Figure 1. We will begin by deriving a very general expression before returning to Equations 1 and 2 to determine the specifics.

Transcription

1 Deivation of the Laplacian in Spheical Coodinates fom Fist Pinciples. Fist, let me state that the inspiation to do this came fom David Giffiths Intodction to Electodynamics textbook Chapte 1, Section 4. I have also boowed Fige 1 and the fist two sets of eqations fom that textbook fo efeence and to give s a stating place to bild off of. These ae what I am calling the Fist Pinciples: Fige 1 x sin(θ) cos(φ) y sin(θ) sin(φ) z cos(θ) Eq. 1 x 2 + y 2 + z 2 θ cos 1 z ( x 2 + y 2 + z 2) φ tan 1 ( y x ) Eq. 2 We will begin by deiving a vey geneal expession befoe etning to Eqations 1 and 2 to detemine the specifics. Fist, deive the Laplacian in Catesian Coodinates: Now, sbstitte: ( T x,y,z ) 2 T (x,y,z) 2 T x T y T z 2 Eq. 3 T (x,y,z) G (,θ,φ) Eq. 4 And wok ot each of the Catesian 2 nd Deivatives in the new vaiables. We shall do this one at a time. This can get athe complicated so I will ty and do things piecemeal so the pocess is obvios. I will begin with the x component fist. 1

2 The spheical vaiables of G, which ae, θ, and φ, ae all eqations of x, y, and z, so we mst se the chain le. So: x (G (,θ,φ)) G x + G θ θ x + G φ φ x Eq. 5 2 G x 2 x (G x + G θ θ x + G φ φ x ) Eq. 6 Now distibte: 2 G x 2 x (G x ) + x (G θ θ x ) + x (G φ φ x ) Eq. 7 I will tackle these thee components one at a time, stating with the vaiable: x (G x ) x (G x ) Eq. 8 Hee I se G with a sbscipt to denote the patial deivative of G with espect to. It is a bit of a mixing of notation, bt I find it is easie to see. We will need to se the Podct Rle to wok Eqation 8 ot: x (G x ) G 2 x 2 + x (G ) x Eq. 9 If yo note, the last pat of Eqation 9 contains the patial deivative of G with espect to taken as a patial deivative with espect to x. The oiginal G eqation was: G (,θ,φ) Theefoe, its deivative with espect to mst also be an eqation of the same vaiables, so sing the chain le once again on this last pat: (G ) x (G ) x + (G ) θ θ x + (G ) φ φ x Eq. 10 So Eqation 9 becomes (Wheneve I make an inset sbstittion I will ty to se a colo indicato): x (G x ) G 2 x 2 + x ((G ) x + (G ) θ θ x + (G ) φ This whole expession can be now be sbstitted back into Eqation 7: φ ) Eq. 11 x 2

3 2 G x 2 G 2 x 2 + x ((G ) + x (G θ θ x ) + x (G x + (G ) θ θ x + (G ) φ φ x ) φ φ ) Eq. 12 x We now need to find the expansion of the two emaining pieces in Eqation 12. So fa we have fond 1/6 of the total geneal expession and the magnitde of this convesion begins to become clea. Nevetheless, I shall cay on. Contining my patial deivative shot hand, the θ potion becomes: x (G θ θ x ) x (G θ θ x ) G θ (θ x) x + θ x (G θ) x Eq. 13 And conseqently, as with Eqation 10, the last bit becomes: (G θ ) x (G θ) x + (G θ) θ θ x + (G θ) φ φ x Eq. 14 Ptting Eqation 14 into 13, ptting both (pple) into Eqation 12, and also conveting as mch as possible to my shot hand notation, we now have: Now to finish off with the φ potion: 2 G x 2 (G xx + x (G x + G θ θ x + G φ φ x )) + (G θ θ xx + θ x (G θ x + G θθ θ x + G θφ φ x )) + x (G φ φ ) Eq. 15 x x (G φ φ x ) x (G φ φ x ) G φ φ xx + φ x (G φ) x Eq. 16 As befoe, we need to wok ot that final tem: (G φ ) x (G φ) x + (G φ) θ θ x + (G φ) φ φ x Eq. 17 Ptting Eqations 16 and 17 into Eqation 15, we now have: 3

4 2 G x 2 (G xx + x (G x + G θ θ x + G φ φ x )) + (G θ θ xx + θ x (G θ x + G θθ θ x + G θφ φ x )) + (G φ φ xx + φ x (G φ x + G φθ θ x + G φφ φ x )) Eq. 18 This is 1/3 of the final Geneal Eqation, o soltion to conveting Eqation 3. Aligning the components in the fashion above allows fo some inteesting pattens to emege. In fact we can se these pattens to extapolate the othe two thids of the final eqation by simply being cleve and eplacing vaiables accodingly. This is how I am going to poceed. Howeve, the eade may choose to wok the est of the Geneal Eqation ot in long hand as poof fo themselves. Dealing with the y component of Eqation 3 we have: 2 G y 2 (G yy + y (G y + G θ θ y + G φ φ y )) + (G θ θ yy + θ y (G θ y + G θθ θ y + G θφ φ y )) + (G φ φ yy + φ y (G φ y + G φθ θ y + G φφ φ y )) Eq. 19 And the z component of Eqation 3 becomes: 2 G z 2 (G zz + z (G z + G θ θ z + G φ φ z )) + (G θ θ zz + θ z (G θ z + G θθ θ z + G θφ φ z )) + (G φ φ zz + φ z (G φ z + G φθ θ z + G φφ φ z )) Eq. 20 So finally, we can now wite Eqation 3 in the fom (This is anew new colo coding scheme): 2 G (,θ,φ) 2 G x G y G z 2 (G xx + x (G x + G θ θ x + G φ φ x )) + Eq. 21 (G θ θ xx + θ x (G θ x + G θθ θ x + G θφ φ x )) + (G φ φ xx + φ x (G φ x + G φθ θ x + G φφ φ x )) + (G yy + y (G y + G θ θ y + G φ φ y )) + (G θ θ yy + θ y (G θ y + G θθ θ y + G θφ φ y )) + (G φ φ yy + φ y (G φ y + G φθ θ y + G φφ φ y )) + (G zz + z (G z + G θ θ z + G φ φ z )) + (G θ θ zz + θ z (G θ z + G θθ θ z + G θφ φ z )) + (G φ φ zz + φ z (G φ z + G φθ θ z + G φφ φ z )) Obviosly, at this point some eaangement is called fo. I will take two steps to clean things p. Fist, I will distibte whee I can (Eqation 22), and then I will gop togethe all the patial deivatives of G that ae alike (Eqation 23). Bt fist ecall that when woking with patial deivatives: F ij F ji o j (F i ) i (F j ) 4

5 So: 2 G (,θ,φ) 2 G x G y G z 2 (G xx + G ( x ) 2 + G θ x θ x + G φ x φ x ) + Eq. 22 (G θ θ xx + G θ θ x x + G θθ (θ x ) 2 + G θφ θ x φ x ) + (G φ φ xx + G φ φ x x + G φθ φ x θ x + G φφ (φ x ) 2 ) + (G yy + G ( y ) 2 + G θ y θ y + G φ y φ y ) + (G θ θ yy + G θ θ y y + G θθ (θ y ) 2 + G θφ θ y φ y ) + (G φ φ yy + G φ φ y y + G φθ φ y θ y + G φφ (φ y ) 2 ) + (G zz + G ( z ) 2 + G θ z θ z + G φ z φ z ) + (G θ θ zz + G θ θ z z + G θθ (θ z ) 2 + G θφ θ z φ z ) + (G φ φ zz + G φ φ z z + G φθ φ z θ z + G φφ (φ z ) 2 ) And then: 2 G (,θ,φ) G xx + G yy + G zz + G ( x ) 2 + G ( y ) 2 + G ( z ) 2 + G θ x θ x + G θ y θ y + G θ z θ z + G θ θ x x + G θ θ y y + G θ θ z z + G φ x φ x + G φ y φ y + G φ z φ z + G φ φ x x + G φ φ y y + G φ φ z z + G θ θ xx + G θ θ yy + G θ θ zz + G θθ (θ x ) 2 + G θθ (θ y ) 2 + G θθ (θ z ) 2 + G θφ θ x φ x + G θφ θ y φ y + G θφ θ z φ z + G φθ φ x θ x + G φθ φ y θ y + G φθ φ z θ z + G φ φ xx + G φ φ yy + G φ φ zz G φφ (φ x ) 2 + G φφ (φ y ) 2 + G φφ (φ z ) 2 Eq. 23 A point to note fom Eqation 2, φ is only a fnction of x and y. We can se this fact to eliminate a few of the tems, these being φ deivatives with espect to z, in Eq. 23. The next logical step afte that will be to facto ot the patial deivatives fom the emaining like tems. I have colo coded the like tems in Eqation 24 fo easy viewing: 2 G (,θ,φ) G ( xx + yy + zz ) + G (( x ) 2 + ( y ) 2 + ( z ) 2 ) + 2G θ ( x θ x + y θ y + z θ z ) 2G φ ( x φ x + y φ y ) + G θ (θ xx + θ yy + θ zz ) + G θθ ((θ x ) 2 + (θ y ) 2 + (θ z ) 2 ) + 2G θφ (θ x φ x + θ y φ y ) + G φ (φ xx + φ yy ) + G φφ ((φ x ) 2 + (φ y ) 2 ) Eq. 24 Yo will notice that we have edced the eqation to a somewhat moe manageable state than that of Eqation 21. Eqation 24 is essentially the final fom of the Geneal Laplacian in Spheical Coodinates. Howeve, it s still a lage, ngainly beast and needs to be fthe simplified if we eve hope to se it. The fist step in doing this will be to take the vaios deivates of the spheical tem eqations ", θ, φ", sbstitte them in, and hope that some of this caziness cancels ot. Fist let me identify all of the vaios deivatives we will need to take and pt them back into pope notation: x x θ x θ x y y θ y θ y φ x φ x z z θ z θ z φ y φ y xx 2 x 2 θ xx 2 θ x 2 φ xx 2 φ x 2 yy 2 y 2 θ yy 2 θ y 2 φ yy 2 φ y 2 zz 2 z 2 θ zz 2 θ z 2 5

6 We ll take these in ode one at a time. Fo convenience and space saving, I will make the following tempoay sbstittion in all of the following deivatives: And so: 2 x 2 + y 2 + z 2 x x x ( ) x sin(θ) cos(φ) sin(θ) cos(φ) Eq. 25 It follows then that: y y z z sin(θ) sin(φ) Eq. 25 cos(θ) Eq. 26 And now to diffeentiate these thee eqations again: xx 2 x 2 x ( x x2 ()( 1 ) 2) x2 x2 And: 2 2 sin 2 (θ) cos 2 (φ) 3 1 sin2 (θ) cos 2 (φ) Eq. 27 yy 2 y 2 1 sin2 (θ) sin 2 (φ) Eq. 28 zz 2 z 2 1 cos2 (θ) Eq. 29 Lckily, the eqation has a symmety that we can se to qickly compte the vaios deivatives once we have taken a single o doble deivative with espect to any vaiable. In this case, I choose x and so y and z followed easily. Unfotnately, the θ eqation is not so elegant, paticlaly the second patial deivatives. As a efeshe, let me give yo the geneic fom of the invese cosine deivative: (cos 1 (w)) dw 1 w 2 Also note the following as it will come in handy soon: 6

7 z 2 x 2 + y 2 2 sin 2 (θ) (cos 2 (φ) + sin 2 (φ)) sin(θ) And so: θ x θ x x (cos 1 ( z )) z x 1 z2 z x z 2 2 cos(θ) sin(θ) cos(φ) 2 sin(θ) cos(θ) cos(φ) Eq. 30 The y patial deivative follows easily: θ y θ y y (cos 1 ( z )) z y 1 z2 z y z 2 2 cos(θ) sin(θ) sin(φ) 2 sin(θ) cos(θ) sin(φ) Eq. 31 And fo z : θ z θ z z (cos 1 ( z )) ( z2 ( ) ) 1 z2 ( ( z2 ) ) 1 z2 ( ( z2 ) ) 1 z2 ( z 2 ) 1 z2 ( z2 ) z 2 z2 sin(θ) 2 sin(θ) Eq. 32 Now fo the fn pat. In ode to take the second patial deivatives, we mst se the deaded qotient le: θ xx 2 θ x 2 x ( z x z 2) ( z 2 x )z z x (( z 2) + 2x z2 ) 2 ( z 2 ) 2 ( 2 sin(θ))z z x (( 2 x ) + 2x sin(θ)) sin(θ) 4 ( 2 sin 2 (θ)) 7

8 4 sin(θ) cos(θ) ( 4 sin 2 (θ) cos 2 (φ) cos(θ) ) 2 sin(θ) 4 sin 3 (θ) cos 2 (φ) cos(θ) 6 sin 2 (θ) cos(θ) 2 sin(θ) cos2 (φ) cos(θ) 2 sin(θ) 2 sin2 (θ) cos 2 (φ) cos(θ) 2 sin(θ) cos(θ)(1 cos 2 (φ) 2 sin 2 (θ) cos 2 (φ)) 2 sin(θ) cos(θ)(sin2 (φ) 2 sin 2 (θ) cos 2 (φ)) 2 sin(θ) Eq. 33 As yo can see, the second patial deivatives of θ do not simplify nealy as nicely as the othes we have taken so fa. It is jst something we will have to deal with fo now. θ yy 2 θ y 2 y ( z y z 2) ( z 2 y )z z y (( z 2) + 2y z2 ) 2 ( z 2 ) ( 2 sin(θ))z z y (( 2 y ) + 2y sin(θ)) sin(θ) 4 ( 2 sin 2 (θ)) 4 sin(θ) cos(θ) ( 4 sin 2 (θ) sin 2 (φ) cos(θ) ) 2 sin(θ) 4 sin 3 (θ) sin 2 (φ) cos(θ) 6 sin 2 (θ) cos(θ) 2 sin(θ) sin2 (φ) cos(θ) 2 sin(θ) 2 sin2 (θ) sin 2 (φ) cos(θ) 2 sin(θ) cos(θ)(1 sin 2 (φ) 2 sin 2 (θ) sin 2 (φ)) 2 sin(θ) cos(θ)(cos2 (φ) 2 sin 2 (θ) sin 2 (φ)) 2 sin(θ) Eq. 34 Yo might be able to see it aleady, bt when we pt θ xx and θ yy " togethe late we will be able to cancel A LOT of the tems ot. Bt fo now, let s finish θ zz. At fist glance, it appeas to be anothe qotient le, bt in fact it is not and theefoe mch easie to calclate: θ zz 2 θ z 2 z ( z2 ) + y 2 z ( x2 ) z ( 1 x 2 + y 2 ) 2z 2 x 2 + y 2 2z x2 + y cos(θ) ( sin(θ)) 4 2 cos(θ) sin(θ) 2 Eq. 35 And finally, we get to the φ deivatives. By the way, the geneic fom of the actan deivate is: 8

9 (tan 1 (w)) dw 1 + w 2 φ x φ x x (tan 1 ( y y ( x )) x 2 ) 1 + ( y2 x 2 ) y x 2 + y 2 sin(θ) sin(φ) 2 sin 2 (θ) sin(φ) sin(θ) Eq. 36 And: φ y φ y y (tan 1 ( y 1 x )) ( x ) 1 + ( y2 x 2 ) 1 x + y2 x 1 ( x2 + y 2 ) x x x 2 + y 2 sin(θ) cos(φ) 2 sin 2 (θ) cos(φ) sin(θ) Eq. 37 And now the second patial deivatives: φ xx 2 φ x 2 x ( y x 2 + y 2) x (y(x2 + y 2 ) 1 ) 2xy (x 2 + y 2 Eq. 38 ) 2 I m not going to simplify this one any fthe, and yo will see why in jst a moment: φ yy 2 φ y 2 y ( x x 2 + y 2) y (x(x2 + y 2 ) 1 ) 2xy (x 2 + y 2 Eq. 39 ) 2 That will cancel ot nicely! Now we ae eady to begin o massive sbstittion mission of the deivatives back into Eqation 24. I will se the Eqation 24 colo coding scheme to sbstitte the eqation ot piece meal: G ( xx + yy + zz ) G (1 sin2 (θ) cos 2 (φ) + 1 sin2 (θ) sin 2 (φ) + 1 cos2 (θ) ) G (θ) cos 2 (φ) + sin 2 (θ) sin 2 (φ) + cos 2 (θ) 3) ( (sin2 ) G (2 ) Eq. 40 9

10 G (( x ) 2 + ( y ) 2 + ( z ) 2 ) 2 G 2 (sin2 (θ) cos 2 (φ) + sin 2 (θ) sin 2 (φ) + cos 2 (θ)) 2 G Eq G θ ( x θ x + y θ y + z θ z ) 2 θ (G cos(θ) cos(φ) ) (sin(θ) cos(φ) + sin(θ) sin(φ) 2 θ (G ) (sin(θ) cos(θ) cos2 (φ) 2 θ (G cos(θ) ) (sin(θ) cos(θ) sin(φ) + sin(θ) cos(θ) sin2 (φ) Eqation 42 is the best eslt I seen so fa! Let s keep on going: cos(θ) sin(θ) ) sin(θ) cos(θ) ) sin(θ) cos(θ) ) 0 Eq. 42 2G φ ( x φ x + y φ y ) 2 (G sin(φ) cos(φ) ) ((sin(θ) cos(φ)) ( ) + (sin(θ) sin(φ)) ( φ sin(θ) sin(θ) )) 2 (G sin(φ) cos(φ) sin(θ) sin(φ) cos(φ) ) ((sin(θ) ) ( )) 0 Eq. 43 φ sin(θ) sin(θ) G θ (θ xx + θ yy + θ zz ) G θ (φ) 2 sin 2 (θ) cos 2 (φ)) (cos(θ)(sin2 2 sin(θ) 2 cos(θ) sin(θ) + 2 ) + cos(θ)(cos2 (φ) 2 sin 2 (θ) sin 2 (φ)) 2 sin(θ) G θ (cos(θ) 2 ( sin2 (φ) + cos 2 (φ) 2 sin 2 (θ) cos 2 (φ) 2 sin 2 (θ) sin 2 (φ) + 2 sin 2 (θ) )) sin(θ) G θ (cos(θ) 2 ( 1 2 sin2 (θ) cos 2 (φ) 2 sin 2 (θ) sin 2 (φ) + 2 sin 2 (θ) )) sin(θ) G θ (cos(θ) 2 ( 1 2 sin2 (θ)(cos 2 (φ) + sin 2 (φ)) + 2 sin 2 (θ) )) sin(θ) 10

11 G θ (cos(θ) 2 ( 1 2 sin2 (θ) + 2 sin 2 (θ) )) G sin(θ) θ ( cos(θ) 2 ) Eq. 44 sin(θ) G θθ ((θ x ) 2 + (θ y ) 2 + (θ z ) 2 ) 2 G θ cos(φ) ((cos(θ) 2 2 ) cos(θ) sin(φ) + ( ) 2 + ( sin(θ) 2 ) ) 2 G θ 2 (θ) cos 2 (φ) + cos 2 (θ) sin 2 (φ) + sin 2 (θ) (cos2 2 ) 2 G θ 2 ( 1 2) Eq. 45 2G θφ (θ x φ x + θ y φ y ) 2 G φ θ ((cos(θ) cos(φ) ) ( sin(φ) sin(φ) ) + (cos(θ) ) ( cos(φ) sin(θ) sin(θ) )) 2 G φ θ ((cos(θ) cos(φ) ) ( sin(φ) sin(φ) ) + (cos(θ) ) ( cos(φ) sin(θ) sin(θ) )) 2 G φ θ cos(φ) sin(φ) ((cos(θ) 2 sin 2 ) ( (θ) cos(θ) cos(φ) sin(φ) 2 sin 2 )) 0 Eq. 45 (θ) G φ (φ xx + φ yy ) G φ (( 2xy (x 2 + y 2 ) 2) + ( 2xy (x 2 + y 2 ) 2)) 0 Eq. 47 G φφ ((φ x ) 2 + (φ y ) 2 ) 2 2 G sin(φ) (( φ2 sin(θ) ) + ( cos(φ) 2 sin(θ) ) ) 2 G φ 2 (φ) + cos 2 (φ) (sin2 2 sin 2 ) 2 G (θ) φ 2 ( 1 2 sin 2 ) Eq. 48 (θ) Sbstitting all of these Eqations, 40-48, back into Eqation 24, yo get the ALMOST completed eslts of: 2 G (,θ,φ) G (2 ) + 2 G 2 + G θ ( cos(θ) 2 sin(θ) ) + 2 G θ 2 ( 1 2) + 2 G φ 2 ( 1 2 sin 2 ) Eq. 49 (θ) In fact, this IS the final eslt. Some cleve peson late came along and condensed the expession a little moe. I wold gess if fo nothing else then to make it all fit nicely in a text book! We have come this fa, so no eason to stop now. The fist two tems, ed and ble, ae the eslt of a podct le expansions, as yo can see: G (2 ) + 2 G (2 G ) 11

12 And: G θ ( cos(θ) 2 sin(θ) ) + 2 G θ 2 ( 1 2) ( 1 2 sin (θ) ) θ (sin (θ) G θ ) A wod of cation, occasionally the θ and φ tems ae swapped in the initial stage when deiving Eqations 1. The end eslt is the same, bt the vaiables ae switched. The Laplacian we have deived hee is typically how it is fond in Physics textbooks. So finally, afte 11 pages of wok, we have aived at o answe: 2 G (,θ,φ) 1 G 2 (2 ) + ( 1 2 sin (θ) ) G (sin(θ) θ θ ) + ( 1 2 sin 2 (θ) ) 2 G φ 2 It s petty clea why the deivation of this fomla is left p to the stdent. Yo wold need an entie sbsection in a text book to clealy show the steps. Having seen it magically appea ot of the atho s bag of ticks in a few of my classes with little comment on why it takes this fom, I felt compelled to find ot fo myself. If yo have taken the time to come this fa, thank yo fo eading. -Lyle Anett 12