Chapter 5 Phonons II Chapter 5 Phonons II The Planck Distribution � Last lecture, we looked at how many phonons are �occupied� as a function of the energy of the phonon, and the temperature � This is[.]
Chapter 5: Phonons II The Planck Distribution • Last lecture, we looked at how many phonons are “occupied” as a function of the energy of the phonon, and the temperature x < n >= = − x exp(hω / kT ) − • • • • This is the average number of phonons that are “occupied” or “active” 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) From Statistical Mechanics • • You can also get this result using statistical physics A harmonic oscillator can have the energy levels: ∆E = 0, hf ,2hf ,3hf • The partition function for a single harmonic oscillator is then: Z = exp(0 / kT ) + exp(− hf / kT ) + exp(−2hf / kT ) + Z= − exp(− hf / kT ) • And the average energy is: ∂Z hf exp(− hf / kT ) = (exp(−hf / kT ) − 1) Z ∂ (1 / kT ) (1 − exp(− hf / kT )) hf E= (We will use this later) (exp(hf / kT ) − 1) E =− From Statistical Mechanics • If we think of the energy as coming in “units” of hf, then the average number of units of energy in the oscillator is < n >= • • • E = hf (exp(hf / kT ) − 1) which is what we got before It is important to note that high energy vibrations are suppressed – they are “frozen out” or not active Also, as you lower the temperature, you get the same effect – the number of phonons at a certain energy goes down as you lower the temperature At zero temperature, you get no vibrations, and → What does this look like? • Plotting this as a function of hf/kT, we get 100 for bosons (eg photons, phonons) 80 • So, as you lower the energy, you get more phonons which are occupied at a given temperature (it costs less energy to activate these phonons) As you increase the temperature, you get more phonons occupied as well (there is more thermal energy around to create lattice vibrations) 60 • 40 20 0 10 15 20 25 30 hf/kT Energy increasing Temperature increasing The Einstein model • • • In 1907, Albert Einstein proposed the following model for the lattice vibrations in a solid using the Planck Distribution: He approximated all the atoms to vibrate at the same frequency, so they are identical (we know that this isn’t true, but let’s see what happens) The best way to think of this is as a collection of 3N harmonic oscillators for N atoms (there are ways that each atom can vibrate – x, y, and z) It is commonly called an Einstein solid The Einstein Model • We can now use the Planck Distribution to figure out what is the average level which is occupied for each one of these oscillators at a given temperature < n >= exp(hω / kT ) − • If we know what the average energy level is, then we can calculate the average energy/oscillator: Average energy/oscillator = (average energy level) x (energy in each level) (hω ) =< n > (hω ) = exp(hω / kT ) − The Einstein Model • So, for 3N oscillators, we have a total energy of: Nhω U= exp(hω / kT ) − • We can now calculate the heat capacity by taking the derivative with respect to temperature: Nhω ∂ ∂U CV = = ∂T ∂T exp(hω / kT ) − = Nhω • ∂ ∂T exp(hω / kT ) − The easiest way to this is to change variables The Einstein Model • Let v = ħω/kT Therefore: dv = -(ħω/(kT2))dT Now we can calculate the specific heat: exp(hω / kT ) − hω ∂ hω − exp(v) = Nhω − = Nhω − kT ∂v exp(v) − kT (exp(v) − 1) ∂ CV = Nhω ∂T hω exp(hω / kT ) = Nhω kT (exp(hω / kT ) − 1) hω Cv = Nk kT exp(hω / kT ) (exp(hω / kT ) − 1) Heat Capacity of an Einstein Solid The Einstein Model (High T limit) • • • What does this function look like? Investigate the limits: As T →∞, we have: hω Cv = 3Nk kT exp(hω / kT ) (exp(hω / kT ) − 1) hω = Nk kT + hω / kT (1 + hω / kT − 1) hω ~ Nk kT (hω / kT ) ~ Nk = 3nR Law of Dulong and Petit Heat capacity saturates at high temperatures at 3R/mol The Einstein Model (low T limit) • • What happens at low temperatures? As T → (becomes small), we have: hω Cv = Nk kT hω ~ Nk kT 2 exp(hω / kT ) (exp(hω / kT ) − 1) exp(hω / kT ) (exp(hω / kT )) hω ~ Nk (exp(−hω / kT ) ) kT This goes to zero as ~ (exp(−hω / kT ) ) T goes to zero! (1/exp(large no)) ~ The exp part dominates over the 1/T2 term The Einstein Model (low T limit) • • • How does this compare to experimental data? (T3 relationship) It isn’t a bad approximation, but overall it isn’t perfect The Einstein Model was a huge step forward in the right direction – the first application of statistical physics to problems in solids We know, however, that the lattice can vibrate at different frequencies, so it isn’t very suprising to see that it doesn’t work perfectly 0.3 Heat Capacity • Einstein Model T experimental data 0.2 0.1 0.0 -0.1 Temperature(K) The Einstein Model • • • • • • However, the Einstein Model does work very well for some materials Where we see this? Optical modes of solids (the frequency is almost independent of the wavelength, so this works) Overall, we know that ω(κ) for our dispersion curves – the frequency changes and is not a constant How we use this to describe vibrations in solids? Debye Model Optical branches have a nearly frequency independent dispersion curve Using Dispersion Curves to Calculate the Heat Capacity • • What we need to now is take what we know about the dependance of the phonon frequencies upon the wavevector κ to calculate the heat capacity (this is from dispersion curves) This reduces to summing over all the energies of all the possible phonon modes, multiplied by the Planck Distribution: U = ∑∑ U κ , p =∑∑ < n > hωκ , p κ Sum over all wavevectors p Sum over all polarizations (=3 for 3D) κ p Planck Distribution (no of phonons at a given temp, energy) Energy of a phonon of polarization p, wavevector κ Density of States • Now what we have to is calculate U = ∑∑ κ p hωκ , p exp(hω / kT ) − U = ∑ ∫ dω ( D (ω )) p • p hω exp(hω / kT ) − Where what we have done is replace the sum over κ by an integral If we this, we now have to use what is called the density of states, D(ω) This is the number of modes per unit frequency range Density of States • • What is the density of states, physically? It is defined as being: dN D(ω ) = dω • No of modes Frequency Why we have to use this? For photons, for example, you would expect that you can have an infinite amount of modes at each energy However, for solids this isn’t the case –there is a limit on the number of possible wavelengths that you can have due to the fact that we are dealing with a discrete system – the atomic spacing places a limit on the number of possible modes Density of states There are only so many wavelengths that are allowed, for example, in a solid n = mode n = mode n = N1/3 mode Because of the discrete nature of the system, there are only so many allowed modes of vibration/energy unit (because there are only so many wavelengths allowed) We have to calculate this for all the possible modes in 3D