 In our previous lecture, we discussed about the various conservation principles and let us briefly recapitulate what we discussed. So, we started with the Reynolds transport theorem. Then I mean we first derived the Reynolds transport theorem. I am just recapitulating what we have done. So, we will not get into the details. So, in the Reynolds transport theorem what we essentially did is we expressed the change in the system in terms of change in the control volume. We applied the Reynolds transport theorem to the principle of conservation of mass that gave rise to the continuity equation. Then we started considering the Reynolds transport theorem for momentum conservation and typically linear momentum conservation. So, we have not entered into the linear momentum conservation as yet but we tried to basically express various forces acting on the control volume. So, we tried to express the forces in terms of the surface force and the body force. So, we define what is the traction vector and we express the traction vector in terms of the various stress tensor components by using the Cauchy's theorem. So, that is what we did and we will take it up from there. So, it is important to understand that we are going to apply the conservation of linear momentum by expressing the rate of change of linear momentum for the system by relating that for the system with that for the control volume. Once we do that question remains that what happens with the angular momentum because when we write the equations of conservation very common equation that we write is equation of linear momentum conservation where does or where goes the consideration for angular momentum conservation. So, we will consider that in a moment. So, if you look here you will see that we are considering a rectangular element in the rectangular element the dimensions along x and y are delta x and delta y and what we are trying to see is a basically rotation of a 2 dimensional element. So, the 2 dimensional element is in the x y plane and we are interested to characterize the rotation of this element with respect to the z axis. So, when we want to do that we will express various forces acting on the element. So, what are the forces acting on the element? The forces acting on the element are surface forces and body forces. This is not a complete free body diagram we have not shown the body force simply because we want to find out the moment of all forces with respect to the centre of mass of this element and because I mean the body force will pass to the centre of mass in this particular case it does not have any net moment with respect to the centre of mass. So, therefore we are if we consider an axis a centroidal axis which is basically coincident with the centre of mass in this case because it is considered to be a homogeneous body. So, this may be considered as an element of fluid or an element of solid. So far we have not distinguished between fluid and solid. So, let us try to identify various surface forces we have already discussed that the body force has no contribution towards creating a net moment with respect to the centroidal axis but surface forces will have. So, what are the surface forces? So, if you consider the right surface for example you have a normal stress which is tau 11 the first index just for recapitulation first index is the direction normal of the surface. So, tau 11 here means that the direction normal of the surface is 1 that is x 1 direction or x direction and the second index 1 indicates that the force itself is along x 1 direction. So, this is tau 11 is force per unit area. So, the total force is tau 11 times the area area is delta y times the unit width along the z axis. So, tau 11 into delta y into 1 similarly you have the shear stress which is tau 12 the first index 1 for the direction normal along x 1 the second index 2 for the force itself being directed sorry I will put the cursor properly the force index force itself being directed along the x 2 direction. So, that makes it tau 12. So, tau 12 into delta y. So, similarly on the opposite phase you will see similar expressions like tau 11, tau 12 only they are all in the negative x 1 and negative x 2 direction. The reason is that the outward normal to the left phase is actually minus x 1 not plus x 1. So, similarly we consider the upper phase upper phase the normal stress is tau 22 and the shear component of stress is tau 21 and the lower phase the normal component is tau 22 and the shear is tau 21 same thing but their positive senses are just opposite because their direction normal 1 is positive x 2 and another is negative x 2. So, now we are interested to know what is the net moment because of with respect to the centroidal axis of the element. So, the net moment usually if you see here the moment is contributed by the forces tau 12 into delta y and tau 12 into delta y they are acting in the anti-parallel direction. So, they are from giving rise to a couple moment. So, the couple moment is given by tau 12 into delta y times the distance between these 2 forces that is the couple arm that is delta x. So, tau 12 into delta y into delta x and this is a anticlockwise moment. So, we consider the anticlockwise moment to be positive then the tau 21 into delta x on the upper phase and tau 21 into delta x in the negative x direction in the lower phase that will form another couple moment. So, that is tau 21 into delta x times the distance between these 2 forces that is delta y. So, this is acting in the clockwise direction because this is acting in the clockwise direction it is negative. So, minus tau 21 into delta x into delta y. So, this is the net couple moment the normal stresses tau 11 tau 22 these all these forces due to these stresses will pass through the centroidal axis and therefore will not give rise to any couple moment. So, net couple moment is the moment of inertia with respect to the same centroidal axis times the angular acceleration. So, the moment of inertia is proportional to the mass which is rho into delta x into delta y that is the into 1 1 is the width that is the mass into delta x square plus delta y square into some constant that constant for a rectangular element is 1 by 12. But I mean let us not so much bother about the constant the some constant say k into delta x square plus delta y square times the angular acceleration alpha. Now, if you take the limit as delta x delta y tends to 0 see you look at this expression carefully in both sides if delta x delta y tends to 0 delta x into delta y will be cancelled because they are tending to 0 they are not equal to 0. So, you can cancel that from both sides and then in addition in the right hand side you can substitute delta x and delta y both as tending to 0 in the limit. So, you will get tau 12 equal to tau 21. So, we can generalize that by considering all possible rotations and write tau ij is equal to tau ji. So, we write this expression not explicitly in the equation of motion that we usually describe but we write it implicitly because in the Cauchy's theorem when we write t is equal to tau i tau ji into nj or tau ji into eta j if you consider the previous slide. So, t is equal to tau ji into eta j this is what we had derived in the previous lecture. Now, we will substitute in place of tau ji we will write tau ij. So, we will write t with the ith component of t basically is equal to tau ij into eta j remember that because j is a repeated variable there is a summation over j from j is equal to 1 2 3. So, eta 1 is the x component of the unit normal vector of the surface eta 2 is the y component and eta 3 is the z component. So, to summarize the contribution of angular momentum we are not writing an explicit separate equation for angular momentum in terms of differential equation but in the differential equation of linear momentum we are going to substitute tau ij equal to tau ji that is where we are incorporating the consideration for angular momentum. Now, a very important question is tau ji equal to tau ij always valid. Now, this is a very critical question because see while deriving this tau ij equal to tau ji we considered a body force but no body couple. So, no there is we even if we consider a body force the body force will not give rise to a net moment with respect to the centroidal axis but if you have a body couple that is a some couple which is acting over the entire volume of the body then that will give rise to a moment a couple moment and that moment should add to this. Therefore, in presence of a body couple you will not have tau ij is equal to tau ji but normally we write tau ij equal to tau ji for fluids the reason is fluids will normally not sustain body couples but if fluids contain particular type of matters like if the fluids contain engulfed particles which by virtue of their own rotation can introduce body couple in the fluid then it is possible that the fluid can fluid the equation of motion for the fluid has to be written by taking the body couple into account and that kind of fluid is there in the in the in the study of fluid mechanics and that type of fluid one such example is micro polar fluid. So, we will not be discussing about micro polar fluids in details because I mean we do not have enough scope in this fundamental course to talk about that but these are special types of fluids you know which can sustain couple body couples. So, it is possible that for such a fluid you do not have tau ij is equal to tau ji and you write separate differential equations for angular momentum conservation just like we write differential equations for linear momentum conservation we need to write separate differential equations for angular momentum conservation. Now, micro polar fluids are very much important in the context of microfluidics research because there are certain examples when micro polar fluids are used to model some biological fluids like for example there are instances when micro polar fluids have been used to model blood and you know biological fluids are very important in microfluidic applications. Therefore, I mean we cannot rule out those possibilities where you will have tau ij is not equal to tau ji but if the fluid is not sustaining any body couple it is true that tau ij is equal to tau ji and we will consider that kind of example for the remaining part of this course. Now, maybe for little bit of the remaining part of the exercise we will go to the board and do it because some of the algebra is little bit involved and that is not worked out in details in the slides. So, what we will do is that as I told you earlier that we will like switch over from the board to the slide whenever necessary and for some detail working out we will use the board rather than the slide. So, let us say that so what formalism we have come across is that we have surface force, we have body force. The surface force we are writing in terms of what? So, the other ith component of the surface force what we are doing if we have a control volume like this you consider a small area dA, the surface force per unit area is ti along the ith direction and that you multiply with dA and integrate this over the entire control surface. So, now this ti what we can write? We can write as the Cauchy's theorem gives it tau ji into eta j but from angular momentum conservation since tau ji equal to tau ij we can write this as tau ij eta j dA. Now we will try to express this in terms of dot product of 2 vectors why we are interested to express this in terms of dot product of 2 vectors is because we can then express it in terms of a volume integral by using the divergence theorem that is what is our motivation. See I will try to give you a big picture in all the derivations using Reynolds transport theorem we have some surface integrals and some volume integrals. You will find that the terms in the Reynolds transport theorem in general forget about the complexity at the end it will be either surface integral or volume integral. Now to derive the differential equation from the integral form what we do we convert the surface integral into volume integral by using the divergence theorem. So, the first requirement is express the surface integral in terms of a dot product. So, that we will be in a position to convert it into volume integral then all terms will be converted into volume integral and you will get integral of something into dv equal to 0 because that dv the choice of the control volume is arbitrary. So, that something integral of which is 0 that something is 0 that will give rise to the differential equation. So, that is the basic philosophy by which you derive the differential form of the conservation equation from the integral form of the conservation equation. So, we will be expressing this in terms of what we will be expressing this in terms of a dot product. So, what we do is that see let us maybe use a different color to highlight this. So, what this is? So, this is tau i1 eta 1 plus tau i2 eta 2 plus tau i3 eta 3 right. So, that expression this one we can write tau i1 into i cap plus tau i2 into j cap plus tau i3 into k cap dot eta 1 i cap plus eta 2 j cap plus eta 3 k cap where ijk are unit vectors along x, y, z. So, what we do out of here? We you can see that this is essentially a vector right. It has a 3 components tau i1 tau i2 tau i3 as a some numbers scalar components. So, let us give it this vector a name tau i this is just a name that we give often in the books in some books this is called as stress vector just a name given. So, and this is what? This is the unit vector eta with 3 components. So, this integral is nothing but integral of what tau i dot eta dA. So, this is nothing but divergence of tau i dV integral is over what? The volume that bounds the control surface the volume rather the control surface is the surface that bounds the control volume. So, this is the control volume. So, we have got the surface force the body force. So, the surface force we have written in terms of what? The surface force we have written in terms of a volume integral the body force anyway is normally written in terms of the volume integral. So, the body force if Bi is the body force per unit volume along i. So, what is Bi? Small Bi is the body force per unit volume along i. So, if you just multiply it with dV and integrate it over the control volume then this is the surface force. So, by these 2 we have represented the surface force and the body force. So, we have represented the force in the Reynolds transport theorem for linear momentum conservation. Let us recollect. So, maybe let us go through the slides. So, that we can go a little bit quickly because we have already represented the forces the remaining part is trivial. So, if you look here. So, the resultant force acting on the control volume we have just shown in the board how to do the Reynolds how to find out the resultant force acting on the control volume. So, the resultant force now the right hand side the right hand side is like this is the change within the control volume the first term and this is the outflow-inflow term. So, here again it was the surface integral or area integral we have converted it into volume integral. So, that is how this the last term in the right hand side is divergence of rho ui into v what is ui? ui is the ith component of the velocity v v is the velocity vector it is ith component is ui. So, you have divergence of rho ui v dv. So, how has this term come let us quickly look into the board and figure out that how this how this particular term has come let us switch over to the board. So, you have this particular term is equal to what rho ui is small n that is capital N is m into ui per unit mass is ui this is the last term outflow-inflow term. This v is v relative but we have considered the control volume to be stationary. So, that the relative velocity is same as the absolute velocity. So, here the vector field f is nothing but rho ui into v. So, this for the control surface is as good as integral of divergence of rho ui v dv. So, that is very straight forward this is using the divergence theorem. Let us go get back to the slides. So, as you see here that so the right hand side we have written in terms of the volume integral left hand side is also written in terms of the volume integral because both the forces are expressed in terms of the volume integral. So, the forces one of the forces so one of the forces this is what is written in the right hand side of this equation is actually the left hand side just we have switched the right hand side and the left hand side. So, this is the surface force term del dot tau i that is what we worked out in the board just now. This is the surface force term and this is bi is the body force term. So, everything this term plus this term minus this terms into dv integral of that is 0 because all the terms we have expressed in terms of the volume integral. So, in the volume integral individual terms will be appearing like this is the unsteady term within the changing within the control volume. This is the outflow minus inflow term this is the surface force term and this is the body force term. So, all these terms together multiplied by dv integrated over the control volume is 0. Now, since the choice of the control volume is arbitrary we can write the corresponding differential form like this. Now, we have expressed the divergence in terms of the index notation to write the equation in terms of the index i and j. So, again I will go to the board to explain that let us go to the board to explain that just for those who are like little bit out of touch with this nabla operator. So, this is i del del x plus j del del y plus k del del z. So, in terms of the indices you can write this as i del del x 1 plus j del del x 2 plus k del del x 3 the vector f any vector f is f x i plus f y j plus f z k. So, this is as good as in the index notation f 1 i plus f 2 j plus f 3 k. So, the dot product of these 2. So, what is this del f 1 del x 1 plus del f 2 del x 2 plus del f 3 del x 3. So, here you can see that there is a summation. So, you can write this as del f j del x j I mean this could be instead of j k l m n whatever does not matter because it is the repeated index. So, there is a summation over that. So, when you write del dot tau i. So, you can write this as del tau i j del x j just tau i is like f. Similarly, when you write this expression del dot rho u i v this v has components what this v has components u with a subscript we cannot use the subscript i again because that has already been used to represent the component along i. So, we can use the some other subscript j. So, this is as good as. So, u j is the component of this velocity v. So, this is as good as del del x j of rho u i u j. So, with this sort of notation let us get back to the slides. So, we can see that these two terms one is del del x j of rho u i u j and this term del tau i j del x j. So, we have represented all the terms. Now, I mean can you tell I mean what are the restrictions of this equation like I mean this I always tell in all classes that I teach that is it is important to know the equations which you need which you require to solve problems and we will solve problems using these equations. But it is more important to know what are the restrictions under which or assumptions under which those equations are used. So, if you recall that when we derived starting from the Reynolds transport theorem we made two important assumptions that the control volume is stationary and the control volume is non deformable. But you can see later on that if the control volume was not stationary and if the control volume is non deformable that will what effectively it will do it will just give rise to some extra forcing terms. For example, if the control volume is not stationary if the control volume itself is accelerating or rotating we will talk about the rotating control volume in the context of microfluidics the lab on a CD that you have the device a rotating device in which you have microfluidic channels. So, a rotating control volume is a very practical example that we will consider in microfluidics and then we will revisit these equations. You will see that for those cases what you essentially require is basically you have to account for some pseudo forces which you experience in the accelerating reference frame those are so called inertia forces. So, that will give rise to extra forces in the forcing term. So, if you have a deformable control volume and if you have a accelerating control volume all that will boil down to some modification of the force term you will have additional force terms. But if you take all those forces into consideration then this equation is very generic it does not have any restriction provided that whatever is the additional force to take care of the special nature of the control volume that you take care of by modifying this bi term. If you do that then this is pretty general and this should be true for all fluids. So, this is a very important equation because here we get an equation with which we can start as an equation of motion for all fluids does not matter whether it is a special type of fluid that will satisfy certain law or it is some other special type of fluid that will satisfy a different law. Now before proceeding further we will look into the equation try to see that if we are solving this equation then what are the number of equations and the number of unknowns. So, if you look at this equation so let us note down the number of equations and the number of unknowns may be let us do that in the board. So what are the equations that we have derived so far first the continuity equation. So, let us write that this is the continuity equation the momentum equation also known as the Navier or the Cauchy equation this equation Navier equation or Cauchy momentum equation I mean again like these equations are given such names to honour the famous scientist or mathematicians who have contributed very significantly towards the development of the modern fluid mechanics. Now let us look into the number of equations and number of unknowns. So, let us list in general rho may be a variable so you can consider rho as a variable. Now you can have a special case when rho is not a variable so but that special case we will consider later on but we consider the general case when rho can be a variable then u1, u2, u3 the velocity components and the tau ij's how many tau ij's are unknown see ideally i is from 1 to 3 and j is from 1 to 3 there are 9 components but because tau ij is equal to tau ji you have basically 6 unknowns. So, this is a symmetric tensor with tau ij is equal to tau ji so there are 6 such so 6 this is 1 these are 3 so in totality you have 10 so you have 10 unknowns how many independent equations you have this is one equation and then 3 components this is just the ith component so i equal to 1 i equal to 2 and i equal to 3 so 3 components so 3 equations here and 1 equation here so 4 equations or 4 independent equations with 10 unknowns so it is an enclosed system of equations and unknowns you have to match the equations with the unknowns that means you require a closer model that means you require a sort of modeling a mathematical paradigm that will invoke more equations that relate some of these may be tau ij with some of these. Now question is with what tau ij is related that is where now you have to appeal to the special case of a fluid when you consider a fluid tau ij is dependent on what? See a great distinction between a fluid and a solid though I must give you a caution that there are very weak distinctions between fluid and a solid for some substances which are partly fluidic and partly solidic type of behaviour so what I am giving you is not a strict definition of what should be the characteristic of a fluid or what should be the characteristic of a solid but in general think of some limiting cases if you think of a elastic solid in elastic solid the stress is related to strain right and we know that the stress within proportional limit the stress is linearly proportional to the strain right it is not within elastic limits because elastic limits may be beyond the limit of linear proportionality there can be non-linear elastic behaviour so it is better to say within proportional limits you have stress is proportional to strain right so that is the Hooke's law. In general stress is related to strain it may be linearly proportional it may be non-linear whatever for fluids if you want to relate stress with strain the problem is that fluids are what fluids are continuously deforming even the under the action of a very small shear stress so fluids can be non-deforming or under rest only when it is subjected to a pure normal stress and that pure normal stress that in that situation that pure normal stress can be described by a hydrostatic state of stress expressed in terms of the pressure of the fluid but if the fluid is continuously deforming then what happens if the fluid is continuously deforming and if you relate the stress with the strain then the strain will be tending to infinity as you allow infinite time so a time factor will come into the picture but in reality because the fluid is continuously deforming it is not the strain itself that is important but the rate of strain that is important that is how fast or how slow the fluid is getting strained or the fluid is getting deformed so for fluids normally you relate stress with the rate of strain or rate of deformation so when we write tau ij this tau ij we are relating to what this tau ij we are relating to one situation where the stress in the limit boils down to the hydrostatic condition that even if the fluid is at rest some stress is shown if the fluid is at rest what stress appears that is the normal stress because of pressure so we represent that I will explain how why it is represented like that first this delta ij this delta ij is called as Kronecker delta I have not yet written the full tau ij expression so please keep some space for writing that this is Kronecker delta so delta ij is equal to 1 if j is equal to i and is equal to 0 otherwise maybe it is not capturing the frame so I will write a little bit on the other side so delta ij is equal to 1 if j is equal to i and is equal to 0 otherwise so you can see that tau ij when is it normal stress and when it is is it shear stress it is normal stress only when j is equal to i and then only it will be minus p and if j is not equal to i that is shear stress pressure cannot give rise to a shear component of stress pressure can give rise to only normal component of stress and why minus minus is because the positive normal component of stress is by sin convention considered as tensile whereas pressure by definition pressure means it is acting compressively so that is what is pressure so minus p delta ij this is the hydrostatic component that means if the fluid is at rest this is what is the component of stress that is acting one very important point is there is a distinction between something called as mechanical pressure and something called as thermodynamic pressure so here we are trying to express this through thermodynamic pressure we will see when thermodynamic pressure is equal to the mechanical pressure the thermodynamic pressure is the pressure that will satisfy the thermodynamic equation of state for example for an ideal gas it is pb equal to nrt that p we are talking about as thermodynamic pressure whereas mechanical pressure is simply a manifestation of the average of the normal component of stress negative average of the negative of the average of the normal components of stress in the 3 directions so tau minus of tau 1 1 plus tau 2 2 plus tau 3 3 by 3 so this is the hydrostatic but very importantly even if the fluid is under dynamic condition this is acting but there is something beyond this that is acting when the fluid is under dynamic condition so that is called as tau ij deviatoric so the deviatoric component of stress is acting when the fluid is under deformation when the fluid is under motion under motion fluid under motion means fluid under continuous deformation now what will be the relationship between tau ij deviatoric and the rate of deformation or the rate of strain see this is the part of the stress that is responsible for deformation therefore for fluid this must be related to the rate of deformation now the rate of deformation if you if you recall the kinematics of fluid flow you can recall that the rates the deformation the linear deformation angular deformation volumetric deformation all those are expressed in terms of some gradient of velocity right the linear strain the angular strain the volumetric strain all these are expressed as collection of terms with some gradient of velocity so the velocity gradient is an implicit indicator of the deformation but we have to isolate a deformation from rigid body motion because rigid body motion is also represented by these types of terms right so to do that we can write this as half of see this is nothing but very trivial we can write as half of a plus b plus a minus b right so here this is what is a and this is what is b you can look into the special features of this this term remains the same if you swap i and j right the first term which I have shown by this curly bracket this term remains the same if you swap i and j therefore this is the symmetric tensor right on the other hand the next term it will become minus of this if you swap i and j so that is a skew symmetric tensor or an anti symmetric tensor so this is something which is very obvious because every tensor can be written as a sum of a symmetric and a skew symmetric tensor just like in matrix algebra you have learnt that every matrix can be written as a sum of a symmetric and a skew symmetric matrix and you have to remember that matrices are nothing but some systematic ways of representing tensors of say a 3x3 matrix is a systematic way of representing a second order tensor so matrix algebra will follow very close to the tensor analysis so with this understanding now let us try to look into the physics of these terms so what this term relates to this term with a half at including a half I mean you forget about the half this del ui del xj plus del uj del xi is an indicator of the deformation if i is not equal to j it talks about angular deformation if i is equal to j it talks about linear deformation so this is deformation. So remember that the deformation can be both linear and angular many times we have the perception that deformation means it has to be angular deformation not really deformation could be related to both angular deformation as well as linear deformation and this is what and this is related to what this is related to rotation of the fluid so the rotation of the fluid you relate to what the rotation of the fluid you relate to what you do not relate the rotation of the fluid with the stress because rotation is a rigid body type of behavior. A rigid body will not be subjected to stress, it is a deformable body that is subjected to stress. So rotation is not something about which we will bother for the time being. So tau ij will be related to deformation. So when we say that it will be related to deformation, the deformation, how many indices does it require? So if you want to represent this term, let us call it eij. So eij is half of del ui del xj plus del uj del xi. Maybe I will write again in a separate place because the computer monitor is blocking it. So eij is equal to half of del ui del xj plus del uj del xi. So now what we can see that tau ij is a second order tensor. It is related to another second order tensor which is the deformation tensor and a second order tensor should now be expressed as a function of another second order tensor basically. So one second order tensor is the stress. It should be expressed as a function of the rate of deformation tensor which is a second order tensor. Now question is how they are related? There can be any arbitrary relationship between the stress and the deformation. But the stress and the deformation may be also mapped linearly. So the linear relationship between the stress and the deformation exists for some special types of fluids which are called as Newtonian fluids. So for Newtonian fluids the stress is linearly related to the rate of deformation which we simply call as deformation for fluid because for fluid by deformation we always mean the rate. So stress is related to the rate of deformation for fluids for special types of fluids called as Newtonian fluids the stress is linearly related to the rate of deformation. So there must be a linear transformation that maps the stress with the rate of deformation. So what maps the stress with the rate of deformation in terms of tensor theory it is a transformation it is a mapping that maps a second order tensor on to a second order tensor. We have earlier seen through Cauchy's theorem that a second order tensor is such which maps a vector on to a vector like the if you recall that the Cauchy's theorem when you couple it with the angular momentum conservation principle it is originally tau ji but if you consider tau ij equal to tau ji it becomes this. So you will get tau 11, tau 12, tau 13, eta 1, eta 2, eta 3, tau 21, tau 22, tau 23, this is T1, T2, T3 okay. So this is just the matrix representation of the Cauchy's theorem nothing very special. See the power of writing in the index notation is that such a big thing we could condense in the small expression that is the power of writing it in the index notation. So you can see that this is nothing but the stress tensor. What it has done? It has effectively mapped this vector on to another vector, mapped the normal vector to the traction vector. It is a linear transformation that has mapped the normal vector to the traction vector. So it is a second order tensor that has mapped a vector on to a vector. Vector is a first order tensor. Similarly you now require a second order tensor instead of this now you have tau ij related to the deformation on this side there must be a mapping that maps this second order tensor. This is again a second order tensor. This is also a second order tensor. There must be something that maps the second order tensor on to a second order tensor that is nothing but a fourth order tensor. Just like a second order tensor maps a vector on to a vector, a fourth order tensor maps a second order tensor on to a second order tensor. So we can write how many indices we will require for a fourth order tensor, we will require 4 indices. For a second order tensor, we require 2 indices i and j. So for a fourth order tensor, we require 4 indices. So tau ij, we can write a fourth order tensor Cijkl into Ekl, where we have defined what is E. So this is the fourth order tensor. This is the generalized Newton's law of viscosity. But the problem is that this is too cumbersome to be used for solving a practical problem in fluid mechanics. The reason being that how many such constants you will require, the i will be from 1 to 3, j will be from 1 to 3, k will be from 1 to 3, l will be from 1 to 3. So 3 into 3 into 3 into 3 total 81 constants. Of course, you can immediately reduce those 81 constants to 36 constants by using the symmetry that is tau ij is equal to tau ji. And even those are very large number of constants. So you remember that in practical examples like when we deal with water or air, we do not require 36 or 81 constants to specify the property of water or air. That means things can be boiled down to something which is somewhat more simplistic than this by appealing to some special considerations. What are those special considerations? Special considerations are isotropic fluid and homogeneous fluid. We have already considered this as a Newtonian fluid. So Newtonian fluid is a fluid that will obey the Newtonian's law of viscosity. So if we have isotropic homogeneous and Newtonian fluid, we will be in a position to show that this 81 number of constants will boil down to 2 effective constants, 2 material constants, 2 properties of the fluid, 2 independent properties of the fluid. So briefly what is isotropy? Isotropy means that it is direction independent. That means if you have a fixed set of coordinate axis to represent a particular property, then if you have a rotation of the coordinate axis, so that the orientation is changed, you will get, you will still have the same representation. The representation does not depend on the rotation of the coordinate axis. Homogeneous means it is spatially uniform. That is the property does not vary as you go from spatially one position to another. So you can use one particular material property for the entire volume of material or maybe one sub-volume of the entire volume of material. You do not have to consider point to point variations that is inhomogeneity. So isotropy and homogeneity are two very common characteristics which again are not satisfied for all engineering fluids but some common engineering fluids which we encounter not just in the classical fluid mechanics but also in microfluidics based applications. So we will be seeing in the next class that how we relate this isotropy and homogeneity with this but remember that this is not the complete tau ij. This is tau ij deviatoric. So the total tau ij is the hydrostatic tau ij plus the deviatoric tau ij. And in the next class we will see that how we can relate the deviatoric tau ij with e ij with e kl basically and that we will do for homogeneous and isotropic fluid. So that we will be able to express tau ij in terms of the pressure and the velocity gradients that will give rise to a constitutive law which we can use for homogeneous isotropic and Newtonian fluids. And that we will substitute in the Navier's equation and we will invoke another hypothesis which we call as Stokes hypothesis. So in the next class our agenda will be to discuss about how the generalized form of the Newton's law of viscosity boils down to a special form for homogeneous isotropic fluids and then how that can be incorporated in the Navier's equation in addition to the considerations of Stokes hypothesis and that will eventually lead us to the derivation of the celebrated Navier's Stokes equation. So that we will take up in the next lecture. Thank you.