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G 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
(357)

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,
(358)

where n is the number density defined by n = ∫ δfpdv. Using the relation between the particle distribution function and guiding-center distribution function, equation (358) is written as

∫ nU = v δf (x − ρ,v )dv. ⊥ ⊥ g
(359)

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).
(360)

Plugging this expression into Eq. (359), we obtain

∫ ∫ nU ⊥ ≈ v⊥δfg(x,v)dv− v⊥ ρ⋅∇ δfg(x,v)dv
(361)

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. (361) is reduced to

∫ nU = − v ρ ⋅∇δf (x,v)dv ⊥ ⊥ g
(362)

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

∫ nU ⊥ = v⊥v-×-e∥ ⋅∇ δfg(x,v)dv ∫ (Ω ) = v e∥× ∇ δf (x,v ) ⋅v dv. ∫ ⊥ Ω g ⊥ = v⊥H ⋅v ⊥dv, (363)
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. (363). In terms of velocity space cylindrical coordinates (v∥,v⊥,α), v⊥ is written as
v ⊥ = v⊥ (ˆx cosα+ ˆy sinα ),
(364)

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,
(365)

where Hx and Hy are independent of α. Using these in Eq. (363), we obtain

∫ nU ⊥ = v⊥ (xˆcosα+ ˆy sinα )v⊥ (Hx cosα + Hy sinα)dv ∫ = v2⊥ [ˆx(Hx cos2α+ Hy sin αcosα) +yˆ(Hx cosα sinα + Hy sin2α)]dv. (366)
Using dv = v⊥dv∥dv⊥dα, the above equation is written as
∫ ∫ ∫ nU = ∞ dv ∞ v dv 2πv2[ˆx(H cos2α + H sinα cosα)+ ˆy(H cosαsin α +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⊥(ˆxHx π+ ˆyHy π)] ∫−∞ ∫ 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∥- = m Ω × ∇ δp⊥, (367)
where
∫ ∞ ∫ ∞ 2 δp⊥ ≡ dv∥ v⊥dv⊥δfg(x,v)mv⊥-2π ∫−∞ 0 2 mv2⊥- = δfg(x,v ) 2 dv, (368)
is the perpendicular pressure carried by δfg(x,v). The flow given by Eq. (367) is called the diamagnetic flow.

 

 

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