The Laplacian in Spherical Polar Coordinates

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1 Univesity of Connecticut Chemisty Education Mateials Depatment of Chemisty The Laplacian in Spheical Pola Coodinates Cal W. David Univesity of Connecticut, Recommended Citation David, Cal W., "The Laplacian in Spheical Pola Coodinates" 007. Chemisty Education Mateials. Pape This Aticle is bought to you fo fee open access by the Depatment of Chemisty at It has been accepted fo inclusion in Chemisty Education Mateials by an authoized administato of Fo moe infomation, please contact

2 The Laplacian in Spheical Pola Coödinates C. W. David Depatment of Chemisty Univesity of Connecticut Stos, Connecticut Dated: Febuay 6, 007 I. SYNOPSIS In teating the Hydogen Atom s electon quantum mechanically, we nomally convet the Hamiltonian fom its Catesian to its Spheical Pola fom, since the poblem is vaiable sepaable in the latte s coödinate system. This eading teats the bute-foce method of effecting the tansfomation of the kinetic enegy opeato, nomally called the Laplacian, fom one to the othe coödinate systems. II. PRELIMINARY DEFINITIONS We stat with the pimitive definitions x = sin θ cos φ z = cos θ thei inveses = x y z θ = cos 1 z = z cos1 x y z φ = tan 1 y x y = sin θ sin φ attempt to wite using the chain ule x = x θ,φ θ x θ,φ φ x φ,θ y = z = y θ,φ z θ,φ θ y θ,φ θ z θ,φ φ y φ,θ φ z φ,θ III. PRELIMINARY PARTIAL DERIVATIVES The needed above patial deivatives ae: = sin θ cos φ 3.1 x Typeset by REVTEX = sin θ sin φ 3. y = cos θ 3.3 z we have as a stating point fo doing the θ tems,

3 d cos θ = sin θdθ = dz 1 z d = dz 1 z d = dz z 1 xdx ydy zdz 3.4 so that, fo example when dy = dz = 0 we have which is sin θdθ = cos θ so that sin θdθ = z x dx sin θ cos φdx = 1 cos ϑ 3 dz θ cos θ cos φ = x θ cos θ sin φ = y but, fo the z-equation, we have which is so one has sin θdθ = sin θdθ = sin θdθ = dz z 1 zdz 1 z 3 dz = z 3 dz 1 z 3 dz = sin θ 3 dz θ = sin θ z Next, we have as an example tan φ = sin φ cos φ = tan1 y x taking the patial deivatives on both sides, we obtain so 1 dsin φ sin φ cos ϑ cos dcos φ φ 1 sin φ cos dφ = dy φ x y x dx o 1 cos dφ = dy φ x y x dx so, afte multiplying acoss by cos φ leads to at constant x at constant y IV. φ = cos φ y sin θ φ = sin φ x sin θ φ = z THE FIRST PARTIAL DERIVATIVE TERMS Given these esults above we wite z = cos θ y = sin θ sin φ x = sin θ cos φ sin θ cos θ sin φ cos θ cos φ θ 4.1 cos φ θ sin θ φ 4. θ sin φ sin θ φ 4.3 V. GATHERING TERMS TO FORM THE LAPLACIAN Fom Equation 4.1 we fom

4 while fom Equation 4. we obtain z = cos θ [ cos θ sin θ θ sin θ cos θ sin θ θ θ y = sin θ sin φ cos θ sin φ [ sin θ sin φ cos θ sin φ [sin θ sin φ cos θ sin φ θ cos φ [sin θ sin φ cos θ sin φ sin θ φ fom Equation 4.3 we obtain [ sin θ cos φ x = sin θ cos φ cos θ cos φ cos θ cos φ [sin θ cos φ cos θ cos φ θ sin φ [sin θ cos φ cos θ cos φ sin θ φ θ cos φ sin θ φ θ cos φ sin θ θ cos φ sin θ θ sin φ sin θ θ sin φ sin θ θ sin φ sin θ φ φ φ φ φ Exping, we have while fo the y-equation we have z = cos θ cos θ sin θ sin θ cos θ θ θ sin θ sin θ cos θ sin θ cos θ sin θ θ θ θ 5.4 y = sin θ sin φ 5.5 [ cos θ sin φ cos θ sin φ sin θ sin φ θ 5.6 θ [ sin θ sin φ cos φ cos φ sin θ φ 5.7 sin θ φ cos θ sin φ [cos θ sin φ sin θ sin φ 5.8 θ [ cos θ sin φ sin θ sin φ cos θ sin φ θ θ 5.9 [ cos θ sin φ cos φ cos θ cos φ sin θ φ 5.10 sin θ φθ cos φ [sin θ cos φ sin θ sin φ 5.11 sin θ φ [ cos φ cos θ cos φ cos θ sin φ sin θ θ 5.1 θφ [ cos φ sin φ cos φ cos φ sin θ sin θ φ sin θ φ 5.13

5 finally = sin θ cos φ sin θ cos φ x [ cos θ cos φ cos θ cos φ sin θ cos φ θ θ [ sin φ sin φ sin θ cos φ sin θ φ sin θ φ cos θ cos φ [cos θ cos φ sin θ cos φ θ [ cos θ cos φ sin θ cos φ cos θ cos φ θ θ [ cos θ cos φ sin φ sin φ sin θ φ sin θ φθ sin φ [sin θ sin φ sin θ cos φ sin θ φ [ sin φ cos θ sin φ cos θ cos φ sin θ θ θφ [ sin φ cos φ sin φ sin θ sin θ φ sin θ φ Now, one by one, we exp completely each of these thee tems. We have, fo the y-equation: z = cos θ 5. cos θ sin θ 5.3 θ sin θ cos θ 5.4 θ sin θ 5.5 sin θ cos θ 5.6 θ sin θ cos θ 5.7 θ sin θ θ 5.8 y = sin θ sin φ 5.9 sin θ cos θ sin φ θ cos θ sin θ sin φ 5.31 θ sin φ cos φ φ cos φ sin φ 5.33 φ cos θ sin φ cos θ sin θ sin φ 5.35 θ sin θ cos θ sin φ 5.36 θ cos θ sin φ 5.9 θ 5.37 cos θ cos φ sin φ sin 5.38 θ φ cos θ cos φ sin φ 5.39 sin θ φθ cos φ cos φ sin φ 5.41 φ cos φ cos θ 5.4 sin θ θ cos θ cos φ sin φ sin θ θφ cos φ sin φ 5.13 sin 5.44 θ φ cos φ sin θ φ 5.45

6 finally, fo the x-equation, we have x = sin θ cos φ 5.46 sin θ cos θ cos φ θ sin θ cos θ cos φ θ cos φ sin φ φ 5.49 sin φ cos φ 5.50 φ cos θ cos φ sin θ cos θ cos φ 5.5 θ sin θ cos θ cos φ θ cos θ cos φ θ 5.54 cos θ cos φ cos φ sin φ 5.16 sin θ φ 5.55 sin φ cos φ cos θ 5.56 sin θ φθ sin φ sin φ cos φ φ 5.58 cos θ sin φ sin θ θ cos θ sin φ cos φ sin θ θφ 5.60 sin φ cos φ 5.0 sin θ φ 5.61 sin φ sin θ φ 5.6 Gatheing tems as coefficients of patial deivatives, we obtain fom Equations 5., cos θ sin θ sin φ sin θ cos φ fom Equations 5.3, 5.6, 5.30, 5.36, 5.4, 5.47, 5.53, 5.59 cos θ sin θ sin θ cos θ θ sin θ cos θ sin φ sin θ cos θ sin φ cos φ cos θ sin θ while we obtain fom Equations 5.8, 5.37, 5.54: sin θ cos θ cos φ sin θ cos θ cos φ cos θ sin φ sin θ cos θ sin θ θ 5.63 θ sin θ Fom Equations 5.5, 5.34, 5.40, 5.51, 5.57, cos θ sin φ sin θ cos θ sin φ cos φ cos θ cos φ sin φ cos θ cos φ 1 θ Fom Equations 5.3, 5.38, 5.44, 5.49, we obtain sin φ cos φ φ cos θ cos φ sin φ sin cos θ cos φ sin φ cos φ sin φ cos θ cos φ θ sin θ cos θ cos φ sin φ sin φ cos φ sin θ sin θ Fom Equations we obtain φ cos φ sin θ sin φ sin θ 1 sin θ φ 5.67

7 The mixed deivatives yield, fist, fom Equations 5.33, 5.41, 5.50, 5.58 leading to cos φ sin φ cos φ sin φ sin φ cos φ sin φ cos φ φ Fom Equations 5.4, 5.7, 5.35, , 5.48 θ sin θ cos θ cos φ sin θ cos θ sin θ cos θ cos θ sin θ sin φ cos θ sin θ sin φ sin θ cos θ cos φ Fom Equations cos θ cos φ sin φ cos φ sin φ φθ sin θ sin θ sin φ cos φ cos θ cos θ sin φ cos φ sin θ sin θ Gatheing togethe the non-vanishing tems, we obtain 1 θ cos θ sin θ θ 1 sin θ φ VI. MAPLE EQUIVALENT A. Example 1 which is one of the two classic foms fo. The othe is 1 1 sin θ sin θ sin θ θ θ φ Hee is a set of Maple instuctions adjusted fom the -dimensional code [1 fo ou 3-dimensional case, which will get you the same esult: estat; f:=g,theta,phi; tx := sintheta*cosphi*difff,costheta*cosphi/*difff,theta -sinphi/*sintheta*difff,phi; tx:=exp sintheta*cosphi*difftx,costheta*cosphi/*difftx,theta -sinphi/*sintheta*difftx,phi; ty := sintheta*sinphi*difff,costheta*sinphi/*difff,theta cosphi/*sintheta*difff,phi; ty:=expsintheta*sinphi*diffty,costheta*sinphi/ *diffty,thetacosphi/*sintheta*diffty,phi; tz := costheta*difff, -sintheta/*difff,theta; tz := expcostheta*difftz,-sintheta/*difftz,theta; del := txtytz: del := algsubs costheta^=1-sintheta^, del : del := expalgsubs cosphi^=1-sinphi^, del ; Hee is anothe vesion of the same thing: B. Example

8 > #CARTESIAN TO SPHERICAL POLAR > estat; > withplots: Waning, the name changecoods has been edefined > uu:=usqtx^y^z^,accosz/sqtx^y^z^,actany,x; uu := u x y z z, accos, actany, x x y z > ux:=diffuu,x: > uy:=diffuu,y: > uz:=diffuu,z: > uxx:=diffux,x: > uyy:=diffuy,y: > uzz:=diffuz,z: > Lapu:=simplifyuxxuyyuzz: > assume,positive; > Lapu:=simplifysubsx=*sintheta*cosphi, > y=*sintheta*sinphi, > z = *costheta, > actansintheta*sinphi,sintheta*cosphi=phi, > accoscostheta=theta, > Lapu,tig: > Lapu := subsactansintheta*sinphi,sintheta*cosphi=phi, > accoscostheta=theta, > Lapu: > Lapu := algsubs-1costheta^=-sintheta^,lapu: > Lapu:=expLapu; Lapu := D u, θ, φ sinθ cosθ D, u, θ, φ D 3, 3u, θ, φ sinθ 3/ sinθ D 1u, θ, φ D 1, 1 u, θ, φ It takes some getting used to Maple notation to see that this is the expected esult. ae bette ways to cay out the tansfomation fom Catesian to Spheical Pola indeed any othogonal coödinate system. VII. COMMENTS The eade should be awae that the bute foce methods used hee ae pimitive in the exteme, that thee [1 Mathias Kawski, kawski/maple/maple.html