2 κ ω 2 ( ) ( ce ω 2 ) ω 2 ω 2 ce ω 2 ω 2 ce ω 2. For our experiment (verify this!) ω pe ω ce, ω <<ω pe,ω ce We can <. Once again if the field is large enough When the ratio of κ < the integral can be re-written as (31) φ( k) = qe iωt 2π 2 ε iπ 2 H ( 1) k αρ cos k z dk and 21

22 α = κ ( ) > Note for our conditions: κ ω 2 ce ω 2 ω 2 ω 2 ce (what is this ratio?) 2 ω Ho is a new function called a Hankel function. Guess what these are also orthogonal functions and are combinations of Bessel functions. cos nπ (32) H 1 n ( x) = J n ( x) + in n (x) ; N n (x) = ( )J n x sin nπ ( ) J n ( x) ( ) where x = αρk Also H 1 n ( x) = i ( ) n+1 eix x n m= i m ( ) m m! 2x ( n + m)! ( n m)!, n The N s are called Neumann functions. Note for n= you must use a limiting procedure to get N! But for n = N ( ). where γ is the x value for which N(x)= the first zero of the ( x) = 2 ln x + γ ln2 π Bessel function Jo occurs at x=2.448 thus N ( x) 2 ln x π ( ) From the definition equation (31) the integral becomes (33) φ ( ρ,z) = iqeiωt J k αρ 4πε κ ( ) + in ( k αρ) cosk zdk These two integrals can also be looked up: N ( ax)cos( bx) dx = and J ( ax)cos( bx) dx = 1 b 2 a 2 1 a 2 b 2 ; <a<b, zero otherwise ; <b<a, zero otherwise (except when a=b) This has solutions depending on the ratio αρ > z, αρ < z. Since the magnitude of α = ω 2 ce ω 2 and in our experiment B=1G, f = 1 Mhz α Then αρ < z 22

23 (34) φ ( ρ,z) = iqeiωt 4πε We can simplify this i = qeiωt 1 z 2 α 2 ρ 2 4πε z 2 κ ρ 2 (35) φ ( ρ, z) = qeiωt 1 qe iωt = 1 4πε z 2 κ + ρ 2 4πε κ z 2 κ + ρ 2 κ This is similar to the potential of a charge in vacuum φ( r,z) = q 1 4πε ρ 2 + z 2 but the plasma dielectric function modifies the r and z dependence. What we are measuring in the LAPTAG experiment is the electric field. Since as we saw before E = φ we have to take derivative. Remember we are in cylindrical coordinates so : φ = φ r ˆr + 1 φ ˆφ + φ r φ z ẑ Since our answer does not depend on the angle phi we just have to take two simple derivatives: qe φ ( ρ, z) iωt 1 = ˆρ + 4πε κ ρ z 2 + ρ 2 z κ We are interested in the radial component of E so 1 ẑ z 2 + ρ 2 κ (36) E r = qe iωt 1 4πε κ + ρ 2 κ 2 z2 2ρ qe iωt ρ = 4πε κ 2 κ z 2 + ρ 2 κ 3 κ

24 This is what we measure. The electric field as a function of radius at a fixed axial distance is shown in the figure below: E(r) at fixed axial (Z) location Let us graph this for our experiment: B=1G, fwave = 7 MHz and the plasma density 2X1 8 cm -3 (late afterglow) we plot the electric field as a function of radius 1 cm from a point source antenna If we plot the electric field in the r-z plane then we get the figure above The is actually a plot of the log of the electric field because it drops off so rapidly from the exciter. If you neglect the dielectric function it goes as 1/r2 but with the dielectric it is worse.. The y axis here is along z. r is in cm as well as z. Given the expression Eq 36 for the electric field we can ask Is there a location at which the electric field becomes infinite? This would occur when the denominator of equation 35 goes to zero. In nature, of course, things don t become infinitely large. We left out collisions of the electrons with neutral particles for example in our derivation. These and other collisions would add another term to the denominator which is imaginary. This prevents the denominator from ever becoming zero. The condition is: z 2 + ρ 2 z2 κ = This is satisfied when : = ρ 2 κ. 24

25 Define an angle theta such that tan 2 θ = ρ z Our condition is now: (37) tan 2 θ = κ Theta defines what is called the resonance cone angle. There is a cone in the cylindrical geometry of ρ,z ( ) (38) tan 2 θ = κ 2 ω 1 pe ω 2 ω 2 ce = 1 ω 2 pe ω 2 The cone angle depends on the density, the magnetic field and the frequency of the wave. For our conditions ω < ω ce,ω pe and tan 2 θ tanθ f 1 ω 2 pe f pe ω 2 ce ω 2 pe ω f ce = ω 2 pe ω 2 ce ω 2 pe ω 2 1 = ω 2 ω 2 pe + 1 ω 2 ce For the general case prove that (39) sin 2 θ c = ω 2 ( ω 2 pe + ω 2 ce ω 2 ) ω 2 peω 2 ce For conditions similar to those in the Laptag device the figure below shows the resonance cone angle for fixed density and magnetic field 25

26 26