Figure 1: black: Ė_{M}, red: W_{L}, blue: W_{n}, orange: W_{i}, yellow: W_{n} + W_{i}, for τ′_{AD} = 1 and ζ = 10^{−9} (top), 10^{−5} (middle), and 10^{−3} (bottom), just like in the 2019 paper.

Basic equations

\frac{∂A}{∂t} = u_{i} × B − ημ_{0}J,  (1)

ρ_{i} \frac{Du_{i}}{D_{i}t} = J×B−∇p_{i}+∇·(2νρ_{i}S_{i})−ρ(ρ_{i}γ+ζ)(u_{i}−u),

(2)

ρ\frac{Du}{Dt} = ρf −∇p+∇·(2νρS)+ρ_{i}(ργ +αρ_{i})(u_{i} −u),

(3)

\frac{D ln ρ_{i}}{D_{i}t} = −∇ · u_{i} + ζρ/ρ_{i} − αρ_{i},  (4)
\frac{D ln ρ}{Dt} = −∇ · u − ζ + αρ_{i}^{2}/ρ,  (5)

Frictional heating

H_{i} = +ρ(ρ_{i}γ + ζ)(u_{i} − u) · u_{i},  (6)
H_{n} = −ρ_{i}(ργ + αρ_{i})(u_{i} − u) · u,  (7)

Figure 2: Same as Figure 1, but with 10 times larger γ, so τ′_{AD} = 10.

So that

H_{i} + H_{n} = ρ(ρ_{i}γ + ζ)(u_{i} − u)^{2} + R, (8)

where

R = α(ρ_{i} − ρ_{i0})^{2} − ζ(ρ − ρ_{0})^{2}  (9)

is a residual that vanishes if ρ_{i} and ρ are equal to their iniial values, ρ_{i0} and ρ_{0}, respectively.

In the 1-fluid approximation, we have, instead, ϵ_{AD} = (τ_{AD}/ρ_{0})⟨(J × B)^{2}⟩.

1 Results

Ė_{M} < 0, so magnetic energy is lost and goes via W_{L} into kinetic energy of the ions, which is then dissipated via frictional heating.

The oscillations are relaxation oscillations owing to the initial pressure imbalance. Neutral and charged particles oscillate in antiphase.

Figure 3: Heating profile of (a) ions (black) and neutrals (blue) in the 2-fluid description, and (b) in the 1-fluid description, for τ′_{AD} = 10, ζ = 10^{−3}, α = 10^{5}, γ = 10^{3}, ρ_{i0} = 10^{−4}, and ρ_{0} = 1.