magnetostatics and the round wire REVIEWED

Magnetostatics and the Round Wire PhL 11.6.13 (11.15.13) This was my breakthrough calculation showing how, when you compute A for a round wire including the surface current, you get an A which is linear in μ1 outside the wire, but linear in μ2 inside the wire. This was always a mystery to me, though I knew it somehow had to come out that way. It was not until I was aware of the surface current that I could get this result. The order of this file is a little deceiving. I really did Section 5 first to discover the surface current (first time ever I did that), then I used that in the 3D analysis of the sections 1-4 to compute A, then B, then H. I was truly amazed when H was correct. In the last section I use the 2D method to compute A and get the same answer apart from constant Λ terms. All the material here is now in Appendix B. There I did the 2D method, and left the 3D method as a reader exercise with waypoints in the last section of App B. Those waypoints are derived in a second short document that should not get lost. So I think this document can be marked REVIEWED. For me it as pretty significant on my stumbling path. 1. Setup of the Problem of finding A for a round wire. 1 2. Evaluation of the two integrals 3 3. Install integral results into the A formula 6 4. Verify the Az results. 7 5. The Surface Current 7 6. Surface current for a non-round wire. 10 7. Redo this all using 2D propagator 14 1. Setup of the Problem of finding A for a round wire. Here is my picture of interest, end view of the round wire I have my reliable B&B open to page 143. They quickly arrive at (using my μ scale) 2A(x) = -μJ(x) B&B (5.46) magnetostatics where μ is that appropriate for point x in space. Now if I regard the surface current K discussed below to be part of region 1, I can write 2A(x) = -μ1Jm(x) x in region 1 2A(x) = -μ2Jc(x) x in region 2 Now I can add these equations to get an equation for region 1+2: 2A(x) = -μ1Jm(x) -μ2Jc(x) // compare to (B.6.1) which is 2D version where the currents don't exist in the same region, but that is fine. We then find A(x) = (1/4π) ∫dV' [μ1Jm(x') +μ2Jc(x')] (1/R) R = |x-x'| and now I think we are getting somewhere. This is the generalization of A = (1/4π) ∫dV (μJ) (1/R) B&B (5.47) so my situation. Now I can rewrite the above as A(x) = (1/4π) ∫dS' [μ1K(x')] (1/R) + (1/4π) ∫dV' [μ2Jc(x')] (1/R) In our case, everything is in the z direction. I should maybe have said that earlier. But in Cartesians the above is OK as stated. So Az(x) = (1/4π) ∫dS' [μ1Kz(x')] (1/R) + (1/4π) ∫dV' [μ2Jcz(x')] (1/R) Now we know that (round wire, DC ) ( the K result comes from below) Jcz(x') = Jcz = I/(πa2) Kz(x') = Kz = (1-μ2/μ1) Ht1 = (1-μ2/μ1) (1/2πa) I Then we have for all of region 1 and 2 Az(x) = (1/4π) μ1Kz ∫dS' (1/R) + (1/4π) μ2Jcz ∫dV' (1/R) // compare (B.6.4) where ∫dS' is over the entire wire surface, and ∫dV' is over the entire wire volume. Installing the above values we rewrite as Az(x) = (1/4π) μ1(1-μ2/μ1) (1/2πa) I ∫dS' (1/R) + (1/4π) μ2 I/(πa2) ∫dV' (1/R) or Az(x) = (1/4π) (μ1-μ2) (1/2πa) I ∫dS' (1/R) + (1/4π) μ2 I/(πa2) ∫dV' (1/R) = (1/4π) (1/2πa) I { (μ1-μ2) ∫dS' (1/R) + (2/a) μ2∫dV' (1/R) } = (1/4π) (1/2πa) I { (μ1-μ2) I1 + (2/a) μ2I2 } // defines two integrals 2. Evaluation of the two integrals The Integrals (this is all just in cross section) I1 ≡ ∫surfdS' (1/R) = ∫(adθ') !Syntax Error, Idz' (1/R) I2 ≡ ∫dV' (1/R) = ∫(r'dθ') ∫dr'!Syntax Error, Idz' (1/R) We now access information in some other docs. In cyl coords we know that |x -x'|2 = R2 = r2 + r'2 - 2rr'cos(θ-θ') + (z-z')2 If we choose to observe things from z = 0 we get R2 = r2 + r'2 - 2rr'cos(θ-θ') + z'2 = s2 + z'2 s2 = (x-x')2+ (y-y')2 R = Then we have I1 ≡ ∫(adθ') !Syntax Error, Idz' (1/ ) = a !Syntax Error, I dθ' !Syntax Error, Idz' (1/ ) I2 ≡ ∫(r'dθ') ∫dr'!Syntax Error, Idz' (1/ ) = !Syntax Error, I r'dr' !Syntax Error, I dθ' !Syntax Error, Idz' (1/ ) and we have the same dθ'dz' integral in both cases. I show in appendix C that !Syntax Error, Idz' (1/ ) = -2ln(s/Λ) = -2ln(/Λ) First work on I2: I2 = !Syntax Error, I r'dr' !Syntax Error, Idθ' [-2ln(/Λ) ] = - 2 !Syntax Error, Ir'dr' [!Syntax Error, Idθ' ln[(r'2+r2- 2rr' cosθ')/Λ2] ] where I just drop the Λ stuff since it just makes a constant which we drop in B = curl A. So I2 = - 2!Syntax Error, Ir'dr' !Syntax Error, Idθ' ln(r'2+r2- 2rr' cosθ') I then show in verifications that !Syntax Error, Idθ' ln(r'2+r2- 2rr' cosθ') = π ln { [(r'2+r2) + | r'2- r2|] /2 } so that I2 = - 2π !Syntax Error, Ir'dr' ln { [(r'2+r2) + | r'2- r2|] /2 } I show in verifications that !Syntax Error, Ir'dr' ln { [(r'2+r2) + | r'2- r2|] /2 } = a2 ln(r) for r > a which means r' < r !Syntax Error, Ir'dr' ln { [(r'2+r2) + | r'2- r2|] /2 } = (1/2)r2 for r <a where we always drop constants since they don't affect B = curl A. So we conclude that I2 = - 2π [ a2ln(r) θ(r>a) + (1/2)r2θ(r<a) ] Now what about the I1 integral? I1 ≡ ∫(adθ') !Syntax Error, Idz' (1/ ) = a !Syntax Error, I dθ' !Syntax Error, Idz' (1/ ) = 2a !Syntax Error, I dθ' !Syntax Error, Idz' (1/ ) As before, the dz' integral gives -2ln(/Λ) so we then have I1 = 2a !Syntax Error, I dθ' [-2ln(/Λ)] = -2a !Syntax Error, I dθ' ln[(r'2+r2- 2rr' cosθ')/Λ2] = -2a !Syntax Error, I dθ' ln[(r'2+r2- 2rr' cosθ')] Now I showed above that !Syntax Error, Idθ' ln(r'2+r2- 2rr' cosθ') = π ln { [(r'2+r2) + | r'2- r2|] /2 } so it seems then that we have I1 = -2πa ln { [(r'2+r2) + | r'2- r2|] /2 } Now the I1 is in our surface integral and there we have r' = a, so this says I1 = -2πa ln { [(a2+r2) + | a2- r2|] /2 } So this has two regions as well. If a > r then I1 = -2πa ln { [(a2+r2) + (a2- r2)] /2 } = -2πa ln(a2) r < a I1 = -2πa ln { [(a2+r2) - (a2- r2)] /2 } = -2πa ln(r2) r > a So I find that I1 = -2πa { ln(a2)θ(r<a) + ln(r2)θ(r>a) } so backtrack now to Az(r) = (1/4π) (1/2πa) I { (μ1-μ2) I1 + (2/a) μ2I2 } where we have so found above that I1 = -2πa { ln(a2)θ(r<a) + ln(r2)θ(r>a) } I2 = - 2π [ a2ln(r) θ(r>a) + (1/2)r2θ(r<a) ] Rewrite I2 so same θ: I1 = -2πa [ ln(a2)θ(r<a) + ln(r2)θ(r>a) ] I2 = - 2π [(1/2)r2θ(r<a) + a2ln(r) θ(r>a) ] 3. Install integral results into the A formula We then get two results for Az(r) depending on the size of r. Installing Az(r) = (1/4π) (1/2πa) I { (μ1-μ2) I1 + (2/a) μ2I2 } For r > a (outside) we get Az(r) = (1/4π) (1/2πa) I { (μ1-μ2) [-2πa ln(r2)] + (2/a) μ2[- 2π a2ln(r)] } = - (1/4π) (1/2πa) I 2 a { (μ1-μ2) [π ln(r2)] + 2πμ2[ln(r)] } = - (1/4π) (1/2πa) I 2 a { (μ1-μ2) [2π ln(r)] + 2πμ2[ln(r)] } = - (1/2) (1/2πa) I 2 a { (μ1-μ2) ln(r) + μ2ln(r)} = - (1/2π) I { (μ1ln(r)} = - (1/2π) I μ1ln(r) r > a outside On the other hand, for r < a we get Az(r) = (1/4π) (1/2πa) I { (μ1-μ2) [-2πa ln(a2)] + (2/a) μ2[- 2π(1/2)r2] } = -(1/2) (1/2πa) I { (μ1-μ2) [a ln(a2)] + (2/a) μ2[(1/2)r2] } = -(1/2) (1/2πa) I { (μ1-μ2) [a ln(a2)] + (1/a) μ2[r2] } = -(1/2) (1/2πa) I { (μ1-μ2) [a ln(a2)] + a μ2[r2/a2] } = -(1/2) (1/2π) I { (μ1-μ2) [ ln(a2)] + μ2[r2/a2] } Now I guess drop the constant term to get = - (1/4π) I μ2[r2/a2] r < a inside So I seem to obtain these expressions for Az(r) by evaluating the Helm integral thing Az(r) = - (1/2π) I μ1ln(r) r > a outside // same as (B.6.17) Az(r) = - (1/4π) I μ2[r2/a2] r < a inside // non-constant part of (B.6.18) I may have errors, but I obtain here an important general result of great interest to me: in the outside formula, you see μ1 appear, whereas in the inside formula you see μ2 appear. 4. Verify the Az results. Now let's try to get B from B = curl A. It happens that B = [ -∂rAz] B = -∂rAz So we compute B1t = (1/2π) I μ1 (1/r) r > a outside B2t = (1/4π) I μ2 2r/a2 = (1/2π) I μ2 (r/a2) r < a inside Then we compute H H1t = (1/2π) I (1/r) r > a outside H2t = (1/2π) I (r/a2) r < a inside But these are just the results you get from curl H = J Ampere's Law: H1t 2πr = I outside H1r 2πr = I (πr2/πa2) = I(r2/a2) inside So FINALLY I see how the 3D Helmholtz type integral solution gives the desired results, and you must include the surface current indeed! So when you compute the results of this problem using the vector potential A = (1/4π) ∫dV (μJ) (1/R) the computation is NON TRIVIAL and you see all the details above. 5. The Surface Current The results of this section were used in Section 1 above. I am going to draw a picture which shows how the B field lines up the mag dipoles to create a surface current. This clearly shows the mechanism which causes a surface current in the opposite direction to J. What is the size of this surface current? Here is a blowup of the rightmost point in the left drawing: so there is a surface current K going away from the observer inside the loop Apply our integral form curl H = ∂tD + J H ds = ∫S [∂tD+J] dA (1.1.23) Now H does not "see" the surface current, so J is only the conduction current. This is a mag static problem so that ∂tD = 0 . We are going to find of course that Ht2 = Ht0 at the boundary We confirm this by directly computing H from Ampere's Law (current is uniform) 2πr Ht2 = (πr2/πa2) I = (r2/a2) I => Ht2 = (1/2πa2) I r r < a Outside the conductor, since H does not "see" K we get 2πr Ht0 = I => Ht0 = (1/2π) I r-1 r > a and the values match at r = a where we have Ht0 = Ht1 = Ht2 = (1/2πa) I. Now how do we compute the surface current K? consider curl B = με ∂tE + μJ B ds = μ ∫S [ε ∂t E + J] dA (1.1.24) curl B = μJ B ds = ∫S μ J dA (1.1.24) where J includes the volume conduction current and the surface current Jm = curl M. Again ∂t E = 0. Write the above as B ds = ∫S μ [Jc + Jm] dA In the limit that s→ 0 the finite volume Jc makes no contribution. The Jm surface current is located entirely in the μ0 region 0, as you would say for a surface charge. Thus ∫S μ [Jc + Jm] dA ≈ μ0 K L Meanwhile B ds = (Bt0)L - (Bt1)L Then we find that μ0K = Bt0 - Bt2 = μ0Ht0 - μ2Ht2 = (μ0-μ2) Ht0 and then the surface current density is, K = (1-μ2/μ0) Ht0 // this derivation is in Appendix B now which goes away if μ2 = μ0. Mystery regarding integral the curl equations Go back to the claim that B "sees" all the currents: curl B = με ∂tE + μJc + μJm (*) // wrong! Divide by μ on the left to get curl H = ε ∂tE + Jc + Jm But this says that H sees Jm which we know is wrong. What gives? Well, what gives is that I have the curl B equation written incorrectly. Start with the curl H equation curl H = ε ∂tE + Jc curl μ0H = μ0ε ∂tE + μ0Jc Then write μ0H = B-μ0M to get curl (B-μ0M) = μ0ε ∂tE + μ0Jc or curl B = μ0ε ∂tE + μ0Jc + μ0 curl M or curl B = μ0ε ∂tE + μ0Jc + μ0 Jm // right This is the corrected version of (*) shown above. Then go do the integral form (1/μ0)curl B = ε ∂tE + Jc + Jm (1/μ0) B ds = ∫S [ε ∂tE + Jc + Jm] dA NOW we can apply this thing to get (1/μ0) [(Bt0)L - (Bt1)L] = KL and then K = (1/μ0) [Bt0 - Bt1] = (1/μ0) [μ0Ht0 - μ1Ht1] = (1/μ0) [μ0 - μ1] Ht1 = (1-μ1/μ0) Ht1 as found above, but not much more clearly! I need to go fix Chapter 1 once again ******* to get all this stuff right. But hold off on that until we finish with the current problem. 6. Surface current for a non-round wire. When I get to Chapter 4, I will have this volume of current in the arbitrarily shaped conductor, Jz1(x,y,z) = b1(x,y) i1(z) Question: Can you compute the surface current knowing the volume current? Consider curl H = ∂tD + J H ds = ∫S [∂tD+J] dA (1.1.23) In a DC situation or perhaps the AC situation with E = 0 inside the conductor, this says H ds = ∫S J dA This would get you the line integral of Ht around a path just inside the surface. If not a round wire, we have no obvious way to recover Ht from this formula at some point on the boundary. So consider instead curl H = J ∂xHy(x,y) - ∂yHx(x,y) = Jz(x,y) This is a 2D PDE where the source is given. Is this a standard equation of some sort? Hold that thought and try this. Apply curl to both sides curl curl H = curl J grad div H - 2H = curl J Now if we always work inside the conductor where there is some fixed μ, then this says (div B = 0) 2H = - curl J and then I want to write my standard Helmholtz solution H(x) = . (A.1.2) Now if J = Jz only, we can say curl F = (∂xFy- ∂yFx) + (∂yFz- ∂zFy) + (∂zFx- ∂xFz) Therefore curl J = (∂xJy- ∂yJx) + (∂yJz- ∂zJy) + (∂zJx- ∂xJz) Now as x' moves around in the above integration (over the entire conductor), we can take our primed unit vectors be same as unprimed -- they are just constants. We also assume J is only a function of x and y, and furthermore we only have Jz . Then Then we can say curl J(x) = (∂yJz(x,y)) + (- ∂xJz(x,y)) curl' J(x') = (∂y'Jz'(x',y')) + (- ∂x'Jz'(x',y')) Then the integral becomes Hx(x,y) = (1/4π) ∫d3x' ∂y'Jz'(x',y')/R R = |x - x'| Hy(x,y) = - (1/4π) ∫d3x' ∂x'Jz'(x',y')/R So we have at least made some progress. In theory we can compute H just inside the boundary using these expressions. We could do the z' integral in our usual manner perhaps. !Syntax Error, Idz' (1/ ) = -2ln(s/Λ) = -2ln(/Λ) But instead of doing that, suppose we do parts in each equation above, and (Jz1/R) = 0 at ≡∞ due to the 1/R. Then meanwhile R2 = (x-x')2 + (y-y')2 +(z-z')2 2RdR = 2(y'-y)dy' => ∂y'R-1 = -R-2∂y'R = -R-2 (y'-y)/R = -R-3(y'-y) So then maybe we get Hx(x,y) = - (1/4π) ∫d3x' Jz'(x',y')[ ∂y'R-1] = (1/4π) ∫d3x' Jz'(x',y') (y'-y)/R3 So then Hx(x,y) =(1/4π) ∫d3x' Jz'(x',y') (y'-y)/R3 Hy(x,y) = - (1/4π) ∫d3x' Jz'(x',y') (x'-x)/R3 so these are some kind of moments of Jz/R3.Now what about the s' integral? !Syntax Error, Idz' (s2+z'2)-3/2 = 1/s2 says Maple, Then we seem to have Hx(x,y) = (1/4π) ∫dx'dy' Jz(x',y') (y'-y)/s2 Hy(x,y) = - (1/4π) ∫dx'dy' Jz(x',y') (x'-x)/s2 s ≡ No matter how I rewrite this, it is always going to require some integral over the volume current density and won't have some trivial form such as Jz near the boundary. So in Chapter 4 we are going to have to "carry" a whole extra term to handle the boundary current if the μ's are not equal. So go look now at Section 4.3 where I now add in a magnetization current Az1(x) = ∫ [Jz1(x') + Jz1m] dx'dy'dz' R = |x - x'| (4.3.1) We then make the same assumption of separation of variables to write Jz1(x,y,z) = b1(x,y) i1(z) A/m2 1/m2 A (4.3.2) where i1 is scaled such that !Syntax Error, Idx dy b1(x,y) = 1 . (4.3.3) and then I have to add Jz1m(x,y,z) = b1m(x,y) i1(z) A/m2 1/m2 A (4.3.2) !Syntax Error, Idx dy b1m(x,y) = λ1 . (4.3.3) where λ is some new constant! In (4.3.2) we have already set the sale of i1 so it is fixed. I could calculate λ for a round wire I suppose. It is dimensionless and is a function of the conductor shape I think. Each conductor has a different λ so they don't cancel when you form W(z). Well this was a first cut at this subject. 7. Redo this all using 2D propagator We start with a 2D magnetostatics equation which is (1.5.4) with β2 = 0, 22DA(x) = -μ1Jm(x) -μ2Jc(x) The particular solution is obtained using the 2D propagator g = (1/2π) ln(1/R) , A = (1/2π) ∫dV' [μ1Jm(x') + μ2Jc(x')] ln(1/R) R = |x-x'| = (1/2π) { ∫dS' [μ1Km(x')] ln(1/R) + ∫dV'[ μ2Jc(x')] ln(1/R) } so Az = (1/2π) { ∫dS' [μ1Kz(x')] ln(1/R) + ∫dV'[ μ2Jcz(x')] ln(1/R) } = -(1/2π) { ∫dS' [μ1Kz(x')] ln(R) + ∫dV'[ μ2Jcz(x')] ln(R) } = -(1/4π) { μ1∫dS' [Kz(x')] ln(R2) + μ2∫dV'[Jcz(x')] ln(R2) } At this point we insert the uniform prescribed current Jcz and our previously computed surface current Kz, Jcz(x') = Jcz = I/(πa2) Kz(x') = Kz = (1-μ2/μ1) (1/2πa) I with the result that Az = -(1/4π) { μ1(1-μ2/μ1) (1/2πa) I ∫dS' ln(R2) + μ2 I/(πa2)∫dV' ln(R2) } = -[I/(4π2a2)] { μ1-μ2) (a/2) ∫dS' ln(R2) + μ2∫dV' ln(R2) } The volume integral (per unit length in the z direction, so really an area integral) ∫dV' ln(R2) = !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln [r'2 +r2-2rr' cos(θ-θ')] = 2 !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln [r'2 +r2-2rr' cos(θ-θ')] = 2 !Syntax Error, Ir' dr' !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] Let us define this x integral as Q(r',r) ≡ !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] so then ∫dV' ln(R2) = 2 !Syntax Error, Ir' dr' Q(r',r) Meanwhile, our surface integral (really a line integral) of interest is ∫dS' ln(R2) = !Syntax Error, I[a dθ'] ln [r'2 +r2-2rr' cos(θ-θ')] |r'=a = 2a !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] |r'=a = 2a Q(a,r) so that both integrals of interest require computation of Q(r',r), which we replicate here, Q(r',r) ≡ !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] This integral may be evaluated using GR7 ***** with a = r'2 +r2 and b = -2rr' and a2-b2 = (r'2-r2)2 so that = | r'2-r2 | .The condition a > |b| > 0 is met since (r±r')2 > 0 => r2+r'2 > ±2rr' which says a > ±b so a > |b|. Thus, Q(r',r) = !Syntax Error, Idx ln [r'2 +r2-2rr' cosx)] = π ln [ ] = = 2π This our two integrals of interest are ∫dS' ln(R2) = 2a Q(a,r) = 4πa ∫dV' ln(R2) = 2 !Syntax Error, Ir' dr' Q(r',r) = 4π !Syntax Error, Ir' dr' where we still have a dr' integral to carry out. For r > a we know that r > r' since r' integrates over the inside of the wire. Then ∫dV' ln(R2) = 4π!Syntax Error, Ir' dr' ln(r) = 4π ln(r) [ a2/2 ] = 2πa2 ln(r) r > a On the other hand, for r < a we have to break the integral into two parts : ∫dV' ln(R2) = 4π !Syntax Error, Ir' dr' ln(r) + 4π !Syntax Error, Ir' dr' ln(r') = 4πln(r) (r2/2) + 4π [(1/2)x2(lnx-1/2) ]|ar = 2π r2 ln(r) + 4π [(1/2)a2(lna-1/2) - (1/2)r2(lnr-1/2)] = 2π r2 ln(r) + 2π [a2(lna-1/2) - r2(lnr-1/2)] = 2π r2 ln(r) + 2π a2lna - πa2 - 2π r2lnr + πr2 = 2π a2lna + π(r2-a2) r < a The result then is the following, where below it we repeat the earlier surface integral result, ∫dV' ln(R2) = ∫dS' ln(R2) = 4πa We can now install these integrals into our Az expression *** to get Az = -[I/(4π2a2)] { μ1-μ2) (a/2) ∫dS' ln(R2) + μ2∫dV' ln(R2) } = -[I/(4π2a2)] { μ1-μ2) (a/2) 4πa + μ2} We can then write these out for the two separate cases Az(r>a) = -[I/(4π2a2)] { μ1-μ2) (a/2) 4πa ln(r) + μ2 2πa2 ln(r) } = -[I/(4π2a2)] { μ1-μ2) 2πa2 ln(r) + μ2 2πa2 ln(r) } = -[I/(2π)] { μ1-μ2) ln(r) + μ2 ln(r) } = -[I/(2π)] { μ1ln(r) } Az(r<a) = -[I/(4π2a2)] { μ1-μ2) (a/2) 4πa ln(a) + μ2( 2πa2ln(a) + π(r2-a2) ) } = -[I/(4πa2)] { μ1-μ2) 2a2 ln(a) + μ22a2ln(a) + μ2(r2-a2) ) } = -[I/(4πa2)] { μ1 2a2 ln(a) + μ2(r2-a2) ) } = -[I/(2π)] { μ1 ln(a) + μ2(r2-a2)/(2a2) ) } At the boundary r = 0 both expressions give Az(r=a) = -[I/(2π)] { μ1ln(a) } The next step is to compute the magnetic field B = curl A given our function Az(r). We find B = curl A = - ∂rAz(r) First, for r > a we get -∂rAz(r) = ∂r[I/(2π) { μ1ln(r) }] = (I μ1/2π) (1/r) r > a then for r < a, -∂rAz(r) = I/(2π) ∂r { μ1 ln(a) + μ2(r2-a2)/(2a2) ) } = (I μ2/2π) μ2 r/a2 Thus B = (I μ1/2π) (1/r) for r > a which is in the medium with μ1 B = (I μ2/2π) (r/a2) for r < a which is in the medium with μ2 Using B = μH in each region we then get H = (I /2π) (1/r) for r > a => Hθ = -(I/2πr) H = (I /2π) (r/a2) for r < a => Hθ = - (Ir/2πa2) Amazingly, these results agree with the elementary calculation results given in ***.