Statistical Mechanics - 6CCP3212 - King’s College London - November 2023 PDF

6CCP3212 Statistical Mechanics Lecturer: Prof. Eugene A. Lim Office : S7.09 King’s College London Department of Physics November 29, 2023 the historic entr...

6CCP3212 Statistical Mechanics Lecturer: Prof. Eugene A. Lim Office : S7.09 King’s College London Department of Physics November 29, 2023 the historic entry, “practically zero.” The notebook further records that the helium level stood quite still. The experiment continued into the late afternoon. At the end of the day, Kamerlingh Onnes finished with an intriguing notebook entry: “Dorsman [who had controlled and meas- ured the temperatures] really had to hurry to make the ob- servations.” The temperature had been surprisingly hard to control. “Just before the lowest temperature [about 1.8 K] was reached, the boiling suddenly stopped and was replaced by evaporation in which the liquid visibly shrank. So, a remark- ably strong evaporation at the surface.” Without realizing it, the Leiden team had also observed the superfluid transition of liquid helium at 2.2 K. Two different quantum transitions had been seen for the first time, in one lab on one and the same day! Three weeks later, Kamerlingh Onnes reported his re- sults at the April meeting of the KNAW.7 For the resistance of ultrapure mercury, he told the audience, his model had yielded three predictions: (1) at 4.3 K the resistance should be much smaller than at 14 K, but still measurable with his equipment; (2) it should not yet be independent of tempera- ture; and (3) at very low temperatures it should become zero within the limits of experimental accuracy. Those predictions, Kamerlingh Onnes concluded, had been completely con- firmed by the experiment. For the next experiment, on 23 May, the voltage resolu- tion of the measurement system had been improved to about 30 nV. The ratio R(T)/R0 at 3 K turned out to be less than 10−7. (The normalizing parameter R0 was the calculated resistance Figure 4. Historic plot of resistance (ohms) versus temper- of crystalline mercury extrapolated to 0 °C.) And that aston- ature (kelvin) for mercury from the 26 October 1911 experi- ishingly small upper sensitivity limit held when T was low- ment shows the superconducting transition at 4.20 K. ered to 1.5 K. The team, having explored temperatures from Within 0.01 K, the resistance jumps from unmeasurably 4.3 K down to 3.0 K, then went back up to higher tempera- small (less than 10–6 Ω) to 0.1 Ω. (From ref. 9.) tures. The notebook entry in midafternoon reads: “At 4.00 [K] not yet anything to notice of rising resistance. At 4.05 [K] not the cryostat—just in case the helium transfer worked. yet either. At 4.12 [K] resistance begins to appear.” The mercury resistor was constructed by connecting That entry contradicts the oft-told anecdote about the key seven U-shaped glass capillaries in series, each containing a role of a “blue boy”—an apprentice from the instrument- small mercury reservoir to prevent the wire from breaking maker’s school Kamerlingh Onnes had founded. (The appel- during cooldown. The electrical connections were made by lation refers to the blue uniforms the boys wore.) As the story four platinum feedthroughs with thin copper wires leading to goes, the blue boy’s sleepy inattention that afternoon had the measuring equipment outside the cryostat. Kamerlingh let the helium boil, thus raising the mercury above its 4.2-K Onnes followed young Holst’s suggestion to solidify the mer- transition temperature and signaling the new state—by its cury in the capillaries by cooling them with liquid nitrogen. reversion to normal conductivity—with a dramatic swing of the galvanometer. The first mercury experiment The experiment was done with increasing rather than To learn what happened on 8 April 1911, we just have to fol- decreasing temperatures because that way the temperature low the notes in notebook 56. The experiment was started at changed slowly and the measurements could be done under 7 am, and Kamerlingh Onnes arrived when helium circula- more controlled conditions. Kamerlingh Onnes reported to tion began at 11:20am. The resistance of the mercury fell with the KNAW that slightly above 4.2 K the resistance was still the falling temperature. After a half hour, the gold resistor found to be only 10−5R0, but within the next 0.1 K it increased was at 140 K, and soon after noon the gas thermometer de- by a factor of almost 400. noted 5 K. The valve worked “very sensitively.” Half an hour later, enough liquid helium had been transferred to test the Something new, puzzling, and useful functioning of the stirrer and to measure the very small evap- So abrupt an increase was very much faster than Kamerlingh oration heat of helium. 1 Onnes’s model could account for.8 He used the remainder of The team established that the liquid helium did not con- his report to explain how useful that abrupt vanishing of the  Acknowledgments As a statistical mechanical nincompoop, I have benefited from being able to learn from the many books and online lecture notes that exist, of which I have freely stolen and borrowed from for this set of lecture notes. In particular, I would like to thank David Tong for his excellent lecture notes that I have learned greatly from. Finally, I would like to thank Sophie for her patience and support. Figure on the previous page shows the historic plot of the discovery of superconductivity, by Heike Kamerlingh Onnes in 1911. The figure shows the resistance (Ohms) vs temperature (K). At 4.2 K, there is a sudden drop in the resistance from 0.1 Ohms to 10−6 Ohms, a phase transition signaling the onset of superconductivity. 2 Contents 1 Introduction and Review of Classical Thermodynamics 7 1.1 The problem of counting..................................... 7 1.2 The Laws of Thermodynamics.................................. 8 1.2.1 Zeroth Law and the equation of state......................... 9 1.2.2 1st Law of Thermodynamics.............................. 10 1.2.3 2nd Law of Thermodynamics.............................. 12 1.3 Thermodynamic potentials................................... 14 1.3.1 Free energies and thermodynamic potentials...................... 14 1.3.2 The Maxwell Relations................................. 17 1.3.3 Heat capacity and specific heat............................. 17 1.4 The chemical potential...................................... 18 1.5 Intensive/extensive variables and conjugacy.......................... 19 2 Statistical Ensembles 22 2.1 Phase space of microstates.................................... 22 2.2 The microcanonical ensemble.................................. 23 2.2.1 Statistical equilibrium and the 2nd Law........................ 25 2.2.2 Temperature....................................... 27 2.2.3 An example : Schottky Defects............................. 28 2.2.4 Heat bath......................................... 30 2.2.5 *Recurrence time..................................... 31 2.3 The canonical ensemble..................................... 32 2.3.1 The Boltzmann Distribution.............................. 34 2.3.2 The partition function.................................. 35 2.3.3 Entropy of canonical ensembles............................. 38 2.3.4 An example : Paramagnetism.............................. 39 2.4 The grand canonical ensemble.................................. 42 2.5 Some final remarks........................................ 44 2.5.1 Ensemble Averages of energy and particle number.................. 44 2.5.2 Discrete to continuous distributions.......................... 45 3 Classical Gas 47 3.1 Ideal gas.............................................. 47 3.1.1 Equipartition Theorem.................................. 49 3.1.2 An example : diatomic gasses.............................. 50 3.2 Maxwell-Boltzmann distribution................................ 52 3.3 Distinguishable vs indistinguishable particles......................... 54 3  3.4 Ideal gas as a grand canonical ensemble............................ 55 3.5 Non-ideal gas........................................... 56 3.5.1 Interacting gasses..................................... 57 3.5.2 Van der Waals forces................................... 58 4 Quantum Gas 63 4.1 Quantum states vs classical particles.............................. 63 4.1.1 Density of states..................................... 63 4.1.2 Relativistic vs non-relativistic particles......................... 65 4.1.3 Many particles vs many quantum states........................ 65 4.1.4 Indistiguishability part deux.............................. 66 4.2 Bosons............................................... 68 4.2.1 The Blackbody Radiation................................ 69 4.2.2 Phonons in solids : the Debye Model.......................... 74 4.2.3 Mode freeze-out..................................... 77 4.3 Fermions.............................................. 79 4.3.1 Non-relativistic fermion gas............................... 79 4.3.2 The classical limit : high temperature and low density................ 80 4.3.3 Low temperature fermions : degenerate fermion gas................. 82 4.3.4 Slightly less degenerate fermion gas........................... 85 5 Phase Equilibrium and Phase Transitions 88 5.1 Bose-Einstein Condensates : your first phase transition.................... 88 5.1.1 BEC phase diagram................................... 91 5.1.2 Heat capacities of phase transitions........................... 92 5.1.3 Real BEC......................................... 93 5.2 Overview of phases of matter.................................. 93 5.2.1 Phase equilibrium.................................... 95 5.2.2 The Clausius-Clapyron equation............................ 96 5.2.3 The Van der Waals model of phase transition..................... 98 5.2.4 Universality, order parameters and critical exponents................. 100 5.3 Phase transitions......................................... 102 5.3.1 Landau Theory...................................... 102 5.3.2 2nd order phase transition................................ 103 5.3.3 1st order phase transition................................ 105 5.4 The Ising model......................................... 107 5.4.1 Mean field theory..................................... 110 5.4.2 Connection to Landau Theory............................. 113 5.4.3 The missing physics of fluctuations........................... 115 A Some Mathematical Formulas 118 A.1 Stirling’s Approximation..................................... 118 A.2 Some useful integrals....................................... 118 4  What did we just sign you up for? This is a 3rd year required course in statistical mechanics taught at the Department of Physics, King’s College London. It is designed to be taught over a period of 10 weeks, for a total of 40 contact hours. I expect that we will not be able to cover the material presented in this lecture notes completely given the time – in this case I will assign some reading. Of course, if you can read ahead yourself, that’s even better. This set of notes is not meant to be complete – I have provided some recommended textbooks to go along with it. Additional material will also appear on the class website from time to time. Apart from the lectures themselves, and these notes, there will be 5 homework sets. Some of the homework problems will be alluded to in these lecture notes – you should absolutely do them as they will help you cover the gaps that inevitably exist. Like all physics courses, the key to mastery is to practice solving a lot of problems. So in addition to the homework sets, I will also provide some additional problems to help you along. Finally, a confession: this is the first time I am teaching statistical mechanics. Actually, it’s worse: I have never taken a statistical mechanics course in my life. So caveat emptor. 5  Recommended Books * is highly recommended for this course. Online resources will be posted on the class webpage. F. Reif, *Fundamentals of statistical and thermal physics., McGraw-Hill, 1965. This is a solid, rigorous book, and would make a good introductory standard text. It covers most of the basics very well, though due to its dated nature it falters when come to more advanced topics. Be careful when you use this though as it defines work as done by instead of on the system so there are a lot of minus sign differences when compared to these notes. F. Mandl, *Statistical Physics, 2nd Ed., Wiley 1997. This is another intro level book which takes a more verbose approach to the basics, but does not go very deep. It is a good companion to Reif. D. Chandler, Introduction to Modern Statistical Mechanics, Oxford University Press 1987. A good modern introduction, which covers most of the material of this lecture. It has a nice extended multi-chapter discussion on the basics of the Ising model without going too crazy with the math. L. Landau and E. M. Lifshitz, Statistical Physics, Part 1, 3rd Ed, Elsevier 1980. Ignore all the fear mongering about the “difficulty” of the Landau and Lifshitz 10-volume treatise of modern physics, this book is complete, clear and full of brilliant insight from one of the founders of modern statistical mechanics. It is aimed at a high level, but you will profit immeasurably even if you just dip into it. D. Goodstein, States of Matter, Dover 1985. This book is aimed at people who have already done a beginning statistical mechanics course, and would like to dig deeper into the more advanced topics. It is a very readable book for those who wanted to pursue some of the topics taught in this lecture further. Bonus : Goodstein is a good writer, albeit with a slightly warped sense of humour. (Solid-liquid transition is like pregnancy, really, David?) R. K. Pathria, Statistical Mechanics, Elsevier 1997. A PhD level book which covers the basics in detail with a very modern and excellent choice of advanced material. A good book to refer to if you want to investigate with further depth the topics taught in this module. D. Tong, *Lectures on Statistical Physics, University of Cambridge Part II Mathematical Tripos http : //www.damtp.cam.ac.uk/user/tong/statphys.html. Not a book, but might as well be. This is an excellent set of lecture notes (one of the many) by David Tong. It is aimed at 3rd year undergraduates, although at a slightly more mathematically inclined crowd. 6 Chapter 1 Introduction and Review of Classical Thermodynamics..we must learn how to count the number of states it is possible for a system to have or, more precisely, how to avoid having to count that number. D. Goodstein 1.1 The problem of counting The universe is big. By that I don’t mean in size (though it is quite large in size) – I mean it contains a lot of things. A small cup of tea contains about 5 moles of water, and that’s 3 × 1024 molecules of water. The Earth has about 7.5 × 1018 grains of sand. Anfield stadium at Liverpool can hold 5.4 × 104 football fans. Our Milky Way galaxy contains about 3 × 1011 stars. This lecture room contains about 6 × 1027 molecules of air (O2 , N2 and CO2 ). The permanent magnet that is on your refrigerator door contains 1018 little magnetic dipoles that arrange themselves to give you its collective magnetic property. Each of these individual things obey some equation of motion – laws that tells them how to move or behave. Stars will move according to the Newton’s law of motion, gas particles will move according to the laws of quantum mechanics, football fans will move depending on whether Liverpool is leading or losing, magnetic dipoles will flip depending on the presence of external magnetic fields governed by the Maxwell laws of electromagnetism etc. Of course, you have spent the last two years studying these laws (and perhaps a lifetime being a Liverpool football fan). We can perhaps solve the system when there are 2 or 3 things, but how on Earth are we suppose to solve a system of 1023 things? We can try to use a computer. Scientists have tried to simulate the motion of the “entire universe”, in order to understand how the galaxies and stars come to be distributed around the cosmos the way they are today. These simulations employs a lot of CPUs, of the order 106 (another big number!). The record holder so far (2017) is the TianNu simulation, which simulated 3 × 109 particles with 3.3 × 106 CPUs for 52 hours. Not even close to getting the behaviour of your cup of tea correct. Obviously, we do know a lot about how the universe actually works. We know that to make a good cup of tea, you need to boil water so that it is hot, and then dunk your favourite tea bag in. We know that we need to jiggle the tea bag, so that the flavour can diffuse into the water. Not only that, we can 7 calculate these processes – we know exactly how long the kettle takes to boil if we know how much heat we are adding to it. We can calculate how quickly the tea will diffuse into the water. In words, we mostly understand how the macroscopic physics of the world works. We don’t care about how each molecule in the cup of water behaves, all we need is its temperature. So instead of trying to solve the impossible 10N particle system, we can write down macroscopic laws – the Laws of Thermodynamics is the canonical example of such a set of laws, for example, although not the only one. Nevertheless, there are physical processes that defy our ability to attack it with macroscopic methods. Sometimes, the microphysics actually matter. For example, you are told to make sure your water is boiling before dunking your teabag in. How does water boil? Why does water boil? So, we are back to the problem of trying to count these things. This is where statistical mechanics come in. Statistical mechanics is the branch of physics of which we try to explain the macrophysics of a system with many things from its fundamental microphysics, without actually trying to solve the entire system of equations describing all the particles. But more than that, it is a study of how, when we put a lot of things together, interesting physics that are not manifestly “fundamental” can appear. Famously, we can derive the Laws of Thermodynamics by considering 1023 particles, not by considering 3 particles1. The history of the development of physics is rife with examples where deeper insight is gained when considering a phenomena from both its microphysics, its macrophysics and the interplay between them. The goal of this course is to teach you the basic principles of statistical mechanics. We will then use this new knowledge to dip our toes into some more advanced topics, at least in a introductory and not-so-crazy-deep way, to give you a sense of the wide applications of statistical mechanical techniques. Hopefully this will provide you with a base to explore further. 1.2 The Laws of Thermodynamics In the days of yore, when people do not quite understand the microscopic structure of matter, they did a bunch of experiments and came up with all sorts of definitions to figure out how matter behave. Furthermore, their experiments are blunt – they can only measure macroscopic quantities such as pres- sure P , volume V , work W etc. Despite these difficulties, remarkably they managed to come up with the correct laws that govern matter given these quantities, and these laws are known as the Laws of Thermodynamics. To specify these laws, we first introduce the idea of the thermodynamical state. A state is simply a set of variables that completely describe the condition of the system. For example, the state of a particle can be described by its position x and momentum p. We cannot of course describe a system by specifying all the positions and momenta of all particles as we have discussed earlier, but we can still assign some macroscopic variables to it. The simplest way of describing a thermodynamical state is to specify its pressure P and volume V – these are known as state variables, and the space of all possible (P, V ) is called the state space. Of course, a thermodynamical system can have more than 2 state variables, for example the tem- perature T , number of particles N , its entropy S (of which we will have plenty to say later) etc. An important point, which is still not obvious, is that the state variables are not independent variables. Think of the state variables as coordinates and the state space as the space where these coordinates span. The number of independent variables define the dimensions of the state space. Like regular coordinates (x, y), we can perform a change of coordinates to (x0 , y 0 ) in any way we like. As we will see below, P , V and T are not independent variables in general. Furthermore, we also need a couple of definitions, which you might be familiar with. The first definition comes from the idea of that we can insulate a system from all external influences so that no 1 Sometimes, such “macro” from “micro” concepts are called by the sexy moniker emergent. 8 energy or particles are exchanged between the system and the rest of the universe. The special insulator that achieves this is called an adiabatic wall, and a system which is enclosed in adiabatic walls is called an isolated system. Two systems separated by adiabatic walls will not interact with each other. Conversely, if two systems are allowed to remain in thermal contact and hence exchange thermal energy but not exchange particles, they are said to be separated by diathermal walls. Systems which can exchange thermal energy and do work but not exchange particles with the environment are called closed systems. If they can exchanged particles, then the systems are opened2. The second definition is that of equilibrium – an isolated system left alone for some (usually long) time will relax to a state where there is no macroscopic changes. This end state is called the equilibrium state. Unlike the first definition which are statements about the engineering of walls, this idea is actually a conjecture – why would a system wants to evolve into its equilibrium state? Indeed, this is a puzzle of which the study of statistical mechanics sought to shed light on, so we will come back to this later. For now, we take this as a given (as you have been taught for many years). Finally, before we continue, we can define the idea of functions of state, or state functions. Func- tions of state are any mathematical relations which relate the state variables of a system in thermodynamic equilibrium3. Given these, we can state the laws. 1.2.1 Zeroth Law and the equation of state The Zeroth Law states that if systems A and B are in equilibrium with system C, then they are also in equilibrium with each other. Now you might have learned in your Thermal Physics course that this implies that the temperatures for A and B are equal with the temperature for C. While this is true, it seems like a tautology: isn’t the being in equilibrium means that your temperatures are the same? Actually it is not: notice that we have specified the Zeroth law without specifying the notion of temperature – in fact it is the zeroth law that allows us to define a notion of temperature. To see this, let’s consider systems which can be described by their state variables pressure P and V. As we discussed earlier, in general a system can be described by more than 2 such state variables but let’s keep things simple for now. Consider systems A and C in thermodynamic equilibrium. This means that their thermodynamic variables PA , PC , VA , and VC must be related in some way, so this means there must exist some function FAC (PA , VA ; PC , VC ) = 0. (1.1) Equation Eq. (1.1) is a functions of state. Note that is is a constraints – it tells us that only three of the four possible thermodynamic quantities are free. Why are we allowed to write down such a function? We use the notion of equilibrium we described above, i.e. the state does not change with time, so dFAC /dt = 0, and hence FAC = const, and we can set the constant to zero WLOG. We further assume that we can solve Eq. (1.1) to obtain VC as a function of PA , VA an PC , i.e. VC = fAC (PA , VA ; PC ). (1.2) Similarly since systems B and C are also in equilibrium, we can find a similar relation VC = fBC (PB , VB ; PC ), (1.3) 2 Note that sometimes people say “closed” when they really mean “isolated” – and this include an older version of these notes! To get ahead of ourselves, entropy of a closed system can decrease but that of isolated systems cannot. 3 At the risk of getting ahead of ourselves, as a well known example, you recall that an ideal gas in equilibrium can be described by the equation of state P V = N kb T , where kb is the Boltzmann constant. This is a function of state. 9 which implies that fAC (PA , VA ; PC ) − fBC (PB , VB ; PC ) = 0. (1.4) But now, invoking the zeroth law, A and B is also in equilibrium, so there must exist an equivalent function of state FAB (PA , VA ; PB , VB ) = 0. (1.5) Since Eq. (1.4) must be the same as Eq. (1.5), it means that PC in the former must appear in such a way that it cancels. In other words there must exists some quantity which is common to both system A and B such that fAC ↔ TA and fBC ↔ TB , i.e. fAC (PA , VA ) − fBC (PB , VB ) = 0 ⇒ TA (PA , VA ) = TB (PB , VB ). (1.6) This quantity T , as you all must have guessed, is the temperature, and the equation which relates the thermodynamic quantities P and V (and possibly other quantities) to temperature T = T (P, V ) is called an equation of state. As an example, you have met the equation of state of an ideal gas PV T = (1.7) N kb where N is the number of particles and kb = 1.38064852 × 10−23 m2 kg s−2 K−1 is the Boltzmann constant. Note that since the equation of state relate the temperature T to the state variables P , V , N and possibly others, we can use it to replace a state one of the state variable with T. In other words, for an ideal gas P , V , T and N are not independent variables but is related by the equation of state Eq. (1.7) as we discussed earlier. Having said all that, we will come back when we discuss statistical mechanics proper and show you a much more remarkable definition of temperature. 1.2.2 1st Law of Thermodynamics The first law of thermodynamics is nothing but a statement on the principle of conservation of energy. Consider an isolated system A, then if we do work W on it, then its internal energy (or simply just the energy) E will change by ∆E = W. (1.8) Notice that the change in internal energy E is independent of what kind of work done to the system (you can stir it, compress it with a piston etc). Similarly, how the system “accommodate” this additional energy is not specified – it could be that the particles move faster, or they vibrate more vigorously if they are di-atomic etc. In any case, the state of the system will change. Hence, one can imagine the existence of a function of state for energy, E(P, V ). Suppose now that the system is not isolated, this means that not all the work done affects the internal energy, but “leaks” in or out, then we can write ∆E = W + Q (1.9) where Q is called heat. When Q = 0, the process is adiabatic. Note that despite the temptation, it is incorrect to write E = W + Q Wrong! (1.10) This is wrong because while E(P, V ) is a function of state (i.e. E depends on the state of the system), neither W nor Q are functions of state, i.e. you can’t write them as W (P, V ) nor Q(P, V ). The reason is obvious when you think a bit more about it : W is work done on the system, and Q is the energy that 10 got transfered in or out of the system via its diathermal walls, neither which has anything to do with the actual state of the system. At this point, we should now kill a pernicious misconception that is taught in high skool that seems to pervade into college: heat is not a form of energy, but a form of energy transfer. Indeed, both work and heat are different forms of energy transfer. Work usually refers to energy transfer through observable macroscopic degrees of freedom – piston moving, stirring with a spoon, magnetic work done by turning on an external magnetic field etc. Heat, on the other hand, refers to energy transfer through microscopic degrees of freedom (the kinetic energy or vibrational individual particles). This is the “word” version of why equations like E = W + Q is wrong – we cannot subdivide up the internal energy E into “heat” energy and “work” energy. However, since heat and work are energy transfers, it makes sense to think of them as infinitisimal changes to the internal energy. We can then rewrite Eq. (1.9) in differential form known as the 1st Law of Thermodynamics, dE = d¯Q + d¯W (1.11) where we have written the differentials d¯Q and d¯W instead of dQ and dW to emphasise the fact that Q and W are not functions of state, and hence not expressible as Q(P, V ) or W (P, V ). Such differentials are called inexact differentials. Exact vs Inexact Differentials Consider a differential dF , dF = A(x, y, z)dx + B(x, y, z)dy + C(x, y, z)dz. (1.12) If furthermore, there exists some scalar F (x, y, z) such that       ∂F ∂F ∂F A(x, y, z) = , B(x, y, z) = , C(x, y, z) = , (1.13) ∂x y,z ∂y x,z ∂z x,y then we say that dF is an exact differential. The implication is that the differential Rx dF can be integrated to obtain a value x10 dF = F (x0 , y0 , z0 ) − F (x1 , y1 , z1 ). In other words, the integral only depends on the beginning and end point of the boundaries and do not depend on paths. For example, consider the exact differential dF = xdy + ydx, then R x0 ,y0 x1 ,y1 dF = x0 y0 − x1 y1 where F (x, y) = xy. It then simply follows that the loop integral of an exact differential is always identically zero I dF = 0. (1.14) On the other hand, if there exists no scalar function F (x, y, z), then the differential Eq. (1.12) is called inexact. Inexact integrals depend on the paths of the integration. In the context of classical thermodynamics, if dF is exact then F is a function of state, otherwise F is not. Since dE is an exact differential, we can express it as     ∂E ∂E dE = dP + dV, (1.15) ∂P V ∂V P which is a trivial consequence of the fact that E(P, V ) is a function of state. Integrating this equation between two states (P1 , V1 ) and (P0 , V0 ) gives us the total internal energy change E(P1 , V1 ) − E(P0 , V0 ), which is independent of paths. 11  Let’s now consider the inexact differential d¯W , where you have learned from your thermal physics course can be given by d¯W = −P dV (1.16) where P is the pressure and dV is the infinitisimal volume change4. Since P, V are our state variables, it’s clear that we cannot find a scalar function W (P, V ) such that dW = −P dV (you can try). There is a negative sign because we have defined d¯W to be work done on the system, so if a positive change in V means that system has expanded and done work on the environment. We can still integrate this of course, but the integral will now depends on the path of the process (Fig. 1.1). Figure 1.1: The work done ∆W between two states (P0 , V0 ) and (P1 , V1 ) is dependent on the path. Indeed, if we make a loop, say flip the direction of the arrows of path a or b in Fig. 1.1, then the loop integral I I d¯W = − P dV 6= 0, (1.17) a calculation of which you have done many times in your thermal physics course, e.g. if the loop is clockwise, then the system converts heat to work and is an engine. 1.2.3 2nd Law of Thermodynamics Since the 1st Law is just a statement on the conservation of energy, it does not tell us how a state will evolve. A system in equilibrium will remain so if we leave it alone, but it will not remain in equilibrium if we do something to do it – we can do work on it, apply heat to it, remove adiabatic walls etc. If we are gentle and careful however, we can change the state of the system slowly, making sure that while the state is changing (e.g. its P and V is changing with time), it remains in equilibrium. Hence the state travels around its state space, remaining in equilibrium at every point – such a process is called reversible. As its name implies, it means that we can happily (and gently) reverse the process, going back to its original state. On the other hand, in general, changing the state of a system will move it far away from equilibrium. Once this occurs, we cannot undo this process – such processes are called irreversible. Of course, such a definition seems like tautology: “irreversible processes are those that are not reversible” doesn’t really explain any physics. To truly understand this, we would need to understand how states evolve from non-equilibrium towards equilibrium – we will discuss this when we have the tools of statistical mechanics. For now, let’s consider a hand-wavy argument. Consider a thermally isolated system, with two compartments V1 and V2 separated by a physical adiabatic partition in the middle which can be opened 4 Be warned that the sign convention we have used is not universal! Some references uses d¯W = P dV , so W defines work is done by the system. 12 or closed (see Fig. 1.2). Compartment V1 is initially filled with an ideal gas at temperature T and pressure P , while compartment V2 is empty. Now we play a game : without explicitly tracking the motion of each particle, we are asked to guess which compartment all the particles are, and we get a point for each correct answer. Of course, initially, we will ace the game: all the particles are in container V1 and we get full marks. However, if we then open up the adiabatic partition betweem the two compartments, the gas will now expand and fill up both compartments. The gas then fills up the entire container and hence is changing, but as we noted earlier, it will evolves towards equilibrium after some time. Playing the same game again, it is clear the game is now harder, since for each particle, we have roughly a 50% chance of getting the answer correct. In other words, our knowledge of where the individual particles have decreased. This decrease in our knowledge, or equivalently, the increase in the disorder, means that we cannot reconstruct for free the original state since we have “forgotten” some information. Such a process is hence irreversible. Of course, this does not mean that we can no longer go back to the original state, to do that we have to painstakingly hunt down each particle and move them physically back to V1 – i.e. doing work5. Figure 1.2: A thermally isolated container with two compartments, separated by an adiabatic wall which can be opened or closed. Figure 1.3: A gas inside a piston is allowed to expand as we add heat d¯Q to it such that it is reversible. We can quantify this loss of knowledge, or increase in the disorder, with a quantity S, called entropy. In an isolated system as shown in Fig. 1.2, this implies that ∆S ≥ 0 for thermally isolated systems. (1.18) On the other hand, consider a system which is not isolated and undergoes a reversible infinitisimal process which it absorbs some quantity of heat d¯Q. For example, a gas inside a piston expands by dV via the transfer of an infinitisimal quantity of heat d¯Q (see Fig. 1.3), keeping the temperature T fixed. We then define the infinitisimal change in entropy d¯Q dS = for reversible processes. (1.19) T 5 You might ask: how do we regain knowledge by doing work? As it turns out, there is a very deep connection between information and work – it can be shown that possession of a bit of information can be converted into work via the Szilard’s Engine. We won’t discuss this, but it is well worth checking it out yourself. 13 Combined, the pair of equations Eq. (1.18) and Eq. (1.19) is the 2nd Law of Thermodynamics. Notice that in Eq. (1.19), we have written dS instead of d¯S – in other words, dS = d¯Q/T is an exact differential. This means that the entropy S is a function of state – a much more useful thing. But wait! How do you prove that this is true? As it turns out, in “pure” classical thermodynamics, there is no satisfactory answer. However, in statistical mechanics, there is a very clear answer – we will learn that we have gotten the whole definition the wrong way round as Eq. (1.19) is not a definition of S but instead can be thought of as a definition of T in a reversible system. Entropy S is defined to be S ≡ kb ln Ω(E, V, N,... ) (1.20) where Ω is called the statistical weight of the system and is a function of all the independent state variables of the system, and hence S is a function of state. Anyhow, we have gotten ahead of ourselves, and this will be clear when we tackle statistical mechanics proper in the next chapter. Since S is a function of state and hence its integral only depends on the boundaries, then its loop integral vanishes identically viz. d¯Q I I dS = = 0 for reversible processes. (1.21) T Eq. (1.21) is known as the Clausius Equality. As you will be asked to show in a Homework problem, the Carnot cycle which you have studied in great detail in your Thermal Physics course obeys this equality as it is a reversible cycle. What is the relationship between d¯Q and dS in the case of irreversible processes? Consider a non- isolated system undergoing some process where we inject some quantity of heat d¯Q at some temperature T , evolving in time t. If the process is reversible, then the following holds according to Eq. (1.19) d¯Q dS =T. (1.22) dt dt However, if the process is irreversible, then according to Eq. (1.18), its entropy can only increase regardless of whether we inject any heat or not, and hence for a general process d¯Q dS Tvib all degrees of freedom become active. Not only that, the “activation” is sudden – there is no active rotation degrees of freedom until Trot is hit and then all the particles began to rotate. We will come back to discuss this in section 4.2.3 after we have build up some quantum mechanical muscle in the next chapter. 51 Figure 3.2: The heat capacity CV of a diatomic gas as a function of temperature. Instead of the classically predicted CV = (7/2)kb , CV stays constant at (3/2)kb at low temperatures, until some fixed temperature Trot where it jumps to (5/2)kb. It stay at that value until Tvib is reached, where it then jumps to the predicted value of (7/2)kb. (Figure stolen from Wikipedia.) 3.2 Maxwell-Boltzmann distribution Before we continue on with new things, let’s take a moment to rederive an old friend, the Maxwell- Boltzmann distribution of molecular speeds of a gas at temperature T. Assuming that the gas is held at fixed temperature in a diathermal box (i.e. no particles are exchanged), so we are dealing with a canonical ensemble. As we have learned from section 2.3.1, the probability of finding a microstate of energy E is given by the Boltzmann distribution Eq. (2.44). Now instead of considering a gas, we consider the canonical ensemble of single particle of mass m. The kinetic energy for this particle is E = mv 2 /2 with v ≡ |v| being the absolute velocity of this particle as usual. The probability of finding a particle at absolute velocity v is then given by (following Eq. (2.44)) 2 f (v)e−mv /2kb T P (v) = (3.29) Z with the partition function being the sum over all possible states as usual. So the probability of finding the particle between the velocities v and v + dv is the integral 1 2 P (v)dv = f (v)e−mv /2kb T dv. (3.30) Z Our goal is now to compute f (v). To do this, we again deploy the omnipotent partition function (in integral form) 1 Z Z Z = d x d3 p e−E/kb T 3 (2π~)3 Z V 2 = 3 d3 p e−mv /2kb T , (3.31) (2π~) where we have done the volume integral as usual. We now need to convert the 3-momentum integral d3 p into an integral over the absolute velocity – for every value of v, there exist many possible combinations 52  √ √ of v = (vx , vy , vz ) such that v = |v|, e.g. v = (v, 0, 0), v = (0, v, 0), v = (0, v/ 2, v/ 2) etc. This traces out a locus of a 2-sphere with radius v, and thus d3 p = m3 d3 v = m3 v 2 dvdΩ. R RR The solid angle integral dΩ = sin θdθdφ = 4π as usual, so Eq. (3.31) becomes 4m3 πV Z 2 Z= dv v 2 e−mv /2kb T. (3.32) (2π~)3 This means that the Maxwell-Boltzmann distribution must be f (v) ∝ v 2 , or 2 2 f (v)e−mv /2kb T dv = Av 2 e−mv /2kb T dv , (3.33) The normalization constant A can be computed by insisting that the total probability Z ∞ 2 Av 2 e−mv /2kb T dv = 1 , (3.34) 0 which gets us r  3/2 2 m A=. (3.35) π kb T Putting everything together, we get the famoous Maxwell-Boltzmann distribution as promised r  3/2 −mv 2 /2kb T 2 m 2 f (v)e = v 2 e−mv /2kb T. (3.36) π kb T Figure 3.3: The Maxwell-Boltzmann distribution. Notice this is the velocity distribution for a single particle. However, if we assume that the particles Q are non-interacting, so that the partition function simply multiplies Z, it’s easy to show that the entire calculation follows through, and Eq. (3.36) remains the same. In other words, the distribution do not depend on density. Of course, real gasses’ particles interact with one another, and as we will see later in section 3.5, the partition function no longer multiplies and distribution will be dependent on the density. Nevertheless, the Maxwell-Boltzmann distribution is an excellent model for low density systems. Let’s take a moment to reacquaint ourselves with this distribution (Fig. 3.3). For a given fixed m and T , the distribution of the velocities at small v is dominated by the v 2 term since the exponent 2 limv→0 e−v → 0 so it rises until some maximum vmax which can be found by setting d h 2 i f (v)e−mv /2kb T =0, (3.37) dv vmax p which you can show in a homework problem to be vmax = 2kb T /m. 53  Notice that there the probability drops to zero at v = 0 – the probability of finding a particle that is standing still is effectively zero. At high v on the other hand, the exponential starts to dominate, and we experience an exponential tail off. This tail off has a “half-life” that scales as T. So the higher that temperature, the slower the tail-off, i.e. the particles are more likely to be at higher velocities comparatively which is exactly as one intuitively would have guessed. Like all distribution functions, we can use it to compute averages of quantites that are functions of v. For example, the mean velocity of the particle is given by Z ∞ 2 hvi = dv vf (v)e−mv /2kb T r0 8kb T =. (3.38) πm We can also calculate the mean kinetic energy of the system of particles using Gauss’ integrals (see Appendix A.2) 1 hEi = mhv 2 i 2 Z ∞ 1 2 = m dv v 2 f (v)e−mv /2kb T 2 0 3kb T = , (3.39) 2 which is exactly the same answer we have gotten from equipartition arguments in section 3.1.1 for particles with 3 translational degrees of freedom – not surprising because we are using the same distribution. The Maxwell-Boltzmann distribution is historically the first statistical law of physics – it was derived by Maxwell in 1859 from the principles of the Kinetic Theory of Gas which postulates that gas can be described as a large number of particles in random motion. This theory, which you have studied in your Thermal Physics course, is the progenitor of statistical mechanics, and is still widely applicable in the description from the universe from the evolution of the universe to describing fluids in your bath tub. 3.3 Distinguishable vs indistinguishable particles We now have a confession to make. The partition function for the ideal gas we have derived in section 3.1, Eq. (3.9),  3N/2 mkb T Z =VN , (3.40) 2π~2 is actually incorrect. The reason is that we are overcounting the microstates in the following way. When we derived Eq. (3.9), we have simply multiplied the partition function of N different single particle partition functions. This procedure makes the extremely important (and incorrect) assumption that these N particles are distinguishable, i.e. there are labels on them numbering them from 1 to N. But in reality, at the quantum mechanical level, the particles of a real gas is indistinguishable – e.g. every hydrogen molecule is exactly the same as every other hydrogen molecule. This point is illustrated in Fig. 3.4. In general, for N indistinguishable particles, we will overcount the number of microstates N ! times, so the partition function should be modified to 3N/2 VN  Z mkb T Z̃ → =. (3.41) N! N! 2π~2 We emphasise that particles that can be labeled in some way – say same-colored billiard balls – are not indistinguishable. Indistinguishable particles cannot be labeled at all. For example, in Fig. 3.4, the right hand set is not “6 identical sets”, but they are exactly the same set. 54 Figure 3.4: Indistinguishable vs distinguishable particles: if 3 particles are distinguishable then we can label them (left plot, blue, green and red), we can arrange it in 3! = 6 different configurations or microstates. However, if they are indistinguishable we cannot label them (right plot, all pink), there is only one configuration or microstate. Fortunately, many of the results that we have computed such as that of the Maxwell-Boltzmann distribution in section 3.2 does not change bacause the factorial factor only changes the normalization. On the other hand, any quantity that directly requires the actual number of microstates may require modification. The most important of this is of course, the entropy, Eq. (2.70) S = kb (ln Z̃ + β Ē). (3.42) Now the mean energy does not change ∂ ln Z̃ Ē = − ∂β ∂(− ln N ! + ln Z) = − ∂β ∂ ln Z = − , (3.43) ∂β since the log makes the N ! factor additive and the partial then gets rid of it. This is a mnemonic to remember when figuring out which equations are affected. On the other hand, the first ln Z̃ term now gets this extra term S = kb (ln Z̃ + β Ē) = kb (ln Z − ln N ! + β Ē). (3.44) Hence the entropy is smaller of ln N ! – this is not surprising since in reality there are less possible configurations. Finally, you might wonder about the case of the paramagnet we studied in section 2.3.4 – aren’t electrons also indistinguishable particles? You would be right, but in that case, the electrons are held in place in a lattice – hence we can label them with the lattice coordinates, rendering them distinguishable. 3.4 Ideal gas as a grand canonical ensemble In the section on the ideal gas section 3.1, we have considered the canonical ensemble, which means the particle number N is constant. In general, this can change – for example, in the atmosphere of the sun, the number density of the hydrogen atoms changes as nuclear processes fuse them into helium. 55  As usual, we begin with the grand canonical partition function Eq. (2.102), X Z(µ, T, V ) = e−β(Er −µNr ) r X = e−β(Er −µNr ) Er ,Nr X = Z̃(Nr , T, V )eβµNr , (3.45) Nr where Z̃ is the canonical partition function of the ideal gas (taking into account indistinguishable parti- cles), Eq. (3.41), so we can just copy down the answer X V Nr Z(µ, T, V ) = eβµNr , (3.46) Nr !λ3Nr Nr where we have used the thermal de Broglie wavelength Eq. (3.12). This terrible looking sum can actually be easily computed in the limit of very large N , using the series n=∞ X xn x2 ex = =1+x+ +... (3.47) n=0 n! 2! so it sums to our final expression for grand partition function of the ideal gas,  βµ  Ve Z(µ, T, V ) = exp. (3.48) λ3 With Eq. (3.48), we can then roll out all the thermodynamical quantities as we have seen in section 2.4. The mean particle number is 1 ∂ ln Z V eβµ hN i = = , (3.49) β ∂µ λ3 so the chemical potential of the ideal gas is given by hN iλ3   1 µ = ln. (3.50) β V Now recall that λ3 is roughly the de Broglie wavelength, so is the length scale when quantum effects are important. Meanwhile, V /N is the “available” volume for each particle, so to trust our classical calculation of the ideal gas, λ3  V /N – this means N λ3 /v  1 – i.e. µ < 0. In other words, the chemical potential for an ideal gas is negative. There is nothing wrong with it, though in general µ can be both positive and negative, depending on the system. To derive the equation of state, we use the Landau potential, from Eq. (2.114) and Eq. (1.80), Ψ = −kb T ln Z = −P V (3.51) but ln Z = V eβµ λ−3 = hN i from Eq. (3.49), and the equation of state immediately follows. 3.5 Non-ideal gas We have so far considered ideal gasses, of which the particles do not interact. They are very good approximations for situations when the gas density is low. In this section, we will discuss how we can model “real gasses”, which can have various forms of interactions. 56 3.5.1 Interacting gasses To begin, let’s consider a system of N indistinguishable particles of mass m which interact with one another. As should be familiar by now, we want to construct its partition function. The Hamiltonian (or energy) of this system is X p2 X i H= + U (rij ) , (3.52) i 2m i>j | {z } | {z } position only momenta only where U (x1 , x2 ) is the interacting potential between any two particles with position xi and xj , which depends on the distance between the particles rij ≡ |xi − xj |. (3.53) Notice that the Hamiltonian Eq. (3.52) is a sum of terms that depends on momentum and terms that depend on position, and we sum over all i and j with the condition i > j so that we avoid overcounting. The partition function is in general an integral over all the momenta and coordinates, 1 1 Z Y Z Y Z= d3 pi d3 xj e−βH. (3.54) (2π~)3N N ! i j d3 xi and d3 pi appear so often that it is very convenient to introduce Q Q The product integral measures i i a shorthand Y Y d3 xi ≡ Dx , d3 pi ≡ Dp. (3.55) i i Now since we can split the Hamiltionian into momentum only and position only parts, we can separate the integrals (using the shorthand U (rij ) = Uij ) Z  Z  1 1 −β i p2i /2m P P −β j>k Ujk Z = Dp e × Dx e N ! (2π~)3N Z  1 1 P = 3N Dx e−β j>k Ujk. (3.56) N! λ We have used the ideal gas result Eq. (3.9), but without the “trivial” position integral as it is no longer trivial in the presence of the potential U (rij ). Now the Helmholtz free energy for the ideal gas is Eq. (2.74) Fid = −kb T ln Zid VN   = −kb T ln λ3N N !     1 N = −kb T ln + ln V. (3.57) λ3N N ! The Helmholtz free energy for the interacting gas is then F = −kb T ln Z     1 Z P −β Ujk = −kb T ln + ln Dx e j>k λ3N N !   1 Z P −β j>k Ujk = Fid − kb T ln Dx e VN = Fid − kb T × I , (3.58) where we have expressed the remaining integral as   1 Z P −β j>k Ujk I ≡ ln Dx e. (3.59) VN 57 This integral encodes the change to the system due to the presence of the interactions. In general, it is an incredibly difficult integral to perform since all the particles may interact with every other particles. To proceed, we want to understand the nature of the interactions, so let’s look at Uij closely. 3.5.2 Van der Waals forces Consider two particles (say the molecules of some gas), the Hamiltonian is (p21 + p22 )/2m + U (r12 ). r12 = |x1 − x2 | ≡ r is the distance between the particles. Physically, we know that the particles cannot be infinitisimally close to each other, since Pauli’s exclusions principle will prevent them from occupying the same space. Suppose r0 is roughly the radius of the particle , then we can roughly model this behavior as U (r) → ∞ for r < r0 , (3.60) i.e. it is infinitely expensive energetically for the particles to overlap r < r0 – hence making the potential repulsive at short distances. Such an infinite core is called a hard core. Meanwhile, from experiments, we notice that there exists a very weak molecular attractive force between molecules that are neither ionic or covalent, called Van der Waal’s forces. This force vanishes at as r → ∞, so we can model it as a potential  r 6 0 U = −U0 , (3.61) r which is sometimes called the London potential. Van der Waal’s forces, as you studied in your high skool days, describe the very weak but present forces between the molecules of a gas. The origins of the Van der Waals forces are quantum mechanical – technically, they are a sum of all vacuum polarizations multipoles between molecules. Combining Eq. (3.60) and Eq. (3.61), we get ( ∞ , r < r0 U (r) = r0 6
. (3.62) −U0 r , r > r0 This potential is sketched in Fig. 3.5. Figure 3.5: The potential for the Van der Waal’s forces which consists of a hard core model with weak attractive London interaction. Given this potential, let’s see how we can perform the integral Eq. (3.59) using a very clever trick. The exponent e−βU contains all the possible interactions between all the particles, but we can make use of the fact that since we are dealing with gasses with low densities, the most likely interaction is a single interaction between two particles, followed by the less likely two simultaneous interactions either between 3 particles or between two pairs of particles, and even rarer n > 3 simultaneous interactions. So what we want is some kind of formal expansion of the integrand which will allow us arrange the integration along such n-interactions. We implement this trick by introducing the following change of variables f (rij ) ≡ e−βU (rij ) − 1. (3.63) 58 This is a very clever function3. Notice that for any pair of particles labeled i and j, if they are very weakly interacting when they are far away (at rij → ∞), then since Uij → 0, f (rij ) = e0 − 1 → 0. If they are close rij → 0 and strongly interacting, then Uij → ∞, and f (rij ) → −1. This means that when we do RQ 3 the i d xi integral over all the coordinates, they will only contribute when around the domain where xi ≈ xj – a fact we will make use of very soon. Using this function, we can then expand the exponent as (using the shorthand fij ≡ f (rij )) P Y e−β j>k Ujk = (1 + fjk ) j>k X X X = 1+ fjk + fjk flm + fjk flm fqw +... j>k j>k,l>m j>k,l>m,q>w = 1 + (f12 + f13 + · · · + f23 + f24... ) +(f12 f23 + f13 f34 +... ) + (f13 f23 f34 +... ) +.... (3.64) Thus one can interpret fij as the contribution to the partition function of the interaction between particle P i and j. The linear terms f12 + f23 + · · · = j>k fjk are the interactions of two particles, and as the sum makes clear, there are N (N − 1)/2 pairs of these. The next to leading order term f12 f23 + f13 f34 + · · · = P j>k,l>m fjk flm are the interactions of either 3 particles or simultaneous interactions of two pairs of particles (e.g. f12 f34 ) etc. Such an expansion is called a cluster expansion and is a very useful device which you will see again and again in any calculations involving the partition functions (e.g. in quantum field theory). Using our intuition that the most dominant interactions will be for a single pair of particles4 , we will drop all the higher order f terms and keep just the linear term. The integral I then is   1 Z P −β j>k U (rjk ) I = ln Dx e VN   1 Z Y = ln  N Dx (1 + fjk ) V j>k    1 Z X = ln  N Dx 1 + fjk +...  (3.65) V j>k The “1” integral is trivial Dx = V N. Our introduction of the f expansion has cleverly changed our R integrand from a crazy product of many exponents into a sum of linear fjk terms – where each fjk is the contribution of the interaction of a single pair. Better still, each pair interaction is exactly identical as any other pairs, and the sum gives us N (N − 1)/2 such pairs in the entire integrand, with the total ! N (N − 1) Z X Z Y 3 Dx fjk → d xi fjk , 2 i j>k for any choice of j and k. Since N is very large, we can simply approximate N (N − 1) ≈ N 2. Now each interaction, fjk = f (rjk ) only depends on the coordinates of two particles, we can trivially integrate all the coordinates xi except for when i = j, k giving us a factor of V N −2. Putting everything back together, the integral I is N2  Z  3 3 I = ln 1 + d x j d x f k jk. (3.66) 2V 2 Since we can always fix the coordinate of one of the particle, say the k particle, and express the j particle R∞ as distance from k, r = |xj − xk | then d3 xj d3 xk → V 0 d3 r, where the factor of V comes from the R R 3 It’s called the Mayer f function for those who likes fancy names. 4 Ifyou have spotted that we have brazenly ignored simultaneous interactions of 2 pairs of particles, e.g. f12 f34 , then well done. This is an important point we will come back to a bit – see footnote 5. 59 fact that we integrate over all possible locations for the k particle , we can further reduce I into a single integral N2 ∞ 3  Z  I = ln 1 + d r f (r). (3.67) 2V 0 We can proceed with this integral if we like, but now is a worthwhile time to stop doing math, and think about physics a bit. As we said earlier, f (r) only contributes to the integral when the particles are near each other. Looking at the potential in fig. 3.5, it’s clear that f only contribute when r. r0 , so the volume integral d3 rf (r) ∼ −r03 (recall that f (r) → −1 at small distances). This means that (ignoring R signs) N2 N 2 r03 Z d3 r f (r) ∼. (3.68) 2V V Now N/V is the number density of the gas, and since r03 is roughly the volume of the particle or molecule, 1/r03 must be the number density of the particles if we pack them together touching each other! Thus the combination N r03 1, (3.69) V must be true – it’s just the statement that particles of a gas are not packed edge-to-edge. But, we further assert that the combination N 2 r03 1, (3.70) V is also true. This is less obvious – why are we allowed to do this? Indeed, at first glance, it seems wrong. For example, a mol of hydrogen occupy 22.4 m3 as usual, while molecular hydrogen has the radius r0 = 1.2 × 10−10 m. N = 6.02 × 1023 for a mol, so N V /r03 = 4.6 × 10−8. But then N 2 V /r03 ∼ 1015 , which is very large! What gives? The “formal” answer is that we assert that we are only doing this integral for small quantities of gas, i.e. when N  108. The reason we are allowed to do this is because we know that the free energy is extensive, i.e. they are additive Ftotal = F1 + F2 , as we have seen in section 1.5. Thus, once we have computed F for a small quantity of gas, it’s easy to simply sum them up5. With this further assumption, we can then use the expansion ln(1 + x) ≈ x for x  1 to finally write down the integral we have to do N2 Z I= d3 r f (r). (3.71) 2V This is now looking much more promising! Recalling that the f function has the property f (r) → −1 at r. r0 while f (r) → 0 at large r, we break the integral into two parts N2 Z I = d3 r f (r) 2V Z r0 Z ∞ N2  = d3 r f (r) + d3 r f (r). (3.72) 2V 0 r0 For the Van der Waals potential Eq. (3.62), the 1st integrand above is f (r < r0 ) = −1 since U (r < r0 ) → ∞, we get Z r0 Z r0 4πr03 d3 r f (r) = 4πr2 (−1)dr = −. (3.73) 0 0 3 5 We have to assume that the subsystems do not interact of course, but this is explained by something we have swept under the rug when we do the expansion in Eq. (3.64). Some of you might have noticed that some 2nd order terms (e.g. f12 f34 ) interactions are actually between 2 pairs of interactions. Now, these are rare as we asserted, but in a sufficiently large volume of gas, they can occur. Since F is extensive, we are in danger of undercounting the number of interactions. The solution is simply to restrict the volume (and hence the number of particles N ) such that we are safe to assume that there is only a single interaction in the entire volume at any one time. This restriction is equivalent to our assumption that N 2 r03 /V  1. 60 Meanwhile, the third term f (r  r0 ) = e−βU (rr0 ) − 1, and assuming (see below) βU is small in this range, we can expand e−βU − 1 ≈ (1 − βU +... ) − 1 = −βU +... where the dots denote terms of O(β 2 U 2 ) and higher. The 3rd integral then becomes Z ∞  r 6 0 4πr03 βU0 4πr2 dr = βU0. (3.74) r0 r 3 Putting everything together, the final integral is then N 2 4πr03 I= (βU0 − 1). (3.75) 2V 3 Inserting this back into Eq. (3.58), we obtain the final answer for the Helmholtz free energy N 2 kb T 2πr03 F = Fid − (βU0 − 1). (3.76) V 3 To find the equation of state, we use Eq. (1.32), P = −(∂F/∂V )T ,    N kb T N a P = 1− −b , (3.77) V V kb T where we have defined the coefficients 2πr03 U0 2πr03 a≡ , b≡. (3.78) 3 3 Rearranging Eq. (3.77), −1  N2 N2     V N V kb T = P + 2a 1+ b ≈ P + 2a −b (3.79) N V V V N where we have expanded the last term, again using the fact that N b/V ∼ N r03 /V  1 to Taylor expand the second bracket. Eq. (3.79) is the famous Van Der Waals equation of state, which you may have encountered in high skool. We have glossed over a few things in the derivation, so let’s make some parenthical remarks here: High temperatures: In deriving Eq. (3.74), we have assumed that βU0 = U0 /(kb T )  1. This means that the temperature T  U0 /kb. As an example, for Argon gas interactions, U0 /kb ∼ 400 Kelvins. R∞ Short range forces only: When we integrate the potential in Eq. (3.74), 0 r2 (r0 /r)6 dr, the London potential scales as 1/r6 , while the integral scales as r3. This means that when we take the upper limit r = ∞, it is zero and hence the integral converges. However, it is clear that if the potential has been longer range, i.e. U (r) ∝ r−n where n ≤ 3, then the integral diverges. For example, the Coulomb potential scales like 1/r so this derivation will not work. Low densities: As we have explained earlier, we have used the fact the following low density as- sumption N 1  3. (3.80) V r0 This suggests that we can use the parameter N/V an expansion parameter, where higher order effects are proportional to O(N/V )n. This allows us to motivate the following virial expansion of the equation of state "  2  3 # P N N N N = 1 + B1 (T ) + B2 (T ) + B3 (T ) +... , (3.81) kb T V V V V 61  where Bi (T ) are known as virial coefficients. For the Van Der Waals gas above, we see that B1 (T ) = −(a/kb T − b). These coefficients can be calculated theoretically, for a given potential, by taking into account higher order interactions. For those interested can check out Goodstein, section 4.4. In practice though, the virial coefficients are usually measured from experiments. The Lennard-Jones Potential. In the literature, especially in chemistry texts, instead of the hard core + London potential we used Eq. (3.62), the so-called Lennard-Jones potential or the 12-6 potential, is often used    σ 12  σ 6 U = 4 − , (3.82) r r where  and σ are the Lennard-Jones coefficient and effective radius of the molecules respectively. These coefficients are usually measured experimentally. You will show in a homework problem we will recover the same results in this section using this potential. 62 Chapter 4 Quantum Gas So far, we have considered classical systems. Even so, the universe is actually quantum, and hence in a couple of places, we have leaned on quantum mechanics to motivate some otherwise unexplained assertions. Firstly, we have argued in section 2.5.2 that the correct normalization factor when we compute the continuous partition function is 1/(2π~) per spatial dimension. Secondly, in section 3.3, we have argued that quantum mechanically particles are indistinguishable. Nevertheless, in those sections, we have not fully embraced quantum mechanics. In this chapter, we take the plunge into the quantum world. 4.1 Quantum states vs classical particles Consider a quantum particle of mass m in a 3 dimensional box with equal sides a, and volume V = a3 , and periodic boundary conditions. Extending the 1-D derivation in section 2.5.2 to 3-D, the eigenfunction fo the time-independent Schrodinger’s equation is the plane wave solution in 3-D, ψ(x) = Aeik·x (4.1) with the wavenumbers k = (kx , ky , kz ) quantized as 2πni ki = , ni = 0, ±1, ±2,... , (4.2) a so the energy of the particle is ~2 k 2 4π 2 ~2 2 2 2 q En = = (n + n + n ) , k ≡ kx2 + ky2 + kz2. (4.3) 2m 2ma2 x y z The quantum numbers ni labels the quantum state of the particle. 4.1.1 Density of states So far this is simply a review of quantum mechanics. Bringing in the machinery of statistical mechanics, we make the connection that each quantum state of the particle represents a possible microstate, and hence the partition function for a single quantum mechanical particle is then given by Eq. (2.43) X Z1 = e−βEn , (4.4) n where the sum is over all possible quantum numbers n ≡ (nx , ny , nz ). (4.5) 63 Recall that the lengthscale when quantum mechanics is important is given by the de Broglie wavelength, or as we have discussed in section 3.1, the thermal de Broglie wavelength Eq. (3.12) s 2π~2 λ≡. (4.6) mkb T As long as the box size is much, much greater than this wavelength, a  λ, the number of possible states n is super-duper large, so we can approximate the sum over all n as an integral X Z → dnx dny dnz. (4.7) n Following the arguments in section 2.5.2, the measure dni is a dni = dki , (4.8) 2π so ∞ 4πV Z Z Z V 3 dnx dny dnz = d k= k 2 dk (4.9) (2π)3 (2π)3 0 where we have used isotropy to turn the dkx dky dkz integral into a volume integral over the 2-sphere. The partition function Eq. (4.4) is then Z ∞ 4πV Z1 = k 2 e−βEn (k) dk , (4.10) (2π)3 0 where we have explicitly specified the k dependence of En to be clear. We can easily convert the dk measure into an energy measure dE by using Eq. (4.3) ~2 k dE = dk , (4.11) m so ∞ 4πV 2mE m −βEn Z Z1 = e dE (2π)3 0 ~2 k~2 Z ∞r 4πV 2mE m −βEn = e dE (2π)3 0 ~2 ~2 √ Z ∞ 4 2πV = m3/2 E 1/2 e−βEn dE (2π~)3 0 Z ∞ ≡ e−βE g(E)dE , (4.12) 0 where the measure √ 4 2πV 3/2 1/2 g(E)dE = m E dE , (4.13) (2π~)3 quantifies the density of possible microstates of a single particle between E and E + dE. This measure is called the density of states. Since the factor e−βEn just came along for the ride, in general we can use this measure for any distribution f (E) – we simply replace the sum over quantum states n with the density of states X Z ∞ f (E) → g(E)f (E)dE. n 0 Notice that we have obtained the correct normalization factor (2π~)−3 for the partition function, which is not surprising since we followed the same arguments. Also not surprising is that the quantum single particle partition function Eq. (4.12) is the same as its “classical” counterpart Eq. (3.4) – which we have derived with quantum mechanical input to “make it right”. 64 4.1.2 Relativistic vs non-relativistic particles So far we have worked with non-relativistic particles, where the energy E = ~2 k 2 /2m. But of course, Einstein told us that, when a particle is moving a close to the speed of light, the energy is given by the relativistic formula E 2 = ~2 k 2 c2 + m2 c4. (4.14) The measure in this case is then EdE = ~2 kc2 dk. (4.15) You will show in a homework problem that, this leads to the following relativistic version of the density of states 4πV E p 2 g(E)dE = E − m2 c4 dE. (4.16) (2π~)3 c3 For massless particles m = 0, say like the photon, this becomes 4πV E 2 g(E)dE = dE. (4.17) (2π~)3 c3 We should come clean and point out that we have sort off cheated by using a result we have derived using the Schrodinger’s equation (valid only for non-relativistic particles) to obtain a relation for relativistic particles. In this case, this is actually fine, because the eigenfunctions are plane waves, which also happens to be solutions to the relativistic equations (e.g. the free Klein-Gordon equation). 4.1.3 Many particles vs many quantum states What about a system of many quantum particles? As usual, we will take advantage of the fact that the partition function of a non-interacting composite system multiply, Y Z= Z1 , Q but we can choose how to label the microstates we take the product over. Since particles are in general not conserved and can be created/destroyed, the product is “over all possible configurations over all possible particle numbers”. In other words, we are really working with a grand canonical ensemble Z → Z. When we previously studied the grand canonical ensemble in section 2.4, we introduced a new pair of conjugate variables: the total particle number N and the chemical potential µ per particle. While we can do the same here, we can actually do better! We take advantage of the fact that, for a particle with a definite quantum state n, we already know the entire solution for its evolution, i.e. every particle with quantum number n is exactly described by the wavefunction Eq. (4.1) with the energy Eq. (4.3) – they are completely indistinguishable, so it doesn’t make sense to label them individually. So instead of making a single particle a subsystem, and then taking the product of all such single particle subsystems to construct the grand partition function, we choose to partition the system up into subsyste