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h1. Last Time


In a microcanonical ensemble, <math>N, V,</math> and <math>E</math> fixed, and in a canonical ensemble, <math>N, V,</math> and <math>T</math> are constant.  Constants were derived last time, and they were shown to be independent of boundary conditions.  

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The term <math>\beta</math> was determined, and an expression for <math>P_v</math> for canonical ensembles was found. A result was that <math>\beta = \frac{1}{kT}</math>.  Students are advised to be familiar with this type of derivation for the exam.  

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By considering a state function, <math>f = \ln Q</math>, an expression for entropy was found.  By letting the system undergo reversible change <math>dV</math> while coupling to a heat bath, we determined <math>dW_{\mbox{rev}}</math> and then <math>S</math>.  A derivation involved taking the total derivative, adding and rearranging terms, and looking at the total differential of the function.

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Entropy was explained in terms of information theory.  It is proportional to the number of optimal binary questions needed to be asked to determine what state a system is in.

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A case was considered wherein the temperature approaches zero.  This means that <math>\beta \rightarrow \infty</math>.  Below is an expression of the probability and the limit as temperature approaches zero.

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<math>P_{ij} = \frac{e^{-\beta \Delta E_i}}{\sum_v e^{-\beta \Delta E_i}}</math>

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<math>P_{0,GS} = \lim_{\beta \to \infty} \frac{e^{-\beta \Delta E_0}}{\sum_{ij} e^{-\beta \Delta E_i}}</math>

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<math>P_{0,GS} = \frac{1}{\Omega (E_0)}</math>

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Consider the degeneracy of the ground state, <math>\Omega (E_0) = 1</math>.  The probability of being in the ground state is one and the probability of being in any other state is zero.  This means that the value of entropy is zero.  The system is ordered in the ground state.  There is no possible variation, and there is no need to ask any questions about what state the system is in.  There is no ignorance.  This provides a microscopic interpretation of the third law, which states that the entropy of a system approaches a minimum value as the temperature drops to zero.  As the temperature increases, there are more states available, and there is a need to ask questions about what state the system is in.

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<math>P_{GS} = 1</math>

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<math>P_{if \ne GS} = 0</math>

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<math>S = 0</math>

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h2. Summary

Microcanonical Ensemble
* <math>N, V,</math> and <math>E</math> fixed

Canonical Ensemble
* <math>N, V,</math> and <math>T</math> fixed 
* more common

Determined <math>\beta</math> and an expression for <math>P_v</math> for canonical ensembles
* <math>\beta = \frac{1}{kT}</math>
* Students advised to be familiar with this type of derivation for exam

Determined entropy for canonical ensembles. 
* Allow system to undergo reversible change <math>dV</math> while coupling to a heat bath
* Determined <math>dW_{rev}</math> and then <math>S</math>.

[see lecture slides]

Limiting cases:
* <math>T \to 0</math> (<math>\beta \to \infty</math>) 
** <math>P_{ij} = \frac{e^{-\beta \Delta E_i}}{\sum_v e^{-\beta \Delta E_i}}</math>
*** <math>P_{0,GS} = \lim_{\beta \to \infty} \frac{e^{-\beta \Delta E_0}}{\sum_{ij} e^{-\beta \Delta E_i}}</math>
*** <math>P_{0,GS} = \frac{1}{\Omega (E_0)}</math> 
*** <math>\Omega (E_0) = 1</math>
**** <math>P_{GS} = 1</math>
**** <math>P_{ij,GS} = 0</math>
**** <math>S = 0</math>
***** This means that the system is in the ground state. There is no need to ask any questions to know what state the syste is in.
***** The system is ordered in ground state. No ignorance.

Microscopic interpretation of 3rd law

h1. Real Example

Consider a real example where degeneracy is a function of energy.  Consider the graph below.  The degeneracy of the ground state energy is proportional to the number of particles, which is large, approaching <math>10^{23}</math>.

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!Degeneracy_and_energy.PNG!

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<math>\Omega (E_0) \prop N</math>

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<math>N \gg 1</math>

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Entropy is no longer zero; it is something finite.  Below are expressions.

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<math> S = - k_B \sum_{ij} P_{ij} \ln P_{ij} </math>

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<math> S = - k_B \ln \frac{1}{\Omega} </math>

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<math>S = k_B \ln \Omega (E_0)</math>

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h2. Numeric Example 1: Only one state is filled; S  0, complete order; No questions necessary

Consider a numeric example in which only one state is filled.  Calculate the entropy and find it is very small.  The third law is more or less valid.  Compare to entropies calculated at the macroscopic scale.  Sum over the states.

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<math>k_B = 1.38 \cdot 10^{-23}</math>

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<math>S \approx 1.38 \cdot 10^{-23} \ln (10^{23})</math>

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<math>S \approx 10^{-21} \frac{J}{K}</math>

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<math>\mbox{macroscopic scale}</math>

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<math>S \approx k_B \ln (10)^N</math>

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<math>S=3.2 \frac{J}{K}</math>

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h2. Numeric Example 2: All probabilities equal; S  max, maximal disorder; max number of questions

Consider the opposite extreme in which the temperature approaches infinity.  This means that <math>\beta</math> approaches zero.  An exponential term written below is equal to zero for all states.  This means that the probability of any state is equal to one.

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<math>e^{-\beta \Delta E_i} = 1</math>

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This case can be compared to the dice; there are equal probabilities of all outcomes.  The entropy is maximum.  There is maximum disorder.  In this case, there is a maximum number of questions that must be asked to find out what state a system is in.  The ability to access more states increases complexity.  Degeneracy goes up (?).  It is hard to find what state a system is in.

h2. Numeric Example 3: Not all probabilities equal; S  intermediate; need to ask some questions

Consider the case when temperature is greater than zero but does not approach infinity.  Bias the system for some states.  Some states are more likely than others.  There is a non-uniform distribution of probabilities.  Ask binary questions to find what state a system is in.  There are different probabilities of accessing states at different energies.  A plot is below wherein <math>E</math> is the energy needed to produce a state. Consider a solid at high temperature.  There are many more states associated with a liquid state, but it is possible to be in an ordered state for a brief period.

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!Plots_of_probability_versus_energy_associated_with_different_T.PNG!

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h2. Example: Binary Solid

Consider an example of a binary solid.  When the temperature is low, the system is more likely to reside in the ground state.  At intermediate temperatures, the probability of being in a disordered state, <math>P_{\mbox{disordered}}</math>, becomes more significant.  The energy of a disordered state is larger than an ordered state, <math>E_{\mbox{disordered}}</math>, but there is a higher degeneracy.  At high temperature, the probability of the liquid state is imporant.  The degeneracy is much high of a liquid.  There is an entropy driven process of accessing more states.  Higher energies are entropy stabilized.

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!Binary_Solid.PNG!


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<math>\Omega(E_{\mbox{liquid}}) \gg \Omega(E_{\mbox{ordered, disordered})</math>

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<math>\Omega(\mbox{E_{\mbox{dis}}) \gg \Omega(E_0)</math>

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h2. Summary

Consider the degeneracy to be a function of energy
* Degeneracy of ground state: <math>\Omega ( E_0)</math>
* With large degeneracy, <math>\Omega</math>, and many non-interacting particles, some states dominate
* Expression of entropy
** <math>S = - k_B \sum_{ij} P_{ij} \ln P_{ij}</math> 
** <math>S = - k_B \ln \frac{1}{\Omega}</math>
** <math>S = k_B \ln \Omega (E_0)</math>

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Example 1
* Only one state is filled
* <math>S = 0</math>
* complete order
* no questions necessary
* calculation
** <math>k_B = 1.38 \cdot 10^{-23}</math> 
** <math>S \approx 1.38 \cdot 10^{-23} \ln (10^{23})</math>
** <math>S \approx 10^{-21} \frac{J}{K}</math>
* comparison calculation
** <math>S \approx k_B \ln (10)^N</math>
** <math>S=3.2 \frac{J}{K}</math>
* On macroscale, the 3rd law is more or less valid.

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Example 2
* Temperature approaches infinity
* ( <math>\beta \to 0</math> )
* For all states: <math>e^{-\beta \Delta E_i} = 1</math>
* Probability, <math>P_{ij}</math>, is the same for all states
* Entropy: <math>S = -k_B \sum_{ij} P_{ij} \ln P_{ij}</math> 
** Maximum entropy or disorder in this case
** Maximum number of smart questions need to be asked to determine the state

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Example 3
* Consider a temperature between zero and infinity.
** The probability distribution versus energy is non-zero.
** There is a need to ask some binary questions to find what state the system is in
** The term <math>E</math> is the energy needed to produce the state
** When the temperature is small, only the ordered states are accessible
** When the temperature is large, all states are accessible

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Example of binary solid
* <math>T</math> low: <math>P_{\mbox{ordered}} > P_{\mbox{disordered,liquid}}</math>
** System mostly resides in ground state
* <math>T</math> intermediate: <math>P_{\mbox{disordered}}</math> becomes more significant. 
** <math>E_{\mbox{dis}} > E_{\mbox{ordered}}</math> but there is higher degeneracy <math>\Omega (E_{\mbox{disordered}}) >> \Omega ( E_0)</math>
* <math>T</math> high: <math>P_{\mbox{liquid}}</math> becomes important 
** <math>\Omega (E_{\mbox{liquid}}) >> \Omega ( E_{\mbox{ordered,disordered}})</math>

h1. Important Relations with Thermodynamics

Find important relations with thermodynamics.  Calculate the average energy, <math>\overline{E}</math>, and pressure, <math>\overline{p}</math>.  Fix <math>N, V,</math> and <math>T</math> as in a canonical ensemble.  An expression of entropy can be rearranged to relate the entropy to the Hemholtz free energy, which is the characteristic potential for fixed <math>N, V,</math> and <math>T</math>

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<math>S = k_B \ln Q + \frac{\overline{E}}{T}</math>

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<math>-k_B \ln Q = \overline{E} - TS</math>

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<math>-k_B \ln Q = F</math>

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Consider the differential form of the Hemholtz free energy.  The conjugates <math>S, P,</math> and <math>\mu</math> are determined in thermodynamic equilibrium.

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<math>F = -k_B T \ln Q </math>

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<math>\mbox{Entropy}</math>

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<math>S = - \left (\frac{\partial F}{\partial T} \right )_{V, N}</math>

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<math>S = k_B \ln Q + k_B T \left (\frac{\partial \ln Q}{\partial T} \right )_{V, N}</math>

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<math>\mbox{Pressure}</math>

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<math>P = - \left (\frac{\partial F}{\partial V} \right )_{T, N}</math>

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<math>P = k_B T \left (\frac{\partial \ln Q}{\partial V} \right )_{T, N}</math>

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<math>\mbox{Chemical Potential}</math>

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<math>\mu = \left (\frac{\partial F}{\partial N} \right )_{T, V}</math>

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<math>\mu = - k_B T \left (\frac{\partial \ln Q}{\partial N} \right )_{T, V}</math>

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Calculate the partition function through first-principles model.  Solve the Schrondinger equation to find energy levels and determine thermodynamic variables.  There is a simple relation between the characteristic potential and the partition function.  All thermodynamic relations can be found.  Remember expressions of average energy and pressure.

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<math>\overline E = \sum E_v P_v</math>

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<math>\overline p = \sum_v p_v P_v</math>

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<math>\overline p =- \sum_v \left (\frac{\partial E_v}{\partial v} \right ) P_v</math>

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h2. Summary

Consider <math>N, V,</math> and <math>T</math> fixed as in the canonical ensemble
* There is an expression relating <math>\overline E</math> and <math>S</math>, and terms can be rearranged to relate the partition function, <math>Q</math>, to Hemholtz free energy, <math>F</math>.
** <math> S = <math>k_B \ln Q + \frac{\overline E}{T}</math>
** <math> - k_B T \ln Q = \overline E - TS</math>
** <math> - k_B T \ln Q = F</math> 

Put the Hemholtz free energy, which is the characteristic potential for <math>T, V,</math> and <math>N</math> boundary conditions, in differential form and determine thermodynamic relations with conjugates <math>S, P,</math> and <math>\mu</math>
* <math>F = F(T,V,N)</math>
* <math>dF = -SdT - PdV + \mu dN</math>
** Entropy
*** <math>S = - \left (  \frac{\partial F}{\partial T} \right )_{V,N}</math>
*** <math>S = k_B \ln Q + k_B T \left (  \frac{\partial \ln Q}{\partial T} \right )_{V,N}</math>
** Pressure
*** <math>P = - \left (  \frac{\partial F}{\partial V} \right )_{T,N}</math>
*** <math> P=k_B T \left (  \frac{\partial \ln Q}{\partial V} \right )_{T,N} </math>
** Chemical Potential
*** <math>\mu = - \left (  \frac{\partial F}{\partial N} \right )_{T,V}</math>
*** <math> \mu=-k_B T  \left (  \frac{\partial \ln Q}{\partial N} \right )_{T,V} </math>

Derive all thermodynamic relations.

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Remember expression of average energy and pressure
* <math>\overline E = \sum E_v P_v</math>
*<math>\overline p = \sum_v p_v P_v = \sum_v (\frac{\partial E_v}{\partial v} P_v</math>