 Welcome everybody to this fourth lecture in the series on thermodynamics. And today my focus is on the calculation of phase diagrams using computer methods. And at some stage in your metallurgical or materials career, you will come across the need to do a computer calculation of phase diagram, and you can do that in two ways you can just use it blindly. You can use a computer program, or you can use it, knowing how it works. So my goal today is to explain the essence of how these computer programs work, because as you will see towards the end of the lecture. There are some issues, which you need to think about when applying these methods. We've considered thermodynamic models and the simplest one here was the ideal solution where there is no topic of mixing, but simply a configurational entropy of mixing. And then we had the regular solution model where we still assume that the atoms are mixed at random. Even though there is a finite access free energy of mixing so this access free energy consists of an enthalpy term and entropy terms which are not from the configuration of entropy but for example thermal entropy and so forth. So this necessarily tells you that the atoms may not be randomly mixed at low temperatures but nevertheless we assume that there's an ideal mixture of atoms, a random mixture of atoms. Okay, so to summarize, you know the ideal solution assumes a random distribution of atoms. And we know that, you know, most solutions are not random, and therefore there's an access gives free energy term which defines our regular solution model. And that tend to be symmetrical about a concentration of a half. Now, of course, these models we have considered essentially for binary systems, although you can generalize the regular solution models to multi component systems, for example, all grand and stuff and certain have done that work. But we are interested actually in a more much more complex set of allies that are commercially useful. Okay, so the binary allies and so forth are very important when you're doing research. When you're doing research to discover make some discoveries, but the real allies have to satisfy a basket of properties and therefore you will use quite a few alloying additions inside your material, whether it's nickel base or copper base or whatever. And you know you have this whole palette of elements to work with some of which are incredibly expensive and very rare so you wouldn't use much of that, and others, which are not. The problem is really complicated if you want to create a model, which will be able to deal with such generality, and with many components and many possible phases when these elements combine. Now, the fundamental property that we need by measurement is the heat capacity. And in these generalized models of solutions, we express the heat capacity as an empirical polynomial. So this particular polynomial is found to work extremely well in representing the heat capacity data. And supposing that it doesn't work very well over the entire temperature range, then that's not a problem in a computer you can stitch this with another factor with another such equation which will work in a different domain of temperature. So you represent the heat capacity empirically, because it actually varies in a complicated way with temperature, and you know, you can't have that complication when you're dealing with generic calculations. So this is a polynomial that we use to represent heat capacity as a function of temperature, and these are just fitting constants, and you can use several of these polynomials to represent the entire temperature range. And therefore you then have the capacity to work out the enthalpy change if the temperature changes, and similarly the entropy change. And therefore, you know if I integrate this equation in this manner. I get a general equation for the free energy where again the bees are the empirical constants. And here we have the temperature terms which you obtain by integrating this in this manner and this manner. So this is how you would represent free energy inside computer programs like thermo calc, empty data, fax age, and so forth. So the models that we have considered so far, which are derived are just too simple to represent the complexity of real data and that is why we use these empirical polynomials. So the computer can access these thermodynamic data from these equations and from the coefficients. Now in a binary solution, we have this access free energy of mixing and the contribution from the configuration entropy. Now we've got to design of a computer method so that if some thermodynamic data somewhere need to be changed, it doesn't affect all the other data. So the access free energy term here is represented by another polynomial of this kind. But this is the concentration of a times the concentration of be multiplied by some coefficient here, which is, which also is multiplied by the difference in concentration. And this I means you can extend this polynomial to however many terms you like to represent the access free energy data accurately so for example here I've expanded it to I equals zero and I equals one. So in this term you might recognize the first term, which is basically the regular solution term xA xB times a regular solution constant so you can identify LAB zero with the regular solution constant, and this is an additional term because this might not adequately represent the variation of the access free energy with the concentration and you can add more terms to this polynomial. So in future work, you know if you discover a different value of LAB one, then you can just modify this term you don't need to modify the other terms in that respect. So it is designed so that you don't have to rework everything when you have some new and perhaps more accurate data. So this is a binary solution. Now, when it comes to turner is there are fewer thermodynamic data available. So one trick is to take, supposing we have a ternary of A, B and C, then we take the three binaries here, and simply combine them so this equation here, this part of the equation is no different from what I illustrated here. So that's for a binary and similarly for the different binaries so we have three binary phase diagrams, and we are going to combine these terms to represent the access free energy of a ternary system. And again, you know, if, if some terms are modified you don't need to modify all the others. Now obviously, this means that when we go away from the binary systems into into a ternary system, we are using some sort of an approximation. Yeah, because we are simply looking at the end members of the ternary system. But nevertheless, it's a good way to do things when you don't have any data for ternary interactions here we only have interactions between C and A, B and C and A and B. So this is the first, first approach when data lacking. And it indicates if your calculation for the ternary system doesn't turn out to be correct that indicates that you need to generate data, because the whole calculation relies on experimental data. Now, nowadays there are actually for, for the popular systems, for example steels and nickel base allies, there are ternary interaction data available. So here is how you'd represent those data you would have this additional term, which has all three concentrations here A, B and C. And these are now our ternary interaction. These are now the ternary interaction parameters here and show for yourself that if I set one of these. For example, see the concentration of C to zero, then this reverts back to the original formulation for the binary system. Okay. So these are ternary system interactions which can be incorporated. And similarly, these polynomials can be generated for any number of components, assuming that the thermodynamic data are available to derive these quantities. Actually, you know, the computer calculation of phase diagrams is one of the best examples of extremely systematic work, which was started by Kubaszewski and Evans in London, Kauffman in Boston, and Hillard and his colleagues in Sweden. There might be others, but I'm just quoting the ones that I know. And they got together to collect and assess data. Okay. And then you can have different computer programs to access those data and do calculations of thermodynamic properties and phase diagrams. The process has been so systematic over more than half a century that the databases are huge now and commercially available, routinely commercially available, even in industry. Furthermore, there are systematic journals in which such data are published. For example, the CalFAT journal computer. I forget what CalFAT stays for, but it's about computer calculation of phase diagrams. And there's another one called phase equilibria and so on where such data will be published and assessed by the whole community. When you do your experimental work, you have to ask yourself a question. Can the experimental data that I've generated, which are much more than in your publication, can those be accessed by future researchers? Very important question. If I wanted the experimental data of Newton, I can get them. But for a modern scientist, I may not find anything more than a publication in the future. So archiving of experimental data helps enormously in progressing science. Now I'm going to show you some of the tricks you can do with these calculations. And I'll illustrate this by a ternary phase diagram here, which is between iron, chromium and nickel. And these are the two phase fields where you see the straight tie lines here. And these triangles are tie triangles. So at the corners of these tie triangles, the pair of phases at the end of each side of the triangle are in equilibrium. And this is a single phase field. Now this is an iron-chrome nickel system. But supposing I have iron-chrome carbon. And carbon can diffuse much faster than iron and chromium. So it is possible to get a transformation in which the carbon diffuses, but the iron and chromium do not. In which case, you will get what's known as constrained equilibrium, that the carbon can reach a uniform chemical potential, but the iron and chromium may not. In these calculations, you can suppress some of these solutes from actually moving about so that the ratio of chromium to iron remains constant everywhere, even though carbon is moving. When you run such a calculation, you have to select the phases that you think might actually be occurring in your system. So in this particular calculation, which is for a low carbon, 0.14 carbon steel, this smattering of silicon and manganese and nickel, molybdenum and chromium. My metallurgical experience tells me that these phases should be included in the calculation. Okay. If you are in any doubt, you can include all the phases in the database in your calculation. But I know that I know from my experimental work that these are the phases that will form. So I've limited the calculation to these. And then you specify the chemical composition in rate percent or mole fraction, etc. So BCCA2 is just right and FCA1 is austenite. This is M23 C6 carbide. M stands for metal atoms because we have several metal atoms. And the carbide is not a pure carbide of chromium and carbon, but you can have a mixture of atoms on the metal sub lattice. Okay, this is M6C carbide. When you run the calculation, you have to select a temperature and a pressure, because the free energy is a function of both. And in this case, I'm working at atmospheric pressure. Okay, so it's about 100,000 Pascals, which is atmospheric pressure. This is M6173 Kelvin, which is basically I'm doing this calculation for tempering of martensite in that steel, and I want to temperate it by 400 degrees centigrade. This now is the output, okay, which you don't get from just looking at the phase diagram. So we have carbon concentration, its chemical potential and its activity in the system. And when the calculation is complete, this is the information that is absolutely relevant from a phase diagram point of view. We can't plot this like a normal phase diagram because this had many, many components, but the only information you are looking from in a phase diagram is this, that what is the proportion of the three different phases. This gives you the proportion of the three different phases out of the total that you've added into the calculation. And what it's saying is that at this temperature, I only have three, these three phases stable. In these phases, the exact composition is given in these data. So for example, cement diet has that much iron, a quarter mole fraction of carbon, which is what we expect because it's F3C so this should be 0.75. And we have other elements yet, for example, manganese, nickel, and chromium inside the cement diet. If you add all those up, it'll come to about 0.75. So, just well, just well, can you mute yourself please. Okay, I've done it. So, so we have the chemical compositions of the individual phases, and we have the quantities of the individual phases as well. Now I'm using here moles but you can you can use whatever you like, you know, wait percent, etc. Now, as I raise the temperature. Okay, to do something like I think this would come to over 800 degrees centigrade. So, austenite becomes the stable phase, ferrite disappears and M23C6 disappears, but I have a two phase field, consisting of austenite and cementite, you don't see that in a normal iron carbon phase diagram. And when the carbon concentration is, is above the eutectoid concentration, bear in mind, our carbon concentration here is just 0.15 weight percent, and yet we have a two phase field here. So that that would be because it's a multi component system. In, in doing these calculations, if I just go back one slide, supposing that I know that M23C6 does not form at 400 degrees centigrade, because the diffusion of chromium atoms and so forth is is very slow. During the time scale of the tempering experiment. But nevertheless equilibrium says it should form. But I want to see the situation in which it doesn't form, then I can just remove M23C6 from the calculation, and then see how these results change. So you have the possibility of examining the sequence of stability of the phases in your system. As you progressively eliminate eliminate one phase, which phase becomes stable and so forth. So cementite might form first because it only requires a shorter institution of carbon and M23C6 might be suppressed. So at some temperature you know I might incorrectly predict that I get M23C6 during my experiment which doesn't correspond to equilibrium. If I held it at 400 degrees centigrade for a very long time, I would get M23C6. But from a practical point of view when you're studying the tempering of martensite, you might want to suppress the substitutionally alloyed carbides because they simply will not grow at a reasonable rate at 400 degrees centigrade. That you can do in these calculations very easily by going back to the choice that you had here and just remove these substitutional solids from the calculations. So you can study the sequence of matter stable precipitates that will form in your alloy. And you can also say that look if I put this tempered martensite into service at 400 degrees centigrade for 40 years, for example in in one part of a power station. Then I expect to see significant changes in the structure with the possibility of M23C6 carbide precipitation. And what will be the consequence of that? Okay, so if you are going to put the material into service at a high temperature, then it's very important to look at all the equilibrium phases to decide on whether they are going to form within the time scale of your application or not. So you know this basically completes the story on phase diagram calculation and notice that I haven't actually drawn any diagram, all right, because you don't need to the numbers say it all. However, you might want to plot the volume fraction of cement type that you get as a function of temperature that's perfectly all right. But there's no point in plotting many dimensional phase diagrams when you're doing multi component systems. Now I'm going to show you how even these calculations help you to discover new things. Okay. I'm going to go into this transformation here, which is a bay night where a plate of martensite effectively forms, but the temperature is high enough so the carbon escapes from that plate, and eventually precipitates as cementite here. So this is the structure that we call a bay night. And you can suppress this part of the reaction by adding, for example, silicon, which retards the precipitation of cement time. Okay, so let's assume that we've had this plate of martensite, and the plate of the carbon ending up in the residual austenite. Then we expect that the carbon concentration inside this plate will be very small. So let's assume we are transforming at around 400 degrees centigrade. Then, if you look over here, the concentration of carbon in the ferrite that is in equilibrium with austenite will be very very small. Right. So let's let's say 0.0 to 8% carbon. Because carbon prefers to be in the austenite than in the ferrite. So the equilibrium, the iron carbon equilibrium phase diagram. And notice that I'm not plotting the whole diagram because I only have two phases here. So I don't have cementite here, for example. That's perfectly okay. I explained to you in the last lecture that we can extrapolate these phase boundaries to whatever temperature that we like using the thermodynamic method. Since I have only two phases, I don't need to worry about anything else. And these are the only two phase boundaries I'm interested in. Let's see. We can measure the carbon concentration inside this plate using the atom probe, which I introduced in the last lecture. You know, it's basically a technique for looking at atoms and for identifying the atoms using a time of light my spectrometer. And this is the structure that I'm talking about you have plates of ferrite with carbon enriched austenite. Back in 1981, when we did the measurements, we found that the concentration of carbon in the bayonetic ferrite was much, much higher than equilibrium. Right. Now, with the atom probe available on in those days, we assumed that this carbon is not actually in solid solution, but is segregated to dislocations which would be a natural thing to assume. Because carbon does not want to stay there and this heat treatment was certainly long enough to allow the carbon to move away. But by segregating to defects, it becomes stabilized within the ferrite. So we have the carbon studies and these two in particular by Perry Loma and so forth, and cover Lara back in 2012 and 2013. You can collect many, many more atoms to build a picture. And here is such a picture. I told you in the last lecture that the specimen is in the form of a thin needle. So what you're looking at here is a thin needle and each dot is an atom. In this case, we are only using the carbon atoms to form this image. Okay, so you can select the atoms from which you want to form the image. So this, this is austenite. These are dislocations where carbon has segregated to dislocations. Okay. So that is, as I explained, just explained, I expected the carbon to be at dislocations. However, you can see that there's a lot of carbon, which is not at dislocations. It's actually in solid solution. And you know, the concentration is quite large that remains in solid solution, even though the sample will have been held at 200 degrees centigrade for 10 days. It's quite bizarre, you know, the carbon does not escape into the austenite. So you can see the concentration here. It's about 10 days. It's really orders of magnitude higher than the equilibrium solubility. So the question is, why is this right, but just to summarize, there is a huge excess of carbon in the ferrite, and it stays in the ferrite in spite of the heat treatment that we've given it. In fact, in steels like this, the austenite is actually harder than the ferrite. But here, the ferrite is harder than the austenite because the carbon is remaining in the ferrite. A large part of the carbon is remaining in the ferrite. So what is the reason for this? There was it really, really puzzled puzzled me for a long time. This is how a displacive transformation happens. This is one of the deformations involved in the creation of a plate of martensite. So all of these atoms, red and blue are austenite, but I marked these as red because within these two unit cells of austenite, I can identify a body centered tetragonal cell of austenite. So it's just relabeling the unit cell of austenite into a body centered tetragonal. And that gives us a clue on how the transformation happens because if I compress along this axis and uniformly expand along these two axes, then that's known as the Bain strain and that converts that tetragonal cell, which has a very large C over a ratio of about 1.24 into a body centered tetragonal cell with a much smaller C over a ratio if a material contains carbon or a body centered cubic cell if it doesn't contain carbon. The point that I want to make here about this Bain strain is that all of the carbon atoms that will be located in the austenite at the octahedral interstices end up on one set of interstices in the martensite and therefore the martensite becomes tetragonal. So when the transformation actually happens, it will lead to tetragonal bainitic ferrite, even if that is not favored by ordering. In other words, if you're transforming at a high temperature, then you do not favor the ordering of carbon atoms on one set of cell edges. So if you're currently thinking, maybe, you know, if I go back, we are looking at the wrong equilibrium here. Okay, this is between body centered cubic ferrite and face centered cubic austenite. Maybe we should be looking at the equilibrium between tetragonal ferrite and austenite. And of course there is no phase diagram for that. So how do you cope? Well, we have to create data, okay, for the tetragonal form. How do you create data? There's a method known as first principles calculations where, you know, you set up a unit cell and in this case it might be a tetragonal cell. You can put the energy of that cell, the cohesive energy of that cell and compare that with a cubic cell. And therefore you get a term which you can then substitute into the phase diagram calculations. So this is the result here. This is now no longer an iron carbon phase diagram between cubic ferrite and austenite, but it's between body centered tetragonal ferrite and austenite. And you can see that the solubility has increased dramatically. Okay. This is the corresponding magnified plot of the cubic ferrite solubility when it's in contact with austenite. So this could be the reason why the carbon remains inside the bainitic ferrite until you temper for long enough for precipitation reactions to happen. But the sort of values that we have here are consistent with the excess carbon that was measured, although these calculations are only for iron carbon, whereas that particular alloy was a more complex alloy. Okay, so let's think about this. So we predict therefore that bainitic ferrite should be tetragonal. And sure enough, when we do measurements in an X-ray synchrotron, we find that it is, it is tetragonal. This is the C-O-ray ratio. And what we are doing here is we are heating up the sample to high temperatures. So the C-O-ray ratio remains roughly the same. And then of course when reactions start to happen, tempering reactions, you get a drop in the tetragonality. And there are now many other experiments which confirm that the bainitic ferrite in that sample is tetragonal. And indeed, this is a paper from IIT Kharagpur by someone that you already know, I think Shiv Brotsingh and his student. I don't know what these yellow lines are, but never mind. They're yellow, so it doesn't matter. Okay, so they concluded that by taking tetragonal ferrite for bainite, they could get better agreement in the interpretation of the dielectric data than by taking cubic. And others have now done experiments using neutron diffraction, lattice imaging, and so forth, which confirmed that this is, this is the tetragonal structure. Now, just recently, maybe a few months ago, there was a paper published using positron analysis, which says that, look, another reason why the carbon remains is that the carbon could be associated with the vacancies in the iron lattice, in other words, with the host lattice. Okay. Now in the last lecture, I showed you how not the last lecture but one of the lectures, I explained to you how to work out the equilibrium number of defects and I was doing this in the context of saying graphene and nanotubes will not scale up in strength. So, if Delta G is the enthalpy of a defect, you know, the excess energy I need to put into create a defect, then I can write that the total free energy change is this, and set that to zero. And I get the equilibrium number of defects. This is the number of defects over the total number of entities, and simply exponential of minus Delta G over KT. Now I want to treat this problem where I've lost an iron atom from the middle of this cell, and this carbon atom which was in the octahedral site, exactly on that face drops down a little bit it doesn't go into the middle. Okay, for various reasons I won't go into, but it drops down a little bit here. We can use the same formalism that I explained earlier to work out the equilibrium number of these defects. That means the, there is clearly a binding energy between a host vacancy. That means an iron vacancy and a carbon atom, and that might reduce the rate at which, you know it's logical to assume that will reduce the rate of diffusion into the austenite. And that is what this paper, a few months ago in science reports concluded that the real reason why it remains there is because of this association. And what they forgot, or what they didn't do is how many carbon atoms are actually associated with host vacancies. And there's a paper going back from 1988 by McClellan and colleagues. Extremely nice thermodynamic theory okay thermodynamic theory, which works out the fraction of carbon atoms that will be associated with host vacancies in ferrite. Right I'm going to show you the result without explaining too much, but you'll see that the pattern of the calculation is the same as we did for the vacancies. So this is the ratio of carbon host vacancy pairs to the total iron atoms. Okay, so you can think about this as concentration of carbon vacancy pairs. And this vacancy is a host vacancy. This is a coordination number, the concentration of vacancies. And we've already know how to work that out. And this term you're familiar with you know this is the energy of the carbon energy in creating a carbon host vacancy pair. And there are some other terms, you don't need to go into them in more detail but again you see this term here. When you do this calculation. You see that 10 fraction 10 to the minus seven of the carbon atoms will be associated with host vacancies, and they reach the logical conclusion that the association of carbon atoms with host vacancies has no perceptible influence on the diffusion of carbon or of iron. And, you know, even if you multiply this by a factor of 1000, it doesn't matter, it will be still be only 10 to the minus three of the total number of carbon atoms a fraction 10 to the minus three, which will be associated with host vacancy. This is not the explanation for why the carbon remains inside the host lattice. But notice this is, this is really nice because what we've done here is we've combined detailed thermodynamics, you know, which was done back in 1988 to correct this interpretation of the positron data which was not, not particularly quantitative. Now the second example I'm going to show you is more recent that, and it's to illustrate that sometimes the thermodynamic data that we use in these computer programs may actually be wrong. Right. But you discover that only by looking at something that you're interested in. I was interested in, in this commonly assumed sequence of precipitation when we temper martensite that we first get epsilon carbide doesn't completely relieve the carbon from the martensite. And then the epsilon transforms into what's known as eta carbide, and then into Chi carbide, and finally the stable cementite. Okay. So this is the commonly assumed tempering reaction in steels that we start off with some transition carbides, and with sufficient tempering we end up with a mixture of cementite and ferrite and these transition carbide I'm giving you the approximate compositions here. But what we are doing is we are getting towards towards F3C. Now, there was some thermodynamic data published in 2014 by Naragi and Ogren and colleagues, which, which indicated that the iron carbon phase diagram should be like this. Here we have ferrite and cementite stable and that's what we normally assume to cover the whole of this temperature range. But according to those data, below about this temperature 450 degrees centigrade, let's say, the cementite should transform to chi and below that the chi should transform to eta. And you know this this idea was actually mentioned a long time ago by Chipman that this might be the case. So what this would say is that look if I took a mixture of ferrite and cementite and I cooled it, then I should end up with ferrite and chi, and that simply hasn't been observed. And it could be that at these temperatures, the kinetics of this reaction are too slow, because we also have only a small amount of carbon in the ferrite. So, we looked at this in detail, and some of the justification that the authors use for creating these data came from first principles calculations. When I looked at the whole series of first principles calculations available for the energy of formation of these carbides. There was just too much variation to be able to conclude what is more stable and what is not stable. And of course you know these first principles calculations only refer to a phase in isolation. They're not looking at the phase, which is inside ferrite or inside austenite. So calculations are for zero Kelvin and zero pressure. But nevertheless, you know if you don't have anything then this is a good way to proceed. So, I looked at a large quantity of experimental data and came up with this plot. Now on the horizontal axis we have what's known as the Larson Miller parameter which is basically a combination of time and temperature to express the strength of a heat treatment, a tempering heat treatment. And this is the bulk carbon concentration of an iron carbon alloy. And in all cases, cementite is more stable at high tempering times than any of the transition carbides. And these particular steels are very interesting because they have very low carbon concentration 0.03 weight percent, and you do not get any transition carbides but you get cementite. And it's important that transition carbides only form when the driving force is large, because you know they are they are not stable carbides, and therefore the driving force relative to cementite precipitation will be small. The driving force for cementite precipitation is small, they may not form at all. So in these alloys you do not find any any transition carbides at all. And I happen to have an alloy of pure iron 1.61 weight percent carbon which I made something like 30 years ago. My students to do a long term tempering experiment. And sure enough, it was just cementite forming even when we have a very large carbon concentration so the driving force is more than sufficient to form transient carbides and if they're stable they would remain, but all of these data indicate that cementite is more stable. Now, why were these data not wrong because you know we don't have good experimental data for these transition carbides thermodynamic data. So, what they did was basically said look I can write the heat capacity of kai carbide as that of cementite plus graphite required to change the composition of cementite into this. And you know that's fine that's that's an approximation, but the experiments indicate that it's not actually possible to get a stable phase field of alpha and Kai in that phase diagram. Now this was important for something that I was working in, and therefore I discovered that this is probably not the case that we have other stable phase fields below alpha and cementite. Sometimes you have no choice at all, but to do first principles calculations. Okay, because it isn't possible even in principle to make a thermodynamic measurement and this, this is one such case. So this is the unit cell of cementite projected onto one of its author on big faces. So these fractions here just indicate the heights of atoms above the basal plane. And the carbon atoms, and the carbon atoms lie on mirror planes inside the unit cell and how one third of the iron atoms also lie on mirror planes, but the others are at general positions with no particular symmetry. So there are two kinds of iron atoms here, those located on mirror planes and those located where the only symmetry is a one fold axis. So, when doing what we want to do is to find the thermodynamic properties of silicon inside cementite. And silicon has such a low solubility in cementite that it's not possible to do equilibrium measurements. In principle, you can't do that, but I know that when cementite forms at a very low temperature, the silicon is inherited by the cementite and therefore changes the thermodynamic properties of the cementite. To do calculations I need that information, because we use silicon quite a lot to stop the precipitation of cementite and we're good to know how much silicon you need to add. You take the unit cell of cementite and you calculate its energy of formation using first principles methods. And then you substitute a silicon atom into one of the mirror sites in this case, and the general position in this case, and you find the change in energy. And the difference here gives you information which can feed into the phase diagram calculation to work things out. So first principles calculations are at their best when there is nothing else that you can do. Okay, because here, you know, these are again calculations done for zero Kelvin and zero pressure, but there is no choice. This is the only way that we can get the necessary thermodynamic information. So nowadays, you know, the thermodynamic databases include results from first principles calculations. All right, so first principles calculations basically look at electron theory. And from that, work out energetics. I'm going to finish with an unsolved problem. So this is an iron 28% manganese alloy. And when I, when I heat it it undergoes a transformation below 200 degrees centigrade to austenite. And then when I cool it, it undergoes a transformation to martensite, which is epsilon martensite that means the hexagonal form of martensite. When we do a phase diagram calculation for iron 20 manganese, it indicates that there is no epsilon, right? Alpha naught is just an ordered form of iron manganese, right? So according to the phase diagram, there is no epsilon martensite, but we know that we get it, right? So that could be some sort of kinetic effect, but let's think. Okay, now if I suppress the ferrite from the calculation, right? I don't allow alpha or alpha naught to fall. Then I predict that, yes, I can get epsilon martensite. So if the thermodynamic data are right, then what this says is that the epsilon is not a stable phase, right? So you may ask, why don't I do an experiment where I hold the mixture of austenite and epsilon martensite here for a very long time, you know, the course of the PhD project and see whether I get ferrite for me. Well, it's impossible because just to give you an example, at 200 degrees centigrade, the diffusion distance of an iron atom in 10 days is 10 to the minus 17 meters. So it would take centuries for me to do that experiment, right? So I don't know. I don't know whether the thermodynamic data are not right or whether epsilon is simply a metastable phase, but there's another really interesting conclusion that I reached. And we are trying to design an alloy in which we can form epsilon at a higher temperature. So substantial quantities of epsilon. It's not proving to be easy at all. Basically, there has never ever been an observation of the epsilon phase, hexagonal martensite in iron, except by martensitic transformation. There's no equiaxed form of epsilon and so forth. And I really, really would like to create an alloy where I can get grains of epsilon forming without applying any pressure. So this is ambient conditions.