Lecture 4 Notes, Electromagnetic Theory I Dr. Christopher S. Baird University of Massachusetts Lowell
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1 Lecture 4 Notes, Electromgnetic Theory I Dr. Christopher S. Bird University of Msschusetts Lowell 1. Orthogonl Functions nd Expnsions - In the intervl (, ) of the vrile x, set of rel or complex functions U n (x) where n = 1, 2,... re orthogonl if: U n * xu m xdx=, m n - When m = n, the integrl is nonzero. The functions re orthonorml if normlized to one: U n * xu m xdx= nm - An ritrry, integrle function f(x) cn e expnded in series of the orthonorml functions U n (x) ccording to: N f x= n U n x n=1 - To find the expnsion coefficients n, we multiply oth sides y the function U * M(x), integrte, nd use the orthonormlity property: N f xu * m x= n U n xu * m x n=1 N f xu m * xdx= n=1 n U n xu * m xdx f xu * m xdx= n nm N n=1 f xu m * xdx= m - Interchnge the ritrry lel m for n nd get the finl form: N f (x)= n=1 n U n ( x) where n = f ( x)u n * (x)dx Finite Series Expnsion
2 - If the functions form complete set, nd ll functions tht re useful in physics do, then the series expnsion ecomes more ccurte representtion of the function f(x) s more terms in the series re kept. The most ccurte is the infinite series: f (x)= n=1 n U n ( x) where n = f ( x)u n * (x)dx Infinite Series Expnsion - If the intervl (, ) is expnded to e infinite, thn the orthogonl functions ecome continuum of functions, the index vrile n ecomes continuous vrile k, nd the orthogonlity condition ecomes normlized to the Dirc delt function: U k *xu k ' x dx= k k ' f (x)= A(k )U k (x)dk where A(k)= f (x)u n * ( x)dx Infinite Continuous Expnsion 2. Fourier Series - The most commonly used orthogonl functions re sines nd cosines, constituting Fourier series. - Strt with generl expnsion in terms of sines nd cosines over the intervl (-/2, /2): f x= A n cosk n x B n sink n x n= - In order for the series to e vlid representtion of the function in the intervl, the series must e periodic outside the intervl, so tht f(-/2) = f(/2). Using this requirement leds to: A n cosk n /2 cos k n / 2B n sin k n /2 sin k n /2= n= - This must e true independent of A n nd B n, so tht the coefficients must e zero: sin k n /2 sin k n / 2= sin k n /2= k n /2=n k n = 2n The series now ecomes: f x= n= An cos 2 n x B sin n 2 n x
3 / 2 / 2 To find the coefficients, multiply oth sides y cos 2 m x where m is n integer nd integrte over the intervl (-/2, /2): dx f x cos 2 m x = n= / 2 A n /2 cos 2 m x cos 2 n x / 2 dxb n /2 cos 2 m x sin 2 n x dx - The first integrl on the right is zero, except when m = n due to orthogonlity, nd the second integrl is lwys zero. / 2 / 2 dx f xcos 2 m x 2 / 2 =A m cos 2 m x / 2 dx - Use integrtion y prts to solve the integrl on the right nd finlly fter releling: A n = 2 / 2 / 2 f xcos 2 n x dx - The sme pproch is repeted, multiplying oth sides y sin 2 m x B n = 2 / 2 f xsin 2 n x / 2 dx nd integrting: - In summry, ny function on the intervl (-/2, /2) cn e expnded in Fourier series: f (x)= n= An cos ( 2πn x ) +B sin n ( 2π n x ) where A n = 2 / 2 /2 f ( x)cos( 2 π n x ) dx nd B = 2 / 2 n /2 f ( x)sin( 2πn x ) dx - A more useful form of the Fourier series is in terms of complex exponentils: f (x)= 1 n= A n e i (2π n x /) where A n = 1 / 2 / 2 f ( x')e i (2 πn x '/ ) dx' - Becuse the summtion index now spns negtive infinity to positive infinity insted of just zero to positive infinity, oth terms e +ikx nd e -ikx hve een comined into one term. Sometimes it is more useful to leves the terms seprte. - If the intervl ecomes infinite ( ), the series ecomes n integrl of continuum of functions. There is now no restriction on k n, so it is just k: f x= Ak e i k x dk
4 - Multiply oth sides y e i k ' x nd integrte: f xe i k ' x dx= Ak e i k ' x e i k x dx dk - Using the orthogonlity of complex exponentils: A k= 1 2 f x e i k x dx - More often, the constnt A n is redefined to f (x)= 1 A(k)e i k x dk where A(k)= 1 2π 2π e i(k k ') x dx=2πδ(k k ') 1 2 A n to mke the equtions symmetric: f (x)e i k x dx Fourier Integrl 3. Seprtion of Vriles: Lplce Eqution in Rectngulr Coordintes - Often differentil eqution cn e roken into set of independent equtions. - The Lplce eqution 2 = is used when chrge-free region is ounded y oundry where the potentil is known. In rectngulr coordintes: 2 x 2 2 y 2 2 z = 2 - Try solution of the form: x, y, z= X xy yz z Y ( y)z (z) 2 X ( x) x 2 + X (x) Z ( z) 2 Y ( y) y 2 + X (x)y ( y) 2 Z ( z) z 2 = - Divide ech side y X x Y yz z 1 d 2 X x 1 d 2 Y y 1 d 2 Z z = X x d x 2 Y y d y 2 Z z d z 2 - Totl derivtives hve replced prtil derivtives ecuse the functions re of only one vrile. - This eqution must hold for ll possile vlues of the independent coordintes, therefore the terms must e independent. Ech cn e set to n ritrry constnt: 1 d 2 X x = 2, X x d x 2 - Simplify the equtions: 1 d 2 Y y = 2, Y y d y 2 1 d 2 Z z = 2 where 2 2 = 2 Z z d z 2
5 d 2 X x d x 2 = 2 X x, - And find the solutions: d 2 Y y d y 2 = 2 Y y, d 2 Z z = 2 Z z d z 2 X ( x)=a e i α x +B e i α x, Y ( y)=c e i β y +D e i β y, Z ( z)=f e γ z +G e γ z X ( x)=a+b x if α=, Y ( y)=c +D y if β=, Z ( z)=f +G z if γ= - The prticulr solution for non-zero constnts is: x, y, z= X xy yz z Φ( x, y, z)=(a αβ e i α x +B αβ e i α x )(C αβ e i β y + D αβ e iβ y )(F αβ e γ z +G αβ e γ z ) if α nd β - The prticulr solutions for when the constnts my e zero re: Φ( x, y, z)=(a β + B β x)(c β e i β y + D β e i β y )(F β e β z +G β e β z ) if α = nd β Φ( x, y, z)=(a α e i α x +B α e i α x )(C α +D α y)(f α e αz +G α e α z ) if α nd β = Φ( x, y, z)=(a + B x)(c +D y)( F γ= +G z) if α = nd β = - The generl solution is the sum of ll possile prticulr solutions: Φ( x, y, z)= α β ( A αβ e i α x +B αβ e i α x )(C αβ e iβ y +D αβ e i β y )( F αβ e γ z +G αβ e γ z ) + β + α ( A β +B β x)(c β e i β y +D β e iβ y )(F β e β z +G β e β z ) ( A α e i α x +B α e i α x )(C α +D α y)(f α e α z +G α e α z ) +(A + B x)(c +D y)(f +G z) - This is the most generl solution possile. Most geometries re simple enough tht most of these terms drop out. - Let us now discuss differentil equtions in generl. - A derivtive eqution such s d y =2 d x only specifies the solution up to n ritrry constnt. In this cse y = 2x + C. We must hve n dditionl piece of informtion, oundry condition, to specify unique solution to the derivtive eqution. In this cse, if we know y(1) = 4, then we find the unique solution to e y = 2x A differentil eqution is just comintion of derivtives nd functions. - For every derivtive in differentil eqution, the solution will hve one integrtion constnt, nd there must e one oundry condition to specify it. - The Lplce eqution in three-dimensions hs second-order derivtive (which is relly just derivtive pplied twice) in ech dimension, for totl of six derivtives. This mens tht for prticulr solution, there will e six constnts, nd we need six oundry conditions to specify these constnts. The six oundry conditions re just the vlue of the potentil on the six sides of the ox contining the volume of interest. - Looking t the most common prticulr solution to Lplce's eqution in three-dimensionl
6 rectngulr coordintes, it would seem tht we hve more thn six constnts: Φ( x, y, z)=(a αβ e i α x +B αβ e i α x )(C αβ e i β y + D αβ e iβ y )(F αβ e γ z +G αβ e γ z ) - But in relity, mny of the constnts cn e comined nd lso γ is function of the others constnts: Φ( x, y, z)= A αβ (e i α x +B αβ e i α x )(e i β y + D αβ e iβ y )(e α 2 +β 2z +G αβ e α2 +β 2z ) - It is good prctice when pproching prolem to write down the numer of constnts, identify them, nd write ll of the oundry conditions. 4. Exmple of Rectngulr Boundry Conditions for Chrge-Free Regions - Consider for simplicity ox with one corner t the origin nd the opposite corner t the point (,, c) in the (x, y, z) dimensions, where the potentil is everywhere zero on the surfce of the ox except t the z = c surfce where: x, y, z=c=sin x sin y - The other five oundry conditions re the five other sides of the ox held t zero. - A creful nlysis leds us to see tht ll the zeropotentil sides force ll prticulr solutions to dispper except the most common ones: Φ( x, y, z)= α β ( A αβ e i α x +B αβ e i α x )(C αβ e iβ y +D αβ e i β y )( F αβ e γ z +G αβ e γ z ) - Now we pply ll oundry conditions one y one: x=, y, z= = α β ( A αβ +B αβ )(C αβ e iβ y +D αβ e i β y )(F αβ e γ z +G αβ e γ z ) - This is only true for ll y nd z if A αβ +B αβ =, or B αβ = A αβ. - The solution is now: Φ( x, y, z)= Φ( x, y, z)= α β α β A αβ (e i α x e i α x )(C αβ e iβ y +D αβ e i β y )( F αβ e γ z +G αβ e γ z ) A αβ sin (α x)(c αβ e i β y + D αβ e iβ y )(F αβ e γ z +G αβ e γ z ) - Next pply the oundry condition x=, y, z= :
7 = α β A αβ sin(α)(c αβ e i β y +D αβ e i β y )( F αβ e γ z +G αβ e γ z ) - This is only true for ll y nd z if =n where n=,1,2,... so tht the solution ecomes: Φ( x, y, z)= n β A nβ sin ( nπ x ) (C nβ ei β y +D nβ e i β y )(F nβ e γ z +G nβ e γ z ) - The exct sme process occurs in the y dimension, yielding: Φ( x, y, z)= n m A nm sin ( nπ x - With α nd β determined, γ hs een determined: ) sin ( mπ y ) (F nm eγ z +G nm e γ z ) where n, m=,1,2,... = 2 2 = n2 2 m2 2 = n 2 / 2 m 2 / Next pply the oundry condition x, y, z== : = n m Φ( x, y, z)= A n m sin ( n π x n m ) sin ( m π y ) (F +G ) n m n m nd thus G n m = F n m, yielding: A nm sin ( nπ x ) sin ( mπ y ) sinh (π n2 / 2 +m 2 / 2 z ) - Apply the lst oundry condition: x, y, z=c=sin x sin y sin ( π x ) sin ( π y ) = n m A n m sin ( nπ x ) sin ( mπ y ) sinh(π n2 / 2 +m 2 / 2 c) - The only term of the series expnsion tht is needed to represent the left side of the equlity is the n = 1, m = 1 term, reducing the eqution to: 1= A 1,1 sinh (π 1/ 2 +1/ 2 c) which uniquely determines the finl coefficient: 1 A 1,1 = sinh 1/ 2 1/ 2 c - The finl solution to this exmple is:
8 sin Φ( x, y, z)= ( π x ) sin ( π y ) sinh (π 1/ 2 +1/ 2 z ) sinh (π 1/ 2 +1/ 2 c ) - We cn mke the cse more generl y supplying the ritrry oundry condition: x, y, z=c=v x, y - Then the previous nlysis still follows, except now the lst oundry condition requires: V ( x, y)= n m A n m sin ( nπ x ) sin ( m π y ) sinh (π n2 / 2 +m 2 / 2 c ) - The oundry vlue function V is eing expnded in Fourier series with coefficients: A n m sinh (π n 2 / 2 +m 2 / 2 c ) - As ws done with the generl Fourier series, we solve for the coefficients y multiplying oth sides y sines of x nd y nd integrting, so tht the orthogonlity picks out coefficients: A n m sinh (π n 2 / 2 +m 2 / 2 c )= 4 A n m = 4 sinh (π n 2 / 2 +m 2 / 2 c) - So tht the generl solution is: Φ( x, y, z)= where A n m = n m dx dx dyv ( x, y)sin( nπ x ) ( sin m π y ) dyv ( x, y)sin( nπ x ) ( sin m π y ) A n, m sin ( n π x ) sin ( mπ y ) sinh (π n2 / 2 +m 2 / 2 z ) 4 sinh (π n 2 / 2 +m 2 / 2 c) dx dyv ( x, y)sin( nπ x ) ( sin m π y )
9 5. The Lplce Eqution in Polr Coordintes - If the oundry conditions of chrge-free region exhiit uniformity in one dimension nd circulr shpe in the other two dimensions symmetry, it is much more nturl mthemticlly to use polr coordintes rther thn rectngulr coordintes. - If the oundry conditions re uniform in the z-dimension, the three-dimensionl cylindriclcoordintes prolem reduces to two-dimensionl polr-coordintes prolem. - Just like ws done for rectngulr coordintes, seprtion of vriles nd Fourier series cn e used to solve the Lplce eqution in polr coordintes. - The two-dimensionl Lplce eqution in polr coordintes is given y: 2 = = - We use the seprtion of vriles pproch y trying solution of the form:,=r - Sustituting it in: R R 1 2 = Multiply y R 2 nd ring one term to the right: R R = The two terms re seprtely functions of two independent vriles nd must hold for ll vlues of the two independent vriles, so they must e relted y constnt, which we cll υ 2. R R =2 nd 2 = Put ech in n instructive form: R =2 R nd 2 = The generl solution for is now pprent: R = nd = A e i i B e - So tht the generl solution to the Lplce eqution in polr coordintes for is:,=r
10 ,= A e i B e i - If =, the differentil equtions ecome: R = nd 2 2 = with the generl solutions for = : R = ln nd A B - So tht, finlly, the most generl solution ecomes: Φ(ρ,ϕ)=( + lnρ)(a + B ϕ)+ ( ν ρ ν + ν ρ ν )( A ν e i ν ϕ +B ν e i νϕ ) ν - The coefficients re now determined y pplying oundry conditions nd recognizing the result s Fourier series expnsion. - Becuse the z dimension is uniform nd is therefore ignored, we hve solved second-order differentil in two dimensions, so there should e four totl integrtion constnts nd four oundry conditions. If we tke the most common prticulr solution nd comine constnts, we find this to e the cse: P,= e i B e i - Note tht υ is only n integer if the region of interest where we re trying to solve for the potentil includes ll possile zimuthl ngles, or in other words if there is no physicl oundry condition t some fixed zimuthl ngle. Otherwise, υ is not n integer. This mens tht the requirement tht the potentil e single-vlued, Φ(ρ,ϕ)=Φ(ρ,ϕ+2π), is only vlid if the full sweep of possile zimuthl ngles is included in the region of interest. 6. The Lplce Eqution with z-uniform Cylindricl-Shell Boundries - Consider cylindricl shell of rdius ρ with potentil of V(φ) on the shell. We re seeking the potentil inside the shell. - Becuse the region of interest involves the full sweep of φ, to keep φ single-vlued we must require: z Φ(ρ, ϕ)=φ(ρ, ϕ+2π) - This condition is not true for geometries for which the region where vlid potentil is required does not spn ll possile vlues of φ. -In this geometry, the single-vlued requirement leds to: ln A B A e i B e i, ρ y x
11 = ln A B 2 A e i 2 B e i 2, - Becuse the vriles re independent, every term in the series must mtch, leding to: B = s well s: e i i 2 =e which is only possile if ν=m where m is n integer: m=,1,2,... - The solution now ecomes:,= ln m=1 m m m m A m e i m B m e i m - Becuse the region of interest includes the origin, the potentil must e finite t the origin. Thus m = nd =, to keep those terms from lowing up t the origin. This reduces the solution to:, = m A m e i m B m e i m where the term hs now een included in the sum. m= - The index m cn e mde to run from negtive infinity to positive infinity, thus covering oth terms:,= A m m e i m m= - Now pply the oundry condition: =, =V V = A m m e i m m= - Multiply oth side y complex exponentil, integrte, nd use the orthogonlity of exponentils to find:,= A m m e i m where A m = 1 m= 2π ρ - Or, written in more intuitive form: π m π V (ϕ ')e i(mϕ ') d ϕ' Φ(ρ,ϕ)= A m= m( ρ m ρ ) e i mϕ where A m = 1 π V (ϕ')e i(mϕ' ) d ϕ' 2π π - Although this exmple only considered one cylindricl shell, the sme pproch is used to solve for oundry conditions on multiple cylindricl shells.
12 7. The Lplce Eqution with Intersecting Plnes - Next consider two plnes tht intersect, one on the x-xis, the other plne t n ngle β from the first plne, nd oth uniform in the z dimension, with oundry conditions:,==v,,==v, =, =V 1 y (ρ, φ) - This is lso pplicle to the edge of more complex prolem, where the chrges nd other surfces re fr enough wy tht they only come into ply in tht they crete the oundry condition t ρ, thus the Lplce eqution still pplies. - As efore, inclusion of the origin in the region of interest leds to υ =, including = :,= A B A e i B e i, - Apply the oundry condition,==v V =A A B, - Which is only possile if A =V, A = B so tht the solution now ecomes:,=v B B sin, - Apply the oundry condition,==v : V =V B B sin, =B B sin, β x - This must e vlid for ll ρ, requiring B = nd ν=m π/β where m=,1,2,..., yielding:,=v B m m / sinm / m=1 - Note tht ν is not equl to n integer in this cse, ut is equl to =m / ecuse we hve physicl oundries t certin zimuthl ngles. - Apply the finl oundry condition, which is typiclly dependent on the chrge fr wy: =, =V 1 m/ V 1 =V B m sin m / m=1
13 m/ V 1 V = B m sin m / m= - Multiply oth sides y sin n / nd integrte oth sides from to β: d V 1 V sin n /= m= m B m / d sin m /sin n / - Due to orthogonlity, the integrl on the right is lwys zero, except when m = n: n / d V 1 V sin n /= B n d sin 2 n/ - The integrl on the right evlutes to /2 : B n = 2 n / d V 1 V sin n / - The finl solution then tkes the form: Φ(ρ,ϕ)=V + B m ρ m π/β sin(m π ϕ/β), B m = 2 m= β ρ mπ/ β β d ϕ(v 1 (ϕ) V )sin (m π ϕ/β) - Let us get sense of the generl ehvior of the potentil ner the edge. In the series expnsion, we keep only the fist non-zero term, m = 1: Φ(ρ, ϕ)=v +B 1 ρ π/β sin(π ϕ/β) ner ρ= - The electric field, E=, is clculted in polr coordintes: E= [ ρ ρ + ϕ 1 ρ ϕ] [ V + B 1 ρ π/β sin (π ϕ/β)] E= π B 1 ρ π/β 1 ( ρ sin(π ϕ/β)+ ϕ cos(π ϕ/β)) β - The surfce chrge densities t ϕ= nd ϕ=β re: =[ E n] surfce = E, = σ(ρ)= ϵ π B 1 ρ π/β 1 β
14 Numericl Plots of surfce chrge densities for vrious ngles, higher densities in drker lue. 8. Finite Element Anlysis - Often the Poisson eqution cnnot e solved nlyticlly. Insted we must solve the prolem numericlly. Finite Element Anlysis is one useful numericl method. - This pproch hs three foundtionl ides: 1. Expnd the electric potentil's solution into series sum over set of simple orthogonl functions ϕ i so tht we cn clculte the derivtives of the known functions insted of the unknown electric potentil. Note tht this is simple one-dimensionl set of functions, ut tht it spns two or three-dimensionl spce. 2. Set up the prolem so tht it ecomes liner lger prolem, ecuse computers cn clculte this type of prolem reltively quickly. 3. Choose expnsion functions tht re loclized to different points in spce ( finite elements ) so tht the mtrix ends up sprse nd cn e clculted much more quickly. - Consider the Poisson eqution in two dimensions vlid in region ounded y known Dirichlet oundry conditions: 2 ψ= g where g=ρ/ϵ - Bring everything to the sme side: 2 ψ+g= - Multiply everything y the test function so tht we hve it there to work with: ϕ i 2 ψ+ϕ i g =
15 - Integrte this eqution over the entire surfce region S so tht we hve something tht looks like Green's first identity: S [ϕ i 2 ψ+ϕ i g]d= - Green's first identity for two-dimensions is: S (ϕ 2 ψ+ ϕ ψ)d = C ϕ ψ n d l - We re ssuming Dirichlet oundry conditions so tht the derivtive of the potentil long the oundry (Neumnn oundry conditions) cn e set to zero mking the right side of this eqution go wy. S (ϕ 2 ψ+ ϕ ψ)d = S ϕ 2 ψ d = S ϕ ψ d - Apply this to the Poisson eqution to otin: S [ ϕ i ψ+ϕ i g]d= S ϕ i ψ d= S ϕ i g d - Now expnd the electric potentil into sum of these expnsion functions weighted y unknown coefficients A j : ψ= j A j ϕ j - Note tht the index j represents the set of functions spnning oth x nd y dimensions. - Apply this to the Poisson eqution: S ϕ i [ j A j ϕ j ] d= S ϕ i g d j A j[ S ϕ i ϕ j d ] = S ϕ i g d - At this point, the fctor in squre rckets depends only on our choice of expnsion functions. - If we choose simple set of expnsion functions, we cn nlyticlly preclculte the fctors in the rcket nd then just tret them s numers computtionlly. - The key to this pproch is tht we choose expnsion functions tht re only non-zero in smll finite region, so tht the surfce integrl over these functions reduces to n integrl over smll region. If this region is smll enough, the source g is ssumed to e constnt cross this region nd cn come out of the integrl:
16 A j j[ S ϕ i ϕ j d ] =g i S ϕ i d - Here g i is the chrge density t the loction in spce where function ϕ i is non-zero. - The right-hnd side is just set of numers tht cn e clculte eforehnd so tht the prolem hs een reduced to liner lger prolem where A k re the unknowns. j K ij A j =G i where K ij = S ϕ i ϕ j d nd G i =g i S ϕ i d - In mtrix nottion this ecomes: K A=G - The solution is: A=K 1 G - The core computtionl tsk of solving the Poisson eqution hs een reduced to finding the inverse of mtrix, which computers cn do very efficiently. - Note tht the potentil is known long the oundries nd this must e included s well. - The Finite Element method is therefore summrized s follows: 1. The humn chooses set of loclized expnsion functions ϕ i 2. The humn nlyticlly clcultes ll K ij = S ϕ i ϕ j d nd S ϕ i d 3. The computer finds the inverse K The computer clcultes ll G i =g i S ϕ i d 5. The computer clcultes A=K 1 G 6. The computer clcultes the finl solution to the electric potentil ψ= j A j ϕ j
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