[MUSIC] So as I just stated, we need to classify all these cases of mechanical coupling between fluids and solids. And we need a tool to do this. The tool we shall use is called Dimensional Analysis. It is widely used in fluid mechanics, not so much in solid mechanics. What is it exactly? Let us consider as an example flow around a sphere, pure fluid mechanics. The configuration may be defined, for instance, by giving the upstream velocity U. The density of the fluid rho, and the viscosity mu. The diameter of this sphere is going to be L. All of these are dimensional quantities in the sense that their magnitude must refer to a scale of units. For instance, the velocity must be given in meters per second or kilometers per hours, or miles per days, as you want. But you will always need length and time units to express it. Out of these four dimensional quantities, we may define a ratio called the Reynolds number. Reynolds equals rho UL over mu, which is dimensionless. It is dimensionless in the sense that we do not need any scale of unit to express it. It is just a number. As you know from your fluid mechanics courses, this particular dimensionless number is efficient in in classifying the topology of the flow around a sphere. At low values of the Reynolds number, the fluid's creeping around the sphere, while at high Reynolds number, it is detached as sketched here. So dimensional analysis is a tool that can be used to classify. Certainly, we should go together a bit further in the use of dimensional and dimensionless quantities and the fundamentals of what is called dimensional analysis. [MUSIC] Dimensional analysis refers to physical laws. What do I mean by physical law? I just mean a relation between quantities that I use to represent the system I am considering. This, in general, would be dimensional quantities such as length, frequencies, forces, and so on. Such a physical load would read f (x1, x2,...,xN) = 0 in a general form. For instance, the physical law would give the frequency of oscillation of a pendulum as a function of its size and of gravity. Here, let us just take as a principle that a physical law should only relate dimensionless quantities. Why only dimensionless quantities? Because, of course, if I change the units in which I express my dimensional parameters in the form above, that magnitude is going to change, and the law will not be satisfied. If I switch from meters to inches, for instance, the law is not valid anymore, and this is not acceptable. So I write this law F(X1, X2,...,XP) = 0, which are all dimensionless. As an illustration, let us consider again the flow on a sphere. Imagine that we want to find the physical law that relates the force exerted by the fluid, namely the drag, to the parameters that I gave before. All these parameters are dimensional. The law reads in a general form, f (D, U, rho, mu, L) = 0, where D is the drag. As I've stated before, we should look for laws relating dimensionless parameters, not these dimensional ones. By combining them, I can find two dimensionless ones which are right here, and the relation takes this form. What are these parameters? The second one is the well-known Reynolds number that I mentioned before. The first one is the drag divided by rho U squared L squared. This is usually referred to as the drag coefficient. So the law of drag is actually the low between the drag coefficient and the Reynolds number. Do we know anything on this law? Yes, certainly. Here is what this law looks like from experimental data. So what have we gained here by using large F instead of small f, dimensionless in place of dimensional? Essentially, we have considerably reduced the dimension of the parameter space to explore. It had a dimension 5, and now only 2. This much easier. So this seems efficient. But we need some general guidelines on how to go from dimensional to dimensionless. [MUSIC] First, we need to define what we mean by dimensional variable. Let x be a physical variable. We are going to say that its dimension exponents, in terms of length, mass, and time, are respectively alpha, beta, and gamma. This means that when we give a value for x, we use a combination of length, mass, and time units at these exponents. We note this x brackets is L to the alpha, M to the beta and T to the gamma. For instance, if we consider gravity, g, it's value is about 9.81 meter per second squared, so that its dimension exponents are 1, 0 and -2 in length, mass, and time respectively. Why? Because meter per second squared is a length divided by the square of the time. The question now is to go from the dimensional quantities that we consider for a given problem, the small x1 to small xN, to the dimensionless quantities, large X1 to large XP. These X1 to XP have dimensions 0, 0, and 0 as they are dimensionless. At this stage, we don't know how many need to be used, and we don't know what they are. In the previous case, I just exhibited the drag coefficient and the Reynolds number and it happens that these two are found to be related in experiments. So when you want to go from considering dimensional quantities to dimensionless ones, what can you do? There is a rather general theorem called the Pi Theorem or the Vaschy-Buckingham Theorem, which tells you how many dimensionless quantities you need to look for. This theorem states that the number of dimensionless quantities, P, is equal to that of the dimensional ones, N, minus R. What is R? It is the rank of the matrix of dimension exponents. This matrix is formed by the columns of the dimension exponents of all variables, as you can see here. Remember that the rank of a matrix is a number of independent lines or columns that you can find. [MUSIC] Let us give an example. In the case of drag on a sphere, here's what we have. The number of dimensional quantities, N, is five, D, U, rho, mu, and L. The matrix reads as follows. The drag is a force which is the length to the 1, mass to the 1, and time to the -2. The velocity now is a length to the 1, time to the -1, and so on for the density, the viscosity and the diameter. The matrix has a rank of three, as we can find three independent vectors in its columns. So R equals 3. Using the Pi Theorem, we know that P equals n minus R, so 5-3, which is 2. We have therefore to find two dimensionless quantities only, and the two I have defined before, the drag coefficient and the Reynolds number can be used. Note that Pi Theorem does not tell me what the dimensionless variables are, just how many of them I should use. In practice, I can choose any that I want, provided they are independent. What is the good choice of dimensionless parameters? There is no good choice, there are just efficient choices that allow me to express the physical law in a convenient manner. We shall have plenty of example of this. There are still some very useful guidelines to build appropriate dimensionless numbers The main one is the physical meaning of these quantities. As they are dimensionless, we can always see each of them as the ratio of two dimensional quantities, for instance, two lengths, two times, two forces, and so on, for instance, the drag coefficient of the sphere that we have been talking about. It may be understood as the drag force divided by or scaled by the sum of dynamic pressure rho U square over the cross flow area L square. When drag is caused by pressure difference between the upstream and downstream, as it is for higher Reynolds number flow, then we have CD of order one. That makes sense. [MUSIC] More generally, as our physical laws relate dimensionless parameters, they relate ratios of times, of forces, of lengths. This means that if all lengths or all times in the system are rescaled, the same laws apply. Actually, this was introduced by a French philosopher named Henri Bergson in the early 20th century. He stated that if all motions were to happen twice as fast, nothing would have to be changed in our formulas. He was a precursor of dimensional analysis. Note that he was a Nobel Prize, but in literature, not in physics. So to summarize, we were looking for a tool that could be used to classify the huge variety of cases of interactions. We knew that dimensional analysis has been used to do this in fluid mechanics. What we know now is that dimensional analysis allows giving you guidelines in the construction of physical laws between parameters. It gives you the number of dimensionless parameters to expect from the set of dimensional parameters that you choose to take into account. And these dimensionless parameters have a very strong physical meaning, as they are related to ratios between quantities of interest. So dimensional analysis seems particularly adapted to the classification of complex phenomena. So let us apply this powerful tool to fluid solid interactions and see what happens. [MUSIC]