 Welcome to dealing with materials data, this is a course on collection analysis and interpretation of data from material science and engineering. So we are looking at the module of probability distributions and in this session we are going to talk about the nucleation process and how it is related to the Poisson distribution. Nucleation is a process which is very very important in material science and engineering. Here is an example of an under cooled melt and out of which nuclei is forming, there are actually several nuclei that have formed and they are going to grow and that is how this under cooled melt is going to transform into a complete solid. What we are interested is in the process of the formation of this nuclei and we want to have the right statistics to describe this process. So in order to describe the process again we have to understand exactly what is happening physically and once we know that we can translate this problem in mathematical terms to the correct statistics and then solve the problem. So let us find out what this is all about. So when does melt turn into a solid, we all have seen this that you keep water inside the freezer it turns into ice, you keep ice outside in the room it turns into water. So there is a driving force for water to be in liquid form or in solid form and that depends on temperature. So as a function of temperature if you actually plot what is known as Gibbs free energy then you can define what is this transformation temperature in this case it is a melting temperature that is decided by the free energy. So above this temperature we find that the liquid free energy is lower as compared to the solid free energy. So above this temperature the system would prefer to be a liquid, below this temperature we see that the solid has lower free energy than the liquid so the system would prefer to be a solid. So the melting temperature is that temperature at which these solid and liquid have the same free energy so it is basically the intersection point of the free energies as a function of temperature and at the point of intersection is where we define the transition temperature. In this case because it is solid liquid transition I am calling it as a melting temperature or freezing temperature. You can also think of other transformations and this definition is more general than for this specific scenario. So in every case you will find that the driving force for transformation is given by these free energy changes the system is trying to minimize the free energy. So it would like to be on this line and then on this line it does not want to continue here because there is a another state that is available with lower free energy so the system would prefer to shift to that. So if you are heating the system the solid will turn into a melt and if you are cooling the liquid will turn into a solid. So we are looking at this process of cooling the liquid and it is turning into solid. But you can notice that there is no driving force at all for transformation at the melting temperature because the two free energies are equal. In other words the thermodynamic definition of melting point says that if you keep a solid and its melt at this temperature it will have a tendency neither to melt nor to solidify. So there will be no change in the solid melt interface at all it will neither move towards the solid nor move towards the melt and it will remain where it is. And in fact if you go slightly lower also it is not going to be favorable for the liquid to turn into solid unless it reaches a critical value in terms of temperature or critical under cooling as we call it. So we call it under cooling because you have to take the liquid below the melting temperature to some extent before the liquid turns into solid. You would expect this to be a spontaneous process but it does not happen at this temperature you have to be below. How much below your below the melting temperature is what is known as the under cooling. And a critical under cooling is needed for the liquid to turn into solid or some solid region like this to form in the melt. Why is that? So that is what is explained in this figure. So you have the liquid free energy, you have the solid free energy this is the melting point and you need to be in temperature below the melting temperature to some value and this difference is what is known as under cooling. And this under cooling has to reach a critical value that is because at this temperature for example it is favorable let us consider this point at this temperature for example it is favorable for the liquid to turn into solid because solid has lower free energy liquid has higher free energy. But if it forms this nucleus from the melt there is the nucleus so that will have the solid free energy and so much of melt has turned into solid so there will be so much of gain in terms of free energy. But there is also this interface that it has to form between the melt and the solid. This interface costs energy for the system so unless the gain in energy can compensate for this the system is not going to prefer to form this solid from the melt. So the critical under cooling is that temperature where you see that this gain in free energy is more than sufficient to pay for the interface cost in that case because it will gain so much energy and it will pay for this and it will still have gain in energy it is going to grow so that is when this actually is called as nucleus. So there is a critical under cooling that has to be achieved so that you will achieve critical change in free energy which will compensate for the cost you have to pay for making this interface so till that is reached you will not have it. And in all of this we are assuming that the solid is forming in the melt homogeneously that is it is not for example forming at the mould walls or it is not forming on any impurities or any other particles that are then in the melt. If suppose there is a particle in the melt and the nucleation happens on the particle there the particle melt interface will already have an interfacial energy. If suppose that is more than the solid melt interfacial energy that will have its own driving force so the critical free energy change might not be valid in such cases so you have to also incorporate the interfacial energy associated with the melt particle interface or melt impurity interface or the mould wall melt interfacial energy and so on. So that is why we are not considering this case we are considering nucleation from the melt and such nucleation is known as homogeneous nucleation. So what is the critical nucleus size and so that can again be derived. If suppose I assume that the nucleus is spherical particle then you can see that the surface area will go as square of the radius of the particle and interfacial energy will be proportional to the surface area so interfacial energy will go as r squared. On the other hand volume of radius r of sphere will go as r cube and the free energy decrease that it gets because of turning so much of melt into a solid will go as r cube. For very small r r squared will still be dominant and so r cube will not be favorable but as the particle size becomes larger and larger you will see that this is the point at which the contribution that comes from the volume free energy will more than compensate for what it has to pay in terms of the interfacial free energy. So beyond this point so this is the change over point at which you will have the critical nucleus size. So if a particle of this size exists it is at a size where further increase in size is going to decrease the overall system free energy. If it is anywhere here any decrease in the solid size is what is going to decrease the overall system free energy. So in these stages it is not preferred to have these solid particles anything here is neither stable nor unstable and beyond this these particles become stable. So this is the critical size of the nucleus and this is the critical dg star as we call it. So this is obtained by writing analytical expressions assuming that this are spheres typically or you can assume any shape as long as you describe in terms of the shape how the interfacial free energy and volume free energy change and you have to minimize this quantity. So take the first derivative equate it to 0 and the solution you get is r star. You can show that r star is 2 times gamma by delta g. So that delta gv is the overall volume free energy gain that you get by turning the melt into solid and gamma is the interfacial energy. So it also comes from what is known as Gibson equation which says that r is if suppose r is radius then 2 gamma by r is the excess free energy associated with that interface. So you have to compensate for it so delta gv happens to be 2 gamma by r. So in any case so you can see that there is a critical size and beyond that size the nucleus will grow and this process of turning the melt into a solid by formation of such a stable particles is what is known as nucleation specifically we are looking at homogeneous nucleation. Now the rate of homogeneous nucleation J in meter cube per second units is the average number of stable nuclei in a given volume in a given amount of time. Now the formation of nucleus nuclei is a stochastic process. What gives rise to such particles in the first place is the thermal fluctuations. Because of thermal fluctuations different sized solid like particles are always forming in the melt. They are forming in fact because they are based on thermal fluctuations they are forming at all temperatures. But if suppose your temperature happens to be lower than the melting temperature then when the system accesses those states it realizes that it can decrease its free energy. But if suppose it accesses these sizes then at these sizes it is going to find that it is not favorable so they are going to shrink back. But if suppose it happens to access this size then it knows that it had already decreased the system energy. So it will prefer those states and so it will continue to do that. So this is what is happening. And so these thermal fluctuations that give rise to such regions of solids if the size of these particles happen to be smaller than the stable size then they are called embryos. They keep forming and they keep dissolving back into the belt. And they are called nuclei if they are above the critical size and so they can grow. So homogeneous nucleation is what we have assumed we have not assumed that the nucleation is on mold wall or impurity or like that. So in the case of homogeneous nucleation if that is the rate of nucleation how do we get this rate? So the idea so in principle you have to consider the formation of different sized particles and their dissolution and or building up one by one you know the particle and it is growing in size. And at every stage it has definite probability to shrink and some finite probability to grow and you can track this and find out when it actually goes over the critical size that is one approach. But there is another idea which says that because of thermal fluctuations what is the probability of finding a particle of exact critical size r star and now to this solid if you add one atom or molecule it will become a nuclei. So if you know the probability or frequency of incorporation of one atom or molecule into critical nucleus and that is of some f 0 then the thermal fluctuations which give rise to these solid particles in the melt that is described by the Boltzmann distribution. Like I said earlier we are also going to talk about Boltzmann distribution and the other statistical distributions like Fermi Dirac and Bose-Einstein at some point. But this distribution is described by exponential minus del G by KT where K is the Boltzmann constant T is the absolute temperature, del G is the free energy of the nucleus of critical size. In our schematic this is del G and this is r star. So if you have exponential minus del G by KT gives the probability or the occurrence of nuclei of solids of this particular size and you multiply by the frequency so that will give you the J. Now suppose J is unchanging that there is constant under cooling so this is again another approximation this is not true always because if you form a solid and if there is some latent heat change then the temperatures can change and so on and so forth. So there are lots of things that can happen so the under cooling might not remain constant. But if you assume that under cooling remains constant so the homogeneous nucleation rate remains a constant then we are asking the question what is the probability of M nucleation events in a volume V of liquid in time T that is the question we are asking. So the number of nuclei N will be T times V times J because J is a per meter cube per second so if you multiply by the volume and time you will get the number of nuclei that will nucleate in a volume V at in time T and we are trying to get this probability so this will be given by probability distribution which is known as Poisson distribution. So Poisson distribution is also distribution that you will find for example for accidents and errors and things like that so we will also look at one such example later. So for nucleation for homogeneous nucleation so the number of nuclei that you would find and the probability of M nucleation events of having N nuclei is given by N power M by M factorial exponential minus N so this is the probability distribution function for the Poisson distribution. So this is the probability distribution for Poisson distribution function it so happens that the expectation value for the Poisson distribution and the variance both happen to be N and Poisson can be thought of as a limiting case for binomial for the probability tending towards 0 and for large M the Poisson distribution actually approaches normal distribution. So the Poisson distribution using r poise is the command and d poise p poise q poise and r poise as usual they give probability density cumulative distribution function quantile function and r poise actually gives you the random variance of Poisson distribution. Let us assume M to be 20 and try to do these commands using r so M is 20 and again we are going to make 3 plots and those are for the density cumulative distribution and the quantile function and the Poisson distribution requires just the parameter M. So here is the probability distribution, here is the cumulative distribution and here is the quantile function. Of course you can also get the random variates and the way to get the random variates is just to say r poise so I want 20 random variates and v is the parameter so you can give so you have 12, 16, 24 etc of the random variates which are picked from the Poisson distribution. So let us go back so this is how you deal with by now I am assuming that you are comfortable with these commands because this is a third or fourth time we are using similar commands and so let us also look at another problem. So we looked at the nucleation problem and we found that nucleation events can be described by the Poisson distribution for homogeneous nucleation assuming constant nucleation rate. There is also another problem set of problems where Poisson distribution is useful or it can be used in approximation for a binomial distribution and here is one example. Suppose a steel company produces 1000 sheets of a given dimension in a day and if the sheets contain some longitudinal cracks of more than a critical size they will be rejected and for a given chemistry if you make sheets out of this composition we know that the probability of sheets having flaws greater than a critical size is about 3 percent let us say. Now if you pick 100 sheets of samples randomly from these 1000 sheets that you make what is the probability that 3 sheets will have flaws greater than the critical size right. So this is the question and of course you can think of it as a binomial distribution problem and you can also calculate it as a approximate by a Poisson. So let us do that so what do we have we have 100 and the probability is 0.03 and so if you have 100 then at least 3 you expected to fail and you are looking for probability for when you test 100 3 of them will be failing and so the Poisson distribution for 3 failures and for the probability the mean 3 right in this case this also happens to be 3 is 0.224. Now you can also calculate this quantity as a binomial what is the binomial that is here right you can calculate the total is 100 and you are looking for 3 out of it with a probability of 0.03 so that is 0.227. So you can see that the difference between the 2 is at the third decimal place some 0.03 or something so you can actually calculate the difference okay so it is 0.003 so you can see and you can see here also. So the binomial distribution as p tends to 0 you can think of it as approaching the Poisson distribution so because of which for small p you can use Poisson distribution to calculate these probabilities. So this is one of the properties of the Poisson and we will also look at some other properties later after we do the normal distribution. So we have found that this is the last of the discrete distributions we have looked at Poisson distribution and this distribution has lots of use it is very important in studies like nucleation and it is also useful for finding out defects in the manufacturing process for example and it can also be used for approximating the binomial for very small values of p the probability that goes into the binomial distribution function. So to summarize we have looked at probability distributions specifically we have looked at discrete probability distributions we have looked at Bernoulli trials binomial negative binomial hyper geometric and Poisson distributions and we have looked at at least two very important cases where these distributions are of use one of them is a microscopy technique to get the composition of very small volumes so that is the atom probe technique where we found that negative binomial hyper geometric and things like that are useful and where we also did some error analysis along with the distribution so it is an interesting mix of problem and we have now also looked at Poisson distribution and how it is useful for understanding things like nucleation. So with this we are going to conclude the discrete distribution scenario there is one more uniform distribution which can be either discrete or continuous so we will come back to it at some later point and now we are going to move to continuous probability distributions of which the most important is of course normal so we are going to start with normal in the next session. Thank you.