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Dimensional analysis and reciprocal space...

The simplest reciprocal space formula is gd=λL, where d is a vector and g is a bivector

$unitA = {a0 L^2, {ax L, ay L }, axy } ; unitB = {b0 L^2 , ... sp; } ; (* try scalars dimensionless, vectors in L, bivectors bxy in L^2 in an L^2 space *)$

$unitA = {a0 , {ax/L , ay/L }, axy/L^2 } ; unitB = {b0 , {bx/L , by/L }, bxy/L^2 } ; (* try scalars dimensionless, vectors bx in L, bivectors bxy/L^2in a dimensionless space*)$

$unitA = {a0/L , {ax/L^2 , ay/L^2 }, axy/L^3 } ; unitB = {b0/L , {bx/L^2 , by/L^2 }, bxy/L^3 } ; (* try scalars dimensionless, vectors bx in L, bivectors bxy/L^3 in a 1/L space*)$

$( {{a0 L^2}, {{ax L, ay L}}, {axy}} )$

$( {{a0 L^2 + b0 L^2}, {{ax L + bx L, ay L + by L}}, {axy + bxy}} )$

$( {{-axy bxy + ax bx L^2 + ay by L^2 + a0 b0 L^4}, {{-ay bxy L + axy by L + ax b0 L^ ... ax bxy L + ay b0 L^3 + a0 by L^3}}, {axy b0 L^2 - ay bx L^2 + a0 bxy L^2 + ax by L^2}} )$

$unitA = {a0, {ax M, ay M}, axy/M^2M^2} ; unitB = {b0, {bx M, by M}, bxy/M^2 M^2} ; (* ... unless we use vectors in meters with bivectors normalized by unit area like bxy/L^2 *)$

Mul2[unitA, unitB]//MatrixForm (* now multivector multiplication mixes areas with scalars in even order, but otherwise works ... *)

$( {{a0 b0 - axy bxy + ax bx M^2 + ay by M^2}, {{ax b0 M + a0 bx M - ay bxy M + axy by M, ay b0 M - axy bx M + ax bxy M + a0 by M}}, {axy b0 + a0 bxy - ay bx M^2 + ax by M^2}} )$

$Simplify[Mul2[{0, {0, 0}, 1}, avec]/aCell]//MatrixForm (* new reciprocal basis vector a^* = (i2 a)/a . b *)$

( {{0}, {{ay/(ax bx + ay by), -ax/(ax bx + ay by)}}, {0}} )

$Simplify[Mul2[{0, {0, 0}, 1}, bvec]/aCell]//MatrixForm (* new reciprocal basis vector b^* = (i2 b)/a . b *)$

( {{0}, {{by/(ax bx + ay by), -bx/(ax bx + ay by)}}, {0}} )

Created by Mathematica (April 17, 2005)