Nuclear Fusion and Astrophysical Plasmas

Question 1

What is plasma?

Answer 1

Ionized gas. ‘Fourth state of matter’. Higher enthalpy.

Quasi-neutral gas of charged and neutral particles which exhibits collective behaviour

Question 2

Ideal gas law (and assumptions)

Answer 2

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

Question 3

Maxwell-Boltzmann distributions

Answer 3

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

Question 4

Examples of plasmas

Answer 4

• Crookes Tube (Gas discharge in rarefied gas)
• The Sun
• Nebulae
• Fusion plasmas
• Lightning
• Flame
• Aurora
• Sterilization cold plasmas
• Plasma etching

Question 5

Difference between plasma and equilibrium gas

Answer 5

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)

Question 6

Pros/cons to controlled fusion

Answer 6

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

Question 7

Relevant fusion reaction

Answer 7

²₁D + ³₁T → ⁴₂He + ¹₀n + 17.6 MeV

Question 8

How to aquire fusion fuel

Answer 8

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.

Question 9

Energy released in reaction

Answer 9

Find mass deficit, convert to energy, scale up to amount of fuel

Question 10

Q

Answer 10

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

Question 11

Temp. eV conversion

Answer 11

T = kTₖ/e (eV)
k is boltzmann
Tₖ is temp in kelvin

Question 12

Energy distribution in fusion products

Answer 12

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

Question 13

Reaction rate

Answer 13

R = n₁n₂σv

n₁n₂, densities of reactants
σ cross-section (includes coulomb barrier, nuclear dynamics, etc)
v relative velocity

Question 14

How is energy used once released in fusion

Answer 14

• Charged particles: keep heat in plasma
• neutrons: breeding tritium, heat exchange
• Gammas: (lossy)

Question 15

Conditions for fusion

Answer 15

• High temperature to overcome electrostatic repulsion of nuclei:
  ⇒ Efficient DT fusion requires temperature of order 10keV
  ⇒ Fully-ionised plasma, 10 keV&raquo_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

Question 16

Two main types of fusion reactor

Answer 16

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

Question 17

What is the Lawson criterion

Answer 17

Quantifies condition under which efficient production of fusion energy is possible

Essentially compares generated fusion power with additional power required to heat plasma

Question 18

Derive the Lawson criterion

Answer 18

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

Question 19

Power loss in a plasma

Answer 19

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

Question 20

Ratio of fusion to heating power

Answer 20

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

Question 21

Ignition condition

Answer 21

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

Question 22

Cyclotron frequency and radius

Answer 22

Derive using centipetal and Lorentz forces
ω꜀ = |eB/m|

r_L = |m/vₚₑᵣₚ/eB|
Larmor radius

Question 23

Main field components in a tokamak

Answer 23

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

Question 24

Plasma in the sun

Answer 24

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.

Question 25

Planetary magnetospheres

Answer 25

• 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

Question 26

Outer heliosphere

Answer 26

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

Question 27

What causes aurorae?

Answer 27

Interaction of energetic magnetospheric and solar wind charged particles with Earth’s neutral atmosphere

Question 28

Average energy per particle

Answer 28

N particles of mass m and velocity u

⟨E⟩ = 1/2N Σ(1toN) mᵢuᵢ²

Question 29

Average kinetic energy in a plasma

Answer 29

⟨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

Question 30

Saha Equation (degree of ionisation)

Answer 30

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

Question 31

Derive plasma oscillation

Answer 31

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’

Question 32

Electric field shielding frequency

Answer 32

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)

Question 33

Frequency criterion for plasma

Answer 33

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 34

Question

Answer 34

Answer

Question 35

Derive Debye length

Answer 35

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

Question 36

How to show Deybe shielding

Answer 36

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

Question 37

Potential for test charge in a plasma

Answer 37

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

Question 38

Plasma parameter

Answer 38

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

Question 39

Force between two charged particles

Answer 39

F₁₂ = -q₁q₂/4πε₀r² r̂₁₂

Question 40

Collision time, cross section, change in momentum, and collision frequency for an electron-ion collision

Answer 40

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

Question 41

Heat and collision freq.

Answer 41

hotter plasmas have lower collision frequencies as faster particles are less likely to collide.

Question 42

Condition for magnetic confinement

Answer 42

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

Question 43

Plasma Beta

Answer 43

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

Question 44

Summary of plasma criteria and characteristics

Answer 44

• Criterion #1: τ꜀ ≫ τₚ or τ꜀/τₚ ≫ 1
• Criterion #2: λ_D ≪ L
• Magnetised plasma (optional): L ≫ r_L
• Collisional if L&raquo_space; λₑᵢ
• Collisionless if L &laquo_space;λₑᵢ

Question 45

How does the B-field affect particle motion?

Answer 45

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.

Question 46

Properties of the cyclotron frequency

Answer 46

• 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.

Question 47

Derive Larmor radius

Answer 47

v x B is centripetal
– mv⊥²/r = qv × B
= q v⊥ B

=> r_L = mv⊥/|q|B = v⊥/ω꜀

Question 48

Gyrofrequency

Answer 48

f꜀ = ω꜀/2π

Question 49

Helical motion in a uiform B-field

Answer 49

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∥

Question 50

Pitch angle of helix

Answer 50

α = tan⁻¹ (v⊥/v∥)

Question 51

Magnetic moment of a charged particle in motion

Answer 51

µ = 1/2 mv⊥²/B

Question 52

E × B drift

Answer 52

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.

Question 53

Particle motion in uniform electric and magnetic field

Answer 53

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.