¹⁴The last expression is obtained by substituting (6.69) in J₃ = ∘I2-∕J2 and using Taylor’s series expansion about tr(ϵ) = 0, to obtain

----------- ∘ --- ∘ 2 2 ∘ --------- ∘ --------- J3 = I2= J1 --tr(B-)≈ 3-+-4tr(ϵ)≈ 1 + 2tr(ϵ) ≈ 1+ tr(ϵ). J2 2J2 3 - 2tr(ϵ)