Fundamentals of Plasma Physics

17.7 The strongly coupled regime: crystallization of a dusty plasma 499

In region 1,ψ >> 1 so that the ion density is given in principle by Eq.(17.72) which
would give a right hand side scaling as

√

ψ.This ion term would dominate the electron and
dust terms since the latter are always less than unity in this region. Thus,keeping just the
ion term on the right hand side, Poisson’s equation in region 1 reduces to

1

r ̄^2

∂

∂ ̄r

(

̄r^2

∂ψ

∂ ̄r

)

︸ ︷︷ ︸

vacuum

≃

2

√

πψ

ψ

︸︷︷ ︸

ion

; (17.78)

however, the ion term above can be neglected compared to the vacuum term√ because
ψ >> 1 sinceψ >> 1 .Thus Poisson’s equation can be approximated in region 1 by
just the vacuum term. This assumption is quite good near the dust grainsurface whereψis
indeed very large compared to unity, but becomes marginal whenψapproaches unity at the
outer limit of region 1.
In region 2, the ions have a Boltzmann distribution soni/ne 0 =1+ψ.SinceTi/Te<<
1 , the normalized electron density isne/ne 0 = exp(−ψTi/Te)≃ 1 .BecauseZψ>> ̄ 1 ,
the dust density is nearly zero in this region.
In region 3, Eq.(17.74) approximates to

1

̄r^2

∂

∂ ̄r

(

r ̄^2

∂ψ

∂r ̄

)

︸ ︷︷ ︸

vacuum

≃ 1+ψ

︸︷︷︸

ions

−(1−α)

(

1 −

ψTi

Te

)

︸ ︷︷ ︸

electrons

−

(

1 −Zψ ̄

)

α

︸ ︷︷ ︸

dust

≃ (1+αZ ̄)ψ (17.79)

whereTi/Te<< 1 has been used. For finiteα, the dust term dominates becauseαZ>> ̄ 1.
The system has two boundary conditions, one at the dust grain surface and the other at
infinity. The former occurs in region 1 and is set by the radial electric field on the dust grain
surface. This boundary condition is obtained from Gauss’ law which relates∂ψ/∂r ̄at the
dust grain surface to the dust grain charge and radius. The boundary condition atinfinity
can be considered as the larger ̄limit for region 3 and requiresψto vanish as ̄rgoes to
infinity.
If the same form of Poisson’s equation were to characterize regions 1, 2, and 3, then
the system would be governed by a single second-order ordinary differential equation and
the two boundary conditions described in the previous paragraph would suffice to give
a unique solution. However, the equations in regions 1, 2, and 3 differ from each other
and so must be solved separately with appropriate matching at the two interfaces between
these three regions. The criteria for matching at the interfaces, determined by integrating
Poisson’s equation twice across each interface, are that bothψand its radial derivative must
be continuous at each interface. An important feature of this analysis is thatthe locations
of the two interfaces are unknowns to be determined by solving the matching problem.
Since regions 1 and 2 are of finite extent, both decaying and non-decayingψsolutions are
allowed in these regions. However, only a decaying solution is allowed inregion 3.
Region 1, which is governed by Eq.(17.75), has vacuum-like solutions of the form
ψ∼c ̄r−^1 +dwherecanddare constants. The constantcis chosen to give the correct
radial electric field at the surface of the dust grain test charge anddis chosen so thatψ=1
atr ̄iwhich is the as-yet undetermined location of the interface between regions1 and 2.