EM Boundary Value Problems

Transcription

1 EM Bounday Value Poblems 10/ 9 11/ By Ilekta chistidi & Lee, Seung-Hyun A. Geneal Desciption : Maxwell Equations & Loentz Foce We want to find the equations of motion of chaged paticles. The way to do this: - given chages q i with position i (t) and velocity v i (t), the electic and magnetic field E(,t) and B(,t) can be computed - with E and B known, the foce F can be computed - given F, the futue motion can be pedicted This poblem tuns out to be NON-LINEAR, fo the motion of chages depend on the Loenz foce, which depends on pesent E and B, which depend on pevious q i position, which depend on pevious foce and so on, i.e. the histoy of chaged-paticle position and velocity matte as well. Theefoe, the inteaction of chages and fields makes the poblem too complicated, and in ode to solve it we need BOUNDARY CONDITIONS. Example: Chage q at =0. q Then, the electic and magnetic field E = + E 3 o B = B o confim Maxwell's equations fo any E o and B o, so the solution is not unique. But, if we impose the B.C. E and B 0 fo 0 we have a unique solution, because we demand that E B 0. 0, 0 The solution to an EM poblem is staightfowad fo the case in which the chage distibution is eveywhee specified, fo then, the electic field ( E ) and magnetic field( B ) ae diectly computed.

2 1. The elationship between souce chages and the electic field Coulomb s integal law ρ( )( ) E( ) = dv (1) 3 The pai of diffeential equations E = 4πρ (Gauss Law) () 1 B E = (3) c t Scala potential & Gadient opeation ρ( ) Φ( ) = v d v 1 A E = Φ c t (4) (5). The elationship between souce cuents and the magnetic field Biot-Savat integal law 1 J ( ) ( ) B( ) = dv 3 c (6) The pai of diffeential equations 4π 1 E B = J (7) c c t B = (8) Vecto potential & Cul opeation 1 J ( ) A ( ) = dv c (9) B = A (10)

3 3. The Loentz Foce v F = q( E + B) (11) c This equation is known as the Loentz Foce on a moving chage and is valid fo time-vaying as well as fo steady-state fields. Physics = Pediction Investigation fo the tajectoy of a paticle Given chaged paticles E, B Given E(, t), B(, t) fo a paticle q(, v; t) F B. The Equations of Laplace & Poisson Often, we ae confonted with situation whee we do not know, a pioi, the chage distibutions, and consequently we cannot diectly detemine E,Φ. An impotant example of this is when we have a system of conductos whose elative potentials ae known, but the chage densities on the conducto suface ae not. In this instance, as we will see, thee exists a solution, poducing well-behaved electic fields and chage densities. Ou explicit fomula fo E and Φ demand that we know the chage distibutions on the conducto sufaces. It tuns out that we employ two citeia simultaneously to solve two poblems. Fo example, although each of two linea algebaic equations may contain two unknowns and neithe can be solved alone fo both unknowns, the two in combination can be solved.

4 In the case of electostatics, two elations that can be solved simultaneously ae as follows: Fom () and (5) fo static fields ( = 0) : t E and E = Φ = 4πρ These two equations may be futhe combined into one equation-namely, Φ = 4πρ (1) : Poisson s Equation If ρ = 0 in some egion of space, then Φ = (13) : Laplace Equation Φ( ) 0 o at least not diveging, is condition fo finite total enegy of chage system, Ty Φ = const. Φ = x, y, z and Φ = x ( x y ) = 0, ( x + y z ) = 0 Fo steady cuent, 4π B = J c B is not a consevative field. Theefoe it makes no sense to intoduce a magnetic scala potential in the same sense as we intoduced an electostatic potential. Nonetheless, thee usually ae cetain egions of space whee B = 0. Fo these egions we ae pemitted to intoduce a scala potential function just so long as the space is simply connected. A simply connected egion is one fo which any closed cuve constucted theein can be shunk down continuously to a point without the cuve leaving the egion. y

5 Assuming then that thee exists a simply connected volume V, we define a magnetic scala potential function Φ m in V, such that 4π B = Φm (14) c Substituting this in (8) gives Φ m = (15) 1. Polynomial of Degee N Fo the Laplace equation (13), let s conside a homogeneous polynomial; l m n x y z and l + m + n = N polynomial of degee N P = homogeneous polynomial of degee N Non-intege powes excluded because they lead to singula behavio if ( ) is applied enough times. Insist that Φ and all it s deivatives ae bounded. Counting homogeneous polynomial Suppose thee ae N + objects, and if we divide these objects into N and, then the possible numbe of distibution is N + ( N + )! ( N + )( N + 1) = = N N!! And Laplace equation gives the homogeneous polynomial of degee N. Then, the numbe of independent polynomial of degee N which obey Laplace equation will be ( N + )( N + 1) N( N 1) = N + 1 ( ) Theefoe, P N ( x, y, z) = 0 gives N + 1 polynomials.

6 . Two-dimensional vecto space analogue It follows fom the pevious analysis that the geneal poblem to be solved is: ˆ ( E, B) (, J) ( E, B) ˆ 1 Ο = ρ = Ο ( ρ, J) whee Ô is some opeato ("Maxwell opeato"), i.e. the invese of this opeato has to be found. We can bette peceive this poblem by using a two-dimensional vecto space, whee Ô is a x matix M. We will investigate the case whee M -1 does not exist (M non-invetible), because the opposite case is tivial. We conside the bounday-value poblem. Fo given matix M and vectos u,w, we can wite as follows; Mu = w (16) Suppose M = and u o = then, Mu o = 0 and (16) does not have a unique solution, since we can add to evey solution u a tem µ u0 and yet (16) is valid. Example : M, u Φ : homogeneous Laplace equation 3. Uniqueness of solutions Assuming that a mathematical solution to (16) indeed exists. This is not obvious, although physically we know that if we have a system of conductos, with an abitay chage o potential on each, then thee will exist some electostatic field solution in the space. How can we get unique solution to (16)? Let a 1 u =, then u is othogonal to a 1

7 Theoem : If Poof : Let s suppose a M = w, then u is a unique solution. a a α α' M = w, then M a α α' = 0 α α α' α' = 1 µ 1 α α = α' α' So, by imposing a condition on u (potential), namely that u is othogonal to u o, we assued the uniqueness of the solutions. β Unde this condition, the only possible choice fo w (souces) is w = β Example Let s conside Poisson s equation (1) Φ = 4πρ and we want unique solution, 3 1 Φ( ) = d ρ ( ) : a solution but unique? If we conside Φ = 4πρ, then ( Φ Φ) = 0 must ule out solutions to this. ( ) If we assume Φ is C (all deivatives ae continuous), then Φ Φ = homogeneous polynomial ( x, y, z) None of these polynomials vanish as except fo Φ Φ = const All divege as. only possible Φ Φ = const Unde these assumptions, Φ indeed is unique. Φ = 4πρ, Φ 0 as with bounday condition is invetible

8 Futhe equiement on souce ; ( ) 0 ρ as Ty contay ρ ( ) = const. = ρ0 Note that if we did not limit the chage density ρ 0, we would have multiple solutions of the equation Φ = 4πρ Fo example Φ( z) = πρoz and 4πρo x + y + z Φ = 3 ae both solutions, but they clealy violate the b.c.. Thus, by ou (wong) choice of souce, we violated the b.c. and lost the uniqueness of the solution. Resolve these issues by taking a limit; ρ0 ρ = 0 o ( < R) ( > R) 4 E Q 3 E = Q( R) 3 fo ( > R) 0 = πρ = ( ) fo ( < R) 3 Fo well-posed poblem Φ = 4πρ need B.C. on Φ and confinement conditions on ρ

9 [Remak] We shall show that a potential distibution obeying Poisson's equation is completely specified within a volume V if the potential is specified ove the sufaces bounding that volume. Such a uniqueness theoem is useful fo two easons: (a) It tells us that if we have found such a solution to Poisson's equation, whethe by mathematical analysis o physical insight, then we have found the only solution; and (b) it tells us what bounday conditions ae appopiate to uniquely specify a solution. If thee is no chage pesent in the volume of inteest, then the theoem states the uniqueness of solutions to Laplace's equation. we assume that the solution is not unique- that two solutions, a and b, exist, satisfying the same bounday conditions- and then show that this is impossible. The pesumably diffeent solutions a and b must satisfy Poisson's equation with the same chage distibution and must satisfy the same bounday conditions. It follows that with d defined as the diffeence in the two potentials, d = a - b, A simple agument now shows that the only way d can both satisfy Laplace's equation and be zeo on all of the bounding sufaces is fo it to be zeo. Fist, it is agued that d cannot possess a maximum o minimum at any point within V. With the help of Fig.1, visualize the negative of the gadient of d, a field line, as it passes though some point o. Because the field is solenoidal (divegence fee), such a field line cannot stat o stop

10 within V. Futhe, the field defines a potential. Hence, as one poceeds along the field line in the diection of the negative gadient, the potential has to decease until the field line eaches one of the sufaces S i bounding V. Similaly, in the opposite diection, the potential has to incease until anothe one of the sufaces is eached. Accodingly, all maximum and minimum values of d ( ) have to be located on the sufaces. Figue.1 Field line oiginating on one pat of bounding suface and teminating on anothe afte passing though the point o The diffeence potential at any inteio point cannot assume a value lage than o smalle than the lagest o smallest value of the potential on the sufaces. But the sufaces ae themselves at zeo potential. It follows that the diffeence potential is zeo eveywhee in V and that a = b. Theefoe, only one solution exists to the bounday value poblem stated with (1). 4. Paticula and Homogeneous Solutions to Poisson's and Laplace's Equations Suppose we want to analyze an electoquasistatic situation as shown in Fig.. A chage distibution () is specified in the pat of space of inteest, designated by the volume V. This egion is bounded by pefect conductos of specified shape and location. Known potentials ae applied to these conductos and the enclosing suface, which may be at infinity.

11 Figue. Volume of inteest in which thee can be a distibution of chage density. To illustate bounding sufaces on which potential is constained, n isolated sufaces and one enclosing suface ae shown. In the space between the conductos, the potential function obeys Poisson's equation. A paticula solution of this equation within the pescibed volume V is given by the supeposition integal. This potential obeys Poisson's equation at each point within the volume V. Since we do not evaluate this equation outside the volume V, the integation ove the souces called fo in (1) need include no souces othe than those within the volume V. This makes it clea that the paticula solution is not unique, because the addition to the potential made by integating ove abitay chages outside the volume V will only give ise to a potential, the Laplacian deivative of which is zeo within the volume V. Is (1) the complete solution? Because it is not unique, the answe must be, suely not. Futhe, it is clea that no infomation as to the position and shape of the conductos is built into this solution. Hence, the electic field obtained as the negative gadient of the potential p of (1) will, in geneal, possess a finite tangential component on the sufaces of the electodes. On the othe hand, the conductos have suface chage distibutions which adjust themselves so as to cause the net electic field on the sufaces of the conductos to have vanishing tangential electic field components. The distibution of these suface chages is not known at the outset and hence cannot be included in the integal (1).

12 A way out of this dilemma is as follows: The potential distibution we seek within the space not occupied by the conductos is the esult of two chage distibutions. Fist is the pescibed volume chage distibution leading to the potential function p, and second is the chage distibuted on the conducto sufaces. The potential function poduced by the suface chages must obey the souce-fee Poisson's equation in the space V of inteest. Let us denote this solution to the homogeneous fom of Poisson's equation by the potential function h. Then, in the volume V, h must satisfy Laplace's equation. The supeposition pinciple then makes it possible to wite the total potential as The poblem of finding the complete field distibution now educes to that of finding a solution such that the net potential of (3) has the pescibed potentials v i on the sufaces S i. Now p is known and can be evaluated on the suface S i. Evaluation of (3) on S i gives so that the homogeneous solution is pescibed on the boundaies S i. Hence, the detemination of an electoquasistatic field with pescibed potentials on the boundaies is educed to finding the solution to Laplace's equation, (), that satisfies the bounday condition given by (5). Supeposition to Satisfy Bounday Supeposition will often be used in anothe way to satisfy bounday conditions. Suppose that thee is no chage density in the volume V, and again the potentials on each of the n sufaces S j ae v j. Then The solution is boken into a supeposition of solutions j that meet the equied condition on the j-th suface but ae zeo on all of the othes.

13 Each tem is a solution to Laplace's equation, (6), so the sum is as well.