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
\[\frac{1}{2} m v_{th} ^2 \approx \frac{1}{2} kT \rightarrow v_{th} \approx \sqrt{ \frac{kT}{m}}\]Since electrons are much lighter than ions (\( m_e = m_i / 1800 \) 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
\[n_{e, w} v_e = n_{i, w} v_i\]- \( n_{i,w} \) is the ion density at the wall
- \( n_{e, w} \) is the electron density at the wall
- \( v_e \) is the electron thermal speed at the wall
- \( v_i \) 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 \( n_i \approx n_0 \). Gauss’s law gives the electric field generated by a volume \( A \cdot x \) of electron-free plasma
\[E A = \frac{A \rho x}{\epsilon_0} \quad \rightarrow \quad E = \frac{\rho x}{\epsilon_0}\] \[\rho \approx + n_0 e\] \[V = - \int E \cdot \dd l = - \frac{n_0 e}{2 \epsilon_0} x^2\]The potential energy of electrons reaching the wall is
\[PE = V(-e)= \frac{n_0 e^2 x^2}{2 \epsilon_0}\]The total energy of the electron is
\[KE + PE = \frac{1}{2} k T_e\]where \( T_e \) is the electron temperature. The electron stops when \( KE = 0 \) or
\[\frac{1}{2} k T_e = \frac{n e^2 \lambda _D ^2 }{2 \epsilon_0}\] \[\rightarrow \lambda_D = \sqrt{ \frac{ \epsilon_0 k T_e}{n_0 e^2}}\]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 \( (n_i v_i = n_e v_e) \) flow to the wall? Let’s assume the electrons satisfy a Boltzmann distribution
\[n_e \propto e^{- \varepsilon/kT}\] \[\varepsilon = PE + KE = \frac{1}{2} m_e v_e ^2 + V(-e)\] \[\rightarrow n_e \propto e^{\frac{1}{2} \frac{ m v^2}{kT}} e^{\frac{-V(-e)}{kT}}\] \[\rightarrow n_e = n_0 e^{\frac{V e}{kT}}\]Plugging this into the ambipolar flow condition
\[n_0 v_0 = n_0 e^{\frac{Ve}{kT}} v_e\]In equilibrium the energy is distributed evenly across species. Once again, because the ions are so much heavier, \( v_i \approx v_0 \)
\[kT = m_e v_e ^2 = m_i v_0 ^2\] \[\rightarrow \frac{v_0}{v_e} = \sqrt{\frac{m_e}{m_i}} = e^{\frac{Ve}{kT}}\] \[\rightarrow \frac{Ve}{kT} = - \ln \sqrt{\frac{m_i}{m_e}}\]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
\[\frac{V}{T_e} = - \ln \sqrt{\frac{m_i}{m_e}}\]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.
\[V_{sheath} \approx 4 T_e \qquad (\text{D or H})\]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:
\[W_{impact} \approx 4 Z_i k T_e + \frac{1}{2} k T_i \qquad (\text{D or H})\]Now that we roughly know the total sheath voltage required to maintain ambipolar flow, what do \( n_e \) and \( V \) 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
\[\div \vec E = \frac{e(n_i - n_e)}{\epsilon_0}\] \[E = - \grad V\] \[\rightarrow \nabla ^2 V = - \frac{e(n_i - n_e)}{\epsilon_0}\]Recall that we can relate \( n_e \) to temperature by
\[n_e = n_0 e^{\frac{ Ve}{kT}}\]For the ion density, Mass conservation in the direction of the wall gives
\[n_0 v_0 = n_i v_i\]Conservation of energy of the ions gives
\[\frac{1}{2} m v_i ^2 = - Ve + \frac{1}{2} m_i v_0 ^2\]where \( v_0 \) 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.