Analytical Advances about the apex Field Enhancement Factor of a Hemisphere on a Post Model
UUniversidade Federal da BahiaInstituto de F´ısicaPrograma de P´os-gradua¸c˜ao em F´ısicaAdson Soares de Souza
Analytical Advances about the ApexField Enhancement Factor of aHemisphere on a Post Model
Salvador,2019 a r X i v : . [ phy s i c s . c l a ss - ph ] S e p niversidade Federal da BahiaInstituto de F´ısicaPrograma de P´os-gradua¸c˜ao em F´ısicaAdson Soares de Souza Analytical Advances about the apexField Enhancement Factor of aHemisphere on a Post Model
Dissertation submitted to Programa de P´os-gradua¸c˜ao em F´ısica, Instituto deF´ısica, Universidade Federal da Bahia, as a partial requirement for obtaining aMaster’s Degree in Physics.Supervisor: Prof. Dr. Thiago Albuquerque de Assis.Salvador,2019 bstract
In this dissertation, we analytically study the apex field enhancement factor(FEF), γ a , by constructing a method which consists in minimizing an errorfunction defined as to measure the deviation of the potential at the boundary,yielding approximate axial multipole coefficients of a general axial-symmetricconducting emitter shape, on which the apex FEF can be calculated by sum-ming its respective Legendre series. Such method is analytically applied for aconducting hemisphere on a flat plate, confirming the known result of γ a = 3.Also, it is applied on a hemi-ellipsoid on a plate where the values of the apex FEFare compared with the ones extracted from the analytical expression. Then, themethod is applied for the hemisphere on a cylindrical post (HCP) model. In thiscase, to analytically estimate the apex FEF from first principles is a problem ofconsiderable complexity.Despite the slow convergence of the apex FEF, useful analytical conclusionsare drawn and explored, such as, it is confirmed that all even multipole contri-butions of the HCP model are zero, which in turn leads to restrictions on thecharge density distribution: it will be shown the surface charge density must bean odd function with respect to height in an equivalent system. Also, expres-sions found for the apex FEF depend explicitly on the aspect ratio, that is, theratio of height by base radius. Using the dominant multipole contribution, thedipole, at sufficient large distances, it is shown that, for two interacting emit-ters, as their separation distance increases, the fractional change in apex FEF, δ , decreases following a power law with exponent −
3. The result is extended forconducting emitters having an arbitrary axially-symmetric shape, where it isalso shown δ has a pre-factor depending on geometry, confirming the tendencyobserved in recent analytical and numerical results.1 esumo Nessa disserta¸c˜ao, estudamos analiticamente o fator de amplifica¸c˜ao por efeitode campo eletrost´atico (FEF), γ a , por construir um m´etodo que consiste emminimizar uma fun¸c˜ao de erro definida para medir o desvio do potencial nasuperficie, produzindo coeficientes multipolos axiais aproximados de uma su-perf´ıcie emissora condutora axialmente simetrica, no qual o FEF no ´apice podeser obtido somando a respectiva s´erie de Legendre. Tal m´etodo ´e aplicadoanaliticamente para uma semi-esfera condutora sobre uma placa plana, confir-mando o resultado conhecido γ a = 3. Ademais, o m´etodo foi aplicado a umsemi-elips´oide sobre uma placa, onde os valores do FEF no ´apice s˜ao compara-dos com os valores extra´ıdos a partir da express˜ao anal´ıtica. Em seguida, om´etodo ´e aplicado para o modelo de uma semi-esfera sobre um poste cil´ındrico(modelo HCP). Neste caso, estimar analiticamente o FEF no ´apice a partir deprimeiros princ´ıpios ´e um problema de consider´avel complexidade.Apesar da lenta convergˆencia do FEF no ´apice, conclus˜oes anal´ıticas ´uteiss˜ao tiradas e exploradas, como por exemplo, ´e confirmado que todas as con-tribui¸c˜oes de multipolos pares do modelo HCP s˜ao nulas, o que por sua vezleva a restri¸c˜oes na distribui¸c˜ao de densidade de carga: ser´a mostrado que adensidade de carga superficial deve ser uma fun¸c˜ao ´ımpar em rela¸c˜ao `a alturaem um sistema equivalente. Al´em disso, express˜oes encontradas para o FEF no´apice dependem explicitamente da raz˜ao de aspecto, ou seja, a raz˜ao da alturacom rela¸c˜ao ao raio da base. Utilizando a contribui¸c˜ao de multipolo dominante,o dipolo, a distˆancias suficientemente grandes, mostra-se que, para dois emis-sores interagentes, `a medida que sua distˆancia de separa¸c˜ao aumenta, a varia¸c˜aofracion´aria do FEF no ´apice, δ , diminui seguindo uma lei de potˆencia com ex-poente −
3. O resultado ´e estendido para emissores condutores com geometriaarbitr´aria eixo-sim´etricos, onde tamb´em ´e mostrado que δ tem um pr´e-fatordependendo da geometria, confirmando a tendˆencia observada em resultadosrecentes, anal´ıticos e num´ericos. 2 cknowledgements I’d like to thank a couple people, which, in a way or another, helped me arrivewhere I am right now.Jehovah, the God of the Bible, for everything He has done for me.Professor Thierry Jacques Lemaire, whose lectures (besides really awesome)were the primary reason why I decided to choose the physics course when I wasan undergraduate student, still in engineering.Professors Mario Cezar Ferreira Gomes Bertin, Newton Barros de Oliveira,Gildemar Carneiro dos Santos, Thiago Albuquerque de Assis, and Diego Cata-lano Ferraioli, for the great lectures when I was an undergraduate student.These were really enjoyable classes.Professor Diego Catalano Ferraioli for encouraging me and giving me theinitial kick, in a matter of speaking, for me to start the master program in thefirst place.Professors Julien Chopin and Denis Gilbert Francis David for the awesomelectures while I was in the graduate program, full of really interesting math,calculations, and interesting topics outside the main books.My supervisor, Thiago Albuquerque de Assis, for exposing me to scientificliterature and some open problems, for showing me how things work in thescientific world, for showing me some aspects about what it takes to be a goodscientist, for the (almost always) weekly meetings with insightful talks, andespecially for always being motivating.My dad, for his support and encouragement to write and finish this disser-tation and the research.Also, I’d like to thank CNPq for finantial support.3 ist of abbreviations andsymbols
FEF Field Enhancement FactorHCP Hemisphere on a cylindrical post N The set of natural numbers. That is, N = { , , , . . . } . Z The set of integer numbers. That is, Z = { . . . , − , − , − , , , , , . . . } . h Height of an emitter. R Base radius of an emitter. ν Aspect ratio of an emitter, defined as ν = h/R . (cid:96) Defined as (cid:96) = h − R . V Electrostatic potential. See (2.1). θ Azimuthal angle in spherical coordinates. r Radial distance in spherical coordinates. J Spherical element of area, defined as dS = Jdθdφ . J A Component of element of area, defined as J = J A r sin θ . P l ( x ) Legendre Polynomials. E Magnitude of applied electrostatic field in positive z-direction. F M Magnitude of applied electrostatic field in negative z-direction(that is, F M = − E ).Σ Error function measuring deviation from real potential, de-fined in (2.2). A l Legendre coefficients of the potential, defined in (2.1). A ( n ) l Approximate value of A l , when calculated from linear system(2.6), of dimension ( n + 1) × ( n + 1). Note: by definition, forsome fixed n , it is true that A ( n ) l + n = 0 for all l ∈ N .4 A l Defined as ˜ A l = A l /E .˜ A ( n ) l Defined as ˜ A ( n ) l = A ( n ) l /E . γ a Field Enhancement Factor at the apex of an emitter. γ ( n ) a Approximate value of γ a , by plugging ˜ A ( n ) l at (2.8). G l G-Integrals, as defined in (2.4). I ij I-Integrals, as defined in (2.4). Q l The l -th axial multipole moment. See (2.9) and (2.10). M l The l -th multipole moment of a line of charge. See (4.48). c The distance between the symmetry-axial lines of two emit-ters. δ Fractional reduction in apex FEF.Φ( r ) Newtonian Kernel, defined at (5.26). h ( r ) Resolvent Kernel, defined implicitly by (5.24). See also (5.32). W Interacting first order contributions for the resolvent kernel.5 ontents ν . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.2 Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.2.1 Cylindrical I-Integrals . . . . . . . . . . . . . . . . . . . . 284.2.2 Cylindrical G-Integrals . . . . . . . . . . . . . . . . . . . . 304.2.3 Hemispherical area element . . . . . . . . . . . . . . . . . 314.2.4 Suspended hemispherical G-Integrals . . . . . . . . . . . . 314.2.5 Suspended hemispherical I-Integrals . . . . . . . . . . . . 334.3 HCP Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.4 HCP Multipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.4.1 Line of charge model . . . . . . . . . . . . . . . . . . . . . 354.5 Interacting HCPs . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.5.1 Two interacting HCPs . . . . . . . . . . . . . . . . . . . . 366.5.2 One-dimensional regular array of HCPs . . . . . . . . . . 394.6 HCP FEF for (cid:96) (cid:28) R . . . . . . . . . . . . . . . . . . . . . . . . . 404.7 FEF for HCP Model . . . . . . . . . . . . . . . . . . . . . . . . . 424.8 Numerical Results . . . . . . . . . . . . . . . . . . . . . . . . . . 43 A.1 Cylinder in spherical coordinates . . . . . . . . . . . . . . . . . . 56A.2 General symmetrical shape in spherical coordinates . . . . . . . . 57A.3 Prolate spheroid in spherical coordinates . . . . . . . . . . . . . . 59A.4 Suspended hemisphere in spherical coordinates . . . . . . . . . . 60A.4.1 Element of area . . . . . . . . . . . . . . . . . . . . . . . . 61A.4.2 Approximation: (cid:96) (cid:29) R . . . . . . . . . . . . . . . . . . . . 61A.4.3 Approximation: (cid:96) (cid:28) R . . . . . . . . . . . . . . . . . . . . 62 B Spheroidal Integrals 63
B.1 G-Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63B.2 I-Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
C Suspended Hemispherical Integrals 69
C.1 Exact case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69C.1.1 G-Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 70C.1.2 I-Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 71C.2 Approximation (cid:96) (cid:28) R . . . . . . . . . . . . . . . . . . . . . . . . 72C.2.1 Approximating r . . . . . . . . . . . . . . . . . . . . . . 72C.2.2 Approximating r n . . . . . . . . . . . . . . . . . . . . . . 73C.2.3 Approximating J A . . . . . . . . . . . . . . . . . . . . . . 73C.2.4 Approximating J . . . . . . . . . . . . . . . . . . . . . . . 75C.2.5 Approximating r n J A . . . . . . . . . . . . . . . . . . . . . 76C.2.6 G-Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 77C.2.7 I-Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 797 hapter 1 Introduction
In the context of field emission at low temperatures, often called cold fieldelectron emission (CFE), protrusions in the micro or nano scale over a flat con-ducting plate, experiences an amplification of the external applied electrostaticfield at some parts of its surface [1]. A common characterization parameter ofsuch phenomena is the local field enhancement factor (FEF) at the apex, de-noted by γ a = F a /F M , which is the ratio between the field at the apex of thesurface F a , and the external macroscopic field F M . For an ideal emitter, F M is taken to be F M = V /d , where V is the voltage difference between the twoplates, and d its separation distance. If the distance d is large enough, the apexFEF is known to depend only on the geometry of the protrusion [1].The most common method of measuring emitter characteristics, is by meansof a current-voltage (IV) plot: experimental control over the plate voltage V is assumed, and the emitted current I can be measured, for instance underhigh ultra-vacuum conditions. Field emission can be modeled by a Fowler-Nordheim (FN)-type equation: J = AF M exp( − B/F M ) [2–4], where A and B are parameters which may depend on the macroscopic field F M , J is a currentdensity, which in some contexts is taken as a macroscopic current density J M ,some characteristic current density at a characteristic point J kc , or the localcurrent density J l , depending on what is wanted to be analyzed, or measured.The emission is called orthodox if some conditions are met [5], such as: (1)uniform voltages on the electrodes, (2) the measured current is only due toCFE phenomena, (3) deep tunneling, that is, the Fermi level is “much” lessthan the top of potential barrier in which they tunnel, (4) the work function ofthe emitter is constant across its surface.The parameters A and B are relevant. For instance, one can model thelocal current density by using a one dimensional exact triangular (ET) potentialbarrier. It will depend on the probability of an electron to tunnel the barrier,and, then, J is given by an FN-type equation, where B depends on γ a , and A can be assumed to be a constant dependent on fundamental physical andmathematical constants. To extract the coefficients A and B , one can plotln( I/V ) vs 1 /V , which is known as a FN plot. If the emission is orthodox,8hen, a FN-plot will be a nearly straight line, where the slope will depend on B . This procedure allows the FEF to be measured experimentally, by lookingat the slope of a nearly linear FN type plot [4, 5].As for a technological example: a protrusion might be engineered into aflat surface, with the aim of causing field emission at lower external fields. Forinstance, post-like carbon nanotubes (CNTs) may be used as a single or a largearea emitter. That indeed has several applications, such as, the manufacturingof microwave amplifiers, electron microscopy, field emission displays and otherapplications [6–10]. There has been several works that use a classical modelto estimate the apex FEF of arrays of CNTs [11–13], as well as experimentalmeasurements [14–16].In an attempt to model an isolated post-like protrusion, like a CNT, severalmodels have been made, treating the protrusion as a classical conducting entity,with a specific geometry [1]. Some of them will be listed as follows.The hemisphere on a conducting plate model assumes a hemisphere of radius R above the flat plate, and attempts to calculate the apex FEF. This case canbe analytically solved from first principles [17], yielding a known γ a = 3. Usingit to model the FEF of a general emitter creates some problems: the model failsto capture the height of the emitter.The hemi-ellipsoid on a conducting plate model, as shown in figure (1.1a),is a conducting revolution hemi-ellipsoid of base radius R and height h , placedover a flat conducting plate. It can also be solved analytically using spheroidalcoordinate system [18], thus γ a is known. Such model is more realistic thanthe former, because it can have any height, however, the greater the height, thegreater the Gaussian curvature at the apex, and thus, the greater the chargedensity, which in turn, leads to large (and possibly unrealistic) apex FEFs.The hemisphere on a cylindrical post (HCP) model, as shown in figure (1.1b),is a conducting hemisphere of radius R , over a cylindrical shape of height (cid:96) , overa plate. The overall height of the structure is often called h = R + (cid:96) . Such model,in turn, has a smooth curvature at the apex, and can have any height. On theother hand, this model lacks an analytical solution in the literature, thoughaccurate numerical computations have been done [19].In the case of a CNT, there’s no reason to believe that classical models wouldbe effective, as quantum effects might be relevant. However, a recent paper [20]have numerically explored the possibility for determining a characteristic FEFof a CNT using density functional theory (DFT) [21], a quantum mechanicalmethod, and compared with the the local FEFs of the HCP model. The authorsfound that, depending on how the FEF is defined in the quantum case, thereexists a relation between the FEF calculated from DFT, and the FEF calculatedfrom the classical HCP model. This highlights the relevance of classical models,which will be the focus of this dissertation.It has been focus of several works to find an expression for the FEF in thecontext of the HCP model. Some of them includes: • γ a ≈ ν (Eq. (16) of [1]) • γ a ≈ ν if ν (cid:29) γ a ≈ . .
15 + ν ) . (Eq. (6) of [22]) (Eq. (20) of [1]) • γ a ≈ . . ν ) . (Eq. (2) of [12]) • γ a ≈ .
93 + 0 . ν − . ν (Eq. (22) of [1])In this work, a mathematical method is constructed in order to calculate thepotential over an emitter shape with axial symmetry from first principles, andthe method is applied for the case of a hemisphere on a plate, a hemiellipsoidon a plate, and a HCP on a plate. The method allows analytical conclusions tobe drawn, and these will be explored.In addition, interacting emitters, either in the form of finite arrays, or simplytwo of them separated at a distance c from each other, experiences a reductionin the FEF: It falls from γ a to γ (cid:48) a , where the fractional change in apex FEF, δ = ( γ (cid:48) a − γ a ) /γ a , was attempted to be estimated by several works in differentcontexts. • − δ = exp (cid:0) − . c h h (cid:1) , in ref [23], where h = 2 µm . This formula isdimensionally incorrect, as the argument inside the exponencial is notadimensional. • − δ = exp (cid:0) − . ch (cid:1) , in ref [24]. This fixes the dimensional error fromthe previous formula. However, ref [25] says such formula doesn’t providea good fit for arrays of various geometries. • − δ = exp (cid:16) a (cid:2) ch (cid:3) k (cid:17) , in ref [25]. Parameter k was introduced in order tobetter fit arrays. Furthermore, [25] claims there’s agreement with line ofcharge (LCM) model. • − δ ∼ r/(cid:96) ) (cid:0) (cid:96)c (cid:1) , from [26]. This was obtained by a more analyticalapproach using the floating sphere at emitter plane potential (FSEPP)model, as opposed to curve fitting (as was done with all the others aboveit). The striking contrast is, instead of an exponential behavior, a powerlaw was predicted, with exponent − • Ref [27] numerically estimated δ for pairs of 6 different emitter geometries,and found the c − law. • Ref [28] used method of images to find δ ∼ c − . • Ref [29] used LCM models to find δ ∼ c − . • Ref [30] argues about universality of δ ∼ c − .This work will show the c − law is valid for any axially-symmetric emitter,under an independent approximation (charge density doesn’t change much as c varies), and under approximation c/h (cid:29)
1. Using these approximations, itwill be shown that − δ = K · ( h/c ) , where h is the height of the emitter.Furthermore, it will be shown the pre-factor K is not 1, and in fact, depends10n the aspect ratio, that is, K = K ( ν ), where the functional form of K will begiven.This dissertation is organized as follows: Chapter 02 explains the method, itsconnection with the multipole coefficients, and proves how multipole momentsscales linearly with the external macroscopic field. In chapter 03, the method isapplied to model shapes in which the apex FEF is known: the hemisphere on aplate model, and the hemi-ellipsoid on a plate model. In chapter 04, the methodis applied to the HCP model. It will be shown that even multipoles are zero forthe system HCP emitter on a plate, and some consequences of it. Additionally,an approximate first-order expression for the FEF will be calculated for therange h (cid:28) R . Chapter 05 will prove some theorems for a general emitter, aslong as some conditions are met. Chapter 06 will conclude the dissertation, witha review of the main results obtained, and perspectives of future work. For thesake of transparency, there will be several appendixes, in which some technicalderivations are shown in detail. 11igure 1.1: (a) Hemiellipsoid over a plate model. (b) Hemisphere on a post(HCP) model. 12 hapter 2 Method
In general, it is possible to expand the electrostatic potential by means of spher-ical harmonics. However, if a system has axial symmetry (such as, the HCPmodel), the potential can be written [17] by means of Legendre polynomials, inthe following way: V ( r, θ ) = − E r cos θ + ∞ (cid:88) l =0 A l r − ( l +1) P l (cos θ ) . (2.1)In (2.1), E is the magnitude of the external field, oriented towards thepositive z-axis. In the context of field emission, the applied electrostatic fieldis often considered to be F M (called macroscopic field), oriented towards thenegative z-axis (that is, F M = − E ) (this also implies F = − E ).The problem of interest is to find V ( r, θ ), where, the emitter coincides withan equipotential surface. In order to find the proper coefficients A l , it is imposedthe following boundary condition: V = 0 at the surface S . Based on this, anerror function, to measure how far the solution deviates from the correct one,is defined as the following: Σ( A l ) def = (cid:90) S V ( r, θ ) dS. (2.2)Notice that, if V is smooth, Σ( A l ) = 0 if and only if boundary conditionsare met. Otherwise, Σ( A l ) > A l ) = (cid:90) S E r cos θdS − ∞ (cid:88) i =0 E A i G i + ∞ (cid:88) i =0 ∞ (cid:88) j =0 A i A j I ij , (2.3)where the quantities G i and I ij , which will be called G -integrals and the I -integrals from now on, are defined as: G i def = (cid:90) S r − i P i (cos θ ) cos θdS, I ij def = (cid:90) S r − i − j − P i (cos θ ) P j (cos θ ) dS (2.4)13s for a property visible from the definition, the I integral is symmetric,that is, I ij = I ji .The goal is to find all A l that minimizes Σ( A l ). A necessary condition is: ∂ Σ ∂A l = 0 , for all l ∈ { , , , ... } . (2.5)Imposing (2.5) at (2.3), then: ∂ Σ ∂A l = 0 = ⇒ ∞ (cid:88) k =0 I lk A k = E G l . The above equation can be written in matrix form: I I I I · · · I I I I · · · I I I I · · · I I I I · · · ... ... ... ... . . . ˜ A ˜ A ˜ A ˜ A ... = G G G G ... (2.6)Where ˜ A l = A l /E . If the linear system can be solved, then the values of A l are obtained, and one can find the electrostatic potential (2.1). However, it isreasonable to assume truncating the system yields approximate solutions for thevalues of A l . Truncating the I -Matrix into a n × n matrix, and the G, A -Vectorsinto a n -dimensional vector, then the now finite linear system can be solved forthe A l values. Definition 2.0.1.
I-Matrix, G-Vector, A-Vector, normalized A-Vector are thetruncated matrix filled with the integrals I ij , the truncated vector filled withthe values G l , A l and ˜ A l , respectively. Even if the values I ij and G l doesn’t depend on where the system is truncated(that is, on the value of n ), however, A l does. Thus, if the system is truncatedhaving a matrix of dimensions ( n + 1) × ( n + 1), the A-Values will be called as A ( n ) l . As an example, a good exercise is to solve for n = 0 and n = 1. Truncating at n = 0On that, the matrix is of dimension 1 × I ˜ A (0)0 = G = ⇒ ˜ A (0)0 = G I runcating at n = 1On that, the matrix is of dimension 2 × (cid:20) I I I I (cid:21) (cid:34) ˜ A (1)0 ˜ A (1)1 (cid:35) = (cid:20) G G (cid:21) = ⇒ (cid:34) ˜ A (1)0 ˜ A (1)1 (cid:35) = 1det I × (cid:20) I − I − I I (cid:21) (cid:20) G G (cid:21) Henceforth:˜ A (1)0 = I G − I G I I − I I , ˜ A (1)1 = I G − I G I I − I I (2.7)An important comment is that, A (0)0 (cid:54) = A (1)0 , but, it is possible to write A (1)0 as a function of A (0)0 . For instance, using the fact I = I , then: A (1)0 = A (0)0 I − I G G I − I I The only difficulty in this method, is solving all I -integrals and G -integrals.That can be done analytically depending on the shape to be integrated, but also,by means of numerical integration procedures. In increasing the dimension ofthe linear system by truncating at a high value of n , it is expected to get closerto the real potential. Furthermore, this procedure is valid for any shape S , aslong as it has axial symmetry. Once the A l values are known, one can use it and build the entire potential asin (2.1). Having the potential, the local FEF can be calculated.The Field Enhancement Factor at the apex γ a , is defined as the magnitudeof the local electric field at the apex of the surface E a , divided by the externalfield E . E a = − ∂V∂r (cid:12)(cid:12)(cid:12) θ =0 = E + ∞ (cid:88) l =0 ( l + 1) A l r l +2 Therefore, the FEF: γ a = E a E = 1 + ∞ (cid:88) l =0 ˜ A l l + 1 r l +2 (2.8)In equation (2.8) it was assumed two things: (1) the surface’s tangent planeat the apex coincides with the xy -plane, hence why the field only has a z-direction. (2) The apex of the surface is located at the z -axis. If both conditionsare not met, the full gradient (with radial, polar, and azimuthal terms), mustbe taken, as opposed to an ordinary derivative in r . Thus, the apex field E a r , which, in this case, should be how far the apexof the surface is from the origin. In summary, in both formulas above, the apexis located at cartesian coordinates (0 , , r ). For an sphere of radius R , then, r = R . For a revolution ellipse with radius R and height h , then, r = h . In caseof a HCP surface, r = (cid:96) + R = h . The A l terms of the Legendre expansion are associated with the 2 l -multipolemoments. Thus, what is really happening, is a multipole expansion. The linearsystem (2.6) is choosing the multipole coefficients such that it better meets theboundary conditions. As for an example on the multipole l = 0, the chargemonopole can be calculated by means of Gauss Law taken in a sphere P inwhich S is completely inside the sphere. All one has to do is to derive (2.1) withrespect to r , and then: Q(cid:15) = (cid:90) P E · dS = − (cid:90) π (cid:90) π ∂V∂r r sin θdrdθ = 4 πA Thus, the A term is associated with the zeroth-multipole coefficient (a sim-ple image charge), located at the center of the coordinate system. The term A is associated with a dipole moment, also located at the center. A with aquadrupole moment, and so on.In fact, an axially symmetric multipole expansion (written below) can becompared with the electrostatic potential (2.1), and thus, a relation with A l and Q l is possible: V ( r, θ ) = 14 π(cid:15) ∞ (cid:88) l =0 Q l r l +1 P l (cos θ ) = ⇒ Q l = 4 π(cid:15) A l (2.9)In which, for an axially symmetric shape: Q l = (cid:90) S σ ( r ) r l P l (cos θ ) dS ( r ) (2.10)Where σ is the charge density on the surface, which, is induced by thepresence of the external field E .Furthermore, because (2.6), the values ˜ A l are independent on the externalfield E . It was defined that ˜ A l = A l /E , and, the values of A l are the onesthat actually appear in the electrostatic potential (2.1). Because A l related tothe multipoles Q l by (2.9), then: Q l = 4 π(cid:15) E ˜ A l (2.11)Where ˜ A l depends only on the geometry of the system. In other words, themultipole moments scale linearly with E . If one doubles the external field, allmultipole moments will double. 16 hapter 3 Application of the method
As the method was just explained, one can test it in a simple shape, such as,a hemisphere with radius R centered at the origin, on top of the conductingplane z = 0, and to calculate the apex FEF. By symmetry, this is equivalent ofconsidering a sphere, and no plane, as the potential in the whole space won’tbe changed. Therefore, instead of integrating the hemisphere and the plane(which is infinite in size), the integration will be made on a whole sphere, andno plane, because the area of a sphere is finite, as opposed to a plane. Thesolution will be equivalent to both problems, of course. In addition, about theHCP model, setting h = R , that is, ν = h/R = 1, is equivalent to the sphereproblem. Furthermore, a prolate spheroid with ν = h/R = 1 is also equivalentto the sphere problem. Thus, solving the sphere also gives the results for a HCP ν = 1 and a spheroid of ν = 1. To solve the problem, the A l values are required, and, to calculate those, the G l and I ij values are required. First, the G -integrals: G l = (cid:90) S r − l P l (cos θ ) cos θdS = (cid:90) π (cid:90) π R − l P l (cos θ ) cos θR sin θdθdφ (3.1)If x = cos θ , and using the fact that x = P ( x ) G l = 2 πR − l +2 (cid:90) − xP l ( x ) dx = 2 πR − l +2 (cid:90) − P ( x ) P l ( x ) dx (3.2)By orthogonality of the Legendre polynomials, one can finally get the valueof the G -integrals: G l = 43 πRδ l, (3.3)Where δ i,j = 0 if i (cid:54) = j , and δ i,j = 1 if i = j .17 .1.2 Spherical I-Integrals Now, doing the same with the I-Integrals: I ij = (cid:90) S r i − j − P i (cos θ ) P j (cos θ ) = (cid:90) π (cid:90) π R − i − j − P i (cos θ ) P j (cos θ ) R sin θdθdφ (3.4)Again, doing substitution x = cos θ , and using orthogonality of Legendrepolynomials: I ij = 2 πR − i − j (cid:90) − P i ( x ) P j ( x ) dx = 2 πR i + j j + 1 δ i,j (3.5) The infinite I -matrix is diagonal, more, among the G -Vector, only G is nonzero.Thus, all relevant values are: I = 4 π R , G = 43 πR (3.6)Therefore, solving the system (2.6), one gets: ∞ (cid:88) n =0 I ln ˜ A n = G n = ⇒ I nn ˜ A n = G n = ⇒ I ˜ A = G (3.7)Therefore: ˜ A = G I = R , ˜ A n (cid:54) =1 = 0 (3.8) Writing the expression for the potential at (2.1) considering A n (cid:54) =1 = 0, then: V ( r, θ ) = − E r cos θ + A r cos θ = E (cid:18) r − R r (cid:19) cos θ (3.9)Which coincides with the well-known analytical result for the sphere in anexternal field [17]. The FEF can be calculated by expression (2.8), that is: γ = 1 + ∞ (cid:88) l =0 ˜ A l l + 1 r l +2 = 1 + ˜ A R = 1 + (1 + 1) R R = 1 + 1 + 1 = 3 (3.10)18 .2 Half prolate spheroid over plane The method worked with a hemisphere over a plane: next step would be toconsider a spheroid. A spheroid is by definition an ellipsoid of revolution. Withrespect the z -axis, the spheroid can be elongated or flattened. An elongatedspheroid is called a prolate spheroid, while, a flattened spheroid is called anoblate spheroid.The intersection of the spheroid across the plane z = 0 forms a circle ofradius R . The intersection of the spheroid across the plane y = 0 (or x = 0)forms an ellipse of semi-major axis h . The height of the spheroid is thus h .Given the shape of interest is in the case h ≥ R , by definition, such shape is aprolate spheroid. The aspect ratio of a prolate spheroid is defined as ν = h/R .Again, just like it was done with the sphere, the half prolate spheroid overa conducting plane is equivalent to a full prolate spheroid, with no plane. Theadvantage of the later is clear from the fact that, a plane has infinite surfacearea, while, a spheroid has not, thus, all integrations will be finite. As before,one should calculate the G -vector, the I -matrix, then A -vector, and then, thelocal FEF. The G -integral: G l = (cid:90) S r − l P l (cos θ ) cos θdS (3.11)The G integral was calculated for a sphere, and thus, to avoid confusion, adifferent notation is employed: from now on, ˙ G l (with the dot on top), will bedenoted to exclusively speak about the G integral on a spheroid. In fact, everydotted quantity will be speaking about a spheroid.To solve the integrals, two quantities are required: r and dS , that is, thevalues of r ( θ, φ ) which coincides with a prolate spheroid, and, the area element J ,such that dS = Jdθdφ . A detailed derivation of both can be found in AppendixA, where the final results are in equations (A.10) and (A.13), written below: r = R √ − (cid:15) cos θ , (cid:15) = 1 − R h , J = r sin θ (cid:114) r R (cid:15) cos θ sin θ (3.12)One can identify (cid:15) as the first eccentricity of the ellipse which generated thespheroid. Therefore 0 ≤ (cid:15) <
1, and, if (cid:15) = 0, then, the spheroid collapses into asphere, as R = h .Inserting equation (3.12) into equation (3.11), one can get (B.1) [located inAppendix B], written below:˙ G l = (cid:90) π (cid:90) π r − l P l (cos θ ) cos θ · r sin θ · (cid:114) r R (cid:15) cos θ sin θdθdφ (3.13)19herefore, integrating at φ :˙ G l = 2 π (cid:90) π r − l +2 P l (cos θ ) cos θ sin θ (cid:114) r R (cid:15) cos θ sin θdθ (3.14)Like before, substitution x = cos θ is done, thus arriving at equation (B.10)[located in Appendix B], written below:˙ G l +1 = 2 πR l − (cid:90) − xP l +1 ( x ) (cid:2) − (cid:15) x (cid:3) l − / · (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (3.15)Where G l = 0. Only terms of the form 2 l + 1 contributes with the G -Vector. All others are zero. Integral above can be solved exactly, and it is donein Appendix B, as shown in equation (B.16), written below for convenience:˙ G l +1 = 2 πR l − ∞ (cid:88) n =0 l (cid:88) k =0 ( − n a l +1 , k +1 (cid:18) l − n (cid:19) (cid:15) n A k + n +1) (3.16)Where A n (not to be confused with the coefficients A l in the Legendreexpansion of the potential) is the 2 n -th moment of the square root part of theintegrand, that is: A n = (cid:90) − x n (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (3.17)The moments A n can be solved exactly using hypergeometric functions.Also, a l,k are the coefficients of the Legendre polynomials, that is, P l ( x ) = (cid:80) k a l,k x k . And (cid:0) ·· (cid:1) is the generalized binomial coefficient as defined in (B.12)[located at Appendix B]. The level of sophistication is unnecessary, as suchmoments can be approximated by Taylor expansion. The calculation is done inAppendix B, and the final result can be found in (B.26).Finally, thus, the expression of the G integrals for a prolate spheroid, are:˙ G l +1 ≈ πR l − ∞ (cid:88) n =0 ( − n (cid:18) l − n (cid:19) (cid:15) n l (cid:88) k =0 a l +1 , k +1 (cid:20) n + k + 1 + + (cid:15) n + k + 1 + (cid:21) (3.18)From it, inserting l = 0, one can get G , by using a , = 1 because P ( x ) = x .˙ G = 2 πR ∞ (cid:88) n =0 ( − n (cid:18) l − n (cid:19) (cid:15) n a , (cid:20) n + 1 + 1 + + (cid:15) n + 1 + 1 + (cid:21) (3.19)Furthermore, one can further approximate G l , for instance, if the prolatespheroid is not elongated much, that is, if the (cid:15) ≈
0, then, higher powers of (cid:15) becomes closer to zero, allowing the possibility of truncation of the summationas a tool for approximation. For powers of (cid:15) , that is, n = 0 truncation, yields:˙ G l +1 ≈ πR l − l (cid:88) k =0 a l +1 , k +1 (cid:18) l − (cid:19) k + 1 + (3.20)20runcating at powers of (cid:15) :˙ G l +1 ≈ πR l − l (cid:88) k =0 a l +1 , k +1 (cid:20) k + 1 + − (cid:15) (cid:18) l − (cid:19) k + 2 + (cid:21) (3.21)An example:˙ G = 2 πRa , (cid:20)
10 + 1 + − (cid:15) (cid:18) − (cid:19)
10 + 2 + (cid:21) = 2 πR (cid:20)
23 + (cid:15) (cid:21) (3.22)Which can be re-written:˙ G = 43 πR (cid:20) (cid:15) + O ( (cid:15) ) (cid:21) (3.23)Comparing with (3.6), the G found is exactly the sphere’s G , corrected byan eccentricity. Such expression is suitable for small (cid:15) . The same way that was done with the G -integrals, it will be done with the I -integrals. Inserting the r ( θ ) and J , and integrating on φ , equation (B.27)[located at Appendix B] is obtained, written below:˙ I ij = 2 π (cid:90) π r − i − j P i (cos θ ) P j (cos θ ) sin θ (cid:114) r R (cid:15) cos θ sin θ · dθ (3.24)And then doing substitution x = cos θ , one arrives at expression (B.29),written below:˙ I ij = 2 πR i + j (cid:90) − P i ( x ) P j ( x ) (cid:2) − (cid:15) x (cid:3) i + j (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (3.25)As with the G -integrals, there are parity considerations to make: I ij = 0 iff i + j is odd. And, of course, I ij (cid:54) = 0 iff i + j is even. The integral above has ananalytical solution as shown by equation (B.32), shown below:˙ I ij = 2 πR i + j i (cid:88) u =0 j (cid:88) v =0 i + j (cid:88) n =0 ( − n a i,u a j,v (cid:18) i + j n (cid:19) (cid:15) n A u + v +2 n (3.26)Using (B.25) for the purpose of approximations, one gets (B.33), shownbelow:˙ I ij = 2 πR i + j i (cid:88) u =0 j (cid:88) v =0 i + j (cid:88) n =0 ( − n a i,u a j,v (cid:18) i + j n (cid:19) (cid:15) n (cid:34) n + u + v + + (cid:15) n + i + v + (cid:35) (3.27)21ith above expression, the value of ˙ I :˙ I = 2 πR (cid:88) n =0 (cid:18) n (cid:19) (cid:15) n (cid:20) n + + + (cid:15) n + + (cid:21) (3.28)Re-writing: ˙ I = 2 πR (cid:20)
23 + 25 (cid:15) + 15 (cid:15) + 17 (cid:15) (cid:21) (3.29)Which, neglecting some terms:˙ I = 4 π R (cid:20) (cid:15) + O ( (cid:15) ) (cid:21) (3.30)Which, again, as comparing with (3.6), gives exactly the same for a sphere,except, a correction for the eccentricity. Having expressions for G l and I ij , the values of A l can be found by means of(2.6). Again, G l = 0 iff l is even, and I ij = 0 iff i + j is odd. Therefore, thelinear system (2.6) becomes: I I I I · · · I I I I · · · I I I I · · · I I I I · · · I I I I · · · I I I I · · · I I I I · · · I I I I · · · ... ... ... ... ... ... ... ... . . . ˜ A ˜ A ˜ A ˜ A ˜ A ˜ A ˜ A ˜ A ... = G G G G ... (3.31)The main linear system above is separated into two different linear systems,completely equivalent to the first: a system for odd components, and anotherfor even components, as written below: I I I I · · · I I I I · · · I I I I · · · I I I I · · · ... ... ... ... . . . ˜ A ˜ A ˜ A ˜ A ... = (3.32) I I I I · · · I I I I · · · I I I I · · · I I I I · · · ... ... ... ... . . . ˜ A ˜ A ˜ A ˜ A ... = G G G G ... (3.33)22otice the system (3.32) is a linear system of the form IA = 0. If I isnon invertible (there is, if det I = 0), then there exists non-null solution A .Furthermore, if A is a solution, then cA is also a solution, for any scalar c ∈ R .Thus, there exists infinite solutions, all belonging to the kernel of I . In suchsituation, even valued A wouldn’t be unique, violating the uniqueness theoremof the Laplace Equation. Therefore, such possibility is discarded, even if aformal proof is not offered about if matrix I is indeed singular. Therefore, A l = 0 , l ∈ { , , , , , , ... } .Becaused A l = 0, only A l +1 will be relevant, and they will be given bythe system on (3.33). This means, only odd valued multipoles will contributeto the potential field: dipole, octopole, etc. Even valued multipoles will give azero contribution (that is, image charge, quadrupole moments, etc). All that is missing is to find the values of the A l . Already knowing A = 0, andtruncating (3.33) for a one dimensional system. Therefore:˜ A (1)1 = G I (3.34)Substituting (3.23) and (3.30) at above equation, one gets:˜ A (1)1 = R (cid:15) (cid:15) (3.35)Inserting A (1)1 at (2.8), and because A (1) l = 0 for l >
1, then: γ (1) a = 1 + ∞ (cid:88) l =0 ˜ A (1) l l + 1 r l +2 = 1 + ˜ A (1)1 r = 1 + ˜ A (1)1 r (3.36)Knowing the value of A (1)1 , only r is required, which is the radial position atthe apex of the shape, that is, r = h . Therefore: γ (1) a = 1 + 2 R h (cid:15) (cid:15) (3.37)One can see, above expression only depends on the aspect ratio ν , because (cid:15) can be written in function of ν .It is important to realize at this point all approximations used: An infinitelinear system was truncated to one dimension, such that, only dipole contribu-tions to the local FEF are being computed. More, the eccentricity terms onnumerator and denominator were truncated until (cid:15) order, and, the values of A n where approximated by a second order Taylor expansion. All approxima-tions used are suitable if (cid:15) (cid:28)
1, which is equivalent to R ≈ h , which is equivalentto ν ≈
1. 23 .2.5 Numerical calculations
The method was used to numerically calculate γ ( n ) a (see Fig. (3.1a)) for a prolatespheroid of height and radius h and R , respectively. It was found the error fromthe analytical apex FEF γ a was able to be curved-fitted into an exponentialfunction (see Fig (3.1b)): γ a − γ ( n ) a γ a = A exp ( − bn ) , b > n is the truncation order, and γ ( n ) a is the local FEF calculated at order n . Parameter A should not be confused with the coefficients A l . The γ a usedhere, is the exact theoretical value, which can be found in equation (26) of [1],written below: ν = hR , ξ = (cid:112) ν − , γ a = ξ ν ln ( ν + ξ ) − ξ (3.38)At the table below, it is shown for various aspect ratios the fitting parameters A and b . For instance, when ν = 2 then A ≈ .
723 and b = 0 . /b ≈ . /e . Assuming such behavior continues forward, one can calculatethe order required to numerically calculate the FEF given some precision.Shape A b /bh = 1 . .
332 0 .
821 1 . h = 1 . .
499 0 .
355 2 . h = 1 . .
598 0 . . h = 1 . .
656 0 . . h = 2 0 .
723 0 . . h = 3 0 .
851 0 . . h = 5 0 .
937 0 . . γ ( n ) a vs truncation order n for various aspect ratios. The hori-zontal dashed lines correspond to the analytical values of the apex FEF γ a . (b)Error γ a − γ ( n ) a γ a × n , for the same aspect ratios as shown in (a). Thehorizontal plateau observed for ν = 1 . A n (see Eq. (3.17)) was made with 10 trapezoidalelements. 25 .2.6 Slow convergence of the FEF As Fig. (3.1) shows (and the values of 1 /b ), that, a slow convergence of theFEF is observed as the aspect ratio ν = h/R increases. The reason for this,the hemi-ellipsoid concentrates its charge distribution on the top. Therefore, itcan be approximated as a charge on the top, and its respective image charge(opposite in sign) below the plane. It was shown in (2.9) that the coefficients˜ A l are proportional to the multipole coefficients Q l of the system.For a single charge q located in the z -axis, above the plane at a distance a ,that is, (0 , , a ), then, the axial multipole moments are known to be Q l = qa l .Because of the image charge below, the total multipole is: Q l = qa l + ( − q ) · ( − a ) l = qa l + ( − l +1 qa l (3.39)In other words: Q l = 0 Q l +1 = qa l +1 (3.40)It is clear that Q l grows with the power l . The potential for a multipole Q l is proportional to 1 /r l +1 , and the field as 1 /r l +2 . At the apex, r = h , and a < h but a ≈ h , therefore, for sake of illustration, using expression (2.8), consideringin a rough way that Q l ≈ aq l for all l , then: γ a ≈ ∞ (cid:88) n =0 ( l + 1) qa l h l +2 = 1 + 1 h ∞ (cid:88) n =0 ( l + 1) q (cid:16) ah (cid:17) l (3.41)Because a/h ≈ a/h < n , then, γ ( n ) a → h increases. In general,the larger h is, the more orders are required to correctly approximate γ a .26 hapter 4 Hemisphere on a postmodel
Having shown how the method works for shapes with known FEFs, now, theHCP model will be focused, analytically, using the former method. The HCPmodel consists of a conducting cylinder of radius R and height (cid:96) on top of aconducting plane located at z = 0. On top of the cylinder, there’s a hemisphereof radius R . It would be required to integrate the hemisphere, the cylinder, andthe plane (with a circular hole at the center), however, that won’t be necessary:An equivalent problem is posed by eliminating the plane, and considering anmirror shape down below, as if the plane is a mirror. Thus, now there exists acylinder of radius R , but with height 2 (cid:96) , with (cid:96) of height on the positive z axis,and (cid:96) of height on the negative z-axis, and two hemispheres: one located on topof the cylinder, and the other at the bottom of the cylinder. These are easierto integrate, because their area is finite. This will define a HCP shape of height h = (cid:96) + R , and height to radius ratio of ν = h/R . ν The aspect ratio ν is defined as the ratio between the height and radius of theHCP shape. That is: ν = hR (4.1)It will be motivated in this section, that, it makes sense to have the FEFdepending only on ν .Assume two HCP shapes denoted by ( h , R ) and ( h , R ), such that, bothshapes have an equal aspect ratio ν . This implies: ν = h R = h R = ν (4.2)27he only way to meet such criteria, is having: h = αh , and R = αR , forsome α ∈ R where α >
0. Thus, this represents an scale by α . Therefore, a wayto transform from shape ( h , R ) to shape ( h , R ) where ν ( h , R ) = ν ( h , R ),is by scaling the entire coordinate system ( x, y, z ) → ( αx, αy, αz ).One can verify what happens to the Laplacian operator applied over a genericfunction φ when such scale transformation is applied. By defining r (cid:48) = α r , then: ∂φ∂x (cid:48) = ∂x∂x (cid:48) ∂φ∂x = 1 α ∂φ∂x (4.3)And: ∂ φ∂x (cid:48) = ∂∂x (cid:48) α ∂φ∂x = 1 α ∂x∂x (cid:48) ∂ φ∂x = 1 α ∂φ∂x (4.4)The same can be done for y, z , and, therefore, ∇ φ = α ∇ (cid:48) φ . Thus, ifLaplace equation is satisfied: ∇ φ = 0 ⇐⇒ ∇ (cid:48) φ = 0 (4.5)Therefore, if φ ( x ) is a solution of Laplace equation, so is φ ( αx ). If ψ ( r )be the analytical solution of the HCP shape ( h, R ), that is, ψ satisfies Laplaceequation, and boundary conditions ( h, R ), then, ψ ( α r ) is a solution of Laplaceequation, and satisfies boundary conditions ( αh, αR ).If ψ depends only on ν and the position vector r , as opposed to h and R separately, then it is clear that the scaling conditions are immediately satisfied.This highlights the importance of the aspect ratio ν . It is expected that allquantities ( G -integrals, I -integrals, A -values, and the apex FEF) will dependexplicitly on the aspect ratio ν . The integral to solve: I ij = (cid:90) S r − i − j − P i (cos θ ) P j (cos θ ) dS (4.6)Where S is a cylinder with height 2 (cid:96) and radius R . Recall that r is a radialvector, as in spherical coordinate system. For a cylinder, from Appendix A, itsatisfies equation: r = R sin θ , J = r (4.7)Where J is the area element, that is, dS = Jdθdφ . The limits of integrationare from θ (the top of the cylinder) to π − θ , the bottom of the cylinder, where: a c = cos θ = (cid:96)r , a s = sin θ = Rr , r = (cid:112) R + (cid:96) (4.8)28 detailed derivation is in Appendix A. The integral is thus:¨ I ij = (cid:90) S r − i − j − P i (cos θ ) P j (cos θ ) dS = (cid:90) π (cid:90) π − θ θ r − i − j − P i (cos θ ) P j (cos θ ) r dθdφ (4.9)Thus: ¨ I ij = 2 πR i + j (cid:90) π − θ θ P i (cos θ ) P j (cos θ ) [sin θ ] i + j dθ (4.10)Or, changing x = cos θ :¨ I ij = 2 πR i + j (cid:90) a c − a c P i ( x ) P j ( x ) (cid:104)(cid:112) − x (cid:105) i + j − dx (4.11)First, the function √ − x is always even. Second, if i + j is even, then P i P j is even. Also, if i + j is odd, then P i P j is odd. Because the limits of the integralare symmetrical, then an odd function must be evaluated to zero. Therefore, I ij = 0 if i + j is odd.Now, if one wishes to calculate specific values:¨ I = 2 π (cid:90) π − θ θ dθ = 2 π ( π − θ )¨ I = 2 πR (cid:90) π − θ θ cos θ sin θdθ (4.12)The integrals in θ where used for I and I , because they are easier to solvethan the integrals in the variable x . (cid:90) [sin θ cos θ ] dθ = 14 (cid:90) sin (2 θ ) dθ = 18 (cid:90) [1 − cos (4 θ )] dθ (4.13)Thus: (cid:90) sin θ cos θdθ = 18 θ −
132 sin (4 θ )= 18 θ −
116 sin (2 θ ) cos (2 θ )= 18 θ −
18 sin θ cos θ (cid:0) cos θ − sin θ (cid:1) (4.14)Or, defining r = (cid:96) + R , then:¨ I = 2 πR (cid:20) π −
14 arccos (cid:18) (cid:96)r (cid:19) + 14 R(cid:96)r (cid:18) (cid:96) r − R r (cid:19)(cid:21) (4.15)Or, re-writing both:¨ I = 2 π (cid:20) π − (cid:18) (cid:96)r (cid:19)(cid:21) ¨ I = 2 πR (cid:20) π −
14 arccos (cid:18) (cid:96)r (cid:19) + 14 R(cid:96)(cid:96) + R (cid:96) − R (cid:96) + R (cid:21) (4.16)29f (cid:96) = 0, then ¨ I = ¨ I = 0, because, arccos(0) = π/
2. Expansion of arccosis: arccos( x ) = π − x − x − x
40 + O ( x ) (4.17)Thus, inserting the expansion into (4.16), it becomes, for (cid:96) (cid:28) R : I = 4 π(cid:96)r ¨ I = π R (cid:20)(cid:18) (cid:96)r (cid:19) + R(cid:96)(cid:96) + R (cid:96) − R (cid:96) + R (cid:21) (4.18) The integral to solve: G l = (cid:90) S r − l P l (cos θ ) cos θdS, (4.19)The same recipe will be followed, substituting r and writing dS = r dθdφ .Thus:¨ G l = (cid:90) π (cid:90) π − θ θ r − l P l (cos θ ) cos θr dθdφ = 2 π (cid:90) π − θ θ (cid:18) R sin θ (cid:19) − l +2 P l (cos θ ) cos θdθ (4.20)Therefore, finalizing:¨ G l = 2 πR − l +2 (cid:90) π − θ θ P l (cos θ ) [sin θ ] l − cos θdθ (4.21)As before, applying substitution x = cos θ :¨ G l = 2 πR − l +2 (cid:90) a c − a c x (cid:104)(cid:112) − x (cid:105) l − P l ( x ) dx (4.22)Function √ − x is always even. If l is odd, then xP l is even. If l iseven, then xP l is odd. Because the integral limits are symmetric, if the entireintegrand is odd, then the integral must evaluate to zero, which happens when l is even. Thus, ¨ G l = 0.One can now calculate the value of G .¨ G = 2 πR (cid:90) a c − a c x − x dx = 2 πR [arctanh( x ) − x ] a c − a c (4.23)Where substitution x ← tanh x was done to solve the integral. Therefore:¨ G = 4 πR (cid:20) arctanh (cid:18) (cid:96)r (cid:19) − (cid:96)r (cid:21) (4.24)30f it is the case that (cid:96) (cid:28) R , then a c (cid:28)
1, meaning, the Taylor expandarctanh would be good approximation.arctanh( x ) = x + x x O ( x ) (4.25)Therefore, for (cid:96) (cid:28) R , expression (4.24) can be written:¨ G = 43 πR (cid:34)(cid:18) (cid:96)r (cid:19) + 35 (cid:18) (cid:96)r (cid:19) + · · · (cid:35) (4.26) Expression (A.20), as derived at Appendix A, show the area element of thehemisphere: J = r sin θ (cid:118)(cid:117)(cid:117)(cid:116) (cid:96) r sin θ (cid:32) (cid:96) cos θ (cid:112) R − (cid:96) sin θ (cid:33) (4.27)Where r is shown in expression (A.18) [also in Appendix A], written below: r = (cid:96) cos θ + (cid:112) R − (cid:96) sin θ (4.28)In above equation, the value of (cid:96) is signed: (cid:96) > (cid:96) < J can be approximated as follows: if (cid:96) (cid:29) R (also h (cid:29) R ),as in (A.24), then: J = r sin θ (cid:115) (cid:96) r (cid:96) cos θ sin θR − (cid:96) sin θ , (cid:96)R (cid:29) (cid:96) (cid:28) R , as in (A.26), then: J = r sin θ (cid:114) (cid:96) r sin θ (cid:96)R (cid:28) Recalling dS = Jdθdφ where ( A.
24) shows the value equation for J . Recall theintegral to solve: ˚ G l = (cid:90) S r − l P l (cos θ ) cos θdS Inserting the exact J in expression above, and doing parity considerations,just like it was done with the prolate spheroid, the integral can be simplified asin equation (C.11) [located in Appendix C], written below:˚ G l = 0˚ G l +1 = 4 π (cid:90) (cid:96)/r r − l +1 P l +1 ( x ) xJ A dx (4.31)31here: r = (cid:96) + R , J = r sin θJ A (4.32)Therefore, just like the prolate spheroid, a suspended hemisphere also has G l = 0 if l is even. About solving for the odd values of l , above integralis still too complicated for a direct substitution of r and J . If (cid:96) = 0, thecalculations reduces to the case of the sphere, which was solved exactly usingthis method, and found that a single dipole describes the entire field. It isexpected, that, higher deviations from the spherical shape will cause more andmore contributions of multipole elements. With that in mind, it makes sense toconsider (cid:96) (cid:28) R : in this condition, the dipole is still the dominant coefficient,and higher multipole contributions will probably be small.With that in mind, one can Taylor expand the integral r n P l ( x ) xJ A aroundthe center (cid:96) = 0 until second order, and simplify above integral to the expressionwritten at (C.48) [detailed calculations at Appendix C] case, written below: G l = 4 πR − l +2 (cid:90) (cid:96)/r P l ( x ) xdx + 4 π ( − l + 2) R − l +1 (cid:96) (cid:90) (cid:96)/r P l ( x ) x dx + ( l − l + 2) 4 πR l (cid:96) (cid:90) (cid:96)/r x P l ( x ) dx + 4 lπR l (cid:96) (cid:90) (cid:96)/r xP l ( x ) dx (4.33)These integrals are easier to solve, requiring only the calculations of momentsof Legendre polynomials, and this can be solved exactly. For instance, expression(C.51) [located at Appendix C] is the result of the first moment of Legendrepolynomials P l . A description of how to calculate higher moments can be foundin Appendix C. Nevertheless, the final analytic solutions for a general G l arebig and complicated, and, they do not provide intuitive grounds to base theanalysis. Furthermore, solutions of moments of P i P j , required for I ij , are notknown. Thus, even if one writes complete expressions for G l , the I ij would stillbe lacking, rendering the G l expressions useless.With all of that in mind, one can find specific values of l , which will be donefor G and G . It is already known from expression (4.31) that G = 0. Solving(4.31) for l = 1 yields the expressions (C.55) [located at Appendix C], writtenbelow: G = 43 πR (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + π(cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) − π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 4 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) (4.34)32ust like one could anticipate, if (cid:96) = 0, the value of G becomes the calculatedvalue for the sphere, that is 4 / πR , as can be seen in equation (3.3). The otherterms, are the Taylor corrections for low values of (cid:96) .For the case where (cid:96) (cid:29) R , a calculation of G can also be found in AppendixC. Recall the I-Integral at (2.4), that is:˚ I ij = (cid:90) S r − i − j − P i (cos θ ) P j (cos θ ) dS (4.35)As before, substitute area element dS = Jdθdφ , and thus:˚ I ij = (cid:90) π (cid:90) θ r − i − j − P i (cos θ ) P j (cos θ ) Jdθdφ (4.36)The same procedure that was done for the G l integrals can be done tocalculate I ij . It is shown in expression at Appendix C, (C.16), exact expressionfor the I ij integrals, shown below:˚ I ij = 4 π (cid:90) (cid:96)/r r − i − j P i ( x ) P j ( x ) J A dx, i + j ∈ { , , , , , . . . } (4.37)And I ij = 0 otherwise. Again, doing a Taylor expand around (cid:96) = 0, wheredetailed derivation is at Appendix C, final result can be found in equation(C.56), written below:˚ I ij = 4 πR i + j (cid:90) (cid:96)/r P i ( x ) P j ( x ) dx − ( i + j ) 4 πR i + j +1 (cid:96) (cid:90) (cid:96)/r xP i ( x ) P j ( x ) dx + (cid:2) ( i + j ) − (cid:3) πR i + j +2 (cid:96) (cid:90) (cid:96)/r x P i ( x ) P j ( x ) dx + [( i + j ) + 2] 4 πR i + j +2 (cid:96) (cid:90) (cid:96)/r P i ( x ) P j ( x ) dx (4.38)Given ˚ I = ˚ I = 0 because 1 + 0 is odd, expressions (C.60) and (C.62)[also at Appendix C] show the values of ˚ I and ˚ I , written below:˚ I = 4 π · (cid:20) − (cid:96)r (cid:21) − π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 8 πR (cid:96) (cid:20) − (cid:96)r (cid:21) ˚ I = 4 π R (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) − πR (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 8 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 16 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) (4.39)33t is relevant to notice yet agin again, how they generalize the values foundfor the sphere ( (cid:96) = 0) at equation (3.5), namely: I = 4 π, I = 4 π R (4.40) With all of that, one can come up with the HCP Integrals. The HCP shapeis composed of a hemisphere of radius R on top of a cylinder of length (cid:96) in aplate. It was already discussed that, it was much better to, instead, computethe equivalent problem of a hemisphere of radius R on top of a cylinder of radius2 (cid:96) centered on the z-axis, and another hemisphere of radius R on the bottomof the cylinder. That was the case when it was calculated: ˚ I ij and ˚ G l are theintegrals calculated on both hemispheres, above and below the z-axis, while ¨ I ij and ¨ G l are the integrals for a cylinder of length 2 (cid:96) , (cid:96) above the z-axis, and (cid:96) below the z-axis. Therefore, for a complete HCP shape: G l = ¨ G l + ˚ G l I ij = ¨ I ij + ˚ I ij (4.41)That is true, precisely because, for arbitrary function f is valid: (cid:90) HCP f dS = (cid:90) Suspended Hemispheres f dS + (cid:90) (cid:96) -length Cylinder f dS (4.42) Notice that, it was proved for the exact case, that, ¨ G l = ˚ G l = 0. In addition,¨ I ij = ˚ I ij = 0 for i + j even. Thus, just like the case with the prolate spheroid, theHCP linear system is as described in equation (3.31). It was shown in equation(3.32) that, for such case, A l = 0 for every l , and, by equation (3.33) only odd l at A l contributes to the potential (2.1).Hereby, using similar arguments of the prolate spheroidal case, one can con-clude that only odd multipoles will contribute to the potential of aHCP shape , that is, dipole, octopole, etc.Recalling the multipole moments expression (2.10), if Q l = 0, then, theinduced charge density over the surface σ has to be of a certain way. Q l = (cid:90) S σ ( r ) r l P l (cos θ ) dS ( r ) (4.43)Because the shape is axially symmetric, integration can be done in domains φ ∈ [0 , π ] and θ ∈ [0 , π ], where, the φ integral will yield 2 π , thus: Q l = 2 π (cid:90) π σ ( θ ) r ( θ ) l +2 P l (cos θ ) sin θJ A ( θ ) dθ (4.44)34ubstituting x = cos θ , then:0 = Q l = (cid:90) − σ ( x ) r ( x ) l +2 P l ( x ) J A ( x ) dx (4.45)It was shown for a HCP shape (both cylinder and suspended hemisphere),that r ( x ), J A ( x ) were even functions. Furthermore, P l is even. Notice then,that, if σ ( x ) is odd, then the entire integrand is odd, then all Q l is automaticallyzero. One way for that to happen, is, σ ( x ) being an odd function. Theorem 4.1.
If the induced charge density σ ( x ) is an odd function, then Q l = 0.Evidently, the goal is to prove the converse, that is, if Q l = 0, then σ ( x ) isodd. That can be done, if one has an expression for Q : Q l = (cid:90) − σ ( x ) r ( x ) l +2 P l ( x ) J A ( x ) dx (4.46)Notice that, expression above shows that Q l are proportional to the coeffi-cients of the Legendre expansion of the function σ ( x ) r ( x ) l +2 J A ( x ), that is: r ( x ) l +2 J A ( x ) σ ( x ) = ∞ (cid:88) n =0 Q l l + 1 P l ( x ) (4.47)Therefore, if all Q l = 0, it means that only odd Legendre polynomials con-tribute to the summation, meaning r ( x ) l +2 J A ( x ) σ ( x ) must be an odd function.Because it is already known that r ( x ) and J A ( x ) are even, then, σ ( x ) must beodd. It was thus, proven the following theorem: Theorem 4.2. Q l = 0 for all l ∈ N ∪ { } , if and only if, σ ( x ) + σ ( − x ) = 0.Recall also, that, the HCP shape was considered to be one resembling acapsule, because of the convenience of excluding the plane from the integrals.Therefore, σ ( x ) , x > σ ( x ) , x < E ( r ), normal to the plane.More, if l is even, then the l -multipole of the plane is equal in magnitude,but opposite in sign of the l -multipole of the hemisphere on a post (and thus,both of them sum to zero). And, if l is odd, they are equal in magnitude andin sign. One can seek to explore even more the result of zero even multipoles, and applyit to other models. Let a line of charge along the z-axis, going from − a < z < a a >
0. Such a line has a distribution of linear charge density λ ( z ). Theaxial multipoles of such a system can be calculated: M l = (cid:90) a − a λ ( z ) z l dz (4.48)If one is trying to approximate the potential of a HCP shape by using aline of charge, it is clear that the even multipoles must be zero. Also, fromthe expression above, if λ ( z ) is an odd function, the multipole condition isautomatically satisfied. That happens precisely because z l is even if l is even.Thus, the following is valid: Theorem 4.3. If λ ( z ) is the linear charge density of a line from [ − a, a ] is anodd function, then the multipoles M l = 0.Again, one would like to prove the converse, however, the lack of orthogo-nality of the monomials z l prevents a proof just like the surface charge density σ ( x ) from a HCP shape.Because of this, one is encouraged to choose an odd function when modelingwith a line of charge. For instance, the linear charge density in work [32] waschosen to be an odd function. With two HCP shapes, separated by a distance c , it was proven there’s no imagecharge (because it is the zeroth multipole, and zero is even), thus, the nextmultipole contribution is extremely likely to be of a dipole. If c is large enoughas compared to the dimensions of a HCP shape, the dominant interaction willbe the dipole. With the knowledge of the dipole moment Q and the FEF ofan isolated γ a , it is possible to calculate how much the FEF will fall.The electric field of a dipole: E ( r ) = 3( p · ˆr ) ˆr − p π(cid:15) r (4.49)Therefore, considering the multipoles exactly at the z = 0 at their respectivepositions, which, without loss of generality will be considered to be (0 , ,
0) and( c, , γ (cid:48) a = γ a + E ( c, , h ) · ˆz E (4.50)Where E is the external field. Here, it was explicitly considered only thecontribution in the ˆz direction, because, the electric field at the apex mustparallel to the normal vector of the surface on the apex, which, for a HCP,happens to be ˆz . 36igure 4.1: Two interacting HCP shapesThe distance from the dipole to the tip of the HCP is, r = √ c + h . Fur-thermore, it was shown earlier that all A l (and thus, all multipoles Q l ) scaleslinearly with the external field E . Therefore, the new γ (cid:48) a because the presenceof the other HCP shape, becomes: γ (cid:48) a = γ a + Q E ˆp , ˆr ) ˆr − ˆp π(cid:15) √ c + h · ˆz (4.51)Because p = Q ˆz , then: γ (cid:48) a = γ a + Q E ( ˆp , ˆr ) − π(cid:15) √ c + h (4.52)Where: cos( ˆr , ˆz ) = hr = h √ h + c , sin( ˆr , ˆz ) = cr = c √ h + c (4.53)Thus: γ (cid:48) a = γ a + Q E π(cid:15) ( c + h ) / (cid:20) h h + c − (cid:21) (4.54)Or, equivalently, using equation (2.9) connecting Q and A , instead, onecan write in terms of ˜ A : γ (cid:48) a = γ a + ˜ A ( c + h ) / (cid:20) h h + c − (cid:21) (4.55)Which can also be written: δ = γ (cid:48) a − γ a γ a = − γ a ˜ A ( c + h ) / (cid:20) − h h + c (cid:21) (4.56)37herefore, δ decreases by the third power of c , until a certain value c k , andit begins to increase again. Such value can be calculated from expression above:3 h h + c k − ⇐⇒ c k = h √ c > h √ ⇒ δ < c = h √ ⇒ δ = 0 c < h √ ⇒ δ > h is comparable to the distance between the posts d , then thevariation in FEF begins to decrease, until zero is reached. It is important tonotice this treatment considered the interaction of them to be identical to twoindependent dipoles.It is not the case that at h √ h √
2. In fact, it is not even clear if the independent approxi-mation is valid for d ≈ h . Thus, what was done must be regarded as a goodapproximation only for large distances c .As for the pre-factor K in terms of the normalized distance s = c/h , onecan get: δ = − γ a ˜ A h (1 + ( c/h ) ) / ≈ − γ a ˜ A h s if ch (cid:29) γ a depends only on the aspect ratio term by term, then:˜ A l h l +2 = f l ( ν ) = ⇒ ˜ A l = h l +2 f l ( ν ) (4.60)Where f l is some function. This implies ˜ A l (and hence all multipole moments Q l , because of (2.11)) do not depend on the aspect ratio, rather, they need tobe normalized by h l +2 . Considering the special case l = 1, the dipole term:˜ A = h f ( ν ). Making the substitution into (4.59), then: δ ≈ − γ a s f ( ν ) = ⇒ K ≈ ˜ A h γ a = f ( ν ) γ a (4.61)Therefore, if δ is put in term of the normalized distance s = c/h , then thepre-factor K depends only on the aspect ratio, K = K ( ν ). Furthermore, K (cid:54) = 1,in contrast with some references [23–25] where the pre-factor was absent in thecurve-fitting functions (that is, equivalently, K = 1 was assumed).38 .5.2 One-dimensional regular array of HCPs Let an infinite number of HCP shapes located at positions ( nc, , n ∈ Z . That is, the HCPs are distributed in a single axis. The aim is tocalculate the fractional change in the apex FEF, δ . From (4.56), the result willbe: − δ = γ (cid:48) a − γ a γ a = (cid:88) n ∈ Z ∗ γ a ˜ A ( n c + h ) / (cid:20) − h h + n c (cid:21) (4.62)Where it was denoted Z ∗ = Z − { } . A reasonable first step would be tosimplify the expression in the summand. Notice that, as n grows larger:3 h h + n c → , ˜ A ( n c + h ) / → ˜ A n c (4.63)Therefore, the sum becomes: − δ = 2 ∞ (cid:88) n =1 γ a ˜ A n c = 2 ˜ A γ a c ∞ (cid:88) n =1 n = 2 ˜ A γ a c ζ (3) (4.64)Where ζ ( n ) is the Zeta-Riemann function, and ζ (3) ≈ . E γ a is the electric field on the top of the HCP shape.The dependence continues to be − δ = c . In other words: − δ = 2 ˜ A γ a c ζ (3) = 2 Q ζ (3)4 π(cid:15) E γ a c (4.65)In order to investigate what happens in the borders, a possible model couldbe a semi-infinit e array: Let an infinite number of HCP shapes located atpositions ( nc, ,
0) where n ∈ N ∪ { } . In that case: − δ = − γ (cid:48) a − γ a γ a = (cid:88) n ∈ N γ a ˜ A ( n c + h ) / (cid:20) − h h + n c (cid:21) ≈ (cid:88) n ∈ N γ a ˜ A ( n c + h ) / ≈ (cid:88) n ∈ N γ a ˜ A n c = ˜ A γ a c ∞ (cid:88) n =1 n = ˜ A γ a c ζ (3) (4.66)Therefore, the fractional reduction in apex FEF, is: − δ = − γ (cid:48) a − γ a γ a ≈ ˜ A γ a c ζ (3) (4.67)39his result shows that, δ in an infinite array is twice as more as the δ ina semi-infinite array, meaning, apex fields on the emitters of the borders arelarger than the emitters located on the the middle of the linear array. Thisis consistent with [31], where it was reported that the emitters at the bordersdegrades faster. (cid:96) (cid:28) R Hereby, for (cid:96) (cid:28) R , for G , equations (4.26) and (4.34) can be summed, writtenbelow: G = 43 πR (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + π(cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) (hemisphere) − π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 4 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) (hemisphere)+ 43 πR (cid:18) (cid:96)r (cid:19) (cylinder) (4.68)Thefore, one term of the suspended hemispherical G and cylindrical G inte-grals cancel out. G = 43 πR + π(cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) − π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 4 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) (4.69)At first order of (cid:96) , the correction of G from the sphere case is merely π(cid:96) .The relevant value now, would be I . Thus, one can sum (4.18) with (4.39),then: I = 4 π R (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) − πR (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 8 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 16 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + π R (cid:18) (cid:96)r (cid:19) + π R R(cid:96)(cid:96) + R (cid:96) − R (cid:96) + R (4.70)Truncating the linear system (3.33) at 1 ×
1, then: A = G I (4.71)40here, all it is required to do, is to insert (4.70) and (4.69) at equationabove. In order to simplify a little, terms of (cid:96) can be ignored. Therefore: A (1)1 ≈ πR + π(cid:96) π R − πR (cid:96) + π R (cid:96)r = R R + (cid:96) − (cid:96)R + (cid:96)r (4.72)Using (2.8), the FEF for becomes: γ (1) a = 1 + 2 A (1)1 ( R + (cid:96) ) = 1 + 2 R ( R + (cid:96) ) · R + (cid:96) − (cid:96)R + (cid:96)r , (cid:96) (cid:28) R (4.73)One can see, it depends on the ratios h/R = ν , or (cid:96)/R = ( h − R ) /R = ν − h increases, γ a → Q l ≈ qa l for some a < h , but a ≈ h , thus the FEF convergesslowly, as shown in equation (3.41) which presents a rough estimation, writtenbelow: γ a ≈ h ∞ (cid:88) n =0 ( l + 1) q (cid:16) ah (cid:17) l (4.74)41 .7 FEF for HCP Model For a HCP shape of radius R and total height h = R + (cid:96) , where R > (cid:96) >
0, the first step is to calculate the integrals (2.4), written below: G i = (cid:90) S r − i P i (cos θ ) cos θdS, I ij = (cid:90) S r − i − j − P i (cos θ ) P j (cos θ ) dS (4.75)Above surface integrals simplifies to ordinary integrals as stated in equations(4.22), (4.11), (4.31), (4.37), (4.41) with aid of (4.27), (4.32) (4.28) where inte-gration limits can be found from (4.32) and (4.8). All required equations aresummarized below: I ij = ˚ I ij + ¨ I ij G l = ˚ G l + ¨ G l (4.76)¨ G l = 2 πR − l +2 (cid:90) (cid:96)/r − (cid:96)/r x (cid:104)(cid:112) − x (cid:105) l − P l ( x ) dx (4.77)¨ I ij = 2 πR i + j (cid:90) (cid:96)/r − (cid:96)/r P i ( x ) P j ( x ) (cid:104)(cid:112) − x (cid:105) i + j − dx (4.78)˚ G l +1 = 4 π (cid:90) (cid:96)/r r − l +1 P l +1 ( x ) xJ A dx (4.79)˚ I ij = 4 π (cid:90) (cid:96)/r r − i − j P i ( x ) P j ( x ) J A ( x ) dx, i + j even (4.80) J A ( x ) = (cid:118)(cid:117)(cid:117)(cid:116) (cid:96) r (1 − x ) (cid:32) (cid:96)x (cid:112) R − (cid:96) (1 − x ) (cid:33) (4.81) r ( x ) = (cid:96)x + (cid:112) R − (cid:96) (1 − x ) (4.82) r = (cid:112) R + (cid:96) (4.83)Where it was proved:¨ G l = ˚ G l = G l = 0 , ∀ l ∈ N ∪ { } ¨ I ij = ˚ I ij = I ij = 0 , ∀ i, j s.t. i + j ∈ { , , , , , . . . } (4.84)Once I ij and G l are all known, solve the linear system (3.33), written below: I I I I · · · I I I I · · · I I I I · · · I I I I · · · ... ... ... ... . . . ˜ A ˜ A ˜ A ˜ A ... = G G G G ... (4.85)42nce the values ˜ A l are solved, one can find the potential in the overall spaceby inserting the values at (2.1), written below: V ( r, θ ) = − E r cos θ + E ∞ (cid:88) l =0 ˜ A l r − ( l +1) P l (cos θ ) . (4.86)Where, the FEF is expressed in equation (2.8), where the apex is located at r = h , written below: γ a = 1 + ∞ (cid:88) l =0 ˜ A l l + 1 h l +2 (4.87) Calculations have been done using the method, and truncating the linear system(2.6) until n , for aspect ratios ν = h/R = 1 . ν = 2. For comparison,the apex FEF was obtained numerically [Thiago Albuquerque de Assis, privatecommunication] by using the minimal domain size (MDS) method [19].For ν = 1 . ν = 2, the apex FEF values were found to be γ MDS ≈ . γ MDS ≈ . γ ( n ) a from themethod does converge to γ MDS , as shown in the plots on Fig (4.1).43igure 4.2: (a) The FEF γ ( n ) a is plotted against the truncation n . The dashedline correspond to γ MDS . (b) Error γ ( n ) a − γ MDS γ MDS × n .44 hapter 5 General shapes
The fact that even multopole moments were zero for the hemisphere, hemi-ellipsoid on a plate, and the HCP shape, is no coincidence. It turns out, allshapes which are axially symmetric, and have mirror symmetry in the xy-plane,will have all even multipoles zero. In this chapter, this will be proved, and theconsequences explored.
Theorem 5.1.
Let an emitter shape S under a external electrostatic field E .Let S be described in spherical coordinates by r ( θ, φ ). If a shape S has axialsymmetry, that is r ( θ, φ ) = r ( θ ), independent of φ , and, if S has mirror sym-metry in the xy-plane, that is r ( θ ) = r ( π − θ ), then, under an uniform appliedfield E , the combined system will have all even multipoles zero. Proof.
To prove the stated theorem, it is suficient to sshow that r ( x ) is even, J A ( x ) is even, and show the respective I and G integrals are zero. Doingconversion x = cos θ , then, r ( x ) = r ( − x ), because cos( π − θ ) = − cos θ = − x .Therefore, r ( x ) is an even function.A detailed derivation of J can be found in the Appendix A. Defining J = J A r sin θ , then: J A ( θ ) = (cid:115) (cid:18) drdθ (cid:19) (5.1)Again, doing x = cos θ : J A ( x ) = (cid:115) (cid:18) drdx dxdθ (cid:19) = (cid:115) (cid:18) − drdx (cid:112) − x (cid:19) (5.2)That is: J A ( x ) = (cid:115) − x ) (cid:18) drdx (cid:19) (5.3)45f dr/dx is either odd or even, then J A ( x ) is even function. Because r ( x ) iseven, then r (cid:48) ( x ) has definite parity, namely: r ( x ) = r ( − x ) = ⇒ r (cid:48) ( x ) = − r (cid:48) ( − x ) (5.4)Thus, if r ( x ) is even, then r (cid:48) ( x ) is odd, then r (cid:48) ( x ) is even, then J A ( x ) iseven.About the IG integrals, doing substitution x = cos θ , and recall θ ∈ [0 , π ]and φ ∈ [0 , π ]. Because S has axial symmetry, nothing does depend on φ , thusthe φ integral evaluates to 2 π . One is left with: G l = (cid:90) S r − l P l (cos θ ) cos θdS = 2 π (cid:90) − r − l +2 ( x ) P l ( x ) xJ A ( x ) dxI ij = (cid:90) S r − i − j − P i (cos θ ) P j (cos θ ) dS = 2 π (cid:90) − r i + j ( x ) P i ( x ) P j ( x ) J A ( x ) dx (5.5)Because r ( x ) is even, then r n ( x ) is even, for any n ∈ N . Therefore, becausethe symmetrical intervals [ − , P n ( x ) has definite parity, theentire integrand on both integrals have definite parity. If integrand is odd,integral is zero. Then G l = 0. And, if i + j is odd, then I ij = 0. Thelinear system (2.6) does reduces to the same as equation (3.32) and (3.33), andtherefore, A l = 0. Because A l is proportional with the axial multipoles as inequation (2.9), then, Q l = 0, finishing the proof. Theorem 5.2.
Let an emitter shape S under an external electrostatic field E .Let S be described in spherical coordinates by r ( θ, φ ). If a shape S has axialsymmetry, that is r ( θ, φ ) = r ( θ ), independent of φ , then the local potential isaxially symmetric, that is, V ( r, θ, φ ) = V ( r, θ ). Proof.
In general, the potential can be written: V ( r, θ, φ ) = E r cos θ + ∞ (cid:88) l =0 l (cid:88) m = − l B lm r − ( l +1) Y lm ( θ, φ ) (5.6)At the surface, V = 0. Because the shape is axially symmetric, then: V ( r ( θ ) , θ, φ ) = 0 = E r cos θ + ∞ (cid:88) l =0 l (cid:88) m = − l B lm r ( θ ) − ( l +1) Y lm ( θ, φ ) (5.7)Because Y lm ( θ, φ ) = N lm P lm (cos θ ) e imφ , then: − E r cos θ = ∞ (cid:88) l =0 r ( θ ) − ( l +1) l (cid:88) m = − l N lm B lm P lm (cos θ ) e imφ , ∀ θ ∈ [0 , π ] (5.8)46herefore, if we integrate both sides on φ : − πE r cos θ = ∞ (cid:88) l =0 r ( θ ) − ( l +1) l (cid:88) m = − l N lm B lm P lm (cos θ ) (cid:90) π e imφ dφ (5.9)Using (cid:82) π e imφ dφ = 2 πδ m , then: − E r cos θ = ∞ (cid:88) l =0 r ( θ ) − ( l +1) N l B l P l (cos θ ) , ∀ θ ∈ [0 , π ] (5.10)Comparing (5.8) with (5.10), we conclude N lm = 0 if m (cid:54) = 0, because r ( θ )is an arbitrary function. This means, potential (5.6) reduces to (2.1), meaning, V ( r, θ, φ ) = V ( r, θ ), thus, the entire system is axially symmetric. Such resultalso justifies the usage of (2.1). Corollary 5.2.1.
Let an emitter shape S under an external electrostatic field E . Let S be described in spherical coordinates by r ( θ, φ ). If a shape S hasaxial symmetry, that is r ( θ, φ ) = r ( θ ), independent of φ , then the surface chargedensity σ ( θ, φ ) distribution has axial symmetry σ ( θ, φ ) = σ ( θ ). Proof. If V is axially symmetric, E = −∇ V also is, and σ = (cid:15) E · n , where n isthe normal unit vector of S. Therefore, σ must be axially symmetric. Theorem 5.3.
Let an emitter shape S under an external electrostatic field E . Let S be described in spherical coordinates by r ( θ, φ ). If a shape S hasaxial symmetry, that is r ( θ, φ ) = r ( θ ), independent of φ , and, if S has mirrorsymmetry in the xy-plane, that is r ( θ ) = r ( π − θ ), then the surface charge density σ ( θ, φ ) distribution both has axial symmetry σ ( θ, φ ) = σ ( θ ), and, oppositemirror symmetry: σ ( θ ) = − σ ( π − θ ). Proof.
By theorem (5.2.1), because S is axially symmetric, then σ is also axiallysymmetric. This means, the multipole moments Q l can be integrated directlyon φ yielding 2 π : Q l = (cid:90) S σ ( θ ) r l P l (cos θ ) dS = 2 π (cid:90) − σ ( x ) r ( x ) l +2 P l ( x ) J A ( x ) dx (5.11)In which substitution x = cos θ was done. Notice that, expression aboveshows that Q l are proportional to the coefficients of the Legendre expansion ofthe function σ ( x ) r ( x ) l +2 J A ( x ), that is: r ( x ) l +2 J A ( x ) σ ( x ) = ∞ (cid:88) n =0 Q l l + 1 P l ( x ) (5.12)By previous theorem, Q l = 0, it means that only odd Legendre polynomi-als contribute to the summation, meaning r ( x ) l +2 J A ( x ) σ ( x ) must be an odd47unction. Because it is already known that r ( x ) and J A ( x ) are even (see proofof previous theorem), then, σ ( x ) is odd, that is σ ( x ) = − σ ( − x ). Because σ ( x )is odd, and − x = − cos θ = cos( π − θ ), then, σ ( π − θ ) = − σ ( θ ). Theorem 5.4.
The potential V of an axially-symmetric mirror-symmetric shape S may be approximated with arbitrary precision by a line of charge in thesymmetry-axis along interval [ − a, a ] , a >
0, with linear charge density λ ( z ). Proof.
The multipole moments of an axially symmetric surface, with a surfacecharge density σ ( θ ), can be calculated as: Q l = (cid:90) S σ ( θ ) r l P l (cos θ ) dS = 2 π (cid:90) − σ ( x ) r ( x ) l +2 P l ( x ) J A ( x ) dx (5.13)The multipole moments of a line of charge from z ∈ [ − a, a ] with linear chargedensity λ ( z ) can be calculated as: M l = (cid:90) a − a λ ( z ) z l dz = (cid:90) − λ ( ax )( ax ) l adx = 2 a l +1 (cid:90) λ ( ax ) x l dx (5.14)Choosing λ such that M l = Q l for all l , means the response from the line ofcharge is identical to the response of the surface S .Let λ ( ax ) = Λ( x ). Under a polynomial approximation:Λ( z ) = (cid:88) n a n x n = ⇒ m l = (cid:90) Λ( x ) x l dx = n − (cid:88) k =0 k + l + 1 a k (5.15)In other words: m m m ... m n = / / / · · · /n / / / / · · · / ( n + 1)1 / / / / · · · / ( n + 2)... ... ... ... . . . ... n n +1 1 n +2 1 n +3 · · · n − a a a ... a n (5.16)Such matrix, is known as the Hilbert Matrix, and it is known to be an ill-posed problem. Nevertheless, solving for larger and larger n should yield moreaccurate Λ. Therefore, in theory, one can find Λ with arbitrary precision. Let two shapes S and S , axially symmetric, mirror symmetric on the xy-plane.By Theorem (5.1), the monopole charge is zero, because monopole is the zerothmultipole, and zero is even. Therefore, the next contributing multipole factor,is the dipole. If these two shapes are in a distance big enough as compared48ith their own lengths, then, the leading multipole contribution is the dipole.By theorem (5.3), it was proven that σ must have xy-symmetry. However, it ispossible to choose σ such that dipole is zero (thus, the next contributing term,would be the octopole). So far, nothing prevents a special shape from havingzero (or close to zero) dipole moments, and high octopole moments.Consider S and S emitter shapes with height h (cid:48) and h (cid:48)(cid:48) , axial dipole mo-ments Q (cid:48) and Q (cid:48)(cid:48) , both nonzero, and FEFs γ (cid:48) a and γ (cid:48)(cid:48) a . Therefore, if both shapesinteract independently of each other, then, they will feel each other’s electricdipole. It will be assumed shape S is centered at position (0 , ,
0) and shape S is at position ( c, , E ( r ) = 3( p · ˆr ) ˆr − p π(cid:15) r , γ (new) a = γ a + E ( d, , h ) · ˆz E (5.17)Thus: γ (cid:48) (new) a = γ (cid:48) a + Q (cid:48)(cid:48) E π(cid:15) ( c + h (cid:48) ) / (cid:20) h (cid:48) h (cid:48) + c − (cid:21) γ (cid:48)(cid:48) (new) a = γ (cid:48)(cid:48) a + Q (cid:48) E π(cid:15) ( c + h (cid:48)(cid:48) ) / (cid:20) h (cid:48)(cid:48) h (cid:48)(cid:48) + c − (cid:21) (5.18)In other words: δ (cid:48) = γ (cid:48) (new) a − γ (cid:48) a γ (cid:48) a = − γ (cid:48) a ˜ A (cid:48)(cid:48) ( c + h (cid:48) ) / (cid:20) − h (cid:48) h (cid:48) + c (cid:21) δ (cid:48)(cid:48) = γ (cid:48)(cid:48) (new) a − γ (cid:48)(cid:48) a γ (cid:48)(cid:48) a = − γ (cid:48)(cid:48) a ˜ A (cid:48) ( c + h (cid:48)(cid:48) ) / (cid:20) − h (cid:48)(cid:48) h (cid:48)(cid:48) + c (cid:21) (5.19)Therefore, the fractional change in the apex FEF, δ , falls as − δ ∼ Kc − forthe general shapes. More, the pre-factor K depends only on the geometry ofthe emitter shape, since γ a and A depends only on geometry.It is worth pointing out that, if γ a depends only on the aspect ratio termby term, then A (in fact, all A l ) will not depend on the aspect ratio, rather,will depend on h and R in general. This comes from Eq. (2.8), in which, byhypothesis, A /h depends only on ν .˜ A h = f ( ν ) = ⇒ ˜ A = h f ( ν ) (5.20)This can be seen, for instance, in the spheroidal case in Eq. (3.35), andeven in the HCP case in Eq. (4.72). Also, it can be seen from (3.40), where a is hypothesized to be much more sensitive on h than the base radius R of agiven shape. This also shows that the multipole moments Q l are not expectedto depend only on the aspect ratio ν .It was assumed that electrostatic interactions are negligible (independentshapes), that is, the interaction of both shapes is such, that, the surface chargedistribution of them is not disturbed. A natural question that rises, is if such ap-proximation is indeed valid, and when the charge distributions begin to change.That will be done in the next section. 49 .3 Potential Theory The only way to check electrostatic interactions, is by fully solving Laplaceequation. That will be done here. The system to be solved: ∇ V = 0 , V ( r (cid:48) ) = 0 , ∀ r (cid:48) ∈ S, −∇ V ( r (cid:48) ) = E , if | r (cid:48) | → ∞ (5.21)This one is slightly harder to be solved analytically by the methods of po-tential theorem. However, notice above problem is equivalent to: ∇ ˜ V = 0 , ˜ V ( r (cid:48) ) = E r cos θ, ∀ r (cid:48) ∈ S, (5.22)Where solution is V ( r, θ ) = − E r cos θ + ˜ V ( r, θ ). This indeed solves bound-ary conditions because: V ( r, θ ) | S = − E r cos θ + ˜ V ( r, θ ) (cid:12)(cid:12)(cid:12) S = − E r ( θ ) cos θ + E r ( θ ) cos θ = 0 (5.23)The problem (5.22) can be solved by a Fredholm integral equation of thesecond kind [33], written below: E r cos θ = 12 h ( r ) + (cid:90) S h ( r (cid:48) ) ∂ Φ ∂ν (cid:48) ( r − r (cid:48) ) dS ( r (cid:48) ) (5.24)Where the solution is given by: ψ ( r ) = − (cid:90) S h ( r (cid:48) ) ∂ Φ ∂ν (cid:48) ( r − r (cid:48) ) dS ( r (cid:48) ) , Φ( r ) = 14 π | r | = 14 πr (5.25)If one solves (5.24) for h ( r ) (this function should not be confused with theheight of a emitter, a mere number), one can plug h ( r ) in (5.25), and the solution ψ is known. The single and double layer kernels for three dimensional systemsare: Φ( r ) = 14 π | r | = 14 πr , ∇ Φ = − π r | r | (5.26)Integral Equation (5.24) can be solved by means of Liouville–Neumann se-ries, which basically consists in noticing h appears both outside and both insidethe integral. Thus, isolating h : h ( r ) = 2 E r cos θ − (cid:90) S h ( r (cid:48) ) ∂ Φ ∂ν (cid:48) ( r − r (cid:48) ) dS ( r (cid:48) )= 2 E r cos θ − (cid:90) S h ( r (cid:48) ) r − r (cid:48) | r − r (cid:48) | · n (cid:48) dS ( r (cid:48) ) (5.27)We substitute above equation, in the h inside the integral: h ( r ) = 2 E r cos θ − (cid:90) S (cid:20) E r (cid:48) cos θ (cid:48) − (cid:90) S h ( r (cid:48)(cid:48) ) ∂ Φ ∂ν (cid:48)(cid:48) ( r (cid:48) − r (cid:48)(cid:48) ) dS ( r (cid:48)(cid:48) ) (cid:21) ∂ Φ ∂ν (cid:48) ( r − r (cid:48) ) dS ( r (cid:48) )(5.28)50r: h ( r ) = 2 E r cos θ − (cid:90) S E r (cid:48) cos θ (cid:48) ∂ Φ ∂ν (cid:48) ( r − r (cid:48) ) dS ( r (cid:48) )+ 4 (cid:90) S (cid:90) S (cid:48) h ( r (cid:48)(cid:48) ) ∂ Φ ∂ν (cid:48)(cid:48) ( r (cid:48) − r (cid:48)(cid:48) ) ∂ Φ ∂ν (cid:48) ( r − r (cid:48) ) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) ) (5.29)Or: h ( r ) = 2 E r cos θ − (cid:90) S E r (cid:48) cos θ (cid:48) r − r (cid:48) | r − r (cid:48) | · n (cid:48) dS ( r (cid:48) )+ 4 (cid:90) S (cid:90) S (cid:48) h ( r (cid:48)(cid:48) ) (cid:20) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:21) (cid:20) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:21) dS (cid:48) ( r (cid:48)(cid:48) ) dS ( r (cid:48) ) (5.30)Plugging h again, we get: h ( r ) = 2 E r cos θ − (cid:90) S E r (cid:48) cos θ (cid:48) r − r (cid:48) | r − r (cid:48) | · n (cid:48) dS ( r (cid:48) )+ 4 (cid:90) S (cid:90) S (cid:48) E r cos θ (cid:20) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:21) (cid:20) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:21) dS (cid:48) ( r (cid:48)(cid:48) ) dS ( r (cid:48) ) − (cid:90) S (cid:90) S (cid:48) (cid:90) S (cid:48)(cid:48) h ( r (cid:48)(cid:48) ) (cid:20) r (cid:48)(cid:48) − r (cid:48)(cid:48)(cid:48) | r (cid:48)(cid:48) − r (cid:48)(cid:48)(cid:48) | · n (cid:48)(cid:48)(cid:48) (cid:21) (cid:20) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:21) (cid:20) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:21) dS (cid:48) ( r (cid:48)(cid:48)(cid:48) ) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) )(5.31)Plugging h again, and doing so iteratively, one arrives at Liouville–Neumannseries, and a complete solution for h . h ( r ) = E r cos θ + ∞ (cid:88) n =1 ( − n (cid:90) S (1) (cid:90) S (2) · · · (cid:90) S ( n ) [ · · · ] n (cid:89) k =1 dS ( r ( k ) ) (5.32)It is important to say, no attempt have been made to prove that above seriesconverges for some shape. Now, it is easy to find the contribution of interacting systems, by simply con-sidering S = S ∪ S , where S is a shape, and S is another shape. Let h bethe resolvent kernel of the isolated shape S , and h for the isolated shape S .Their combined resolvent kernel of S will be, as written by a Liouville-Neumannseries: h ( r ) = h ( r ) + h ( r )+ 2 (cid:90) S (cid:90) S E r (cid:48) cos θ (cid:48) (cid:20) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:21) (cid:20) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:21) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) )+ 2 (cid:90) S (cid:90) S E r (cid:48) cos θ (cid:48) (cid:20) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:21) (cid:20) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:21) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) )+ · · · (5.33)51otice that, if both shapes are separated by a distance c , then, about the firstinteracting integral W , calculated at r ∈ S , because, ultimately, objective isthe apex FEF. | W | = (cid:12)(cid:12)(cid:12)(cid:12) (cid:90) S (cid:90) S E r (cid:48) cos θ (cid:48) (cid:20) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:21) (cid:20) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:21) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:90) S (cid:90) S E z (cid:48) (cid:12)(cid:12)(cid:12)(cid:12) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:12)(cid:12)(cid:12)(cid:12) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) ) ≤ (cid:90) S (cid:90) S E z (cid:48) (cid:12)(cid:12)(cid:12)(cid:12) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) r − r (cid:48) | r − r (cid:48) | (cid:12)(cid:12)(cid:12)(cid:12) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) ) ≤ (cid:90) S (cid:90) S E z (cid:48) sup (cid:26)(cid:12)(cid:12)(cid:12)(cid:12) | r (cid:48) − r (cid:48)(cid:48) | (cid:12)(cid:12)(cid:12)(cid:12) , ∀ r (cid:48) ∈ S , ∀ r (cid:48)(cid:48) ∈ S (cid:27) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) )= 2 E z (cid:90) S (cid:90) S (cid:20) s + ( c − t ) (cid:21) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) )= 2 E z s + ( c − t ) ) (cid:90) S (cid:90) S dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) )= 2 E z A A ( s + ( c − t ) ) (5.34)In this case, A and A (not to be confused with coefficients A l of theLegendre expansion of the potential) are the surface areas of the shapes S and S respectively. In addition, z is the maximum z distance of shape S . Also,it was considered that:1 s + ( c − t ) = sup (cid:26)(cid:12)(cid:12)(cid:12)(cid:12) | r (cid:48) − r (cid:48)(cid:48) | (cid:12)(cid:12)(cid:12)(cid:12) , ∀ r (cid:48) ∈ S , ∀ r (cid:48)(cid:48) ∈ S (cid:27) (5.35)The same thing can be done for W : | W | = (cid:12)(cid:12)(cid:12)(cid:12) (cid:90) S (cid:90) S E r (cid:48) cos θ (cid:48) (cid:20) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:21) (cid:20) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:21) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:90) S (cid:90) S E z (cid:48) (cid:12)(cid:12)(cid:12)(cid:12) r (cid:48) − r (cid:48)(cid:48) | r (cid:48) − r (cid:48)(cid:48) | · n (cid:48)(cid:48) (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:12)(cid:12)(cid:12)(cid:12) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) ) ≤ (cid:90) S (cid:90) S E z (cid:48) sup (cid:26)(cid:12)(cid:12)(cid:12)(cid:12) | r (cid:48) − r (cid:48)(cid:48) | (cid:12)(cid:12)(cid:12)(cid:12)(cid:27) (cid:12)(cid:12)(cid:12)(cid:12) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:12)(cid:12)(cid:12)(cid:12) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) )= 2 E z (cid:20) s + ( c − t ) (cid:21) (cid:90) S (cid:90) S (cid:12)(cid:12)(cid:12)(cid:12) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:12)(cid:12)(cid:12)(cid:12) dS ( r (cid:48)(cid:48) ) dS ( r (cid:48) )= 2 E z s + ( c − t ) (cid:90) S | k | dS ( r (cid:48) )= 2 | k | E z A s + ( c − t ) (5.36)52 useful result is Gauss’ lemma, which states: k (cid:48) = (cid:90) S (cid:20) r − r (cid:48) | r − r (cid:48) | · n (cid:48) (cid:21) dS ( r (cid:48) ) = (cid:90) S ∂ Φ ∂n ( r − r (cid:48) ) dS ( r (cid:48) ) (5.37)Where, if S = ∂V , then: k (cid:48) = 0 , if r ∈ Vk (cid:48) = − / , if r ∈ ∂Vk (cid:48) = − , if r ∈ V c (5.38)However, because the integrand is an absolute value, the sign of ( r − r (cid:48) ) · n (cid:48) is relevant. If there’s no change in sign in the entire integration domain, thenthe integral will yield | k | = | k (cid:48) | , as in Eq (5.3.18). If there’s change in sign,then, nothing can be said.If r is taken immediately above the protrusion, then r ∈ V c . However, thereis a change in sign, as ( r − r (cid:48) ) · n (cid:48) > r (cid:48) is taken at the apex, and ( r − r (cid:48) ) · n (cid:48) < r (cid:48) is taken at the tip of the lower hemisphere. Because of that, one cannot useGauss’ lemma in this case, and show that k = 0. In this case, | k | is non-zero.If r is taken immediately below the protrusion (that is, r ∈ V ), or if r isexactly at the apex (that is, r ∈ ∂V ), intuitively there’s no sign change, andGauss’ lemma can be used, yielding | k | (cid:54) = 0 as well.Therefore, first order electrostatic interaction term is: | W | ≤ | W | + | W | ≤ E z A A ( s + ( c − t ) ) + 2 | k | E z A s + ( c − t ) ∼ c + | k | c (5.39)If k = 0, the resolvent falls at least as c − . In this case, because electric dipoleapproximation is of δ ∼ c − , and electrostatic interactions become importantwith c − , thus, expressions (5.2.3) are expected to hold.But, if k (cid:54) = 0, it was shown the resolvent falls at least as c − , therefore,nothing can be told about if independent dipole-dipole is a good approximation,precisely because all expressions obtained for W are upper bounds.53 hapter 6 Conclusion
The electrostatic potential was assumed to be of the form of Eq. (2.1), whichis true for a surface S with axial symmetry (see theorem (5.2)). Then, an errorfunction Σ was defined in order to minimize the errors from boundary conditions.A necessary condition was found at Eq. (2.6), on which, the parameters ˜ A l , G l and I ij were related, and only depended on the boundary S . Several analyticalconclusions could be drawn from this set up.The coefficients A l related to the multipole moments Q l of the system bymeans of equation (2.11), in which, was shown that for a general axially sym-metric shape, all multipole moments scale linearly with the applied externalelectrostatic field E . That is, if one doubles the field, all multipole momentswill double, as expected.At the case of a hemisphere on a plate, the system (2.6) was solved exactly,giving the known potential (3.9) for a sphere, which a known apex FEF γ a = 3(3.10).Then, it was applied to the hemi-ellipsoid model on a plate. It was shownby equation (3.32) that all even multipole contributions of the system are zero.It was possible to obtain an expression for the FEF for h (cid:28) R at equation(3.37). Numerical calculations have been done, by truncating the linear systemand attempting to approximate the values of ˜ A l . Unfortunately, it was foundthat convergence is slow.The method was then applied to the hemisphere on a cylindrical post (HCP)model. As in the case of the prolate spheroid, it was shown by the same way,that all even multipole contributions (image charge, quadrupole, etc) of a HCPshape over applied field are zero, thus, only odd multipoles contribute (dipole,octopole, etc). This had several consequences: it gave information of how theinduced surface charge density must behave, in particular, it was shown intheorem (4.2) that σ ( x ) + σ ( − x ) = 0, that is, the charge density must be anodd function. Both multipole and σ restrictions was generalized to shapes withaxial symmetry, and mirror symmetry in the xy-plane, as given by theorems(5.1) and (5.3).Another consequence is that other models attempting to approximate a HCP54hape must comply the conditions over the multipoles, which imposes restric-tions. For example, for a line of charge in the interval z ∈ [ − a, a ] with linearcharge density λ ( z ), it was shown in theorem (4.3) some encouragement tochoose an odd function λ ( z ) as well. By means of theorem (5.3), this holds alsowhen modeling general shapes (as enunciated on the theorem) with a line ofcharge, as was done in [32].Furthermore, theorem (5.4) was proven for a general emitter shape, showingthat any axially symmetric shape can be approximated by the line of chargemodel, with arbitrary precision. A relation between the surface charge density σ and the linear charge density λ is shown in the theorem. This means, if onecan find λ , one can also find σ , and vice versa.In a HCP shape, because image charges are zero, the next non-zero contri-bution is of a dipole. It was therefore considered two HCP shapes, interactingvia primitive dipole, and it was shown that the FEF decays by third power lawwith respect to the distances of the hemispheres (4.56), in agreement of numer-ical simulations [27] and analytical models [28]. The same thing happens withone dimensional arrays as calculated at (4.65). Such result was also generalizedfor two general axial symmetric and mirror symmetric shapes (not necessarilyidentical), where the fractional change in the apex FEF of both shapes wascalculated at equation (5.19). Such equation shows that, δ ∼ Kc − , where c isthe distance between the emitters. It was also shown the pre-factor K does de-pend on the geometry of the emitter shapes, confirming the tendency in recentanalytical and numerical results [27, 28]. The functional form of K was shownin Eq. (4.61).It was shown by means of potential theory that the next contribution in thedouble layer resolvent kernel is bounded by equation (5.39). This result showsthe resolvent falls in at least c − , and therefore, nothing can be said about if theindependent dipole-dipole approximation is a valid approximation, or the rangewhere it is valid. However, other layers could have been chosen (say, simplelayer, or, some other), and the analysis would be different. Furthermore, noproof was given that the resolvent converges, thus, such result should be usedwith caution.As for perspectives from future work, the method could be extended fora general emitter (not necessarily axial symmetric). The potential would bewritten as in equation (5.6). It is possible that some sort of IG integrals couldbe found to minimize the potential on the surface. The entire treatment wouldbe more complicated, but, it is possible that some interesting theorems can beproved. This could be investigated further. In addition, investigation about therole of potential theory in the context of field emission could be more explored.55 ppendix A Shapes in sphericalcoordinates
A.1 Cylinder in spherical coordinates
The equation of a cylinder is x + y = R , and we know that, the radialcoordinate is r = x + y + z = R + z . However, because z = r cos θ , wehave: r = R + r cos θ . We can isolate r , and get: r (1 − cos θ ) = R .Therefore, for the cylinder: r = R sin θ That could be done quite obviously from another way: considering a triangle,where hypothenus is a vector from origin to a point in the cylinder, in such away that θ is the azimuthal angle from the spherical coordinate system. Fromit, we can extract: cos θ = zr , sin θ = Rr , tan θ = Rz We have in spherical coordinates: x = r cos φ sin θ = R sin θ cos φ sin θ = R cos φ The same can be done for y and z . And thus, we’ll have a vector: x = ( r cos φ sin θ, r sin φ sin θ, r cos θ ) = (cid:18) R cos φ, R sin φ, R tan θ (cid:19) . With it, we can figure out the normal vector of the cylinder: ∂ x ∂φ × ∂ x ∂θ = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˆx ˆy ˆz − R sin φ R cos φ
00 0 − R sin θ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − R φ sin θ − R φ sin θ = − R sin θ cos φ sin φ dS = Jdφdθ . That is: J = (cid:13)(cid:13)(cid:13)(cid:13) ∂ x ∂φ × ∂ x ∂θ (cid:13)(cid:13)(cid:13)(cid:13) = R sin θ = r Now we have all data necessary to integrate. We now seek the limits ofintegration. At the top of the cylinder, r = (cid:96) + R . Henceforth:cos θ = (cid:96)r = (cid:96) √ (cid:96) + R = 1 (cid:113) (cid:0) R(cid:96) (cid:1) = a c sin θ = Rr = R √ (cid:96) + R = 1 (cid:113) (cid:0) (cid:96)R (cid:1) = a s And, lastly: tan θ = R(cid:96) , θ = (cid:96)R A.2 General symmetrical shape in spherical co-ordinates
For a general element of area, we need the equation of the shape itself, that is, r = r ( θ, φ ). Because the shape is axially-symmetric, r doesn’t depend on φ , andtherefore, r = r ( θ ).We’ll now calculate the element of area. For that, we need to calculateseveral derivatives. Let us begin: ∂x∂θ = ∂r∂θ cos φ sin θ + r cos φ cos θ = xr ∂r∂θ + z cos φ∂y∂θ = ∂r∂θ sin φ sin θ + r sin φ cos θ = yr ∂r∂θ + z sin φ∂z∂θ = ∂r∂θ cos θ − r sin θ = zr ∂r∂θ − r sin θ And now, derivating with respect with φ . ∂x∂φ = − r sin φ sin θ = − y∂y∂φ = r cos φ sin θ = + x∂z∂φ = 0 = 057he normal vector: ∂ x ∂φ × ∂ x ∂θ = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ˆx ˆy ˆz − y x xr ∂r∂θ + z cos φ yr ∂r∂θ + z sin φ zr ∂r∂θ − r sin θ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Thus: ∂ x ∂φ × ∂ x ∂θ = x (cid:0) zr ∂r∂θ − r sin θ (cid:1) y (cid:0) zr ∂r∂θ − r sin θ (cid:1) − y (cid:0) yr ∂r∂θ + z sin φ (cid:1) − x (cid:0) xr ∂r∂θ + z cos φ (cid:1) (A.1)And, therefore, finally, the element of area is: J = (cid:0) x + y (cid:1) (cid:18) zr ∂r∂θ − r sin θ (cid:19) + (cid:20) y (cid:18) yr ∂r∂θ + z sin φ (cid:19) + x (cid:18) xr ∂r∂θ + z cos φ (cid:19)(cid:21) = (cid:0) x + y (cid:1) (cid:18) zr ∂r∂θ − r sin θ (cid:19) + y (cid:18) yr ∂r∂θ + z sin φ (cid:19) + x (cid:18) xr ∂r∂θ + z cos φ (cid:19) + 2 xy (cid:18) yr ∂r∂θ + z sin φ (cid:19) (cid:18) xr ∂r∂θ + z cos φ (cid:19) (A.2)We can expand the terms even more, and then, group them in powers of z ,and factor out what we can: J = z (cid:34) ( x + y ) (cid:18) r ∂r∂θ (cid:19) + ( x cos φ + y sin φ ) (cid:35) + z (cid:20) − x + y ) sin θ ∂r∂θ + (cid:18) r ∂r∂θ (cid:19) (cid:0) sin φ (cid:0) y + 2 x y (cid:1) + cos φ (cid:0) x + 2 xy (cid:1)(cid:1)(cid:21) + (cid:34) ( x + y ) r sin θ + (cid:18) r ∂r∂θ (cid:19) ( x + y ) (cid:35) (A.3)And then we can finally arrive at: J = z (cid:34) ( x + y ) (cid:18) r ∂r∂θ (cid:19) + ( x cos φ + y sin φ ) (cid:35) + 2 z ( x + y ) (cid:20) − (cid:18) sin θ ∂r∂θ (cid:19) + (cid:18) r ∂r∂θ (cid:19) ( x cos φ + y sin φ ) (cid:21) + (cid:0) x + y (cid:1) (cid:34) (cid:18) r ∂r∂θ (cid:19) (cid:35) (A.4)Now, we’ll substitute x, y, z by its values: x = r cos φ sin θ, y = r sin φ sin θ, z =58 cos θ . And for that, some quantities might be useful: x + y = r sin θ (cid:0) cos θ + sin θ (cid:1) = r sin θx cos φ + y sin φ = r cos φ sin θ + r sin φ sin θ = r sin θ ( x cos φ + y sin φ ) = ( r sin θ ) = r sin θ (A.5)Now, we substitute, and, doing some more calculations, we arrive at thisexpression: J = r sin θ (cid:34) (cid:18) r ∂r∂θ (cid:19) (cid:35) = ⇒ J = r sin θ (cid:115) (cid:18) r ∂r∂θ (cid:19) (A.6)Or, because r doesn’t depend on φ , that is, depends exclusively on θ , then: J = r sin θ (cid:115) (cid:18) r drdθ (cid:19) (A.7) A.3 Prolate spheroid in spherical coordinates
For a prolate spheroid (elongated ellipsoid of revolution), with radius R andheight h , obeys the quadric equation: x R + y R + z h = 1 (A.8)Because r = x + y + z , it becomes: x + y + R h z = R x + y + z + (cid:18) R h − (cid:19) z = R r + (cid:18) R h − (cid:19) z = R r (cid:20) (cid:18) R h − (cid:19) cos θ (cid:21) = R (A.9)Where it was used z = r cos θ . The equation of the prolate spheroid becomes: r = R √ − (cid:15) cos θ , (cid:15) = 1 − R h (A.10)In here, (cid:15) is the eccentricity of the revolution ellipse. Acceptable values of (cid:15) are: 0 ≤ (cid:15) <
1, where (cid:15) = 0 iff R = h , that is, the problem of the sphere whichwas solved. The derivative with respect to theta, becomes: drdθ = − R(cid:15) cos θ sin θ [1 − (cid:15) cos θ ] / (A.11)59hus: drdθ = − r R (cid:15) cos θ sin θ (A.12)Using (A.7), we find for the spheroid: J = r sin θ (cid:114) r R (cid:15) cos θ sin θ (A.13) A.4 Suspended hemisphere in spherical coordi-nates
The suspended hemisphere is much more complicated than the cylinder. Thehemisphere is suspended over the cylinder, thus its center is located at (0 , , ± (cid:96) ).The equation of both hemispheres are: x + y + ( z ∓ (cid:96) ) = R . Again, we recall: x + y + z = r , in spherical coordinates. Then: x + y + ( z ∓ (cid:96) ) = R x + y + z ∓ z(cid:96) + (cid:96) = R r ∓ z(cid:96) + (cid:96) = R Therefore, using z = r cos θ , we get: r = R − (cid:96) ± r(cid:96) cos θ . The sameresult can be reached by a triangle, except this time, unlike the cylinder case,our triangle is no longer a right triangle. Yet, by cosine law, we can get at thesame result.Such a triangle becomes a right triangle in the limits of integration: φ ∈ [0 , π ] and θ ∈ [0 , θ ]. At θ = 0 we have r = (cid:96) + R = h . At θ = θ , we have r = (cid:96) + R . And, the θ can be calculated using such right triangle:sin θ = Rr , cos θ = (cid:96)r , tan θ = R(cid:96) (A.14)The equation we got, r = R − (cid:96) ± r(cid:96) cos θ , is a quadratic equation, andcan be solved analytically: r = ± (cid:96) cos θ ± (cid:112) R − (cid:96) sin θ (A.15)Where, the first ± is due to the location of the sphere (either upwards, ordownwards), while the second ± is due to the ±√ ∆ of the quadratic equation.Therefore, these are four equations, where only two of them are correct. To findout, we look at extreme values of θ , namely, 0 and 2 π . θ = 0 = ⇒ r = ± (cid:96) ± Rθ = π = ⇒ r = ∓ (cid:96) ± R (A.16)60e know that, in both cases, the correct expression should be r = (cid:96) + R .Therefore, we pick ++. That leads us to: r = (cid:96) cos θ + (cid:112) R − (cid:96) sin θ, θ ≤ π , (cid:96) ≥ , (Upper Hemisphere) r = − (cid:96) cos θ + (cid:112) R − (cid:96) sin θ, θ ≥ π , (cid:96) ≥ , (Lower Hemisphere)(A.17)Another way to interpret such result, is to have one unique formula: r = (cid:96) cos θ + (cid:112) R − (cid:96) sin θ, θ ∈ [0 , π ] , (cid:96) ≥ , (Upper Hemisphere) (cid:96) ≤ , (Lower Hemisphere)(A.18) A.4.1 Element of area
We already have equation (A.7). Now, all we are lacking to do, is to evaluate thederivative. We could choose either to deal with unsigned (cid:96) as in equation (A.17,or with signed (cid:96) , as in equation (A.18). Notice that, dealing with unsigned (cid:96) , twoexpressions would be required for the derivative, while with signed (cid:96) expression,only one equation would be required. Choosing the signed expression, andcalculating the derivative, one can get: ∂r∂θ = − (cid:96) sin θ (cid:34) (cid:96) cos θ (cid:112) R − (cid:96) sin θ (cid:35) (A.19)All we have to do now, is to insert this expression into the J we had, and,then, finally: J = r sin θ (cid:118)(cid:117)(cid:117)(cid:116) (cid:96) r sin θ (cid:32) (cid:96) cos θ (cid:112) R − (cid:96) sin θ (cid:33) (A.20)Don’t forget that, r still depends on θ by (A.18), so, there is still one sub-stitution left. But, we’re going to leave it at that. A.4.2 Approximation: (cid:96) (cid:29) R As we have noticed, the expression for the suspended hemispherical area elementis quite complicated, and we seek to simplify doing approximations of these kind.For that, we seek in understanding how our variables, (cid:96), sin θ, cos θ r behaves.For that, we seek our attention to (A.18). Notice that: r = (cid:96) cos θ (cid:124) (cid:123)(cid:122) (cid:125) Grows as O ( (cid:96) ) + (cid:112) R − (cid:96) sin θ (cid:124) (cid:123)(cid:122) (cid:125) Grows as O (1) (A.21)61urthermore, with aid of (A.14), we can come up with a more general rela-tionship, which is valid for all (cid:96) (not only (cid:96) (cid:29) R ), which is:0 ≤ sin θ ≤ Rr , (cid:96)r ≤ cos θ ≤ (cid:96) (cid:29) R , sine will keep itselfvery close to zero all the time (because r ≈ h ), and cosine will keep itself veryclose to one all the time.This can be identified geometrically, from the triangle. We can do the samething with the square root in expression (A.20), in which we can expand, inorder to determine the assymptoptics. J = r sin θ (cid:118)(cid:117)(cid:117)(cid:117)(cid:117)(cid:116) (cid:124)(cid:123)(cid:122)(cid:125) J ∈ O ( (cid:96) ) + (cid:96) r sin θ (cid:124) (cid:123)(cid:122) (cid:125) J ∈ O (1) + (cid:96) r (cid:96) cos θ sin θR − (cid:96) sin θ (cid:124) (cid:123)(cid:122) (cid:125) J ∈ O ( (cid:96) ) + (cid:96) r (cid:96) cos θ sin θ (cid:112) R − (cid:96) sin θ (cid:124) (cid:123)(cid:122) (cid:125) J ∈ O ( (cid:96) ) (A.23)Now, we pick the highest asymptoptics, that is, O ( (cid:96) ), and neglect all others.We, thus, have: J = r sin θ (cid:115) (cid:96) r (cid:96) cos θ sin θR − h sin θ , (cid:96) (cid:29) R (A.24)Now we have a much simpler area element J : There’s no longer a squareroot inside a square root, and a few other benefits. A.4.3 Approximation: (cid:96) (cid:28) R We can’t use the same asymptoptics as equation ( A. (cid:96) ≈ r , which is clearly not true anymore under this approximation regime. Now,we have r ≈ R . In addition, from (A.22), we conclude that cosine and sine willvary freely as θ changes from 0 to θ , precisely because θ is an angle close to π/ J = r sin θ (cid:118)(cid:117)(cid:117)(cid:117)(cid:117)(cid:116) (cid:124)(cid:123)(cid:122)(cid:125) J ∈ O (1) + (cid:96) r sin θ (cid:124) (cid:123)(cid:122) (cid:125) J ∈ O ( (cid:96) ) + (cid:96) r (cid:96) cos θ sin θR − (cid:96) sin θ (cid:124) (cid:123)(cid:122) (cid:125) J ∈ O ( (cid:96) ) + (cid:96) r (cid:96) cos θ sin θ (cid:112) R − (cid:96) sin θ (cid:124) (cid:123)(cid:122) (cid:125) J ∈ O ( (cid:96) / ) (A.25)Considering only the O (1) term, we recover J = r sin θ , the same for ansphere, especially if one considers r ≈ R . Henceforth, we’ll also pick the term O ( (cid:96) ). Therefore: J = r sin θ (cid:114) (cid:96) r sin θ, (cid:96) (cid:28) R (A.26)62 ppendix B Spheroidal Integrals
B.1 G-Integrals
Recalling (2.4): G l = (cid:90) S r − l P l (cos θ ) cos θdS Having the area element in (A.13), together with (A.10), the G-integral canbe written as˙ G l = (cid:90) π (cid:90) π r − l P l (cos θ ) cos θ · r sin θ · (cid:114) r R (cid:15) cos θ sin θdθdφ (B.1)Therefore, integrating at φ :˙ G l = 2 π (cid:90) π r − l +2 P l (cos θ ) cos θ sin θ (cid:114) r R (cid:15) cos θ sin θdθ (B.2)The substitution x = cos θ yields: dx = − sin θ , where θ ∈ [0 , π ] = ⇒ x ∈ [ − , G l = 2 π (cid:90) − r − l +2 P l ( x ) x (cid:114) r R (cid:15) x (1 − x ) dx (B.3)It is important to notice something:˙ G l = 2 π (cid:90) − r − l +2 (cid:124) (cid:123)(cid:122) (cid:125) Even P l ( x ) x (cid:124) (cid:123)(cid:122) (cid:125) Even if l ∈{ , , ,... } (cid:114) r R (cid:15) x (1 − x ) (cid:124) (cid:123)(cid:122) (cid:125) Even dx (B.4)Here we used the fact, that, r ( x ) and its derivative are even: r ( x ) = r ( − x ) , (cid:18) r drdθ (cid:19) ( x ) = (cid:18) r drdθ (cid:19) ( − x ) (B.5)63s can be computed directly from (A.10) and (A.12). Because the integraldomain is symmetric (from -1 to 1), an odd integrand will yield zero, which willhappen with l ∈ { , , , , ... } . When integrand is odd, we guarantee nonzeroresult, due to symmetric domain (unless the entire integrand is identically zero,which is not the case). ˙ G l = 0 , if l ∈ { , , , , , . . . } ˙ G l (cid:54) = 0 , if l ∈ { , , , , , . . . } (B.6)Using (A.10), we get:˙ G l = 2 πR − l +2 (cid:90) − xP l ( x )[1 − (cid:15) x ] − l +22 (cid:114) (cid:15) x (1 − x )1 − (cid:15) x dx (B.7)It only makes sense to calculate G l +1 , for l ∈ { , , , . . . } . With that inmind, consider:˙ G l +1 = 2 π (cid:90) − r − (2 l +1)+2 · xP l +1 ( x ) · (cid:114) r R (cid:15) x (1 − x ) dx ˙ G l +1 = 2 π (cid:90) − r − l +1 · xP l +1 ( x ) · (cid:114) r R (cid:15) x (1 − x ) dx (B.8)Again using (A.10), we get:˙ G l +1 = 2 πR − l +1 (cid:90) − xP l +1 ( x )[1 − (cid:15) x ] − l +1 / · (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (B.9)Which can also be written as:˙ G l +1 = 2 πR l − (cid:90) − xP l +1 ( x ) (cid:2) − (cid:15) x (cid:3) l − / · (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (B.10)As the case with the Legendre polynomials, we can write the power in termsof a binomial expansion, except this one will have infinite terms. The more (cid:15) approaches one, the more terms will be needed to approximate it. (cid:2) − (cid:15) x (cid:3) l − / = ∞ (cid:88) n =0 (cid:18) l − n (cid:19) (cid:0) − (cid:15) x (cid:1) n (B.11)Where we define the generalized binomial coefficients as: (cid:18) αk (cid:19) := α ( α − α − · · · ( α − k + 1) k ! . (B.12)Rewriting (B.10), we get:˙ G l +1 = 2 πR l − (cid:90) − xP l +1 ( x ) · ∞ (cid:88) k =0 (cid:18) l − n (cid:19) ( − n (cid:15) n x n · (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (B.13)64aking the proper substitutions, and writing the Legendre polynomials as: P l ( x ) = l (cid:88) k =0 a lk x k (B.14)Then: P l +1 ( x ) = l +1 (cid:88) k =0 a l +1 ,k x k = l (cid:88) k =0 a l +1 , k +1 x k +1 Therefore:˙ G l +1 = 2 πR l − (cid:90) − x l (cid:88) k =0 a l +1 , k +1 x k +1 · ∞ (cid:88) n =0 (cid:18) l − n (cid:19) ( − n (cid:15) n x n · (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx ˙ G l +1 = 2 πR l − l (cid:88) k =0 a l +1 , k +1 ∞ (cid:88) n =0 ( − n (cid:18) l − n (cid:19) (cid:15) n (cid:90) − x k +2 x n · (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx ˙ G l +1 = 2 πR l − ∞ (cid:88) n =0 l (cid:88) k =0 ( − n a l +1 , k +1 (cid:18) l − n (cid:19) (cid:15) n (cid:90) − x k + n +1) · (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (B.15)Finally:˙ G l +1 = 2 πR l − ∞ (cid:88) n =0 l (cid:88) k =0 ( − n a l +1 , k +1 (cid:18) l − n (cid:19) (cid:15) n A k + n +1) (B.16)Where A n is the nth moment of f in the [ − , A n = (cid:90) − x n f ( x ) dx = (cid:90) − x n (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (B.17)It is possible to write A n exactly in terms of hypergeometric functions,but, seeking simplicity, we’ll content ourselves with the second order Taylorexpansion, which we already computed. There are a few important properties: A n +1 = 0 by parity. Furthermore, because f ( x ) > , ∀ x ∈ [ − , x < , ∀ x ∈ [ − , x n +2 < x n , ∀ x ∈ [ − , x n +2 f ( x ) < x n f ( x ) , ∀ x ∈ [ − , A n +2 < A n , ∀ x ∈ [ − ,
1] (B.18)65ewriting (B.16) as ˙ G l +2 = πR l − (cid:80) ∞ n =0 a n , one gets: a n = l (cid:88) k =0 ( − n a l +1 , k +1 (cid:18) l − n (cid:19) (cid:15) n A k + n +1) a n = ( − n (cid:18) l − n (cid:19) (cid:15) n l (cid:88) k =0 a l +1 , k +1 A k + n +1) a n +1 = ( − n +1 (cid:18) l − n (cid:19) (cid:15) n l (cid:88) k =0 a l +1 , k +1 A k + n +1) (B.19)Therefore: a n +1 a n = ( − n +1 (cid:0) l − n (cid:1) (cid:15) n +2 (cid:80) lk =0 a l +1 , k +1 A k + n +2) ( − n (cid:0) l − n (cid:1) (cid:15) n (cid:80) lk =0 a l +1 , k +1 A k + n +1) (B.20)Thus: a n +1 a n = (cid:20) − ln + 1 + nn + 1 (cid:21) (cid:15) (cid:80) lk =0 a l +1 , k +1 A k + n +2) (cid:80) lk =0 a l +1 , k +1 A k + n +1) (B.21)Thus, if n + 1 / > l , then a n +1 < a n . More, if both factors are close toone, then, we get a n +1 ≈ (cid:15) a n . This is particularly relevant, if one is interestedusing the summation for numerical computation.What is left, is to Taylor expand f ( x ). f ( x ) = (cid:112) g ( x ) , g ( x ) = h ( x ) q ( x ) , h ( x ) = (cid:15) x (1 − x ) q ( x ) = (1 − (cid:15) x ) (B.22)Then: f (cid:48) ( x ) = g (cid:48) ( x )2 f ( x ) , h (cid:48) ( x ) = 2 (cid:15) x (1 − x ) q (cid:48) ( x ) = − (cid:15) x (1 − (cid:15) x ) f (cid:48)(cid:48) ( x ) = g (cid:48)(cid:48) ( x )2 f ( x ) − g (cid:48) ( x ) f (cid:48) ( x )2 f ( x ) , h (cid:48)(cid:48) ( x ) = 2 (cid:15) (1 − x ) q (cid:48)(cid:48) ( x ) = − (cid:15) (1 − (cid:15) x ) g (cid:48) ( x ) = h (cid:48) ( x ) q ( x ) − h ( x ) q (cid:48) ( x ) q ( x ) q ( x ) g (cid:48)(cid:48) ( x ) = h (cid:48)(cid:48) ( x ) q ( x ) − h ( x ) q (cid:48)(cid:48) ( x ) − h (cid:48) ( x ) q (cid:48) ( x ) + 2 h ( x ) q (cid:48) ( x ) q ( x ) (B.23)With that, we find: f (0) = 1 , f ( ±
1) = 1 , f (cid:48) (0) = 0 , f (cid:48)(cid:48) (0) = (cid:15) Because we have f (0) = f ( ±
1) = 1 and f (cid:48) (0) = 0 and f (cid:48)(cid:48) (0) >
0, there mustnecessarily exist at least two points of maximum inside [ − , x m and − x m , where − < x m <
1, such that f (cid:48) ( x m ) = 0 and f (cid:48)(cid:48) ( x m ) < f (cid:48) ( x ) = 0 yields a polynomial equation of 7th degree, in which all theroots can be found analytically. The value ± x m is important, because, if wehave f ( x m ), we’d have an upper bound for f in the [ − ,
1] domain, enabling usto estimate the error we’re committing by integrating a Taylor expansion, and,by allowing us to come up with better trying functions. It turns out, not mucherror is done by Taylor expanding, and, in general, the greater (cid:15) , the greaterthe error. Therefore, we’ll consider: f ( x ) ≈ (cid:15) x (B.24)Therefore, integrating, A n is found: A n ≈ (cid:90) − x n (cid:20) (cid:15) x (cid:21) dx = 1 n + + (cid:15) n + (B.25)Therefore, while (B.16) is the exact expression, a reasonable approximationon top of it can be:˙ G l +1 ≈ πR l − ∞ (cid:88) n =0 l (cid:88) k =0 ( − n a l +1 , k +1 (cid:18) l − n (cid:19) (cid:15) n (cid:20) n + k + 1 + + (cid:15) n + k + 1 + (cid:21) (B.26) B.2 I-Integrals
Recalling (2.4): I ij = (cid:90) S r − ( i + j +2) P i (cos θ ) P j (cos θ ) dS Inserting (A.13) at equation above:˙ I ij = (cid:90) π (cid:90) π r − ( i + j +2) P i (cos θ ) P j (cos θ ) · r sin θ (cid:114) r R (cid:15) cos θ sin θ · dφdθ Integrating in φ :˙ I ij = 2 π (cid:90) π r − i − j P i (cos θ ) P j (cos θ ) sin θ (cid:114) r R (cid:15) cos θ sin θ · dθ (B.27)Making the substitution x = cos θ :˙ I ij = 2 π (cid:90) − r − i − j P i ( x ) P j ( x ) (cid:114) r R (cid:15) x (1 − x ) · dx By the same parity arguments as shown in equation (B.4) and (B.5), weconclude: ˙ I ij = 0 iff P i P j is odd iff i + j is odd. And, thus, ˙ I ij (cid:54) = 0 iff i + j is even. Therefore, one only needs ˙ I ij iff i + j = 2 n for some integer n . Let j = 2 n − i , and thus, I i, n − i (cid:54) = 0. 67ubstituing x = cos θ into the radial equation (A.10), one gets r ( x ), andinserting at equation above, one gets: I ij = 2 π (cid:90) − (cid:20) R √ − (cid:15) x (cid:21) − i − j P i ( x ) P j ( x ) (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx Thus: I ij = 2 πR i + j (cid:90) − P i ( x ) P j ( x ) (cid:104)(cid:112) − (cid:15) x (cid:105) i + j (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (B.28)Because i + j must be even for I ij (cid:54) = 0, then, i + j is an integer. Meaning,the square root in the middle of above expression vanishes, as indicated byexpression below: I ij = 2 πR i + j (cid:90) − P i ( x ) P j ( x ) (cid:2) − (cid:15) x (cid:3) i + j (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (B.29)As it was done in the case of G -integrals, using (B.11) and (B.14), to find: I ij = 2 πR i + j (cid:90) − i (cid:88) u =0 a u,i x u j (cid:88) v =0 a v,j x v i + j (cid:88) n =0 (cid:18) i + j n (cid:19) ( − (cid:15) x ) n (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx, (B.30)Unlike G -integrals, the expansion by the binomial theorem doesn’t yield aninfinite series, precisely because the exponent is an integer. I ij = 2 πR i + j i (cid:88) u =0 j (cid:88) v =0 a u,i a v,j i + j (cid:88) n =0 (cid:18) i + j n (cid:19) ( − (cid:15) ) n (cid:90) − x u + v +2 n (cid:115) (cid:15) x (1 − x )(1 − (cid:15) x ) dx (B.31)Therefore, for i + j even, the exact expression is: I ij = 2 πR i + j i (cid:88) u =0 j (cid:88) v =0 i + j (cid:88) n =0 ( − n a u,i a v,j (cid:18) i + j n (cid:19) (cid:15) n A u + v +2 n (B.32)We can approximate, using (B.25), for i + j even, and get: I ij = 2 πR i + j i (cid:88) u =0 j (cid:88) v =0 i + j (cid:88) n =0 ( − n a u,i a v,j (cid:18) i + j n (cid:19) (cid:15) n (cid:20) n + u + v + + (cid:15) n + u + v + (cid:21) (B.33)For i + j odd, we know I ij = 0. 68 ppendix C Suspended HemisphericalIntegrals
C.1 Exact case
From (A.18), recall: r = (cid:96) cos θ + (cid:112) R − (cid:96) sin θ From (A.20), recall the element of area: J = r sin θ (cid:118)(cid:117)(cid:117)(cid:116) (cid:96) r sin θ (cid:32) (cid:96) cos θ (cid:112) R − (cid:96) sin θ (cid:33) Both equations are valid for signed (cid:96) , depending if one is referring to theupper hemisphere or lower hemisphere. With substitution x = cos θ , just like itwas done with the prolate spheroid, we’ll get: r = (cid:96)x + (cid:112) R − (cid:96) (1 − x ) r = (cid:96) x + R − (cid:96) (1 − x ) + 2 (cid:96)x (cid:112) R − (cid:96) (1 − x ) r = ( R − (cid:96) ) + 2 (cid:96) x + 2 (cid:96)x (cid:112) R − (cid:96) (1 − x ) (C.1)Like the prolate spheroid, r ( x ) is an even function, because, if x >
0, thencos θ >
0, then we are dealing with the upper hemisphere, then (cid:96) >
0. For x < (cid:96) <
0. Therefore, r ( x ) = r ( − x ).The same substitution can be done with J . J = r sin θ (cid:118)(cid:117)(cid:117)(cid:116) (cid:96) r (1 − x ) (cid:32) (cid:96)x (cid:112) R − (cid:96) (1 − x ) (cid:33) θ was left on purpose precisely because dx = − sin θdθ , thus, it willvanish when a change of variables is done in the integral. Thus, making thesubstitution into the r inside the square root, we get: J = r sin θ (cid:118)(cid:117)(cid:117)(cid:117)(cid:116) (cid:96) (1 − x ) (cid:104) (cid:96)x + (cid:112) R − (cid:96) (1 − x ) (cid:105) (cid:32) (cid:96)x (cid:112) R − (cid:96) (1 − x ) (cid:33) (C.2)For the substitution of r , we motivate the following definition: J A = (cid:118)(cid:117)(cid:117)(cid:117)(cid:116) (cid:96) (1 − x ) (cid:104) (cid:96)x + (cid:112) R − (cid:96) (1 − x ) (cid:105) (cid:32) (cid:96)x (cid:112) R − (cid:96) (1 − x ) (cid:33) (C.3)Notice J A is also an even function, as J A ( x ) = J A ( − x ), recalling that, if x > (cid:96) >
0, and, if x < (cid:96) < J = r J A sin θ = J A sin θ (cid:104) (cid:96)x + (cid:112) R − (cid:96) (1 − x ) (cid:105) (C.4)Notice that, as (cid:96) → J A →
1, meaning, J → r sin θ , the sphere case. C.1.1 G-Integrals
Recalling (2.4): G l = (cid:90) S r − l P l (cos θ ) cos θdS Just like it was done previously with other shapes, it is done here. Unlike thespherical or spheroidal case, integration limits happens at the upper hemisphereis θ ∈ [0 , θ ]. For the upper hemisphere, we have: G l = (cid:90) θ (cid:90) π r − l P l (cos θ ) cos θr sin θJ A dθdφ (C.5)Therefore, integrating in φ , we get: G l = 2 π (cid:90) θ r − l +2 P l (cos θ ) cos θ sin θJ A dθ (C.6)One shouldn’t forget there also exists the bottom hemisphere. The totalintegration limits are θ ∈ [0 , θ ] ∪ [ π, π − θ ]. G l = 2 π (cid:32)(cid:90) θ + (cid:90) ππ − θ (cid:33) r − l +2 P l (cos θ ) cos θ sin θJ A dθ (C.7)70ow, substitution x = cos θ can be done. Then, dx = − sin θdθ . Also, recallintegration limits from (A.14), thus we get: G l = 2 π (cid:90) (cid:96)/r r − l +2 P l ( x ) xJ A dx + 2 π (cid:90) − (cid:96)/r − r − l +2 P l ( x ) xJ A dx (C.8)Recalling P n ( − x ) = ( − n P n ( x ), we can do a substitution x = − y in thesecond integral of expression (C.8). Thus: G l = 2 π (cid:90) (cid:96)/r r − l +2 P l ( x ) xJ A ( x ) dx + 2 π (cid:90) (cid:96)/r ( − l r − l +2 P l ( y )( − y ) J A ( − y ) dy (C.9)Therefore, using J A ( x ) = J A ( − x ): G l = 2 π (cid:90) (cid:96)/r r − l +2 P l ( x ) xJ A ( x ) dx + 2 π ( − l +1 (cid:90) (cid:96)/r r − l +2 P l ( x ) xJ A ( x ) dx (C.10)Thus: G l = 0 G l +1 = 4 π (cid:90) (cid:96)/r r − l +1 P l +1 ( x ) xJ A dx (C.11) C.1.2 I-Integrals
Having found an integral expression for G l , we seek attention to I ij . Again,looking at (2.4): I ij = (cid:90) S r − i − j − P i (cos θ ) P j (cos θ ) dS Which becomes: I ij = 2 π (cid:32)(cid:90) θ + (cid:90) ππ − θ (cid:33) r − i − j P i (cos θ ) P j (cos θ ) sin θJ A dθ (C.12)Substitution x = cos θ yields: I ij = 2 π (cid:90) (cid:96)/r r − i − j P i ( x ) P j ( x ) J A dx + 2 π (cid:90) − (cid:96)/r − r − i − j P i ( x ) P j ( x ) J A dx (C.13)Making transformation x ← − x in the second integral: I ij = 2 π (cid:90) (cid:96)/r r − i − j P i ( x ) P j ( x ) J A dx + 2 π (cid:90) (cid:96)/r r − i − j ( − i P i ( x )( − j P j ( x ) J A dx (C.14)Or: I ij = 2 π (cid:90) (cid:96)/r r − i − j P i ( x ) P j ( x ) J A dx + 2 π ( − i + j (cid:90) (cid:96)/r r − i − j P i ( x ) P j ( x ) J A dx (C.15)71hich simplifies to: I ij = 4 π (cid:90) (cid:96)/r r − i − j P i ( x ) P j ( x ) J A dx, i + j ∈ { , , , , , . . . } (C.16)And I ij = 0 otherwise. C.2 Approximation (cid:96) (cid:28) R It would only be a matter of substituting J A from equation (C.3) into (C.11) or(C.16) However, J A form is too complicated to yield a direct integration, thus,we seek to approximate it. C.2.1 Approximating r We need to find an approximation for r and for J . One possibility, is to Taylorexpand with respect to (cid:96) , centered at (cid:96) = 0. Doing that for r , one can get: r = ( R − (cid:96) ) + 2 (cid:96) x + 2 (cid:96)x (cid:112) R − (cid:96) (1 − x ) dr d(cid:96) = − (cid:96) + 4 (cid:96)x + 2 x (cid:112) R − (cid:96) (1 − x ) + 12 2 x(cid:96) · ( − (cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) dr d(cid:96) = − (cid:96) + 4 (cid:96)x + 2 x (cid:112) R − (cid:96) (1 − x ) − x(cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) dr d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 2 x √ R = 2 xR (C.17)Now we find the second derivative: d r d(cid:96) = dd(cid:96) (cid:34) (cid:96) + 4 (cid:96)x + 2 x (cid:112) R − (cid:96) (1 − x ) − x(cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) (cid:35) d r d(cid:96) = − x + 12 2 x · ( − (cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) − x(cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) − x(cid:96) (1 − x ) · − ( − (cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) d r d(cid:96) = − x − x(cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) − x(cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) − x(cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) d r d(cid:96) = − x − x(cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) − x(cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) (C.18)Therefore, we conclude: r ( x ) (cid:12)(cid:12) (cid:96) =0 = R , dr d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 2 xR, d r d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = − x (C.19)And thus, the Taylor approximation for r ( x ) is: r ( x ) = R + 2 xR(cid:96) + (cid:0) x − (cid:1) (cid:96) + O ( (cid:96) ) (C.20)72 .2.2 Approximating r n Recall: r = (cid:96)x + (cid:112) R − (cid:96) (1 − x ) (C.21)The first derivative and second derivative: dr n d(cid:96) = nr n − drd(cid:96)d r n d(cid:96) = n ( n − r n − (cid:18) drd(cid:96) (cid:19) + nr n − d rd(cid:96) (C.22)Thus, the following derivatives are required: drd(cid:96) = x − (cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) d rd(cid:96) = − − x (cid:112) R − (cid:96) (1 − x ) − (cid:96) (1 − x ) [ R − (cid:96) (1 − x )] / (C.23)Therefore, evaluating them at (cid:96) = 0, yields: drd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = x, d rd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = − − x R (C.24)Thus: dr n d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = nxR n − , d r n d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = n ( n − R n − x − nR n − − x R (C.25)Re-writing: dr n d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = nxR n − d r n d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = nR n − (cid:0) nx − (cid:1) (C.26) C.2.3 Approximating J A If a generic function f : R → R is written as: f ( x ) = (cid:112) g ( x ), then: f (cid:48) ( x ) = 12 g (cid:48) ( x ) (cid:112) g ( x ) = 12 g (cid:48) ( x ) f ( x )Looking at J A from (C.3), we notice J A as a function of (cid:96) has similar behavior73s the above function. We define: c ( (cid:96) ) = (cid:112) R − (cid:96) (1 − x ) a ( (cid:96) ) = (cid:96) (1 − x ) (cid:104) (cid:96)x + (cid:112) R − (cid:96) (1 − x ) (cid:105) = (cid:96) (1 − x )[ (cid:96)x + c ( (cid:96) )] b ( (cid:96) ) = (cid:96)x (cid:112) R − (cid:96) (1 − x ) = (cid:96)xc ( (cid:96) ) J A = (cid:113) a ( (cid:96) ) (1 + b ( (cid:96) )) (C.27)Therefore, the first derivative: dJ A d(cid:96) = 12 J A (cid:20) dad(cid:96) (1 + b ( (cid:96) )) + 2 a ( (cid:96) ) dbd(cid:96) (1 + b ( (cid:96) )) (cid:21) dJ A d(cid:96) = 12 J A (1 + b ( (cid:96) )) (cid:20) dad(cid:96) (1 + b ( (cid:96) )) + 2 a ( (cid:96) ) dbd(cid:96) (cid:21) (C.28)The second derivative: d J A d(cid:96) = − J A dJ A d(cid:96) (1 + b ( (cid:96) )) (cid:20) dad(cid:96) (1 + b ( (cid:96) )) + 2 a ( (cid:96) ) dbd(cid:96) (cid:21) + 12 J A dbd(cid:96) (cid:20) dad(cid:96) (1 + b ( (cid:96) )) + 2 a ( (cid:96) ) dbd(cid:96) (cid:21) + 12 J A (1 + b ( (cid:96) )) (cid:20) d ad(cid:96) (1 + b ( (cid:96) )) + dad(cid:96) dbd(cid:96) + 2 dad(cid:96) dbd(cid:96) + 2 a ( (cid:96) ) d bd(cid:96) (cid:21) (C.29)Now, we find the derivatives of a , b , and c , starting with c : dcd(cid:96) = 12 ( − (cid:96) (1 − x ) (cid:112) R − (cid:96) (1 − x ) = − (cid:96) (1 − x ) c ( (cid:96) ) d cd(cid:96) = dd(cid:96) (cid:20) − (cid:96) (1 − x ) c ( (cid:96) ) (cid:21) = − − x c + (cid:96) (1 − x ) c dcd(cid:96)d cd(cid:96) = − − x c + (cid:96) (1 − x ) c (cid:96) (1 − x ) c = − − x c + (cid:96) (1 − x ) c (C.30)Then, derivative of a : dad(cid:96) = 2 (cid:96) (1 − x )[ (cid:96)x + c ( (cid:96) )] − (cid:96) (1 − x )[ (cid:96)x + c ( (cid:96) )] (cid:18) x − (cid:96) (1 − x ) c ( (cid:96) ) (cid:19) dad(cid:96) = 2 (cid:96) (1 − x )[ (cid:96)x + c ( (cid:96) )] − (cid:96) (1 − x )[ (cid:96)x + c ( (cid:96) )] (cid:18) x − (cid:96) (1 − x ) c ( (cid:96) ) (cid:19) (C.31)74hen, derivative of b : dbd(cid:96) = dd(cid:96) (cid:20) (cid:96)xc (cid:21) = xc − (cid:96)xc · (cid:96) (1 − x ) cdbd(cid:96) = xc − (cid:96) x (1 − x ) c ,d bd(cid:96) = − xc dcd(cid:96) + 2 (cid:96)x (1 − x ) c + 4 (cid:96) x (1 − x ) c dcd(cid:96) (C.32)Considering that c (0) = R , then: dcd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 0 , d cd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = − − x c = − − x R ,dad(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 0 , d ad(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 2(1 − x ) c = 2(1 − x ) R ,dbd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = xc = xR , d bd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 0 (C.33)Therefore, evaluating a (0) = b (0) = 0, one can get: dJ A d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 0 , d J A d(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = d ad(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 2(1 − x ) R (C.34)Thus, the Taylor expansion for J A is: J A = 1 + 2(1 − x ) R (cid:96) + O ( (cid:96) ) (C.35) C.2.4 Approximating J Recall we have defined: J = r sin θJ A . Therefore: dJd(cid:96) = dr d(cid:96) sin θJ A + r sin θ dJ A d(cid:96) (C.36)Therefore: dJd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 2 xR sin θJ A + R sin θ · dJd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 2 xR sin θJ A (C.37)Doing the same for second derivative: dJd(cid:96) = d r d(cid:96) sin θJ A + dr d(cid:96) dJ A d(cid:96) sin θ + dr d(cid:96) dJ A d(cid:96) sin θ + r sin θ d J A d(cid:96) (C.38)75herefore: d Jd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 2 J A sin θ + 2 xR · · sin θ + 2 xR · · sin θ + R sin θ · − x ) R d Jd(cid:96) (cid:12)(cid:12)(cid:12)(cid:12) (cid:96) =0 = 2 J A sin θ + R sin θ · − x ) R (C.39)We now write J up to second order: J = (cid:2) R + 2 xR(cid:96) + 2(1 − x ) (cid:96) + O ( (cid:96) ) (cid:3) sin θ (C.40) C.2.5 Approximating r n J A This approximation is fairly important, because it is precisely the integrand ofthe G l and I ij integrals. We proceed by the same way: dd(cid:96) [ r n J A ] (cid:96) =0 = (cid:20) dr n d(cid:96) J A + r n dJ A d(cid:96) (cid:21) (cid:96) =0 (C.41)First, notice r n | (cid:96) =0 = R n , precisely because r | (cid:96) =0 = R . That can also beseen by direct substitution of (cid:96) = 0 in the expression for r n . Now, we seek tocalculate the first derivative. We’ve already calculated the derivatives, whichcan be seen from equations (C.34) and (C.26). dd(cid:96) [ r n J A ] (cid:96) =0 = nxR n − (C.42)Now, to second order: dd(cid:96) [ r n J A ] (cid:96) =0 = (cid:20) d r n d(cid:96) J A + 2 dr n d(cid:96) J A d(cid:96) + r n d J A d(cid:96) (cid:21) (cid:96) =0 (C.43)Again, using equations (C.34) and (C.26), we arrive at: d d(cid:96) [ r n J A ] (cid:96) =0 = nR n − (cid:0) nx − (cid:1) + R n − x ) R d d(cid:96) [ r n J A ] (cid:96) =0 = R n − (cid:2) n (cid:0) nx − (cid:1) + 2(1 − x ) (cid:3) d d(cid:96) [ r n J A ] (cid:96) =0 = R n − (cid:2) n x − n + 2 − x (cid:3) d d(cid:96) [ r n J A ] (cid:96) =0 = R n − (cid:2)(cid:0) n − (cid:1) x − ( n − (cid:3) (C.44)Therefore, the final answer: r n J A = R n + nxR n − (cid:96) + (cid:2)(cid:0) n − (cid:1) x − ( n − (cid:3) R n − (cid:96) + O ( (cid:96) ) (C.45)76 .2.6 G-Integrals To solve the integrals, recall equation (C.11), written below: G l +1 = 4 π (cid:90) (cid:96)/r r − l +1 P l +1 ( x ) xJ A dx (C.46)Our goal would be to Taylor expand the integrand r − l +1 P l +1 ( x ) xJ A ( x ).Because the expansion is around (cid:96) = 0, the terms xP l ( x ) won’t contribute,because they do not depend explicitly on (cid:96) . The remaining terms r − l +2 J A isa case of (C.45), for n = − l + 2. Therefore, making the substitutions, one canget: G l = 4 π (cid:90) (cid:96)/r P l ( x ) x (cid:8) R n + nxR n − (cid:96) + (cid:2)(cid:0) n − (cid:1) x − ( n − (cid:3) R n − (cid:96) (cid:9) dx (C.47)Therefore, one is left with the following integrals to compute: G l = 4 πR − l +2 (cid:90) (cid:96)/r P l ( x ) xdx + 4 π ( − l + 2) R − l +1 (cid:96) (cid:90) (cid:96)/r P l ( x ) x dx + ( l − l + 2) 4 πR l (cid:96) (cid:90) (cid:96)/r x P l ( x ) dx + 4 lπR l (cid:96) (cid:90) (cid:96)/r xP l ( x ) dx (C.48) Moments of Legendre Polynomials
Making reference of (C.48), one requires to calculate first, second and thirdmoments of the legencre polynomials, that is: (cid:90) x m P n ( x ) dx, m ∈ { , , } (C.49)A recurrency relation can be useful in computing integrals:(2 n +1) P n ( x ) = ddx [ P n +1 ( x ) − P n − ( x )] = ⇒ (cid:90) P n ( x ) dx = P n +1 ( x ) − P n − ( x )2 n + 1 + C (C.50)To calculate first moments of the Legendre polynomials, all one has to do,77s to integrate by parts: (cid:90) xP n ( x ) dx = x (cid:90) P n ( x ) dx − (cid:90) P n +1 ( x ) − P n − ( x )2 n + 1 dx (cid:90) xP n ( x ) dx = x P n +1 ( x ) − P n − ( x )2 n + 1 − n + 1 (cid:20) P n +2 ( x ) − P n ( x )2( n + 1) + 1 − P n ( x ) − P n − ( x )2( n −
1) + 1 (cid:21) + C (cid:90) xP n ( x ) dx = x P n +1 ( x ) − P n − ( x )2 n + 1 − n + 1 (cid:20) P n +2 ( x ) − P n ( x )2 n + 3 − P n ( x ) − P n − ( x )2 n − (cid:21) + C (C.51)Once the first moment is calculated, it can be used to calculate the secondmoment, again using integration by parts: (cid:90) x P n ( x ) dx = x (cid:90) P n ( x ) dx − (cid:90) x P n +1 ( x ) − P n − ( x )2 n + 1 dx (cid:90) x P n ( x ) dx = x (cid:90) P n ( x ) dx − n + 1 (cid:20)(cid:90) xP n +1 ( x ) dx − (cid:90) xP n − ( x ) dx (cid:21) (C.52)The third moment can be found by the same procedure. An analytical exactsolution for any moment can be found, though, the expressions start to grow insize pretty quickly. Value of G Given a general expression will be needlessly complicated, we’ll focus on l = 1,that is, G . The Legendre polynomials for these orders are P ( x ) = x . We’llproceed by calculating moments of P . (cid:90) xP ( x ) dx = (cid:90) x · xdx = x C (cid:90) x P ( x ) dx = (cid:90) x · xdx = x C (cid:90) x P ( x ) dx = (cid:90) x · xdx = x C (C.53)Using expression (C.48), we find for G : G = 4 πR x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r + 4 π(cid:96) x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r − πR (cid:96) x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r − πR (cid:96) x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r G = 4 πR (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 4 π(cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) − πR (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 4 πR (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) (C.54)78r, rewriting: G = 43 πR (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + π(cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) − π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 4 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) (C.55) C.2.7 I-Integrals
Substituting r − i − j J A as in equation (C.45) into (C.16) for n = − i − j , one findsthe expression for I ij , valid for i + j even, is: I ij = 4 πR i + j (cid:90) (cid:96)/r P i ( x ) P j ( x ) dx − ( i + j ) 4 πR i + j +1 (cid:96) (cid:90) (cid:96)/r xP i ( x ) P j ( x ) dx + (cid:2) ( i + j ) − (cid:3) πR i + j +2 (cid:96) (cid:90) (cid:96)/r x P i ( x ) P j ( x ) dx + [( i + j ) + 2] 4 πR i + j +2 (cid:96) (cid:90) (cid:96)/r P i ( x ) P j ( x ) dx (C.56)The required Legendre moments are more complicated to calculate explicitly: (cid:90) x m P i ( x ) P j ( x ) dx (C.57)Thus, we’ll directly evaluate I and I . We already know that I = I =0, because 1 + 0 is odd. Moments as before can be calculated easily: (cid:90) xP ( x ) P ( x ) dx = (cid:90) xdx = x C (cid:90) x P ( x ) P ( x ) dx = (cid:90) x dx = x C (cid:90) xP ( x ) P ( x ) dx = (cid:90) x · x dx = x C (cid:90) x P ( x ) P ( x ) dx = (cid:90) x · x dx = x C (C.58)Using expression (C.56) to calculate I , one gets: I = 4 π · | (cid:96)/r − − · πR (cid:96) x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r + 2 · πR (cid:96) · | (cid:96)/r (C.59)79hich evaluates into: I = 4 π · (cid:20) − (cid:96)r (cid:21) − π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 8 πR (cid:96) (cid:20) − (cid:96)r (cid:21) (C.60)The same thing can be done for I : I = 4 πR x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r − · πR (cid:96) x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r + 2 · πR (cid:96) x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r + 4 · πR (cid:96) x (cid:12)(cid:12)(cid:12)(cid:12) (cid:96)/r (C.61)Which evaluates into: I = 4 π R (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) − πR (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 8 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) + 16 π R (cid:96) (cid:34) − (cid:18) (cid:96)r (cid:19) (cid:35) (C.62)80 ibliography [1] R. G. Forbes, C. J. Edgcombe, U. Valdr`e, Some comments on models forfield enhancement . Ultramicroscopy , 57 (2003)[2] R. H. Fowler, L. Nordheim, Electron emission in intense electric fields . Proc.R. Soc. A , 173 (1928)[3] R. G. Forbes, J. H. B. Deane,
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