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Graduate Course The Saha Equation + Debye Length • • • • • • The chemical potential Chemistry (this is the route to the Saha equation! The Saha equation Debye Length Plasma Parameter Plasma Frequency Chemical Reactions Let us consider a chemical reaction, for example hydrogen and oxygen reacting to form water vapour: 2H 2 + O2 ⇔ 2H 2O −2H 2 − O2 + 2H 2O = 0 In a isolated box, atoms can move between molecules during a reaction, but the number of atoms is conserved. Let the coefficient of the Bi molecule be bi, where we take the products to have positive b. € ( for example, the first molecule in the reaction is H2, thus B1 is H2, b1=-2). Thus a general chemical reaction can be written m ∑b B = 0 i i=1 i Conservation of atoms Let there be Ni molecules of type Bi. During the chemical reaction the number of molecules of a particular type can change, but the atoms have to go somewhere! dN i = λbi € where λ is some constant that is the same for all molecules Conditions for equilibrium For a single component system the chemical equilibrium means that the particle number is constant. This implies a constant chemical potential The condition for chemical equilibrium for a multi-component system can be written as ∑ µ dN i i =0 i i.e. € € ∑b µ i i i =0 Reactions at constant temperature and volume ∂F µi = ∂N i V ,T ,N j≠i Recall F is the free energy So our condition for equilibrium now reads € ∂F =0 ∑ bi ∂N i V ,T ,N j≠i i F = −k BT ln Z N where, as we know, € Chemical equilibrium states can be calculated if we know the partition functions of the reactants and products. € Reactions in ideal gases For systems of weakly interacting indistinguishable particles (like ideal gases) we know that ZN = (Z ) sp N! ∏ (Z sp ) N = Ni i ∏N ! i Zsp is the partition function of a single particle i F = −k BT ( N i ln Z sp − ln N i!) € = −k BT ( N i ln Z spi − N i ln N i + N i ) ∂ µi = −k BT N i ln Z spi − N i ln N i + N i ) ( ∂N i µi = −k BT (ln Z spi − ln N i ) € Law of mass action Thus our condition for chemical equilibrium now becomes ∑ b (ln Z i spi − ln N i ) = 0 i Or to put it in a really compact form: € ∏ (Z i bi bi ) = (N ) ∏ spi i i This is known as the law of mass action. If we know the partition functions, we€know the concentrations of reactants and products. Chemistry Without going into detail, we know that the partition functions must be some function of volume and temperature. Therefore the law of mass action states: N1b1 N 2b2 N 3b3 ...N ibi = K(T ,V ) where K is known as the ‘equilibrium constant’ of the reaction. € Plasma Physics • Rather than do any chemistry, let us do some physics that uses the same ideas - plasma physics. • The ionization of atoms looks similar to a chemical reaction. • Let us consider the simplest case: the ionization of atomic hydrogen. H ⇔ p + + e− p + + e− − H = 0 • We will assume that the density is low enough, and the temperature high enough, that the electrostatic energy can be ignored € compared with the thermal € energy. Thus we assume a very weakly interacting particles (ideal gas). Saha Equation (1) bi bi (Z ) = (N ) ∏ spi ∏ i i i be− = b p+ = 1 , bH = −1 € € N e− N p + NH = Z sp − Z sp + e p ZH But in a hydrogen plasma, the number of electrons is the number of protons, so N e2− Z spe− Z sp p+ € = NH ZH So now we need to work out the partition functions…. Partition functions (1) • To work out how ionized the gas is, we need to work out the partition functions of electrons, protons, and hydrogen atoms • In order to do this, we need to decide what we mean by zero energy • We will take zero energy to be free electrons/protons at rest • Note that a static recombined atom in the ground state has an energy of -13.6eV (the Rydberg) on this energy scale Partition functions (2) The electrons and protons look like free gas atoms…the partition function is just the total volume divided by the volume of the particle Z sp − 2 πme k BT 3/2 = 2 ×V 2 h Z sp + 2 πm p k BT 3/2 = 2 ×V 2 h e spin p The volume occupied by a particle is given by the deBroglie wavelength € So do the hydrogen atoms, but a static H atom has an energy of -Ry compared with a static electron/protron: € +Ry 2 πm H k BT 3/2 Z sp = 2 × 2 × exp V 2 H h k BT Saha Equation (2) N e2− Z spe− Z sp p+ = NH ZH 3/2 3/2 2 π m k T 2 πme k BT p B 2V 2V 2 2 N e2− h h €= Ry 2 πm H k BT 3/2 NH 4 exp V 2 h k BT assume mp=mH, then € −Ry 2 πme k BT 3/2 N e2− = V exp 2 NH k BT h Saha Equation Solving Saha for Hydrogen Define the degree of ionization, ξ which is defined in terms of the total number of atoms (ionized +unionized), N0, such that € N e = N p = ξN 0 N H = (1− ξ )N 0 so the Saha equation can be written as 3/2 ξ € V −Ry 2 πme k BT = exp = f (N 0 ,V,T ) 2 (1− ξ ) N 0 k BT h 2 ξ= € f2+4f − f 2 Typical Numbers for Hydrogen As an example, take a density of atoms/ions of 1020 m-3 (typical Tokamak plasma) T (eV) f(N0,V,T) ξ 0.6 2x10E-3 4.4E-02 0.8 0.9 0.6 1.0 37.4 0.975 Plasmas are ‘easy’ to make • Notice that the hydrogen is 97% ionized at a temperature of 1.0 eV • But the ionization energy was 13.6 eV • The plasma is created at a ‘low’ temperature. Why? • The answer is in the statistical weight…ionization occurs by those few high energy electrons on the tail of the distribution function • The continuum is ‘heavy’ compared with the recombined atoms – explains why it is slightly easier to ionize as the density decreases Multi-electron atoms (1) Let AZ represent the ion A with Z out of a total of Z’ electrons removed. A0 ⇔ e− + A1 .................... AZ ⇔ e− + AZ +1 .................... AZ ' −1 ⇔ e− + AZ ' Z sp(e− ) Z sp(Z +1) € Z sp(Z ) N e− N (Z +1) € = N (Z ) Can no longer assume N e− = N (Z +1) Multi-electron atoms (2) Z' N e− = N1 + 2N 2 + ...+ ZN Z + ...+ Z ' (N Z ' ) = ∑ iN i i=0 € IZ Z sp(Z +1) g(Z +1) = exp Z sp(Z ) gZ k BT Procedure: (1) For given ion density guess Ne. (2) Use this guess to € all the ion density ratios computationally. (3) Check if determine this set of ratios of ion densities is consistent with the initially guessed electron density. (4) If not adjust guess. (5) Iterate until convergence achieved. Example Aluminum at a density of 1020 cm-3 Debye Shielding (1) + - + - - + - + - + + - + Excess of electrons close to test charge, deficit of ions close to test charge. + - - + - + + + - - + + - Debye Shielding (2) Assume potential at a distance r from the test charge is φ (r) eφ (r) eφ (r) ne (r) = n0 exp ≈ n0 1+ kT kT € Therefore there is an excess of electrons of order eφ (r) n (r) ≈ n0 kT + e € By similar reasoning there is a deficit of ions at the same position € eφ (r) n (r) ≈ −n0 kT − i Debye Shielding (3) Therefore the excess charge density at the point r is given by eφ (r) ρ (r) = e(n (r) − n (r)) ≈ −2n0 kT − i + e For self consistency the potential itself is related to the charge density by Poisson’s equation € 2 1 ∂ ∂φ ρ (r) 2n e ∇ 2φ (r) = 2 r 2 = − = 0 φ (r) r ∂r ∂r ε 0 ε 0 kT € Debye Shielding: solution 2r A φ (r) = exp− r λD where the Debye length is given by € ε 0 kT λD = ne2 2r Q € φ (r) = 4 πε r exp− λ 0 D Plasma Parameter A ‘good’ plasma is one that has a large number of particles within a Debye Sphere: 4 N D = n 0 πλ3D >> 1 3 Also note that the Debye length is the typical distance that a thermal electron travels during a plasma period… € ε 0 kT kT ε 0 m vth λD = = = 2 2 ne m ne ω p € Plasma Oscillations • Ions are heavy compared with electrons - assume they are stationary • Displace the electrons from the ions and ‘let go’ - the resulting oscillation occurs at what is known as the plasma frequency • Note we assume no collisions take place • We also assume that the electrons are cold - their motion is due to the electrostatic restoring force – we ignore thermal motion Plasma Frequency l Apply Gauss’ law: 2 nel x 2 El = ε0 nex E= ε0 2 + + + - + - + - + - + - + - - - + + + + - + - + - + - + - - - - + + + - + - - + + + + - - + - + - + - + overall neutral + - + - + - + - - - - + + + - + - - - + + + + - + - + - + - + - - - - + - + - + - Ex E 2 + - x Ex = 0 (electrons flow to left) € d x ne x Equation of motion ∴ F = m 2 = −eE = − dt ε0 2 ne 2 ω Oscillation at frequency p = ε0 m € Typical plasma frequencies Useful aide memoire : fp ~ 9000 n 1/2 (n in cm-3). Ionosphere , n ~ 104 cm-3 fp ~ 1 MHz Tokamak n ~ 1012 cm-3 fp ~ 10 GHz Laser plasma n ~ 1021 cm-3 fp ~ 1 THz Note about plasma frequency 1/2 ε 0 k BT k BT 1 vth λD = = ≈ 2 m ωp ωp ne € Note that the Debye length represents the distance travelled by a typical thermal electron during the oscillation period of one plasma wave. Summary • All equilibrium chemistry is in the partition functions of the reactants and products • The density of reactants and products is determined by the law of mass action ∏ (Z ) = ∏ (N ) • Ionization is a little like a chemical reaction - applying the law of mass action leads to the Saha equation. bi bi spi i € i i −Ry 2 πme k BT 3/2 N e2− = V exp 2 NH h k BT • The ground state of the atom dominates the partition function € Summary • Charges in a plasma are shielded over a distance of order the Debye length ε 0 kT λD = ne2 • Electrostatic waves occur in plasmas, in a cold plasma they are independent of wavelength, and oscillate with a frequency given by € ne2 ω = ε0 m 2 p € Graduate Course Opacity and X-ray Spectra • Einstein A and B Coefficients • Introduction to Opacity • Various forms of equilibrium Blackbody Radiation • In undergrad classes we studied blackbody radiation – i.e. a thermal distribution of photons. • But how do photons get into thermal equilibrium? To put it another way, why does a sodium or hydrogen lamp emit a line spectrum, but a tungsten filament a ‘blackbody’ continuum? Or when do you see line radiation from a plasma, and when a continuum? The Blackbody/Line Spectrum Problem • To answer this riddle we will use a circular argument (!) • We will first assume Blackbody radiation can exist • We will use this fact to determine the Einstein A and B coefficients (as in Atomic Physics) • We will then explain the difference between the line radiation source and the blackbody source Reminder about Blackbody Radiation V dω 3 ρ(ω )dω = 2 × 2 3 ω 2π c [exp(ω / k BT ) − 1] ω 3 dω ρ(ω )dω = C [exp(ω / kBT ) −1] where V C= 2 3 π c Einstein A and B Coefficients (1) • Put an atom at temperature T in a blackbody radiation bath also of temperature T • Assume the atom only has two levels, populated by the Boltzmann distribution • Equate the transition rates between the levels (i.e. stimulated absorption = stimulated emission + spontaneous emission) Einstein A and B Coefficients (2) N2 atoms, energy ε2 Degeneracy g2 ε 2 − ε1 = ω 0 N1 atoms, energy ε1 Degeneracy g1 −ω N2 g2 0 = exp N1 g1 k T B A21N2 ρ(ω 0 ) N2 B2 1 ρ(ω 0 ) N1 B1 2 N2 ( A2 1 + ρ (ω 0 ) B2 1) = N1 ρ (ω 0 ) B1 2 Einstein A and B Coefficients (3) g2 −ω 0 −ω 0 3 3 A 1 − exp + CB ω exp = CB ω 2 1 2 1 0 1 2 0 k T k T g1 B B This equation must hold at all temperatures. Thus if T is very high we find B2 1 g1 = B1 2 g2 Whereas if T is very low we see g1 A2 1 = Cω 30 B1 2 = Cω 30 B2 1 g2 Line Radiation The line emission has a finite width - perhaps due to Doppler broadening, pressure broadening, or natural lifetime. Gaussian Line Profile 1.2 1 0.8 0.6 0.4 0.2 0 Broadening of the line could come about due to the thermal motion of the atoms. ω0 1 Frequency ∞ ∫ φ (ω )dω = 1 0 Optically thin source Gaussian Line Profile 1.2 1 ∞ 0.8 ∫ φ (ω )dω = 1 0.6 0.4 0 0.2 0 1 ω0 Frequency If the source is not very thick, the photons have little chance of being reabsorbed on the way out, we see line radiation. An optically thick source (1) Gaussian Line Profile 1.2 1 0.8 0.6 0.4 0.2 0 ω0 1 Frequency Gaussian Line Profile 1.2 1 0.8 0.6 L 0.4 0.2 0 1 ω0 Frequency Absorption length L ~ 1/κω Where κω is the absorption coefficient An optically thick source (2) • Lots of photons are emitted at line centre, but the absorption depth is also short - we only see photons from roughly the last absorption depth • Not so many photons are emitted from the wings of the line, but their chance of being reabsorbed is corresponding less - i.e. the absorption depth is much longer • These two effects cancel out, and the number of photons we see does not depend on the line shape! • Observed energy density ~ ρ ω = N2 A2 1φ ω ω × L ( ) ( ) An optically thick source (3) N 2 A21φ (ω )ω jω ρ(ω ) = = κω κω N 2 A21φ (ω )ω N 2 A21 ρ(ω ) = = N1B12φ (ω )ω N1B12 ω ρ(ω ) = Cω exp − kB T 3 Energy density looks similar to Blackbody Radiation in the high temperature limit. € An optically thick source (4) We forgot stimulated emission:- κ ω = ( N1 B1 2 − N2 B2 1)φ (ω )ω 1 jω ρ(ω ) = ρ BB (T ) = Cω = ω κω exp − 1 kB T 3 ρ BB (T ) is the energy density of a blackbody at temperature T The optically thin to thick transition (1) δρ(ω ) = [ jω − κ ω ρ (ω )]δx δρ(ω ) = [ ρ BB (T ) − ρ (ω )]κ ωδx Integrating over a length x:- ρ(ω , x ) ∫ 0 δx x δρ(ω ) = − ∫κ ωδx ρ(ω ) − ρ BB (T ) 0 which results in ρ(ω , x ) = ρ BB (T )[1− exp (−κ ω x )] The optically thin to thick transition (2) Tau = 0.1 Tau = 0.2 Tau = 0.5 Tau = 1.0 Tau = 2.0 Tau = 5.0 Tau = 20.0 1.2 1 Intensity 0.8 ρ(ω, x ) = ρ BB (T) if κ ω x >> 1 0.6 0.4 € ρ(ω, x ) = jω = N 2 A21φ (ω )ω if κ ω x << 1 0.2 0 Frequency € τ=κx at the center of the line (Doppler Profile) Hohlraums Cavity at temperature T Hohlraums are used to ensure that the surface of the material (which loses the photons) stays at the same temperature as the bulk of the material. Non-LTE Physics • Radiation escapes from the plasma - therefore the downward and upward rates are no longer in balance • There is competition between collisional and radiative processes Complete Equilibrium 2 σ2,1 Β1,2 - e e- Α2,1 hν hν Β2,1 σ1,2 1 N1ρ(ν1,2)B1,2+N1NeC1,2=N2ρ(ν1,2)B2,1+N2A2,1+N2NeC2,1 Detailed Balance (Collisions) 2 σ2,1 eeσ1,2 N1NeC1,2=N2NeC2,1 ΔE C2,1 g2 = exp− C1,2 g1 k BT Collisional equilibrium occurs when the electron density is high and the radiative rates can be€neglected Coronal Equilibrium Assume weak radiation field, and density so low that A2,1 >> NeC2,1 then we have ‘coronal equilibrium’ such that N1NeC1,2 = N2A2,1 so N2/N1 = (NeC1,2)/A2,1 Note the upper state population is now proportional to the electron density, and is clearly below the value given by Boltzmann’s law Example Aluminium - 400eV 0.001 Hy-2/Hy-1 1 0-5 1 0-7 1 0-9 1 0-11 1 014 1 016 1 018 Ne 1 020 1 022 Density dependence of ratios of ionization stages • Saha (LTE) predicts weak dependence on density • Collisional ionization is balanced by three-body recombination (if these are in balance we have Saha) • At lower densities collisional ionization is balanced by radiative recombination, and ionization is density independent Weak density dependence is used as a temperature diagnostic Aluminium at 101 9cm-3 Aluminium - 400eV 1 05 100 1 04 Hy/13+ Hy/13+ 1000 100 10 1 10 1 014 1 016 1 018 Ne 1 020 1 022 0.1 100 1000 Temperature (eV) Use ratios of high lying optically thin lines in LTE with next ionization stage to find ionization ratios and hence temperatures. Thermal contact between levels in weak radiation field If the plasma is thin, then radiation can escape. Under these conditions N1NeC1,2 = N2NeC2,1+N2A2,1 For one level to be in ‘thermal contact’ with the next level we need Ne >> A2,1/C2,1 In practice the numbers work out to be Ne >> 1.6x1012 x (Tkelvin)1/2 (ΔEev)3 m-3 Temperature Diagnosis • High lying states are in good ‘thermal contact’ with the ground state of the next ionization stage. • Therefore the ratio of lines from optically thin high lying states from different ion states is a good measure of the ratio of the ground state populations of successive ionization stages. • This ratio is strongly dependent upon temperature, but only weakly dependent on density - thus we have a temperature diagnostic. Temperature Diagnosis continued… Consider the intensity of radiation from level n in a hydrogenic ion to the ground state. I H (n,1) = AH (n,1)N H (n) But the level is in good thermal contact with the next ionization stage… χ H (n,∞) NB gB = exp− € N H (n) g H (n) k T B Similarly for a high lying level, n’, in a helium-like ion € χ He (n ' ,∞) N H (1) g H (1) = exp− ' ' N He (n ) g He (n ) k T B Temperature Diagnosis continued… Hence the ratio of the intensity of the lines is given by χ H (n,∞) − χ He (n ' ,∞) I H (n,1) AH (n,1)g H (n)g H (1)N B = exp ' ' ' I He (n ,1) AHe (n ,1)g He (n )g B N H (1) k BT € But the exponential is very close to unity (as the energy difference is so small – e.g. for Al n=4 of H, it is 144 eV, but for n=4 of He-like, is 123.3 eV, just over 20 eV different) Thus the intensity of the lines is a good measure of the ionization stage ratios, and thus a good measure of temperature Typical Spectrum Density Diagnosis • An ion in a plasma experiences an electric field due to the neighbouring electrons and ions • This field shifts the energy levels – the shift being some function of the field and therefore distance to the perturbing charge • This shifts the transition energy of the radiation • As there are many ions, there are a range of fields being experienced, leading to a broadening of the observed spectral line • The simplest example is Stark broadening of emission from hydrogenic ions, where linewidth ~ ne2/3 Stark Broadening • Shift of Degenerate levels in Hydrogen is proportional to electric field. • Field (and range of fields) scales as 1/r2. • Electron density scale as r-3. • Therefore linewidth scales as ne2/3 Line-Width Calculation Notice a full calculation has a line width that roughly scales with density as we predict. Summary • Blackbody radiation arises because in a thick source the probability of being absorbed is proportional to the probability of being emitted • In general plasmas are in collisional-radiative equilibrium • At low densities coronal equilibrium is appropriate. Ionization ratios are density independent, but upper state levels are proportional to density • Density insensitivity allows us to use high lying lines as temperature diagnostics (as they are in good thermal contact with the next ionization stage) • Line widths can be used as a density diagnostic