Derivation of Eq. (122), not ﬁnished

From the deﬁnition of μ, we obtain

2 2 ∂μ-= − v⊥--∂B0-+ -1--∂v⊥-= −-μ-∂B0-+ ∂v-⊥ ⋅ v⊥ ∂x 2B20 ∂x 2B0 ∂x B0 ∂x ∂x B0

(425)

Using

∂v⊥- ∂[v-−-v∥e∥]- ∂e∥ ∂v∥ ∂x = ∂x = − v∥ ∂x − ∂x e∥,

(426)

expression (425) is written as

∂μ- μ-∂B0- ∂e-∥ v⊥- ∂x = − B0 ∂x − v∥ ∂x ⋅B0 ,

(427)

which agrees with Eq. (10) in Frieman-Chen’s paper[3].

1 ∫ 2π ∂μ --- dαv⋅ --- 2π 0∫ 2π ∂x( ) = -1- dαv⋅ − μ-∂B0-− v∥∂e∥ ⋅ v⊥ 2π 0 B0 ∂x ∂x B0 =?0

∫ 2π [ ( )] 1-- dαv ⋅ v × -∂- e∥ ⋅ ∂δG0 2π 0 ∂x Ω ∂X =

Using the fact that δG₀ is independent of α, the left-hand side of Eq. (122) is written as

⟨v⋅[λB1 + λB2 ]δG0⟩α 1 ∫ 2π { [ ∂ ( e∥)] ∂δG ∂μ ∂δG ∂α ∂δG } = --- dαv ⋅ v × --- -- ⋅---0-+ ------0-+ ------0- 2π ∫02π { [ ∂x( Ω )] ∂X ∂x ∂μ } ∂x ∂α = -1- dαv ⋅ v × ∂-- e∥ ⋅ ∂δG0-+ ∂μ-∂δG0 2π 0 ∂x Ω ∂X ∂x ∂μ