Well friends. Today I would like to discuss
the energy flow with an electrostatic wave.
We shall discuss current density due to a
plane monochromatic wave, then current density
due to a wave whose amplitude is a function
of space and time, and then deduce an expression
for energy stored in the electric field of
a wave. And deduce the expression for energy
flow per unit area per unit time with the
electro static wave, and we shall also discuss
the group velocity.
Well, this is based on this book interaction
of electromagnetic waves with electron beams
and plasmas by Professor C.S. Liu and myself.
However, you can find discussion of today’s
theme in any book in plasma physics.
Well, let me begin with the response of a
plasma to a monochromatic plane electrostatic
wave whose electric field is E is equal to
A to be simple or a specific lets the wave
is longitudinally polarized in the z direction,
and amplitude is constant in the time variation
is i omega t, and space variation is like
k z. This is the electric field of an electrostatic
wave, we have already solved the equation
of motion for electrons including the pressure
term, and we deduced an expression for electron
velocity from there we wrote that the perturbed
velocity of electrons due to this waves is
equal to e E upon m i omega square minus k
square v thermal square into omega.
If thermal velocity is not there this is the
thermal velocity of electrons then, this is
the same expression for drift velocity as
due to an electromagnetic wave, but the electrostatic
wave is difference because of the thermal
effects the denominator is modified and from
there if you write down the perturbed current
density due to the wave which is a product
of electron charge equilibrium electron density,
and perturbed velocity or drift velocity of
electrons.
If you substitute this there then, this is
expressible as sigma into E, and sigma turns
out to be minus n 0 e square and omega in
the denominator upon m i omega square minus
k square v thermal square in this derivation
I have ignored the effect of collisions. If
collisions are finite then wherever omega
occurs you have to replace omega by omega
plus i nu then this equation is valid for
coalitional plasma also. Usually when we are
talking of Linear waves omega is around omega
p, and collision frequency is orders of magnitude
is smaller than omega p, hence this effect
is small though it may be significant, but
it is small. This is the current density due
to an electrostatic wave whose amplitude is
constant in time as well as in space. In order
to deduce an expression for energy flow with
the wave, we like to examine the current density
of plasma response to a wave whose amplitude
has a slow dependence, but significant dependence
on time and position. So, let me go over to
that stage.
Now, I would like to discuss the electron
response when wave amplitude is a function
of time and space. I think we shall learn
something very significant from this deduction.
I am considering that my system has an electric
field which is directly along z because it
is a longitudinal electrostatic wave propagating
on z axis, but amplitude I will permit to
have a dependence of time and z Exponential
minus i omega t minus k z. So, besides having
fast phase dependence like this the way has
a slow dependence on time and amplitude in
terms of it is amplitude. Let us see what
the consequence of this. If you write down
the equation of motion is if I divide the
equation of motion by mass it reads as delta
v by delta t is equal to minus e E upon m
minus pressure term which reads as v thermal
square which is t e upon m upon n 0 and gradient
of n.
Here I have used the linearized equation of
motion. Where v dot Del v term has been dropped
means the products of perturbed quantities
have be dropped. This is what we call as a
linear zed equation of motion that we had
written last time. And actually this is for
the perturbed velocity and perturbed density.
So, I have written like this. And the equation
of continuity for the perturbed density is
delta n 1 by delta t plus divergence of n0
v1 is equal to 0. I wish to solve these equations
when is not a constant, but a slowly varying
function of t and z. I will slowly following
a procedure of iteration solve these equations
and derive some a result and then I will point
out a general procedure which is very simple,
but in order to appreciate that simple procedure.
I need to do some deduction from these equations.
I will be little slow and elaborate in deriving
or rather in obtaining the solution to these
equations.
What I do I express my perturbed electron
velocity n terms of some quantity a and I
am taking all quantities vector quantities
along z because it is a longitudinal wave.
Suppose this is slowly varying amplitude of
velocity which is upon t and z exponential
minus i omega t minus k z. I have written
the past time and z dependence the same way
as the electric field because there is a source
there is a response, but I am permitting my
amplitude to have some dependence on t and
z. Then I can write down my perturbed density
also as say b which is a function of t and
z exponential minus i omega t minus k z. Now
what I do, I substitute these two expressions
in the equation of motion continuity and treat
delta a by delta t to be small as compared
to minus i omega and delta b by delta t also
to be small.
So, I am presuming that delta a by delta t
is less than omega into a and delta b by delta
t less than omega v. This is called iteration
approximation. What I am going to do, first
of all I will ignore the dependence of 
a on time and b on time and similarly I am
assuming there z derivative also to be small
as compared to k, means delta a by delta z
is less than k a and delta b by delta z less
than k b. This is the approximation I am going
to make. What I do first of all when I substitute
this I ignore the derivatives of a and b in
the equation of motion continuity then those
equations read like this. Let me actually
substitute this and see how those equations
look when you substitute this expansion for
v 1 in the equation of motion you get minus
i omega a plus delta a by delta t. This is
d delta a by delta t term exponential factor
is common and will cancel out in both sides
is equal to minus e into a minus the pressure
term which is v thermal square upon n 0 delta
z of n 1.
So, when you m also is here. When you evaluate
that term there are two terms i k into b plus
delta b upon delta t. This is the term and
this is term. These two terms together are
called perturbed terms and I can call them
if I take this term on the right hand side
and add these two terms together I will call
them as S 1. Then this equation becomes, let
me define S 1 is equal to minus delta a by
delta t minus b thermally square upon n 0
delta v by delta t. These two terms must I
ignore and obtain the value of in terms of
A; obviously, there is b here.
So, I should write down the equation of continuity
as well and the equation of continuity is,
if you substitute the expansion for density
it gives you minus i omega b plus delta b
by delta t plus n 0 divergence of delta z
of v 1 which gives me i k a plus delta a by
delta z is equal to 0. It looks little complicated,
but simple substitution of v 1 and n 1 in
the equation of motion and continuity. Here
also there are two small terms delta v by
delta t and delta a by delta z. I can take
these two terms on the right hand side and
define the sum of these 2 as S 2 which is
equal to minus delta b by delta t, because
I take on the right hand side and then this
becomes and there is another term minus n
0 delta a by delta z.
This is the second term S 2 that I get what
you should have is in terms of S 1 and S 2
the equation of motion then is minus i omega
a plus i k v thermal square upon n 0 into
b is equal to minus e A upon m plus S 1. The
equation of motion takes this form and the
equation of continuity this one, when I transfer
these partial derivative terms from the right
hand side takes the form i k. This term I
will write first n 0 a then minus i omega
b is equal to S 2. What I am suggesting is
that first ignore S 1 and S 2, when you ignore
S 1 and S 2, you can solve these equations
to obtain b is equal to from this equation
k upon omega into n 0 a and when you use this
value of b in this equation and ignore this
one you will get a is equal to plus e A omega
upon m i omega square minus k square v thermally
square.
This 
is the same velocity. If you multiply the
exponential term as we had obtained in the
local approximation when the amplitude was
a constant and this is the density perturbation
when you multiply the phase term. So, as before,
we obtain the same thing when you ignore this
S 1 and S2, but one use this value of b and
a in S 2 and S 1. S 2 is put this value here
and this value of a here and solve these two
equations you can get in the new case when
S1 and S2 are retained. This is when S 1 and
S 2 are ignored. If S 1 is equal to 0 and
S 2 is equal to 0 these are the values of
b and a. Then I am suggesting that now evaluate
S 1 and S 2 using these values of b and a
and then solve these two equations returning
these two terms and you will get the following.
I just give the result, this algebra is simple
and I am pretty sure you can easily simplify
this. The net result is this.
So, to the next approximation when you evaluate
a and b. T hey turn out to be a is equal to
e omega a upon m i omega square minus k square
v thermal square where v thermal is the electron
thermal velocity this term as before. Now
the additional terms are these minus e upon
m omega square minus k square v thermally
square multiplied by I will write down this
in different ink because these are the 2 important
contributions. Because of S 1 and S 2 which
is omega square plus k square v thermally
square into delta a by delta t plus twice
k omega v thermally square delta a by delta
z. This is an important addition to the solution,
this is if you ignore these two terms this
last two terms then a is the same as due to
a uniform amplitude wave. And these are the
contributions that take into account that
time variation of amplitude and space variation
of amplitude. These two terms together. They
are important terms and we shall learn in
little while the role played them in energy
flow.
Well, if you know the velocity the current
density can be written and current density
J z in this case will be minus e into n 0
into velocity. Velocity is a into exponential
minus i omega t minus k z. This is the value
of J z and when you substitute the value of
a here you get an expression which is equal
to minus n 0 e square just multiply n 0 e
here. Because minus n 0 e square upon m i
omega square minus k square v thermally square
into omega this omega into a into this quantity
is simply e E z. This is first term and second
term is plus because plus n 0 e square upon
m omega square minus k square v thermally
square multiplied by these two terms in the
bracket. Let me write them with red ink omega
square plus k square v thermally square delta
a by delta t plus twice k omega v thermally
square delta a by delta z into exponential
term this exponential. So, I do not have room
here to write exponential. So, I will write
e and then put dot there this exponential
factor phase factor is to be there.
Now the beauty here is that if you call the
coefficient of e z as sigma then, you can
easily see that this quantity this term which
is multiplying delta a by delta t is simply
delta sigma by delta omega with an iota and
similarly this quantity is exactly same as
delta sigma by delta k you can just verify.
Let me write this, I will define this coefficient
from here through here this quantity is conductivity.
If I called sigma then these other coefficients
can be easily shown and you must show yourself
lf must check yourself.
That J can be written as sigma E z plus i
delta sigma upon delta omega into delta A
by delta t exponential minus i omega t minus
k z, and another term plus i delta sigma by
delta k, actually this negative sign delta
A by delta z exponential minus i omega t minus
k z.
This is a very simple comprehensive way of
writing an expression for J, because only
in terms of a single quantity called conductivity
and it is derivatives. I had been able to
write J due to an electrostatic due to an
electrostatic wave whose amplitude is a function
of time and z. Of course, I have assumed that
a depend slowly on time and slowly on z. This
is the approximations I have made, but in
that limit this is good. Well this was a very
special case of a wave when there was no static
magnetic field when I did not consider collisions
and so, on. But in general I like to point
out a general procedure. whenever you have
to you are encountered with waves of time
varying amplitude and position what to do
if you know the response to a monochromatic
wave whose amplitude is constant you can easily
deduce the response due to time varying amplitude
and time varying and the space varying amplitude
the procedure is as follows.
You know if you have electric field is equal
to say some function of t and z exponential
minus I omega t minus k z, suppose this is
my electric field. If I differentiate this
partially with time I will get delta E by
delta t how much that would be. That would
be simply first of all I will differentiate
this quantities will gives you minus i omega
into this A will be there and then I will
differentiate this. We will get delta A by
delta t and of course, z unit vector is there
and exponential minus i omega t minus k z
is there. What I am saying, that if the wave
has constant amplitude then this term is 0.
So, delta t is simply equal to minus i omega.
But when the wave has a time varying amplitude
then delta t is not equal to this, but minus
i omega plus delta t has to be there means
omega has to be replaced by or minus i omega
has to be replaced by minus i omega plus delta
t.
If you compare this case with the case when
there is no time variation of amplitude. Then
this term is not there. The difference in
the two cases is that minus i omega has to
change by this or simply if I divide this
equation by minus i, then this says that omega
just to replace by omega plus i delta t. Because
if I divide this equation by minus i this
becomes omega goes to omega plus i delta t.
And this delta t operator has to operate not
on the whole field, but only on the amplitude.
In general your sigma the conductivity is
a function of omega and k. But you should
do, replace omega by omega plus i delta t
and similarly if you differentiate this with
respect to z you can show that k has to go
from k minus i delta z. If you can make these
two changes in sigma and operate this sigma
over the amplitude of the electric field you
will get the right expression.
So, this expression that I have written here
complicated expression can be deduced from
J is equal to sigma e in a simple way if I
substitute sigma omega by this and k by this
let us see how do we do it.
We begin with J is equal to sigma which 
is a function of frequency and k in general
and this direction is z into a exponential
minus i omega t minus k z, but what I am saying
that I am going to replace this omega by let
me write down J z as how much then sigma which
is a omega plus i delta t and k is to be replaced
by k minus i delta z operating over a exponential
factor minus i omega t minus k z. Well, if
this is a slow dependence as compared to omega
as we have been assuming then you carry out
the Taylor expansion. Any function, if there
is a small increment in the variable can be
expanded by using Taylor series. This expression
becomes simply sigma at omega and k plus there
is an increment here. So, it becomes delta
sigma upon delta omega into I delta t and
because of this will be a term minus i delta
sigma upon delta k into delta z and these
are operating over a into the exponential
term. This same factor phase factor.
You may note that this is exactly same as
before. Rather than going over the entire
calculation and in this is valid in general.
Even if there is a magnetic field using this
is valid, this is a very general thing. This
procedure is very similar to the procedure
adopted by Schrödinger when he deduced the
wave equation Schrödinger equation from de
Broglie’s hypothesis. Essentially, he said
that omega should be replaced by an operator
and you should also replace k by an operator.
The same way I am suggesting here that omega
can be replaced by because it is not operating
over the entire electric field, it is operating
this you know over the amplitude. I am replacing
omega by omega plus I delta t because the
time dependence of the phase term was already
incorporated in deducing sigma. It is only
the time dependence of amplitude that was
ignored. I have to replace this by omega by
omega plus i delta t and so on.
This is a very standard procedure in all areas
of physics whenever you are dealing with waves
or fields whose amplitudes vary with time
and position slowly. This is a 1 important
thing that I wanted to mention to you. Well,
now this question of simple interpretation
I think if time permits, then I would like
to say something more significant from in
a different way you can arrive at the same
expression. Let me go a step further, let
me write this expression sigma e bring this
a in the interior, sigma into e is z rather
this becomes i delta sigma upon delta omega
into delta a by delta z delta t minus i delta
sigma upon delta k and delta a by delta z
and obviously, in these two terms I am multiplying
an exponential term minus i omega t minus
k z. I have to interpret this. Well, I am
going to make an approximation here. As we
have already seen the sigma that we have derived
here is almost an imaginary quantity. Because
the expression for sigma if you look is has
a i there. It is imaginary. However, if we
include collisions then it may have been small
real part because collisions are a small quantity.
I can write down conductivity a sum a small
real part and a very large imaginary part.
where sigma i is much bigger than sigma r
or magnitude bigger, but let me retain this
small sigma r to have a physical interpretation.
Now for electrostatic waves there is no magnetic
field in the wave. Convention point in vector
is 0 e cross h is 0 because h is 0. Now, let
us understand the implication of J from the
last Maxwell equation if curl of h is 0 then
J plus delta D by delta t is 0. The last Maxwell
equation becomes this if h is 0. I want to
multiply this equation by dot E and take the
time average. So, multiply this equation by
dot E and let us see and take time average.
Let us see what I get, please remember J is
always a real quantity though I have written
in a complex way, but whatever expression
I have written real part of that expression
is J.
So, J is a physical quantity and electric
field also is a physical quantity though written
in terms of a complex expression just takes
the real part. So, I am going to use this
identity that real part of A into real part
of B is equal to half real part of A dot B
take complex conjugate of any quantity either
A or B plus a dot b. This is the identity
that we had proved earlier and you can verify
even now. Take A as small a 1 plus I a 2 and
B as small b 1 plus I b 2 where a 1 b 1 a
2 b 2 are all real you can verify this easily.
Now electric field changes with time exponential
minus i omega t minus k z first time dependence.
Similarly, J also has the similar time dependence.
So, if I replace this a by J and b by e, but
I get this term will have 2 omega frequency
variation and time average will be 0 and this
term will have the exponential phase terms
will cancel because this is a complex conjugate.
When I multiply this equation by dot E only
these terms survives this goes this goes away
and let us see what the consequence is.
So, time average of this is 0. Let me just
mention this. This is time average is 0, you
can forget this term is 0 rather time average
is not exactly 0 and how about this term.
This is the only thing I have to retain a
dot b star by 2 and take the real part. I
will write down term by term J has three terms
and this may also have two terms. Let me simplify
these, I think I want to be slow on this because
this derivation is very fundamental to plasma
physics. I want you to follow it with greater.
Let me write down J dot E a star and I take
the real part of this quantity and divide
by 2. This is what I am going to evaluate
if I do this the first term J is equal to
sigma e. It becomes real part half of sigma
into e dot e star which is the same thing
as modulus of A square which is a real quantity.
This real part means real part of sigma will
survive. Second term was plus i delta sigma
by delta omega into delta A by delta t.
If you recall I am multiplying this by E star
they are on the same direction, just multiply
by a star. A could be complex in general and
similarly, there is another term which is
minus i delta sigma by delta k delta A by
delta z into A star, this is the meaning of
real part of J dot E star by 2 means the time
average value of J dot E is this the quantity.
As, I mentioned sigma here is largely sigma
i. I replace by i times sigma i. This becomes
means if I take the real part of 
this quantity it becomes half sigma r modulus
of a square because this is real. So, real
part will be sigma r only. Only real part
will survive. And sigma i is because sigma
in these expressions is nearly equal to i
times sigma i. If I substitute this sigma
in here and in there then, this becomes also
real quantity i and i will become minus. This
becomes a real quantity; similarly this also
becomes a real quantity and then I have to
take the real part of this quantity, which
is you can easily show is equal to minus delta
sigma by delta omega sigma i rather delta
omega into 1 by 4 delta t of A A star.
And similarly this term becomes when I plus
delta sigma i upon delta k 1 by 4 delta 2
of a star. You, may wonder why I am getting
4 not 2. The reason is that I am getting a
term here. I think one should understand this
with clarity that real part of A star delta
A by delta t, I have to evaluate which is
the same thing as half this number a star
delta a by delta t plus. It is complex conjugate.
Any complex numbers real part is half of that
number plus it is complex conjugate. Complex
conjugate of this quantity means replace this
A star by A and replace the un-starred quantity
by the starred quantity. This should go away
and this should be there this should not be
there and this can be easily shown is equal
to half delta t of A star same thing. I have
used this is identity you can identical. This
is a very important expression. It has three
terms the one which involves the real part
of conductivity it may be tiny, but it may
this term may be comparable to this term because
a is A very slowly varying function of time.
Though sigma i may be large, but because the
slow dependence of A star on time this term
may be these terms may be comparable. Let
us understand now the physical implication
of the fourth Maxwell equation that I have
written as J plus delta d by delta t equal
to 0. J dot e star i have obtained I have
yet to obtain the other term d delta d by
delta t dot E star.
I am always keeping in focus J plus delta
d by delta t is equal to 0. This term is how
much D is epsilon 0 into delta e by delta
t which is equal to minus i omega into E plus
delta A by delta t exponential minus i omega
t minus k z and when you take E dot of this
quantity and time average this quantity. If
I take E dot and this becomes real, but this
is imaginary, real part of this is 0. If I
take half real part of E dot delta D by delta
t and take star here and then this term does
not contribute. This is the only term that
contributes and that gives you epsilon 0 is
there; 2 is there and this becomes A star
dot delta A by delta t and I take the real
part of this quantity. Real part of this quantity
has to be taken, which is, the same thing
as epsilon 0 by 4 delta of modulus of a square
by delta t. J dot to be obtained earlier E
dot delta D by delta t we have obtained now
add the two terms and put them equal to 0,
what do you get essentially this expression.
If you just club these terms together what
you get is this half sigma r A square plus
epsilon 0 by 4 this term one comes from here
minus delta of sigma i upon delta omega into
1 upon epsilon 0 into delta of a square upon
delta t because of J dot E. The second term
and E. This term is coming this one and then
you get a last term plus 1 by 4 delta sigma
i by delta k into delta of modulus of A square
delta 2 is equal to 0. This has a simple interpretation
because when there is conductivity real part
means the current density has a component
in phase with the electric field there is
power dissipation. This is the power dissipation
per unit volume. Then, this term should be
something like rate of increase of energy
density of the electric field. This factor
multiplied by modulus of a square can be interpreted
as energy density of the electric field and
this term should be then expressible or can
be interpreted as the pointing representing
divergence of pointing flux.
This equation has lot of physical implication.
Let us try to appreciate in plasma physics
we do not talk that much in terms of conductivity.
As, we talk in terms of effective plasma permittivity,
but the two are unrelated as I defined earlier.
Effective plasma permittivity, we define as
1 plus i sigma upon omega epsilon 0 and sigma
is almost imaginary at least in these terms
that are appearing here the time derivative
amplitude and space derivative amplitude in
those terms. This is nearly equal to 1 minus
sigma i upon omega epsilon 0, from here, I
can write down sigma i is equal to omega epsilon
0 into 1 minus epsilon effective. This is
a better way of expressing that equation rather
than in terms of sigma. I would like to write
that equation in terms of effective plasma
permittivity and if you do this, the equation
becomes rather simple. Half sigma r modulus
of A square plus delta t of I will call a
term w E plus delta z of w E into v g equal
to 0. That equation can be put in this form;
however, you may wonder there was delta k
of epsilon effective. Where that is and what
is W E. you just switch off this sigma i in
that expression you will find it.
If W E turns out to be 1 by 4 epsilon 0 into
delta omega of omega epsilon effective into
modulus of A square or E star. Normally, a
half vector comes in here, but because we
have taken time average, half comes because
of time average also. This is the energy stored
in the electric field. This is called energy
stored in the electric field.
Energy stored in the electric field of the
wave per unit volume and what I have done
regarding there was a term like delta sigma
i by delta k what did I do with that. It was
replaced by because sigma i. when, I wrote
down in terms of epsilon effective. It became
is equal to minus omega epsilon 0 because
they were constant. I am differentiating this
expression partially to k. So, omega can be
treated to be constant and then delta epsilon
effective upon delta k. But one must remember
one thing that epsilon effective is a function
of omega and k and for electrostatic wave
this is always equal to 0. If I differentiate
this function any increment in this function
due to increase in omega in change in k would
of this form delta epsilon effective upon
delta omega into increment in omega plus delta
epsilon effective upon delta k into increment
in k this should be 0 because this is 0 always
at omega k, as well as, at omega plus delta
omega and k plus delta k also this is 0.
So, from here I can obtain the value of this
expression in terms of delta omega of this
quantity you may just check and then you can
write down the J dot E equation the way I
had written and I have called delta omega
by delta k or delta k as v g group velocity.
This can be written as delta omega by delta
k.
Let me write that J dot E equation again,
which was half sigma r A square plus delta
delta t of W E plus delta delta z of W E into
v g is equal 0. Now simple interpretation
would say that, this is the amount of heat
dissipation, the power lost by the wave per
unit volume. This is the increase in energy
density per unit volume. If in certain region
suppose consider a region between this is
z and z plus dz, if in certain volume if I
multiply by the volume of this region dz each
term by dz, and area I am taking unit area
of cross section of this plane as unity. Then
the volume element will be dz, multiply this
by dz, then the first term gives you the energy
dissipation per unit volume. This gives you
the increase in energy per unit time in that
region.
Energy is being lost here more over energy
is increasing here. It means energy should
be coming in here from somewhere. So, more
energy should pour in and less should get
out only then increase in energy density and
dissipation energy is possible. This quantity
must represent the energy flow. Because divergence
of energy flow is represents like in pointing
theorem also. This is some sort of a pointing
vector, less quantity more quantity of this
comes in less gets out only then there is
an increase in energy density of the electric
field and increase and power loss there. The
energy flow density or pointing vector for
the electrostatic wave would be the pointing
vector would be S. I am talking of average
pointing vector is equal to W E into v g,
the direction of energy flow will be in the
z direction in this case because my wave was
taken to be in the two direction.
This is equivalent of pointing vector for
electromagnetic wave this is 
e cross h where as for electrostatic waves
for which magnetic field is not there is no
magnetic perturbation time average pointing
vector is W E into v g, where W is the energy
density of electric field per unit volume
is energy stored in the electric field per
unit volume. I have given you a derivation
of two expressions one of W E in a dispersive
medium and second in the derivation of time
average pointing vector. One think, I would
like to mention to you. For electrostatic
waves the expression for conductivity was
different then the expression for conductivity
for electromagnetic waves. The thing was for
an electromagnetic wave, if the direction
of propagation is like this the electric field
was in the transverse direction. So, there
was no compression of charges this is E M
wave.
Whereas for electrostatic wave your electric
field is oscillating in the same direction
as k this is electrostatic wave and there
is a compression rare faction of charge and
hence pressure term is significant. Here,
there is no pressure term there. We had seen
that in the case of electromagnetic waves
conductivity does not involve any temperature,
any thermal effects temperature comes through
collision collisions only, but not through
pressure, but here in case of electrostatic
wave’s temperature comes or thermal velocity
comes through pressure term. It is a separate
dependence. The two conductivities are different
and hence epsilon effectives are also different,
and it will be useful for you to obtain the
energy density for 
a plasma wave and energy density to a sound
wave and also to calculate the group velocity.
You can easily verify that for a plasma wave
group velocity is equal to thermal velocity
square upon v phase.
