Section 06_Thermal_Properties

# Section 06_Thermal_Properties - Physics 927 E.Y.Tsymbal...

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Physics 927 E.Y.Tsymbal Section 6: Thermal properties Heat Capacity There are two contributions to thermal properties of solids: one comes from phonons (or lattice vibrations) and another from electrons . This section is devoted to the thermal properties of solids due to lattice vibrations (the contribution from electrons in metals will be considered separately). First, we consider the heat capacity of the specific heat . The heat capacity C is defined as the heat ' Q which is required to raise the temperature by ' T , i.e. Q C T ' ' . (6.1) If the process is carried out at constant volume V , then ' Q = ' E , where ' E is the increase in internal energy of the system. The heat capacity at constant volume C V is therefore given by V V E C T w § · ¨ ¸ w © ¹ . (6.2) The contribution of the phonons to the heat capacity of the crystal is called the lattice heat capacity . The total energy of the phonons at temperature T in a crystal can be written as the sum of the energies over all phonon modes, so that ( ) p p p E n Z ¦ q q q = , (6.3) where p n q is the thermal equilibrium occupancy of phonons of wavevector q and mode p ( p = 1 3 s , where s is the number of atoms in a unit cell). The angular brackets denote the average in thermal equilibrium. Note that we assumed here that the zero-point energy is chosen as the origin of the energy, so that the ground energy lies at zero. Now we calculate this average. Consider a harmonic oscillator in a thermal bath. The probability to find this oscillator in an excited state, which is characterized by a particular energy E n is given by the Boltzmann distribution: / 0 B n k T n P P e Z ± = , (6.4) where the constant P 0 is determined from the normalization condition 0 1 n n P f ¦ , (6.5) so that 1 / 0 0 B n k T n P e Z ± f ± § ¨ © ¹ ¦ = · ¸ . (6.6) The average excitation number of the oscillator is given by 1

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Physics 927 E.Y.Tsymbal / 0 / 0 0 B B n k T n n n k T n n ne n nP e Z Z f ± f f ± ¦ ¦ ¦ = = . (6.7) The summation in the enumerator can be performed using the known property of geometrical progression: 0 1 n n x f ¦ . (6.8) Using this property we find: ² ³ 2 0 0 1 1 1 f f ± ± ¦ ¦ n n n n d d nx x x x dx dx x x x , (6.9) where . Thus we obtain / kT x e Z ± = / 1 1 1 (1 ) 1 1 B k T x n x x e Z ± ± ± = ± . (6.10) The distribution given by Eq. (6.10) is known as the Planck distribution. Coming back to the expression for the total energy of the phonons, we find that ( )/ ( ) 1 p B p k T p E e Z Z ± ¦ q q q = = . (6.11) Usually it is convenient to replace the summation over q by an integral over frequency. In order to do this we need to introduce the density of modes or the density of states D p ( Z ). D p ( Z ) d Z represents the number of modes of a given number s in the frequency range ( Z , Z + d Z ). Then the energy is / ( ) 1 B p k T p E d D e Z Z Z Z ± ¦ ³ = = (6.12) The lattice heat capacity can be found by differentiation of this equation with respect to temperature, so that 2 / / 2 ( ) ( 1 B B k T B V B p k T p e k T E C k d D T e Z Z Z Z Z § · ¨ ¸ w © ¹ w ± ¦ ³ = = = ) . (6.13) We see that the central problem is to find the density of states D p ( Z ), the number of modes per unit frequency range.
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