Debye Length
Adrian DownJanuary 22, 2007
1 Particle number density
1.1 Review
Previously, we used Boltzmann’s law to express the number density
n
of aspecies of particle in thermal equilibrium subject to a potential
φ
,
n
(
r
) =
n
0
e
−
qφ
(
r
)
kT
In the quasinuetral situtaion, the number of ions
n
i
and electrons
n
e
areapproximately equal,
n
i
≈
n
e
≈
n
0
. However, the distribution of thesediﬀerent kinds of particles are not necessarily unifrom in space in the quasineutral situtation, and so charge gradients and ﬁelds can still exist withinthe plasma.
1.2 Example: Gaussian potential
Consider a Gaussian potential
φ
acting on a population of electrons. In thiscase, the charge is
q
=
−
e
. The distribution of electrons is thus,
n
e
(
r
) =
n
0
e
+
eφkT
The sign of the exponent indicates that there is an enhancement of the number of electrons in the region of the potential. This can be seen physicallyfrom the electric ﬁeld, which is the gradient of the potential. In all space,the electric ﬁeld points outwards from the peak of the potential, so the electrons, having negative charge, are drawn towards the potential, opposite of the electric ﬁeld.1
2 Derivation and deﬁnition of the Debye length
2.1 Poisson’s equation
The previous example is not very physical. As soon as the distribution startsto evolve, the change in the density
n
changes the potential
φ
. To ﬁnd thepotential, we need Poisson’s equation,
φ
·
E
=
ρ
0
E
=
−
φ
⇒−
2
φ
=
ρ
0
ρ
is the charge density, which must take into account all species of particle.Assuming the charge of the electrons and ions to be equal and opposite,
ρ
=
s
n
s
q
s
=
n
i
e
−
n
e
e
If
n
e
and
n
i
are equal and uniform, then there is complete cancellation. Ingeneral though, this is not the case.
Note.
In an inhomogenious medium, a tensor must be used to represent theproperties of the dielectric,
·←→
·
E
D
=
ρ
0
2.2 Introduction of a test charge
We can consider a test change
s
placed in the plasma. Using the expressionsfor the number density derived previously to express the number density,
ρ
=
n
i
e
−
n
e
e
+
s
=
en
0
e
−
eφkT i
−
e
eφkT e
+
s
Note.
We allow for the case that
T
e
=
T
i
.In the onedimensional case, Poisson’s equation with the inclusion of thetest charge is,
−
∂
2
φ
(
x
)
∂x
2
=
n
0
e
0
e
−
eφ
(
x
)
kT i
−
e
eφ
(
x
)
kT e
+
s
2
2.3 Linearization
Finding an analytic solution to this equation is nontrivial. The physicalsolution is to make an approximation and solve the equation in the resultinglimit. To linearize the equation, make the approximation that
eφkT
1, sothat the exponents can be expanded. This approximation corresponds to
kT
eφ
, meaning that most particles in the plasma are “freestreaming”,unaﬀected by the potential
φ
.For small
x
, the exponential can be Taylor expanded,
e
x
≈
1 +
x x
1
⇒
e
−
eφkT i
≈
1
−
eφkT
i
e
eφkT e
≈
1 +
eφkT
e
The linearize form of Poisson’s equation is,
−
∂
2
φ∂x
2
=
n
0
e
0
1
−
eφkT
i
−
1
−
eφkT
e
+
s
=
n
0
e
2
0
1
kT
i
+1
kT
i
φ
+
s
The coeﬃcient of the
φ
term can be considered a constant. By comparingwith the left side, it must have units of length squared. The Debye length isdeﬁned as the square root of this constant. We can simplify the expressionfurther by deﬁning an eﬀective temperature,1
T
eﬀ
=1
T
e
+1
T
i
The Debye length is then,
Deﬁnition
(Debye length
λ
D
)
.
λ
D
=
0
kT
eﬀ
n
0
e
2
Note.
In convenient units, the Debye length can be written,
λ
D
= 7
.
4
T
(eV)
n
(cm
−
3
)m3
3 Interpretation of
λ
D
: shielding
3.1 Introduction
Shielding in a plasma is often due primarily to the electrons in the plasma.Consider electrons impinging on a surface. The ﬂux of these electrons is
φ
=
nv
th
. The thermal velocity can be approximated by kinetic theory,modulo a constant factor,
kT
∼
m
v
2
⇒
v
th
∼
kT m
Because of the dependance on the mass, the thermal velocity of the heavier ions is often much less than that of the electrons. Hence if there is a wallin a container, the ﬂux of electrons onto that wall will often be much higherthan that of ions. The accumulation of negative charge will form a sheath.This eﬀect is known as sheath theory, and it is presented in chapter 8 of thetext.
3.2 Diﬀerential equations
3.2.1 Possion’s equation
Poisson’s equation governs the formation and evolution of such a sheath of charge,
∂
2
φ∂x
2
=1
λ
2
D
φ
+
s
The presence of the constant factor
s
does not change that the solution tothis equation is exponential,
φ
(
x
) =
Ae

x

λD
+
Be
−
x

λD
where
A
and
B
are normalization constants. As a boundary condition, werequire that
φ
→
0 as
x
→∞
, so
A
= 0.
3.2.2 Electric ﬁeld
We can obtain an expression for the potential
φ
from the electric ﬁeld
E
. Todo so, we assume that the test charge is actually a sheet that is inﬁnite in4