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Wall-supported Plasma

Consider a "hot" plasma surrounded by a "cold" wall. What are the properties of the plasma (both electrons and ions) at the wall? The speed of a particle headed towards the wall is

12mvth2≈12kT→vth≈kTm \frac{1}{2} m v_{th} ^2 \approx \frac{1}{2} kT \rightarrow v_{th} \approx \sqrt{ \frac{kT}{m}}

Since electrons are much lighter than ions (me=mi/1800 m_e = m_i / 1800 for hydrogen), they are moving much faster and will leave the plasma at a much higher rate than the ions. The outflow of negative charge causes a positive buildup in the plasma, slowing the electron loss until the electrons and ions leave at the same rate

ne,wve=ni,wvi n_{e, w} v_e = n_{i, w} v_i

  • ni,w n_{i,w} is the ion density at the wall
  • ne,w n_{e, w} is the electron density at the wall
  • ve v_e is the electron thermal speed at the wall
  • vi v_i is the ion speed into the wall

Now, what would be the thickness of electron-free plasma needed to stop a thermal electron? Near the wall itself, since the electron mass is so much smaller than the ion mass, we can estimate ni≈n0 n_i \approx n_0 . Gauss's law gives the electric field generated by a volume A⋅x A \cdot x of electron-free plasma

EA=Aρxϵ0→E=ρxϵ0 E A = \frac{A \rho x}{\epsilon_0} \quad \rightarrow \quad E = \frac{\rho x}{\epsilon_0}

ρ≈+n0e \rho \approx + n_0 e

V=−∫E⋅dl=−n0e2ϵ0x2 V = - \int E \cdot \dd l = - \frac{n_0 e}{2 \epsilon_0} x^2

The potential energy of electrons reaching the wall is

PE=V(−e)=n0e2x22ϵ0 PE = V(-e)= \frac{n_0 e^2 x^2}{2 \epsilon_0}

The total energy of the electron is

KE+PE=12kTe KE + PE = \frac{1}{2} k T_e

where Te T_e is the electron temperature. The electron stops when KE=0 KE = 0 or

12kTe=ne2λD22ϵ0 \frac{1}{2} k T_e = \frac{n e^2 \lambda _D ^2 }{2 \epsilon_0}

→λD=ϵ0kTen0e2 \rightarrow \lambda_D = \sqrt{ \frac{ \epsilon_0 k T_e}{n_0 e^2}}

Cool, so the sheath will be somewhere on the order of the Debye length.

What voltage drop between the plasma bulk and the wall is necessary to maintain ambipolar (nivi=neve) (n_i v_i = n_e v_e) flow to the wall? Let's assume the electrons satisfy a Boltzmann distribution

ne∝e−ε/kT n_e \propto e^{- \varepsilon/kT}

ε=PE+KE=12meve2+V(−e) \varepsilon = PE + KE = \frac{1}{2} m_e v_e ^2 + V(-e)

→ne∝e12mv2kTe−V(−e)kT \rightarrow n_e \propto e^{\frac{1}{2} \frac{ m v^2}{kT}} e^{\frac{-V(-e)}{kT}}

→ne=n0eVekT \rightarrow n_e = n_0 e^{\frac{V e}{kT}}

Plugging this into the ambipolar flow condition

n0v0=n0eVekTve n_0 v_0 = n_0 e^{\frac{Ve}{kT}} v_e

In equilibrium the energy is distributed evenly across species. Once again, because the ions are so much heavier, vi≈v0 v_i \approx v_0

kT=meve2=miv02 kT = m_e v_e ^2 = m_i v_0 ^2

→v0ve=memi=eVekT \rightarrow \frac{v_0}{v_e} = \sqrt{\frac{m_e}{m_i}} = e^{\frac{Ve}{kT}}

→VekT=−ln⁡mime \rightarrow \frac{Ve}{kT} = - \ln \sqrt{\frac{m_i}{m_e}}

If we measure temperature in electron volts (because of course we do), then Boltzmann's constant is equal to the electric charge, so the ratio of the sheath voltage to the wall temperature is given by

VTe=−ln⁡mime \frac{V}{T_e} = - \ln \sqrt{\frac{m_i}{m_e}}

For a specific species (ion-to-electron mass ratio) we can calculate this ratio numerically. For D it is about -4.1, for H it is -3.7, so for a D-H plasma it will be about 4.

Vsheath≈4Te(D or H) V_{sheath} \approx 4 T_e \qquad (\text{D or H})

Since the sheath voltage is negative, the ions which are able to overcome the positive charge buildup will be accelerated towards the wall. The final energy of the ions will be:

Wimpact≈4ZikTe+12kTi(D or H) W_{impact} \approx 4 Z_i k T_e + \frac{1}{2} k T_i \qquad (\text{D or H})

Now that we roughly know the total sheath voltage required to maintain ambipolar flow, what do ne n_e and V V look like as we move through the sheath towards the wall?

The voltage will be the gradient of the electric field, which will come from a difference in number density between the ions and electrons. Gauss's law says that

∇⋅E=e(ni−ne)ϵ0 \div \vec E = \frac{e(n_i - n_e)}{\epsilon_0}

E=−∇V E = - \grad V

→∇2V=−e(ni−ne)ϵ0 \rightarrow \nabla ^2 V = - \frac{e(n_i - n_e)}{\epsilon_0}

Recall that we can relate ne n_e to temperature by

ne=n0eVekT n_e = n_0 e^{\frac{ Ve}{kT}}

For the ion density, Mass conservation in the direction of the wall gives

n0v0=nivi n_0 v_0 = n_i v_i

Conservation of energy of the ions gives

12mvi2=−Ve+12miv02 \frac{1}{2} m v_i ^2 = - Ve + \frac{1}{2} m_i v_0 ^2

where v0 v_0 is the thermal speed of the ions in the plasma bulk. With that, all we need are boundary conditions at the wall and in the plasma bulk.