 So, we will start today's lecture with this wrapping up what we did in the last class which was the problem of the Rayleigh Taylor instability, right. So, the Rayleigh Taylor instability we went through the math, the algebra and found that when you impose the condition that you want a non-zero solution, the relationship between the growth rate sigma and the wave number in terms of all the other parameters is given by this relationship, okay. And what we were trying to basically summarize is that if we do not have any surface tension gamma is 0, then sigma squared varies linearly with alpha, okay. That is what we see because this goes off, gamma is 0 and this is a constant and then system is unstable for rho 2-rho 1 being greater than 0 because here sigma squared is positive always. If it is unstable that means when you have the heavier fluid on top, lighter fluid at the bottom is unstable system that is what this says but this is unstable for all wave numbers, okay. This is unstable for all wave numbers alpha because sigma squared is always positive and the more the alpha, the more the other thing. So, there is no selection of a particular wave number that is going to happen, okay. The higher the wave number, the more is a growth rate whereas if gamma is not equal to 0, then system is unstable for sigma squared sorry is unstable for sigma squared greater than 0, okay. The system is unstable when the sigma squared is positive because if sigma squared is positive, sigma is going to have a plus the square root of that or minus the square root of that. So, there is one component which is going to be growing, okay and even if one component grows that means it is unstable. So, the system is unstable for sigma squared greater than 0 and when does that happen? This happens if rho 2-rho 1 g is greater than gamma alpha squared, okay. So, it means for large or sorry for low alpha or large wavelength system is unstable because alpha is low this condition is going to be satisfied, okay and for the reverse that is low wavelength system is stable, okay. So, what that means is the surface tension actually is going to have a stabilizing influence, okay because it is associated with this negative sign here, okay. That is something which I want you to keep in mind and if you were to now plot for rho 2-rho 1 greater than 0, if we plot for rho 2-rho 1 greater than 0 maybe I just plot sigma squared versus alpha clearly for alpha equal to 0 sigma squared is 0 and there is some value of alpha depending upon the gamma for which again it is 0. So, there is an interval here in which it is going to be going up and coming down, okay. For alpha large system is stable that means sigma squared is negative, okay and in between for low alpha system is unstable sigma squared is positive, okay. So, there is a change of the stability and what this means is there is some kind of a maxima which you are going to see in this dispersion curve, okay. What does this maxima correspond to? This tells me the wave number which is going to be the one which is fastest growing, the one which is going to actually grow fastest. So, these wave numbers are also unstable but what you are going to actually see in actual experiment is going to be a pattern which is going to be dominated by this wave number alpha, okay. So, this is similar to what you saw in the Rayleigh-Bernard convection problem where we found the alpha by the point where the growth rate was having a maximum or the Rayleigh number was having a minimum, okay. So, I just wanted to point out this analogy here. Yeah, Jason, is there a problem? Yeah, I was just wondering the way alpha is now affecting the negative value of alpha and the sigma squared becomes negative. You should take a negative value of alpha. Yeah, but we are going to look at, yeah. So, if you are talking about you are saying alpha is negative, I am just wondering if we have actually made an assumption of alpha being positive anywhere in the derivation alpha x, e power i alpha x, x is what we have, no, you know, exponential and the alpha that we have is actually square root of alpha x squared plus alpha y squared. So, alpha is positive. So, this is a 2-dimensional problem and alpha is nothing but square root of alpha x squared plus alpha y squared, okay and this is a positive quantity. Yeah, exactly. So, alpha is positive. So, we are looking only at the positive half of the plane. So, the point I am trying to make here is when you include surface tension, there is something like a selection of a particular pattern, particular wave number, okay, which is what you are going to see and that is occurring naturally in the system because of the physics, okay. So, I think we will just stop the Rayleigh Taylor discussion with this. What I want to do is go on to the next problem which is the problem of the capillary jet instability which we have mentioned a couple of times in the class before, okay. So, basically for large alpha, the question is for large alpha sigma squared is negative but remember, yes indeed sigma squared is negative. So, as far as sigma is concerned, you will get a purely imaginary number that means there is no real part. The real part is 0. You will get plus or minus i multiplied by some number, okay. So, you have a pure, that means the real part is 0. So, you are on the boundary of stability, the threshold. That is the reason I did not discuss this case where you have, so you really cannot tell if it is stable or unstable, okay. If you know for sure it is positive, you know for sure it is negative, then you can make this conclusion. So, I am focusing only here where I know for sure it is unstable, okay. Here it is purely imaginary. So, it is marginally stable or neutrally stable. If you give a disturbance, it is neither going to grow nor it is going to decay, okay. The real part is 0. So, whatever disturbance you are going to give is just going to stay as it is, okay. So, basically this portion of the curve you really cannot. So, what you need to do is go for higher order terms to understand what happens. So, you can actually conclude about stability or instability only if the real part is either negative or positive. If it is 0, you really cannot tell anything. What you need to do is go for higher order terms, okay. So, that is the limitation of this linear stability analysis. In a sense you can only tell you for sure stable or unstable depending on whether it is negative or positive, the real part. System is stable, yeah, yeah. Now, for low wavelength here, low wavelength is, oh, you are objecting to this. You are objecting to this statement here. Oh, yeah. So, maybe if I write this, it is marginally stable. Maybe on the basis of this analysis, this is what I have to say. Maybe it is neutrally stable is what I am going to write. It is neutrally stable. Actually, we cannot really make any conclusions. It is inconclusive. I think that is a good point. So, what this means is the lesser you write, the less chance of you making a mistake. The more you write, the more chance of you making a mistake, right. I think that is an important lesson. More than whether it is stable or unstable. We will talk about this capillary jet instability, okay. And this particular problem we have analyzed, I mean we have mentioned in the past, but was analyzed long time ago by Rayleigh. I think we need to give him enough respect. So, we will call him Lord Rayleigh, okay. And we are going to basically do this analysis. Again, it is going to be a very simplistic analysis with a lot of assumptions. But then there is a lot of rich information which comes out of it. So, you can use that as a basis for doing more complicated analysis by relaxing some of the assumptions, okay. That is the idea. So, what is this capillary jet instability problem? It is one, supposing you have a jet of liquid falling vertically down. Let us say that is the vertical direction. What we expect is because of the gravitational force, the jet is going to accelerate, okay. And because it accelerates, it is going to constrict. Eventually what you see is this guy is going to pinch off and you are going to get drops, okay. That is an experiment which you will see every day in the morning when you open the tap in your bathroom, okay. So, the idea is, is it possible for us to make a prediction of the size of these drops? How does this happen? What is it that is causing this thing to break? Because theoretically it can keep on shrinking and then keep on thinning down and it can go on forever. The velocity will keep on increasing. So, what is it that is causing it to break into drops? It is clearly the surface tension effect, okay. And that is the reason this is called the capillary jet instability because capillary is basically associated with surface tension, okay. So, again here, surface tension is important, has to be included in the model, right. Otherwise, you will not be able to actually get this breakup. The other thing we are going to do is we are going to do something similar to what we did for the Rayleigh-Taylor problem. We are going to assume that the jet is inviscid because viscosity is not really the one which is causing the jet breakup. Viscosity is only going to possibly change the rate of growth. It is not going to decide whether it is positive or negative. It is going to make the growth rate small if viscosity is high. So, if you have very viscous fluid like honey and you just drag and drop honey from a height, the honey is not going to break, okay. Because it is going to break after a very, very long time. So, the time for breakup, the growth of the disturbance is going to be very large, okay. Or the growth rate is very low, okay. So, viscosity is not really going to decide this. It is only going to make it break further away, okay. What we are interested in is when it breaks, what is the size of the drop? Supposing that is the question you are asking, then you can actually neglect the effect of viscosity, okay. Anyway, we will neglect it and if you think it is important, we can always include it later. So, we will follow lot Rayleigh and we will make another simplification which is we will consider the jet to be stationary or a horizontal circular, I will just call it a thread. So, this is very hypothetical situation again, which means you just imagine something like a circular jet with no velocity. It is just a blob of liquid which is in the form of a circular cylinder and a stationary, okay. That is your geometry. So, the question now is, is this, if you had such a liquid thread, would it break up into drops because of surface tension, okay. And there is no acceleration like we had here. So, that would clearly depend upon the size because the surface tension force is going to depend upon the size of this thread. So, the smaller the thread, the more is going to be the effect of the surface tension force because the surface area to volume ratio is higher, okay. So, the question we are asking is, will this thread break up and disintegrate into drops, okay. And the next question would be what would be the size of the drops that we get. So, what are the advantages of making these assumptions that I have? So, what have I done? I have made a lot of assumptions, one inviscid, no viscosity. I am also saying that this jet is not moving, it is just sitting there, okay. Which means I go back to this problem, which was similar to my Rayleigh-Binard problem where my base state was 0 velocity. My Rayleigh-Taylor problem where the base state was 0 velocity. Here again, my base state is 0 velocity. That just helps me do the analysis, okay and get some insight. So, if you are not happy with any of these assumptions, then you have to go and become the homework problem, right. So, that is what we will do. We assume that this is the base state. Well, you do not assume the base state. The base state is the trivial solution u equal to 0. All 3 components are 0, okay. Liquid is not moving and clearly corresponding to this base state, there is going to be a pressure distribution, okay. And what is going to be the pressure distribution? When you have a flat interface or not a flat interface, a circular interface, there is going to be a difference in the pressures. The pressure here is not going to be the same as the pressure in the atmosphere, okay. Let us say the ambient fluid is atmospheric and you have P atmosphere here and this is P because there is only one pressure which is that of the liquid. That is the pressure of the liquid. So, clearly P-P atmosphere equals gamma divided by r, okay. This 1 by r turns out to be del dot n. We will see that when we do this analysis. So, this is actually gamma del dot n. Del dot n turns out to be 1 by r, okay. So, that is my base state. There is a pressure difference. There is a pressure jump across this interface, okay and that is because the liquid is actually curved. So, if you look at the cross section, the cross section is actually circular, okay. And in order for you to maintain this circular shape, you need to have P greater than P atmosphere. It is different. So, in order to analyze this problem, use the same approach as what we did earlier. Write down the governing equations. We have the base state and we need to do the linearization and go ahead with the solution. Again, same business of trying to decompose into wave numbers. What we will do is we are going to assume that this thread is infinitely long, okay. The thread is infinitely long and so then I can actually decompose it in terms of some kind of Fourier mode, okay. The other thing that we can do is make a further assumption of there being axis symmetry in the problem, okay. That is there is no variation in the theta direction, okay. That is just to make the algebra easy. In fact, people have worked with by including the theta direction also and there is a result which you know you can get again analytically. So, we further assume, what do we assume? Theta symmetry, okay and infinite extent in the z direction. So, it is going is a very infinitely long thread because if I do not assume infinite in the z direction, I need to put boundary conditions and I will be at a loss or answer that question, okay. So, this basically gives you some insight. So, we write the equation of continuity which is 1 by r under these assumptions, this becomes this plus d by dz of uz equals to 0. That is my equation of continuity and then I have rho d by dt of ur plus ur du r by dr plus uz du r by dz equals minus dp by dr, okay. And then similarly I have the viscous term does not show up because I am assuming it is inviscid, okay and let us say I am just doing this thing without any gravitational effect because I do not think gravity is the one which I am interested in studying. If I want to study the effect of gravity, I do the vertical jet problem because that is the one which is actually going to cause the break up. What I am focusing on is, I am focusing on how the surface tension is going to actually cause the break up, okay. So, gravity is neglected and then I have rho t, okay. Remember what we have done is neglect, think, gravity and viscosity. I need to keep surface tension but that is going to come in the boundary condition. So, you can already see a little bit of what is going to happen. So, those are my governing equations subject to the boundary conditions which are my kinematic boundary condition and my normal stress boundary condition, okay. Those are the 2 conditions which I need to invoke just like we did for the Rayleigh Taylor problem at the interface here, okay. So, we will do that but first let me just deal with the equations here. u r is of the form u r ss plus epsilon u r tilde, okay. That is my steady state, that is my perturbation which is order epsilon. What I want to do is u r ss of course is 0 because nothing is moving, the jet is stationary, okay and similarly you can put for the user also. So, if you put that the equation of continuity for the perturbations, this becomes tilde plus this is of order epsilon, okay. Do the same thing for the Navier-Stokes equation. What do you get? d by dt of u r tilde plus epsilon times that plus epsilon times u r tilde times d by dr of epsilon u r tilde plus epsilon uz tilde d by dz of tilde times epsilon again equals minus dp by dr of steady state plus epsilon p tilde, okay. These terms again the convective terms are of order epsilon squared and therefore these guys drop off and order epsilon. The other thing which I want to mention here is dpss by dr is 0. What we are talking about is the pressure inside the liquid in the radial direction is going to be 0 because if it is not 0 there is going to be some kind of a convection, okay. I assume there is no convection or velocity in the radial direction therefore dpss by dr has to be 0. What that means is the pressure is uniform in the cross section but there is a pressure jump across the interface because p1-p2 p atmosphere is gamma divided by r. So, in the cross section the pressure is uniform, okay. So, what this means is pss is independent of r and so at order epsilon my linearized equation is this. You can do the same thing for the other direction. Similarly, you get rho dou uz tilde by dou t equals-dp tilde by dz, okay. Again in the steady state the gradient is 0 for the pressure in the axial direction. So, as far as the equations are concerned these are the equations, okay. But then I need to solve this subject to some boundary conditions and what we do is we do not worry about boundary conditions in the axial direction. Why? Because it is infinitely long and so we are going to give periodic perturbations in that direction. What we need to do is worry about the boundary conditions in the radial direction and that involves the normal stress boundary condition and the kinematic boundary condition. What I want to do is I want to talk about, so when I am doing this linearized analysis I am going to give a perturbation and what this perturbation is going to do is it is going to be of some arbitrary perturbation. I am drawing it periodically here but I mean it is some arbitrary perturbation here, okay. As a result of which the surface can deflect. So, when I am going to giving a disturbance the I am not restricting my surface to be circular and conserve volume, okay. So, as a result of the disturbance the surface deforms. The interface is going to deform, okay. That is the general situation and that is what we are interested in. I have to include this in my model otherwise the guy is not going to break, right. I mean I keep my interface flat it is going to remain a circle for the ever. So, I need to include this thing and see how this guy is going to behave. And if your radius of the unperturbed surface is A then r equals A is the base state, okay. And what we can do is the perturbed state is r equals A multiplied by 1 plus epsilon times f of z. So, I am giving a perturbation. The perturbation is the form of f z some arbitrary function of z, okay. And it is a very small perturbation and that I am indicating by epsilon here. Same thing as what we did for the Rayleigh-Teller problem, okay. Only thing is the Rayleigh-Teller problem I had h which was function of x and y, okay. But now to make my life simple I am just saying that things are not changing in the theta direction. That is the reason I not included my theta dependency here, okay. Because it is axisymmetric theta does not show up. Only it varies only in the z direction. Just keeps the algebra a bit simple. But at the end of the day you saw when we included two directions x and y. The two wave numbers which I can actually combine most of the time, right. I got alpha x square plus alpha y square and I just said that was equal to some alpha square, okay. So mathematically only it becomes slightly different. Otherwise the analysis is the same, okay. What we want to do is we want to get the normal stress boundary condition. The normal stress boundary condition is what I wrote earlier which is p1 minus p2 equals gamma multiplied by del dot n. That is my normal stress boundary condition. And this is going to be valid always in the sense p is actual pressure. In fact, I should not write p2. This is actually p ambient, right. That is constant. And what I want to do now is calculate this del dot n which is my curvature. And but I need to find del dot n for this deformed surface because that is the gentle case. So how do you find n? Before finding del dot n I need to find n. n remember is gradient of f divided by absolute value of the gradient of f. That is n, okay. And f is r minus a times 1 plus epsilon f of z, okay. This is f equal to 0. So what is the gradient of f? Er minus a epsilon f prime, f prime is df by dz divided by 1 plus, no, that is no. This is the gradient of f. I am just differentiating this with respect to r. That is associated with 1 Er, differentiate with respect to z. I get a epsilon f prime and that is associated with dz, okay. And therefore n is going to be, so that is my unit normal vector, okay. And what I need to do is calculate del dot n. I am not sure if we did this problem already. Maybe we did. Remember somebody was talking about 2 curvatures and that is what I want to show today, okay. We need to calculate del dot n. How do you calculate del dot n? It is Er by r d by dr of r plus ez d by dz, okay. Operating on n, which is, so what I need to do is operate, I am doing the dot product. Er and ez are perpendicular to each other. Is that a problem? I think so fine. Taking the dot product, so Er dotted with Er is 1. Er dotted with ez is 0. And I need to concentrate this term with that term and this with this, okay. Er and ez are perpendicular. What I also want you to recognize is that this particular term is a function only of z. It does not contain r. The entire thing is independent of r, okay. So for all practical purposes, when I am differentiating with respect to r, this guy is going to give me 0. But then I will have, I have to use the product rule. r multiplied by that, right. I have to use the product rule. So what I will get is 1 by r times this with that 1 plus a square epsilon squared f prime squared. That is what this is going to give me. 1 by r times this and Er dotted with Er is 1 and that is d by dr of r, which is 1, okay. This is just mathematics. So you do not have to worry too much about it. But now we are going to do this with that. But now you remember that this is a function of z. So you have to use some quotient rule, like we did the last time and maybe what I will do is just write this plus d by dz of minus a epsilon f prime divided by square root of 1 plus a square epsilon square f prime squared, okay. So use the quotient rule now and it is going to be similar to what we did. I would like to actually write the answer but I am not sure of the sign minus or plus. So I just have to do it again, okay. I am looking at the second term over there. It is d by dz. We can just write the second term. Denominator times derivative of the numerator, which is I think it is minus plus a epsilon f prime numerator times the derivative of the denominator, right, which is 1 by 2 times f double prime. So you bring that to this side and you can do the simplification, okay. Since I have done this problem before, I think this is what you will get. That is what you get. It is a minus sign. This is exactly what we got last time. You take the denominator here, multiply it and this will cancel off with this and you have this multiplied by 1 contributing. So that is the second term and therefore del dot n is equal to 1 by r times 1 plus a square epsilon square f prime square to the power half minus a epsilon f double prime divided by, that is what we get. Now I want you to focus on the fact that the curvature term is actually made up of 2 terms, okay. And I want to give you some physical significance to these 2 terms. The significance is that this particular term is what I would call a radial curvature and this is an axial curvature, okay. And just to explain what these things mean, look at this deformed thread. So you have a curvature along the plane of the board, okay. The curvature along the plane of the board, that is the rz plane. This is z and this is r. rz plane is my axial curvature. And this is my f of z, remember. This is f of z. So this is associated with my f double prime, okay. In the r theta plane, the r theta plane is actually perpendicular to this, okay. It is circular. Why do I say circular? Because I assume theta symmetry. If you want, you can assume non-circular but basically in r theta plane, this is the shape. So there is a curvature of the cross section, okay. And this curvature is the radial curvature. So there are essentially 2 effects. One is a radial curvature. One is the axial curvature and both of them together give you the actual curvature which you have to include in the problem, okay. So well, we have got this particular thing, del dot n. What we have to do is we have to use this in our boundary condition, the normal stress boundary condition. And in the normal stress boundary condition, we have to again do a bird of patient series analysis, get the term of order epsilon to the power 0, get the term of order epsilon to the power 1, okay. And because the equations of r to the order of epsilon, the boundary conditions also have to be of the order of epsilon. The way I am going to convert this to order epsilon is just by doing a binomial series expansion. And take this to the numerator, take it to the power minus half and do a power series expansion. Do the same thing there and find out what is the term of order epsilon, what is the term of order epsilon to the power 0, okay. So once we do that, at least the normal stress boundary condition is clear, okay. The normal stress boundary condition, remember is P1-P2 equals gamma del dot n. And I am going to write this as P1ss-, sorry, epsilon P1 tilde-P atmosphere. P1 is P1ss-epsilon P1 tilde. P2 is P atmosphere, outside liquid, okay. Equals gamma my surface tension times this guy-, okay. So do your binomial series expansion, 1-1-half of a squared epsilon squared f prime squared, etc., etc. right? Is that right? Yeah. But the r can change with z, right? Yeah that is correct. That is what we are what we are going to get? Let us do this then you will know, yeah, it means that is what we are going to find out, at order epsilon we will find out what it is, what it is at order epsilon to the power 0 and what it is at order epsilon to the power 1, okay, so we will find that out when we do this. So you are talking about this term, right, you are talking about this term, you are telling me whether it is going to be always of order, whether it is going to be always equal to 1 by r, yeah, but the r can change with z, yeah. So I am getting 1 upon r into some extra term, so does it mean that extra term is equal to 1? No, I do not know, it does not mean that extra term is equal to 1. The radial curvature you are saying is equal to 1 by r and no, this is the radial curvature, okay and let us do like this, let me finish my, I will answer your question, but let me finish this analysis and then we will come to your question. So how does this binomial series thing work? 1 by 1 plus nx minus a epsilon times f double prime times 1 minus 3 by 2 times a square epsilon square f prime square plus etc, okay, is this right? Yeah, so now I have forgotten something here, this r itself is varying with z, I need to include the fact that r is actually 1 plus a epsilon f of z because I am going to be evaluating this along the boundary, at the boundary r is not equal to a, for the water surface r is actually 1 plus a epsilon f of z, okay, so this has to be, I need to do a binomial series expansion of this as well, take it to the top. So this is going to now give me a into, you are right, yeah, yeah, a into 1 plus epsilon, yeah, thanks, so this is going to be gamma times by a times 1 minus epsilon f to the power minus 1 times 1 minus half times a square epsilon square f prime square, I do not worry about this term because this is a higher order term, okay, when I multiply this and this is going to give me minus a epsilon f double prime, the rest of the terms are higher order terms. So, I get gamma by a 1 plus epsilon f times 1 minus this thing, I am not sure if there is a sin problem, where is the 1 plus, yeah, that is good, it should be plus because this should be minus, yeah, now everything is fine, I was wondering if I made a mistake again, so this gives me gamma by a minus gamma by a f epsilon minus a epsilon f double prime, so I do not know if this answers your question that the radial curvature is actually gamma by a, I mean at order epsilon, it can be written as gamma by a minus gamma by a f epsilon, so this is your radial curvature when you do the binomial series expansion, okay. So, what I want to do is, I want to look at the left hand side and the right hand side, these terms are of order epsilon to the power 0, this is going to balance this term which is of order epsilon to the power 0, this P1 tilde is going to balance the order epsilon terms, so what this means is the boundary conditions, the normal stress boundary condition gives me P1 ss minus P atmosphere equals gamma divided by a and P1 tilde is equal to minus gamma by a f minus, I think the gamma multiplying this also, right, the gamma multiplying this, gamma a f double prime, yeah. That is basically what your normal stress boundary condition is at order epsilon, so this is how your perturbation pressure is going to vary, okay, so all I have done is written the normal stress boundary condition to order epsilon, I mean whenever you have any term like something having an epsilon, a function of epsilon in the denominator or a sine or a cosine term, you are going to do a Taylor series expansion and then reduce it to order epsilon, a power series in epsilon, that is what we have done, okay. And this is remember fine because in the base state that is what I expect and for the perturb state this is what I expect, okay.