transverse isotropic

Detail for Lai p 325: transverse isotropic tensor PhL 5.21.12 In these notes I take a quick shot (2 hours) at this problem and then give up for now. This needs more advanced methods than I am using right now and it is not in the main path really. _________________________________________________________________________________ Now suppose you have a case of what I might call cylindrical symmetry. The axisym axis is taken to be the z axis. Recall from above C'abcd = Faa'Fbb' Fcc'Fdd' Ca'b'c'd' where this time Fab rather than being a reflection in the x,y,or z plane, is a reflection in a plane that is rotated by angle β from the x=0 plan, and we have this symmetry for any β. This probably means the material is isotropic in the x-y direction. How do you reflect in a plane at angle β? If x is a point in space, then let xt = (x,y). Then we have [Rz(β) xt] → - [Rz(β) xt] // This is 2D parity. z → z // no change in z Therefore I would say that Fab = Rz(β)ab where this is the 3D matrix. So the condition on C then is Cabcd(β) = Rz(β)aa' Rz(β)bb' Rz(β)cc' Rz(β)dd' Ca'b'c'd' We know this C matrix will be a subset of the one shown above which is 5.49.3 p 325, since the one above applies for just the two reflections (x and y) which occur for β = 0 and β = π/2. But how on earth do we make sense out of the general case above? Since there are only 9 independent C elements here, we could put 9 distinct elements on the LHS and then we have 9 specific equations. For example C1111(β) = Rz(β)1a' Rz(β)1b' Rz(β)1c' Rz(β)1d' Ca'b'c'd' C1122(β) = Rz(β)1a' Rz(β)1b' Rz(β)2c' Rz(β)2d' Ca'b'c'd' etc. Is this not then 9 equations in 9 non-vanishing unknowns, and that for each value of β ? That seems to be too many conditions somehow, but they must not all be unique. Let's try this directly with the Cab matrix instead? But it is not a tensor so don't know what to do. Two tricks might help here: (1) try specific angles like β = π/4 (2) anytime an index is 3, one R factor goes away. The 2D matrix for π/4 can be taken to be (1/) . Derivative attempt. (small β) Cabcd(β) = Rz(β)aa' Rz(β)bb' Rz(β)cc' Rz(β)dd' Ca'b'c'd' C3333(β) = Rz(β)3a' Rz(β)3b' Rz(β)3c' Rz(β)3d' Ca'b'c'd' = C3333 // learn nothing Next try C3331(β) = Rz(β)3a' Rz(β)3b' Rz(β)3c' Rz(β)1d' Ca'b'c'd' = C3331 = Rz(β)1d' C333d' = cosβ C3331 + sinβ C3332 = C3331 But how can this be true? C3331= cosβ C3331 + sinβ C3332 for any β The only way is if all three vanish, and we know that is the case from our theorem above. So let's only consider the non-zero entries in 5.49.3 and not waste time on the others. Try things with two 3's: Cab33(β) = Rz(β)aa' Rz(β)bb' Rz(β)3c' Rz(β)3d' Ca'b'c'd' = Rz(β)aa' Rz(β)bb' Ca'b'33 where a and b are both not 3. But even this is a painful problem. Basically this is some kind of "ill proposed problem". I just don't grok it in terms of why there should be a unique solution, etc etc. My method is too weak. Idea: what about using the rotation generators and just look near β = 0. Then Rz() = Jz = R() = exp(-iJ) Rz(β) = 1 -iβJ3 (J3)jk = -i 3jk. Rz(β)jk = δjk -iβ(-iε3jk) = (δjk - βε3jk) At least this is something I can write down, all this last stuff is valid for small β only. Then Cabcd(β) = ( δaa' - βε3aa') ( δbb' - βε3bb') ( δcc' - βε3cc')' ( δdd' - βε3dd') Ca'b'c'd' = β4 (ε3aa' ε3bb' ε3cc' ε3dd' Ca'b'c'd') – β3(ε3aa' ε3bb' ε3cc' Ca'b'c'd + ε3aa' ε3bb' ε3dd' Ca'b'cd' + ε3aa' ε3cc' ε3dd' Ca'bc'd' + ε3bb' ε3cc' ε3dd' Cab'c'd') + β2 (there are 4,2 = 4!/2!2! = 4!/4 = 3! = 6 terms here ) + β (there are 4,1 = 4!/3! = 4 terms here) + Cabcd Now if the LHS is supposed to be independent of β, we should be able to say ∂nCabcd(β) = 0 for any positive integer n. So lets try n = 4 and see what we get ε3aa' ε3bb' ε3cc' ε3dd' Ca'b'c'd' = 0 This seems to be a set of 24 = 16 conditions. For example, for abcd = 2222 we get ε321 ε321 ε321 ε321 C1111 = 0 which says C1111 = 0 which is wrong! Something is wrong somewhere. Well, my approximation has omitted many β4 terms so it is all garbage. But what about the first derivative, since then I will have all those terms. Cabcd(β) = ( δaa' - βε3aa') ( δbb' - βε3bb') ( δcc' - βε3cc')' ( δdd' - βε3dd') Ca'b'c'd' ≈ - β [ε3aa' Ca'bcd + ε3bb' Cab'cd + ε3cc' Cabc'd + ε3dd' Cabcd' ] So I think at least this is a valid set of conditions: ε3aa' Ca'bcd + ε3bb' Cab'cd + ε3cc' Cabc'd + ε3dd' Cabcd' But this is 4! 4! = 242 equations! I am still no closer. Question: why is there even a solution to this problem? Why can you satisfy this equation for all β ??? Cabcd = Rz(β)aa' Rz(β)bb' Rz(β)cc' Rz(β)dd' Ca'b'c'd' If this equation is really true, then this must be true Rz(β)aa' Rz(β)bb' Rz(β)cc' Rz(β)dd' = δaa' δbb' δcc' δdd' // wrong This is wrong because we know that only 21 of 81 elements are independent so there could be cancellations. Try making symmetrized forms of the C matrix. C(L)abcd = (Cabcd + Cbacd)/2 C(LR)abcd = (C(L)abcd + C(L)abdc)/2 C(S) = (C(LR)abcd + C(LR)dcba)/2 Looking at the web, I see that this is a much harder general problem than I want to do right now, but it is a good one.