Poisson's equation

`BoltzmannElectrons`

Boltzmann plasma electrons for positive-ion extraction.

density: $\rho_e = -\rho_{e0} \exp\left(\frac{\phi - U_p}{T_e}\right)$
neutral: $\rho_{e0} = \sum_\mathrm{beams} \frac{J}{v_0}, \qquad v_0 = \sqrt{\frac{2 E e}{m}}$

Assumes

isothermal electrons in equilibrium with the potential: no inertia, flow or sheath beyond the Boltzmann factor
$\rho_{e0}$ given, or by default the charge density of the beams at their start, so the plasma is neutral there (ParticleBeams add nothing)
$T_e$ and $U_p$ in eV and V; the species has the sign $s = -1$

Thermal positive plasma ions for negative-ion extraction.

density: $\rho_p = \rho_{p0} \exp\left(-\frac{\phi - U_p}{T_p}\right)$
neutral: $\rho_{p0} = \sum_\mathrm{beams} \frac{J}{v_0}, \qquad v_0 = \sqrt{\frac{2 E e}{m}}$
both signs: $\rho = s\, \rho_0\, e^{-s (\phi - U_p) / T}, \qquad s = +1 \text{ (ions)}, \ -1 \text{ (electrons)}$

Assumes

the plasma's positive ions, repelled by the extraction field; the negative ions and the co-extracted electrons are traced beams
isothermal ions in equilibrium with the potential, $T_p > 0$
$\rho_{p0}$ given, or by default the negative charge density of the beams at their start

Solve laplace(phi) = -rho / epsilon_0 (plus a Boltzmann plasma species) on a mesh.

poisson: $\nabla^2 \phi = -\frac{\rho_b + \rho_p}{\varepsilon_0}, \qquad \nabla^2 = \partial_x^2 + \frac{1}{r}\partial_r (r\, \partial_r) \text{ (cylindrical, } \partial_\theta = 0)$
shortley weller: $h^2 \partial_x^2 \phi \approx \frac{2\, \phi_-}{(\alpha + \beta)\alpha} + \frac{2\, \phi_+}{(\alpha + \beta)\beta} - \frac{2\, \phi_0}{\alpha \beta}$
radial: $h^2 \frac{1}{r}\partial_r (r\, \partial_r \phi) \approx \frac{2/\alpha - 1/j}{\alpha + \beta}\, \phi_- + \frac{2/\beta + 1/j}{\alpha + \beta}\, \phi_+ - \frac{2}{\alpha \beta}\, \phi_0, \quad j \ge 1$
axis: $h^2 \frac{1}{r}\partial_r (r\, \partial_r \phi)\Big|_{r = 0} \approx 4 (\phi_1 - \phi_0)$
neumann: $\phi_{-1} = \phi_1 + 2 h g: \qquad h^2 \partial_x^2 \phi \approx \frac{2 (\phi_1 - \phi_0)}{\beta^2} + \frac{2 h g}{\beta}$
rhs: $b = -\frac{h^2 \rho}{\varepsilon_0} - \sum_\text{fixed neighbours} w\, \phi_\text{fixed}$

Order

2

Assumes

$\alpha$ , $\beta$ the distances to the lower and upper neighbour or electrode surface in units of $h$ (1 away from electrodes); for $\alpha = \beta = 1$ the 5-point stencil
$g$ the outward normal derivative of a Neumann face [V/m]
the linear system solved by SuperLU (scipy.sparse.linalg.splu), its factorisation cached per set of fixed nodes
the order is in $h$ , for a smooth potential away from electrode corners

References

functionsource

Return the potential (nx, ny) for the charge density rho [C/m^3].

residual: $R(u) = A u - b + s\, a\, e^{-s (u - U_p) / T}, \qquad a = \frac{\rho_0 h^2}{\varepsilon_0}$
newton: $\left(A - \operatorname{diag}\frac{a\, e^{-s (u - U_p) / T}}{T}\right) \delta u = -R(u), \qquad u \leftarrow u + \delta u$
stop: $\max |\delta u| < 10^{-6} \max\left(\operatorname{ptp}(\phi_\text{fixed}),\ T,\ 1\ \mathrm{V}\right)$

Assumes

$s = \pm 1$ the sign of the Boltzmann species; the exponent capped at 50 to keep a bad first guess finite
started from the vacuum solution or the previous potential; undamped, at most 100 steps, the last iterate returned without a warning
without a plasma species one linear solve; an initial plasma region is held at $U_p$ as extra fixed nodes

Built from