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Today, we will continue our discussion of Vlasov theory of plasma wave, and we shall
discuss Landau damping and growth of waves. First, we will drive a dispersion relation
for plasma wave, then deduce the damping rate, and in case of a drifting Maxwellian distribution
function or bump in tail, we will examine the possibility of wave amplification giving
rise to instability. We will also use to discuss the growth of ion acoustic wave by a drifting
electron distribution function; and we discuss the consequence of such instabilities.
The references for today presentation are the same as before as in the last lecture;
three books by Stix, Krall and Trivelpiece and by F. F. Chen. .
Let me recapitulate what we were doing last time. We were examining the possibility of
having a plasma with equilibrium distribution function F 0. This was the equilibrium distribution
function, perturbed by a small perturbation f 1 self consistently with the creation of
an electric field, which we wrote down as having E z is equal to minus e sorry minus
delta phi by delta z, and we were considering that f 1 was uniform in x and y. So, it was
a function of t, function of z, and function of v x, v y, v z in general; phi was a function
of e and z. So, a small initial perturbation n, the distribution function and self consistent
electric field creation in the system; how about this perturbed perturbation will evolve
in time? Will it dump out or will it continue to grow? Let there was a issue I would I was
think to address. For this purpose, we had introduced fourier
transforms, and we introduced a quantity called fourier transform of f 1 f 1 k that was in
z. So, this was I had written as 1 upon under root 2 pi and integral f 1 exponential of
minus i k z dz minus infinity to infinity. This was the fourier transform of f 1. And
fourier transform of phi, I will call as phi k is equal to 1 upon root 2 pi integral minus
infinity to infinity potential multiplied by minus i k z d z.
So, what we did was that, we had a linear linearized equation, which was of this form
delta f 1 by delta t plus v delta delta z of f 1 is equal to minus e delta delta z of
phi upon electron mass minus e is the electron charge multiplied by delta f 0 by delta v
z and when I multiply this equation by on both sides by exponential minus i k z dz and
carry out the integration over z from minus infinity to infinity. I got an equation for
f 1 k and that equation was, let me write down that equation f 1 k was, delta f 1 k
delta t plus i k v z f 1 k is equal to minus e upon m i k phi k delta f 0 by delta v z.
So, this is the equation governing the fourier transform of perturbed distribution function.
And this phi k is related to f 1 k through the poisons equation, which was phi k is equal
to e upon k square epsilon 0, where epsilon 0 is the free space permittivity and triple
integral f 1 k dv x dv y dv z. You may note one thing in here, that I can
easily multiply this equation by dv x dv y and integrate over v x and v y. because there
is no derivative or no other coefficient having v x and v y so, it is very simple. So, I can
easily multiply this equation by dv x and dv y and integrate after all and I can let
me define a quantity f 1 k as f 1 k dv x dv y double integral. So, f 1 k dv x dv y then,
this whole double integral will reduce to simply f 1 k. So, I can write down this is
simply e upon k square epsilon 0 f 1 k into dv z and this f 1 k will depend only on obviously
is on time k and v z, that is all dependence. So, in order to obtain an equation for f 1
k, I multiply this equation by dv x dv y and integrate and the f 0 when I integrate over
dv x dv y, I define capital F 0 as f 0 dv x dv y. So, this is called one-dimensional
distribution function and F 0 now depends only on v z, because v x v y dependence have
been integrated over. So, my equation governing F 1 would be F 1 k would be delta delta t
of F 1 k plus i k v z F 1 k is equal to minus e upon m i k phi k delta F 0 by delta v z.
So, it becomes a one-dimensional equation as for as, f is concerned its velocity phase
dependence is concerned.
Now, to get read of this time dependence, we carry out a Laplace transform. Laplace
transform has to be carefully done. So, I write down F 1 k, which is a function of time,
besides other dependence, it depends on v z also. I write down this quantity as F 1
k, let me m omega. Now, this a function of v z of course is there, exponential of minus
i omega t d omega, where F 1 k omega is called the fourier transform of F 1 k. So, I will
write down F 1 k omega is equal to F 1 k, which is a function of t and v z, exponential
of i omega t dt sorry this is d t, this goes from 0 to infinity and this 1 upon 2 pi in
here What I am saying here is, this Fourier transform
I define as the time integral of my distribution function multiplied by exponential i omega
t. And this F 1 k may have a dependence on time which may go as exponential of say gamma
t. Now, if x this as a exponential of I gamma t or gamma t of this form, then the integration
over omega obviously, I expect this goes goes from minus infinity to plus infinity but then
omega should be allow to have a imaginary part large enough. So that, the exponential
decay part of this is more dominant than the exponential growth part of this. Otherwise,
this will over flow because in this integral if omega has a large imaginary part, it does
not have a large imaginary part but this has then a time goes on this integral will overflow.
So, this integral should not be over flowed and hence this omega should be allowed to
have a imaginary part, which is substantially large, because any imaginary part will omega
will be minus. So, suppose I write here this omega as omega real plus i times omega imaginary
then, this term will goes as exponential minus omega i into t. What I want, that this integral
should not diverge. So, whenever this perturbation has a time dependence of this form. The time
dependence of this exponential part should be more dominant than this one, so that this
integral converges otherwise, this will not converge.
So means, this is defined only for in that for those values of omega for imaginary part
is substantially bigger than gamma. That is that is a important. So, when I carry out
this limit, I must be careful, this goes from minus infinity to infinity. but it should
have a substantially large imaginary part plus i times some imaginary parts suppose
alpha plus i times alpha alpha is like give any quantity greater than gamma. That is important.
So, if I have to carry out this, if I write down this suppose, I plot real part of omega
here and imaginary part of omega here and I find that the Fourier transform has or suppose,
this is the value or these are the values any values suppose, this is the value imaginary
part omega F 1 k suppose has a dependence on time which is equivalent to imaginary part
omega equal to so much like gamma. Then, my line of integration should be here. This is
the line of integration of omega integral from minus infinity to plus infinity but at
a height alpha. This is the primary the thing means, any singularity in the Laplace transform
should be below the line of integration, this what I have to say.
So, when we carry out the inverse Laplace transform, we should be careful about it,
but I think, we will refer to this refer this discussion to a little later in stage. Let
me first go back to our Fourier transformed equation. What I do, multiply the equation
equation for F 1 k by exponential of i omega t dt and carryout the time integral. Let us
see, the first term in the equation is delta delta t of F 1 k multiply this by f i omega
t dt and integrate. let us see, what do I get, I am doing this from 0 to infinity. This
quantity will be first integral of this quantity will be F 1 k and F 1 k as I mentioned depends
on time and v z this has to be obtained at 2 limits sorry multiplied by exponential of
i omega t. The limits are from 0 to infinity minus differential of this and integral of
this. So, it becomes F 1 k and differential of this is i omega exponential of i omega
t dt 0 to infinity. As, I mentioned to you omega has a substantial
large imaginary part. So, when t goes to infinity, this vanishes. So, at the upper limit this
entire quantity goes to 0. Only at the lower value this will survive and at that this quantity
is one and this is F 1 k 0 v z. How about this integral, if I take I omega outside then,
this is simply the Laplace transform of F 1 k. So then, this integral is simply sum
of two terms. This integral delta delta t of F 1 k exponential of i omega t d t, which
was from 0 to infinity is equal to minus F 1 k at 0 v z minus second integral was simply
F 1 k omega into i omega. And then the equation becomes simple, its simply gives you F 1 k
omega is equal to simply simplify this, e upon m omega minus k v z k, here phi k there
into delta f 0 by delta v z. So, we got a simple expression for f 1 k.
Now, let me go over to the expression for phi k, I have a so, if I multiply phi k equation
also by exponential i omega t dt and integrate, I will get phi k omega is equal to e upon
k square epsilon 0 multiplied by dv z integral from minus infinity to infinity multiplied
by F 1 sorry this is capital F 1 this is capital F 1 capital F 1 k omega multi[ply]- this is
all. I think I forgot to write this term, this is also a term there. So, when I multiply
this take this on the right hand side it becomes plus actually, when this goes on the right
hand side it becomes minus 1 up[on]- yeah this becomes F 1 k 0 v z divided by omega
minus k v z into I. This term, I have taken into account. I had
forgotten to take this into consideration so, the I should add. Now, these 2 equations
this is phi k omega also here. My suggestion is that or rather what I want to do now. I
would like to use this value of F 1 k omega here and I will get two terms, one on the
right hand side will contain a phi k omega that I will combine with this term and I will
get another term having this integral dv z of this quantity.
So, let me write down phi k omega I get, is equal to some numerator divided by the denominator.
Numerator turns out to be an integral, which is if I simplify this is e upon k square epsilon
0 into i k common and then you will get a integral of this form F 1 k at 0 v z dv z
upon v z minus omega by k minus infinity to plus infinity and the denominator of this
is it is 1 minus, let me just see the denominator is rather simple. Let me write down this expression.
This turns out to be equal to omega p square upon k square into 1 upon n 0 I would say
here n 0 the number of electrons per unit volume into minus infinity to infinity delta
F 0 upon delta v z upon v z minus omega by k.
So, you get the Fourier Laplace transform of wave potential as a ratio of two quantities
N and D and here omega p I have defined as n 0 e square omega p square is n 0 e square
upon m epsilon 0. but because there was no n 0 explicitly there. So, I have multiplied
denominator numerator by n 0 and this is how I have written. But obviously, 1 n 0 is contained
in the distribution function F 0. The important contribution of landau lies in understanding
this. The thing is that, whenever you carry out the Laplace transform, you have to integrate
to obtain phi from phi k omega, I have to carry out the inverse Laplace transform or
integration over D omega from minus infinity to plus infinity above all the poles, and
that is a complicated integral. Landau says that, if D equal to 0, hence the singularities.
Now obviously, this is integral, this can be 0, if it is 0 for some values of omega
after omega is a range and complex frequencies. So, for some values of omega this will be
0. Those values of omega for which D is 0 are
called poles of phi k omega and the maximum contribution to an integral will arrives from
those poles. but unfortunately your Laplace transform is not defined for those frequencies
of complex omegas at which this becomes 0. So, then you have to carry out what we call
as the analytic continuation of the of this integral means, of this Laplace transform
has to be this is essentially quantity defined in complex omega plane. So, you want to define
Laplace transform in a domain where similarities lie. So, let me draw a graph here. Suppose,
I plot here real part of omega and imaginary part of omega here, D the denominator of the
Laplace transform suppose becomes 0 at some point here, it may become 0 somewhere here.
Then, your integration has to be carried out on this line; this is the contour for omega.
Integration has to be carried out here, this is contour one.
Landau says that, if there are no poles between here and this line. So, if I choose this suppose,
this is another path of integration, this is path two because my please always keep
this in view. I want to sorry I want phi k at time t that I write in terms of phi k omega
exponential minus i omega t d omega ranges minus infinity plus epsilon sorry plus i alpha
to infinity plus i alpha this length is i alpha from here to here and this is 1 upon
2 pi. So, in order to carry out this integration over this line in the complex plane like this,
landau says that it will be much easier to carry out a integration of this over a another
path, which is far below the imaginary omega equal to 0 axis. This is line over a which
omega imaginary omega is 0. So, go underneath because imaginary omega will be highly negative.
So, when omega is a large in negative imaginary part, this quantity will vanish. So, only
contribution will come from the poles. So, landau suggested that rather than carrying
out integration over here, it is more convenient to carryout integration over here because
only the contribution of poles will come. but then your function is not defined below
the singular line similarity line because as I mention to you otherwise the Laplace
integral overflow. So, in order to avoid the over flow of the Laplace integral, you defined
your Laplace transform only in upper half upper plane above the similarity. The issue
was that, how will you generalize your function. Now, let me define what is analytic continuation?
Suppose, in my complex plane, some function is defined some in some domain, this is a
function F 1 is defined and another function is defined F 2 in some other domain, this
is the F 2 function is defined here. but if in this overlap region F 1 is equal to F 2,
then we say that F 2 is the analytical continuation of F 1. This overlap region may be a simple
surface or single line a single curve it mean, it need not be a big finite area region the
area of this may be negligibly small, it could be simple joining line. So, continuation of
F 1 to F 2, F 2 is called analytic continuation of F 1 if, in the overlap region the two functions
are equal or at the boundary joining the two the two functions are equal then this is called
analytic continuation. And what is a function, when it is analytic continuation.
A function is called analytic, when the derivative of the function you can define suppose, there
is a function g a function of z. You call this analytic, when g has a real part function
of x and y suppose, I z I write as x plus i y then, this function can be written as
u of x y and v of x y. u and v are called the real and imaginary parts of g function
and x and y are called the real part imaginary part of z function, of z variable. And there
is a condition that, when delta u y some derivative of this with respect to x or y is related
to some derivative of y or v with respect to other variable means, the differential
coefficient of g is uniquely defined irrespective of suppose, there is a function is defined
at this point here and its defined its neighborhood, when you go from 1 point to another point
either from here to here, or from here to here, or from here to here, or from here to
here it gives the same limit then the function is called analytic
So, when the derivative does not depend on, how z goes to 0, you know we define normally
g at z plus dz minus g at z divided by dz and limit dz going to 0, this is called g
prime. but dz can go to 0 in a horizontal line or a vertical line but when you give
an increment from here to here, or here to here if the rate of increment is same, then
only the function is called analytic. Means, it is independent of whether, you make dx
equal to 0 or dy equal to 0. so on, which root you move, then the function is called
analytic. So, thing is that, if my function is analytic in some domain, the analytic continuation
can be achieved by detouring the path of after all the Fourier transform that we obtained
here, are in the form of integral over v z. So, Landau suggested that, why do not we treat
v z is a complex variable. Normally, the integral in these n and d the
two numerator and denominator of phi k omega. They contain integrals over v z, he suggested
that treat v z is a complex variable and presume that, those functions n and d functions, that
you have written, if they are analytic and certain region and if you can avoid the singularity,
then in all the domains where singularity is do not exists, the function will be analytically
continuous. So, what he suggested was this. That if, I have to carry out suppose, I look
at this n expression. N is equal to e upon k square epsilon 0 i k minus infinity to infinity
F 1 k 0 v z dv z. This is the initial perturbation upon v z minus omega by k. This is some function
of omega by k. So, what Landau suggested that, in order to carry out this integration, what
you are having is this.
Suppose, I plot real part of v, this integration is over real v z but he says that in order
to look for analytical continuation of Laplace transform, I treat v z as a complex variable.
So, real part of v z if I plot here, imaginary part of v z I plot here, what do I get. If
I move on this horizontal line, my integration has to be turn on the horizontal line, but
this omega by k with finite imaginary part omega, I can have a pole here or I can have
a pole underneath. Suppose, pole is here then he says that, this integration over this line
is the same as over this line, this is suppose, he call as path 3, this is path 4.
The integral along this line and integral along this line, they will differ only by
the contribution of the singularity or rigidly of this pole. And, if you have a pole underneath
here, Landau suggested, then your path integral should be like, this go over this line go
underneath and surround this here and so line integral this is was a suggestion, that the line integral from here to here is
the same as line integral over this path plus contribution from the singularity. And you
will have the contribution singularity here as well, if the contribution this singularity
lies underneath the imaginary v z line, then you detour your contour like this.
Then, n defined by this integral, this the formed contour will be analytically continuous
and your Laplace transform is then valid in the domain where imaginary part of omega is
negative and far negative. So, just by detouring this, he could show that for imaginary part
of omega higher values certainly, this is the same, because if the singularity lies
here then and your function is defined there is no problem. The problem arises, when the
pole lies here so, he says deform the contour. So, what he wrote was this and the same thing
was done for D, D was written as 1 minus omega p square denominator was k square n 0 delta
F 0 by delta v z upon v z minus omega by k dv z. This is equal to 0 are the poles, when
I put this equal to 0, they are the poles. So, they say that, if imaginary omega is greater
than 0 means, the poles lie here then the zeroes of D will be given by this integral.
Then, we are carrying over the integration over the real line. Whereas, if your poles
lie imaginary part of omega is less than 0, then you have to detour the contour and this
is 1 minus omega p square k square n 0 minus infinity to infinity delta F 0 by delta v
z upon v z minus omega by k dv z, a contribution from this singularity has to added. So, he
added a term here, this turns out to be equal to, when you add a contribution to singularity,
it is 2 pi i times the residue and it turns out to be equal to 2 pi i times omega p square
upon k square n into delta F 0 by delta v z at v z is equal to omega by k. This is the
additional term, that he added. So, Landau says that, whenever imaginary part of omega
is less than 0, in that case because I wanted to really bring my contour to carryout inverse
Laplace transform to imaginary omega being very negative in that case, he says that this
detour this contour by this and contour defined like this will have will be analytically continuous.
And then, he says that these two can be written together in the form of a single, make a single
statement by using what we call as the principle value. You define principle value of a integral
through an isolated value, let me just define this.
The principle value of an integral is defined as the well suppose, there is a pole somewhere through an isolated
singular point called a pole is the average of line integrals above and below the pole. So, you are assume two line
integrals, there is a pole somewhere just above, just below carryout the line integrals
here and there and average some average of the two will be called the principle value.
And if, you write this in the terms of principle value, then denominator of this Fourier Laplace
transform will be 0 correspond to 1 minus omega p square upon k square and p means principle
value of this integral minus infinity plus infinity, that is along the v z axis real
v z axis delta F 0 by delta v z divided by v z minus omega by k dv z minus i pi omega
p square by k square n 0, n 0 is here as well, multiplied by delta F 0 delta v z at v z is
equal to omega by k and put this is equal to 0. And this is the dispersant relation,
because the values of omega which satisfy this equation, in general those values are
complex and your phi k at time t will go as exponential of minus i omega t where omega
are the values which satisfy this equation, let me call them as omega prime values.
So, omega prime are the roots of this equation. So, the entire problem of examining the growth
or decay of perturbations voice down to obtaining the zeroes of this equation. This equation
is an integral equation, the first integral or there is only one integral has a velocity
space integral from minus infinity to plus infinity but this omega you to allow complex
values to omega and principle value, I will just defined as the average of line integrals
just above the pole and just below the pole. And this is the additional contribution that
gives rise to damping of waves. If, this were not there as of did actually did this theory
of plasma waves prior to Landau and he did not get in damping because he did not treat
this singularity carefully. So, when you put this is equal to 0, you do not get any damping
of wave. but this is the additional term that, you have to keep in view that gives rise to
damping of the wave. Well, I think this entire mathematics can be put in a much simpler way. The entire
exercise was, actually I carried out just to show you that, if you consider initial
value problem, then you have to carry out the analytic continuation of the function
and in that in search of a analytic continuation Landau detour the contour of v z integral
treating v z is a complex variable, but once you have recognize that, that whenever you
are encountered with the v z integral or integration velocity space, you will follow the Landau
prescription then, the rest of the procedures very trivial very simple. And, I would like
to rework out the solution of equation.
What we were really having is, let me write down the equation again. Delta f by delta
t plus v dot del f minus e rather plus e delta delta z of phi delta f 0 by delta v z equal
to 0. This was the one-dimensional, this was the equation I can write down this z component
here and this z here forget this. So, this was my equation, which I wanted to solve and
I can solve this very easily saying that, f is equal to f 0 plus f 1. I linearized this
and my equation became delta f 1 by delta t plus v z delta f by delta z, this is the
f 1 is equal to minus e delta delta z of phi delta f 0 by delta v z, this is all.
So, rather than doing any Laplace or Fourier transform, I consider, suppose my phi is equal
to a exponential minus i omega t minus k z. I presume a Fourier component Fourier Laplace
component of phi and I want to find out what is the response of my f 1 in quasi steady
state. Then, I will say that my f 1 should also have a similar dependence on t and z.
So, I say that f 1 should also have as a exponential minus i omega t minus k z. Substitute this
back in here replace delta delta t by minus i omega delta delta z by i k and this equation
becomes simple.
F 1 becomes equal to e k phi upon m omega minus k v z into delta F 0 by delta v z simple.
How about the density perturbation, n 1 is equal to f 1 dv x dv y dv z triple integral.
Substitute this in here, this becomes e k phi upon m outside, I can take minus k common
also, take minus here and k outside here, in the interior you get, delta F 0 upon delta
v z upon v z minus omega upon k and this is dv z d x dv x dv y triple integral. The denominator
does not depend on v v x v y that derivatives also not respect to v z, it is simply with
respect to v z. So, what I can do to be.
For a special case, when I take f 0 to be maxwellian, which is equal to n 0 pi to the
power minus 3 half v thermal to the power minus 3 exponential minus v x square, plus
v y square, plus v z square divided by 2 t e. This is the distribution function f 0 for
a maxwellian distribution function. My suggestion is that, when you do this, then this when
you carryout delta delta v z this factor exponential factor is can be taken out of the differential
operator and you can carry out the integration over v x and v y. When you do this, then limits
on all components of v x v y v z are from minus infinity to plus infinity. So, on carrying
out v x v y integrations you will just turns out to be.
And recognizing that, minus infinity to infinity exponential minus x square dx is equal to
root pi. If you recognize this, then this integral turns out to be n 1 is equal to minus
e k phi upon m k, you will get n 0 there upon under root pi you, get rest of things cancel
out and 1 v thermal you get and you get here delta F 0 by delta v z upon v z minus omega
by k and dv z minus infinity to infinity.
And, if you substitute for, well what is F 0. F 0 is called one-dimensional distribution
function. Actually, I made a mistake, this n 0 is not there, that is involved in F 0
this is equal to n 0. Actually, all these factors are not there they have in a part
of F 0. So, let me remove them here, they are not there n 0 upon under root pi into
v thermal exponential minus v v z square upon v thermally square, this is called one dimensional
distribution function. Now, when I substitute this back in here, this integral takes a simple
form and let me write down this n 1 is equal to e phi upon m k will cancel out, you will
get n 0 outside n 0 negative sign is already there, pi to the power half, then you have
v thermal here and if I differentiate this function, you will get twice v z upon v thermally
square into exponential. So, here you get, with a negative sign. So,
minus is here, 2 will be there and v thermal q will be there and you get here v z exponential
minus v z square upon v thermally square divided by v z minus omega by k dv z. Its goes from
minus infinity to infinity. People normally define v z by v thermal as a new variable.
Let me call this as p so, v z upon v thermal let me call a new variable call p and let
me define omega by k v thermal as a quantity called xi. Then, this integral takes the following
form, n 1 turns out to be exactly equal to, this is equal to 2 n 0 e phi upon m pi under
root into v thermal square and you get this integral of the form p exponential minus p
square upon p minus xi d p minus infinity to infinity.
I define a function called z function of xi, identically equal to 1 upon under root pi
exponential minus p square upon p minus xi d p. This is called plasma dispersant function.
Limits are from minus infinity to infinity and in carrying out this integration one should
be careful does this is not only principle value, this has to add a term corresponding
to the landau prescription. So, the entire understanding of or analysis
will Landau is contained in the in the interpretation of this integral, not only principle value
but the additional imaginary term that I had added because of the going round the singularity.
So, now this p can be replaced by p minus xi plus xi and when p minus xi term cancels
with this. This gives you root pi so, it becomes a simple expression and this xi can be taken
out and you get this entire expression in a simple way, neat way.
And let me write down this, it becomes n 1 is equal to twice n 0 e upon m v thermally
square into phi here, 1 plus xi z function of xi. This is a very important deduction.
The perturbed electron density due to the plasma wave is related to this expression.
This is the expression for it and I would like to substitute this in the Poisson equation.
The Poisson equation is del square phi is equal to minus up rather plus n 1 upon epsilon
0. Replace this by minus k square so, you will get phi is equal to minus n 1 upon k
square epsilon 0 and when I substitute this here, you will get this is equal to 2 omega
p square upon k square v thermally square with a negative sign into 1 plus xi z of xi.
And this entire quantity can be written as minus chi e into phi.
So, chi that is called the electron susceptibility is twice omega p square upon k square v thermally
square into 1 plus xi and plasma dispersant function of xi. And this equation gives you,
the dispersant relation for plasma waves 1 plus chi equal to 0. So, this is a very important
thing that, 1 plus chi e is equal to 0 is the dispersant relation for plasma waves.
We do not have time today to discuss these implications. I think we need to need one
more lecture to discuss in these implications. I think, we shall continue next time in discussing
the dispersant relation and obtaining the damping etcetera damping rate of the wave.
Probably, I like to close at this point but let me remind you that, the entire contribution
of Landau really lies reinterpreting the integral over a v dv z in the dispersant relation.
And when you properly carryout the integration following Landaus contour Landaus prescription,
you always get an additional term that contains the derivative of the distribution functions
delta F 0 by delta v z. As, I mention to you, physically we were expecting
that we will get growth or damping of the wave, when the distribution function has a
positive or negative slope at the phase velocity of the wave when v z is equal to omega by
k and the same thing is contained explicitly in Landaus prescription. I think, we shall
discuss this in our next lecture. Thank you.