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Transcript

Statistical Mechanics quantum statistics Rayleigh-Jeans formula Planck radiation law “Very strange people, physicists - in my experience the ones who aren't dead are in some way very ill.”—Douglas Adams Error in syllabus: skip section 8 (in addition to sections 7 and 11) “Testable” sections from chapter 9: 1 through 6, 9, 10. 9.4 Quantum Statistics Here we deal with ideal particles whose wave functions overlap. We introduce quantum physics because of this overlap. Remember: n ε = gε f ε The function f(ε) for quantum statistics depends on whether or not the particles obey the Pauli exclusion principle. “The wierd thing about the half-integral spin particles (also known as fermions) is that when you rotate one of them by 360 degrees, it's wavefunction changes sign. For integral spin particles (also known as bosons), the wavefunction is unchanged.” –Phil Fraundorf of UMSL, discussing why Balinese candle dancers have understood quantum mechanics for centuries. Recall that electrons and other particles with half-integral spin (1/2, 3/2, 5/2, etc.) are fermions and obey the Pauli exclusion principle. The wave function of a system of fermions is antisymmetric because it changes sign upon the exchange of any pair of fermions. We will find that fermions follow Fermi-Dirac statistics. Recall also that photons and other particles with integral spin (0, 1, 2, etc.) are bosons and are not subject to the Pauli exclusion principle. The wave function of a system of bosons is symmetric because it sign remains unchanged upon the exchange of any pair of bosons. We will find that bosons follow Bose-Einstein statistics Let’s refresh our memory on symmetric and antisymmetric wave functions. Much of this discussion (everything inside the yellow boxes) is a repeat of lecture 24. Consider two identical particles (1 and 2) which may exist in two different states (a and b). I = a (1) b (2) II = a (2) b (1) If the particles are indistinguishable, then we cannot tell whether the system is in state I or II, and, because both states are equally likely, we write the system wave function as a linear combination of I and II. If the particles are bosons, the system wave function is symmetric: 1 B = a (1) b (2) + a (2) b (1) = S 2 If the particles are fermions, the wave function is antisymmetric: 1 F = a (1) b (2) - a (2) b (1) = A 2 What happens if we try to put both particles 1 and 2 in the same state? If the particles are distinguishable, we can simply write M = a (1) a (2) . The subscript M indicates Maxwell-Boltzmann statistics,* because the particles are distinguishable. The probability density for distinguishable particles is M*M = a (1)* a (2)* a (1) a (2) . *“Huh? I thought you said Maxwell-Boltzmann statistics is for classical (not quantum) particles. How come the wave functions?” Be quiet! Actually, recall that you can always use QM. You usually don’t use QM unless you have to. In this case, it is useful for comparing MB results with quantum statistics. For bosons B = 2 a (1) a (2) . with a probability density B *B = 2 a (1)* a (2)* a (1) a (2) B *B = 2 M*M . In other words, if the particles are bosons, they are twice as likely to be in the same state as distinguishable particles! On the other hand, if the particles are fermions, 1 F = a (1) a (2) - a (2) a (1) , 2 with a probability density F * F = 0 . If the particles are fermions, it is impossible for both particles to be found in the same state. In general, the presence of a boson in a particular quantum state increases the probability that other bosons will be found in the same state… …but the presence of a fermion in a particular quantum state prevents other fermions from being in that state. We are now (almost) ready to write down the distribution functions for bosons and fermions. Remember, the distribution function gives the probability that a state of energy is occupied by a particle. For bosons, we use a function called the Bose-Einstein (BE) distribution function. In 1924, Indian physicist S. N. Bose submitted a paper using statistical mechanics and the idea of light quanta to explain Planck’s radiation law. The paper was rejected. Einstein translated the paper, got it accepted, and extended Bose’s ideas from photons to other particles. Bose’s paper removed any final objections to the photon theory of light. Bose For fermions, we use a function called the Fermi-Dirac (FD) distribution function. In 1926, Enrico Fermi and Paul Dirac independently realized that the Pauli exclusion principle leads to a different kind of statistics for fermions (including electrons). Dirac shared the 1933 Nobel prize with Schrödinger. Dirac was a brilliant mathematician. Sometimes it seems like all the great mathematicians of the 1930’s worked on problems involving quantum mechanics. Dirac Dirac was famous for saying exactly what he meant, and no more (typical mathematician?). Once when someone, making polite conversation at dinner, commented that it was windy, Dirac excused himself, left the table and went to the door, looked out, returned to the table and replied that indeed it was windy.* When Dirac won the 1933 Nobel prize, he decided to turn it down because he hated publicity. When it was pointed out he would receive far more publicity by turning it down, he changed his mind. *http://www-groups.dcs.st-and.ac.uk/history/Mathematicians/Dirac.html For bosons, the distribution function is fBE (ε) = the difference 1 e e ε / kT -1 . For fermions, the distribution function is fFD (ε) = 1 e e ε / kT +1 . These were derived in an appendix in the previous edition of Beiser, but are just given as truth here. Remember, bosons are particles with integral spins. There is no limit on the number of bosons which may occupy any particular state (typically the ground state). Examples are photons in a cavity, phonons, and liquid 4He. Also remember, fermions are particles with half integral spin, with only one particle per state n, ℓ, mℓ, ms. The +1 in the denominator of f(ε) means that the f(ε) is never greater than 1. Examples are free electrons in metals and nuclei in collapsed stars. The Fermi-Dirac Distribution Function fFD (ε) = 1 e eε / kT +1 The Fermi energy εF is the energy at which fFD = 1/2. From the above equation we see that εF = -kT, and we can write fFD as fFD (ε) = 1 e ε - ε F / kT +1 blowing up the size a bit so you can see better On the next slide is a plot of the Fermi-Dirac distribution function at T=0, 150, 300, and 1000K. T=0 T = 150 T = 300 T = 1000 F = 3 eV Here’s a comparison of our three distribution functions. Bosons “like” to be in the same energy state, so you can cram many of them in together. Fermions don’t “like” to be in the same energy state, so the probatility is the least. When bosons get very cold, they all fall into the same (lowest) energy state. Their wave packets merge and form a single wave packet. If the bosons are atoms, the individual atoms lose their identity and form a “super atom.” Cornell and Wieman of U. Colorado and Ketterle won the 2001 Nobel prize for making a Bose-Einstein condensate out of 2000 rubidium atoms. They had to cool the atoms to 0.000001 K to do this. Formation of the BEC… The BEC is a new state of matter. BEC : ordinary matter :: laser : light bulb light. Practical applications of BEC: none yet. (Who knows…) Study table 9.1, page 310. Good for multiple choice questions. MB BE FD system? … … … particle? classical bosons fermions properties … … … examples … … … dist. funct. … … … dist. properties … … … 9.5 Rayleigh-Jeans Formula “OK, this quantum statistics stuff has some pretty math in it, but what good is it?” We’ll find we need to use Fermi-Dirac statistics to “explain” properties of metals and semiconductors. Bose-Einstein statistics “explains” blackbody radiation. We talked about blackbody radiation in chapter 2 (lecture 5). You should review those notes. A blackbody is the best possible absorber / emitter of radiation. The spectrum of blackbody radiation depends only on the temperature of the blackbody. This is nice because you don't need to worry about the details of the composition of the blackbody. A box with a hole in it makes a good laboratory approximation to a blackbody emitter. Plotted to the right are a couple of spectra for blackbodies at different temperatures. John William Strutt (Lord Rayleigh) tried to calculate this classically. To calculate the spectrum of radiation from a blackbody, consider a box with a hole in it. Radiation can get in the hole. It then reflects back and forth inside the cavity. a spherical cavity turns out to be the best device for experimentally realizing blackbody radiation We've discussed this problem before in several different contexts. If the radiation interferes destructively, it will never get out. We only see radiation (waves) getting out that were able to set up standing waves inside the cavity. Our job is thus to determine what kinds of standing waves can exist in the cavity, and what their energies are. It is not difficult, but it is somewhat tedious, to calculate the number of standing waves allowed for any particular frequency interval. This is the result: g f df = 8L3 c3 f 2 df , where g(f)df is the number of standing waves ("modes") between f and f+df, and L3 is the cavity volume. Note that the number of modes is proportional to f2. The details of this derivation are worthwhile if you are going to study solid state physics at a more advanced level. I won't test you on the derivation in this class. You can calculate the density of standing waves by dividing the above result by the volume, L3. To get the energy density (energy per unit volume), simply multiply g(f) by the energy of an oscillator of frequency f (we assume the waves originate in oscillators in the cavity walls). Remember, kT/2 goes with each degree of freedom, and an oscillator has two degrees of freedom, so multiply by kT. The spectral energy density is thus u f df = 8 f 2 kT c 3 df . This is called the Rayleigh-Jeans formula because British mathematician and astronomer James Jeans corrected a small mathematical error in Rayleigh’s work. That’s 1 point off for a math error. We’ve already seen (chapter 2) how well this works. The formula leads to the "ultraviolet catastrophe." It predicts increasing energy density with increasing f. It doesn't match experiment at all. It doesn't even come close. Something is seriously wrong. theory experiment Rayleigh produced a lifetime of brilliant work. He won the Nobel prize in 1904 for discovering the inert gas argon. However, as Beiser points out, Rayleigh’s most important contribution to physics may have been this disastrous theory of blackbody radiation. Rayleigh died never really accepting the quantum theory of light. “An important scientific innovation rarely makes its way by gradually winning over and converting its opponents…What does happen is that its opponents gradually die out”—Max Planck. 9.6 Planck Radiation Law In the last section we derived the Rayleigh-Jeans formula for the spectral energy density for blackbody radiation. We found that it failed miserably. What's wrong with the Rayleigh-Jeans formula? Rayleigh divided up the energy into kT/2 for each degree of freedom. This is fine for a continuous energy distribution. We now know that it doesn't work for harmonic oscillators, where energy is quantized in units of hf. Therefore, our energy density was wrong. Our density of standing wave modes, g f df = 8L3 c3 f 2 df , as calculated above is OK. To get the correct energy density of photons in the cavity, we multiply our density of modes by the energy of each mode and the probability as a function of T that a mode is occupied. u f df = energy of mode × density of modes × probability that mode is "occupied" u f df = hf G f fBE f df u f df = 8h c 3 f3 e hf / kT -1 df Warning—two different f ’s; might be better to use for frequency here. The last equation on the previous slide is called the Planck radiation formula, and it works very well. Plug it into Mathcad and try it some time. Note that was set equal to 0 in Bose-Einstein distribution function, which was used in the equation above. is a constant which, mathematically, “says” that the number of particles is conserved. If we don't require conservation of particles, we set =0. Wien's displacement law You can calculate the wavelength at the peak of the energy density by expressing u as a function of wavelength and taking the derivative. u f df = 8h c 3 f3 e hf / kT -1 df Use c = f to write u = u() and set 0 = du /d to find for maximum. The result is Wien's displacement law: λ max T = 2.898×10-3 m K . at which u is maximum! not maximum in spectrum! Wien's displacement law tells which way the peak in the blackbody spectrum shifts as the temperature changes. You can write a formula for u(f) as we did, or for u(). The plots look similar: But the Rayleigh-Jeans formula blows up as f and as 0 (can be visually confusing!). Also, the peak in the spectrum shifts towards higher f as T increases, but to lower . In these plots, B corresponds to our u and corresponds to our f. Homework problem 9.31 The brightest part of the spectrum of the star Sirius is located at a wavelength of 290 nm. What is the surface temperature of Sirius? λ max T = 2.898×10-3 m K 2.898×10-3 m K T= λ max 2.898×10-3 m K T= 290×10-9 m T = 9990 K Aren’t stars hotter than this? Yes—interior temperatures, and temperatures a short distance above the surface, reach millions of degrees. Applications: Portable, variable temperature blackbody radiation source for precision calibration of radiation thermometers. Portable thermal imager. Anybody here have children? Whey you do, they come with a sickness guarantee. They are guaranteed to get sick. Then you have to take their temperature. Sure. Ever tried to get a screaming 2year old to hold a thermometer in his (her) mouth? Taking a sick toddler’s temperature is a 2-parent job. Enough said. Or you can get one of these. Worth every penny of the $35-40 it costs. No, I am not affiliated with any thermometer company! The Stefan-Boltzmann Law The total energy density inside the cavity is found by integrating the energy density over all frequencies. utotal = 0 8h f3 c3 ehf / kT -1 The result is proportional to T4. df = a T 4 . a constant The Stefan-Boltzmann law says R=eσT4, where R is the radiated energy per second per unit area, and e is the emissivity of the radiating surface, and σ is a constant (see Beiser for the value of σ). Homework problem 9.26 If a certain blackbody radiates 1 kW when its temperature is 500 ºC, at what rate will it radiate when its temperature is 750 ºC? R = eσ T 4 at 500 o C : some of the other at problems for this material are more complex, but are still fair game for the exam T R 2 = R1 T 4 2 4 1 R1 = eσ T14 750 o C : R 2 = eσ T24 R 2 eσ T24 = R1 eσ T14 don’t forget! 750 +273 T2 = R1 = 1 kW T1 500 +273 4 R 2 = 3.05 kW . 4 Global Warming? It is a fact that human activity has put enormous amounts of “extra” CO2 into the atmosphere since the beginning of the industrial revolution. It is also a fact that CO2 is a greenhouse gas. There exists extensive data which suggests the earth is warming. Much of the data is subject to interpretation. (i.e., one might say “yes, warming,” or one might say “coincidence, not proof.” My sense is that “most” scientists believe something is truly going on. Some very emphatically. Others less so. What will be the result of increasing earth temperature? We don’t know. There will be change. Perhaps dramatic. Perhaps not in the direction we might predict. How do you feel about change? Global warming is intimately connected to blackbody radiation and the Stefan-Boltzmann law. Opinion: those running our government need to learn more physics. Consider designing a cryostat or furnace. What things would you worry about? What would you do to make the cryostat or furnace work well?