Concepts for specific heat

Transcription

1 Concepts for specific heat Andreas Wacker, Matematisk Fysik, Lunds Universitet November 8, 1 1 Introduction In this notes I want to briefly eplain general results for the internal energy and the specific heat for simple classical and quantum systems. The focus is on two model systems namely a and a Free particle with energy E = p m Harmonic oscillator with energy E = p m + 1 mω ( Applications to solid state physics are briefly discussed. The central input is that thermodynamics tells us that in thermal equilibrium the probability P i to find a systems in state i with energy E i is given by the Boltzmann distribution (or canonical distribution (1 P i = e βe i Z with β = 1 k B T and the partition function Z = i e βe i In particular, we find that for T =, the system is always in its ground state with lowest energy. In the following, the two indefinite integrals are frequently used. d e α = Classical Physics π α and d e α = π 4α 3 for a >.1 Free particles We start with a free particle in one dimension, which can take arbitrary momentum p. In this case the sum over i becomes an integral of p and P(p δp is the probability to find the system in the interval δp around p. Then the partition function reads mπ Z 1 = dp e βp /m = β The epectation value for the kinetic energy is = 1Z1 m dp p β /m m e βp = mπ 1 1 π(m 3 = 1 m 4β 3 β = k BT

2 Specific heat, Andreas Wacker, Lund University, November 8, 1 For three dimensions, we have ( ( ( Z 3 = d 3 p e βp /m = dp e βp /m dp y e βp y/m dp z e βp z/m = Z1 3 and = m with ( ( = 1Z3 p dp m m e βp /m p p y p + + z m m m ( dp y e βp y/m } {{ } =Z 1 dp z e βp z/m } {{ } =Z 1 and similarly for p y/m and p z/m. Adding up these contributions provides = 3 k BT m This can be directly etended to N particles and we find = k BT In thermal equilibrium a system on N free particles in D dimensions has the internal energy E = ND k BT (3 which is precisely k B T/ for each degree of freedom Then the specific heat (at constant volume is given by which holds, e.g. for ideal monoatomic gases. c v = de dt = NDk B (4. Harmonic oscillator In this case the energy depends both on p and and we have to integrate over both variables in order to get the correct distribution. We find Z = dp d e β(p /m+mω mπ π / = β βmω and p m + mω = k B T Thus the average energy is just twice as large as for the free particle, as the energy is evenly distributed between the kinetic and potential energy. (This only holds for quadratic potentials! In thermal equilibrium each oscillation mode has the internal energy E = k B T (5 The atoms in a crystal have 3N atoms different oscillation modes as given by the vibration(phonon spectrum. Thus we epect the specific heat per atom as c V N atoms = 3k B

3 Specific heat, Andreas Wacker, Lund University, November 8, 1 3 This is called the Dulong-Petit law, which is valid for solids at sufficiently high temperatures (as we will see below, room temperature is at the borderline for many substances. For iron, the atomic mass is m = kg = kg we thus estimate a specific heat of 3k B /m = 447 J/(K kg. This is actually a surprisingly good approimation for the eperimental value of about 44 J/(K kg at room temperature..3 Comment on phase space Tacitly, it was assumed in the preceding subsections, that the probability should be evaluated for the variables momentum p i and positions i. If one would use the energy, to describe the states instead, we would find Z = de e βe = 1/β and E = de Ee βe /Z = k B T, for all systems. Thus one wonders, why the choice of space and momentum is the right one? The reason lies in classical mechanics, where one obtains one set of (generalized momentum and position for each degree of freedom. Thus these variables (as well as other pairs obtained by canonical transformations stand out. 1 3 Quantum Physics 3.1 Harmonic oscillator Quantum physics tells us the the one-dimensional harmonic oscillator with angular frequency ω has discrete energies E n = (n + 1/ ω with n =, 1,,... We find the partition function Z = e βen = n= eβ ω/ 1 e β ω and the average energy E n = 1 Z E n e βen = 1 Z n= Z(β β =... = ω + ω e β ω 1 This provides the specific heat C V = d dt E β ( ω e β ω n = k B (e β ω 1 The results is plotted in Fig. 1. In the case β ω 1 (i.e. k B T ω this becomes vanishingly small, while for β ω 1 (i.e. k B T ω we obtain the classical result k B. One says, that the degree of freedom freezes in around a temperature where k B T = ω. As the energies of the phonons for solids are typically some tens of mev (while k B T = 5 mev at room temperature, we epect modifications from the Dulong-Petit law addressed above. These become even more prominent, when lowering the temperature. The low-energy phonon spectrum is given by the acoustic phonons with a spectrum ω(q = c q, where c is the sound velocity (actually these are three branches with different and direction-dependent velocities. Assuming for simplicity, that all modes with ω(q < k B T provide k B to the specific heat, while those with higher frequency do not participate at all, be obtain that the specific heat is just given by 3k B times the number of q values satisfying q < k B T/( c. This number is proportional to the volume of a sphere, i.e., proportional to T 3. Thus the phonon part for the specific heat vanishes at T 3 if the temperature approaches the absolute zero, which is known as the Debye law. 1 Actually these canonical variables are the ones which also satisfy the canonical commutation relations [ˆp i, ˆ j ] = δ ij i in quantum mechanics.

4 Specific heat, Andreas Wacker, Lund University, November 8, 1 4 c v ( k B 1,8,6,4,,1, k B T (hω Figure 1: Specific heat of a phonon mode with angular frequency ω. 3. Single particle in a bo For a bo of length L with infinite potentials outside, the eigenstates of the stationary Schrödinger equation are ( nπ Ψ n ( L sin with energy E n = π n for n = 1,,... L ml Thus we get the partition functions Z = n e βen. Now we set α = β π /ml. In the limit of α 1 (i.e. L is much larger than the thermal wavelength / πmk B T we may replace the sum by an integral, as e αn changes only weakly between consecutive values of n. We obtain Z d e α = π 4α and E n 1 Z dα e α = k BT This recovers the classical result. On the other hand for α 1, i.e. for high temperatures, the probabilities for P n are vanishingly small for n and P 1 = 1. This we find E n E 1 = π ml In this case the specific heat becomes zero, in contrast to the result k B / for high temperatures. Together with the preceding subsection this can be summarized as Quantum physics provides discrete energy levels. If the spacing of these levels is small compared to k B T, the classical result for the specific heat is recovered for a single particle. Otherwise, if the k B T becomes smaller than spacing between the ground and first ecited quantum level, the contribution to the specific heat vanishes. One says that the degree of freedom freezes in. 3.3 Many particles in a finite bo Net to the eistence of discrete energy levels, the Pauli principle is another important consequence of quantum physics. It states that for systems of identical particles, each must be Note that the are some factors π changed, in order to match other conventions

5 Specific heat, Andreas Wacker, Lund University, November 8, 1 5 placed in a different level. Thus the lowest energy for N electrons in a one dimensional bo is obtained by putting them into the levels n = 1,,...N/, where the electron spin allows to put two electron in each orbital level. The energy of the last filled level is denoted as the Fermi level E F = π N in one dimension 8mL which is actually a function of the one-dimensional electron density N/L. In the same spirit the three-dimensional case is treated. For a cube with length L the energy levels are E n,n y,n z = ( π ml n + n y + nz. Here one may use all combinations n,n y,n z 1 with n +n y+n z n ma, and their number can be estimated by πn 3 ma/6 Question 1: Derive this number! Taking into account spin, this allows to accomodate N = πn 3 ma/3 electrons and we find ( E F = 3π N /3 m L 3 for a cube with length L in three dimensions For an electron density /(3Å 3 we obtain E F = 6 ev, a typical value for a Fermi energy in a metal. In order to be thermally ecited the electrons have to occupy levels above this Fermi energy. As E F is huge compared to the thermal energy of k B T = 5 mev, we find that the vast majority of electron is frozen in and does not contribute to the specific heat. Thus c v is much smaller than the value N3k B / epected for a classical gas in three dimensions. In order to estimate the magnitude we assume, that only the states with E n,n y,n z E F k B T contribute to the specific heat by thermal ecitations (i.e. being partially ecited to levels above E F. Their number is N ecitable = L3 (m 3/ [ ] E 3/ 3π 3 F (E F k B T 3/ L3 (m 3/ E F k π 3 B T = 3N k BT E F Assuming further that the ecitably states contribute with the classical specific heat of free particles we find 3 c V N ecitable k B = 9 kb T E F More detailed calculations provide an additional factor π /9, see [1]. The main point is that the specific heat is strongly reduced by the ratio 3k B T/E F (i.e. 1% at room temperature and that it is proportional to the temperature. The Pauli principle restricts the possible ecitations with low energy and thus the heat capacity of degenerate systems is strongly reduced compared to the classical result. 4 References [1] C. Kittel, Introduction to Solid States Physics (seventh edition (John Wiley & Sons inc, New York, 1996.