Tip:
Highlight text to annotate it
X
We will continue our discussion on surface gravity waves. Today, on the 09 lecture we
will talk about the pressure fluctuation in the surface gravity waves and show that the
energy flux and propagation is related to the group velocity.
We will use the group velocity to characterize waves as either dispersive or non-dispersive
waves. If the group velocity of all the constituents is same then, we will notice that they are
called non-dispersive system and in contrast, if the group velocity keeps changing with
wave number we have a dispersive wave system. This is a very interesting topic in various
aspects of applied physics and mathematics. We will talk in great detail about the dispersive
waves. When it comes to surface gravity waves, if
it is formed in a confined region, then we would also notice that we can obtain standing
waves. We will discuss in detail and show how the dispersion relation becomes degenerate
in this case. This will be followed by our study over dynamics
of waves, over water with varying depth. We will notice that the waves those are formed
will vary greatly in terms of wavelength and frequencies. We will be able to show you that
whatever may be the wavelength, we can write out a conservation equation for wavelength.
Talking about ray - optics was developed for studying. Optics - we will talk about dispersion
and ray paths for dispersive ways. One of you at the end of the class ask me little
more about deep water waves and shallow water waves, what are the particle paths, there
were some questions regarding that. So, I thought, I will just simply go through it
once again.
We look at these surface gravity waves as a consequence of the linearised irrotational
approximation and then the flow field is defined uniquely by only one unknown that is the velocity
potential phi, which is given like this. So, this is your a travelling wave solution, this
have been obtained by phi. You can successively take derivative with respect to x and z to
get the velocity components. From the velocity components you can equate it to its Lagrangian
description, which upon integration gives you the departure of the particle paths from
its equilibrium condition, it is given by x naught and z naught.
So take a look at this, this is the way it looks like. Then once you eliminate this x
naught and z naught and t you end up by getting an equation of an ellipse here.
This is the general form of the semi major axis, once again there is no square here;
that is a mistake; I will correct it once again. So basically what happens is that is
a generic expression for the semi-major and semi-minor axes.
However, if you look at specifically for deep water waves then H would approximately be
10 into infinity of very large. Then if you write these in terms of the exponential form
you can see that for H very large this quantity sine hyperbolic k into z plus z naught plus
H by sin kH. This part would go to this value and so the other one would be that cosine
hyperbolic k into z naught plus H by sine kH. So, what happens as a consequence, A1
is equal to B1.
Well, in case if you have forgotten, probably none of you have. If I plot let us say this
hyperbolic functions, then what I find is that cosine hyperbolic begins with 1 and then
it goes to infinity; that is your cosine hyperbolic x. If you look at the sine hyperbolic x, it
goes like this, so that is your sine hyperbolic x; that also asymptotes to infinite value.
Whereas, the tan hyperbolic k actually this goes like this.
Now, you can very clearly see that when x goes to infinity both of their asymptotes
becomes together and you get this A1 equal to B1. So, what you get in a sense is a circular
particle trajectory.
If I have to plot these particle trajectories this is the bottom of the bed, then say this
is the surface gravity wave, then what you are going to see is the clockwise trajectory.
As you keep coming down, this is your datum for the z axis, so as you keep on coming down
you are going to see circles of smaller and smaller radius that is because of this factor
coming down e to the power kz naught will keep on coming down, because z naught is negative
there, as you go down it you will get this. This is how you get and that is what we also
wrote here. All particles from the same vertical column will be in phase, the trajectory defined
by these particles will describe smaller and smaller circles for deep water wave. Same
way, you can look at the shallow water limit for which the H itself will go to 0 and then
you can very clearly see that sine hyperbolic component will go to 0. Well, the cosine hyperbolic
component goes to 1. That is the consequence of that figure that we have drawn.
What happens, you can see this is your A1 and this is your B1. What you are seeing is
A1 cosine hyperbolic part would be much greater than B1. What we are going to see in this
case is, of course we are going to see elliptic trajectories with the semi major axis elongated as compared to semi minor axis.
You also note that the semi major axis this part would not change very much with z naught,
because they will be in the close vicinity of one itself.
So what happens as a consequence is here as you have seen the amplitude actually keeps
coming down and it goes to 0 like this, this is what you get. In this case, if I similarly
try to look at envelop of the semi major and semi minor axis I will find that A1 would
virtually remain same. As I keep on going down this will become thinner and thinner,
and so and so forth. This is what you are going to see in the shallow water wave.
Now, what is a deep water wave? That is what we need to find out. We can draw information
from these trigonometric hyperbolic functions. Whenever, you have kH greater than say 1.75
you will notice that tan hyperbolic will rapidly approach one. You look at your dispersion
relation that is where you would see the tan hyperbolic part comes and it is a kind of
saturates to that. There you would see that this will be kind of value, so your kH greater
than 1.75 is a kind of a threshold below which you can talk about intermediate range.
If it is far below then you will say it is shallow water, but anything above 1.75 you
could classify it as a deep water wave, this is something that you would know. Now, a bit
of trivia or information, if you look at ocean waves, if you are very close to the continent
what is called as a continental shelf then there the depth of water is roughly around
100 meters or so. But, if you go to the open ocean little further inside the deep sea,
there it would be something of the order of few kilometers, something like 4 kilometers
or there about right. This surface gravity wave that we are talking
about usually corresponds roughly about lambda is of the order of 150 meter. If you talk
about h of the order of 100 meters even in the continental shelf, we are not talking
about very close to the beach but little further, may be few 100 meters away from the beach,
then you would be seeing this kind of depth and there the wave of this kind is qualified
as a deep water wave. For all practical purposes the waves that
you would see in the sea excluding the one very close to the vicinity of the beach front
can be classified as a deep water wave; this is what I told you. Probably your exposure
in optics and acoustics you have talked about longitudinal wave and transverse wave, but
this is a mechanical wave and it has got this interesting property that is the trajectories
are not necessarily defining a vertical or a horizontal motion, but by the combination
of the two you get either circles or ellipses. One thing now what we could do is we could
also work out how the pressure varies. So for the pressure variation you can use what
we have already derived in the Bernoullis equation.
What was Bernoullis equation? Of course, take a look at it, it is an unsteady flow, so you
will have this term and then we will have p by rho plus this plus a constant which we
can conveniently set the datum to 0. Now, what you can do is you can define the pressure
fluctuation, what do I mean by pressure fluctuation? It is the departure of the pressure from the
undisturbed value of the surface.
Basically, we will be talking about p prime is the pressure fluctuation that should be
the actual pressures that we have written in the Bernoullis equation. Then we will be
doing this, because minus rho z was the actual pressure undisturbed case from hydrostatics;
that is what you get. So this is your pressure fluctuation, it is easy for you to see that
from here that this two together is going to give us p prime.
So that is what we are going to get p prime by rho is equal to minus del phi del t, so
I will write this as this. I have obtained the expression of phi, you can calculate this
as rho a omega square by k and cosine hyperbolic, this into the phase part. If I use the expression
that dispersion relation, it tells us omega square equal to gk tan hyperbolic kH. I could
simplify and then I could write this as rho ga, and cosine hyperbolic k z plus H, all
divided by cosine hyperbolic kH, of course this remains the same.
Having obtained the pressure fluctuation we would like to talk about something which is
of immediate interest to us, namely how actually the energy propagates? Why we are trying to
talk about this is simply for the reason that we have defined a new quantity - group velocity.
I do not know how many of you are very intimately aware of this quantity, I find that main stream
literature on the book they do talk about group velocity, but do not emphasize its physical
importance that is why I would like to emphasize it a little more.
Now, having obtained this expression what I could do is I could try to calculate the
energy and the rate at which the energy flux is propagating in the fluid that is the very
important consideration.
So, let us talk little bit about energy consideration. If we try to talk about energy, I think we
have briefly talked upon it. We have gone through this expression for dispersion relation,
calculated the group velocity. We have obtained the limiting values for the deep water waves,
shallow water waves. What we noticed that the deep water waves goes at half the phase
speed, while the shallow water waves go at the speed of the phase.
Now, we did talk about this energy flux and I did make this observation. Now I think that
a deeper appreciation of that last bullet would be in order for us to appreciate what
really is the true role of group velocity. Unfortunately, this has been lost not only
in the precision of the physics of the problem but also in competition. This plays a major
role that is what we are going to see as we go long.
So, having obtained this solutions given in terms of phi and the velocities, one thing
of course we can talk about is the kinetic energy, which I will call let us say E of
k. How will I define this kinetic energy? Well, at any location I will calculate the
kinetic energy say per unit volume. Then what I am going to do is I am going to average
it over a wavelength and integrate it across the whole depth.
Basically, we are talking about a surface gravity wave like this, so I will look at
it and integrate the energy from the top all the way to the bed. So, basically we will
also be integrating it from minus h to 0, this will be taken about this.
So, you can very clearly see that this dx integral is actually taking the average over
A1 wavelength. We can say that this is the kinetic energy integrated over the depth and
averaged over a wavelength, so there you have it. We have this expression for u and w, we
can plug it in there. I will take the liberty of omitting the steps and will tell you that
you would get the x and z integrals, they are decoupled and one of which the x integral
will give in terms of a times cosine cos kx minus omega t or sine that has been squared,
so that would give a kind of a expression rho g into eta square bar.
So this is like mean square wave elevation that is your eta square bar x. So, I have
eta as the wave elevation, I will square it and I average it over 1 wavelength that is
what we are writing. Since eta is a cos kx minus omega t you can work it out, it is rather
easy, what is this going to be? 1 over lambda eta square dx that is the definition of eta
bar square eta square bar. If I do that I will plug it in here and then I will work
it out, I am going to get a square by 2, we are very straight forward.
Now that is story for kinetic energy, we are looking at very genuine case of lossless wave
propagation and that would allow us to calculate the potential energy. These are the major
players in defining the energy of the system because there are no losses, no heat addition
and etcetera. We can talk about the dynamics in terms of kinetic and potential energy what
will be it? Well, very simply stated, this will be actually the work done to deform a
phase surface interface from its undisturbed position n. So that is easily obtained again, I will call
that as Ep, what would that be? Well, once again I will write the pressure, work done
that I will find out for the deformed case going from the bottom minus H to eta - the
deformed phase, rho gz will be the pressure that will act over small element dz, I am
integrating over the whole thing, as we have done it before we will integrate over x and
make an average. It is the same context like kinetic energy
was integrated across the depth and averaged over 1 wavelength, so potential energy is
also are written in the same bases. It is the same thing integrated over depth and averaged,
so there you have it. Once again, these are simple algebra you can
do it by yourself, what you would find? You will find that it should be also equal to
half rho g, you can anticipate that. You are going to get a steady state where you are
going to get a constant amplitude waves, what does it mean?
Like what we talked about earlier, it would mean a perfect balance transfer of energy
from the potential to kinetic like your simple pendulum. It is a non dissipative system that
is what you are seeing here also.
No price for guessing that the kinetic energy and potential energy should look like the
same. Having obtained this what we could do is we can talk about the total energy of the
system, well that would be sum of these two, you find that should be equal to rho g square
bar, that should be equal to half rho g a square.
Next what we are going to look at is, this is the total energy of the system, if I integrate
it over the whole depth that is what we are going to get. So, if that is the energy we
want to find out when the wave is propagating - travelling what is the rate at which that
corresponding energy is going as a flux? If I want to calculate the rate of transmission
of energy
across any vertical plane, rate of transmission of energy is basically the flux and this is
done; what we mean by energy flux? This energy flux which I will indicate by F tilde that
would be nothing but what would it be the pressure velocity work.
Some of you would be familiar, some of you may find it little tricky, but that is essentially
the thing. If I look at dz element, p into dz is the force acting and if I am looking
at the rate at which it is going, the work done would be that force and displacement
is dx. If I am trying to calculate its weight that will be dx dt. So that dx dt - that rate
is given by this and this is your force p times dz, so that is you flux; that is the
way the work is done per unit time for the displacement dx.
So, work it out. Now, I have just erased, but you can look at your class notes that
p prime, this p is the pressure so I can write it in terms of the fluctuation. So, what I
would get is F tilde would be equal to p prime minus rho gz time u dz, so it has two components.
One would of course be coming from the fluctuating pressure part and the other part . So this
you understand, so this is going from minus h to whatever eta or 0, I mean probably take
the corresponding linearised version and we will do that.
So, you can set the limit here minus h to 0 and this is the other part, it could be
minus h to 0, we have rho guz dz. Now as we have done before we have now reported quantities
which are averaged over 1 wavelength, so we retain that spirit and define F. So, that
is basically flux averaged over 1 wavelength. If I look at that expression that would be nothing but, F would be nothing
but, I will integrate F tilde dx and this is what we mean that bring us like quantities.
What we have talked about the energy? We are correspondingly finding out the flux which
is averaged over 1 wavelength. What do you find when I am trying to average this quantity
over 1 wavelength? What will happen x dependence comes from here in u and u expression that
you saw was in terms of cosine kx minus omega t? If I took an average of cosine function
over 1 wavelength then that will give me 0, so this part is not going to contribute because
the average in process in x direction on u over 1 wavelength, it will yield a null value.
So, what you are getting here is you are getting this expression 0 to lambda and the z integral
minus h to 0 and we are going to get p prime u dz dx. So, you have the expression in front
of you and we can plug those values of the quantities we already have.
We have the expression for p prime that you look at it today, you would note that for
F we have written the flux, flux would then have the x dependent part that would have
cosine square kx minus omega t; this is your x integral. So that part is multiplied by
all those associated quantities rho a square omega cube by k sine hyperbolic square kH;
we will be performing the integrals of minus H to 0; we will have cosine hyperbolic square
k times z plus H. So, this times dz there are this two quantities,
I have decoupled the x and z integral. Once again you can figure it out, this will be
half, well of course this is a square, bring it here that will give you a square by 2.
In place of omega cube I can use the dispersion relation; I can take omega square and put
gk. If I do that I am omitting steps, you can
do it by yourself or if you are lazy you can look at the book, there you will find that
this expression is given by
as shown . So essentially, after all that what you are seeing that F is equal to this;
your E you have just now derived, the energy total energy and this expression is Vg.
So, what you are seeing? What we expected to see was that the energy is being transported
across any vertical plane is given by the energy that is being transported times a velocity
and that typical velocity happens to be the group velocity, not c. So this is the physical
implication of group velocity, it basically tells you the speed at which the energy propagates.
This we have shown it for specific example of surface gravity waves, but if you also
recall when we looked at the superposition of two neighbors, there also we see that the
amplitude propagates are Vg and V related with that amplitude square to energy. So you
have seen the same thing, this is rather very important we have given a physical system
and we are asked to compute how the energy is propagating by that system, then we need
to know the correct velocity scale. See what happens, many times we are so much
preoccupied which our knowledge of simple linearised waves where Vg happens to be equal
to c and then you do not have to distinguish between Vg and c. You always draw a general
conclusion out of a mistaken notion that Vg is always going to be c and that is not the
case.
So, what happens is in many of the physical systems we will be looking at, we would be
finding out the speed at which the energy propagates. If we are able to calculate that
carefully; I mean that is a very important concept; tell you what; this was there in
an original paper by Hamilton. You have heard of Hamilton in mechanics let us say Irish
mathematician. Hamilton actually in one of his paper talked
about energy propagation speed, but it is probably Rayleigh who did a very good job
and laid the foundation of group velocity. He was the first to tell that group velocity
is different from phase speed. Now as I was telling, you if you are looking at 1d convection
equation with c positive, you have substituted trail solution. A trail solution would have
amplitude, a phase part and then plugged that in over there, you get this dispersion relation.
If you take the derivative of omega with respect to k it gives you the Vg. In this case, the
phase or the group velocities are the same; this is one of the properties of non-dispersive
system. Whereas, in large number of situation where you come across that is including the
surface gravity wave what we see? We see that Vg is not equal to c because the phase speed
itself is a function of k.
This part is non zero that makes Vg different for different k. What happens in consequence
of that? An initial impulse is given to the system, fragments itself into various components,
this will be travelling at their own velocity and you are going to see the different k component
going at a different speed. Now, what you notice that the phase is defined
by phase speed c, but if the energy is propagating at a speed which is greater than the phase
speed then what will happen? You would see that small wavelets riding on the overall
wave packet. They will actually build up in the front because they are going at a faster
speed compare to the phase, you will see them building up in the front of the group. Because,
the energy is not there what will happens? They will disappear from the back.
This is something we will have to realize that if the group velocity is more than c
then you will see the wavelength building up in the front, disappearing at the back
and vice versa. When the group velocity is less than phase speed then you will see that
the energy is lagging behind, while the system has the propensity to show that the phase
can go ahead; that is what will happen. You would see that the wavelets will fall back,
because there is no any of the energy to support them. So, they will keep disappearing in the
front. Next time, when you are near a pool of water,
you decide to drop a stone and you can see which case depends on the depth of the pool
that is, you are looking at on that type of wave that you can create. So, that is why
we raise this question, take a look at it. Now, one thing I wanted to tell you is that
so far we have been talking about waves which are going in the positive x direction that
is why the phase was defined as kx minus omega t.
Now, suppose we are try to discern what is happening with gravity waves which are propagating
in the upstream direction, then its general description would be given by the similar
expression but you see the xt dependence on the phase is interface by a plus sign, this
we will indicate that the wave is propagating upstream. What happens, you go through the
same analysis that you have done? You would find the dispersion relation a p h with a
negative sign, it is just that difference. Now, what do you find? Calculate the group
velocity by taking it is derivative with respect to k. You would find the magnitude will remains
the same but it appears with opposite sign and there is nothing wrong in that simple
observation.
However, that is where something happens interesting, if we have two waves one is going in the positive
direction, the other one is going in the negative direction, if they decide to superpose, then
what happens? Well once again, simple trigonometry will tell you this. Here is an interesting
scenario, the system depends on x and t, how? They are disjoint, they appear differently.
We also notice some very interesting property that around kx equal to plus minus is odd
multiplier of pi by 2, you will have eta equal to 0.
If that is the case then these locations are the nodes with 0 displacements. What happens
here? I would get a kind of a space dependence that is given by cos kx, I have that but the
time dependence is a simple harmonic, so what happens? All the point would go up and down
in the same phase that is what we meant is here by the surface particles will oscillate
in unison and the wave will not propagate. I am particularly bringing it to your attention,
because if you look at that book by kundu and cohen, I think that is where they have
made a mistake by claiming that there is a group velocity that is given by either the
upstream propagating or the downstream propagating path. But that is not true; the energy does
not propagate, because you do not have a space time dependence appearing together. You need
to have them in a composite function, which will be appearing together like what we have
seen k x plus minus omega t. If I have a limited body of water like lakes
or ponds then the standing waves get created, why because, if you recall the d'alembert
solution we talked about the Cauchy problem, we took the interval going from minus infinity
to plus infinity and if I created a disturbance somewhere then it can go in both the directions.
Now, if I intervene by putting in some kind of a boundary what will happen? Those waves
will go there and reflect back. Then the incident and the reflected wave can create standing
waves and what happens? The dynamics would be given by this because the nodes will be
fixed at the shores and that will determine the fundamental wavelength that is the longest
lambda by 2 that you can get between the two shores, you will get half the wave; that is
what you can get. From this, we also summarize that k and omega
are not related. So, we do not have a dispersion relation, they are independently appearing
their and as I said that books tell you that dispersion relation remains the same, but
it isn't true. Here, no energy is transported by the wave which is in the intervening media,
standing waves has this property. Now, there are some other interesting properties of surface
gravity waves. If I create an impulse on the surface of the liquid then what will happen?
Simultaneously many harmonics will be created.
This is the case of a dispersive system, what will happen is that different k component
will travel at different speed, what you would see as a consequence, if I give you a compact
disturbance source with time? It will enlarge the region over which the disturbance is left.
Now, you have the expression for Vg. You would take a look and you would notice that depending
on the value of k you are going to see different parts going at different rate. What would
you see at the leading edge? If you look at the expression for Vg you would notice that
the smallest k travels at a higher speed. So, near the leading edge of the disturbance
packet you would see the longer wavelength disturbances.
Initially, I may create disturbance which have a compact basis but then with time it
will not only stretch. You will also see the longer wavelength to the right in the system
that is a generic statement, but further refinement occurs when we start looking at the dynamics
of the disturbances as seen. We are looking at those disturbances forming over constant
depth or a varying depth. In fact, this kind of variation of depth can add to very interesting
situation.
I can show you this picture and this is what? I was telling you probably this is not drawn
very carefully and that shows up what should have actually happened in this figure. The
longer wavelength part should be on this side. This is the disturbance, this is the x axis
and the time is increasing like this . So, with the mean passage of time the disturbance
is stretched over larger and larger x. This part would be dominated by longer wavelength
smaller k. So, this is not drawn very carefully as I should have done it, now what happens?
We have seen that it is no more a monochromatic wave.
So, what is happening is, we are going to see some kind of a displacement, which you
would like to say that is a slowly varying function of x and t, and slowly its changing.
The same way the phase would be written in terms of theta, which is also a function of
x and t, how different is this theta of x and t from its slowly varying function? This
will still be written here like kx minus omega t, but the important point to recall here
is that k and omega they are changing with x.
They are no more in a fixed entity like what we have talked about when we looked at monochromatic
wave, so what happen? If I have a theta which we wrote like kx minus omega t then well I
could take a derivative of this partial with sorry d theta dx, a slowly varying function
I would just write it as k. That is what we have here that k of x comma t would be obtained
by looking at the variation of phase function with x. The same way, I could differentiate
this with respect to time and that will approximately give me minus omega.
This is all happening because we are talking about slow variation. If that is the way you
look at the displacement, find its variation with x and t, talk about a local k and local
omega, then you can further differentiate this one with respect to t, this one with
respect to x and you see the right hand side will add up to 0 .
So, you see a very interesting property that for such slowly varying waves the del k del
t plus del omega del x goes to 0. Since, this del omega del x we could write with the use
of chain rule in terms of d omega d k into del k del x. This d omega dk is a group velocity,
so what we are noticing that for slowly varying waves is that k is governed by this equation.
This is like your convection equation that we had obtained for displacement of other
physical variable. So, similar such thing is being noticed here
for the wave number k provided, we track it at the value of Vg and see if we are going
at the speed Vg then we would be always looking at the same k, because this is nothing but
the total derivative d dt of k. This tells us that if we decide to track the same value
of k then we should be fixing our gauge and the gauge should move at Vg.
This kind of confusion occurs if you have noticed some of these waves in pond or confined
space. Then, most of the time what we try to do? We try to follow a crest or a trough
something of that kind, then what happens? You see the following things; one we have
already said that as it goes outwards what will happen? You are going to go over to longer
over longer length, how? Then all of a sudden when you have energy cannot keep phase with
the speed Vg that it has, and then it will disappear that is what you see.
However, if you want to follow the same wavelength component then you really must move your gauge
with this velocity Vg. Now, if I go back to my xt plane then I should be looking at two
sets of lines; one is given by dx dt equal to c, those rays will tell me along which
c is constant. There is the other ray path that will draw that it is given by dx by dt,
given by Vg is the line along with k remains constant.
What we find that at time t equal to 0 I have created some band of disturbances in this
range of x with the passage of time that band widens because of dispersion effect and then
what we can see? We can see that the ray path given by dx dt equal to Vg that defines where
I would see the actual disturbances, because that is the speed at which energy is going.
So, the extreme of that Vg range will determine how this envelope is going to enlarge and
this thin lines tells you along which your phase is constant. If I am looking at surface
gravity waves what happens is, along this line k and Vg remains constant. These are
the ones those are indicated by the thick straight lines. The thin lines and paths of
wave crests those appeared from the back because of why? We are looking at let us say either
of the case; suppose we are looking at deep water wave then c is of course greater than
Vg and for deep shallow water c will be equal to Vg.
So if we are looking at deep water wave c will be greater than Vg, then what will happen?
You will see that crest will appear from the back and then they will disappear in the front
that is determined by the range of Vg - admissible Vg. This range of k is of course decided by
the initial impulse, how I have distributed the initial energy across different k that
I can get it from the initial condition. What happens later is a story that is dictated
up on by the dispersion, as we note here the ray paths open up with time and this is a
very important observation. If I create a disturbance at one point with time what happens?
You are going to see constant itself although we are talking about one d wave propagation,
but here x is more like your arm in a cylindrical quadrant system. So, the same energy is going
to be distributed over larger and larger area. So what happens? The amplitude must come down,
when you see such waves amplitude is coming down it is not a viscous effect, it is not
due to loss that the amplitude is coming down, it is because of distributing energy over
larger area. That is another nice property of dispersion and what happens, because of
dispersion amplitude keeps coming down as time progresses. We will have a better and
better support for your theory, because amplitude comes down and that was our starting assumption
for linearization. So, I will stop here and we will continue
may be couple of more lectures and talk about certain very interesting properties of this.