Better B field expressions REVIEWED

B field presentation PhL 9.20.14 Summary of this Doc: Here I set out to simplify the B field presentation in Appendix D. To do this, two new items have to be computed, since they appear in the curl E equations: ∂rfm = β' ∂xfm(x) ∂r(rEθ) = (β' ∂x) [ (x/β') hm(x) ] = ∂x (x hm(x)) I show that ∂x (x hm(x)) = x em - m gm where new em ≡ [ + ] ∂xfm = fm [ (m/x) - ] I then end up with Br(r,m) = - (1/4) (a/ω) ηm I Rdc { r-1m (β') fm + k2 hm } Bθ(r,m) = (j/4)(a/ω) ηm I Rdc {k2 gm - β'2fm [ (m/x) - Jm+1(x)/Jm(x)] } Bz(r,m) = (j/4)(a/ω) ηm I Rdc {k β' em } and this is how the results appear in new (D.4.13.), but in a different order. I then start the task of using Maple to verify the above solutions, but that task is carried out in the main text and not below. I used an intermediate file "new section D_4 D_5 and D_6 on B fields.doc" along with the Maple file " B field verify 9_20.mws ". I am not happy with the way I present the B fields in Appendix D. The E fields are so compact, Second summary of the E field solutions : Rdc = β'2 = β2 - k2 (D.2.33) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (ak) gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (ak) hm hm = [ - ] Can I come up with something compact like this for the B fields? After all, it is just this: -jω B = curl E -jωB(r,m) = [ r-1jmEz +jkEθ] + [-jkEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] (D.4.6) Things are pretty simple except for these two objects: ∂rEz ∂r(rEθ) Basically we have to compute x = β'r ∂rfm = β' ∂xfm(x) ∂r(rEθ) = (β' ∂x) [ (x/β') hm(x) ] = ∂x (x hm(x)) Let's try to get Maple to compute these two objects. Let df = ∂xfm(x) and Maple says The first and third term I recognize as + h (x) The first item ∂xfm(x) is pretty simple f'(x) = ∂xfm(x) = Jm'(x) [ - ] ∂rfm = β' Jm'(x) [ - ] and this is just some other function of x and xa for me to add to the list. The second item is ∂x (x hm(x)) = x h'm(x) + hm(x) and h'm(x) = - Now we then get x h'm(x) + hm(x) = x [ - ] + - = - I do know that from Schaum p 137 24.19 and 24.20 that x J'm+1 = x Jm - (m+1)Jm+1 x J'm-1 = (m-1)Jm-1 - x Jm Then = = = = Then x h'm(x) + hm(x) = - = x [ + ] - m [ + ] = x [ + ] - m gm which is pretty simple. What single letter should I use for this sum. It is sort of f-Plus. Perhaps e. Here is Maple code to confirm the above. First just enter Then we want to show that ∂x (x hm(x)) = x em - m gm So that was not so bad. Meanwhile The last line looks like df =(m/x) f m - Jm+1(x) [ - ] = (m/x) fm - fm = fm [ (m/x) - ] Let's try to verify that: Here then are my results: First, ∂xfm(x) = fm [ (m/x) - ] ∂x (x hm(x)) = x em - m gm Then ∂rfm = β' ∂xfm(x) = β' fm [ (m/x) - ] We are then going to get ∂rEz = (1/4) ηm I Rdc (aβ') ∂rfm = (1/4) ηm I Rdc (aβ') β' fm [ (m/x) - ] = (1/4) ηm I Rdc (a) β'2 fm [ (m/x) - ] dim(RHS) = volts = wrong Next ∂r(rEθ) = ∂r(r(1/4) ηm I Rdc (ak) hm) = (1/4) ηm I Rdc (ak) ∂r(r hm) = (1/4) ηm I Rdc (ak) ∂x(x hm) = (1/4) ηm I Rdc (ak) [ x em - m gm] To summarize, our two items needed are ∂rEz = (1/4) ηm I Rdc (a) β'2 fm [ (m/x) - ] dim(RHS) = volts = wrong! ∂r(rEθ) = (1/4) ηm I Rdc (ak) [ x em - m gm] We can then state our curl stuff: B(r,m) = (j/ω){ [ r-1jmEz +jkEθ] + [-jkEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] } (D.4.6) Br(r,m) = (j/ω)[ r-1jmEz +jkEθ] = (j/ω)[ r-1jm (1/4) ηm I Rdc (aβ') fm +jk(1/4) ηm I Rdc (ak) hm] = (1/4) ηm I Rdc (j/ω) [ r-1jm (aβ') fm +jk (ak) hm] = (1/4) ηm I Rdc (j/ω) ja [ r-1m (β') fm + k2hm] = - (1/4) ηm I Rdc (a/ω) [ r-1m (β') fm + k2 hm] which is a pretty simple result really! Check dimensions dim(RHS) = volts/m * m-sec * 1/m2 = volts-sec/m2 OK Let's try the next one Bθ(r,m) = (j/ω)[ -jkEr - ∂rEz ] = (j/ω)[ -jk(j/4) ηm I Rdc (ak) gm - (1/4) ηm I Rdc (a) β'2 fm [ (m/x) - ] ] = (1/4)(j/ω) ηm I Rdc { -jk(j) (ak) gm - (a) β'2 fm [ (m/x) - ] } = (1/4)(j/ω) ηm I Rdc {k (ak) gm - (a) β'2 fm [ (m/x) - ] } = (1/4)(j/ω) ηm I Rdc {ak2 gm - a β'2 fm [ (m/x) - ] } = (j/4)(a/ω) ηm I Rdc {k2 gm - β'2fm [ (m/x) - Jm+1(x)/Jm(x)] } This is a little messier, but still not too bad. Finally, Bz(r,m) = (j/ω) [ r-1∂r(rEθ) - r-1jmEr ] = (j/ω) [ r-1(1/4) ηm I Rdc (ak) [ x em - m gm] - r-1jm(j/4) ηm I Rdc (ak) gm ] = (1/4) (j/ω) ηm I Rdc [ r-1 (ak) [ x em - m gm] - r-1jm (j) (ak) gm ] = (j/4) (a/ω) ηm I Rdc [ r-1 (k) [ x em - m gm] - r-1jm (j) (k) gm ] = (j/4) (a/ω) ηm I Rdckr-1 [ [ x em - m gm] +m gm ] = (j/4) (a/ω) ηm I Rdckr-1 [ x em - m gm +m gm ] = (j/4) (a/ω) ηm I Rdck r-1 [ x em] with a lucky cancellation at the very end! I will of course have to check all these things somehow. Results so far are Br(r,m) = - (1/4) (a/ω) ηm I Rdc { r-1m (β') fm + k2 hm } Bθ(r,m) = (j/4)(a/ω) ηm I Rdc {k2 gm - β'2fm [ (m/x) - Jm+1(x)/Jm(x)] } Bz(r,m) = (j/4)(a/ω) ηm I Rdc {k r-1 x em } = (j/4)(a/ω) ηm I Rdc {k β' em } I will of course have to check these results carefully, but the presentation is compact! This is very much better than what I have in Appendix D right now. Dimensions all OK. Phase 2. Let's assume the above are correct and see if I can make Maple verify them! To simplify things, lets just delete the following factor which is common to all Ei and all Bi fields (1/4) ηm I Rdc a Then we can say Ez(r,m) = β' fm Er(r,m) = j k gm Eθ(r,m) = k hm -jωBr(r,m) = j ( r-1m (β') fm + k2 hm ) -jωBθ(r,m) = (k2 gm - β'2fm [ (m/x) - Jm+1(x)/Jm(x)] ) -jωBz(r,m) = (k r-1 x em ) = kr-1β'r em = k β' em These six fields should satisfy the curl E equation, and it is now our task to check -jω B = curl E [ r-1jmEz +jkEθ] + [-jkEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] So it is my task to show these three things: -jωBr(r,m) = [curl E]r -jωBθ(r,m) = [curl E]θ -jωBz(r,m) = [curl E]z In Maple lets use -jωBr = mjwBr and so on Then the equations are mjwBr = curlEr mjwBth = curlEth mjwBz = curlEz