Chapter 5 Phonons II Thermal Properties Chapter 5 Phonons II Thermal Properties Thermal Properties of Phonons � As mentioned before, now we are going to look at how what we know about phonons will lea[.]
Chapter 5: Phonons II Thermal Properties Thermal Properties of Phonons • As mentioned before, now we are going to look at how what we know about phonons will lead us to a description of the heat capacity as a function of C/T temperature • What we have to is establish the rules we need to count how many phonons are “active” at a certain temperature, and then figure out how much energy goes into each Slope = β Y-intercept = γ T2 CV = γT + βT3 Electron part Lattice part Historical Perspective • 1819: Dulong and Petit observed that the specific heat at constant volume for most solids was 3R (~25 J/(K mol)) near room temperature ∂U CV = = 3R ∂T V • • How can we explain this? We need to use statistical physics: the principle of equipartition of energy (R = gas constant = 8.314 J/(K mol)) (Note: this is only high-T case At low temp, C goes to zero Why?) Equipartition Theorem • Equipartition theorem: At temperature T, the average energy of any quadratic degree of freedom is ½ kT • Degree of Freedom: way that energy can be distributed Moment of intertia Rotational motion Vibrational motion Angular Velocity ½ Iωx2 , ½ Iωy2, ½ Iωz ½ ksx2 + ẵ mv2 Spring constant ã So, each harmonic oscillator has degrees of freedom, and an average thermal energy of kT Boltzmann’s constant = 1.38 x 10-23 J/K Equipartition Theorem • • • Therefore, if every atom can move in the x, y, and z direction like a spring, then there are harmonic oscillators for every atom The total energy for the system is then = 3NkT (N= total no of atoms) The heat capacity for these oscillating atoms is then: ∂ ∂U CV = = (3 NkT ) = Nk = 3nR ∂T V ∂T • Or, per mol, CV = 3R No of moles Low-T Heat Capacity • • • • • Upon cooling to low temperature, scientists found that this law was no longer valid It also wasn’t true for some materials, like diamond (why?) Also, it should be pointed out that the shape of the curves look different for different materials Can we use what we know about phonons to calculate the heat capacity? Some of our heat capacity goes to the electrons, and other sources, but in most materials the lattice vibrations absorb most of the energy Answer: You need to use quantum statistics to describe this properly Bosons and fermions • • • Before we need to talk about how many phonons there are at a certain temperature, we need to discuss some terms in quantum statistics Particles are divided into categories: Fermions: particles that cannot be in the same energy level (eg Electrons These particles have ½ integer spin (electrons have s = ½, for example)) Bosons: particles that can be in the same energy state (eg Photons, phonons These particles have integer spin (photons have S = 0, for example)) It should be obvious that you can have more than one photon, or lattice vibration at the same energy This isn’t so obvious for the case of electrons (Chapter 6) Electrons can only be “paired up” if their spins are pointing in the opposite direction (and thus they have slightly different energies) The Planck Distribution • • • • Max Planck (German physicist, late 19th century) first came up with the idea of quantized energy His first work was done with photons, particles of light Blackbody Radiation: an empty cavity, except for photons inside, is at a certain temperature T These photons are in equilibrium with the temp of the cavity So, if you poked a hole in it and looked at the radiation given off by the cavity, you would see light of a characteristic wavelength (most of the photons have a certain wavelength, which corresponds to the temp of the cavity) Blackbody radiation • The higher the temperature of the blackbody, the higher the energy of the characteristic photons, and the higher the frequency (lower the wavelength) Infrared Visible Frequency Increasing • UV This explains why stars have certain colours (red stars are cooler than white stars) hc λmaxT = 4.96k Analogy • • • • There is an analogy between photons in a cavity and phonons in a solid: EM Waves Lattice vibrations Photons in a cavity Phonons in a solid It turns out that we can use the same rules to treat both photons and phonons That is why it is said that phonons follow a Planck Distribution of energies Planck figured out how these light wave energy levels were populated as a function of temperature (by considering them to be harmonic oscillators) We are going to use the same rules to see how many phonons are occupied at a given temperature (approximate each phonon to be a harmonic oscillator – same energy levels) Planck Distribution • For a set of identical harmonic oscillators in equilibrium, the ratio of the no of oscillators in the (n+1)th state to the nth state is: Level n+1 Level n N n +1 = exp(−∆E / kT ) Nn Energy difference between n+1 and n (which is = ħω) • Series of identical harmonic oscillators Energy levels = ½ (n+1)ħω (n is integer) So, the fraction of the total no of oscillators in the nth quantum state is: No of phonons in energy level n Total no of phonons Nn = ∞ ∑N s exp(−ε n / kT ) ∞ ∑ exp(−ε s / kT ) Sum is over all possible energy levels ε0, ε1, etc Planck Distribution • • The distributions of these energy levels is called the Planck Distribution What does this look like? phonons phonons phonons phonon phonons e-ħω/kT Prob of being “occupied” n=1 n=2 Energy n=3 So, at any given energy εs = s ħω = s hf (above the zero energy level n=0) Planck Distribution • Let’s rewrite our equation to match the textbook: This is small as n gets large Fraction of phonons = at energy n Nn = ∞ ∑N s exp(−ε n / kT ) ∞ ∑ exp(−ε s / kT ) = exp(−nhω / kT ) ∞ ∑ exp(−shω / kT ) A constant • In order to calculate the heat capacity, we need to know how many phonons there are at each energy level on the average So, what is the average energy excitation level at a temp T? Average for Planck Distribution • • This is like an expectation value in quantum mechanics Eg What is the average size of a 1s hydrogen orbital? < r >= ∫ψ 1s * rψ 1s dV • So, by analogy, our average energy level which is occupied is: ∞ < n >= ∑ n exp(−nhω / kT ) ∞ ∑ exp(−nhω / kT ) How can we this sum? • We know that ∞ ∑0 x = − x s • So, replacing x for e we have: ∞ ∑0 e = − e s • And therefore ∞ ∑ exp(−shω / kT ) = − e s =0 (denominator) − hω / kT How can we this sum? • We can use this result, and a trick, to calculate the numerator: ∞ ∂ x ∂ s s ∑0 sx = ∂x ∑0 x = ∂x − x = (1 − x) ∞ • So therefore: exp(−hω / kT ) ∑0 s exp(−shω / kT ) = (1 − exp(−hω / kT ))2 ∞ The Planck Distribution • Putting it all together, we have: x < n >= = − x exp(hω / kT ) − • • • • • This is the average energy level, or the average number of phonons (since this is equal to n) that are occupied at a temperature T with frequency ω What does this mean? High T limit: as T →∞, = 1/(exp(ħω/kT)-1) ~ 1/(1+ (ħω/kT) -1) ~ kT/ ħω (this is a large number) Low T limit: as T → 0, ~ exp(- ħω/kT) ~ (ground state) This is a handy number, because it tells us how many phonons are at each frequency (we will need this to calculate the total phonon energy)