Lecture 11. Applying Boltzmann Statistics. Heat capacities. Elasticity of a Polymer. C V of molecules for real!! When equipartition fails

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1 Lecture 11 Applyig Boltzma Statistics Elasticity of a Polymer Heat capacities C V of molecules for real!! Whe equipartitio fails Readig for this Lecture: Elemets Ch 8 Readig for Lecture 12: Elemets Ch 9 Lecture 11, p 1

2 Last time: Boltzma Distributio If we have a system that is coupled to a heat reservoir at temperature T: The etropy of the reservoir decreases whe the small system extracts eergy E from it. Therefore, this will be less likely (fewer microstates). The probability for the small system to be i a particular state with eergy E is give by the Boltzma factor: Z = e E where, to make P tot = 1. P Z is called the partitio fuctio. E / kt = e Z / kt Lecture 11, p 2

3 Two-state Systems i Geeral Cosider a two-state system with a eergy differece E betwee the two states. E 2 E How do the occupatio probabilities of the states vary with T? E kt 1 e P1 = E P2 = E 1 e kt 1+ e kt 0.6 The low eergy state is preferetially P 1 occupied at low T, but the states 0.4 approach equal occupacy at high T. P 2 Probabilities E kt/ E This behavior will be exactly the same for every two-state system with the same E. Lecture 11, p 3

4 Summary: Collectio of Spis We used the Boltzma factor (ad rememberig that the sum of the probabilities is always 1) to tell us the probabilities of each of the two eergy states of a sigle magetic momet i a magetic field. P up B E dow = + µb E up = µb e e = ; P = e + e e + e µ B / kt µ B / kt µ B / kt µ B / kt dow µ B / kt µ B / kt I a collectio, the average umber poitig up (or dow) is just N times the probability: N up = NP up, ad N dow = NP M = µ dow ( Nup Ndow ) = Nµ ( Pup Pdow ) µ B / kt µ B / kt e e = Nµ / / Usig these averages, we ca calculate e µ B kt e µ B kt + macroscopic properties: = Nµ tah ( µ B / kt ) total magetic momet, M iteral eergy, U heat capacity, C B etropy, S Lecture 11, p 4

5 Supplemet: Iteral Eergy of a Collectio of Spis Recall how to calculate the iteral eergy U: U = N u E u +N d E d = -(N u -N d )µb = -NµB tah(µb/kt) What does this look like as a fuctio of T? B E up = -E dow = -µb E dow E up U/µBN kt/µb Low T (kt << µb): Boltzma factor ~ 0. All spis are stuck i low eergy state. U = NE up = -µbn, idepedet of T High T (kt >> µb): Boltzma factor approaches 1. Almost equal umbers i the up ad dow states. U (N/2)(E up +E dow ) = 0, idepedet of T Lecture 11, p 5

6 Supplemet: Heat Capacity of a Collectio of Spis We ow have U(T), for fixed B, so we ca get the heat capacity, C B (at costat B), by takig U/ T. B= cost 2 2 sech / U µ B CB = = Nk B kt T kt ( µ ) For kt << µb, C B vaishes, because all are stuck i groud state. C B /kn For kt >> µb, C vaishes, because the probabilities of the two states each approach 0.5, ad cease to deped o T kt/µb A collectio of 2-state spis does ot behave aythig like a ideal gas. Lecture 11, p 6

7 O the importace of polymers Polymers play a major role i society. I 1930, Wallace Carothers (PhD UIUC, 1924) et. al at DuPot ivet eopree. I 1935 Carothers goes o to ivet ylo the miracle fiber (but commits suicide i 1937, just before it s importace is realized). I WWII, ylo productio was directed to makig parachute caopies. Rubber also played a major role i WWII. You eed rubber for tires, gas masks, plae gaskets, etc. I 1941 our access to 90% of the rubber-producig coutries was cut off by the Japaese attack o Pearl Harbor. What to do? Make sythetic rubber. Who did it first? Carl Speed Marvel, UIUC! Today, plastics are used for may, may, may thigs. Other polymers of ote: cellulose, proteis, DNA, Lecture 11, p 7

8 Statistical Mechaics of a Polymer A polymer is a molecular chai (e.g., rubber), cosistig of may parts liked together. The joits are flexible. Here we cosider a simple (i.e., crude) model of a polymer, to uderstad oe aspect of some of them. Cosider a weight hagig from a chai. Each lik has legth a, ad ca oly poit up or dow. Thus, it s a system cotaiig 2-state compoets. This is similar to the spi problem. Each lik has two eergy states: E = 2aw L a The reaso is that whe a lik flips from dow to up, the weight rises by 2a. (We igore the weight of the chai itself.) I the molecular versio of this experimet, the weight is replaced by a atomic force istrumet. Weight w = mg Lecture 11, p 8

9 Act 1 Suppose our polymer has 30 segmets, each of legth a. Each segmet ca be orieted up or dow. 1) What is the chai legth of the miimum etropy state? a) L = 30a b) L = 0 c) 0 < L < 30a 2) What is the miimum etropy of the chai? w = mg a) σ mi = 0 b) σ mi = 1 c) σ mi = l30 Lecture 11, p 9

10 Solutio Suppose our polymer has 30 segmets, each of legth a. Each segmet ca be orieted up or dow. 1) What is the chai legth of the miimum etropy state? a) L = 30a b) L = 0 c) 0 < L < 30a The miimum etropy state has the fewest microstates or arragemets of the liks. 2) What is the miimum etropy of the chai? w = mg a) σ mi = 0 b) σ mi = 1 c) σ mi = l30 The miimum etropy state (with L=30a) has oly oe microstate. Lecture 11, p 10

11 The Equilibrium Legth The solutio is mathematically the same as the spi system with the substitutio µb wa. The average legth of the rubber bad is (compare with magetizatio result): L Na tah wa = kt The average eergy is: E = -w L N = # segmets, a = segmet legth w = mg As the polymer stretches, its etropy decreases, ad the reservoir s etropy icreases (because U R icreases). The maximum total etropy occurs at a itermediate legth (ot at L=0 or L=Na), where the two effects cacel. Questio: What happes whe you heat the rubber bad? Lecture 11, p 11

12 Act 2 Suppose we rapidly stretch the rubber bad. 1) The etropy of the segmet cofiguratios will a) decrease b) remai the same c) icrease 2) The temperature will a) decrease b) remai the same c) icrease Lecture 11, p 12

13 Solutio Suppose we rapidly stretch the rubber bad. 1) The etropy of the segmet cofiguratios will a) decrease b) remai the same c) icrease The miimum etropy state has the fewest microstates or arragemets of the liks, i.e., the exteded chai. 2) The temperature will a) decrease b) remai the same c) icrease Lecture 11, p 13

14 Solutio Suppose we rapidly stretch the rubber bad. 1) The etropy of the segmet cofiguratios will a) decrease b) remai the same c) icrease The miimum etropy state has the fewest microstates or arragemets of the liks, i.e., the exteded chai. 2) The temperature will a) decrease b) remai the same c) icrease Why is that??? Lecture 11, p 14

15 Act 2 Discussio Because we are stretchig the bad rapidly, this is a example of a adiabatic process: Q = 0 = U - W o U = W o Stretchig the bad does work o it, so U icreases. The liks themselves have o U, so the eergy goes ito the usual kietic eergy (vibratioal) modes T icreases. This is similar to adiabatically compressig a ideal gas (cf. firestarter demo). We ofte used dw o = -pdv. You could redo all the thermal physics, istead for elastic materials, usig dw o = +F dl. Or for batteries, usig dw o = +V dq, Or for magets Lecture 11, p 15

16 Heat Capacity & Harmoic Oscillators We ca use the Boltzma factor to calculate the average thermal eergy, <E>, per particle ad the iteral eergy, U, of a system. We will cosider a collectio of harmoic oscillators. The math is simple (eve I ca do it!), ad It s a good approximatio to reality, ot oly for mechaical oscillatios, but also for electromagetic radiatio. Start with the Boltzma probability distributio: We eed to calculate the partitio fuctio: To do the sum, remember the eergy levels of the harmoic oscillator: E = ε. Equally spaced: Defie: The: x Z ε / kt = e = x = P = (1 e ) e e ε / kt = 0 = 0 ε / kt ε / kt 1 1 x P Z E e = Z = e / kt E It s just a geometric series. / kt E = ε 0 Lecture 11, p 16

17 Heat Capacity & Harmoic Oscillators (2) The ratio ε/kt is importat. Let s look at the probability for a oscillator to have eergy E, for various values of that ratio. P 1 P kt=0.3ε 0.5 kt=ε kt=10ε kt=10hf P kt=100ε kt=100hf The importat feature: At low temperatures, oly a few states have sigificat probability. Lecture 11, p 17

18 Heat Capacity & Harmoic Oscillators (3) Let s calculate the average oscillator eergy, ad the the heat capacity. E = ε, ad P = (1 e ) e so E = 0 = E P = e ε ε / kt ε / kt ε / kt 1 See supplemetal slide for the algebra. At high T (whe kt >> ε), e ε/kt 1 + ε/kt: <E> kt, equipartitio!! <E> 3ε 3 2.5ε 2.5 2ε 2 1.5ε 1.5 <E>=kT, (equipartitio) Actual <E> 0.5ε 1ε 1 At low T (whe kt << ε), e ε/kt >> 1: <E> εe -ε/kt << kt. 0.5ε ε 1ε 1.5ε 2ε 2.5ε 3ε kt Equipartitio requires that kt is much larger tha the eergy level spacig, so that there are may states with E < kt. Lecture 11, p 18

19 Heat Capacity & Harmoic Oscillators (4) Calculate the heat capacity by takig the derivative : C Nk, whe kt 2 ε / kt d E ε e = N = Nk dt k T ε Nk kt ε ε / kt ( e 1) 2 2 ε / kt e Nk, whe kt At what temperature is equipartitio reached? To aswer this, we eed to kow how big ε is. We use a fact from QM (P214): ε = hf, where h is Plack s costat = J-s f is the oscillator frequecy. For typical vibratios i molecules ad solids, kt = hf i the rage 40 K to 4,000 K. ε Nk C 0.5ε Vibratios are froze out at low T Equipartitio holds at high T kt Lecture 11, p 19

20 Act 3 Very sesitive mass measuremets (10-18 g sesitivity) ca be made with aocatilevers, like the oe show. This catilever vibrates with a frequecy, f = 127 MHz. FYI: h = J-s ad k = J/K 1) What is the spacig, ε, betwee this oscillator s eergy levels? Li, et al., Nature Naotechology 2, p114 (2007) a) ε = J b) ε = J c) ε = J 2) At what approximate temperature, T, will equipartitio fail for this oscillator? a) T = K b) T = K c) T = 295 K Lecture 11, p 20

21 Solutio Very sesitive mass measuremets (10-18 g sesitivity) ca be made with aocatilevers, like the oe show. This catilever vibrates with a frequecy, f = 127 MHz. FYI: h = J-s ad k = J/K 1) What is the spacig, ε, betwee this oscillator s eergy levels? Li, et al., Nature Naotechology 2, p114 (2007) a) ε = J b) ε = J c) ε = J ε = hf = ( J-s) ( Hz) = J 2) At what approximate temperature, T, will equipartitio fail for this oscillator? a) T = K b) T = K c) T = 295 K Lecture 11, p 21

Solutio Very sesitive mass measuremets (10-18 g sesitivity) ca be made with aocatilevers, like the oe show. This catilever vibrates with a frequecy, f = 127 MHz. FYI: h = 6.6 10-34 J-s ad k = 1.

22 Solutio Very sesitive mass measuremets (10-18 g sesitivity) ca be made with aocatilevers, like the oe show. This catilever vibrates with a frequecy, f = 127 MHz. FYI: h = J-s ad k = J/K 1) What is the spacig, ε, betwee this oscillator s eergy levels? Li, et al., Nature Naotechology 2, p114 (2007) a) ε = J b) ε = J c) ε = J ε = hf = ( J-s) ( Hz) = J 2) At what approximate temperature, T, will equipartitio fail for this oscillator? a) T = K b) T = K c) T = 295 K T = ε/k = ( J)/( J/K) = K Lecture 11, p 22

23 FYI: Recet Physics Milestoe! There has bee a race over the past ~20 years to put a ~macroscopic object ito a quatum superpositio. The first step is gettig the object ito the groud state, below all thermal excitatios. This was achieved for the first time i 2010, usig vibratios i a small drum : Quatum groud state ad sigle-phoo cotrol of a mechaical resoator, A. D. O Coell, et al., Nature 464, (1 April 2010) Quatum mechaics provides a highly accurate descriptio of a wide variety of physical systems. However, a demostratio that quatum mechaics applies equally to macroscopic mechaical systems has bee a log-stadig challege... Here, usig covetioal cryogeic refrigeratio, we show that we ca cool a mechaical mode to its quatum groud state by usig a microwave-frequecy mechaical oscillator a quatum drum... We further show that we ca cotrollably create sigle quatum excitatios (phoos) i the resoator, thus takig the first steps to complete quatum cotrol of a mechaical system. Lecture 11, p 23

24 FYI: Heat Capacity of a Eistei Solid 3N SHO s Cosider a solid as atomic masses coected by sprigs (the atomic bods): Small system (oe atom) Eistei preteds it oscillates idepedetly of other atoms. For high T, Equipartitio Theorem predicts ½ kt for each quadratic term i the eergy: mvx + mvy + mv +κx +κy +κz ) = 3 ( z The eergy ad heat capacity of the etire solid (N atoms) is: kt du U = N< E > = 3NkT CV = = 3Nk dt What about low temperatures? Lecture 11, p 24

25 FYI: Heat Capacity of Eistei solid For a solid with N atoms, total vibratioal eergy is: The heat capacity at costat volume is: C U = 3 N < E > = V (3-D) U T V 3Nε 1 / kt e ε Slope of this U 3NkT T C V ε kt ε kt Classical Result (Lecture 3) 3Nk Few levels populated, heat capacity goes to zero. May levels populated diamod K T Lecture 11, p 25

26 May Modes of Motio? If a molecule has several modes of motio, some may be i equipartitio, while others may be froze out. Cosider a diatomic molecule (H 2 ). It has three quadratic eergy modes: Bod vibratios have a larger ε, correspodig to T ~ 1000 K. Rotatios have a moderate eergy spacigs,correspodig to T ~ 100 K. Traslatios have a cotiuous rage of eergies ever froze out C v 7/2 Nk 5/2 Nk 3/2 Nk Heat Capacity of H 2 Traslatios ever freeze out Rotatios cotribute above T ~ 100 K. Vibratios cotribute above T ~ 1000 K. At T = 300 K, traslatios ad rotatios cotribute to the heat capacity, but bod vibratios do ot. 10K 100K 1000K T Lecture 11, p 26

27 Law of Atmospheres Next Class Thermal Radiatio Next Week: Heat Egies Thermodyamic processes ad etropy Thermodyamic cycles Extractig work from heat Lecture 11, p 27

28 Supplemet: Derivatio of <E> for the Harmoic Oscillator This is always true: E / kt Eβ Z = e = e β kt dz dβ = E e E β, so ( 1/ ) 1 dz e - = EP = E, because P = Z dβ Z E β This is true for the harmoic oscillator: 1 Z = 1 εβ e dz 1 εβ εe = e ε = dβ εβ 1 e 1 e ( ) ( ) ( ) 1 dz ε E = = εβ Z dβ e 1 εβ εβ ( ) 2 2 Lecture 11, p 28

29 Derivatio of Equipartitio for quadratic degrees of freedom To calculate <E>, we must perform a sum: If kt >> ε (the eergy spacig), the we ca tur this sum ito a itegral: q is the variable that determies E (e.g., speed). The oly subtle part is ρ(q). This is the desity of eergy states per uit q, eeded to do the coutig right. For simplicity, we ll assume that 2 ρ is costat. aq 2 kt aq e dq kt Calculate <E>, assumig that E = aq 2 : So, equipartitio follows aturally from simple assumptios, ad we kow whe it fails. See the supplemet for the behavior of liear modes. E E q = EP = 1 = = e ( ) ( ) ρ ( ) E = E q P q q dq 0 2 aq kt dq 2 Lecture 11, p 29

30 Supplemet: Equipartitio for Liear Degrees of Freedom Whe we talk about equipartitio, (<E> = ½kT per mode) we say quadratic, to remid us that the eergy is a quadratic fuctio of the variable (e.g., ½mv 2 ). However, sometimes the eergy is a liear fuctio (e.g., E = mgh). How does equipartitio work i that case? Boltzma tells us the aswer! Let s calculate <E>, assumig that there are lots of states with E < kt, (ecessary for equipartitio), ad that these states are uiformly spaced i y (to simplify the calculatio). Suppose E(y) = ay. ( ) E( y ) ay kt kt 2 E ( y ) e dy aye dy ( kt ) 0 0 a E( y ) ay kt kt E = E y Pdy = = = = kt kt 0 e dy e dy a 0 0 So, each liear mode has twice as much eergy, kt, as each quadratic mode. Lecture 11, p 30

Shedding light on atomic energy levels (segment of Hydrogen spectrum)

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