Nuclear Fusion and Astrophysical Plasmas Flashcards
What is plasma?
Ionized gas. ‘Fourth state of matter’. Higher enthalpy.
Quasi-neutral gas of charged and neutral particles which exhibits collective behaviour
Ideal gas law (and assumptions)
Assume
* Perfectly elastic collisions thermalize gas to equilibrium temperature T
* Otherwise non-interacting
* Independently satisfy two equations of state:
Ideal gas law
PV = Nk₆T
Ideal internal energy
uV = ĉᵥNk₆T
where
ĉᵥ = dimentionless specific heat capacity, e.g. 3/2 for monatomic molecules
(V)olume, (N)o. of particles, (T)emperature, (k₆) Boltzmanns constant, (P)ressure, (u) energy density
Maxwell-Boltzmann distributions
Velocity (1D travel)
f(vᵢ) dvᵢ = √m/2πkT’ exp(-mvᵢ²/2kT) dvᵢ
Speed (energy)
f(v) dv = (m/2πkT)³’² 4πv² exp(-mv²/2kT) dv
Examples of plasmas
- Crookes Tube (Gas discharge in rarefied gas)
- The Sun
- Nebulae
- Fusion plasmas
- Lightning
- Flame
- Aurora
- Sterilization cold plasmas
- Plasma etching
Difference between plasma and equilibrium gas
Gas has binary colissions, plasma has long-range interactions
Fully ionized plasma in equilibrium is like two independent ideal gasses of electrons and ions
Plasmas are good conductors
Electron and ion temps can differ, and distributions can be non-maxwellian (so not in equilibrium)
Pros/cons to controlled fusion
Fossil fuels (coal, gas, oil):
* chemical reactions
* energy density based on molecular bonds (0.1 – 1 eV/nucleon)
* global warming
* finite fuel supply
Nuclear fission:
* nuclear reactions
* energy density based on nuclear binding energy
* safety concerens
* waste disposal
Renewables:
* rely mostly on the sun
* solar, wind, hydro-electric, etc.
* typically low energy density
* dependent on geography and weather
* discontinuous
Nuclear fusion
* nuclear reactions
* energy density based on nuclear binding energy (~10 MeV/nucleon)
* hella fuel available (deuterium/tritium)
* CO₂ neutral
* yields small quantities of radioactive waste.
* no risk of uncontrolled energy release
* fuel is available in all locations on earth
Relevant fusion reaction
²₁D + ³₁T → ⁴₂He + ¹₀n + 17.6 MeV
How to aquire fusion fuel
Deuterium found naturally in water, cheaply obtainable
Virtually no naturally available Tritium
Can be bred from Lithium (found naturally on land)
⁶₃Li + ¹₀n → ⁴₂He + ³₁T + 4.8 MeV
⁷₃Li + ¹₀n → ⁴₂He + ³₁T + ¹₀n - 2.5 MeV
Neutron released in fusion reaction can be used here.
Energy released in reaction
Find mass deficit, convert to energy, scale up to amount of fuel
Q
Fusion power output, characterized by power amplification factor Q
Q = (fusion power)/(heating power)
- Want at least Q >1 (scientific breakeven)
- “Ignition” (self-sustaining reactor) Q ≃ 5
- Practical working reactor (“burning plasma”) at Q ≃> 30
Temp. eV conversion
T = kTₖ/e (eV)
k is boltzmann
Tₖ is temp in kelvin
Energy distribution in fusion products
Energy and momentum conserved
1/2 mₐvₐ² + 1/2 m₆v₆² = Eբᵤₛᵢₒₙ
mₐvₐ + m₆v₆ = 0
Solve to give
Eₐ = 1/2 mₐvₐ² = m₆/mₐ+m₆ Eբᵤₛᵢₒₙ
E₆ = 1/2 m₆v₆² = mₐ/m₆+mₐ Eբᵤₛᵢₒₙ
Reaction rate
R = n₁n₂σv
n₁n₂, densities of reactants
σ cross-section (includes coulomb barrier, nuclear dynamics, etc)
v relative velocity
How is energy used once released in fusion
- Charged particles: keep heat in plasma
- neutrons: breeding tritium, heat exchange
- Gammas: (lossy)
Conditions for fusion
- High temperature to overcome electrostatic repulsion of nuclei:
⇒ Efficient DT fusion requires temperature of order 10keV
⇒ Fully-ionised plasma, 10 keV»_space; hydrogen ionisation energy 13.6 eV - Energy produced by fusion must exceed energy to heat plasma (Q > 1)
- Confinement time high enough to achieve this
⇒ Rection rate produces enough heat to sufficiently heat plasma
for duration of confinement time
Two main types of fusion reactor
Magnetic confinement
* moderate density
* moderate confinement time
* magnetic fields confine plasmas in a torus, where they can undergo fusion
Inertial confinement
* very high density
* very short confinement time
* lasers heat and compress fuel
* direct drive, lasers incident on fuel
* indirect drive, lasers bounce in small metal cylinder containing fuel
What is the Lawson criterion
Quantifies condition under which efficient production of fusion energy is possible
Essentially compares generated fusion power with additional power required to heat plasma
Derive the Lawson criterion
Considering only deuterium - tritium (D-T) fusion, where plasma contains equal number density of D and T ions
nD = nT
Assume fully-ionised plasma
nₑ = nD + nT
Assume all species behave like ideal gasses in thermal equilibrium. Energy density of plasma:
Eₚ = 3/2 (nₑ + nD + nT) k₆T
Two ideal gasses, particles dont collide with singular relative speed. Cross section will generally be a function of relative speed.
Expectation value of rate
R = n₁n₂ ∫ p(v)vσ(v) dv
where p(v) is probability of two particles having relative velocity v, and σ(v) is cross section as a function of relative speed
Assumed thermal distribution, so velocities given by Maxwell-Boltzmann distributions:
p(v) ∼ v² exp(−mv²/2k₆T)
Wont evaluate integral, trust me bro, is often written ⟨vσ⟩ such that
R = n₁n₂⟨vσ⟩
Power output per unit volume, defining Eբ as energy released per fusion reaction. (17.6 MeV for D-T)
P = n₁n₂⟨vσ⟩Eբ
Reaction is confined for time τ_E,
the ‘energy confinement time’
Reaction ‘breaks even’ if energy produced in this time, P τ_E , is equal to energy required to heat reactant plasmas up to reactor temperature.
Define n = nD + nT as total ion density,
Eₚ = 3nkT
Heating constituent plasma from room temperature (T ≈ 300 K) to practical operating point (Tf ≈ 10 keV or 100 ×10⁶ K).
Energy needed to heat up plasma is:
Eₚ = 3nkTf
Inequality between between energy produced and energy required to heat, get basic Lawson Criterion:
P τ_E = nD nT ⟨vσ⟩ > 3nkTf
nτ_E > 12kTf/⟨vσ⟩Ef
Power loss in a plasma
For energy confinement time τ_E
∂W/∂t = - W/τ_E + Pₕₑₐₜ
where W is stored energy density
W = [³⁄₂ nD TD + ³⁄₂ nT TT + ³⁄₂ nₑ Tₑ] k₆V ≃ 3nk₆TV
For steady state want:
∂W/∂t = 0
Ratio of fusion to heating power
If plasma stationary:
W = 3nk₆TV = Pₕₑₐₜτ_E
nk₆T = 1/3V Pₕₑₐₜτ_E
Combine with fusion power (DT reaction)
Pբᵤₛᵢₒₙ = 7.7 n₂₀² Tₖ² V kW
n-T-tau product for DT fusion
Pբᵤₛᵢₒₙ/Pₕₑₐₜ = 0.16 n₂₀² Tₖ τ_E
Will achieve breakeven if n-T-tau > 6
τ_E is in seconds
Tₖ is temperature in keV
n₂₀ is density in units of 10²⁰ m⁻³
Ignition condition
Energy produced by and retained by the fusion reaction is sufficient to heat the plasma.
For DT, only He atoms are confined (neutrons escape the magnetic field) therefore only 20% of total fusion power is available for plasma heating
Cyclotron frequency and radius
Derive using centipetal and Lorentz forces
ω꜀ = |eB/m|
r_L = |m/vₚₑᵣₚ/eB|
Larmor radius
Main field components in a tokamak
Toroidal field (long way round torus) generated by toroidal field coils (solenoid bent into circle)
Poloidal field (short way round torus) generated by plasma current. Current provided by transformer action from central solenoid (poloidal field coils)
Field lines spiral around the chamber
Plasma in the sun
Solar Interior
* Core - hot, very dense plasma
T ≈ 15 million K at centre, n ≈ 10³² m⁻³
* Outer layers – magnetic field generated by “dynamo” due to convective and other motions.
Atmosphere
* Photosphere - T ≈ 6000 K at centre, n ≈ 10²³ m⁻³. Weakly-ionised
“Patchy” magnetic field concentrated into isolated intense tubes of magnetic flux. Largest concentrations seen as sunspots.
* Chromosphere – highly structured. Hotter than photosphere. Temperature rises with height while density falls
* Corona - hot, low-density outer atmosphere T ≈ 1-2 million K, n ≈ 10¹⁴⁻¹⁵ m⁻³. Permeated by magnetic fields.
The Sun’s magnetic field varies on a (roughly) 11 year cycle.
Interaction of solar magnetic field with plasma causes much activity –
notably solar flares and coronal mass ejections
Solar wind is a supersonic flow of plasma at ~500 km s⁻¹
At 1 AU, T ≈ 10⁵ K, n ≈ 5 X 10⁶ m⁻³.
Planetary magnetospheres
- The Earth (and many other planets) are protected from Solar Wind plasma and high energy particles from the Sun by its magnetic field. This forms a cavity in the solar wind called the Magnetosphere – bounded by the
magnetopause. Upstream of the Earth, the Solar Wind is slowed down to subsonic speeds by the Bow Shock - Closer to the Earth is the Ionosphere – the boundary layer between the neutral atmosphere and the plasma magnetosphere. This consists of weakly-ionised plasma, with ionisation by solar UV
Outer heliosphere
The heliosphere forms a cavity within the Intersellar Medium. It is bounded by the heliopause. As the solar wind reaches this boundary, it slows down through a Termination Shock
What causes aurorae?
Interaction of energetic magnetospheric and solar wind charged particles with Earth’s neutral atmosphere
Average energy per particle
N particles of mass m and velocity u
⟨E⟩ = 1/2N Σ(1toN) mᵢuᵢ²
Average kinetic energy in a plasma
⟨E⟩ = ( ∫±∞ 1/2 mu² f(u) du ) / ( ∫±∞ f(u) du )
Integrate numerator by parts, and denominator using
∫±∞ exp(-ax²) dx = √π/a’
Get
⟨E⟩ = 1/2 k₆T in 1D
or
⟨E⟩ = 3/2 k₆T in 3D
Saha Equation (degree of ionisation)
nᵢ/nₙ = 2.405 ×10²¹ T³’²/nᵢ exp(-U/k₆T)
nᵢ is density of ionized atoms,
nₙ is density of neutral atoms,
U is ionization potential energy,
T is plasma temperature,
nᵢ in denominator on right is from recombination
Quasi-neutrality:
nₑ ≃ nᵢ
nₜₒₜₐₗ ≃ 2nᵢ + nₙ
Derive plasma oscillation
Picture two grids of +ve (E₊) and -ve (E₋) particles.
Gaussian box of length ∆x and cross section A
From Gauss’s Law
2E₊A = ρ₊ A∆x/ε₀
where ρ₊ = ne
and n = nₑ + nₚ
Field created
E = E₊ + E₋ = ne∆x/ε₀
Restoring force
F₋ = -eE = - ne²∆x/ε₀
Harmonic oscillator with electronic plasma frequency
ωₚ = √ne²/mₑε₀’
Electric field shielding frequency
Electric fields shielded (EM waves below wont propagate) when oscillating at a frequency less than
fₚ = ωₚ/π
≈ 9000 √nₑ’ Hz (with density in cm⁻³)
Frequency criterion for plasma
Partially ionized gas, collisions are important, plasma oscillations can only develop if mean free time between collisions (τ꜀) is long compared with plasma oscillation period (τₚ)
plasma requires
τ꜀ ≫ τₚ or τ꜀/τₚ ≫ 1
Otherwise, plasma behaves like neutral gas, electric forces have no time to play a siginificant role
Question
Answer
Derive Debye length
Electric field
E = − ne∆x/ε₀
requires work to create
– W = ∫ Fdx = ∫(0to∆x) e E(x) dx
= ne²∆x²/2ε₀
Thermal plasma with temperature Tₑ, particles have average kintetic energy 1/2 k Tₑ
Equate energy scales:
ne²∆x²/2ε₀ ≈ 1/2 k Tₑ
Maximum distance electron can travel (on average)is
∆xₘₐₓ ≈ λ_D
otherwise it is stopped by the electric field
Debye Length:
λ_D = √ε₀kTₑ/ne²’
Gas is a plasma if length scales L are larger than the Debye length
λ_D ≪ L
How to show Deybe shielding
Test charge in plasma with nᵢ = nₑ = n, temperature Tₑ
At t=0, electric potential is free-space
Φ(r) = 1/4πε₀ Q/r
Time progresses, electrons are attracted and ions are repelled
At t≫0, nₑ > n, new potential with charge density
ρ = e(nᵢ - nₑ)
Which can be subbed into Poisson
electron number density is given by Boltzmann Relation
nₑ₍ᵣ₎ = nexp(-qₑΦ(r)/kTₑ)
Sum into Poisson’s equation in spherical coords, then Talor expand for case |eΦ| ≪ kTₑ
e^x = 1+x…
Get
1/r² d/dr (r² dΦ/dr) ≈ − ne²/ε₀kTₑ Φ(r)
= 1/λ_D² Φ(r)
where λD is the Debye shielding length
Potential for test charge in a plasma
Φ(r) = 1/4πε₀ Q/r exp(-r/λ_D)
r → 0, potential is a free charge
r ≫ λ_D, potential falls exponentially, shielded by the plasma
Plasma parameter
Approximate number of particles in a Debye sphere is given by the plasma parameter:
Λ ≈ 4/3 π n λD³
or sometimes
Λ ≈ n λD³
If Λ ≪ 1, sphere is sparsely populated – Strongly Coupled Plasma
Strongly coupled plasmas are cold, and dense. Weakly coupled plasmas are diffuse and hot
Force between two charged particles
F₁₂ = -q₁q₂/4πε₀r² r̂₁₂
Collision time, cross section, change in momentum, and collision frequency for an electron-ion collision
Coulomb force between electron and ion
F꜀ = qₑ²/4πε₀r₀²
where r₀ is closest approach
Force felt for approx. average time τ꜀ ≃ r₀/vₑ
Coulomb collision cross-section
σ꜀ = πr₀²
Change in momentum experienced by electron is
∆(mₑvₑ) ≃ |F꜀τ꜀| = qₑ²/4πε₀vₑr₀
From probability of large vs small angle collisions, change in momentum ∆(mₑvₑ) can be expressed as a fraction of particle momentum ∆(mₑvₑ).
Electron-ion collision frequency is:
νₑᵢ ≃ √2’ωₚ⁴/64πnₑ (kTₑ/mₑ)⁻³’² lnΛ
νₑᵢ ≃ ωₚ/64π lnΛ/Λ
Heat and collision freq.
hotter plasmas have lower collision frequencies as faster particles are less likely to collide.
Condition for magnetic confinement
Length-scale of the system, L
– If L ≫ r_L, plasma is magnetised or magnetically-confined
– If L ≪ r_L, particles essentially unaffected by magnetic field
Plasma Beta
Relative importance of the magnetic field
β = Thermal energy density / Magnetic energy density
= p/(B²/2μ₀) ∼ nkTμ₀/B²
– Low β ≪ 1, magnetic fields drive dynamics
– High β ≫ 1, magnetic field plays minor role in dynamics
Summary of plasma criteria and characteristics
- Criterion #1: τ꜀ ≫ τₚ or τ꜀/τₚ ≫ 1
- Criterion #2: λ_D ≪ L
- Magnetised plasma (optional): L ≫ r_L
- Collisional if L»_space; λₑᵢ
- Collisionless if L «_space;λₑᵢ
How does the B-field affect particle motion?
EoM (Lorentz) for particle with mass m is
m dv/dt = q(E+ v × B)
If particles relativistic, replace m with
m = m₀/(1 – v²/c²)¹’²
where m₀ is particle rest mass.
If particle subject to static and uniform B field ONLY
m dv/dt = qv × B
Dot product with v
v · m dv/dt = v · q(v × B)
m 1/2 d(v · v)/dt = q[v · (v × B)]
RHS is zero, as v⊥B
=> d/dt (mv²/2) = 0
Static magnetic field cant change kinetic energy of a particle, since force always perpendicular to direction of motion.
Consider field lines that are straight and parallel with constant magnetic field strength.
Decompose velocity into parallel and perpendicular:
v = v∥ + v⊥
dv∥/dt + dv⊥/dt = q/m (v⊥ × B)
Split equation:
dv∥/dt = 0 => v∥ = const
dv⊥/dt = q/m (v⊥ × B)
B field has no effect on motion along it (∥), only perpendicular.
Consider
B = (0,0,B₂) = ẑB
m dvₓ/dt = qBvᵧ
m dvᵧ/dt = –qBvₓ
m dv₂/dt = 0
To determine time variation of vₓ and vᵧ, take derivatives
d²vₓ/dt² + ω꜀²vₓ = 0
d²vᵧ/dt² + ω꜀²vᵧ = 0
where ω꜀ = –qB/m
is the gyrofrequency or cyclotron frequency.
Properties of the cyclotron frequency
- Indicative of field strength, and charge and mass of particles.
- Does not depend on kinetic energy.
- For electrons, ω꜀ is positive and electrons rotate in the right-hand sense.
- Plasma can have several cyclotron frequencies.
Derive Larmor radius
v x B is centripetal
– mv⊥²/r = qv × B
= q v⊥ B
=> r_L = mv⊥/|q|B = v⊥/ω꜀
Gyrofrequency
f꜀ = ω꜀/2π
Helical motion in a uiform B-field
d²vₓ/dt² + ω꜀²vₓ = 0
d²vᵧ/dt² + ω꜀²vᵧ = 0
Solutions are
vₓ = v⊥ exp(iω꜀t) = ẋ
vᵧ = m/qB v̇ₓ = ±iv⊥ exp(iω꜀t) = ẏ
where
v⊥= √vₓ² + vᵧ²’
is a constant speed in plane perpendicular to B.
Integrate
x – x₀ = –i v⊥/ω꜀ exp(iω꜀t)
y – y₀ = ±v⊥/ω꜀ exp(iω꜀t)
Take real parts and use Larmor radius
x – x₀ = r_L sin(ω꜀t)
y – y₀ = ±r_L cos(ω꜀t)
These describe a circular orbit about a guiding centre (x₀, y₀).
There is also a velocity v₂ along B which is not effected by B
Overall, get helical motion about a guiding center
r = x̂ x₀ + ŷ y₀ + ẑ (z₀ + v∥t)
Guiding center moves linearly along z with constant velocity v∥
Pitch angle of helix
α = tan⁻¹ (v⊥/v∥)
Magnetic moment of a charged particle in motion
µ = 1/2 mv⊥²/B
E × B drift
Choosing E to lie in x-z plane, so that Eᵧ = 0. EoM is then:
m dv/dt = q(E + v × B)
z-component of velocity is
dv₂/dt = q/m E₂
=> v₂ = qE₂/m t + v₂₀
Straight acceleration along B.
Transverse components are:
dvₓ/dt = q/m Eₓ ± ω꜀vᵧ
dvᵧ/dt = 0 ± ω꜀vₓ
Differentiate with constant E
v̈ₓ = –ω꜀²vₓ
v̈ᵧ = ±ω꜀ (q/m Eₓ ± ω꜀vᵧ)
= –ω꜀² (Eₓ/B + vᵧ)
Can write as
d²/dt² (vᵧ + Eₓ/B) = –ω꜀² (vᵧ + Eₓ/B)
using vᵧ = vᵧ + Eₓ/B.
Solved by
vₓ = v⊥ exp(iω꜀t)
vᵧ = ±i v⊥ exp(iω꜀t) – Eₓ/B
Larmor motion is similar to E = 0, but now with drift v_g of the guiding center in the –y direction.
As +ve charged particle moves parallel to E, gains energy, r_L increases. Opposite in second part of orbit, so radius decreases.
Difference between radius of curvature over orbit gives drift.
Particle motion in uniform electric and magnetic field
If E and B constant, motion is gyro-motion about B + uniform acceleration B (due to E∥) + constant electric drift vE (drift velocity) perpendicular to both E and B
v_E = E×B/B²
Electric drift is independent of q and m, so is same for all species.
