A Review of Appendix B
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Short working note dated 11.26.16 reviewing Appendix C of the 10.24.12 frames doc release. Phil proves the product rule for the time derivative of a tensor AB by taking components, then redoes the vector G rule more simply. He extends it to a rank-2 tensor with epsilon notation, obtaining (∂S T)ab = (∂S' T)ab + ε_ars ω_r T_sb + ε_brs ω_r T_as, matching his earlier result.
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A Review of Appendix C ( as in the 10.24.12 release) PhL 11.26.16
I am looking for a G rule for tensors. I start with a claimed Lemma
Lemma 1: (d[AB]/dt)S = (dA/dt)S B + A (dB/dt)S (C.1)
What does the reader make of the symbol which has suddenly appeared? I could just say
[AB]ij = AiBj
and this is then a rank-2 tensor and that would explain . I write
AB = [AB]ab ea eb (C.4)
where en are the frames doc Frame S basis elements and I point out that ∂S(en) = 0. I claim that
[(d[AB]/dt)S]ab = [(d[AB]ab/dt)S] . (C.3)
I don't prove this, but why not prove it directly in this manner
∂S(AB) = ∂S { [AB]ab ea eb }
= ∂S { (AB)ab } ea eb
Then
[∂S(AB)]ij = ∂S { (AB)ab } [ea eb ]ij
= ∂S { (AB)ab } (ea)i(eb)j
= ∂S { (AB)ab }δa,i δb,j
= ∂S [(AB)ij]
This is really what (C.3) says. You can differentiate and then take component, or you can take component and then differentiate. I have presented this poorly in frames doc I think. Now I could go on with the above
= ∂S [AiBj]
= (∂SAi)Bj + Ai (∂SBj) // just d/dt and product rule
= (∂SA)i Bj + Ai (∂SB)j // using a previous result (d/dt on components)
=[(∂SA) B]ij + [A(∂SB)]ij
= [ (∂SA) B + A(∂SB)]ij
and this I have shown that
[∂S(AB)]ij = [ (∂SA) B + A(∂SB)]ij
and therefore for the tensor itself
∂S(AB) = (∂SA) B + A(∂SB)
and this is the claim of Lemma 1, so I am happy with it.
Just fiddling now:
V = Viei
∂SV = (∂SVi) ei + Vi (∂Sei) = (∂SVi) ei = (∂tVi) ei
On the other hand
∂S'V = (∂S'Vi) ei + Vi (∂S'ei) = (∂tVi)ei + Vi[ – ω x ei]
= (∂tVi)ei - ω x (Viei)
= ∂SV - ω x V
This seems to be a much simpler proof of the G Rule.
Now try this for a tensor
T = Tij eiej
∂ST = (∂tTij) eiej
On the other hand
∂S'T = (∂tTij)eiej + Tij ∂S'(eiej)
Now I need a Lemma which shows this
∂t(AB) = (∂tA) B + A (∂tB)
which I would show by looking at components. Then I can apply this to get
∂S'(eiej) = (∂S'ei) ej + ei (∂S'ej)
= (– ω x ei) ej + ei (– ω x ej)
I don't think I can take this further, so insert into the above to get
∂S'T = (∂tTij)eiej + Tij ∂S'(eiej)
= (∂tTij)eiej + Tij [(– ω x ei) ej + ei (– ω x ej)]
= ∂ST – Tij [(ω x ei) ej + ei ( ω x ej)]
Lets do ε notation to write
(∂S'T)ab = (∂ST)ab - Tij [(ω x ei) ej + ei ( ω x ej)]ab
= (∂ST)ab - Tij [(ω x ei)a(ej)b + (ei)a ( ω x ej)b]
= (∂ST)ab - Tij (ω x ei)aδj,b - Tijδi,a ( ω x ej)b]
= (∂ST)ab - Tib (ω x ei)a - Taj ( ω x ej)b]
= (∂ST)ab - Tib εarsωr(ei)s - Taj εbrsωr(ej)s]
= (∂ST)ab - Tib εarsωrδi,s - Taj εbrsωrδj,s]
= (∂ST)ab - Tsb εarsωr - Tas εbrsωr
= (∂ST)ab - εarsωrTsb - εbrsωrTas
So then
(∂ST)ab = (∂S'T)ab + εarsωrTsb + εbrsωrTas
This agrees with what I already quote
(∂SA)ab = (∂S'A)ab + εarsωrAsb + εbrsωrAas 2