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A Adiabatic electron response

Assume that the total electron density satisfies the Boltzmann distribution on each magnetic surface, i.e.,

( q δΦ) ( q δΦ) ne = Ne exp −-e-- ≈ Ne 1 − -e-- , Te0 Te0
(297)

where Ne is a radial function. Note that this does not imply that the equilibrium density is Ne (it just implies that the total density is Ne at the location where δΦ = 0, which can still be different from the equilibrium density).

Further assume that the magnetic surface average of electron density perturbation (δne = ne −ne0) is zero, i.e.,

⟨ne − ne0⟩f = 0,
(298)

where ne0 is the equilibrium electron density, ⟨…⟩f is the magntic surface averaging operator. Using Eq. (297) in the above condition, we get

( ) Ne = ne0----1-----≈ ne0 1 + qe⟨δΦ-⟩f- . 1 − qe⟨TδeΦ0⟩f- Te0
(299)

Then expression (297) is written as

( ) ( ) ( ) n = n 1+ qe⟨δΦ⟩f- 1 − qeδΦ- ≈ n 1 + qe⟨δΦ⟩f-− qeδΦ- . e e0 Te0 Te0 e0 Te0 Te0
(300)

Then the perturbation δne = ne − ne0 is written as

δne = − n0eqe(δΦ-−-⟨δΦ⟩f). Te0
(301)

This model for electron response is often called adiabatic electron.

A.1 Poisson’s equation with adiabatic electron response

Pluging expression (301) into the Poisson equation (231), we get

2 ∫ qi- ∂Fi0 q2e(δΦ-−-⟨δΦ-⟩f)- ′ − 𝜀0∇ ⊥δΦ− qi mi(δΦ − ⟨δΦ⟩α) ∂𝜀 dv + n0e Te0 = qiδni.
(302)

When solving the Poisson equation, the equation is Fourier expanded into toroidal harmonics. Each harmonic is independent of each other, so that they can be solved independently. For n≠0 harmonics, the ⟨δΦ⟩f terms is zero and thus the the electron term is easy to handle because it’s a local response. There is some difficulty in solving the n = 0 harmonic because the ⟨δΦ⟩f term is nonzero.

I use the following method to obtain ⟨δΦ⟩f.

First slove the n = 0 harmonic of the following equation:

∫ − 𝜀0∇2⊥δΦ′ − qi-qi(δΦ′ − ⟨δΦ ′⟩α)∂Fi0dv = qiδn′i, mi ∂𝜀
(303)

(i.e., Eq. (302) with the electron contribution dropped). Let δΦ′ denote the solution to this equation. Then it can be proved that ⟨δΦ′⟩f is equal to ⟨δΦ⟩f. [Proof:

Taking the flux surface average of Eq. (302), we obtain

⟨ ∫ q ∂F ⟩ − 𝜀0⟨∇2⊥δΦ⟩f − qi -i-(δΦ − ⟨δΦ⟩α)--i0-dv = qi⟨δn′i⟩f, mi ∂ 𝜀 f
(304)

where the adiabatic response disappears. to be continued.

]

Then solving Eq. (302) becomes easier because ⟨δΦ⟩f term is known and can be moved to the right-hand side as a source term.

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