 Today we will start with our discussion on the dynamics of viscous flows. As you remember we were discussing about the transformation of the integral forms of the differential equation, integral forms of the conservation equations to the corresponding differential forms and from that we came up with a differential form for the momentum conservation and we will start with that differential form. So what was the differential form? Let us write it once more. This was the differential form that we derived. We should keep in mind that this here everything is written in the index notation that means there are certain terms where there is an invisible summation. In which term there are summations? There are 2 terms in the left hand side and 2 terms in the right hand side. So out of these terms in which terms you have summations? Second term in the left hand side because you have this repeated index. So this is actually summation of this one with j running from 1 to 3. That means wherever there is such a repeated index this is like a dummy index. In place of j you could write klm whatever because that is not the index that is like free for you. I mean it is summed up. Here also it is j equal to 1 to 3 because it is a repeated index. So whatever index is not a repeated index. Here i is not a repeated index. So i you are free to choose. So i equal to 1 will give the x component of the momentum conservation equation. i equal to 2 will give the y component. i equal to 3 will give the z component. That is the basic way of looking into it. So it is also possible therefore to write in a way in which you write the velocity components uvw. So alternative notations if you write in terms of uvw you write, let us say you write i equal to 1. So it is u1 which is equivalent to u the x component of the velocity. So what is this? Then such a summation so this is the equation written for motion along x. You see the advantage with the index notation is so compact and this when you have expanded it it does not look that compact. Obviously it means the same whenever we will be looking for analytical solutions we will be expanding in this form. But at the same time for the convenience of writing and understanding the index notation many times is quite suitable. Now we will start with this equation. We will use the index notation for deriving the rest of the things. But whatever is our final form again we will come back to the alternative notation. So that you may use those forms for solving the problems. Now we will start with this form with a question what are the assumptions under which this is valid because we may proceed only from that. So we need to understand that what are the assumptions which are inbuilt with this equation. What kind of assumptions were there? Well continuum yes because these are continuum conservation equations. Any more assumption? The reference frame is stationary that was the assumption right. So if you have a moving reference frame it would be easy and straight forward to derive the form because if you recall how did these terms appear. So you had a resultant force in the left hand side of the statement of the Reynolds transport theorem. There you had surface forces and body forces. This term came from surface forces, this term came from body forces. And in the right hand side there was a transient term from which this came and there was an advection term because of the transport of the fluid across the surface of the control volume. From that this term came by converting it into a volume integral. So it is important to keep in mind that what these are. We will see later on that it is equivalent to an equation written in an Eulerian description where the left hand side is like acceleration per unit mass say and right hand side is force per unit mass. So it is as good as like almost like Newton's second law of motion. Now the other question is that which we partially address that what are the physical significances of these terms. And if you look for these terms we will recognize that this right hand side we had recognized. The left hand side we will see that it will come as some function of acceleration or it will be directly related to the acceleration. So acceleration of the flow is related somehow with the transients that is the unsteady effect that is at a particular location if things change with time. And if there is a change because of the change in position from one point to another point where the particle finds a new flow field that is the advective effect. And the combination of that is the total derivative we will be able to express this in terms of the total derivative that we will do subsequently. If you want to write this for a non-inertial reference frame you have to just keep in mind that the left hand side for a non-inertial reference frame is corrected with a correction factor minus equivalent to like a D'Alembert's force like minus mass into acceleration that type of term where that is like a relative acceleration. In that relative acceleration you have terms related to the angular velocity of the reference frame or angular acceleration of the reference frame like that. So you can incorporate those correction terms and that is itself a volume integral and derive it in the same way you will see that you will get some extra terms because of the transformation of coordinates. Otherwise this is quite straight forward and the advantage or the elegance of this form is that it is independent of the nature of the fluid that you are looking for. Till now we have not committed ourselves to whether the fluid is Newtonian, non-Newtonian or whatever it is just for any generic fluid. So this equation is known as Navier's equation of equilibrium. This much we discussed in the last class but this is just a brief recapitulation so that we may start with this. Now our objective will be to see that what equations we have with us. See what is our objective? Our objective is to have sufficient equations so that we may solve for the unknown variables. So we have to identify what are the unknown variables. See if you look into this equation what would be the unknowns? See definitely if you have a forcing on a system you know what is the body force because if you do not know what is the force to which it is subjected you cannot find out how the system evolves under the action of the force. So you know the force. What are the things which are unknown here? So velocity components are unknown here and also the stress tensor components these are not known to you. So the stress tensor components in turn may be expressed in terms of the velocities and that we have to see that is it in terms of the velocities directly or in terms of the gradients of velocities. So we can qualitatively understand from our studies in kinematics of flows that any deformation for a fluid is actually not important in an absolute sense but the rate of deformation is what is important in an absolute sense. So when we say rate of deformation we mean linear deformation, angular deformation, linear deformation may give rise to a volumetric change. So all these are expressed in terms of rates for fluids because the understanding is fluids are anyway continuously deforming. So if we allow more and more time deformation will be more and more that is not a quantification of anything. Quantification is the rate at which that deformation is taking place or the rate of strain so to say. Now that rate of strain should be related to this one but we have to keep in mind that there is a forcing condition, there is a surface forcing condition which should not depend on the rate of deformation. What is that surface forcing condition? So if you have a force on a surface this somehow is a representative of what is the forcing condition on a surface of a fluid element. Now there is a part of that forcing condition which should not be dependent on the rate of deformation and that we should isolate. What is that? That should be the pressure distribution because we have seen that what is the hallmark of a fluid at rest. If there is a fluid at rest that means it is not deforming and we know that fluids are in continuous deformation subjected to shear forces. So that means when fluid is at rest there are 2 important things which are occurring. One is there is no shear force because if there is a shear force it will start deforming except for some special fluids which we have discussed as some special non-Newtonian fluids but in general if you apply a force the fluid will start its deformation. So if it is at rest that means there is no shear force. There is only normal force but a special type of normal force that is not dependent on its deformation because you can also have a normal force which may be related to deformation may not be angular deformation but may be volumetric deformation. So if you have an element you may be able to stretch it or compress it by a normal force. So it is not just true that if you have a normal force then that is not related to deformation. Normal force also may be related to deformation but pressure is going to be such a normal force which is not related to the deformation because it is equally applicable for fluids at rest which are not under deformation. That does not mean that when a fluid is moving it does not have pressure. Simply we can isolate that effect and say that whatever is the remaining effect that is related to the deformation of the fluid. So when a fluid is moving it is equally subjected to pressure but we may isolate that part and say yes the effect due to pressure was something which was not responsible for its deformation and anything else would be responsible for its deformation. Accordingly what we may do is we may decompose this tau ij into 2 parts. One part we say is a hydrostatic component and other part we call as a deviatoric component. So hydrostatic component is something from the name it is clear that as if it talks about a static condition. So it is not related to therefore deformation. So we can broadly say that we are decomposing it into 2 components. One component is not related to deformation and other component is related to deformation. That is the whole idea and when the component that is related to deformation that component should be only a function of pressure because when a fluid is at rest you can only have the description of the equilibrium through the pressure. Now coming to the deviatoric component which will be more interesting for fluid under motion which is under our discussion at the moment. So this deviatoric component should be a function of what? It should be a function of the rate of deformation. So when you write the rate of deformation let us just recall that the rate of deformation in general in the most general form this could be a gradient of velocity where i and j could be different indices. In a special case i and j could be the same index. So we have seen earlier which is quite trivial that you can write this as this writing is nothing very special it is just writing an identity and the whole understanding is that this is like a tensor. You may write it in a matrix form also because you have 2 indices i and j and basically you are decomposing it into a symmetric and a skew symmetric tensor. So the first one is the symmetric one next is the skew symmetric one. Physically this decomposition is very important mathematically may be very very trivial but physical importance is that this is related to the deformation and this is related to the rotation. So when you have a rotational sense you do not directly associate that with deformation because an element may have rotation without deformation that is like a digit body rotation that is a special case. So if you want to relate the deviatoric stress tensor component with something what should be that something I would say that the deformation which is appearing from here should be a natural should be naturally related to the deviatoric stress tensor component right. So that is the first understanding. So we have to first understand that what depends on what because we are trying to develop something which is called as a constitutive relationship or constitutive behavior. What is a constitutive behavior that is if you have think about say Hooke's law in linear elasticity. So you say that stress is linearly proportional to strain within proportional limit of an elastic material. So if you say that stress is linearly proportional to strain that means you are somehow relating a cause with an effect and that comes from the behavior of the material because this you cannot generalize. It is specific to different materials and that material constant will appear in form of a modulus of elasticity. So modulus of elasticity is the gross response of the material against the applied loading condition. In say it is a tensile load and it starts getting elongated. Now the manner in which it is going to get elongated depends on its modulus of elasticity which comes from the inner characteristics of the material. So it is a gross consequence of the material characteristics and that is where it will be different for different materials. So it comes from the constitution of the material that is it is intrinsic to the behavior of the material and that we call as a constitutive behavior. So just like you have for solids constitutive behavior in that way for fluids also you have constitutive behavior in that way and that is how we have Newtonian fluids non-Newtonian fluids like that because different fluids have different constitutive behaviors. So we are interested to have a constitutive behavior or have a functional form of the constitutive behavior. So the functional form in a very qualitative way should be from our discussion that tau ij deviatoric should be a function of the rate of deformation. That much we should appreciate qualitatively okay. Now the question is this function may be of whatever complex nature and important thing is it will depend on what? It will depend on a deformation rate of deformation which is also a second order tensor. So now we are interested to write a function where the dependent variable so to say is a second order tensor, this is a second order tensor and the variable in which or the function on which it is going to depend that is the independent function so to say is also a second order tensor. And we are interested about the functional relationship between these 2. This is not a very simple task. So we will not look into the most general way of characterizing this. There is no very general way of characterizing this. So we have to think about special cases and the special cases are the cases when these are linearly related which is like the Newton's law of viscosity. So we will now consider a Newtonian fluid. So if you have a Newtonian fluid then this will be a linear function. That means the deviatoric stress tensor will be a linear function of the rate of deformation just like Hooke's law for linearly elastic material solid material. So the question is now how we are going to write it? We have to keep in mind that we are now interested to write a transformation that maps a second order tensor onto a second order tensor. Earlier we wrote a transformation which maps a vector onto a vector. If you recall the Cauchy's theorem, the traction vector was mapped in terms of the direction normal or direction cosines of the area which was chosen to evaluate the stress tensor, the traction vector components and that mapping was successful through a second order tensor which is the stress tensor. So the second order tensor was such that it mapped a vector onto a vector that much of mathematical understanding we had. Now we are in a situation where we have to map a second order tensor onto a second order tensor. So there should be some intermediate transformation and that transformation is a fourth order tensor. So fourth order tensor is something which maps a second order tensor onto another second order tensor. So this linear function is written in this form tau ij is equal to a fourth order tensor say Cijkl and a second order tensor. So what is Ekl? Ekl is like okay. So this is Ekl. So E is like the rate of deformation, just this form, just i and j are replaced by k and l. So this is like Eij, whatever is written here okay. Just a symbolic way so that we do not have to write the big thing again and again, that is just a symbolic way of writing it. Now look into this. So what are the repeating indices here? So you can specify, you can fix up i and j whatever you want but you cannot fix up k and l whatever you want because it is summed up over that because these are repeated indices. So you have basically summation over k and summation over l each running from 1 to 3. But the more important thing is that so k and l could be from 1 to 3, i and j also could be from 1 to 3, those are free but you could have i equal to 1, 2 or 3 or j equal to 1, 2 or 3. So how many combinations with Cijkl are possible? In the most general case, 3 into 3 into 3 into 3, 3 to the power 4. So you could have 81 components of Cijkl and that is the most general constitutive behavior for a Newtonian fluid. That means if it has no other special effect, then you would require 81 independent constants to specify its constitutive behavior. That is not again a very simple thing. So of course we are not going into that what would be the picture if we had 81 constants. We will see that those will be reduced to again fewer number of constants even in a more general case because we know tau ij equal to tau ji. So from that symmetry, we can at least reduce a few in number. So we will not go through that route but we will at least keep in mind that mathematically speaking, if you do not use the symmetry of tau ij, you have 81. If you use the symmetry of tau ij, 36. So I will leave it on you as an exercise to show that if you use tau ij equal to tau ji, the 81 constants may be reduced to 36 constants. But the thing is that is also not something which is very very tractable. Think about a problem where you are a practical problem where you require 36 material constant properties to describe a flow. I mean in a very general situation that might be the case but that is not a very common case in engineering or even in science it is not a very common case. So we will do in this course or we will address in this course something which is perhaps the simplest of all these cases but the most commonly encountered one that is we will consider the very special case of a homogeneous and isotropic fluid. So the special case that we consider is homogeneous and isotropic fluid. Homogeneous fluid and isotropic fluid. So till now what assumptions we have made? We have made assumptions first of all stationary reference frame that is with which started our equation. Then a Newtonian fluid special case homogeneous and isotropic Newtonian fluid okay. So now the question is that when you have a homogeneous and isotropic fluid what is the special characteristic that you have? So what should be the dictating factor for a homogeneous and isotropic? So what is homogeneity? So homogeneity means it is same at all points in the flow field that any property is same or position independent at all. So this is related to position independence. So the constitutive behaviour does not vary from one point to another point and isotropy means direction independence. What is the hallmark of that? Say if you have coordinate directions say x and x 1 and y 1 and if you measure or if you try to assess the constitutive behaviour on the basis of that now if you have a transformed coordinate system still orthogonal but let us say x prime and y prime with a rotation. This does not disturb the constitutive behaviour although the orientations of the axis that you have considered for describing the behaviour those have changed but in terms of that new axis the constitutive behaviour still remains the same. The description still remains the same that is called as an isotropy. So we are going to think about a homogeneous and isotropic fluid. So if you have a homogeneous fluid the thing that we will immediately conclude is that Cijkl should be independent of position otherwise if Cijkl is a function of position it loses its homogeneity. What would be the consequence of isotropy? That is something which we need to look into more details mathematically. So for isotropy to say that you want to form something like you have an isotropy and the fourth order tensor Cijkl how could you convert this into a scalar because see let us say our objective is to convert or transform this one into an isotropic scalar okay. Let us say there could be many ways of doing it. Let us say what you are given are vectors. Using vector how can you convert this into a scalar? Forget about isotropy. How can you use vectors to convert it to a scalar? Let us say you have how many vectors would you require first you say. This is just from common index notation understanding you can say 4 because if somehow you can sum it up over these indices then it will have no free indices and that then it will become a scalar. So if you have say 4 vectors a, b, c and d you know vector is characterized by one index. So this say it has a i just for its index notation say this is bj, this is ck, this is dl. Just mathematically now if you have say this a i, bj, ck, dl then and if you write it in this way that means there is an invisible summation over ijkl. So at the end whatever comes out does not have any index and it is a scalar right. So you can employ 4 vectors, 4 independent vectors to convert a fourth order tensor to a scalar that we can understand. But we are interested for a special type of scalar which is an isotropic scalar. Now let us see that what can lead to isotropy if you are dealing with 4 vectors. So how can you say just look into this example. Here we had 2 vectors when we talked about this rotation actually we talked about a vector say maybe one vector oriented along x another vector oriented along y and this vector combination was rotated and after rotating this vector combination the property or the constitutive behavior whatever is described through the new orientation that is same as whatever is the description through the old orientation that is the implication of isotropy. So what is the meaning? The meaning is the meaning is very interesting if you do not change the angle between the 2 vectors then it does not change. So when you what is rotation? Rotation is something which preserves the angle between whatever vectors were there. So as if you are rotating these vectors like a rigid body keeping the angle between the vector same. So if you have vectors a, b, c and d and if you somehow preserve the angle between these 2 taken in pairs say a with b and c with d maybe a with c and b with d and a with d and b with c these are the possible pairings. So with these possible pairings if somehow you keep the angle between these as unaltered then no matter whatever is there orientation in space so long as the relative angle does not get disturbed it should give back the same behavior that is the meaning of isotropy. So if you have vectors a, b, c and d so what kind of combinations you can have taking all those 4 you can have a dot b and that corresponding combination is c dot d then you could have a dot c, b dot d and a dot d, b dot c. Why dot? Because dot directly gives the angle between 2 vectors we were talking about the angle between 2 vectors. So if we consider the dot product that that returns sort of angle between the vectors. So if that is preserved then isotropy is preserved. So why such a combination because you see that this will give rise to a scalar. So if you take these in pairs these are individually scalars so the combinations all combinations of these ones will form a general isotropic scalar involving the 4 vectors a, b, c and d. So this we can write so let us form that scalar. Let us say that that scalar is equal to a dot b, c dot d some multiplier of that say alpha because a scalar multiplier does not change it is any characteristic. So alpha is just an arbitrary scalar multiplier plus beta into a dot c, b dot d and let us say plus gamma a dot d, b dot c. So we have just made a linear combination of these 3. Our objective is to ensure that we get an isotropic scalar. So now what we will do we will write it in terms of the index notations. So if you write it in terms of the index notations how we are going to write it. So we are going to write it in this way. This is the first term that we are writing. Just look into the corresponding terms a dot b may be written as a i b i because it is what a 1 b 1 plus a 2 b 2 plus a 3 b 3 that is the dot product you take the corresponding components and multiply with each other sum it up that is how you get the dot product of 2 vectors. So there is an invisible summation like this and here also it is like this keep in mind i and k are just dummy indices. So in place of i you could write j in place of this you could write whatever. Similarly let us write the other terms plus now you see that we wanted to get it as an isotropic scalar. What is that special isotropic scalar that we are looking for we are looking for c i j k l into a i b j c k d l. That was the form that we are looking for here in the form you do not have the indices a b c and d as different. So if somehow we can convert this indices a of a b c and d as entirely different like one i another j k and l then we would be able to compare this form with this form that we have for our constitutive relationship. So our next objective will be to convert this forms like a i b i c k d k to a form like like a i b j c k d l that form. So how do we do that let us say you want to convert b i to b j that is our objective say because you have a i next you want b j. So you want to convert b i to b j. So there is something very useful in terms of index notation which is called as chronicle delta. So this is given by this is just a notation. So this is defined in this way delta i j such that this is one if i is not equal to j if i is equal to j and 0 if i is not equal to j. So this is just a notation that means if you have only i equal to j this is one otherwise 0. Now let us see let us say we want to write b i delta i j. What is this? See you have to keep in mind that when j is equal to i so rather you may also think it as j equal to i or i equal to j whatever like when they are equal. So think about that only when you have i equal to j then only this is one. So there is only one case when this is equal to one and that is when this i is j right. So this becomes b j because only when i equal to j then these 2 indices are equal that is one otherwise it is 0. So you can see that this is a very simple and elegant way of switching the indices from i to j j to i like that you can play with this. So what we will do? We will use that switching index here. So this b i we will write b j into delta i j. We will keep c k because that is what we want. We want to convert d k to d l. So what we will do? We will have d l into what delta? Delta k l. Similarly let us look into the second term a i. Let us say we keep a i b j is there. We want to change c i to c k. So what will be the transformation? c k into delta i k okay. It is transformation from i to k that is why delta i k. Then we want to change d j to d l. So d l into delta j l. Similarly the third term you have a i you have b j then you want c k. So c j will become c k delta j k and delta j k. d l into delta i l. So if you write in this way see what is our desperate act? The desperate act is we want to write the right hand side just in a form a i b j c k d l okay. So using this simple notation we are able to write it in this way. So let us write that then. So what at the end we can write? We can write that you have now the right hand side which we want to express as say the left hand side as c i j k l into a i b j c k d l. We want to make it an isotropic tensor and the right hand side we have expressed in terms of an isotropic tensor with again you have a i b j c k d l that is like they are common for all terms. Then you have alpha delta i j delta k l plus beta delta j k delta sorry delta i k delta j l and from the third term gamma delta i l delta j k right. So from here it follows that you should have c i j k l is equal to alpha into delta i j delta k l plus beta delta i k delta j l plus gamma delta i l delta j k where delta i j is defined as above. See with this simple exercise we have come up with a very interesting and advantageous conclusion. What is that? We were we started with 81 independent constants in general to describe the material property. The material property is given by this c i j k l. Now how many constants you have? 3, 3 independent constants alpha beta and gamma because delta i j like these are like you know that either 1 or 0. So we can see that and these are position independent. So these are really constants. If these were position dependent this could be also functions of position. So we can say that for a homogeneous isotropic fluid we are now able to come up with 3 independent constants. We may reduce it even to less because we have not yet exploited the symmetry of tau i j that we have never used. We must have tau i j equal to tau j i that we have seen from angular momentum conservation. So the next step is we use the symmetry that is you have tau i j equal to tau j i. So tau i j is what? tau i j is c i j k l e k l that must be equal to c j i k l e k l. So what it follows is that you must have c i j k l equal to c j i k l. So this must be also equal to c j i k l. So how do you write c j i k l just switch i and j here. So that will be equal to alpha delta j i delta k l plus beta delta j k delta i l plus gamma delta j l delta i k. We have just switched i and j. Now you see first of all what is the relationship between delta i j and delta j i? They are the same. Delta is just a symmetric tensor. That means when you require these 2, these 2 get nullified, cancelled. They are the same. For the remaining ones you see delta i k delta j l is this term. One is coming with the coefficient beta in the left hand side. In the right hand side that is coming with the coefficient gamma. That is the only change. And for the other term here, this term is same as like the coefficient of gamma here is same as the coefficient of beta here. So you can just combine that in a simple factorization and write that beta minus gamma into delta i k delta j l minus delta j k delta i l that is equal to 0. Whatever is there inside the bracket, whatever bracket in general this is not 0. Special cases are there but there are certain terms which will not be 0 when the indices are equal. So this in general for all cases since this is not equal to 0, for all cases if you want to make it satisfied, only thing you must have beta equal to gamma. Because what we understand is that the second term in the bracket in some cases it is 0. It is true but in all cases it is not 0. But we must have a constitutive behavior that should be true for all cases. So now you see now we can write c i j k l is equal to alpha into delta i j delta k l plus beta into delta i k delta j l plus beta into delta i l delta j k. So how many independent material constants we have? Two. Keep in mind that there is a exact analogy with mechanics of solids. If you have a solid material, it is a elastic material then how many independent material constants you require to specify its behavior? Two independent constants, modulus of elasticity and Poisson's ratio. You can express the rigidity modulus in terms of any of these two. So you require only two independent material constants. Here also it is totally analogous. So you are requiring two independent material constants to specify its behavior. What are those material constants? That will come from the physical understanding of these terms. So when we want the physical understanding of these terms, we will now write the expression for tau i j. That will give a physical understanding of the term. So tau i j is c i j k l into e k l. So we will multiply this with e k l. So when you multiply this with e k l, so you have alpha delta i j. Let us just write it in an expanded form. Delta k l e k l plus beta delta i k delta j l into e k l plus beta delta i l delta j k into e k l. Let us now simplify these terms. This simplification is easy. It is quite simple. If you consider, let us say this term, delta k l e k l, you tell what should be the corresponding simplification. Let us erase this part and we will write the simplifications here. So delta k l e k l, say we want to write it. How do you write it? You have to keep in mind that this delta is 1 only when l equal to k. So this you can write as e k k or e l l because in all other cases it is 0. What is this actually? This is again not one term. It is e 1 1 plus e 2 2 plus e 3 3 because it is a repeated index. So we have a summation basically. So what is e 1 1 plus e 2 2 plus e 3 3? Let us just write it. e 1 1 is, this is just by using definition of e i j. Then e 2 2 and e 3 3, what is this? This is the divergence of the velocity vector. So we can clearly understand that this has something to do with the compressibility of the fluid, volumetric, rate of volumetric strain of the fluid. So the first term we can understand that it should be related to the rate of volumetric strain of the fluid. So tau i j's first term we have come up with. So tau i j is equal to alpha into delta i j into e k k. Then next terms, let us write the next terms. Next terms you see that what are the indices that you play with here? So what are the repeated things? You have k as repeated and l as repeated. So there are special cases of k and l when this term is non-zero. What is that? When k is equal to i and l equal to j because then only these deltas are non-zero. So the next term will be beta into e i j because this i and j should be something for which delta i k and delta j l are non-zero and that is there when i equal to i becomes k or rather k becomes i and l becomes j because you are writing a summation over k and l. You are trying to find out those values of k and l for which this term is not 0 and that is only k equal to i and l equal to j. The next term, so you have here again you see what are the repeated things? You have k and you have l. So again when l equal to i and k equal to j, so plus beta e j i and the rate of deformation is a symmetric term. So that is the answer we have already seen. So e j i is equal to e i j. That means we are able to write tau i j in a form alpha into delta i j into e k k plus 2 beta e i j. Now the question is how physically we may relate these coefficients alpha and beta? To do that we should understand one very important thing that where from the effects of these come. First you have let us say you have beta. So where from the effect of beta is coming? Let us write what is 2 e i j. So what is 2 e i j? You have 2 into e i j is half of this one. So this is considered as very special case of a one dimensional flow which we did at the very beginning when we are discussing about viscosity that you have only one component of flow say u component of velocity other component was not there. We wrote something very loosely from physical understanding that tau equal to mu d u d y. So when you have say one component only u other component say u 1 only you have and say you are thinking about the gradient along y. So j equal to 2 and other term is not there because only one component of velocity is there. Say you can write this as d u 1 d x 2 that is like d u d y. So you are relating that with some tau. Remember this tau is not the total but tau deviatoric which is related to the deformation. So that is related to this through a coefficient and that we have already seen. Therefore in the general notation this beta must be the viscosity of the fluid mu. So see from mathematics we are converging to the physical understanding and this is very very important. Continuum mechanics is a beautiful subject. It can make you agglomerate physics and mathematics very nicely. So this is mu. This constant is not so well known to us till now whatever we have discussed but it is given a general notation of lambda. This is called as second coefficient of viscosity. We will see later on that what is the significance of this physical significance we will see in details but just a name second coefficient of viscosity. Why? Because already you have a viscosity like mu which already is there. So if you want to have a different coefficient but we have to keep in mind that this is related to sort of angular deformation of the fluid and this is related to volumetric deformation of the fluid. At least qualitatively we can understand the significance of these 2 terms. One term is related to the angular deformation. One term is related to the volumetric deformation. What deformation means? Here type of deformation. Now this is the deviatoric stress tensor only but what about the hydrostatic one? So tau ij there is one tau ij hydrostatic because the total tau ij is tau ij hydrostatic plus tau ij deviatoric. So what is tau ij hydrostatic? That is consider a case when the fluid is in static equilibrium and then even there is a stressing condition because of the pressure distribution. So how do we write it in terms of this tau ij? See you have to keep in mind that it is a pressure effect is like a normal force effect. So it is there only when i equal to j because i equal to j gives the normal components. So it should be something which is 0 if i is not equal to j and which should be 1 if i equal to j multiplied with the magnitude of that. So it should be something multiplied with delta ij that is clear because it should not be non-zero if i is not equal to j. What should be the multiplier? That is the corresponding scalar of it. So what is that? That is because of pressure but you have to keep in mind that pressure by default is compressive in nature whereas our positive sign convention for normal stress is tensile. So it should be minus p. So you come up with the final form where what is the tau ij total? That is tau ij hydrostatic plus tau ij deviatoric. So this is tau ij hydrostatic plus tau ij deviatoric. So what is that? So you have lambda. So first let us write the hydrostatic minus p delta ij plus lambda into e k k into delta ij plus 2 mu e ij that is the tau ij. So we have come up with an expression for constitutive behavior. It has what unknowns? What are the unknowns here? See e ij has unknowns as velocities because it is written in terms of gradients of velocities. So you have velocities and you have pressure. So in the equation of equilibrium see what is our objective at the end to solve for the unknowns. So where we have landed up? We have landed up with an expression for this tau ij which is expressed in terms of unknowns as velocities which are already there plus a new unknown has appeared which is pressure. So in the next class we will see that first of all what are the physical significance of different terms? What is the physical significance of lambda? How it is related to mu and then how we can come up with an equation in the final simplified form where we substitute this constitutive behavior of tau ij in this expression that we will take up in the next class. Thank you.