new appendix B

This has been installed, do not edit. This was appendix C and is now appendix B. Appendix C: The G Rule for a Tensor of Rank n In this section we continue to use these shorthand operator notations, ∂S ≡ (d/dt)S ∂S' ≡ (d/dt)S' ∂t ≡ (d/dt) . (1.31) (C.1) Recall from (1.37) that for a scalar function or for a component of any tensor one has, (∂STijk...) = (∂S'Tijk...) = (∂tTijk...) , (1.37) (C.2) and from (2.9) that ∂S'ei = – ω x ei . (2.9) (C.3) So far we know all about the G Rule for tensors of rank 0 and 1 (scalar and vector), ∂SA = ∂S'A // = ∂tA; A is a scalar function of t (C.2) ∂SA = ∂S'A + ω x A . // A is a vector (2.8) (C.4) What happens for tensors of rank 2 or more? To explore this question, we first collect a few facts. A tensor of rank-n has the following expansion (where q represents the nth letter of the alphabet) T = Σabc..q.Tabc..q (eaebec.....eq) (C.5) Here T has n subscripts and there is a tensor product of n basis vectors. This expansion is a generalization of the expansion of a vector, V = ΣiViei . (C.6) The basis vectors are axis-aligned unit vectors, so we have (en)i = δn,i . (1.3) (C.7) The meaning of the symbol is nothing more than the following, [AB....Q ]abc...q = AaBbCc....Qq . (C.8) This particular tensor AB....Q happens to be the direct product of the n vectors A,B,C...Q, but there are of course general tensors T which cannot be written as such a direct product. As a simple tensor product example, (eaeb)ij = (ea)i (eb)j = δa,i δb,j . (C.9) Here is the expansion of the direct product tensor T = AB, AB = Σab [AB]ab eaeb = Σab AaBb eaeb . (C.10) Before continuing, we develop two very similar Lemmas: Lemma 1: ∂S(AB) = (∂SA) B + A(∂SB) (C.11) where A and B are vectors whose components are functions of time. Proof of Lemma 1: ∂S(AB) = ∂S( Σab AaBb eaeb ) // expansion (C.10) = Σab ∂S(AaBb) eaeb // since ∂Sen = 0 by (1.27) = Σab ∂t(AaBb) eaeb // (C.2) = Σab [(∂tAa)Bb + Aa(∂tBb)] eaeb // regular calculus product rule = Σab [(∂SAa)Bb + Aa(∂SBb)] eaeb // restore ∂S using (C.2) = Σab [(∂SA)aBb + Aa(∂SB)b] eaeb /// commutation rule (1.40b) = Σab [(∂SA)aBb] eaeb + Σab Aa(∂SB)b eaeb // write as two terms = (∂SA) B + A(∂SB) . // expansion (C.10) QED Recall from (1.5) that one can expand a vector A on either the ei or the e'i basis, A = ΣiAiei = ΣiA'ie'i . (1.5) (C.12) Similarly, one can expand the tensor T of (C.5) on the e'i basis to get T = Σabc..q.T'abc..q (e'ae'be'c.....e'q) (C.13) with this special case AB = Σab [AB]'ab e'ae'b = Σab A'aB'b e'ae'b . (C.14) We then have Lemma 2 which is the same as Lemma 1 but with S→S' , Lemma 2: ∂S'(AB) = (∂S'A) B + A(∂S'B) (C.15) The proof exactly follows that of Lemma 1 but we show it anyway: ∂S'(AB) = ∂S'( Σab A'aB'b e'ae'b ) // expansion (C.14) = Σab ∂S'(A'aB'b) e'ae'b // since ∂S'e'n = 0 by (1.26) = Σab ∂t(A'aB'b) e'ae'b // (C.2) = Σab [(∂tA'a)B'b + A'a(∂tB'b)] e'ae'b // regular calculus product rule = Σab [(∂S'A'a)B'b + A'a(∂S'B'b)] e'ae'b // restore ∂S' using (C.2) = Σab [(∂S'A)'aB'b + A'a(∂SB)'b] e'ae'b // commutation rule (1.40b) = Σab [(∂S'A)'aB'b] e'ae'b + Σab A'a(∂SB)'b e'ae'b // write as two terms = (∂S'A) B + A(∂SB) . // expansion (C.14) QED G Rule for a Rank-2 Tensor Recall from (C.5) that, T = ΣabTab (eaeb) . (C.5) (C.16) Then apply ∂S, using the facts ∂Sen = 0 and (C.2), (∂ST) = Σab(∂STab) (eaeb) = Σab(∂tTab) (eaeb). (C.17) On the other hand, (∂S'T) = Σab (∂S'Tab) (eaeb) + ΣabTab ∂S'(eaeb) = Σab (∂tTab) (eaeb) + ΣabTab ∂S'(eaeb) // (C.2) = (∂ST) + ΣabTab ∂S'(eaeb) . // (C.17) (C.18) But, ∂S'(eaeb) = (∂S'ea) eb + ea(∂S'eb) // by Lemma 2 (C.15) = – [ (ω x ea) eb + ea (ω x eb) . // (C.3) (C.19) Inserting this last result into (C.18) and swapping sides then gives the G rule for a rank-2 tensor, (∂ST) = (∂S'T) + ΣabTab [ (ω x ea) eb + ea (ω x eb) ] . (C.20) We can write the G rule for a vector T (rank-1 tensor) in a similar form, (∂ST) = (∂S'T) + ω x T = (∂S'T) + ω x (ΣaTaea) = (∂S'T) + ΣaTa ω x ea (C.21) G Rule for a Rank-n Tensor It is easy to show that the above general pattern applies to tensors of rank 3 and higher, and one ends up with the following lowest tensor G rules, where the tensor rank is shown on the left . 0 (∂ST) = (∂S'T) 1 (∂ST) = (∂S'T) + ΣaTa ω x ea 2 (∂ST) = (∂S'T) + ΣabTab [ (ω x ea) eb + ea (ω x eb) ] 3 (∂ST) = (∂S'T) + ΣabcTabc [ (ωxea)ebec + ea(ωxeb)ec + eaeb(ωxec) ] (C.22) These G rule results may also be expressed in components. For example: 1 (∂ST)i = (∂S'T)i + ΣaTa (ω x ea)i = (∂S'T)i + ΣaTa εirs ωr (ea)s = (∂S'T)i + ΣaTa εirs ωr δa,s = (∂S'T)i + εirs ωrTs . (C.23) 2 (∂ST)ij = (∂S'T)ij + ΣabTab [ (ω x ea) eb + ea (ω x eb) ]ij = (∂S'T)ij + ΣabTab [ (ω x ea)i (eb)j + (ea)i(ω x eb)j ] = (∂S'T)ij + [ Σab Tab (ω x ea)i δb,j + Σab Tab δa,i (ω x eb)j ] = (∂S'T)ij + [ Σa Taj (ω x ea)i + Σb Tib (ω x eb)j ] = (∂S'T)ij + [ Σa Taj εirsωr(ea)s + Σb Tib εjrsωr(eb)s ] = (∂S'T)ij + [ Σa Taj εirsωrδa,s + Σb Tib εjrsωrδb,s ] = (∂S'T)ij + [ Tsj εirsωr+ Tis εjrsωr ] = (∂S'T)ij + εirsωr Tsj + εjrsωrTis . (C.24) We shall do one more case to establish the general pattern, 3 (∂ST)ijk - (∂S'T)ijk = ΣabcTabc [ (ωxea)ebec + ea(ωxeb)ec + eaeb(ωxec) ]ijk = ΣabcTabc [ (ωxea)i(eb)j(ec)k + (ea)i(ωxeb)j(ec)k + (ea)i(eb)j(ωxec)k ] = ΣabcTabc(ωxea)iδb,jδc,k + ΣabcTabcδa,i(ωxeb)jδc,k + ΣabcTabcδa,i δb,j(ωxec)k = ΣaTajk(ωxea)i + ΣbTibk(ωxeb)j + ΣcTijc(ωxec)k = ΣaTajkεirsωr (ea)s + ΣbTibkεjrsωr (eb)s + ΣcTijcεkrsωr (ec)s = ΣaTajkεirsωr δa,s + ΣbTibkεjrsωr δb,s + ΣcTijcεkrsωrδc,s = Tsjkεirsωr+ Tiskεjrsωr + Tijsεkrsωr so we conclude that (∂ST)ijk = (∂S'T)ijk + εirsωrTsjk + εjrsωrTisk + εkrsωrTijs . (C.25) Looking at the last two results we can infer the rank-4 result (∂ST)ijkl = (∂S'T)ijkl + εirsωrTsjkl + εjrsωrTiskl + εkrsωrTijsl + εlrsωrTijks (C.26) Summary of G rule for all ranks (tensor form) (C.27) 0 (∂ST) = (∂S'T) 1 (∂ST) = (∂S'T) + ΣaTa ω x ea 2 (∂ST) = (∂S'T) + ΣabTab [ (ω x ea) eb + ea (ω x eb) ] 3 (∂ST) = (∂S'T) + ΣabcTabc [ (ωxea)ebec + ea(ωxeb)ec + eaeb(ωxec) ] 4 (∂ST) = (∂S'T) + ΣabcdTabcd [ (ωxea)ebeced + ea(ωxeb)eced + eaeb(ωxec)ed + eaebec(ωxed) ] etc. Summary of G rule for all ranks (component form) (C.28) 0 (∂ST) = (∂S'T) 1 (∂ST)i = (∂S'T)i + εirs ωrTs = (∂S'T)i + (ω x T)i 2 (∂ST)ij = (∂S'T)ij + εirsωr Tsj + εjrsωrTis 3 (∂ST)ijk = (∂S'T)ijk + εirsωrTsjk + εjrsωrTisk + εkrsωrTijs 4 (∂ST)ijkl = (∂S'T)ijkl + εirsωrTsjkl + εjrsωrTiskl + εkrsωrTijsl + εlrsωrTijks etc. In the main body of our document we deal only with vector quantities r, v and a , but one can easily imagine "frames of reference problems" which deal also with tensors, such as the inertia tensor.