Green s Function for Potential Field Extrapolation

Transcription

1 Green s Funtion for Potential Field Extrapolation. Soe Preliinaries on the Potential Magneti Field By definition, a potential agneti field is one for whih the eletri urrent density vanishes. That is, J 4 B. ( In addition, the agneti field ust also satisfy Maxwell s equations, in partiular Gauss s law: B. ( Any funtion whih satisfies these two onstraints is a valid potential field. It is often onvenient to write the agneti field in ters of a salar potential, Φ, in the following way: B Φ. (3 This autoatially satisfies the ondition that the urrent density vanishes, sine any salar has the property that the url of the gradient vanishes. Thus one only has to tae into aount the ondition on the divergene, whih an be written B Φ. (4 This is Laplae s equation for the salar potential, whih has been solved (in ters of various speial funtions in soething lie oordinate systes. For this exerise, we will be using only standard Cartesian oordinates. To define a unique solution in a volue, the noral oponent of the agneti field ay be speified on the losed surfae bounding the volue. In soe ases, this volue will be taen as sei-infinite, for exaple, the half-spae above a plane. In this ase, the ondition that the noral oponent of the field be speified at infinity is satisfying by requiring that the agnitude of the field fall off rapidly enough with distane above the plane.

2 . The Green s Funtion For a potential field, the Green s funtion has a reasonably siple for: G x x x r 3, (5 G y y y, (6 r 3 G z z r3, (7 where r (x x +(y y +z, and the boundary is assued to be at z. The field at any point is then onstruted fro B i (x dx dy G i (x,x B z (x,y,. (8 3. Appliation to Disrete Data Now oes the fun part: applying this to B easured at disrete points. It s not as siple as one ight thin, beause one does not want the field due to a series of point soures at eah plae there is a easureent. Instead, require that the vertial field be onstant aross eah pixel. That is, let B z (x,y, ( Θ x x i x ( Θ y y j y ( Θ ( Θ x x i + x y y j + y Use this expression for the field in the integral with the Green s funtion. B l (x,y,z dx dy B z (x,y, G l (x,y,z,x,y dx dy ( Θ x x i x ( Θ ( Θ y y j y ( Θ y y j + y xi + x/ yj + y/ B ij dx dy G l (x,y,z,x,y x i x/ x xi x/ B ij d( x x x i + x/ y j y/ y yj y/ y y j + y/ B ij (9 x x i + x B ij G l (x,y,z,x,y d( ỹg l (x,y,z,x,y

3 3 x xi + x/ B ij d x x x i x/ y yj + y/ y y j y/ dỹ G l (x,y,z,x,y ( with r (x x + (y y + z, tan θ (y y /(x x, and x x x, ỹ y y. Start with the z-oponent, and try to do the integrals expliitly. B z (x,y,z z z x xi + x/ B ij d x x x i x/ y yj + y/ y y j y/ z dỹ ( x + ỹ + z 3/ x xi + x/ ỹ y yj+ y/ B ij d x ( x + z ( x + ỹ + z / y y j y/ x x i x/ x xi + x/ B ij x x i x/ d x ỹ j + y/ ( x + z x + (ỹ j + y/ + z / ỹ j y/ ( x + z x + (ỹ j y/ + z / z { ỹj + y/ x(ỹ j + y/ B ij z(ỹ j + y/ tan z x x xi+ x/ + (ỹ j + y/ + z x x i x/ ỹj y/ x(ỹ j y/ x xi+ x/} z(ỹ j y/ tan z x + (ỹ j y/ + z / x x i x/ B { ij tan ( x i + x/(ỹ j + y/ z ( x i + x/ + (ỹ j + y/ + z tan ( x i x/(ỹ j + y/ z ( x i x/ + (ỹ j + y/ + z + tan ( x i x/(ỹ j y/ z ( x i x/ + (ỹ j y/ + z tan ( x i + x/(ỹ j y/ } z ( ( x i + x/ + (ỹ j y/ + z where x i x x i, ỹ j y y j. This is rather unwieldy, but at least it s an analyti expression that does not involve integrals. Next, try the x-oponent. B x (x,y,z x xi + x/ B ij d x x x i x/ y yj + y/ y y j y/ x dỹ ( x + ỹ + z 3/ y yi + y/ x xj+ x/ B ij dỹ (ỹ + x + z / x x j x/ y y i y/ y yi + y/ B ij dỹ y y i y/ (ỹ + ( x i x/ + z /

4 4 (ỹ + ( x i + x/ + z / { B ij ln ỹ + (ỹ + ( x i x/ + z / y y i + y/ y y i y/ ln ỹ + (ỹ + ( x i + x/ + z / y y i + y/} y y i y/ B ỹ ij + (ỹ ln + ( x i x/ + z / y yi+ y/ ỹ + (ỹ + ( x i + x/ + z / y y i y/ B { ij (ỹj + y/ + (ỹ j + y/ + ( x i x/ + z / ln (ỹ j + y/ + (ỹ j + y/ + ( x i + x/ + z / (ỹj y/ + (ỹ j y/ + ( x i x/ + z / } ln (ỹ j y/ + (ỹ j y/ + ( x i + x/ + z / and this expression will also hold for the y-oponent by appropriately interhanging x and y. ( 4. Solution for a Box Consider now the ase in whih one wishes to deterine the field in the volue < x < a, < y < b, < z <, given the noral oponent of the field on all six of the faes of the box. Following the disussion in Jason (975, onstrut six separate salar potentials, eah of whih has a non-vanishing noral derivative on only one of the walls. Then the salar potential for the solution will be siply the su of the six. Try to onstrut this in suh a way that FFTs an be used, by taing the exponential solutions to be in the diretion noral to the fae on whih the derivative of the potential does not vanish. So, let Φ (x A n (x/a (ny/b h,n a + n b (z whih by onstrution satisfies Φ, and hene B ( Φ. With this definition, Φ x,a a A n sin(x/a (ny/b h a + n b (z,a,n (4 Φ y,b,n n b A n (x/a sin(ny/b h a + n b (z,b (3

5 5 (5 Φ z a + n b A n (x/a (ny/b sinh a + n b (z,n (6 Φ z a + n b A n (x/a (ny/b sinh a + n b (z,n B z (x,y,,n a + n b A n (x/a (ny/b sinh Try to siplify at least the notation soewhat by starting with Φ 3 (x,n a + n / ( x b a { C n + h ( ny b a + n b z + C n h a + n b } a + n b (z thus Φ 3 x,a a a + n / ( x ( ny sin b a b x,a,n { C n + h a + n b z + C } n h a + n b (z (9 Φ 3 n y,b b a + n / ( x ( ny sin b a b y,b,n { C n + h a + n b z + C } n h a + n b (z ( Φ 3 ( x ( ny z, a b,n { C n + sinh a + n b z + C } n sinh a + n b (z B z (x,y, ( x ( ny Cn sinh a b a + n b,n Cn 4 a b ( x ( ny dx dy B z (x,y, ( ab sinh + n a b a b (7 (8

6 B z (x,y, ( x a,n C + n Siilarly, let where and where Φ (x A n A + n Φ (x B n B + n ab sinh,n b sinh 4 ( ny + n a b 6 b a C n + sinh a + n b dx b + n / ( y b { A + n h 4 b sinh,n a sinh + n a b 4 b ( x ( ny dy B z (x,y, ( a b ( nz b + n x + A n h b + n a b dy b dy ( y dz B x (,y,z b a + n / ( x a { B n + h 4 a sinh + n b a 4 + n b a ( y dz B x (a,y,z b ( nz a + n y + B n h a dx a dx ( x dz B y (x,,z a ( x dz B y (x,b,z a } b + n (x a ( nz ( nz } a + n (y b and the vetor potential for the oplete solution is siply Φ Φ + Φ + Φ 3. ( nz ( nz (3 (4 (5 (6 (7 (8 4.. Appliation to Disrete Data The above forulation provides a general solution, given the noral oponent of the field on all six faes of a box. Now onsider the ase in whih the noral field is given at

7 7 disrete (regularly spaed points on eah of the faes, with the goal of onverting as uh as possible of this into Fourier Transfors. There are two ways we an proeed: turn the sine and ine expansions into disrete Fourier transfors by redefining the boundary over a larger area, with the appropriate syetry, or tae the real part of eah Fourier transfor as it is perfored. Start with the latter, sine it will use transfors of saller array sizes, whih will be faster. Note that for the disrete ase, the integrals will be represented as sus, in the following for Lj dx j f(x L j f(x j + f(x j N j L j + f(x j x j N j (9 where x j L j /(N j. Note that this iplies a partiular hoie for the walls of the box, naely that the wall falls on the outerost grid point in eah diension. This assuption is different fro the standard periodi boundary onditions used in fff.pro, for exaple, and also different fro the walls in Yuhong s siulation, in whih the wall is idway between the last grid point and a ghost grid point outside the wall. So, the fator of a half in the first and last grid points represents the fat that only half the pixel is ontained within the walls. To begin with, hange notation one again, this tie using (,, 3 in plae of (x,y,z. With this notation, assue that the volue of interest is x j L j, with j,, 3, and let hene Φ l (x N j N n L j { A l+ n h + n L / ( x j L j L j ( nx L + n x l + A l L n h L j } + n (x l L L l (3 B l± N j N n ( x j { A l+ n sinh N j N n Lj L j A l± n sinh dx j B l± L j ( nx L L l ( rx j L j + n x l + A l L n sinh L j L j ( + n x j ( nx L L j L } + n (x l L L l xl,ll

8 8 A l± n N j N n Lj L j A l± n sinh dx j L N ( + δ r Lj n dx j L L l L j + n L ( rx dx B l± j L j A l± rn sinh L l L j + n L ( rx dx B l± j L j L jl 4 ( + δ r( + δ s A l± rs sinh ( δ ( δ n L j L sinh L l L j + n L Lj L l L j ( nx Lj L ( sx L L ( sx L + n L ( x dx j j ( rx j L j L j ( nx dx ( sx L L L dx j dx B l± ( x j ( nx L j L (3 Note that this expression is not really well defined for n (as is the expression for Φ, but sine this is the Φ onstant ter, and so does not ontribute to the agneti field, I not going to treat it properly. It ay need to be handled differently when oded, however. Also, I a assuing flux balane, so there is no ter of the for Φ B x Using Real Parts This is not as straightforward as I thought, beause one ends up with a issing fator of in the ine ters. Nuerial Reipes has a way (atually, several ways to deal with this, by onstruting fast ine transfors, but for now, proeed with the slow ethod of doubling the diensions. Thus, this setion is not oplete Expanding the Boundary In order to ae use of the standard FFT, and in fat, with the further goal of perforing this with repeated alls to fff.pro, whih assues only the lower boundary is given, and that the boundary onditions are periodi with period L, define a syetrized version of B l in the volue x < L, x < L, x 3 < L 3, and assue that B l vanishes outside of this region. Do this by letting B 3 (L x,x, B 3 (x,x,, (3

9 9 B 3 (L x,x,l 3 B 3 (x,x,l 3, (33 B 3 (x, L x, B 3 (x,x,, (34 B 3 (x, L x,l 3 B 3 (x,x,l 3, (35 and siilarly for B,B. Note that this definition is a little different fro the standard periodi boundary onditions. With this definition, and aing use of the vanishing of B 3 outside the area of interest, evaluate L FFT (B 3 FFT (B 3 L L L L L ( dx B 3 (x,x, exp ix L dx B 3 (x,x, ( x L i sin ( x L ( x + dx B 3 (x,x, i sin L L ( x ( L d B 3 (x,x, exp ix L L L ( x dx B 3 (x,x ( x, i sin L L L + d x B 3 (L x,x, ( x L ( x dx B 3 (x,x ( x, i sin L L + dx B 3 (x,x, L dx B 3 (x,x, L ( x + i sin L ( x L ( x L ( x L i sin ( x L whih has been alulated without aing the hange to disrete data, but presuably ust also hold, provided the FFT is defined properly. A l± n Now go ba to the expressions for the oeffiients in the ine expansion, ( δ ( δ n L j L sinh L l L j ( δ ( δ n sinh L l L j + n L + n L Lj dx j L ( x dx B l (x l j ( nx,l l L j L FFT j (B l (x l,l l (37 where the FFT is taen in the forward diretion to orrespond to IDL s noralization. (36

10 In the disrete version, if B 3 n is given for,n, n,n, then the expanded version of the field is given expliitly by B 3 (N,n B 3 n, (38 B 3,(N n B 3 n, (39 Beause of the expanded diensions for B l, the oeffiients are only eaningful for < N j, n < N. That is, the A l± n are not all independent. In partiular, A l± n A l± (N j,n A l±,(n n. Having deterined the oeffiients in ters of FFTs, next tale the series for the salar potential itself. To do this, it will be onvenient to define a new set of oeffiients, given by αn l + n / { A l+ L j L n h + n x l + A l L j L n h } + n (x l L L j L l (4 for < N j,n < N, and let Now evaluate α l (N,n α l n, (4 α l,(n n α l n. (4 FFT j (αl 4(N j (N N j N j 3 4(N j (N N j 3 4(N j (N N n exp +α l + α l,n (q + N j 3 4(N j (N N + ñ α l,(n ñ } +α l + α,n l (q ip αn l exp exp (N j ip N 3 N j ip exp N j N 3 nn α l n ip exp N j ( q n { N n α l n α l n inq (N ( nq ( nq + i sin N N ( nq ( nq + i sin N N ( nq ( nq } + i sin N N { N ñq N n α l n + i sin ( nq ( nq + i sin N N ( q ñq N

11 N j 3 ip exp 4(N j (N N j } +α l + α,n l (q N j 3 ip exp 4(N j (N N j N 4(N j (N { N j 3 n ( p αn l N j N 4(N j (N { N j n ( p αn l N j { N n N n ( δ n δ nn ( p } + i sin N j ( δ n δ nn N j 3 ( p ( p } + αn l + i sin N j N j N j N 4(N j (N { N j N j + n ( p αn l N j ( α(n l j,n p N 4(N j (N { N j n ( p αn l N j 4(N j (N N j ( nq αn l N ( nq ( δ n δ nn αn l N ( nq N ( nq N ( p + i sin + α,n l + αn l N j j,n (p ( δ n δ nn ( nq N ( p + i sin + α,n l + αn l N j j,n (p p ( + i sin p N j ( δ n δ nn N n ( nq N } + α,n l + αn l j,n (p ( p ( nq N j N p } N j α l n( δ n δ nn ( δ δ Nj (43 whih (basially gives the expression for the salar potential. Slightly redefine α in order to

12 get the various oponents of B. So, for exaple, to get the oponent of the field noral to the wall, let { αn ll A l+ n sinh L j for < N j,n < N, in whih ase + n x l + A l L n sinh L j } + n (x l L L l ( + δ + δ Nj ( + δ n + δ nn (44 B l FFT j (αll. (45 Note that this is only the ontribution to one oponent fro one wall, so still have to su over the other five walls, as well as alulate the other oponents. For a oponent perpendiular to the wall under onsideration, let α lj n L j L j + n / { A l+ L n h L j + n x l + A l L n h L j } + n (x l L L l ( + δ + δ Nj ( + δ n + δ nn (46 for < N j,n < N, and let α l (N,n α l n, (47 α l,(n n α l n. (48 in order to get a sin instead of a ter. Note that now the expression for the field will be B j ifft j (αjl. ( Flux Balane The previous analysis requires that at least the flux through any pair of parallel walls vanish. Consider adding the following salar potential, suggested by Dana Longope, to the solution, to allow flux to exit through any obination of walls. with orresponding field oponents Φ A (x y + A (x z + A 3 (y z (5 B x (A + A x B y (A A 3 y B z (A + A 3 z (5

13 3 Che that the diverene of this vanishes: The flux through the walls is given by B (A + A + (A A 3 + (A + A 3. (5 Ly Lz Φ x dy dz B x x,lx Ly Lz (A + A dy dz x x,lx (A + A L y L z x x,lx (53 Lx Lz Φ y dx dz B y y,ly Lx Lz (A A 3 dx dz y y,ly (A A 3 L x L z y y,ly (54 Lx Ly Φ z dx dy B z z,lz Lx Ly (A + A 3 dx dy z z,lz (A + A 3 L x L y z z,lz (55 In order to balane the flux, don t need all of these ters, so deal with the flux in both the x and y diretions by adding/subtrating a orresponding aount to the flux through the top boundary. Thus, eep only the ters involving A and A 3, thus Φ x A L x L y L z Φ y A 3 L x L y L z (56 This aterial is based upon wor supported by the National Siene Foundation under Grants No and 597. REFERENCES Jason, J. D. 975, Classial Eletrodynais, nd edn. (New Yor: John Wiley & Sons This preprint was prepared with the AAS L A TEX aros v5..