advanced35 min studyLesson 296 in Physics

Two-fluid plasma equations

Take species moments, identify the closure problem, and connect small velocity differences to measurable current.

On this page

What you’ll understand

  • Interpret species continuity and momentum equations as kinetic moments.
  • Compute current and mass-weighted flow separately.
  • Explain why pressure and heat-flux closures are additional physical assumptions.

A plasma can be electrically almost neutral while electrons and ions move differently. A two-fluid description keeps separate densities, velocities, and stresses for each species, coupled through electromagnetic fields and collisions. This preserves more physics than a single bulk velocity, but less than a full distribution function. The reduction is useful only when the unresolved velocity-space information has an explicit and defensible closure.

Moments retain conservation laws

Start with a kinetic distribution fs(x,v,t)f_s(\mathbf x,\mathbf v,t) for species ss. Integrating over velocity gives number density; weighting by velocity gives particle flux and mean velocity. The zeroth kinetic moment yields continuity when ionization, recombination, and other particle sources are absent. The first moment yields species momentum balance. The pressure tensor is built from velocities relative to the species mean, so it represents random-motion stress rather than the bulk momentum flux counted a second time.

The material derivative follows that species' own mean velocity. Electron and ion derivatives need not be the same. Electric forces have opposite signs for their opposite charges; magnetic forces depend on their distinct velocities. The collision force densities exchange momentum between species and sum to zero for internal electron–ion collisions. Neglecting friction separately in one equation while retaining its reaction elsewhere would violate total momentum conservation unless an external momentum reservoir were specified.

∂tns+∇⋅(nsus)=0,msns(∂t+us⋅∇)us=qsns(E+us×B)−∇⋅Ps+Rs\partial_tn_s+\nabla\cdot(n_s\mathbf u_s)=0,\qquad m_sn_s(\partial_t+\mathbf u_s\cdot\nabla)\mathbf u_s=q_sn_s(\mathbf E+\mathbf u_s\times\mathbf B)-\nabla\cdot\mathbf P_s+\mathbf R_s
Number density n_s is m⁻³, velocity u_s is m/s, pressure tensor P_s is Pa, and momentum-exchange force density R_s is N/m³. The exact nonrelativistic moment equations assume number-conserving kinetics with Lorentz forces. They are not closed merely by writing them down.↔ Scroll wide equations horizontally.

Neutral density does not mean zero current

For singly charged ions with ni≃ne=nn_i\simeq n_e=n, current is proportional to the difference of species velocities. Mass flow is instead weighted by species masses. Since ions are heavy, the centre-of-mass flow is usually close to ion flow even when electrons carry a substantial current. Quasineutrality constrains charge density, whereas current describes charge flux. Setting equal densities and then setting equal velocities without a scale argument discards electromagnetic dynamics that the two-fluid model was introduced to retain.

Maxwell's equations supply the fields and their coupling to both species. A quasineutral low-frequency reduction can replace the charge-separation equation while preserving current through Ampère's law in its appropriate ordering. Such a reduction cannot resolve an electron plasma oscillation that fundamentally needs Poisson's equation and electron inertia. Choosing which fields to eliminate is therefore a decision about the phenomena and scales of interest, not merely a way to save algebra.

j=en(ui−ue),ρ=n(mi+me),u=miui+meuemi+me\mathbf j=en(\mathbf u_i-\mathbf u_e),\qquad \rho=n(m_i+m_e),\qquad \mathbf u=\frac{m_i\mathbf u_i+m_e\mathbf u_e}{m_i+m_e}
These expressions assume equal singly charged species densities. Current j is A/m², mass density ρ is kg/m³, and u is mass-weighted velocity. They remain compatible with finite current at vanishing leading-order charge density.↔ Scroll wide equations horizontally.

Closure determines which fluid model you have

Momentum evolution introduces the pressure tensor; pressure evolution introduces heat flux and higher moments. A finite list of exact moment equations therefore has more unknowns than equations. An isothermal closure fixes a temperature and sets scalar pressure proportional to density. An adiabatic scalar closure instead evolves psns−γsp_sn_s^{-\gamma_s} along the species flow when heat exchange and dissipation are negligible. Cold fluid equations discard pressure forces, which is justified by their size for the chosen process, not by the fluid label itself.

Magnetized weakly collisional populations can have different parallel and perpendicular pressures, while strong collisions can support a nearly isotropic local distribution. A scalar closure should therefore be checked against collision, gyro, streaming, and evolution timescales. Landau damping and detailed resonant particle trapping are generally not recovered by a simple local scalar closure. The diagnostic question is what physical information the closure removes and whether the intended calculation depends on that information.

A current with nearly ion-controlled mass flow

A quasineutral hydrogen plasma has n=1018 m−3n=10^{18}\,\mathrm{m^{-3}}, ion velocity 2000x^ m/s2000\hat{\mathbf x}\,\mathrm{m/s}, and electron velocity 1500x^ m/s1500\hat{\mathbf x}\,\mathrm{m/s}. Use e=1.60×10−19 Ce=1.60\times10^{-19}\,\mathrm C, mi=1.67×10−27 kgm_i=1.67\times10^{-27}\,\mathrm{kg}, and me/mi=1/1833m_e/m_i=1/1833. Find current, mass density, and bulk velocity.

  1. The velocity difference is 500x^ m/s500\hat{\mathbf x}\,\mathrm{m/s}. Thus j=(1.60×10−19)(1018)(500)x^=80x^ A/m2\mathbf j=(1.60\times10^{-19})(10^{18})(500)\hat{\mathbf x}=80\hat{\mathbf x}\,\mathrm{A/m^2}. Equal densities remove leading charge imbalance, not this flux difference.

  2. The mass density is 1018(1.67×10−27)(1+1/1833)=1.671×10−9 kg/m310^{18}(1.67\times10^{-27})(1+1/1833)=1.671\times10^{-9}\,\mathrm{kg/m^3}. Electron mass is a small correction to this quantity, despite electrons being essential to the current calculation.

  3. The bulk velocity is 2000−(2000−1500)/(1833+1)=1999.73 m/s2000-(2000-1500)/(1833+1)=1999.73\,\mathrm{m/s} along x. It differs from ion flow by only 0.27 m/s, showing why a bulk momentum equation can approximate ion inertia without requiring electron and ion velocities to be identical.

The result

The current is +80 A/m², the mass density is about 1.671 × 10⁻⁹ kg/m³, and bulk speed is 1999.73 m/s. Charge neutrality and current transport coexist in this example.

Explore the model

Interactive lab

Interactive lab · vectors

Compare species velocity components

Use normalized vectors as separate species-velocity illustrations. Reset and set both controls to angle 0° and magnitude 2; predict x projection 2. Reset and set both controls to angle 0° and magnitude 1.5; predict x projection 1.5. Subtract the two readouts outside the lab. It displays one vector per run and does not evolve coupled fluids or calculate current.

Compare species velocity componentsPlayback 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 forCurrent depends on a velocity difference, while bulk flow requires mass weights; the lab supplies only the vector projections.

Check your understanding

What is still required after taking a finite number of exact moments of the species kinetic equations?

Try it yourself

Locally neglect electron inertia, collisions, and magnetic force along x. Let ne=1018 m−3n_e=10^{18}\,\mathrm{m^{-3}} and ∂pe/∂x=3.20 Pa/m\partial p_e/\partial x=3.20\,\mathrm{Pa/m}. Find the electric field needed for the electron force balance, using e=1.60×10−19 Ce=1.60\times10^{-19}\,\mathrm C. Is this alone a complete equilibrium specification?

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

Give me a hint

Use negative electron charge and include the minus sign on the pressure-gradient force.

Compare with the solution

The electron equation is 0=−eneEx−∂xpe0=-en_eE_x-\partial_xp_e, so Ex=−3.20/0.160=−20 V/mE_x=-3.20/0.160=-20\,\mathrm{V/m}. The electric force on electrons points toward positive x and balances their pressure force toward negative x. This is only a local electron balance. Ion momentum balance, field constraints, boundary conditions, and thermodynamic closure must also be satisfied to construct a full equilibrium. Assigning this electron field does not independently prove that a stationary quasineutral ion population is possible.

Take it with you

  • Two-fluid equations retain separate species motion and stress.
  • Equal electron and ion densities are compatible with finite current.
  • A moment hierarchy needs closure; closure assumptions decide the physics it can describe.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.