 We will then start now to look at a completely different problem, the problem of what is called critical phenomena namely substances, macroscopic objects, large collections of atoms which undergo transitions from one phase to another like liquid to solid or solid to gas and so on and so forth, okay. Now this subject we are going to deal largely with equilibrium phase transitions, critical phenomena because non-equilibrium critical phenomena is an open subject at the moment, there is a lot of work going on and there is no universal agreement on what comes under the umbrella of a non-equilibrium phase transition and what does not, we know it when we see one in that kind of, in that sense. For equilibrium phase transitions or critical phenomena, there is now a well established theory established over the last 50 years or so and it has several features which are of interest to us. In particular, there is going to be connection with the Langevin approach, the Langevin equation and similar notions when we talk about the order parameter and the way the order parameter evolves near critical, the critical region and so on. So let us start now with the most elementary of notions, I have to be critical of this chart starting right away. I am going to zip through the elementary parts of this because it is familiar to all of you but just for the sake of completeness and to refresh your memories as to what we are talking about, we will start with the simplest of notions. First let us look at what is already taught to you in high school namely the fact that you have 3 states of matter as they call them, 3 states of matter. This is not a very happy terminology to start with but we know what they mean, they mean liquid, solid and gas. So the assumption is you have one molecular species, just one kind of molecule with some interaction between the molecules and then depending on the temperature, this substance in sufficiently large collection of these molecules will either be a solid, a crystalline solid when I say solid or a liquid or a gas okay. And we know there are well defined ways of establishing what is the liquid, what is the gas, what is the solid, we will mention some of these things and then one draws what is called a phase diagram. So in such a substance for a given amount of material, the thermodynamic variables are P, V and T okay. As you know from elementary thermodynamics, when you have just a one component substance and a given amount of it, the number does not change, the number of particles then we can forget about the chemical potential, we can forget about concentrations of the different species and so on. And the simplest thermodynamic variables are P, V and T and they are connected by an equation of state. But this equation of state which tells you the value of any one of these variables in terms of the other 2, this equation of state will depend on what the temperature is and what the pressure is etc. It depends on where you are in the space of these 3 variables okay. So one draws typically a P versus T diagram, of course we will also draw V versus T and T versus V versus P and so on, we will do that by and by but one draws first a P, T diagram and then it turns out that at sufficiently high temperatures, the system is in general in the gas phase, at sufficiently very low temperatures it is in the solid phase and it is a liquid somewhere in between okay. So any point on this plane is a state of thermodynamic equilibrium for the substance okay. And then one has typically a figure like this, very schematically, you have a line goes off like that, not necessarily a straight line and the system is in a solid phase here, liquid phase here and a gas phase here and there is very often a triple point here at this place. I am not drawing this to scale at all, I am not telling you what the temperatures are, what are the actual values of these quantities, no such thing, just schematically it looks like this. Now the slope of this graph may be a thousand times the slope of this graph so we are not worried about that. The question is what happens on this line, it is called a coexistence curve or coexistence line, a curve which tells you for a given value of the pressure what is the boiling point so that if you increase the temperature you get into the gas phase or for a fixed temperature if you increase the pressure you cross into a liquid phase, it condenses, the gas condenses. Similarly between the solid and liquid and the solid and the gas here. So what happens here as you go across is boiling or condensation if you come down either freezing or melting as you come down and sublimation across this. So this is called the sublimation curve, this is called the boiling curve and that is called the melting curve. More significantly this thing also tells you for a given pressure what the boiling point is and as you know if you increase the pressure on the gas the boiling point goes up, if you decrease it the boiling point comes down. But very notably this curve just ends in mid air so to speak, just ends at a particular finite point and this is called the critical point. And it is a very special point of great interest and took a long time to unravel its mysteries because in a nutshell thermodynamics fails at that point. So you have to go beyond thermodynamics, you have to go to statistical mechanics but even more than that you have to go to some fairly sophisticated techniques in statistical mechanics to understand what is happening at this point. This curve here keeps going forever or else it branches off into other phase transitions because there could be many, many phases in the solid, not just one kind of crystalline form but many kinds of crystalline forms. It could go from BCC to FCC and so on and so forth. But we are assuming those complications do not exist and keeps going in this fashion. I have drawn a graph where if you increase the temperature, if you increase the pressure the melting point also increases. But that is not necessarily true for water, it goes the other way. So the slope of this curve is actually negative in the case of water which is a very enormous substance and for most materials it is actually positive but with a very high slope, numerical value of the slope. This graph on the other hand always has a positive slope because in all cases if you increase the boiling point of a liquid will increase if you increase the pressure. So it always goes in this fashion. And we also know that there is an equation which tells you that the slope of any of these coexistence curves in the PT plane is given by an equation called the Clausius-Clapeyron equation on this side which tells you how the boiling point or freezing point or whatever it is changes as a function of the pressure, okay. This quantity on the right hand side involves the latent heat divided by the temperature times whatever. So you are used to writing it in elementary thermodynamics as L over T times V2 minus V1 or something like this, right where L is the latent heat, T is the boiling point in this case if you are looking at this graph and V2 minus V1 is the difference in volume, specific volumes in the two phases, okay. Now that is not a very nice way to do this because what V2 is and what V1 is is completely arbitrary. What I call phase 2 and what I call phase 1 is arbitrary but that is subsumed in this L. To go from a liquid to a gas you have to supply the latent heat of boiling and come back the other way you have got to supply, first case latent heat of evaporation, in the other case you have to extract that heat in order to become a liquid here. But this is avoided by writing the proper equation which is the following. It is just one of the Maxwell relations in thermodynamics, okay. You are familiar with what the derivative of the pressure with respect to temperature is for a given volume at volume, constant volume what is this equal to? Are you familiar with the Maxwell relations in thermodynamics, right? Because what you doing? You have to now respond and tell me yes or no, yes. So what is the relation in this case? There seem to be some doubts here otherwise, okay. Let us go, let us take a breather and go back because I think that you should, at all stages you should be able to go back to 1st principles and do everything from the beginning. So digression from the digression, let us go back to thermodynamics, what is meant by thermodynamics. The whole point about thermodynamics is that you have a set of variables which you call thermodynamic variables and then you write an expression for what is called the internal energy in this business. This internal energy you do not in thermodynamics write an equation for it directly. What you say is that if you supply a certain amount of heat DQ to a system and it is an imperfect differential so it is DQ denoted this way and you extract from the system a certain amount of work DW then the difference DQ minus DW this is equal to the change in the internal energy of the system. I will call the internal energy E. Generally one uses this U for this symbol but I prefer to call it E, okay. That is it. This is the law of thermodynamics, law of conservation of energy, the first law of thermodynamics. But you go further and you say no, no, no this thing is part of a bigger system of bigger formalism which says that DE is actually equal to for a reversible thermodynamic process this DQ can be written as with an integrating factor as TDS so it is equal to TDS plus on this side a whole set of generalized forces and corresponding conjugate variables called fluxes. So this can be written as summation over I F I DXI but this is a generalized force. I will give an example right away and this is called a generalized flux. We used exactly this terminology in linear response theory. In fact the formulas that you are going to get, I am going to write a couple of them down are linear response theory formulas for the susceptibility, literally the susceptibility, okay. So it says once you take these integrating factors if you like and multiply the DXIs you end up with the perfect differential DE for the system. Now classic example of this for a single component system is going to be TDS minus PDV plus if the number also changes mu DN that is the chemical potential and that is the change in the number of particles in the system. So now you have a set of thermodynamic variables and these variables can be grouped into this portion, this portion and this portion. This is a generalized force so is P so is mu and this is a generalized flux S, V and N this is the entropy here. But now you immediately see that there is two categories of macro variable, macroscopic variables. There are the so called intensive variables which the generalized forces are. They do not depend on how big the system is, what the size is and so on. And then the extensive variables which depend are proportional to the system size. So I know that at a given pressure and temperature if I double the volume of a gas and double the number of particles the pressure and temperature do not change, nothing changes. So these fluxes are all extensive variables and the forces are intensive variables and you notice that you always multiply this by the differential of that on that side here. From this of course it immediately follows if this is a perfect differential that T equal to Delta E over Delta S keeping V and N etc all these relations follow. But now you can write this not in terms of just D E, you could write an equation for D some thermodynamic potential. You can choose those as independent variables. After all I have S and T sitting here, then I have P and V sitting here and then I have mu and N sitting here and in this system I have, I can draw a diagram like this and I can choose any of these 8 variables, 6 variables, any 3 of them I can choose to express my thermodynamic system. So for instance suppose I chose S, V and N, then as you can see since D E is expressible in terms of D S, D V and D N as a linear combination it implies E is a function of S, V and N. So you have E of S, V and N the internal energy and then you say that in thermodynamic equilibrium this is a minimum, the internal energy is at a minimum that is the state of thermodynamic equilibrium and then you have to ask is it a stable equilibrium, unstable etc and there are convexity conditions which you get as a consequence. But there is no reason why I could have chosen S, V and mu. So I could have chosen S, V and mu and then I could have chosen S, P and mu. So let us write this systematically so that we have S, V and N that is one possibility, S, P and N that is another possibility then S, V and mu that is a possibility, S, P and mu that is a possibility and then I have T, V and N, T, P and N, T, V and mu and T, P and mu that is it. So 2 cubed is equal to 8, these are the 8 possible sets of variables that I can choose to describe the system. Then since I have already got the laws of thermodynamics put in this form it is clear that the thermodynamic potential that is a minimum during internal equilibrium is the internal energy E. I do a Legendre transform on it to go from S to T that is the conjugate variable here by subtracting TS from this potential and I get the Hemmholtz potential. So this is F of T, V and N and then we know that at constant temperature volume and fixed number of particles in thermal equilibrium it is the Hemmholtz free energy that is at a minimum. Similarly I did not like V, I go to P, I choose S, P and N then I can do that by subtracting by adding P, V because this has got a minus P, D, V out here so I can add E plus P, V and I get the enthalpy H of S, P and N which is at a minimum for a process in which you can control the temperature, the pressure, the entropy and the number of particles. Similarly I can go to T, P and N and I get the Gibbs free energy here but I can do any one of them, it does not matter. For instance I could have chosen, I am just trying to figure out, so let us call this Phi, let us call them Phi 1, Phi 2 and Phi 3, no sorry Phi 3, this is arbitrary notation okay, I do not know what the standard notation is. So what was the last one? It was T, P and Mu. I cannot do this, why? It is not a good idea to choose, to describe a thermodynamic system using as a potential a function which is a function of T, P and Mu, why is that? Pardon me? All 3 are intensive, all 3 of these are intensive so there is no information about how big the system is here at all. So this is not going to work and we will see explicitly if you do a Legendre transform on this you will see what happens but we need just one more step before that. You see the first law of the laws of thermodynamics written in this form, I could in some sense consider this also as a generalized force and this is a generalized flux. Then I just put F i d X i for the whole lot, I do not have to take this separately so let me just do that, summation over i F i d X i. With the understanding that there is a generalized force called temperature and there is a corresponding flux called the entropy. Now, extensivity implies that all these generalized fluxes are proportional to E, right? E is proportional to these I mean differential in this form but now we make an assumption that E is a homogeneous function of all the generalized fluxes. So we assume that E is E of X i homogeneous function of degree 1 precisely that is the statement of extensivity. It may not always be valid. For instance if you took cloud in interstellar space which is you know held together by the mutual gravitational attraction of the particles it is not clear that the total energy is proportional to the number of particles because there is a strong attraction and then various other things can happen in between. So there are cases where you have to be careful, this will not work but under normal circumstances this is a homogeneous function of degree 1. What does that actually imply? This immediately implies by Euler's theorem that E must be of the form summation over i X i delta E over delta X i by Euler's theorem with the power 1 here. Had this been of degree R this would have an R, the power R. That is the first, that is the theorem you learn in calculus for homogeneous functions of several variables, right. But we also know from the laws of thermodynamics out here that D E, that delta E over delta X i equal to F i. So together these 2 imply that E equal to summation over i X i F i. So you see although thermodynamics, the laws of thermodynamics gave you statements only about the differential of E of the potentials, not absolute values, the assumption of extensivity which is an extra assumption that has gone in has actually given you an explicit formula here. Now what happens in the case of the single component substance that we have talked about in this instance here, we already see that F 1 is T, F 2 was minus p and F 3 was mu and all we got to do is to put these fellows in and you get an equation which says E equal to T s minus p v plus mu n. That is the Euler's relation in thermodynamics for this single component substance, right. But we also know that E minus T s plus p v must be equal to G because that is what you get by doing a Legendre transform on this to get to this point, to get to a function of T p and n. So we have got one relation here and E minus this thing is equal to G here, right. So this of course immediately implies that G equal to mu n or mu equal to G over n. So the Gibbs free energy has the physical meaning of being, I mean the Gibbs free energy per particle is the chemical potential. So that is the physical meaning of the chemical potential, okay. We will come in a minute to why in phase equilibrium under normal circumstances you say that the chemical potentials have to be equal on either side. This is how you find phase equilibrium. You try to find out the condition for phase equilibrium. If you boil water for instance in a beaker here then the pressure, temperature and the chemical potential in the two phases are equal to each other. We will see in a minute why. Essentially what it says, this Gibbs free energy per particle being chemical potential, it tells you it costs just as much at the boiling point at the instant time of boiling, condition of boiling. It costs you just as much energy to put a molecule, additional molecule in the gas phase as in the liquid phase. That is why you get an equilibrium, okay. So coming back to this, because you have this, if I differentiate this i dE, you see immediately this will also imply that dE equal to summation over i F i dX i plus summation over i X i dF i because E is X i dF i but the laws of thermodynamics tell you that dE is F i dX i already. So this implies that summation over i F i sorry X i dF i equal to 0, whatever it is done. Is that correct? Yeah, that is correct. So while E is X i F i summed over i, dE by the laws of thermodynamics is F i dX i therefore X i dF i must be equal to 0 out here, right. What does it mean in this language? It says 0 equal to S dT minus V dP plus N d mu or in this instance it also means that d mu equal to V over N dP plus minus S over N dT which is equal to little V dP minus little S dT where this is the specific volume, specific entropy namely entropy per particle that is the specific volume per particle, all intensive quantities because you divide it by N. So this implies immediately that mu is a function of P N T and this relation is called the Gibbs-Duhem relation. Now what is going to happen if you try to find a potential here? We have said it is not going to work. It will turn out to be identically 0 because you will end up by subtracting from this minus S T plus P V minus mu N and that by this relation is identically 0 by the relation here, the Euler relation for E, you subtract everything out you get 0. Now that is an indication that is already telling you that that is not a good thermodynamic potential because it is entirely in terms of intensive quantities and won't serve, alright. Now the Maxwell relations tell you the following and this let me, let us take the example here and do this. Let us go back to, let us look at F for instance. So F equal to a function of T, V and N and what is D F? This is found by going from S to T. So we have T D S in the original D E and we are now finding E minus T S which is what this F is. So there is going to be a minus S D T plus P D V because V does not change and then plus T D S mu D N. I want to be careful about this sign, minus P D V, no no, minus P D V plus mu D N I think. So this will imply that S equal to Delta F by Delta T at constant, let us see T D S minus P D V plus mu D N, yeah, it is okay. At V and N and similarly P equal to minus Delta F by Delta V at T and N, constant T and N. So this sort of thing is telling you that because of this relationship, the fact that this is an exact differential, it is telling you that the derivatives of this, of thermodynamic potentials with respect to these variables, the Xi's are other thermodynamic variables. They are the conjugate forces of the generalized fluxes okay. On the other hand, there is nothing to stop you from differentiating a second time in two different orders and we know this is going to be equal to each other. So if I now differentiate this a second time with respect to V, then I get Delta S over Delta V equal to D to F over Delta V Delta T but that must be equal to D to F over Delta T Delta V which must be equal to, pardon me, alright, so Delta S over Delta V equal to minus this guy, that is equal to minus this guy here which is the same as Delta P over Delta T and this is at constant T and that is at constant V. That is called a Maxwell relation. I will let you have fun by finding all the other Maxwell relations. So for our purpose, I mean fortuitously I did not plan it this way. I just took random one but that is precisely the one we need because we are trying to find D P over D T, this slope. So it turns out, I should remember this. So it is the one with F and that is this okay. So we have our first Maxwell relation and we can go back now. So on the boiling curve now with the critical point which we will come to slowly, here is T, here is P. On this boiling curve, the slope of this curve, slope D P over D T is given by Delta S by Delta V. So across this curve, there is a discontinuity in entropy because as you can see the volume for a given amount of material, the entropy of the gas is much higher than the entropy of the liquid, much greater disorder, many more accessible microstates. Similarly as far as the volume is concerned, this specific volume of the gas is far higher than this specific volume of the liquid. So this thing should really be written as Delta S over Delta V. I do not then care which is called phase 1 and which is phase 2. Whatever convention you use, I find Delta S and Delta V with the same convention. So this could be for instance S gas minus S liquid V gas minus V liquid. Now clearly this is positive because the entropy of the gas is definitely higher than that of the liquid. The volume of the gas is definitely higher than that of the liquid in all cases. So this is certainly greater than 0. It is a large denominator. So the slope is generally very, very small, okay. Which explains why this fellow has this behaviour here. On the other hand, if you look at this and apply the same equation between say liquid and the solid, then S liquid is definitely greater than that of the solid. But V liquid need not be greater than V volume of the solid. In fact, in the case of water, it is about 8 percent less. Ice floats on water. So therefore you have a negative slope in this case. The Delta S for liquid is always going to be greater. I mean the S for liquid is always going to be greater than that of a solid unless we are talking about a super fluid. Then of course you have a small entropy system. This is an ordered state in the system. So this slope could be this way or that. And we also see that the liquid is practically incompressible. So Delta V in that case is going to be V liquid minus V solid for a given amount of material is small. But the entropy change is large. So you have got a small denominator, large numerator and therefore the slope is large in magnitude in that case. Otherwise it is small in this case here. Similar arguments for the sublimation curve as well. Now the reason this does not end at a critical point is because we have to ask now what the hell is meant by this critical point. What is meant by it is that as you go along and let us look at the example of the liquid gas of specificity, you need a latent heat to be supplied to the gas in order to make it a liquid to make it a gas to cause it to evaporate. As you go up here down this curve, it turns out that this latent heat starts decreasing and finally vanishes at that point. So you do not need any heat in order to go from one phase to another at this stage. But we will see that the distinction between the two phases disappears. What is the other thing that keeps you distinguishes a liquid from a gas? There is after all a surface and there is a surface tension. That surface tension starts decreasing and at this point the surface tension also vanishes here. So after this, you have got a homogenous mush of some kind. It is a fluid and just call it a fluid phase. Neither liquid nor gas. There is no way of distinguishing. So beyond this point, it is a fluid. They could at this moment, we will see what happens at this critical point. When I do the magnet analogy, then it will become much clearer as to what this flow is, what this critical point does. Now you cannot have that a critical point on this graph because you have a critical point on this graph. It means that I pointed out that every point on this graph, on this plane is a state of thermal equilibrium. So in principle, I could start at this point. I am at a sufficiently low pressure and a high temperature. I am in the gas phase and I go through this process. I take the temperature up at constant pressure and then I take this up here. At constant temperature I increase the pressure, go down in pressure and come down here. At every stage, I am in thermal equilibrium. I have not crossed any coexistence curve. I have continuously moved from the gas phase to the liquid phase without encountering any discontinuity whatsoever. Whereas if I try to do it directly, I would encounter a discontinuity in the volume as well as the entropy. But that is not happening here. So the point about having a critical point, a point where the phase transition line coexistence line ends at some finite point is simply that you can continuously go from one phase to the other and you will not know it but you moved into the other phase alright. Something does happen when you cross this point here or this point but we will not get into that here but it is a continuous path. There is no other distinguishing feature on this phase plane on this plane. But you cannot do that here because if you did then it would imply you can go from a state of isotropic state. The liquid is all directions are exactly the same. It is homogeneous, it is isotropic and so on to a crystalline phase where the symmetry is broken. The rotational symmetry and translational symmetry of a liquid is broken because a crystal has only a smaller group of symmetries. You do not look the same if you rotate it by an arbitrary angle. You have to rotate it by one angle which belongs to the space group so to say and also translation only by lattice vectors will give you the same system again. So the symmetry is lowered here in an ordered phase as much higher here in the disordered phase and you cannot continuously go from one to the other. So graphs like this must either end in other phase, other coexistence lines or they must end in some natural physical boundary like if it is on this side it can hit this point here or it can hit absolute 0 out here. There is no critical point on the liquid gas, crystalline solid liquid phase coexistence curve. Now what happens here at this point? This is a very very hard problem because what essentially happens is that normal thermodynamics fails and fluctuations on all scales, all length scales start playing a role and this took a long time to realize how to handle this, how to find the physical, relevant physical quantities in such a situation. In particular how to explain this apparently universal behaviour of systems at these critical points and this entire problem is an outstanding problem in physics and finally in the 1970s it got resolved with Wilson and Fischer's, Wilson's introduction of the renormalization group. So to understand it properly to calculate critical exponents we need some heavy machinery here but the physics of it can be explained in much simpler terms and we will do that and we will adopt this so called the Ginsburg Landau approach which is a phenomenological approach which does pretty much the same thing provided we know what we are doing and it is a very very physical kind of argument. So let us do that in several slow steps. The first thing I have to do is to ask what happens if I try to draw this graph in the other two planes okay and just to be specific so we do not get confused I am going to focus only on this line liquid and gas, just two phases. We are not going to worry about the solid phase and sublimation curve and so on okay. So now let us draw this T and P and you have a graph like this with a critical point. This was liquid, this was gas. What happens if I draw T here and the volume here? Conventionally one does not draw the volume, one draws the density which is like the reciprocal of the volume for a fixed number of particles but let us do the volume so we can see physically what is happening much more clearly. I pointed out that beyond this curve there is no difference at all and across this curve there is a discontinuity. So it is clear that across this curve the discontinuity means this on this curve the two phases coexist but if they coexist the volumes are very different. The volume of a specific volume of a gas and that of a liquid are very different. So this means that if you solve for V as a function of T it must be double value at a given temperature. At any fixed temperature there must be two volumes if you are on this curve. So this curve would open up into a region and it will look like this, very crudely it will look like this. So let us say that this P is at what I call the critical temperature T C and this is at P C so this is at T C and this is at V C. All the while there is an equation of state of the system which will tell you what one of the variables is given the other two. So if you give me P C and T C I should be able to tell you V C etc. So that is why I am not writing the third variable down. In this graph the assumption is when you are here and you tell me P and T you have also told me V except there are two possibilities for this V but at this point there is just one because at this point the distinction between liquid and gas vanishes and I will call those variables V C P C and T C that is the standard notation here. So it is very easy to see which of these is going to correspond to the gas and which is to the liquid. Well it is not drawn too well but one should really draw one should really draw the density isn't it? Shouldn't it be the density that one, pardon me. When the temperature is increasing the specific volume also increases. There is thermal expansion right. So this is the opposite. So this is not a happy picture but make it wrong. One of these the high density phase in this picture it will look like this. Well actually it is not rho. It is the difference between the two that is being drawn. Rho liquid minus rho gas minus rho liquid or the other way about rho liquid minus rho gas whatever it will become clear as we go along. Whether the liquid has got higher density it does not change as much so it is more likely to be like this here. But what about the isotherm? For a fixed temperature what does P versus V look like and that you are all familiar with P versus V. If it were an ideal gas it would be just hyperbola. P V is a constant time, times temperature and at fixed temperature it is just a constant which is a rectangular hyperbola. That happens at sufficiently high temperatures but as you come down at lower temperatures you have a real gas like a van der Waals gas for instance for which the isotherm typically looks like this. It is actually not drawn to scale better figure would be this. Very high slope here and very small slope here. This is the typical van der Waals kind of isotherm for a real gas and it says that for a given pressure temperature is constant on this isotherm. For a given pressure there are 3 possible values of the volume. It is clearly not true. It also says that one of them has a high slope with respect to V at a constant temperature and the other has a very very small slope. So what is this slope? Remember that Delta P over Delta V at constant T 1 over V times this what is this equal to? Delta V over V sorry V times this what is this guy equal to? This is equal to the bulk modulus right? It is equal to the isothermal bulk modulus and since when you increase pressure the volume decreases minus this or the isothermal compressibility K sub T equal to minus 1 over V Delta V over Delta P at constant temperature. So let us stare at this guy here. You can see that this is a very incompressible part of the isotherm and this is very compressible here. It is also got other problems. Now how will this graph go into at sufficiently high temperatures into that mush above the critical point? The only way it can do it if it does so continuously is that this graph at some point develops an inflection point and after that that inflection point splits into a maximum and a minimum and as you get lower and lower it gets more pronounced. That is the only way it can do it. Now one can couch this in very fancy language but this is the only way this can happen. So there exists an isotherm, a critical isotherm which has an inflection point, this point here and that is PC, this is VC and the corresponding temperature and the corresponding temperature is given by the equation of state. So this is a T equal to PC isotherm. So we have all three diagrams now. There is the PT diagram, there is the VT diagram essentially and the PV diagram out here. What is got to be recognized is that this coexistence line is a coexistence region here, the whole region. But this cannot be a correct picture because it says that if I start, T is fixed, remember there is an isotherm. I take at this point, I start at this point at low pressure and I increase the pressure on the system keeping the temperature constant. Then the system is compelled to move along this graph up to here and then jump to this point. And similarly it has to be moved down along this graph and jump to this point. If I reduce the pressure from the liquid phase, this means there is hysteresis. Your refrigerators won't work with this. We know that there is no hysteresis in the condensation phenomenon, right? So this part of the graph is not correct. There is another problem with this part of the graph and that has to do with the fact that out here, the slope D P over D V is positive. This fellow is positive. That will make the bulk modulus negative which is not possible. It violates thermodynamic stability. If you put pressure on the system, the volume cannot increase, right? It is a statement of thermodynamic stability. It says the thermodynamic potential, corresponding potential is at a minimum, not at a maximum or saddle point. So this fellow has to be positive. It can only happen if this is a negative slope. So this portion of the graph has to be excised anyway. It says it is an unphysical model. But it is even worse because even if you remove this, this is not what happens in a real system. The actual system comes here and at a fixed value of the pressure, it changes to, it condenses to a liquid. On the way back, it does so without hysteresis here. This is called the Maxwell-Thyline construction through the Van der Waals isotherm and it ensures that there is a unique pressure at which condensation or evaporation happens at a given temperature. There is no hysteresis. So the actual isotherm looks like this. There is a flat portion and then comes out and you have a family of such isotherms going all the way up to the critical point here. We are not going to get into that here. There are metastable phases. Actually, yes, okay. So since he is mentioning this, let us take this a little more seriously and see what happens. You see, if I excise these portions all the way through, then they form a curve going through the minima and the maxima looks like this. But I am excising a little more beyond that. And the way this Thyline construction is done, I do not think you can see anything from this figure now, is that this area should be equal to that area. It is called the equal area rule and there are ways of justifying it heuristically. Deriving it rigorously is quite hard, quite hard to do. It is heuristic. It is heuristic. I can give a sort of very crude argument because you can see, you know, by telling which is not a correct argument in terms of the chemical potential but it is not relevant here. So what I want to say is you are removing this portion inside here is an unstable region but you are actually excising a bigger region so which looks like this. So let us draw this figure again. In this figure getting rid of everything else, there is a curve like this and there is a curve out like that, slightly bigger than that and this is V and P. This is the locus of the maxima and then the maxima of these cubic curves in the Van der Waals case whereas the Thyline construction which tells you where the phase transition occurs on either side is a slightly bigger curve than that and the system is unstable here and meta-stable here. So if you rapidly quench it into this state from above then it will phase separate into a portion that is gas and a portion that is liquid. There is a simple lever rule which will tell you how much of it is gas or saturated vapor and how much of it is liquid. Now technically this portion of it it will get through extremely fast because it is thermodynamically unstable and because this portion is meta-stable this is called the spinodal curve this is going to be much slower. In the case of liquid gas it happens pretty fast in any case but in the case of binary metallic alloys for example this might take a long time, it can happen very slowly and then there is a theory, a whole theory for the spinodal decomposition of the curve and so on, right. So it is a separate subject altogether, right. On the other hand we are not concerned with that. The point I want to make is that if I now draw what the actual isotherms are they look like this and this is the point I wanted to make. So here is V, here is P. These fellows come down like this till there is an inflection point and there is nothing beyond that. So gas liquid, gas liquid above the critical point there is nothing, neither gas nor liquid but homogeneous fluid, okay. So that is the PV diagram, the VT diagram and the PT diagram. Now let us look at a magnet okay and we look at the simplest example of a magnet. Something which is got magnetic moments you apply a field and they align in the direction of the field and you switch off the field the magnetization goes to 0, okay. On the other hand we have substances which are ferromagnets where even if you switch off the field you still have a remnant magnetization. So the paramagnet below a certain temperature makes a transition to a ferromagnet and this is the one that has a direct analogy with the liquid gas phase transition. That is what we are going to look at. So let us look at that little more detail. We need a simple model of magnetism. Fortunately for us we have much simpler models of magnetism than we have of fluids because the interaction here is very complicated whereas in magnetism the interaction looks relatively simple. In general it is something called the Heisenberg exchange interaction and many, many simplifications have been done to it and we are going to take the simplest of these things which already captures the essential features of what I want to say, okay. The first point I want to tell you and this is something which we are going to need and what we are going to study later on is that under normal circumstances when a system is in thermal equilibrium, macroscopic system, there are always fluctuations about this equilibrium state but the question is how big are the fluctuations? What is the relative magnitude of these fluctuations relative to say the average value of some variable? If that were not small then thermodynamics would be useless because thermodynamics is a science of averages. You generally expect it to be extremely small, immeasurably small, one part in 10 to the 12 or something like that, right. On the other hand at a phase transition these fluctuations will become fairly large because as we will see the very shape of the graph will change, the central limit theorem will change completely. But first I want to show you since I want to make the self-contained that thermodynamic fluctuations in normal states of equilibrium are really very small for macroscopic objects provided the number of degrees of freedom is sufficiently large. So let us take the simplest of things which you can imagine, the caricature of a caricature, let us take a whole lot of atoms, n of them and let us assume that each of them has only 2 energy states possible, as simple as that. So let us assume that the 2 states, there is an energy 0 and there is an energy epsilon, nothing more than that. And you have n of these particles not interacting with each other so they would not undergo a phase transition and all these particles each of them has energy either 0 or epsilon and it is put in contact with the heat path at temperature, inverse temperature beta. Then the question is what is the average energy, what is the variance of the energy? The trivial calculation. So since each can be either 0 or epsilon, the probability is proportional to the Boltzmann factor, e to the minus beta epsilon, then you have to normalize it. So the so-called partition function is a summation over e to the minus beta e epsilon and since epsilon is 0, you get a 1 and the other one is epsilon so you get 1 plus e to the minus beta epsilon. And the partition function for the full system is just the product of partition functions for all these guys because the number of states is just a product and that is it, that is the partition function. What is the average value of the energy? I called it e but let me call it this to make sure that we understand this. It is the thermal average, the internal energy is the average value of the Hamiltonian. What is this equal to? The simple formula for this is equal to minus delta log Z over delta beta. This is equal to differentiate, you get n, you get 1 plus e to the minus beta epsilon out here and then n epsilon e to the minus beta epsilon times this, which is equal to n epsilon over e to the beta epsilon minus plus 1, that is it, okay. And what is the variance of this energy? Luckily there is a simple formula for it. You can painfully sit down and do the second derivative and things like that but it is actually the derivative of the average energy, the internal energy with respect to beta, okay. So that gives us this is equal to n epsilon over e to the beta epsilon plus 1 the whole square, the minus sign cancels out and then epsilon squared e to the beta epsilon. So the standard deviation is square root of n and this goes away. So the standard deviation of e divided by the average value of e is this guy goes away, the epsilon goes away, that is typical. It says the fluctuation, relative fluctuation in any macroscopic variable will be typically in a state of thermal equilibrium 1 over the square root of the number of degrees of freedom. And if this is of the order of Avogadro's number or 10 to the 24 or something, this is 1 part in 10 to the 12. So that is the reason you can forget about fluctuations in general, okay. But you see if you are in a ferromagnet, a paramagnet where you say that the magnetization is 0 when the field is 0 and you are in a state of permanent magnetization then what has happened is if I plot the magnetization for which let us call it m and the probability that the magnetization is some value m in this fashion then in a paramagnetic phase in the absence of any external field, the magnetization is 0 on the average. But on the average, there is fluctuations about the average. So what you really have is a curve which is extremely sharply peaked about this point, an extremely narrow peak which goes to 0 in the limit when the number of particles becomes infinite. So every time you measure this object, you are going to get on the average, you are going to get 0. On the other hand, if it has a permanent magnetization, it could have either been a magnet pointing up or down if you just take the simplest case of just two possibilities, then you get either a peak here or you get a peak here depending on what your initial condition knows. Now how is a distribution like this going to go to a distribution like that as you cross the curie point as the curie temperature which makes you go from a paramagnet to a ferromagnet. The only way it can do it is as you approach this curie temperature from above and go beyond it left, this distribution must start getting widened out. So it must eventually do this and then it does this and then it fixes at this point, not as a function of time, as a function of the temperature, as I lower the temperature across the critical point, infinitesimally above the critical point it is this, infinitesimally below it is this, which means at the critical point the distribution is extremely broad and this statement fails, the fluctuations become as big as the mean, the standard deviation becomes as big as the mean itself. So if you like, you can say that phase transition with the critical point, at the critical point you have the failure of the central limit theorem, you have the failure of this sharp Gaussian and so on. So that is a probabilistic way of looking at this thing here. It already tells you therefore that thermodynamics is going to fail because the fluctuations have become too large and therefore you expect thermodynamics which deals only with averages to fail, okay. Now the question is how are we going to deal with this situation and that is the task of understanding critical phenomena, okay. Now having come to this point, we will stop here today but next time we will take a simple magnet model and draw the analogs of these three figures for the fluid, develop the magnet fluid analogy and then see what the critical, so-called critical exponents are. We define them, see what they are and then try to see whether we can find the theory which will encompass, which will tell us what critical points behaviour should be like when you look at it in a time dependent way. There is a few more steps to be done before that eventually but that is our goal, okay. Yeah because the whole point here is non-equilibrium. What happens, I mean say it in words here, what happens is that since fluctuations on all length scales will become important, the time scales also will become slower and slower. If you look at large fluctuations, they take longer and longer to heal so something called critical slowing down appears at this, at the critical point which makes for very interesting dynamics.