research36 min studyLesson 301 in Physics

Vlasov–Poisson as an initial-value problem

Normalize a velocity distribution, specify consistent electrostatic initial data, and distinguish characteristic transport from a fluid closure.

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What you’ll understand

  • Distinguish density-normalized distributions from probability densities.
  • Specify the distribution and field constraint for a periodic Vlasov–Poisson problem.
  • Compute density, mean velocity, and a central second moment.

A kinetic plasma state contains more information than density, velocity, and temperature. It specifies how particles populate velocity space at each position, including small resonant populations that a few moments can hide. The collisionless Vlasov equation transports this distribution in the self-consistent mean field. Poisson's equation constrains that field. Solving their initial-value problem requires compatible initial data and boundary conditions, not merely selecting an oscillation frequency from a dispersion relation.

Normalize the object being evolved

Use a one-dimensional spatial model with one retained velocity coordinate. Let F describe number density per unit velocity, so its velocity integral is the physical three-dimensional number density. This reduced distribution can be obtained by integrating a full distribution over the two perpendicular velocities in an unmagnetized longitudinal problem. It is not normalized to one unless divided by local density. Its units differ from those of the original six-dimensional phase-space density, even though both are often called f.

The first velocity moment gives mean flow, while the central second moment gives parallel pressure after multiplication by mass. Subtracting the flow matters: a narrow beam moving rapidly can have large bulk kinetic energy but small pressure. Higher moments measure asymmetry and heat flow that are absent from density and pressure alone. A fluid model retaining only low moments therefore needs a closure; merely taking moments of Vlasov does not determine the unretained moments.

n=∫F dv,u=1n∫vF dv,p∥=m∫(v−u)2F dvn=\int F\,dv,\qquad u=\frac1n\int vF\,dv,\qquad p_\parallel=m\int(v-u)^2F\,dv
F has units s/m⁴ when n is in m⁻³ and v in m/s. The full three-velocity distribution has units s³/m⁶. Parallel pressure is in pascals; this moment is not automatically an isotropic thermodynamic pressure.↔ Scroll wide equations horizontally.

Initial data must obey a field constraint

For electrons with charge minus e and an immobile uniform ion background, the electrostatic acceleration is minus eE divided by electron mass. The density perturbation sets the spatial derivative of E through Poisson's equation. On a periodic domain, integrating that equation requires global neutrality. An arbitrary nonneutral initial distribution is incompatible with a periodic electric field unless the model explicitly changes its background or includes a different boundary condition.

Poisson's equation determines the field only up to a spatially uniform component in this formulation. A common periodic electrostatic convention chooses zero mean E and fixes the additive constant of potential separately. The initial distribution must also remain nonnegative. A formal small perturbation that becomes more negative than the equilibrium tail somewhere is not an admissible particle density. For open boundaries, specify incoming distribution data consistently with the selected electrostatic boundary conditions rather than importing periodic assumptions.

∂tF+v∂xF−emeE∂vF=0,∂xE=eϵ0(ni−∫F dv)\partial_tF+v\partial_xF-\frac e{m_e}E\partial_vF=0,\qquad \partial_xE=\frac e{\epsilon_0}\left(n_i-\int F\,dv\right)
e > 0 is elementary charge in coulombs, me electron mass in kilograms, E in V/m, and ε₀ vacuum permittivity. This one-dimensional collisionless electrostatic model neglects magnetic dynamics and discrete-particle correlations.↔ Scroll wide equations horizontally.

Characteristics preserve fine information

Following a characteristic in position and velocity makes the distribution constant. The characteristic itself depends on the field generated by the whole distribution, so particles do not simply free stream unless that field is absent. The phase-space flow is incompressible: deformation can stretch and fold a distribution without erasing its fine structure. A numerical method that smooths those folds may change entropy even while its density plot looks accurate.

With suitable periodic or decaying boundaries, total particle number and the sum of electron kinetic and electrostatic field energy are conserved. Field energy can decrease while particle kinetic energy increases, so decay of an electric oscillation does not establish collisional dissipation. Retaining the evolving velocity structure is central to interpreting collisionless damping and later echoes. This conservation statement belongs to the ideal continuum model; finite resolution, boundary losses, or added collisions require their own accounting.

x˙=v,v˙=−emeE(x,t),dFdt=0,E=∫dx[∫mev22F dv+ϵ0E22]\dot x=v,\quad \dot v=-\frac e{m_e}E(x,t),\quad \frac{dF}{dt}=0,\qquad \mathcal E=\int dx\left[\int\frac{m_ev^2}{2}F\,dv+\frac{\epsilon_0E^2}{2}\right]
Characteristics use time in seconds. For a unit cross-sectional area, the displayed spatial integral has units J/m². Conservation assumes no boundary power flux, zero-mean periodic E, immobile ions, and sufficiently regular decaying velocity tails.↔ Scroll wide equations horizontally.

Moments of a flat velocity band

A homogeneous species has F = n/(2w) for u − w < v < u + w and zero otherwise. Take n = 6 × 10¹² m⁻³, u = 2 km/s, w = 3 km/s, and mass m = 10⁻²⁶ kg. Find density, mean velocity, and parallel pressure.

  1. The band width is 2w, so integrating its constant height n/(2w) returns n. Symmetry about u makes the mean velocity exactly 2 km/s.

  2. With c = v − u, the central variance is ⟨c2⟩=(2w)−1∫−wwc2 dc=w2/3=3×106 m2/s2\langle c^2\rangle=(2w)^{-1}\int_{-w}^w c^2\,dc=w^2/3=3\times10^6\,\mathrm{m^2/s^2}.

  3. Multiply the variance by mn: p∥=(10−26)(6×1012)(3×106)=1.8×10−7 Pap_\parallel=(10^{-26})(6\times10^{12})(3\times10^6)=1.8\times10^{-7}\,\mathrm{Pa}. Bulk speed is excluded from this pressure moment.

The result

The distribution integrates to 6 × 10¹² m⁻³, has mean velocity 2 km/s, and parallel pressure 1.8 × 10⁻⁷ Pa.

Explore the model

Interactive lab

Interactive lab · vectors

Project before forming a moment

Reset and set BOTH Angle to 0 degrees and Magnitude to 2; predict x projection 2. Then Reset and set BOTH Angle to 60 degrees and Magnitude to 2; predict x projection 1. The static vector illustrates the distinction between speed and a selected velocity component. It contains no particle distribution, pressure integration, or Vlasov–Poisson evolution.

Project before forming a momentPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA kinetic moment uses the specified velocity component and weighting, not simply a particle’s total speed.

Check your understanding

For a density-normalized reduced distribution F(x,v,t), what does ∫F dv equal?

Try it yourself

In a periodic electron–ion model, the initial charge density is ρ(x) = −6ε₀ cos(kx), where the coefficient 6 has units V/m² and k = 2 m⁻¹. Find zero-mean E and zero-mean potential φ with E = −∂xφ.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate Poisson once for E, then once more for potential, keeping the electron charge sign.

Compare with the solution

Poisson gives ∂xE=−6cos⁡(kx) V/m2\partial_xE=-6\cos(kx)\,\mathrm{V/m^2}. Thus E=−(6/k)sin⁡(kx)=−3sin⁡(kx) V/mE=-(6/k)\sin(kx)=-3\sin(kx)\,\mathrm{V/m}. Since E is minus the potential gradient, ϕ=−(6/k2)cos⁡(kx)=−1.5cos⁡(kx) V\phi=-(6/k^2)\cos(kx)=-1.5\cos(kx)\,\mathrm V. Both means vanish over one spatial period. The charge density also has zero mean, so the periodic neutrality condition is satisfied.

Take it with you

  • Distribution normalization fixes both units and the meaning of velocity integrals.
  • A periodic Poisson problem requires neutral charge and a declared mean-field convention.
  • Collisionless characteristic transport preserves fine phase-space information while fields exchange energy with particles.

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