Lect25 - PHYS 172 Modern Mechanics Lecture 25 – The...

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Unformatted text preview: PHYS 172: Modern Mechanics Lecture 25 – The Boltzmann Distribution Fall 2011 1 Read 12.7 Summary: Foundations Einstein Model of Solids (q + N ! 1)! # microstates "#= q !(N ! 1)! (N oscillators, q quanta) Fundamental assumption of statistical mechanics Over time, an isolated system in a given macrostate (total energy) is equally likely to be found in any of its microstates (microscopic 2 distribution of energy). Summary: Entropy and Temperature S ! k ln " 1 dS ! T dEint If the initial state is not the most probable, energy is exchanged until the most probable distribution is reached. 3 Summary: Specific Heat Catom = !Eatom " !T !Esystem N atoms !T Al Pb 4 Today: The Boltzmann Distribution Derivation: Choose Your System Wisely Many Systems or Many Observations? Simple Applications: Probabilities for Atomic Excitations Block of Lead Biological Physics 5 Boltzmann Distribution The Boltzmann distribution comes about by cleverly picking our system. Say that we want to analyze a cup of water. What system do we pick? A tiny volume of the water! Example: a macroscopic volume of water in a glass of water Ωres, Eres Ω, E Etot = Eres + E = constant 6 Recall the Pb Nanoparticle (3 atoms) from Ch 11, p 392-393: A q Ω ln Ω S 4 465 6.20 5 1287 6 3003 Temperature: Tq # 1 "S ! T "Eint $E $q % !& o = $S k $(ln ') B C q T(K) 6.20kB 4-5 58.9 7.16 7.16kB 5 62.5 8.01 8.01kB 5-6 66.6 1 !E ( 3 atoms) 1 8 " 10 #22 C= = = 3.4 " 10 #23 J/K per atom 3 !T 3 (66.6 # 58.9 ) What if we wanted to know what is happening on one of the three atoms? For example, with q=6, how often is atom A found with qA=2? 7 What if we wanted to know what is happening on one of the three atoms? For example, with q=6, how often is atom A found with qA=2? We can simply count the states with 2 quanta on atom A: ! A (qA = 2 ) = (2 + 3 " 1)! =6 2!( 3 " 1)! ! BC (qBC = 6 " 2 ) = ( 4 + 6 " 1)! = 126 4 !(6 " 1)! P= ! A (2 )! BC ( 4 ) 756 = " 0.25 !total (6 ) 3003 ! A (2 )! BC ( 4 ) = 6 " 126 = 756 Pb Nanoparticle (3 atoms) 0.4 Probability 0.3 A B C 0.2 0.1 0 0 1 2 3 quanta on atom A 4 5 8 6 Now, a more interesting question: What is the probability of finding 2 quanta on atom A if it is in contact with a large Pb block? (Assume same T as before.) !E !E # # Probability of !E above ground state ~ " A ( !E ) e kT ! e kT Probability of being in ground state Boltzmann Distribution 1 This is a property of any small system in thermal equilibrium with a large reservoir at a fixed temperature. Probability ratio 0.8 0.6 0.4 0.2 0 0 1 2 3 4 Quanta on Atom A 5 6 7 9 !E !E # # Probability of !E above ground state ~ " A ( !E ) e kT ! e kT Probability of being in ground state Boltzmann Distribution We are now going to show WHY this is true 1 Probability ratio 0.8 0.6 0.4 0.2 0 0 1 2 3 4 Quanta on Atom A 5 6 7 10 VIEWER WARNING! The following slides may contain Derivations or Derivation Byproducts. Do not operate heavy machinery while viewing the following slides. Also, do not go swimming for at least one hour after viewing 11 Probabilities Reservoir A ginormous system with a constant temperature tiny system 12 WARNING! This slide may contain Derivations or Derivation Byproducts. Probabilities What is the probability of finding the system in this state? Reservoir (Ginormous) Etot Reservoir (Ginormous) System (Tiny) 0 (Etot - E) System (Tiny) E 13 WARNING! This slide may contain Derivations or Derivation Byproducts. Probabilities Number of ways of having Esys = E We can find the probability from the number of ways of arranging the energy. !res ( Eres )!( E ) P(E ) = !tot ( Etot ) Reservoir (Ginormous) Total number of ways of arranging energy in combined system: Tot = Res + Sys (Etot - E) System (Tiny) E 14 WARNING! This slide may contain Derivations or Derivation Byproducts. Now Do Some Math... !res ( Eres )!( E ) P(E ) = !tot ( Etot ) We can find the probability from the number of ways of arranging the energy. 15 WARNING! This slide may contain Derivations or Derivation Byproducts. More Math... Eres = Etot − E Use a Taylor Expansion for the Ginormous Reservoir: (E << Eres ) Spot The Temperature! 16 WARNING! This slide may contain Derivations or Derivation Byproducts. More Math... Eres = Etot − E Use a Taylor Expansion for the Ginormous Reservoir: (E << Eres ) Spot The Temperature! 17 WARNING! This slide may contain Derivations or Derivation Byproducts. More Math... Exponentiate Everything... BOLTZMANN FACTOR Very very important! 18 WARNING! This slide may contain Derivations or Derivation Byproducts. VIEWER WARNING! The previous slides did contain Derivations and Derivation Byproducts. So, you should not go swimming for at least one hour after class. The Derivation is now OVER. 19 Boltzmann Distribution The probability of finding energy E in a small system in contact with a large reservoir is E ! We just showed kT P E " # E $e WHY this is true () () The exponential part, e-E/kT, is called the Boltzmann factor. Ω(E) is the number of microstates in the small system at energy E. In many circumstances, Ω(E) changes so slowly compared to e-E/kT that it is essentially constant: P (E ) " const # e ! E kT Ωres, Eres Ω, E 20 Boltzmann Distribution The Boltzmann distribution comes about by cleverly picking our system. Say that we want to analyze a cup of water. What system do we pick? A tiny volume of the water! Example: a macroscopic volume of water in a glass of water Ωres, Eres Ω, E Etot = Eres + E = constant 21 Application: Atomic Excitations How likely is an atom to be in 1st excited, compared to odds of being in ground state? P ( E1 ) e−E1 /kT = −E /kT P ( E0 ) e0 = e−∆E/kT NOTE: kT at room temp = 1/40 eV. For the above atom, odds of being in first excited state are P ( excited state )= 1 1 st e "E 1 / 40 eV = 1 e 40 "E = 1 e 40#4 = 3 #10!70 Typical atomic energy gaps are big compared to room temp. A room-temperature box of neon doesn t glow! (Unless you add energy). 22 Application: Block of Lead ￿ω 8 × 10−22 J = = 57K k k T=300K at room temperature BOLTZMANN FACTOR Very very important! Think of adding q quanta of energy to one lead atom... At room temperature, the Boltzmann factors for exciting vibrations are: Most probable state Watch out! These are relative probabilities 23 Application: Block of Lead T=300K at room temperature Think of adding q quanta of vibrational energy to one lead atom... The actual probabilities use a normalization factor Z (partition function). Most likely to find the vibrations are excited. On average, q=5. 24 Today: The Boltzmann Distribution Derivation: Choose Your System Wisely Many Systems or Many Observations? Simple Applications: Probabilities for Atomic Excitations Block of Lead Biological Physics 25 Next Time: Boltzmann Applications Speed Distribution in a Gas Energy Equipartition and Specific Heat Pressure and the Ideal Gas Law 26 ...
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Lect25 - PHYS 172 Modern Mechanics Lecture 25 – The...

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