section 1J support

Repair of Section 1 (j) PhL 8.17.12 I decided not to put the derivations of (1.38) into frames doc since they are so unimportant and the derivations would be a very boring appendix. But I want the derivations archived in case someone finds an error or claims one. Collection of useful facts: (eix ej)k = εijk (ω)'i = Rijωj (eix e'j)k = εkibRjb (ω)'k εkabRiaRjb = εsijRsk Items 1 and 2 Footnote: Here are details of the above four calculations which the reader is invited to ignore : 1. (da/dt )S = (d [(a)iei] /dt )S = (d(a)i/dt)S ei + (a)i (dei/dt )S = (d(a)i/dt) ei where we use (1.27) . Taking the jth Frame S component of the above, [(da/dt )S]j = {(d(a)i/dt) ei} ej = (d(a)i/dt) δij = (d(a)j/dt) . // item 1 Taking the jth Frame S' component on the other hand gives this: [(da/dt )S]'j = {(d(a)i/dt) ei} e'j = (d(a)i/dt) Rji // item 2 making use of the last line in (1.7). _____________________________________________________________________________ Obscure rotation theorem from matrix Addendum : εabc Rb'b Rc'c = εa'b'c'Ra'a 2. (da/dt )S' = (d [ (a)iei] /dt )S' = (d(a)i/dt)S' ei + (a)i (dei/dt )S' = (d(a)i/dt) ei – (a)i ω x ei using (1.28) in the last step. Taking the jth Frame S component of the above, [(da/dt )S']j = {(d(a)i/dt) ei – (a)i ω x ei } ej . But (ω x ei) ej = ω (eix ej) = ωk (eix ej)k . We compute, (eix ej)k = εkab(ei)a(ej)b = εkabδiaδjb = εkij = εijk (easy to remember) = εjki so that (ω x ei) ej = ωk (eix ej)k = ωk εjki . Therefore [(da/dt )S']j = (d(a)i/dt) δij – (a)i ωk εjki = (d(a)j/dt) - εjki ωk(a)i or [(da/dt )S']j = (d(a)j/dt) - ωk εkij (a)i // item 3 Now taking the jth Frame S' component of our first line we get [(da/dt )S']'j = {(d(a)i/dt) ei – (a)i ω x ei } e'j . But (ω x ei) e'j = ω (eix e'j) = ωk (eix e'j)k . We compute, using the last line of (1.7), (eix e'j)k = εkab(ei)a(e'j)b = εkabδiaRjb = εkibRjb so that (ω x ei) e'j = ωk (eix e'j)k = ωk εkibRjb. Therefore, further using (1.7), [(da/dt )S']'j = (d(a)i/dt) Rji – ωk εkib(a)iRjb // item 4 _____________________________________________________________________________ We can summarize our results to this point: [(da/dt )S]j = (d(a)j/dt) [(da/dt )S]'j = (d(a)i/dt) Rji [(da/dt )S']j = (d(a)j/dt) - εjki ωk(a)i [(da/dt )S']'j = (d(a)i/dt) Rji – ωk εkib(a)i Rjb Now what happens if we redo all of this using the other expansion for a ? ________________________________________________________________ 1. (da/dt )S = (d [(a)'ie'i] /dt )S = (d(a)'i/dt)S e'i + (a)'i (de'i/dt )S = (d(a)'i/dt)S e'i + (a)'i ω x e'i where we use (1.25) . Taking the jth Frame S component of the above, [(da/dt )S]j = { (d(a)'i/dt)S e'i + (a)'i ω x e'i } ej = (d(a)'i/dt)S e'i ej + (a)'i ω (e'i x ej) = (d(a)'i/dt)S Rij – (a)'i ω (ej x e'i) = (d(a)'i/dt)S Rij – (a)'i ωk (ej x e'i)k = (d(a)'i/dt)S Rij – (a)'i ωk εkjbRib We can write this another way as follows, using Rksωs = (ω)'k so Rsk (ω)'s = ωk, ωk εkjbRib = Rsk (ω)'s εkjbRib = (ω)'s [εkjb RskRib] = (ω)'s [εjbk RibRsk] = (ω)'s [εais Raj] Then we have [(da/dt )S]j = (d(a)'i/dt)S Rij – (a)'i (ω)'s [εais Raj] or [(da/dt )S]j = (d(a)'i/dt)S Rij – (a)'i (ω)'k [εmik Rmj] or [(da/dt )S]j = (d(a)'i/dt)S Rij + (a)'i (ω)'k [εkim Rmj] // item 8 Resuming, taking the jth Frame S' component on the other hand gives this: [(da/dt )S]'j = {(d(a)'i/dt)S e'i + (a)'i ω x e'i } e'j = = (d(a)'i/dt)S e'i e'j + (a)'i (ω x e'i) e'j = (d(a)'i/dt)S δij + (a)'i ω (e'ix e'j) = (d(a)'j/dt)S + (a)'i ωk (e'ix e'j)k We compute, (e'ix e'j)k = εkab(e'i)a(e'j)b = εkabRiaRjb = εsijRsk where the last equality follows from an obscure theorem about rotation matrices. Therefore [(da/dt )S]'j = (d(a)'j/dt)S + (a)'i ωk εsijRsk But (ω)'s = Rsk ωk so we can write the last result as [(da/dt )S]'j = (d(a)'j/dt)S + (a)'i (ω)'s εsij or [(da/dt )S]'j = (d(a)'j/dt)S + (a)'i (ω)'k εkij // item 7 ___________________________________________________________________ 2. (da/dt )S' = (d [ (a)'ie'i] /dt )S' = (d(a)'i/dt)S' e'i + (a)'i (de'i/dt )S' = (d(a)'i/dt) e'i using (1.26) in the last step. Taking the jth Frame S component of the above, [(da/dt )S']j = { (d(a)'i/dt) e'i } ej = (d(a)'i/dt) Rij // item 6 Now taking the jth Frame S' component of our first line we get [(da/dt )S']'j = {(d(a)'i/dt) e'i } e'j = (d(a)'i/dt) δij = (d(a)'j/dt) // item 5 The rules to get from first group to the second are S↔S' on both labels and component index types ω → -ω R → R-1 = RT