F. Robicheaux Auburn University Alabama, USArydphy04/lectures/Robi_1.pdf · Plasmas are conductors:...

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Basic Plasma ParametersF. Robicheaux

Auburn UniversityAlabama, USA

ToolsPencil & paper, calculator, PC/laptop, workstation, local cluster, national supercomputer center

Properties:Hot? (need to ionize the atoms/molecules)Density (Sun > 1024 cm-3, space 1 cm-3)Good conductor of electricityMagnetic fields (sometimes)

Examples:StarsIonosphereFusion devicesSolar WindInterstellar Gas

Plasma = ionized gas

Strongly Coupled Plasmas?

JPB 36, 499

Strongly Coupled PlasmasIf the temperature is low enough, PE > KE. Highly correlated motion between the charged particles.

Average volume occupied by one electron: 4πa3/3 = 1/na = (3/4πn)1/3

For ne = 109 cm−3 : a = 6.2 µm << size of plasma

Coulomb coupling parameter: ΓeI <PE>/<KE>Γe = (e2/4πε0a)/kBTe

For Te = 100 K & ne = 109 cm−3 : Γe = 0.027For Te = 1 K & ne = 109 cm−3 : Γe = 2.7

Dimensionless number! Dimensional analysis?

SummaryCoulomb coupling parameter: ΓeI <PE>/<KE>Γe = (e2/4πε0a)/kBTe where a = (3/4πn)1/3

Screening?

+QSlightly more red (- charges) near a +Q charge due to the thermal distribution

n ~ exp[-q V(r)/k T]

Plasmas are conductors: no E-fields?! Charges should be screened by the free charges in the plasma, but nonzero T prevents perfect screening.

Debye length – the distance that a charge screened by a factor of ~ e = 2.718…

Debye length from electrons & ions: 2eeB0D e n 2 / T k ε λ =

Debye Length TheoryNeed to self consistently solve for the potential.The change in electron density due to a potential is

T V(r)/k e n 2-

]}T V(r)/k exp[e - ]T V(r)/k {exp[-e n e )rδρ(

Poisson’s equation for the potential is

V(r) λ1 V(r)

T k εe n 2 rV(r)][

εδρ(r) - rV(r)][

rr1 V(r)

∂∂

=∂∂

Debye Length PhysicsThe solution to this equation is

r)exp(-r/λ V V(r) D

The Coulomb coupling parameter can also be written asΓe = (1/3) (a/λD)2

The coupling is small when there are a large number of electrons within a Debye sphere. (Many charges shift by a small amount.)

Strong coupling occurs when there are few charges within a Debye sphere.

Debye screening length: λD = (ε0 kB Te/2 ne e2)1/2

Langmuir Wave (plasma frequency)Imagine pulling the electrons in a region of the plasma slightly (distance x) to the left. What happens?

The electrons oscillate with the electron plasma frequencyωp = (e2 ne/ε0 me)1/2

For ne = 109 cm−3 : fp = 280 MHzFor ne = 107 cm−3 : fp = 28 MHz

The ions are essentially stationary.

Increasing t2

xεn e ε V

21 E ε V

x n V m 21 KE

– neutral +

+ neutral –

Plasma Frequency Theory, v<<cSolve Maxwell’s equation + Newton’s equation for small changes in density.

t)/m,r(E e- t),r(v]t),r(v[ t

t),r(v

0 t)],r(v t),rn([ t

t),rn(t)/ε,rρ( t),r(E

rrrrrrrrr

=∇⋅+∂

=⋅∇+∂

∂=⋅∇

Use E = E0 + δE, n = ne + δn, ρ = ρ0 - e δn, v = δvne is the background electron number densityassumes the background electron flow is 0

Plasma Frequency Theory, v<<cSolve Maxwell’s equation + Newton’s equation for small changes in density.

t)/m,r(Eδ e- t

t),r(vδ

0 t),r(vδn t

t),rδn(t)/ε,rδn( e- t),r(Eδ

=⋅∇+∂

∂=⋅∇

Use the t derivative of the middle equation and div of last

0 t),rδn( ε mn e

tt),rδn(

∂ rr

Electron plasma frequency: ωp = (e2 ne/ε0 me)1/2

Ion MotionThe large scale fields in the plasma can also give motion to the ions.Regions of slightly higher ion density is not completely screened by electrons. Ions will be pushed out of region of high ion density and pulled into low.

If the modulation in space is sinusoidal waveω = (kB Te qi/e mi)1/2 k (if Te ~ Ti, then v ~ thermal speed)

Ion Acoustic Wave (theory)The ion acoustic wave (for low T) can be found by noting that the electron charge density must almost exactly cancel the ion charge density.

t),r(n t),r(n e

T k - t),r(E

cons t),r(nlneT k t),rV(

)rV( eexp C t),r(neq t),r(n

≅≅

Ion Acoustic Wave (theory)The equations for the ion density and velocity flow:

t),rn( t),rn( e

T k - t),r(E

t)/m,r(E q t),r(v]t),r(v[ t

t),r(v

0 t)],r(v t),rn([ t

t),rn(

rrrrrrrrr

=∇⋅+∂

=⋅∇+∂

Use n = ni + δn, v = δvni is the background ion number densityassumes the background ion flow is 0

Ion Acoustic Wave (theory)The equations for the change in ion density and velocity flow:

t),rδn( n eT k - t),r(E

t)/m,r(E q t

t),r(vδ

0 t),r(vδn t

t),rδn(

=⋅∇+∂

Use the t derivative of the 1st equation and div of 2nd & 3rd

0 t),rδn( m e

q T k t

t),rδn( 2

=∇−∂

∂ rr

This has ignored terms ~ (k λD)2 and (Ti/Te) (ion pressure)

Ion acoustic wave dispersion relation:ω = (kB Te qi/e mi)1/2 k

Thermalization/RandomizationHow does the Maxwell-Boltzmann distribution become established?How does the direction of travel of an electron become randomized due to scattering with the ions?

Two body collisions are all that is needed.

Rate for a collision process is n <v σ>

Collision thermalization time: 1/τ =ne v[e4ln(Λ)/4πε02v4me

2] Λ = 4πε03kBTe λD/e2 ~ 1/θmin>>1For Te = 100 K & ne = 109 cm−3 : ln(Λ) = 6.0 ; τ = 0.064 µsFor Te = 10 K & ne = 109 cm−3 : ln(Λ) = 2.5 ; τ = 0.005 µs

Thermalization/RandomizationThe collision between two charged particles is well studied. The angle through which the particle scatters depends on the charge, reduced mass, relative velocity and impact parameter.

tan(θ/2) = Q e2/(4 π ε0 µ v2 b)

+Qb θ

Rutherford scattering cross section (Q=1):

dσ/d(cosθ) = 2πe4/[(µ 4πε0)2 v4 (1 – cosθ)2]

Angle RandomizationThe electron-ion collisions will give a spread in velocity directions for the electron. In a time δt, the direction spreads by an amount:

<δθ2> = n v δt [e4 ln(Λ)/2πε02 v4 me

2] whereΛ = 4πε0me v2 λD/e2

Why does the Debye length come into this expression?

The plasma modifies the Coulomb potential at large distances. This changes the amount that scatters into small angles (large impact parameter).

The Debye length gives the distance over which the potential is present.

Angle Randomization (Theory)The angle deviation in a time δt diverges if all scattering angles are allowed. Restrict the minimum angle using the Debye length for the maximum impact parameter:tan(θmin/2) ~ θmin/2 = e2/(4 π ε0 µ v2 λD)

The angle deviation is given by

)cos(θ

vm ε π2eδt n v

)cos( d )cos( d

d )]cos( - [1 2δt n v δθmin

2] Λ = 4πε03kBTe λD/e2 ~ 1/θmin>>1

Three Body Recombination (TBR)Two electrons collide in the field of an ion so that one electron loses so much energy to become bound.

Three body recombination rate (e− + e − + A+ e − + A*):Γ = 2 X 10−39 m6 s−1 neni(eV/kBTe)9/2

For Te = 50 K & ne = 109 cm−3 : Γ = 10−4 µs−1

For Te = 10 K & ne = 109 cm−3 : Γ = 0.1 µs−1

For Te = 1 K & ne = 109 cm−3 : Γ = 4000 µs−1

Recombination into states bound by ~4kBTe (size of atom ~ distance between ions at 1 K!)

Three Body Recombination (theory)It is not possible to calculate the TBR rate using pencil & paper. Can calculate how the rate scales with different parameters then use Monte Carlo simulation to determine unknown dimensionless parameters.

Γ= n v σ P = const. X n2 T-9/2

b = size of atom ~ e2/(4 π ε0 kB T)

n = number density

v = velocity ~ T1/2

σ = cross section = π b2 ~ T-2

P = probability for finding another electron ~ n b3 ~ n T-3

TBR (qualified)In most plasmas, the time to scatter the resulting Rydberg atom into the ground state is short compared to 1/Γ. The TBR rate is then the rate for generating ground state atoms.

For ultra-cold plasmas, the time to scatter to the ground state is much longer than the capture time. Need to follow electron-Rydberg collisions.

The capture step itself is the result of MANY collisions. Atoms can re-ionize if BE < 8 kB Te. The many collisions takes time and it could be important if the ion were only briefly exposed to the electrons.

electron-Rydberg collisionsThe electrons cause energy transitions in atoms. The atoms can be l-mixed, de-excited, excited, and ionized.

Again, this problem is too complicated for pencil and paper. But don’t need to solve for all possible incident energies and binding energies. Classically, the Coulomb problem scales (only E0/BE important). Numerically, solve for one binding energy and scale to other BE’s.

l-mixing is much faster than all other processes.

electron-Rydberg collisions (trends)Excitation and ionization requires the incoming electron loses energy; de-excitation more likely when BE >> E0.

The ratio of probability for de-excitation by different energy amounts does not change rapidly as E0 0.

The de-excitation cross section diverges as E0 0 because the dipole can pull electrons in from large distances and there is no energy barrier.

Atoms bound by roughly 5 kB Te are ~ equally likely to be re-ionized or driven to much larger BE.

Classically, the excitation and de-excitation rates are smooth functions that can be fit by simple expressions.

Photon EmissionRydberg atoms can lose energy by emitting a photon.

Accelerating charged particles emit light.

Classical: P = 2 q2 a2/(3 c3 4 π ε0)

Quantum: P = 4 q2 ω2 |pif|2/(3 c3 4 π ε0)

Photon Emission (Simplified)The matrix elements for full quantum calculation of all transitions are extremely tedious.

In many (most?) plasmas, the collisions of the electrons with the atoms cause the angular momentum to completely mix. Only need to calculate radiation from one n-manifold to another.

)n - (n n n1

a π3 3c α 16 A 2

nn fi=→

Radiative decay (ni nf): )n - (n n n1

a π3 3c α 16 A 2

nn fi=→

F. Robicheaux Auburn University Alabama, USArydphy04/lectures/Robi_1.pdf · Plasmas are conductors:...

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