application of curvilinear theory to sphericals

Application of Curvilinear Theory to Spherical Coordinates PhL 9.7.08 Here we treat this Transformation T blindly and just carry out the steps of the M&M curvilinear program as outlined in various other nearby documents. Results are summarized in the last section. This doc is sort of "everything you ever wanted to know about spherical coordinates". Note summary at the end. Contents 1. The Inverse Transformation. 1 2. Compute the inverse T matrix. 1 3. Compute the forward T matrix 2 4. What is the Jacobian? 5 5. What are the basis vectors ei and ei in Cartesian coordinates ? 5 6. What is the metric tensor, what are the Qi, and what is g = det(gij) ? 7 7. What are the mag's of the basis vectors, and what are the unit vectors? 8 8. What are the cross products of the basis vectors (and what is v) ? 8 9. Differential displacements and areas. 9 10. The rotation matrix point of view. 9 12. Derivatives of the unit vectors with respect to coordinates 11 13. The differential operators. 13 14. Summary: 17 (a) The inverse and forward transformations 17 (b) The backwards T matrix Tab ≡ 17 (c) The forward T matrix Tab ≡ 17 (d) The Jacobian J 17 (e) The metric tensor and the Qi 17 (f) The 6 basis vectors in Cartesian coordinates, relations and mag's 18 (g) The unit vectors and their cross products 18 (h) Explicit expressions for the unit vectors and their dot products with the Cartesian ones 18 (i) The differential displacements and areas 19 (j) derivatives of the unit vectors 19 (k) The differential operators, all verified against old printed sheet: 19 Spherical Coordinates Summary Page 20 1. The Inverse Transformation. The inverse coordinate transformation is this: x = x(q) = x(x'). x = rSθCφ y = rSθSφ z = rCθ 2. Compute the inverse T matrix. We use the following ordering for the q-space variables, q1 = θ q2 = φ q3 = r The following derivatives are obtained and we know Tab ≡ , so Tθ1= ∂x/∂θ = rCθCφ Tθ2 = ∂y/∂θ = rCθSφ Tθ3 = ∂z/∂θ = -rSθ Tφ1= ∂x/∂φ = -rSθSφ Tφ2 = ∂y/∂φ = rSθCφ Tφ3 = ∂z/∂φ = 0 Tr1= ∂x/∂r = SθCφ Tr2 = ∂y/∂r = SθSφ Tr3 = ∂z/∂r = Cθ And we get this entered into Maple: , where the columns are second index x,y,z and the rows are first index θ,φ,r, same order as presented above. 3. Compute the forward T matrix The other direction is much harder. We know Tab ≡ so we can start computing things. r2 = x2+ y2 + z2 => rdr = xdx + ydy+zdz T1r = ∂r/∂x = x/r = rSθCφ/r = SθCφ T2r = ∂r/∂x = y/r = rSθSφ/r = SθSφ T3r = ∂r/∂x = z/r = rCθ/r = Cθ The angle ones are not so easy. We can solve our three opening equations for tanθ to get tan2θ = (x2+ y2)/z2 Then differentiate to get (cancel 2's, fiddle around) tanθ sec2θ dθ = (x/z2)dx + (y/z2)dy – [(x2+ y2)/z3]dz Sθ/Cθ3 dθ = (SθCφ/r Cθ2)dx + (SθSφ/r Cθ2)dy – (Sθ2/r Cθ3) dz and then multiply both sides by Cθ3/ Sθ to get dθ = (Cθ Cφ/r)dx + (Cθ Sφ/r)dy - (Sθ/r) dz from which we can read off T1θ = ∂θ/∂x = Cθ Cφ/r T2θ = ∂θ/∂y = Cθ Sφ/r T3θ = ∂θ/∂z = - Sθ/r In a similar manner we can solve for tanφ as tanφ = y/x => sec2φ dφ = dy/x - (y/x2)dx dφ = (Cφ/rCθ)dy - (Sφ/rSθ)dx from which we can read off T1φ = ∂φ/∂x = - Sφ/rSθ T2φ = ∂φ/∂y = Cφ/rCθ T3φ = ∂φ/∂z = 0 So we did all that by brute force. To verify the results, we can take a different tack and use Tab = (T-1)ba So let's enter our matrix T** into Maple and have it compute the inverse TINV where now first index = rows = x,y,z and second index = columns = θ,φ,r. We can quote our manually calculated results from above and paste them into a table, and we see it agrees with TINV. So we can regard Maple as confirming our calculations. Status so far: We now know the following information: The inverse and forward transform are (q1 = θ q2 = φ q3 = r ) x = rSθCφ r2 = x2+ y2 + z2 y = rSθSφ tan2θ = (x2+ y2)/z2 z = rCθ tanφ = y/x The following matrix has as its first column Tθ1 Tφ1 Tr1 , then other columns are 2 and 3 The following matrix has as its first row T1θ T1φ T1r , then other rows are 2 and 3 4. What is the Jacobian? According to our support doc notes we have J= det( ) = det(Tab) and dV = J dV' dxdydz = J dθdφdr So we will have Maple compute det(Tab) where T is the matrix we already entered there. We get: Therefore J = r2sinθ and we then have the very famous result dxdydz = r2sinθ dθdφdr 5. What are the basis vectors ei and ei in Cartesian coordinates ? Recall these "pondering doc" results for Cartesian components of these basis vectors: [ei]n = Tin [ei]n = Tin We can then just read them out of our matrices above: [eθ]1 = Tθ1 = rCθCφ etc So we get [eθ] = which is the first row of matrix T = T*k. Write it this way eθ = rCθCφ + rCθSφ - rSθ eφ = -rSθSφ + rSθCφ er = SθCφ + SθSφ + Cθ Similarly, eφ is the second row, and er is the third row and we wrote these in above Now what about the other ones ei ? [eθ]n = Tθn = (TT)n θ From above and Maple we had: [eθ]n = Tθn = (T-1)nθ = TINVnθ = TINVTθn This just means that the eθ are the columns of our matrix TINV which was this eθ = CθCφ/r + CθSφ/r - Sθ/r eφ = -Sφ/rSθ + Cφ/rSθ er = SθCφ + SθSφ + Cθ compare to our previous results eθ = rCθCφ + rCθSφ - rSθ eφ = -rSθSφ + rSθCφ er = SθCφ + SθSφ + Cθ By observation we see that eθ = eθ / r2 eφ = eφ / r2 er = er and this the linear relationship between ei and ei is what we expect in an orthogonal system. Even were this system non-orthogonal, we would expect things like this to be true: eθ eφ = 0 Let's check this one case eθ eφ = (rCθCφ/r + rCθSφ/r - Sθ/r ) ( -rSθSφ + rSθCφ ) = rCθCφ/r * -rSθSφ + rCθSφ/r * rSθCφ = - CθCφSθSφ + CθSφ SθCφ = 0 But since we have an orthogonal system, we expect this also to be true eθ eφ = 0 but obvious because eθ eφ = eθ eφ/r2 = 0 6. What is the metric tensor, what are the Qi, and what is g = det(gij) ? The metric tensor is gij = ei ei and we know it is diagonal. We just showed above that the off-diagonal element gθφ = eθ eφ = 0. So let's compute the diagonal elements: gθθ = eθ eθ = ( rCθCφ + rCθSφ - rSθ)2 = (rCθCφ)2 + ( rCθSφ)2 + (- rSθ)2 = = r2 ( 1) = r2 = Qθ2 = gθθ gφφ= eφ eφ = (-rSθSφ + rSθCφ)2 = (-rSθSφ )2 + ( rSθCφ)2 = r2Sθ2 = Qφ2 = gφφ grr= er er = ( SθCφ + SθSφ - Cθ)2 = (SθCφ)2 + ( SθSφ )2 + (- Cθ)2 = 1 = Qr2 = grr Because in all three cases ei = const * ei (as observed above), we know at once that all off-diagonal elements are zero, so we know this is an "orthogonal" curvilinear coordinate system. So, the metric tensor is this: g = diag(gθθ, gφφ, grr) = diag(Qθ2, Qφ2, Qr2) = diag(r2, r2Sθ2, 1) and we have these all-important results Qθ = r Qφ = rSθ Qr = 1 which agrees with M&M page 178 equation 5-27. All we have to now is insert these into our various differential operator formulas for orthogonal systems. While here we can compute that g = det(g) = Qθ2Qφ2Qr2 =(r2Sθ)2 so = r2sinθ = J as computed in Sec 4 above! 7. What are the mag's of the basis vectors, and what are the unit vectors? In general we know this ei ei = gii = Qi2 = |ei|2 => |ei| = Qi so we know |eθ| = Qθ = r θ = = eθ/r eθ = rθ = r |eφ| = Qφ = rSθ φ = = eφ/ rSθ eφ = rSθ φ = rSθ |er| = Qr = 1 r = = er er = r = so we have = θ = eθ/Qθ = eθ / r = CθCφ + CθSφ – Sθ =φ = eφ /Qφ = eφ / (rSθ) = -Sφ + Cφ = r = er /Qr = er = SθCφ + SθSφ + Cθ 8. What are the cross products of the basis vectors (and what is v) ? From e3 ≡ e1 x e2 / v we know that eφ x er = v eθ = v eθ/r2 which says (eφ) x (er) = v/r2* ( eθ) (rSθ ) x () = v/r2 (r ) = v /r => x = v/(r2Sθ) But since we know the q-space unit vectors are orthogonal, we conclude that v = r2Sθ Earlier we computed = r2sinθ = J and now we can add v, so we have shown that v = = J = r2sinθ by explicitly computing each quantity. We can finish out our cross product rules by doing cyclic order on the qi as we have set them up: x = x = x = and this agrees with my old hand-written spherical unit vectors page. 9. Differential displacements and areas. The differential displacements are dsθ = dθ dsφ = dφ dsr = dr and the differential "areas" are the following and its cyclics dAθ = dAφr = dsφ x dsr = (dφ ) x (dr ) = dφ dr so get dAθ = dφ dr dAφ = dr dθ dAr = dθ dφ 10. The rotation matrix point of view. The triad of unit vectors ( ) at the location (θ,φ,r) in q-space can be obtained by doing a specific active rotation R = Rz(φ) Ry(θ) on the Cartesian unit vectors. That is to say, we have = Rz(φ) Ry(θ) = Rz(φ) Ry(θ) = Rz(φ) Ry(θ) By operating one's right hand and staring at the picture below, one can easily convince oneself that the above claim is true. Notice that points slightly downward and is tangent to the "only change θ" curve which is a circle but we only show part of this circle. And points right and rear and is tangent to the "only change φ" curve which is a circle. And finally, points in the direction of the "only change r" curve which is a line. These curves are called coordinate lines in M&M Section 5.1. Notice that where they intersect --at the tip of vector r -- they are orthogonal. We can regard this rotation as having the form R = Rz(φ) Ry(θ) Rz(0) so the y-style Euler angles of the rotation are (φ,θ,0). The fact claimed above allows the following rather elegant statement of what these three unit vectors and their dot products are: ( ) = = = Rz(φ) Ry(θ) In other words, the unit vectors are just the columns of the rotation matrix, and the dot products are the matrix elements, as shown. The reader can compare this to our previous computations above, which were = CθCφ + CθSφ – Sθ = () + () + () = -Sφ + Cφ = () + () + () = SθCφ + SθSφ + Cθ = () + () + () Because the matrix is real orthogonal, R-1 = RT, we get the inverse of these equations by reading off the rows instead of the columns: = CθCφ - Sφ + SθCφ = CθSφ + Cφ + SθSφ = – Sθ + Cθ Lest there be any confusion about our active rotation matrices, here they are all with angle θ: Rx() = Ry() = Rz() = 12. Derivatives of the unit vectors with respect to coordinates Derivatives of the unit vectors with respect to angles. The reader needs to constantly stare at the spherical coordinates diagram below while reading this section. Notice that we have included in this figure the unit vector normally associated with cylindrical coordinates. This unit vector lies in the z=constant plane and points from the z axis to the tip of r. Certain unit vector derivatives vanish and we shall get them out of the way. First, if we lengthen the r vector by amount dr, the coordinate triad stays exactly the same, we just "go out to a larger sphere" of radius r+dr, so we have these derivatives at once: ∂/∂r = 0 ∂/∂r = 0 ∂/∂r = 0 ∂/∂r = 0 The following two derivatives are also zero as one can see just looking at the diagram: ∂/∂θ = 0 ∂/∂θ = 0 That is, increasing θ moves the dotted circle down and makes it larger, but and stay the same. Now consider the following detail views. The leftmost view is from the top. The middle view is "from the side" so that all four unit vectors shown are in a plane. The rightmost view just repeats the middle view with a more conventional circle drawing. From the middle figure, we see that is just that has been rotated amount θ. From our experience with rotations, we don't need to go look it all up, we know that = cosθ + sinθ = -sinθ +cosθ We know both signs are plus for from the middle figure, and we know that for r at the equator θ = π/2 we will have = , so the coefficient must be sinθ. Then for we see contributes positively, but contributes negatively, and we know we have to swap the locations of the trig functions, so we get the relation on the right above for . Now getting back to the picture on the left. We know that if we change the azimuthal angle φ by dφ, then we have the following pair of differential relations d = dφ and d = – dφ => ∂/∂φ = and ∂/∂φ = – Normally with the ρ vector going in a circle, we would have dρ = ρdφ but since is a unit vector, ρ=1 and we get the form shown. The second equation of the pair has the reverse sign since a change in points to the center of the circle. A similar situation occurs with the angle θ measured down from vertical. As shown in the right view, although is "reversed", so is the location of the angle, so the equations are exactly the same. d = dθ and d = - dθ => ∂/∂θ = and ∂/∂θ = – Two derivatives are still missing from our list, they are ∂/∂φ and ∂/∂φ . To figure these out, we have to do a small amount of work. Recall three of our equations from above: = cosθ + sinθ = -sinθ +cosθ d = dφ If we change only φ and change it by δφ, we can say d = cosθ d + sinθ d = dφ d = -sinθ d + cosθ d = 0 // 0 since never moves! Multiply the first equation by sinθ and the second by cosθ sinθ d = sinθ cosθ d + sin2θ d = sinθ dφ cosθ d = -sinθ cosθ d + cos2θ d = 0 Add the two equations to find that d = sinθ dφ => ∂/ ∂φ = sinθ We can guess the other equation, but let's just do. Start with our first pair of equations and this time multiply the first by cosθ and the second by sinθ cosθ d = cos2θ d + sinθ cosθ d = cosθ dφ sinθ d = -sin2θ d + sinθ cosθ d = 0 Subtract these to find that d = cosθ dφ => d/∂φ = cosθ All done! We present all the above results in θ,φ,r order As with curvilinear coordinates in general, the orthogonal triad of unit vectors for spherical coordinates changes as the coordinates change, so many unit vector derivatives are non-zero as shown above. With Cartesian coordinates of course all derivatives are zero, for example, ∂/∂y = 0. The triad is in the same orientation at all points in space. 13. The differential operators. We quote the "analysis" set of operators from our "divergence doc" which are all based on having an orthogonal system with orthonormal unit vectors A = [ ∂1(Q2Q3A1) + cyclic] where A = Aii analysis (f) = (1/Qi) (∂if) i analysis A (f) = (Ai/Qi) (∂if) where A = Aii analysis 2(f) = { ∂1{(Q2Q3)/Q1 *(∂1f)} + cyclic } analysis x B = εijk ∂j(QkBk) Qi i analysis = So now we can fill in using 1 = θ 2 = φ 3 = r Q1 = r Q2 = rSθ Q3 = 1 = Q1Q2Q3 = r2Sθ A = [ ∂1(Q2Q3A1) + ∂2(Q3Q1A2) + ∂3(Q1Q2A3)] = [ ∂θ(Q2Q3Aθ) + ∂φ(Q3Q1Aφ) + ∂r(Q1Q2Ar)] = [ ∂θ(rSθ Aθ) + ∂φ(rAφ) + ∂r(r2Sθ Ar)] = [r ∂θ(Sθ Aθ) + r∂φ(Aφ) + Sθ ∂r(r2Ar)] = [ ∂θ(Sθ Aθ) + ∂φ(Aφ) + ∂r(r2Ar) ] or A = ∂r(r2Ar) + ∂θ(Sθ Aθ) + ∂φAφ where we put things in the conventional r,θ,φ order. This agrees with my old copied sheet. (f) = (1/Qi) (∂if) i = (1/Q1) (∂1f) 1 + (1/Q2) (∂2f) 2 + (1/Q3) (∂3f) 3 = (1/r) (∂θf) θ + (1/ rSθ) (∂φf) φ + (1/1) (∂rf) r = (1/r) (∂θf) + (1/ rSθ) (∂φf) + (1/1) (∂rf) = (∂rf) + (1/r) (∂θf) + (1/rSθ) (∂φf) and this agrees as well. Now repeat 1 = θ 2 = φ 3 = r Q1 = r Q2 = rSθ Q3 = 1 = Q1Q2Q3 = r2Sθ A (f) = (Ai/Qi) (∂if) = (A1/Q1) (∂1f) + (A2/Q2) (∂2f) + (A3/Q3) (∂3f) = (A1/r) (∂1f) + (A2/ rSθ) (∂2f) + (A3) (∂3f) = (Aθ/r) (∂θf) + (Aφ/ rSθ) (∂φf) + (Ar) (∂rf) = Ar (∂rf) + (Aθ/r) (∂θf) + (Aφ/ rSθ) (∂φf) and this is one that is not listed on my page, but which started me off on all this stuff! Next, 2(f) = [ ∂1{(Q2Q3)/Q1 *(∂1f)} + ∂2{(Q3Q1)/Q2 *(∂2f)} + ∂3{(Q1Q2)/Q3 *(∂3f)} ] = [ ∂θ{ ( rSθ)/ r *(∂θf)} + ∂φ{( r)/ rSθ *(∂φf) } + ∂r{( r rSθ) *(∂rf) } ] = [ ∂θ{ ( Sθ)*(∂θf) } + (1/Sθ) ∂φ{ *(∂φf) } + Sθ ∂r{( r2) *(∂rf) } ] = [ ∂θ{ ( Sθ)*(∂θf) } + (1/Sθ) ∂φ{ *(∂2f) } + Sθ ∂r{( r2) *(∂rf) } ] = [ ∂θ{ Sθ ∂θf } + (1/Sθ) ∂φ{∂ φf} + Sθ ∂r{ r2 ∂rf } ] = ∂θ(Sθ ∂θf) + ∂φ2f + ∂r (r2 ∂rf) = ∂r (r2 ∂rf) + ∂θ(Sθ ∂θf) + ∂φ2f and this agrees as well with the sheet. Next, x B = εijk ∂j(QkBk) Qi i = [Q1 1 ( ε1jk ∂j(QkBk)) + Q2 2 ( ε2jk ∂j(QkBk)) + Q3 3 ( ε3jk ∂j(QkBk)) ] = [ Q1 1 { ( ε123 ∂2(Q3B3)) + ( ε132 ∂3(Q2B2)) } + Q2 2 { ( ε231 ∂3(Q1B1)) + ( ε213 ∂1(Q3B3)) } + Q3 3 { ( ε312 ∂1(Q2B2)) + ( ε321 ∂2(Q1B1)) } ] = [ Q1 1 { (∂2(Q3B3)) – ∂3(Q2B2)) } + Q2 2 { (∂3(Q1B1)) – ∂1(Q3B3)) } + Q3 3 { (∂1(Q2B2)) –∂2(Q1B1)) } ] 1 = θ 2 = φ 3 = r Q1 = r Q2 = rSθ Q3 = 1 = Q1Q2Q3 = r2Sθ = [ rθ { (∂φ(1Br)) – ∂r(rSθ Bφ)) } + rSθ φ { (∂r(rBθ)) – ∂θ(1Br)) } + 1 r { (∂θ(rSθ Bφ)) –∂φ(rBθ)) } ] OK = [ r { (∂φ(Br)) – Sθ ∂r(rBφ)) } + rSθ { (∂r(rBθ)) – ∂θ(Br)) } + r { (∂θ(Sθ Bφ)) – ∂φ(Bθ)) } ] = { (∂φ(Br)) – Sθ ∂r(rBφ)) } + { (∂r(rBθ)) – ∂θ(Br)) } { (∂θ(Sθ Bφ)) – ∂φ(Bθ)) } = { (∂θ(Sθ Bφ)) – ∂φ(Bθ) } { (∂φ(Br)) – Sθ ∂r(rBφ) } + { (∂r(rBθ)) – ∂θ(Br) } = { ∂θ(Sθ Bφ) – ∂φBθ } { (∂φBr) – ∂r(rBφ) } + { ∂r(rBθ) – ∂θBr } giving the final result x B = {∂θ(Sθ Bφ) – ∂φBθ} + {(∂φBr) – ∂r(rBφ)} + {∂r(rBθ) – ∂θBr } and this too agrees with our sheet. 14. Summary: (a) The inverse and forward transformations x = rSθCφ r2 = x2+ y2 + z2 y = rSθSφ tan2θ = (x2+ y2)/z2 z = rCθ tanφ = y/x and the q-space coordinates are q1 = θ q2 = φ q3 = r (b) The backwards T matrix Tab ≡ where first column = Tθ1 Tφ1 Tr1 , then other columns are 2 and 3 (c) The forward T matrix Tab ≡ where first row = T1θ T1φ T1r , then other rows are 2 and 3 (d) The Jacobian J J= det( ) = det(Tab) = r2Sθ dxdydz = J dθdφdr (e) The metric tensor and the Qi gij = δi,j gii where gθθ = Qθ2 = r2 Qθ = r gφφ = Qφ2= r2Sθ2 Qφ= rSθ grr = Qr2 = 1 Qr = 1 and we find that g = det(g) = Qθ2Qφ2Qr2 =(r2Sθ)2 so = QθQφQr = r2Sθ (f) The 6 basis vectors in Cartesian coordinates, relations and mag's eθ = rCθCφ + rCθSφ - rSθ eθ = eθ / r2 |eθ| = Qθ = r eφ = -rSθSφ + rSθCφ eφ = eφ / r2 |eφ| = Qφ = rSθ er = SθCφ + SθSφ + Cθ er = er |er| = Qr = 1 and we from eφ x er = v eθ = v eθ/r2 we find that v = r2Sθ and thus we have shown that J = = v = r2Sθ => dVq = dθdφdr = dxdydz/J = dxdydz/( r2Sθ) (g) The unit vectors and their cross products = θ = eθ/Qθ = eθ / r = CθCφ + CθSφ - Sθ =φ = eφ /Qφ = eφ / (rSθ) = -Sφ + Cφ = r = er /Qr = er = SθCφ + SθSφ + Cθ x = x = x = (h) Explicit expressions for the unit vectors and their dot products with the Cartesian ones ( ) = = = Rz(φ) Ry(θ) (i) The differential displacements and areas dsθ = dθ dAθ = dφ dr dsφ = dφ dAφ = dr dθ dsr = dr dAr = dθ dφ (j) derivatives of the unit vectors (k) The differential operators, all verified against old printed sheet: (f) = (∂rf) + (1/r) (∂θf) + (1/rSθ) (∂φf) A = ∂r(r2Ar) + ∂θ(Sθ Aθ) + ∂φA 2(f) = (f) = ∂r (r2 ∂rf) + ∂θ(Sθ ∂θf) + ∂φ2f x A = {∂θ(Sθ Aφ) – ∂φAθ} + {(∂φAr) – ∂r(rAφ)} + {∂r(rAθ) – ∂θAr } Our one other result is no longer very interesting because it is obvious once you know the gradient. A (f) = Ar (∂rf) + (Aθ/r) (∂θf) + (Aφ/ rSθ) (∂φf) Spherical Coordinates Summary Page (see application to sphericals .doc) PhL 9.9.08 x = x = rSθCφ r2 = x2+ y2 + z2 x = y = rSθSφ tan2θ = (x2+ y2)/z2 x = z = rCθ tanφ = y/x gθθ = Qθ2 = r2 Qθ = r gφφ = Qφ2= r2Sθ2 Qφ= rSθ J = = v = r2Sθ grr = Qr2 = 1 Qr = 1 dxdydz = r2Sθ drdθdφ = r2dr d(cosθ)dφ ( ) = = = Rz(φ) Ry(θ) (f) = (∂rf) + (1/r) (∂θf) + (1/rSθ) (∂φf) A = ∂r(r2Ar) + ∂θ(Sθ Aθ) + ∂φAφ 2(f) = (f) = ∂r (r2 ∂rf) + ∂θ(Sθ ∂θf) + ∂φ2f x A = {∂θ(Sθ Aφ) – ∂φAθ} + {(∂φAr) – ∂r(rAφ)} + {∂r(rAθ) – ∂θAr }