 And in particular, we had a model, kind of a naive one, that we were studying to try to understand the coexistence of a gas and a liquid, the Van der Waals model. And we noticed that if we consider the pressure-volume equation of state described by the Van der Waals equation, what we'd like to explain is coexistence of a gas and a liquid, which in the pressure-volume plane looks like this. There's some coexistence pressure depending on the temperature. These lines are isotherms when the temperature is fixed. And if we, say, start out with the gas and reduce the volume by compressing it, we reach the point in which the gas can coexist with the liquid, then the pressure stays fixed at a coexistence pressure until the gas is entirely converted to liquid, and then the pressure starts to increase again. So what we actually see in the Van der Waals equation, for the isotherms that have, I should make it exaggerated a little more, for the isotherms with the temperature below the critical temperature is that the pressure as a function of the volume is not monotonic. And we said the right way to interpret that is that this region where dp, dv is positive, where the compressibility is negative indicates an instability. Now what actually happens in those intermediate values of the volume is that there's coexistence between gas and liquid phases described by these two points. In a way we determine the coexistence pressure given this isotherm predicted by our model is using Maxwell's equal area construction. So the area between the line of constant pressure and the isotherm, the area above the horizontal line matches the area below. In terms of how the Helmholtz free energy behaves as a function of volume, the picture looks like this. Actually the slope which minus the slope corresponds to the pressure and so it's always negative. But the Helmholtz free energy as a function of the volume changes from being concave up to concave down. Now what it means is that if we want to minimize the free energy in this region where the straight line which is tangent to the curve at these two points lies below the Helmholtz free energy for the homogeneous phase and the dotted line represents a mixture of these two phases, we can obtain lower Helmholtz free energy by mixing the gas and the liquid in this region than we can by staying in the homogeneous phase. And from that we obtain this equal area prescription. We can also describe what happens when there's coexistence in terms of Gibbs free energy. The Gibbs free energy at a first order phase transition, meaning a transition in which there's coexistence of phases is continuous but its first derivatives are discontinuous. Remember volume is derivative of Gibbs free energy with respect to pressure. And temperature or entropy is minus derivative of Gibbs free energy with respect to temperature at constant pressure. When there's phase coexistence in general, the volume and the entropy are discontinuous. That means the transition has latent heat. We have to add heat to the liquid to boil it and also that the gas and the liquid which coexist with one another have different density. I'll just write different density. The two phases which coexist have different values of the number of particles per volume. So you can think of it this way. If you consider the Gibbs free energy, let's say as a function of temperature, I could draw a similar picture for pressure, that we have a liquid phase which is favored at low temperature because it has the lowest Gibbs free energy and a gas phase which crosses the liquid phase at some point. The system always wants to be in a phase which has minimal Gibbs free energy. So it follows the liquid curve until the curves cross and then follows the gas curve. So the Gibbs free energy itself is continuous through this point, but its first derivatives can jump. And the name first order phase transition really means discontinuous first derivatives of G. So now I want to make a transition as it were to talking about second order transitions. And you might think it reasonable to expect if a first order transition means that first derivatives of the Gibbs free energy are discontinuous, second order would mean that first derivatives are continuous and second derivatives are discontinuous. That's not exactly it because that never happens. You're half right if that's what you think. When we say a phase transition is second order, we mean at the phase transition, first derivatives of G are continuous. So these are transitions that have no latent heat, no entropy discontinuity and no change in volume. The second derivatives of G actually diverge at the phase transition, go to infinity. So for example, we could consider the heat capacity. The heat capacity at constant pressure in particular, which is tau times derivative of entropy with respect to temperature at constant pressure, that's the second derivative of the Gibbs free energy since the entropy is the first derivative, so it's minus tau second derivative of G with respect to tau at constant pressure. In a second order phase transition that behaves like one minus the difference between the temperature and some critical temperature where the phase transition occurs raised to some negative power. So the heat capacity blows up like one over a power of the deviation from the critical temperature. And similarly I can consider the compressibility, which I define as one over V times the derivative of volume with respect to pressure at constant temperature. So remember in the stable phase, derivative of volume with respect to pressure or pressure with respect to volume is negative. So the compressibility is positive. It expresses when we apply pressure the response of the system. It gets a little bit smaller when we increase the pressure. And that too can be viewed as a second derivative of the Gibbs free energy up to some factor in front minus one over volume times second derivative of Gibbs free energy with respect to volume with temperature fixed. And that too blows up like deviation from critical temperature to some power not necessarily the same power, power prime. So an example is we've considered this line of coexistence in the pressure-temperature plane for the gas-liquid transition. That's a line of first order transitions but that line ends at a critical point. As you approach the critical point, the density discontinuity, the difference in density between liquid and gas, goes to zero. And at the critical point, if we go beyond the critical point in temperature, we don't see an abrupt transition at all anymore and no coexistence of phases. This point at the end of the line of first order phase transitions is an example of a second order phase transition. Now this is what we call the critical temperature and critical pressure, critical point in the TP plane where a second order phase transition occurs. So we'd like to understand better the properties of second order phase transitions. And let's try to understand it in the Van der Waals model. So let's imagine we're approaching this critical temperature from above. So in the Van der Waals model, we have isotherms. When the temperature is above the critical temperature, which are monotonic, at the critical temperature, then we make a transition to an isotherm, which is no longer monotonic. So there's a critical isotherm when the temperature is equal to the critical temperature in which pressure as a function of volume has a point of inflection. So DP-DV goes to zero at that point. This is the isotherm where the temperature is equal to the critical temperature. And this is the critical point where DP-DV equals zero. Correspondingly, the compressibility goes to infinity. It's just proportional to the reciprocal of DP-DV. We can also consider approaching the critical point from temperature below the critical temperature. On that case, we have an isotherm which has positive slope in a narrow region. We know what that means. It means that there is base coexistence at a value of the pressure determined by Maxwell's construction, the coexistence pressure. And the discontinuity, well, it involves the difference between the volume here in the gas and the smaller volume here in the liquid. So as we approach the critical temperature from below, these two points get closer and closer together and then they coalesce at the critical temperature where the isotherm becomes this one in which the pressure as a function of volume makes a transition to being monotonic. And it does so by having a single point at which its derivative goes to zero. So I'd like to understand, first of all, if we approach the critical point from below, how that volume discontinuity behaves as the temperature gets closer and closer to the critical temperature from below. So we'll consider tau going to tau c from below tau c. Here we have a picture like this where the two points corresponding to the volume of the liquid in the liquid gas are very close together. And let's try to compute how that discontinuity and volume behaves when the temperature is close to the critical temperature in the Van der Waals model. So it'll be useful to use dimensionless variables like we did when we wrote down the law corresponding states. So I will denote by v hat the volume divided by the critical volume. Critical volume meaning right at this point and this is the critical pressure. This is the isotherm when the temperature is the critical temperature. And p hat is pressure divided by critical pressure. Tau hat is temperature divided by critical temperature. So the dimensionless volume, pressure, and temperature normalized by the critical values. And I'm going to be interested in the case where these are close to one. We're close to the critical point. So it's useful to consider the deviation from one. I'll call that delta p hat. It means the difference between v hat and one. So equivalently, it's the difference between volume and critical volume divided by the critical volume. And same thing for the pressure and the temperature. I'll denote by delta p hat the difference between p hat and one or difference between pressure and critical pressure divided by critical pressure. Delta tau hat is tau hat minus one or tau minus critical temperature divided by critical temperature. So remember, we wrote down the law of corresponding states using these rescaled variables. It follows from van der Waals. We eliminated the other parameters in favor of the critical temperature, volume, and pressure. So that law of corresponding states looks like this. v hat minus the third p hat plus three over v hat squared equals eight-thirds tau hat. And so if we fix p hat and tau hat as we discussed, you can think of that as a cubic equation that we can solve to find the volume in terms of the temperature and the pressure. So after doing a little bit of algebra, I can rewrite that equation of state in terms of these variables delta v hat, delta p hat, and delta tau hat, describing how close we are to the critical values. So we're not going to do the algebra on the board, although it's just a few lines and I put it all in the notes. But obviously it's just algebra. So I think it's legitimate to just write down the answer. The cubic equation to be solved with no further approximations, just taking this equation and making these substitutions, rewriting it in terms of delta v hat, delta p hat, and delta tau hat. It looks like this. Delta v hat cubed equals minus 2 thirds delta p hat plus 8 thirds delta tau hat. That's the constant term on the other side plus minus 7 thirds delta p hat plus 16 thirds delta tau hat that multiplies the linear term in delta v hat. Then there's a quadratic term plus minus 8 thirds delta p hat plus 8 thirds delta tau hat. That's the coefficient of the quadratic term delta v hat squared. And then there's another term which is cubic, which I'm putting over on the other side minus delta p hat delta v hat cubed. So it's just algebra getting from here to here, from those substitutions. Is that a one or two? This? Yes. That's a seven. Okay. So now, though, I want to imagine that we're close to the critical point. That's what we're interested in studying. And that means that these dimensionless variables are all close to one. And so we can consider expanding in a power series in those small quantities. Let's first of all notice that if we're at the critical point, if p is equal to pc and tau is equal to tau c, then that means delta p hat and delta tau hat are both zero. So all the terms on the right-hand side vanish. And this just becomes delta v hat cubed equals zero. So in other words, as we observed last time, the zero of the volume that determines the volume, the root of this equation, becomes triply degenerate at the critical point. That's where these three real roots for the volume coalesce to become degenerate roots. So if we were near the critical point, I'm going to argue that we can neglect, unless we want to calculate higher order corrections and we're going to keep the delta v hat squared and delta v hat cubed term on the right-hand side compared to the first two terms. Now I have a delta v hat cubed term over here, which I'm going to keep, but I'm going to neglect this because it's multiplied by an extra delta p hat, which is small. We're going to consider the approximation in which delta p hat, delta tau hat, and delta v hat are all small compared to one and try to just make a leading approximation under that condition. So now if you come back to this picture, but maybe I'll redraw it in terms of delta p hat and delta v hat, the temperature's just a little bit below the critical temperature. That means there are three roots and looks something like that. And now we want to adjust delta p hat so that we're doing an equal area construction. If we choose this to be the coexistence pressure or the delta p hat corresponding to coexistence, then these areas should be equal according to the Maxwell construction. That's how we actually minimize the Helmholtz free energy. And since this really looks like a cubic function, what it means is that when we choose the coexistence pressure appropriately and we're close to the critical point, these two roots occur at the same absolute value of delta p hat. So once I've adjusted delta p hat to be at the coexistence value for a particular delta tau hat, then the equal area construction guarantees we're going to get a picture like this where the three roots are at plus and minus some value and at zero. And that's how we get these areas to be equal. Well, it's going to look like a cubic when everything's small. And so equal area just means that these two guys have the same absolute value and your third root is zero. So if you look at our equation, that tells us how to ensure we're at the coexistence point for some specified delta tau hat, which is less than zero. We're considering temperature less than critical temperature so that we have three real roots, which means that delta tau hat is negative. And so that our equation, when we neglect these terms because they're higher order, to have three roots for delta v hat, which looks like this picture, this constant term actually has to vanish. Once we've adjusted delta p hat, so that we've satisfied the equal area criterion, this constant in the equation has to be zero. So equal area means it tells us that the condition for coexistence is to choose delta p hat to be equal to four delta tau hat. So constant term in equation of state is zero. So now I know at coexistence what delta p hat is, and that means I can find the coefficient of delta v hat on the right-hand side of the equation when the condition for coexistence is satisfied. So now the equation of state becomes delta v hat cubed. Constant term is going is equal to minus 7 thirds delta p hat plus 16 thirds delta tau hat. Again, doesn't look much like a seven, I guess. Delta v hat plus terms of their higher order that we'll neglect. But now delta p hat is four delta tau hat. So this is minus 28 thirds delta tau hat. If I combine that together with the plus 16 thirds, that gives me minus 12 thirds or minus four tau hat multiplying delta v hat. So minus four delta tau hat delta v hat. And now we can see what the three roots are. Well, there's, by our equal area construction, one of them is that delta v hat equals zero, but that's the unstable one. The gas and liquid correspond to the other two roots. Delta v hat squared equals minus four delta tau hat. That's the condition for a gas and liquid. Remember, this is now positive because delta tau hat is negative. So there are two real solutions, one positive and one negative. The positive one, that's the gas. The negative one is the liquid. So we have delta v hat gas equals two, taking the square root of four, times tau critical minus tau over tau critical because that's delta tau hat. That's minus delta tau hat. That's how I got rid of the minus sign. And then we have the square root of that. And for the liquid, delta v hat for the liquid is the root with the opposite sign. So we've calculated the discontinuity between gas and liquid. The volume of the gas minus the volume of the liquid divided by the critical volume, that's just the difference between this and this. So it's equal to four tau critical minus tau over tau critical to the one-half. So as we approach the critical point from below, the volume discontinuity goes to zero and it goes to zero in a particular way, like the square root of the deviation of the temperature from the critical temperature. This factor four, well to calculate that, we actually use the details of the Van der Waals equation, the corresponding states here, which we derived from the Van der Waals equation last time. So factor four depends on details of Van der Waals equation of state, and therefore we don't really trust the factor of four, because we don't really think the Van der Waals equation is a very accurate description of the gas-liquid transition, not quantitatively. But on the other hand, the fact that the discontinuity behaves like the square root of the deviation from the critical temperature, that's much more general, because that really just followed from the observation, which is true in the Van der Waals equation, but could be true more generally, that the critical point is a cubic inflection point, where we make the transition from having one real root to three real roots. So the fact that it's proportional to tau critical minus tau to the one-half is more general. All we needed was that the critical point is a cubic inflection point. And so we would expect that to be true, even if the Van der Waals equation is not quantitatively correct, as long as our qualitative picture of what's going on at the critical point is correct. And furthermore, that prediction that the discontinuity goes to zero as we approach the critical temperature from below, like tau critical minus tau to the one-half, should apply to lots of different gas-liquid phase transitions. You know, if it applies to water, it should apply to alcohol and ether and so on. So the factor of four was fun to derive, but not of great interest. But volume discontinuity proportional to tau critical minus tau square root is a universal prediction of the model that doesn't depend on the details. So that's nice, since we should expect it to be broadly applicable. A universal just means if we've got the right picture, it should apply to many different critical points. So in particular, to lots of different transitions between gases and liquids and different substances. So now there's good news and bad news. All the way one usually says this is we consider the density rather than the volume. The density is number of particles per volume. So if we keep the number of particles fixed, it's just proportional to one over the volume. So if we consider a small change in density divided by the density, that's the same thing as minus the small change in volume divided by the volume just by differentiating. So the corresponding prediction in our Van der Waals equation for the density discontinuity is rho liquid minus rho gas divided by rho critical is equal to the same thing I wrote down before for tau critical minus tau over tau critical to the one half. So the difference is just that I changed the sign here. The gas has higher volume. The liquid has higher density. So we don't believe the four in general, but we kind of believe the square root and greater generality. Now the good news and bad news is that our idea that there's some kind of universal behavior in a second-order phase transition agrees with experiment. For many different gas-liquid transitions in particular, we find that the density difference between liquid and gas is proportional to tau critical minus tau over tau to a power. And the power, beta, well, it's the same power for many gas-liquid transitions. That's what I mean by universality. Do you mean tau critical? Although, okay, so it's universal, but beta is actually 0.32 when experiments are done. It's 0.32 in lots of different gas-liquid transitions, but it's not one half, it's 0.32. So the idea of universality that we can make general predictions has been indicated that there's something wrong with the way we extracted the quantitative prediction. But nevertheless, let's forge ahead and make some more predictions. Also wrong, but instructive. Let's ask about the compressibility close to the critical point. So suppose we consider, to begin with, approaching the critical point from above this time. So we're approaching the critical isotherm. The critical isotherm is the one with the point where dp dv is equal to 0. That's the critical point. And if the temperature's slightly larger, we have a nearby isotherm, but its slope now over quite goes to 0, though it gets close to 0. That's a isotherm at tau greater than tau critical. And so we can ask, what is the minimum value of the compressibility, while the inverse compressibility dp dv, or the maximum of dv dp, on this isotherm where the temperature is above the critical temperature, but very close to the critical temperature? So what we anticipate is that since that minimum value of dp dv goes to 0 on the critical isotherm, it's going to approach 0 as the temperature approaches the critical temperature from above. And we'd like to know how does it approach 0? It's going to behave like some power of the deviation of the temperature from the critical temperature. What power? So again, we'll compute that in the grandeur wall's model, but mindful, again, of the idea that some of our predictions will depend on the details of that model and others will not. So the question we want to ask is, what is minimum value on some tau greater than tau critical isotherm of dp dv? So let's go back to our approximate equation of state when we're close to the critical point, which I'll just write down again. Delta v hat cubed equals minus two-thirds delta p hat plus eight-thirds delta tau hat. So what we're interested in is how the pressure and volume are related close to the critical point when the temperature is fixed. That tells us the compressibility. The temperature is fixed above the critical temperature, but close to the critical temperature. Sorry, I left out the next term. That's the constant term, and then there's also the we're going to need this too, minus seven-thirds delta p hat plus sixteen-thirds delta tau hat. Delta v hat plus the higher-order terms which I'm going to ignore. So we want to consider very varying p, hence delta p hat and delta v hat with delta tau hat fixed as we're interested in the derivative along an isothermal compressibility. So if I take the variation of both sides, on the left-hand side it's just three delta v hat squared times a d delta v hat. It's a little cumbersome notation, but I mean a small change in delta v hat, d delta v hat. So in this expression, this doesn't vary, but this does. So I have minus two-thirds d delta p hat, and then in this term delta tau hat doesn't vary, but delta v hat and delta p hat do. So when delta p hat varies, I have minus seven-thirds d delta p hat delta v hat, and then when delta v hat varies, I have minus seven-thirds delta p hat, plus 16-thirds delta tau hat, d delta v hat, and the high-order terms, which we're still ignoring. Now where does the minimum of the slope occur? Well, it's kind of obvious that here the slope is zero, so when we're considering an isotherm, which is very close to the critical one, the slope is going to hit its minimum at this same value of the volume, up to higher-order corrections. So the minimum dp dv occurs at delta v hat essentially zero, just like it does on the critical isotherm. So to find minimum, consider delta v hat equals zero. So if delta v hat is equal to zero, the right-hand side of the equation is zero, so I get this relationship between the change in the pressure and the change in the volume. Zero equals minus two-thirds of d delta p hat, and then in this term, this is zero, because we're considering delta v hat to be zero, and the other term is plus minus seven-thirds delta p hat plus sixteen-thirds delta tau hat d delta v hat plus higher-order stuff. So if delta v hat is equal to zero, so maybe you guys can tell me why I wrote this down. I said delta v hat is equal to zero, and that means I can take minus two-thirds delta p hat plus eight-thirds delta tau hat equals zero. Why did I say that? I just have to go back to the equation of state. So in other words, the equation of state tells me how delta v hat is related to the other guys, and so delta v hat equals zero means this is zero, therefore this constant term has to be zero. So now I can substitute in for delta p hat, since this means the delta p hat is equal to four delta tau hat, and I can substitute that in here, and it's very similar to what we did before. This now becomes minus 28-thirds plus 16-thirds, so it's minus 12-thirds minus four minus four delta tau hat multiplying d delta v hat. So now what we found is the relationship between a change in delta p hat and a change in delta v hat along this isotherm when delta v hat is equal to zero, and namely multiplying by three-halves in this equation is equal to minus six delta tau hat. So just remembering, again, what our notation means, delta p hat is deviation of pressure from critical pressure divided by critical pressure, and delta v hat is equal to deviation of volume from critical volume divided by critical volume. So a small change in v hat is the same thing as a small change in v divided by critical v. A small change in p hat is the same thing as a small change in p divided by critical pressure. We're considering integrating along the isotherm, so the derivative of pressure with respect to volume with the temperature fixed is critical pressure divided by critical volume times the derivative of p hat with respect to v hat at constant tau, which we've now seen is minus six delta tau hat. So for the isothermal compressibility, the quantity defined with the sign so it'll be positive minus one over volume derivative of volume with respect to pressure at constant temperature, so we're considering the volume close to the critical volume, so I can replace that by critical volume. And it's convenient to consider the reciprocal because that's the thing which is going to go to zero. The compressibility itself is going to diverge when we get close to the critical point. That's minus v dp dv at constant temperature. That's like six times critical pressure times deviation of temperature from critical temperature divided by critical temperature. That's delta tau hat. So like in our earlier discussion, to get the six, you have to really know the details of the Van der Waals equation, but to get that the inverse of the compressibility diverges like the first power of tau minus tau c, that was more general. The cubic structure of the equation of state. So six depends on microscopic details, or I'll just say on model. Inverse compressibility being proportional to tau minus tau critical is more general. It just follows from the critical point being the cubic inflection point. Alright, so as long as we've come this far, let's ask what happens to the compressibility when we approach the critical point, but this time come up to the critical temperature from below? Well now it's a somewhat different calculation. So kt as tau goes to tau critical from below instead of above. What we found here was derived under the assumption that tau is greater than but close to tau critical. And the reason it's a little different is that now there's a volume discontinuity, and so we want to consider the compressibility for the liquid or the gas, for which delta v is now going to be non-zero. Before we did our calculation when delta v is equal to zero because that's where the slope was minimal when tau was greater than tau c and close to tau c. But now we're not going to want to do that because that's here in the middle of the coexistence region where the phase would be unstable. We actually have a mixture of this phase and this phase. So we need to take into account that delta v is non-zero as we approach the critical point from below. And that's going to make the calculation a little different. The idea is the same, but quantitatively the result will change a little. So remember, as we derived over here from the Maxwell construction, we found the volume discontinuity. Delta v hat was either plus or minus depending on whether we were talking about the liquid or the gas side, minus delta tau hat, which is positive when tau is less than tau c to the 1 hat. We also had that the pressure was equal to 4 delta tau hat at coexistence. So I can use this same equation that I guess I had here. But now delta v hat shouldn't be set equal to zero. It should be set equal to its value on the gas or the liquid side. And when we square it, it's actually the same on either side. So I guess here we have a term which is d delta p hat times a constant. Here we have a term which is d delta p hat times delta v hat. So delta v hat is small. We can neglect this term compared to this one in order to find the lowest order contribution to the impossibility. So I use the same equation, but now substituting in that delta v hat squared is 4 minus delta tau hat. So then what I have is a 3 times 4 times minus delta tau hat d delta v hat. Substituting in what delta v hat squared is here. And then over here I have equals minus 2 thirds d delta p hat. And then this term proportional to d delta v hat, which I'm going to have to keep because it's actually the same order as this term over here. So if delta p hat is 4 delta tau hat, then like before, this is going to be minus 28 thirds plus 16 thirds delta tau hat. And in other words, this last term is minus 4 delta tau hat d delta v hat. So see, we have two terms proportional to d delta v hat. They're both of order delta tau hat. So let's put them over to the same side of the equation. Delta p hat d delta v hat constant temperature. So these both have minus signs. And when I take this minus 4 and I put it over here, this becomes minus 8. This is 3 times 4 minus 8. Then I multiply by 3 halves and that becomes a 12. So this is equal to 12 delta tau hat. Now remember, this is supposed to be negative, but now delta tau hat is negative because the temperature is below the critical temperature. And now our expression for the inverse compressibility. Did I erase what it was before? Why didn't I go back over there? Anyway, I'll just have to remind you what it was before. If I consider minus v dp dv constant temperature, that's the inverse compressibility, before we had 6 times the pressure, before this was 6 delta tau hat but with a minus sign. Now it's 12 delta tau hat. So it's twice as big as before and has the other sign. So whereas before, we had 6 times the critical pressure times the deviation of the critical temperature from the temperature divided by the critical temperature when tau is less than tau c. Now it's twice as big, multiplying tau minus tau c divided by the critical temperature when tau is greater than tau c. So the reason the slope is different is, well, we have this additional contribution of the compressibility coming from the fact that the volume is changing as we get closer to the critical point going from below because these two points are getting closer and closer to the critical point going to coalesce when we get to the critical isotherm. So the 12 like the 6 depends on the model that the compressibility diverges or the inverse compressibility goes to zero with a factor of two difference in slope, that's a general prediction. Again, just coming from the cubic infection point. So there are two general predictions. So the coefficients 6, 12 depend on model, but more general, following just from the cubic infection point, description of this critical point are two things. First of all, that the compressibility, the inverse compressibility, goes like the deviation from the critical temperature to a power, well, I'll write it this way. The compressibility goes like the deviation from the critical temperature to a negative power. Our prediction is gamma is 1 and it blows up like tau minus tau c, coming from either side. But the second thing is that if I look at the inverse compressibility, it goes to zero at the critical point, it approaches it linearly from either side but with a different slope. So here, when we're approaching from below, of course, the slope is negative because it's going down to zero. And here, as we're approaching from above, the slope is positive. But the slope is twice as big here as on the other side. That's another universal prediction. Well, in fact, this observation that we get a different slope, a factor of two difference, that's supported by the experimental data as is the universality of the scaling of the isothermal compressibility as tau approaches tau c with some negative power. But again, our prediction for the power doesn't agree with the data. So universal behavior is seen, but gamma is about 1.3, not 1. So somehow we're missing some of the physics. We got some universal prediction, which was nice, but it's wrong, which is not so nice. Nevertheless, I'm daunted, let's forge ahead and make similar universal predictions in a different setting, just so we see that the idea of getting predictions that don't depend on the details of the model can be applied more generally. So now instead of the gas-liquid transition, let's consider ferromagnetism, because I'd like to make another wrong universal prediction. In the case of a magnetic system, I'd like to consider, quite analogously to what we discussed for the gas-liquid transition, the phase structure when we consider H, the applied magnetic field, which you can think of as being analogous to pressure, M, the magnetization, which is the same thing as the magnetic moment per unit volume, which you can think of as a response to the applied field analogous to volume. When we apply pressure to the fluid, it responds by being compressed. When we apply a magnetic field to a magnetic system, it responds by changing its magnetization. So the magnetization is an analogous to volume. Well, in the case of the fluid, the response to the applied pressure was described by the compressibility. In the case of the magnetic system, we'll consider instead the isothermal susceptibility, the derivative of magnetization with respect to applied field at constant temperature, the magnetic susceptibility. So that's a measure of how stiff the response is, how much the system responds when we apply the field, just like compressibility tells us how much the system responds when we push on it. Now, we have to be a little bit careful about minus signs, which are different here than before. Before, when we talked about work, we said the work done on the system could be written as minus PDV, because if we push on the system against pressure, we're reducing its volume, we're doing work on the system. And in the magnetic case, well, first of all, because I divided out the volume to define the magnetization, I should put back in a factor of volume. Otherwise, I'd be talking about the work per unit volume. But now it would be HDM with a plus sign. So, when we apply the field doing work on the system, it responds by turning on its magnetization lined up with the field. So, there's a large class of magnetic materials for which the response, the way the magnetization behaves as a function of an applied field, looks kind of like this. And the field is weak when H is small. The response is linear. By response, I mean how the magnetization turns on depends on the magnetic field. Now, there's linear response at weak field. When we turn the field off, there's no magnetization. When the field gets strong, the magnetization saturates. That's because the magnetization comes from the little magnetic moments in the material, all the spins, aligning with the magnetic field, and when the material is highly polarized with most of the spins pointing in the same direction, the magnetization can't get bigger. So, it saturates, which is a maximum or a protism maximum at large field or strong field. And that kind of behavior or a material that behaves this way is called a peromagnet. Peromagnet. This is called peromagnetism. The magnetization turns on when you apply the field. When you turn off the field, the magnetization goes away. But there's another type of magnetism called ferromagnetism named after iron, which is an example, where the magnetization in the most probable configuration, the equilibrium configuration as a function of the applied field, this by the way I should have said is an isotherm, so with some fixed value of the temperature, that's how the magnetization depends on the applied field. And a ferromagnet, you get something that looks like this, where right at zero applied field, there is coexistence of faces. In fact, I can get spontaneous magnetization. This is again an isotherm. Start out with a strong field, positive. I'm going to imagine we're talking about what I'll call a uniaxial magnet. So, I'm always considering the magnetic field to be up or down. And similarly, the magnetization is up or down. So, it's just a number which can be positive or negative, depending on whether it's up or down. When the field is large and positive, the magnetization is saturated and positive. We turn off the field slowly, so we stay in the most probable configuration in equilibrium. And when the field goes to zero, there's still a non-zero magnetization. That's ferromagnetism. On the other hand, if I turn on a large negative field, then the magnetization would saturate at a large negative value. And then, when I go along the isotherm, turning off the field and reach zero field, there would be a negative magnetization. So, right here, there's a first-order phase transition. There can be coexistence of faces at zero applied field. You can have a region where the spins are up, which interfaces with another region where the spins are down. They're both stable faces. And so, if you waited a while and you applied... Well, I don't want to go into the story of hysteresis and all that, but at any rate, there's coexistence of phases in the sense that there are two stable phases at the same value of the applied field when h is equal to zero. And what that should remind you of is that the point in the middle of those two points does not exist. The point in the middle is unstable in the same sense that the homogeneous phase, in other words, the homogeneous phase with no magnetization would be unstable. So, the only way to get zero magnetization would be a mixed phase where half of the sample has spin up and half has spin down. That's what I was drawing here. So that should kind of remind you if I turn it on its side so it looks more like this. Of the behavior of pressure versus volume in the case of a first-order gas-liquid transition where there's a discontinuity in the volume along an isotherm at a temperature less than the critical temperature, occurring at some coexistence pressure. So the picture looks like it's reflected. That's because of this minus sign. The fact that the magnetization responds by lining up with the applied field instead of the volume decreasing when you apply pressure. And so if I consider a sequence of isotherms with higher and higher temperature in the gas-liquid transition, the volume discontinuity would get smaller and smaller until finally we would reach a critical isotherm where the slope went to infinity right at zero, something like that. And the same thing happens here. If we lower the temperature, the spontaneous magnet... sorry, if we raise the temperature, maybe I said it wrong before too, if we raise the temperature, then the discontinuity in magnetization gets smaller and smaller. And in fact, there's going to be a critical point, a second-order phase transition where the ferromagnetism goes away, there is no spontaneous magnetization anymore. And right at the critical point, the magnetic susceptibility actually diverges. The slope, the magnetization as a function of field becomes infinite. Just like the compressibility does here. So in the gas-liquid case, we had a coexistence curve, and it could end. And at the end point of that line of first-order phase transitions, we had a second-order transition. And we have a similar phenomenon here with one significant difference. When the temperature is low enough, we have coexistence of phases at exactly zero field. But then, above a critical temperature, there is no spontaneous magnetization. And here, there is spontaneous magnetization. So two phases that are stable at zero applied field when T is less than Tc. And notice that the slope here of the coexistence curve in the H tau plane is zero. And according to the Clausius-Clapeyron relation, that means there's no latent heat. Remember, the slope is the discontinuity in entropy divided by the discontinuity in volume at zero. There's no discontinuity in entropy. There is a discontinuity in magnetization because there's spontaneous magnetization. So delta sigma equals zero, no latent heat. Same entropy in both the positive and negative magnetization phases because really they're just related by symmetry. If I turn my head upside down, that changes spin up to spin down so it has the same entropy. So we'd like to come up with a model kind of analogous to the Van der Waals model for describing the ferromagnetic transition and what happens at the critical point. So first, let's try to understand paramagnetism. So our model of paramagnetism is we don't consider the spins to interact with one another at all. Each one of the spins interacts independently with the applied field. A spin has a magnetic moment. So when we turn on the field, there's an energy splitting between spin up and spin down. I don't know. I guess I'll make the higher energy one spin up. And the splitting, if we apply field, I'm now going to call it B instead of H for a reason you'll see in a minute. B equals applied field. U equals magnetic moment of the spin. So the energy of the magnetic moment interacting with the field is magnetic moment times field. The difference between spin up and spin down in energy, the energy splitting is twice the magnetic moment times the magnitude of the applied field. And so if we're applying the field down, say, so that the, as I drew, the spin wants to align with the field, so spin down has lower energy. We could ask, you know, what's the fraction of all the spins that are up compared to the fraction that are down? The number per unit. No, the number up divided by the total number. So that's just given by a Boltzmann factor. E to the minus the energy difference to mu magnetic field divided by tau. And so the magnetization, which is just the magnetic moment, well, the magnetic moment's just add. So it's the number, total number of spins down minus the total number up times the magnetic moment per spin, which is mu, and some side of find it to be magnetization per volume, I should divide by volume. The number down minus the number up divided by the number down plus the number up. So using that Boltzmann factor and dividing numerator and denominator by the number down, this is just one minus the Boltzmann factor E to the minus 2 mu B over tau divided by 1 plus the Boltzmann factor. So if you multiply numerator and denominator by E to the mu B over tau, you see that it's just a hyperbolic tangent function. This is the hyperbolic tangent of mu B over tau. And the magnetization just goes like, I suppose little n means number of spins per volume. Then the magnetization would just be n times mu times this factor, tangent of mu B over tau. And that just looks like what I was drawing here. So we get magnetization as a function of applied field that looks like that. It saturates when most of the spins are lined up with the field. There's linear response when the field is weak. So in that model, all the spins are responding independently to the applied field. And we'll never get spontaneous magnetization that way. To get spontaneous magnetization, we have to include some interactions among the spins. Just like in the gas-liquid transition, we had to have the gas molecules want to stick together, be attracted at long distances in order to want to clump together to make droplets of liquid. And so here, we're going to want to have an interaction among the spins that make spins want to line up in the same direction. If spin one is pointing up, then the nearby spins are more likely to want to be spin up also. And so let's include that cooperative effect that a spin being up makes its neighbors want to also be spin up in sort of a simple way that's easy to analyze, which is this. Let's say that each spin is responding to a field, which is a combination of the applied field and the field induced by the average of the contributions to the fields of all the other spins. In other words, I will say that each spin sees an effective field, which is the applied field plus some positive number times the magnetization. So this is a contribution to field produced by other spins. So this is called, again, mean field theory. Maybe it's a little clearer now why it's called mean field theory than when I introduced the term before. It means that each spin is responding to the mean or average field produced by all the other spins. And it's not very accurate. I mean, it's not going to give us quantitatively believable results, but it's a model which gives us a way of seeing why ferromagnetism would occur and inferring some of the properties. And so the equation of state that we're going to study is this one, the magnetization behaving this way, but where now we replace B by this effective field, which includes a term depending on M, and that's going to make our equation a more interesting nonlinear one. So I'll just say the magnetization is N mu times hyperbolic tangent mu over tau H plus lambda M. So that's the equation of state relating M, H, and T. And to understand spontaneous magnetization, we want to ask what happens when we turn off the applied field and H is not here. And depending on the value of the temperature, one of two things will happen. We'll either have just one solution for the magnetization. That's the case in which there's no spontaneous magnetization. M equals zero is always a solution. But when the temperature gets low enough, we'll have three solutions. And that's the case in which the spontaneous magnetization will turn on. The stable solutions will have nonzero magnetization. And we'll talk more about that next time.