CONTENTS 1. In this course we will cover more foundational topics such as: These topics may be taught as an independent study sometime next year.

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1 CONTENTS Introduction Prerequisites Knowledge of di erential equations is required. Some knowledge of probabilities, linear algebra, classical and quantum mechanics is a plus Units We will adopt the natural (or Planck) units: ~ = c = G = k = Topics In this course we will cover more foundational topics such as: Thermodynamics, Kinetic theory, Classical Statistical Mechanics Quantum Statistical Mechanics. but will no cover more advanced topics such as: Phase transitions, Renormalization, Quantum gravity, Non-equilibrium processes. These topics may be taught as an independent study sometime next year History A brief history of the material covered in this course can be summarized as follows: (1824) Sadi Carnot. Carnot s engine. (1843) James Joule. Conservation of energy. (1850) Rudolf Clausius. Entropy.

2 CONTENTS 2 (1871) James Clerk Maxwell. Equilibrium thermodynamics. (1872) Ludwig Boltzmann. Kinetic theory. (1876) Williard Gibbs. Statistical mechanics. (1923) Satyendra Nath Bose. Bose-Einstein statistics. (1926) Enrico Fermi and Paul Dirac. Fermi-Dirac statistics. (1927) John von Neumann. Quantum statistical mechanics.

3 Chapter 1 Thermodynamics 1.1 Lecture 1: Preliminaries, Zeroth law, First law Preliminaries Degree of freedom (d.o.f) is a number of parameters necessary to formulate the initial value problem divided by two. For a harmonic oscillator the initial value problem is specified with only two parameters: position coordinate (or angle) and velocity coordinate (or 3

4 CHAPTER 1. THERMODYNAMICS 4 angular velocity) corresponding to only a single d.o.f. In Hamiltonian mechanics the position q and momentum p coordinates come in conjugate pairs (q, p). Thus the number of d.o.f. is the number of such pairs or one half of the total dimensionality of the phase space. Problem: How many d.o.f. in a problem of binary collision of protons in the LHC (Large Hadron Collider)? Solution: Binary collisions involve collisions of only two particles. Each particle is specified by a position and velocity coordinates. This makes the number of d.o.f. (2+2)/2 = 2. Our main objective in this course is to understand the behavior of systems with a very large number of d.o.f. N 1 (e.g. N A = molecules in a box). The task is (in some sense) much more ambitious than the problems we are used to in physics courses where the number of d.o.f. is usually small. For N. 10 analytical methods may be useful; for N computer may work; for N 1google = statistical physics may be the only tool. There are two standard ways to study the large N limit: phenomenological (e.g. thermodynamics) and fundamental (e.g. statistical mechanics). We start with phenomenological approach and later will use a kinetic theory to justify a more fundamental approach. Thermodynamics is a phenomenological theory of systems with many d.o.f. The main idea is that only a small number of measurable parameters (e.g. volume V, pressure P, temperature T, etc.) should be su cient for describing the so called equilibrium states. Equilibrium state is the state whose thermodynamics parameters do not change with time. It is an important experimental fact that in equilibrium all of the parameters are either extensive or intensive. For example, volume is extensive, but pressure is intensive. Extensive parameter is proportional to the amount of substance and intensive parameter is independent on the amount of substance.

5 CHAPTER 1. THERMODYNAMICS 5 Problem: Give an example of neither extensive nor intensive quantity. Explain. Solution: When two wires with electrical resistances R 1 and R 2 are connected in series than the total resistance is an extensive quantity R total = R 1 + R 2, but when the wires are connected in parallel the total resistance is not extensive 1/R total = 1/R 1 +1/R 2. Therefore, electrical resistance is not an extensive nor intensive quantity. Not all of the thermodynamic parameters are independent of each other. Equation of state, f(p, V, T) =0, (1.1) describes the dependence and reduces the number of independent parameters. For example, the equation of state for an ideal gas (su ciently diluted gas) is given by, PV = NkT, (1.2) where N is the number of molecules and k is the Boltzmann s constant. Because of a universal character of ideal gases (1.2) can be used to set a relative temperature scale, but to set an absolute scale and to give a meaning to T =0it is necessary to postulate a Second Law of thermodynamics. For example, measure PV when water boils denote the temperature by T =0, Nk measure PV when water freezes denote the temperature by T =100, use Nk linear extrapolation to assign temperature to arbitrary values of PV. This is Nk the Celsius scale.

6 CHAPTER 1. THERMODYNAMICS 6 It is convenient to think of f(p, V, T) as a 3D function which describes a given thermodynamical system and of (P, V, T) as a point in 3D which describes a given state of the system. The projection of the surface of the equation of state on P V plane is known as the P V diagram. Reversible transformation of di erent types such as 1-2. isothermal (i.e. constant temperature) 2-3. isobaric (i.e. constant pressure) 3-1. isochoric (i.e. constant volume) are described by di erent paths on the P V diagram Zeroth Law Every theory is based on a number of statements which cannot be proved. In mathematics such statements are called axioms, but in physics they go by di erent names: principles, laws, assumptions, etc. With this respect thermodynamics is not an exception as it is based on four laws of thermodynamics numbered from 0 to 3 (for purely historical reasons). Zeroth Law in words: If systems A and C are each in thermal equilibrium with system B, then A is in thermal equilibrium with system C.

7 CHAPTER 1. THERMODYNAMICS 7 Zeroth Law in symbols: A B and C B A C (1.3) where is a relation between systems such that A B reads as A is in a thermal equilibrium with B. Problem: Under assumption that every system is in equilibrium with itself, prove that is an equivalence relation. Solution: Equivalence relation must satisfy three properties: 1) Reflexivity: By assumption, A A. 2) Symmetry: Let A B. From Reflexivity B B. Using Zeroth Law, B B and A B B A. 3) Transitivity: Let A B and B C. From Symmetry C B. From Zeroth Law, A B and C B A C. An important consequence of the equivalence relation ~ is that it divides all of the system into equivalence classes which can be labeled by temperature. Of course, this does not tell us why the temperature should be a real number as opposed to, for example, integers, complex or even more exotic p-adic number. Question To Go: Why temperature is a real number? First Law The Zeroth Law does not tell us why temperature is a real number. Thus additional laws must be postulated before this fact can be established not only experimentally (e.g. using equation of state for ideal gases), but also more theoretically. The next assumption is the First Law which should be viewed as a statement about conservation of energy. First Law in words: The increment in the internal energy U of a system is equal to the difference between the increment of heat Q accumulated by the system and the increment of work W done by it. First Law in symbols: where du is an exact di erential. du = dq dw, (1.4) Exact di erential dx is a di erential whose integral s dx depends only on the limits of integration, but not on the path.

8 CHAPTER 1. THERMODYNAMICS 8 The First law is used to define the state function U which is an extensive quantity: doubling the mass doubles the internal energy. For quasi-static (or su ciently slowly varying) processes the work done by the system dw = PdV such that du = dq PdV. (1.5) Problem: Are the di erentials dq and dw exact or not? Solution: W = s PdV is not a full derivative and therefore depends not only on the initial and final points but also on the path. This can be shown explicitly by integrating along di erent path from (P 1,V 1 ) to (P 2,V 2 ) assuming that P 1 <P 2 and V 1 <V 2 : (P1,V 2 ) (P 1,V 1 ) (P2,V 2 ) (P2,V 1 ) (P2,V 2 ) PdV+ PdV = P 1 (V 2 V 1 ) = P 2 (V 2 V 1 )= PdV+ PdV. (P 1,V 2 ) (P 1,V 1 ) (P 2, V 1 ) Therefore, dw is not an exact di erential and since du is exact due to the First Law dq = du dw is also not exact. It is convenient to think of the conjugate pair P and V as a generalized force (intensive quantity) and generalized displacement (extensive quantity) respectively. Other commonly used conjugate pairs include temperature T and entropy S, chemical potential µ i and number of particles N i of type i, etc. Then the total change in internal energy can be expressed as the sum of the products of generalized forces and generalized displacements: du = TdS PdV + ÿ i µ i dn i. (1.6) In other words, when a given generalized force is not balanced it causes a generalized displacement whose product equals to the energy transfer in or out of the system.