Lecture 6 Notes, Electromagnetic Theory I Dr. Christopher S. Baird University of Massachusetts Lowell

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1 Lecture 6 Notes, Eectrognetic Theory I Dr. Christopher S. Bird University of Msschusetts Lowe. Associted Legendre Poynois - We now return to soving the Lpce eqution in spheric coordintes when there is no ziuth syetry by soving the fu Legendre eqution for = nd : d dx [ x2 d P x dx ] [ 2] 2 x P x = where x=cos - Once this eqution is soved, the gener soution for the Lpce eqution in spheric coordintes wi hve the for: r,, = A r B r A B P = cos A r B r A e i B e i P cos, - Encourged by the Rodrigues' for of the soution to the = Legendre eqution, we try soution of the for P ( x)=( x 2 ) / 2 d g( x) nd substitute in: = x x 2 /2[ d /2 [ d 2 dx dx g x ] 2 d dx g x x2 x 2 /2 d x 2 dx ] g x 2 2 x x2 / 2 d g x dx [ ] x 2 /2 d dx g x 2 x 2 / 2 d dx g x dx g x x2 / 2 - Coect siir ters: = x [ 2 d d 2 dx dx ] g x d d 2 [ ]x g x [ ] 2 dx dx g x - We wnt to ove s uch s possibe inside the derivtives. We cn do this by using the product rue. - The product rue used ties sttes: d dx u v = k=! k! k! [ d k dx ][ u d k k dx ] v k - Let us try to ppy this to rerrnge the first ter in the differenti eqution.

2 - If u= x 2 then d u= 2 x dx, d 2 = 2 dx nd higher derivtives in the expnsion re zero so 2 tht: d dx x2 v = x 2 [ - Moving ters round: x 2 [ d dx v ] 2 x [ d dx v ] [ d 2 d dx v ] = d dx x2 v 2 x[ d - Now if we set v= d 2 x 2 [ d dx d 2 dx 2 g x ] = d dx 2 v ] dx v ] [ d 2 2 g (x) dx this product rue expnsion becoes: dx x2 d 2 dx 2 g x 2 x [ d dx 2 v ] dx g x ] [ d dx g x ] - We cn use this product rue to rerrnge the first ter of the Legendre eqution so it becoes: = d dx [ x2 d 2 dx ] g x 2 d d 2 x g x dx dx g x 2 [ d dx g x ] - Now we wnt to use the product rue expnsion on the second ter of the Legendre eqution. With u=x, du =, higher derivtives re zero nd v= d g( x) dx dx, the gener product rue becoes: d dx [ x d dx ] [ g x =x d dx ] [ g x d dx ] g x - After rerrnging: x[ d dx g x ] = d - Substitute in: dx [ x d dx g x ] [ d dx g x ] = d dx [ x2 d 2 dx [ g x 2 x d 2 dx ] ] g x g x = d dx [ d dx( ( x2 ) d dx g( x) ) + (+) g(x) ] - It is worth noting tht if this is true for =, then it wi utoticy be true for. So

3 we cn set = : = d dx x2 d dx g x g x - Now this is just the = Legendre eqution, which we hve redy soved. We found the soutions to be the ordinry Legendre poynois, P (x), so tht g x =P x. - We now hve the soution to the fu Legendre eqution: P x = x 2 /2 d dx P x - There is n rbitrry phse fctor tht we hve ssued to be one for sipicity, but is conventiony set to, so tht the ssocited Legendre functions becoe: P ( x)=( ) ( x 2 ) /2 d dx P (x) - Or written expicity using Rodrigues' eqution: P ( x)= ( ) 2! ( x2 ) /2 d + dx + ( x2 ) - It is worth noting upon exintion of the eqution = d dx [ d dx x2 d dx ] g x g x tht the highest power of x inside the brckets is, so tht cnnot be greter thn, otherwise the eqution woud be triviy stisfied nd thus there woud be no unique soution. In sury, both nd re integers nd the possibe re -, -(-),...,,..., (-),. - Sever usefu thetic retions invoving the ssocited Legendre functions cn be found (the derivtions re eft to the interested student). P x =!! P x P x = 2 x P x P x x 2 P x = x P x P x x 2 d dx P x = x P x P x - For fixed, the Legendre functions for n orthogon set:

4 P ' x P x dx= 2! 2! ' 2. Spheric Hronics - With the fu Legendre eqution now soved, the gener soution of the Lpce eqution in spheric coordintes hs been found: r,, = A r B r A B P = cos = = = A r B r A e i B e i P cos where P re the ssocited Legendre functions. - Typicy, ost probes require vid soution on the fu rnge of. The singe-vued requireent in this cse forces B =, nd the = ter cn now be incuded in the su with the rest of the ters: r,, = = = A r B r A e i B e i P cos - An ternte wy of presenting this is to et su fro - to nd thus cobine the A nd B ters with the other constnts: r,, = A, r B, r e i P cos = = - The norizing ters re incuded in the undeterined constnts A nd B. When boundry condition is ppied, these constnts becoe specified nd the norizing ters resut ntury. However, to ke the th cener, we cn expicity pu the norizing ters out of the constnts in dvnce. r,, = = = A, r B, r 2! 4! ei P cos - It is now evident tht P (cos θ) ties its noriztion ter fors copete set of orthonor functions in the vribe θ, indexed by for fixed, nd e i φ ties its noriztion ter fors copete set of orthonor functions in the vribe φ, indexed by. The product of both thus cretes orthonor functions tht spns nges of the unit circe. These re known s spheric hronics Y : Y (θ,ϕ)= 2+ 4 π ( )! (+)! ei ϕ P (cosθ) so tht

5 Φ(r,θ, ϕ)= = = ( A, r +B, r )Y (θ,ϕ) - The orthonority of the spheric hronics expicity ens tht: 2 * Y ' ' x, Y x, dx d = ' ' where x=cos or 2π π * Y ' ' (θ,ϕ)y (θ,ϕ)sin θ d θd ϕ=δ ' δ ' - The vidity of this eqution cn be checked triviy by expressing the spheric hronics in ters of its definition nd ppying the orthonority of the ssocited Legendre functions nd the copex exponentis. - Using the definitions of the spheric hronics, the ptient student cn work out the expicit nytic for for ny given nd. The owest-order spheric hronics re especiy sipe nd re typicy tbuted in textbooks. - Becuse the spheric hronics for n orthonor set of equtions, n rbitrry function f, cn be expnded in ters of spheric hronics: f (θ,ϕ)= = = 2π π A Y (θ,ϕ) where A = f (θ, ϕ)y * (θ,ϕ)sin θ d θ d ϕ - There re sever usefu speci cses for spheric hronics tht we shoud keep in ind. - If =, the spheric hronic does not depend on the ziuth nge nd the ssocited Legendre function reduces down to Legendre poynoi: Y, (θ,ϕ)= 2+ 4 π P (cosθ) - If =, the spheric hronic's dependence on por nge becoes especiy sipe, just the sine function to the th power, so there is ony one xiu in this direction: Y, (θ, ϕ)= ( ) (2 +)(2)! e i ϕ sin θ 2! 4π - The spheric hronics with negtive powers of re triviy reted to those with positive powers: Y, (θ,ϕ)=( ) * Y, (θ,ϕ) - Regions of interest incude the positive nd negtive z xis, where the spheric hronics sipify:

6 if Y (,ϕ)={ =} if 2+ 4π if nd Y (π, ϕ)={ 2 + 4π ( ) if =} 3. Boundry Vue Probes in Spheric Coordintes - Consider the probe where there is region of spce without ny chrges, bounded by sphere with rdius r centered t the origin on which there is potenti r=r,, =V,. We wnt to know the potenti inside the sphere. We use the gener soution to the Lpce eqution in spheric coordintes: Φ(r, θ, ϕ)= = = ( A r +B r )Y (θ, ϕ) - The region of interest incudes the origin, thus B = to keep the soution fro bowing up there: Φ(r,θ,ϕ)= = = A r Y (θ, ϕ) - Now ppy the boundry condition: V (θ,ϕ)= = = A r Y (θ,ϕ) - This is n expnsion in spheric hronics, nd the constnts re thus: 2 π π A r = V (θ,ϕ)y * (θ,ϕ)sin θd θd ϕ - The soution is then: Φ(r,θ,ϕ)= = = 2π π A r Y (θ, ϕ) where A =r V (θ,ϕ)y * (θ,ϕ)sin θ d θ d ϕ - Or written in ore intuitive wy by redefining the constnt: Φ(r,θ, ϕ)= = = A ( r r ) 2π π Y (θ,ϕ) where A = V (θ,ϕ)y * (θ, ϕ)sin θd θd ϕ

7 4. The Addition Theore for Spheric Hronics - Consider two coordinte vectors (r, θ, φ) nd (r', θ', φ') which hve the nge γ between the. - The ordinry Legendre poynoi s function of the nge between the, P cos cn be expnded in spheric hronics of the coordinte vectors: P cos = 4 Y * 2 ', ' Y, = - This is known s the ddition theore. - Previousy, we derived the potenti t r in spheric coordintes tht resuts fro unit chrge t soe point r on the z-xis s: r,, = r r = r = r P cos if r r nd r r r = = r P cos if r r - The equtions sti hod when the point r is off the z-xis if the Legendre poynois becoe functions of the nge γ between the observtion point r nd the source point r. r r = r = r P cos if r r nd r r r = = r P cos if r r - This eqution is not terriby usefu becuse the nge γ depends on the spheric coordintes. - We cn use the ddition theore to expnd the ordinry Legendre poynoi so tht the expression is function of spheric coordintes rther thn function of the nge between coordinte vectors: r r =4π = r r =4 π = = = 2+ r r Y * + (θ',ϕ ')Y (θ,ϕ) if r<r nd r r Y * (θ', ϕ ')Y (θ, ϕ) if r>r - This wi be usefu if we wnt to know the potenti produced by utipe point chrges in expicit coordintes. 5. The Lpce Eqution in Cyindric Coordintes - Erier, we soved the Lpce eqution in cyindric coordintes for the speci cse of when the boundry conditions re unifor in the z-diension, nd the probe reduced to por coordintes. - We now revisit the probe for when the boundry conditions re not unifor in the z- diension. - The Lpce eqution in cyindric coordintes is: 2 = z 2 =

8 - Use the ethod of seprtion of vribes by trying soution of the for: r,, z =R Q Z z nd substituting it in: Q Z z R R Z z 2 Q R Q 2 Z z = 2 2 z 2 - Divide by R Q Z z : R R Q 2 Q 2 Z z = 2 2 Z z z 2 - The st ter depends ony on z nd the other ters do not so to be vid for z, they ust be reted by constnt: R R Q - Mutipy the eft eqution by 2 : R 2 Q k 2 = where k 2 = 2 Z z 2 2 Z z z 2 R 2 Q k 2 2 = where k 2 = 2 Z z Q 2 Z z z 2 - The second ter depends now ony on φ nd the rest of the ters do not, so we set it to constnt. A the prti derivtives becoe regur derivtives s the functions re of one vribe ony now: R d d d R d 2 k 2 2 = where 2 = d 2 Q, k 2 = d 2 Z z Q d 2 Z z d z 2 - We put ech eqution in ore intuitive for: d d d R d k 2 2 d 2 R = where 2 Q d 2 = 2 Q, d 2 Z z =k 2 Z z d z 2 - In order to get the ost gener soution, we ust cover four cses: - Cse : If = nd k= then the soutions re: R = A, B, n, Q =C, D, nd Z z = F, G, z - The copete prticur soution is the product of these prts.

9 - Cse 2: If nd k= then the soutions re: R = A, B,, Q =C, e i D, e i nd Z z = F, G, z - The copete prticur soution is the product of these prts. - Cse 3: If = nd k, soutions to the st two equtions re: Q =C,k D, k, Z z = F, k e k z G,k e k z The differenti eqution invoving R hs the se soution s the next cse nd wi be hnded with it. - Cse 4: If nd k, the soutions to the st two equtions re: Q =C, k e i D,k e i Z z = F, k e k z G, k e k z - The differenti eqution invoving R ust be soved by trying series soution. Once found, the gener soution for this cse wi hve the for:,, z = k R, k C, k e i D, k e i F, k e k z G, k e k z - First sipify the eqution by king the substitution: x=k d d R x x d x x d x 2 x 2 R x = - Try soution of the for R x = j x j nd substitute it in: j = j= j x j 2 j 2 2 j x j = j= - Reove the first two ters of the su on the eft nd cobine the reining sus: x x 2 2 x j j 2 j j = j= - Every power of x is independent, thus the coefficient of every power ust vnish: 2 2 =, 2 2 =, j 2 = j j - Fro these it is found tht =±, odd = nd j 2 = j 2 j 2±2 j

10 - Becuse odd powers re zero, we cn iterte over integers nd rewrite the indices: 2 j = 4 j j± 2 j for j=,2,3,... - In sury, one soution to the R eqution is the Besse J function: R, k = J k where J x = 2 j x 2 j, 2 j = 4 j j 2 j nd by convention, = 2 in ters of the G function. j= - One ight think tht the other soution to the R eqution is J -ν, but it turns out tht this is not inery dependent fro J ν when ν is n integer nd we ust find nother soution. - Trditiony, the Neunn function, or Besse function of the second kind is tken s the other inery independent soution. It is defined by: N x = J x cos J x sin x - Usefu properties, incuding the recurrence retions, cn be esiy derived or found in textbooks. - The gener soution for R, in ters of the Besse functions, becoes: R, k = A, k J k B,k N k - The gener soution to the Lpce eqution in cyindric coordintes for cse 4 becoes:,, z = k A, k J k B, k N k C, k e i D, k e i F, k e k z G, k e k z - The ost gener soution to the Lpce eqution in cyindric coordintes is the su of the soutions of the possibe cses: Φ(ρ,ϕ, z)=( A, +B, nρ)(c, + D, ϕ)( F, +G, z) + (A ν, ρ ν + B ν, ρ ν )(C ν, e i ν ϕ +D ν, e i νϕ )(F ν, +G ν, z) ν (A, k J (k ρ)+b, k N (k ρ))(c,k +D, k ϕ)(f,k e k z +G,k e k z ) + k + ν ( A ν, k J ν (k ρ)+b ν,k N ν (k ρ))(c ν,k e i νϕ + D ν, k e i νϕ )(F ν,k e k z +G ν, k e k z ) k - To be usefu, we ust find orthogon sets of functions invoving the Besse functions so tht we cn use the orthogonity condition to invert equtions invoving the boundry conditions. The functions:

11 J x n cn be shown to be orthogon on the interv (, ) for fixed ν nd n =, 2, 3,... where x νn re the roots tht stisfy J x n =. - The orthogonity condition for these functions cn be shown to be: J x n' J x 2 n d = 2 [ J x n ]2 n' n - This cn be used to expnd n rbitrry function in Besse series: f = j= A n J x n where A = 2 n 2 2 J x n f J x n 6. Boundry Vue Probes in Cyindric Coordintes - Consider cyinder with rdius nd height L with its botto centered t the origin. The potenti is everywhere zero on the surfce of the cyinder, except the top where it is potenti V,. This eds to the boundry conditions: =,, z =,,, z= =, nd,, z= L =V, - The gener soution to the Lpce eqution in cyindric coordintes is:,, z = A, B, n C, D, F, G, z A, B, C, e i D, e i F, G, z A, k J k B,k N k C,k D, k F,k e k z G, k e k z k A, k J k B,k N k C,k e i D,k e i F, k e k z G, k e k z k - The region where we need vid soution incudes the origin, so sever of the coefficients ust be zero to keep the soution fro bowing up. This eds to:,, z = C, D, F, G, z C, e i D, e i F, G, z J k C, k D,k F,k e k z G, k e k z k J k C,k e i D,k e i F, k e k z G,k e k z k - The region where we need vid soution spns the fu circe of nges for φ, so we require the function to be singe-vued:,, z =, 2, z

12 - This equity ust hod for vues of of the independent vribes, thus ech ter ust tch. This requireent eds to: D, =, D,k =, nd = where =,,2,3,...,, z = F, G, z C, e i D, e i F, G, z J k F,k e k z G, k e k z k k J k C,k e i D,k e i F, k e k z G,k e k z - Appy the boundry condition:,, z= = =F, C, e i D, e i F, k k J k F,k G, k J k C,k e i D, k e i F, k G, k - This eds to the requireents tht F, =, F, =, G,k = F, k, nd G, k = F,k,, z =G, z C, e i D, e i z k k J k F, k sinh k z J k C,k e i D,k e i sinh k z - A the = ters now hve for for which they cn be cobined with the ters:,, z = C, e i D, e i z J k C, k e i D, k e i sinh k z k - The first ter on the right is just speci cse of the st ter when k = (cose to zero, J hs n syptotic for tht is proportion to ρ ). They cn be cobined:,, z = J k C, k e i D, k e i sinh k z k - Appy the boundry condition: =,, z = : = J k C, k e i D,k e i sinh k z k - This cn ony hod true for vues of the independent vribe nd z if the coefficients re zero:

13 =J k for nd k - There is ony discrete set of soutions to this equtions - its roots. The roots cnnot be found nyticy. When done nuericy, the roots re designted s x for n=, 2, 3... so tht k= x - The soution now becoes:,, z = = n= J x C,n ei D,n e sinh i x z - The fin boundry condition is,, z= L =V, now ppied: V, = = n= J x C, n ei D, n e sinh i x L - This is Fourier series in nd Fourier-Besse series in ρ. We find the coefficients in the usu wy. Mutipy both sides by n exponenti nd integrte: 2 V, e i ' d = = n= J x 2 C, n e i ' D,n e i ' d sinh x L - Use the orthogonity condition e i ' x dx=2,' nd rebe: 2 V, e i d =2 J n= x C sinh,n x L - Mutipy both side by J x ' J x ' 2 nd integrte: V, e i d d =2 - Use the orthogonity condition: J x ' 2 n= J x n' J x ' J x d C sinh, n x L J x n 2 d = 2 [ J x n ]2 n' n V, e i d d =2 2 n= 2 [ J x ]2 n' n C, n sinh x L J x 2 V, e i d d = 2 [ J x n ] 2 C,n sinh x L

14 - Sove for the coefficient: C,n csch x L 2 = 2 [ J x ] 2 d d V, J x e i - In the exct se process the other set of coefficients is found to be: D, n csch x L 2 = d d V, J 2 [ J x n ] 2 x ei * - It is pprent now tht D, n =C,n - In sury, the fin soution is: Φ(ρ,ϕ, z)= where = J ( x ρ n= csch C,n = ( L x ) ) (C, n ei ϕ +C, n 2π π 2 [ J + ( x )] 2 d ϕ * e i ϕ )sinh ( x z ) d ρ ρv (ρ, ϕ) J ( x ρ ) e iϕ