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F Diamagnetic flow **check**

The perturbed distribution function δF given in Eq. () contains two terms. The first term is gyro-phase dependent while the second term is gyro-phase independent. The perpendicular velocity moment of the second term will give rise to the so-called diamagnetic flow. For this case, it is crucial to distinguish between the distribution function in terms of the guiding-center variables, fg(X,v), and that in terms of the particle variables, fp(x,v). In terms of these denotations,  equation () is written as

δFg = -q(δΦ− ⟨δΦ⟩α)∂F0g + δfg. m ∂𝜀
(365)

Next, consider the perpendicular flow U⊥ carried by δfg. This flow is defined by the corresponding distribution function in terms of the particle variables, δfp, via,

∫ nU ⊥ = v⊥δfp(x,v)dv,
(366)

where n is the number density defined by n = ∫ δfpdv. Using the relation between the particle distribution function and guiding-center distribution function, equation (366) is written as

∫ nU ⊥ = v⊥δfg(x − ρ,v )dv.
(367)

Using the Taylor expansion near x, δfg(x −ρ,v) can be approximated as

δfg(x − ρ,v) ≈ δfg(x,v)− ρ ⋅∇δfg(x,v).
(368)

Plugging this expression into Eq. (367), we obtain

∫ ∫ nU ⊥ ≈ v⊥δfg(x,v)dv− v⊥ ρ⋅∇ δfg(x,v)dv
(369)

As mentioned above, δfg(x,v) is independent of the gyro-angle α. It is obvious that the first integration is zero and thus Eq. (369) is reduced to

∫ nU ⊥ = − v⊥ρ ⋅∇δfg(x,v)dv
(370)

Using the definition ρ = −v × e∥∕Ω, the above equation is written

∫ nU = v v-×-e∥⋅∇ δf(x,v)dv ⊥ ∫ ⊥ Ω g ( e∥ ) = v ⊥ Ω × ∇ δfg(x,v) ⋅v⊥dv. ∫ = v ⊥H ⋅v⊥dv, (371)

where H = e∥ Ω ×∇δfg(x,v), which is independent of the gyro-angle α because both e∥(x)∕Ω(x) and δfg(x,v) are independent of α. Next, we try to perform the integration over α in Eq. (371). In terms of velocity space cylindrical coordinates (v∥,v⊥,α), v⊥ is written as

v ⊥ = v⊥ (ˆx cosα+ ˆy sinα ),
(372)

where ˆx and ˆy are two arbitrary unit vectors perpendicular each other and both perpendicular to B0(x). H can be written as

H = Hxˆx + Hyˆy,
(373)

where Hx and Hy are independent of α. Using these in Eq. (371), we obtain

∫ nU ⊥ = v⊥(ˆx cosα + ˆysinα)v⊥(Hx cosα+ Hy sin α)dv ∫ = v2⊥[ˆx(Hx cos2α + Hy sinα cosα)+ ˆy(Hx cosαsinα +Hy sin2α)]dv. (374)

Using dv = v⊥dv∥dv⊥dα, the above equation is written as

∫ ∫ ∫ nU = ∞ dv ∞ v dv 2πv2 [ˆx(H cos2 α+ H sin αcosα)+ ˆy(H cosαsinα + H sin2α)]dα ⊥ − ∞ ∥ 0 ⊥ ⊥ 0 ⊥ x y x y ∫ ∞ ∫ ∞ ∫ 2π = dv∥ v⊥dv ⊥ v2⊥ (ˆxHx cos2α + ˆyHy sin2α)dα ∫−∞∞ ∫0∞ 0 = dv∥ v⊥dv ⊥[v2⊥ (xˆHx π + ˆyHyπ)] ∫− ∞ ∫0 ∞ ∞ 2 = − ∞ dv∥ 0 v⊥dv ⊥[v⊥H π ] ∫ ∞ ∫ ∞ e = dv∥ v⊥dv ⊥[v2⊥ -∥× ∇ δfg(x,v )π ] − ∞ ∫ 0∞ ∫ ∞ Ω = e∥× ∇ dv v dv δf (x,v)v2⊥-2π Ω −∞ ∥ 0 ⊥ ⊥ g 2 e∥-- = mΩ ×∇ δp⊥, (375)

where

∫ ∞ ∫ ∞ 2 δp⊥ ≡ dv∥ v⊥dv ⊥δfg(x,v)mv-⊥ 2π ∫− ∞ 0 2 mv2⊥ = δfg(x,v) 2 dv, (376)

is the perpendicular pressure carried by δfg(x,v). The flow given by Eq. (375) is called the diamagnetic flow.

 

 

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