h1. Fluctuations
There is a spread around the mean for ensembles. Is it enough just to consider the average for one of the fluctuating mechanical variables? A graph of a distribution and an expression of the average of a mechanical variable is below.
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!Probability_versus_mu.png!
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<math>M_{\mbox{thermo}}=\overline{M}</math>
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<math>M_{\mbox{thermo}}=\sum_{\nu} M_{\nu} P_{\nu}</math>
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</center>
As the spread, which is expressed by <math>\sigma</math>, grows larger, the average would not describe the system well. As the spread grows large, the value of <math>M_{\mbox{thermo}}</math> grows questionable. To set the thermodynamic value equal to the average is not a good description. There are so many other energies.
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How large is the spread around the average? Does the spread provide new information?
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<math>\sigma_M = \mbox{second central moment}</math>
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<math>\sigma_M = \sqrt{\overline{ ( M - \overline{M} )^2 }</math>
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h2. Canonical Ensemble
In the canonical ensemble, <math>N, V,</math> and <math>T</math> are constant and <math>E</math> and <math>P</math> are fluctuating mechanical variables. Is the energy, <math>E</math>, really Gaussian? There is a proof in McQuerry. A probability expression and a probability for energy is below. There are many states with the same energy. Look at the probabilities for certain energies. Sum up considering degeneracy. Given a specific set of states means that <math>Q</math> is constant.
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<math>P_{\nu} = \frac{e^{- \beta E_{\nu} }}{Q}</math>
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<math>P(E) = \sum_{E_{\nu} = E} P_{\nu}</math>
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<math>P(E) = \frac{\Omega (E) e^{-\beta E}}{Q}</math>
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<math>P(E) \prop \Omega (E) e^{- \beta E}</math>
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!Probability_versus_energy.PNG!
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The term <math>\Omega (E)</math> increases very rapidly with energy while <math>e^{- \beta E}</math> is decreasing. McQuerry shows more rigorously. The combination of the two curves, <math>\Omega (E)</math> and <math>e^{-\beta E}</math>, is a Gaussian, <math>P(E)</math>.
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Calculate the spread for a certain energy in the canonical ensemble. What is the spread around the average energy? The last two terms are cross terms and are not the same.
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<math>\sigma_E^2=\overline{(E-\overline{E})^2}</math>
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<math>\sigma_E^2=\overline{(E^2-2E\overline{E}+\overline{E}^2)}</math>
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<math>\sigma_E^2=\overline{(\overline{E^2}-2E\overline{E}+\overline{E}^2)}</math>
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<math>\sigma_E^2=\overline{E^2} - 2 \overline{E \overline{E}} + \overline{E}^2</math>
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<math>\sigma_E^2=\overline{E^2} - 2 \overline{E \overline{E}} + \overline{E}^2</math>
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<math>\sigma_E^2=\overline{E^2} - 2 \overline{E}^2 + \overline{E}^2</math>
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<math>\sigma_E^2=\overline{E^2} - \overline{E}^2</math>
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<math>\sigma_E^2=\sum_{\nu} E_{\nu}^2 P_{\nu} - \left ( \sum_{\nu} E_{\nu} P_{\nu} \right )^2</math>
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h2. Summary
h1. Three Step Trick
This series of steps can be applied to various averages.
h2. Step One
Start with the average of a fluctuating variable, <math>{\overline{M}</math>, multiplied by the partition function, <math>Q</math>, applied to particular boundary conditions.
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<math>\overline{E} Q = \sum_{\nu} E_{\nu} e^{- \beta E_{\nu}}</math>
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h2. Step Two
Consider the derivative with respect to the conjugate of fluctuating variable, <math>\beta</math>. The results of differentiating the left- and right-hand side of the equation is below.
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<math>\mbox{left-hand side}</math>
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<math> \frac{\partial}{\partial \beta} \left ( \overline{E} Q \right ) = \frac{\partial \overline{E}}{\partial \beta} Q + \overline{E} \frac{\partial Q}{\partial \beta}</math>
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<math> \frac{\partial}{\partial \beta} \left ( \overline{E} Q \right ) = \frac{\partial \overline{E}}{\partial \beta} Q + \overline{E} \left (- \sum_{\nu} E_{\nu} e^{- \beta E_{\nu}} \right )</math>
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<math>\mbox{right-hand side}</math>
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<math> \frac{\partial}{\partial \beta} \left ( \sum_{\nu} E_{\nu} e^{- \beta E_{\nu}} \right ) = \sum_{\nu} \left ( - E_{\nu}^2 \right ) e^{- \beta E_{\nu}} </math>
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h2. Step Three
Divide by partition function
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<math>\frac{\partial \overline{E}}{\partial \beta} \frac{Q}{Q} + \frac{\overline{E} \left (- \sum_{\nu} E_{\nu} e^{- \beta E_{\nu}} \right )}{Q} = \frac{\sum_{\nu} \left ( - E_{\nu}^2 \right ) e^{- \beta E_{\nu}}}{Q} </math>
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<math>\frac{\partial \overline{E}}{\partial \beta} - \overline{E}^2 = - \overline{E^2} </math>
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<math>\frac{\partial \overline{E}}{\partial \beta} = \overline{E}^2 - \overline{E^2} </math>
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<math>\frac{\partial \overline{E}}{\partial \beta} = - \sigma_E^2 </math>
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<math>\sigma_E^2 = -\frac{\partial \overline{E}}{\partial \beta}</math>
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</center>
h2. Summary
h1. Spontaneous fluctuations in energy
The variance is related to the spontaneous fluctuations in energy. The term on the right-hand side corresponds to a shift in the average energy when perturbing the temperature. The spontaneous fluctuations are related to energy response when fluctuate the temperature. This means that if the system responds much with a change in temperature, there is an easier time fluctuating itself. The spontaneous fluctuations are related to a linear response function. Perturbing the temperature results in a change in the average energy. At a fixed temperature, there is a variation in energy. The extensive variable is fluctuating. The result above is a typical result of fluctuating extensive variables. The calculation of an intensive variable fluctuation is a little more complicated. The same three steps are involved. The calcultion of <math>\sigma_p</math> is a homework problem.
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What is <math>\frac{\partial \overline{E}}{\partial \beta}</math>? Canonical boundary conditions are considered in the last expression.
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<math>\frac{\partial \overline{E}}{\partial \beta} = \frac{\partial \overline{E}}{\partial T} \frac{\partial T}{\partial \beta}</math>
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<math>
\frac{\partial T}{\partial \beta} = \frac{\partial \left ( \frac{1}{k_B \beta}}{\partial \beta}</math>
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<math>\frac{\partial T}{\partial \beta} = - \frac{1}{k_B \beta^2}</math>
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<math>\frac{\partial \overline{E}}{\partial \beta} = -k_B T^2 \frac{\partial \overline{E}}{\partial T}</math>
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<math>\sigma_E^2 = k_B T^2 C_v</math>
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<math>C_v = -T \left ( \frac{\partial^2 F}{\partial T^2 \right )_{V, N}</math>
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The heat capacity is a derivative of energy with respect to the conjugate. It is called susceptibility. Below is the differential form of heat capacity and another expression of heat capacity.
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<math>dF = -SdT - pdV + \mu dN</math>
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<math>C_V = T \left ( \frac{\partial S}{\partial T} \right )_{V, N}</math>
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h2. Summary
h1. Relative Size of Energy Fluctuations
Look at spontaneous fluctuations in systems where certain variables are constant and derive an expression of the fluctuations through properties. Below is a term that is of relevance when considering the relative size of variance. It is the relative deviation from the average.
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<math>\frac{\sigma_E}{E}=\frac{\sqrt{kT^2C_V}}{\overline{E}}</math>
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A good way to understand the order of magnitude of the relative size of fluctuations is to use ideal gas. Expressions of energy and heat capacity are below. There can be a relation between fluctuation size and the number of particles. Consider a mole of particles. The relative size of fluctuations around the gaussian average is very small. This is typical of macroscopic systems/
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<math>\overline{E} \prop Nk_BT</math>
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<math>C_v \prop NK_B</math>
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<math>\frac{\sigma_E}{\overline{E}} \prop \frac{1}{\sqrt{N}}</math>
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<math>N=10^{23}</math>
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<math>\frac{\sigma_{E}}{E} \prop 10^{-11}</math>
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In macroscopic systems, the relative deviations from the mean are very small, which is good. The energy probability distribution of systems with large <math>N</math> is practically a delta function. What happens when <math>N</math> is small? The fluctuations become important when considering microscopic systems. There is a small number of particles in systems at the dimension of a nanometer, and the fluctuations become relevant. Statistical mechanics is applied to macroscopic systems.
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!Probability_versus_E_large_N.PNG!
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h2. Grand Canonical Ensemble
The fixed variables in a grand canonical ensemble are <math>T, V,</math> and <math>\mu</math>, and the fluctuating variables are <math>E, P,</math> and <math>N</math>. Consider the fluctuations of the number of particles. The series of three steps outline above is used between the second and third line.
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<math>\sigma_N^2=\overline{N^2}-\overline{N}^2</math>
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<math>\left ( \sum_{\nu}N_{\nu}^2P_{\nu}-\overline{N}^2 \right )</math>
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<math>\overline{N^2} - \overline{N}^2 = kT \left (\frac{\partial \overline{N}}{\partial \mu} \right )_{V, T}</math>
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The last line looks very similar to what was found before. A relation is used that is also a homework problem.
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<math>\left ( \frac{\partial \mu}{\partial \overline{N}} \right )_{V, T} = - \frac{V^2}{N^2} \left ( \frac{\partial P}{\partial V} \right )_{N, T}</math>
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Use the Gibbs-Duhem relation. Express <math>\sigma_N^2</math> in terms of the compressibility, <math>\mathcal{H}</math>. The variance is related to a susceptibility. It is related to a linear response function.
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<math>SdT - VdP + Nd \mu = 0</math>
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<math>\sigma_N^2 = \frac{N^2 k_B T}{V} \cdot \mathcal{H}</math>
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<math>\mathcal{H} = - \frac{1}{V} \left ( \frac{\partial V}{\partial P} \right )_{N,T}</math>
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<math>\mbox{ideal gas}</math>
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<math>\mathcal{H} = \frac{1}{P}</math>
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<math>\frac{\sigma_N}{N} = \frac{1}{\sqrt{N}}</math>
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h2. Summary
h1. Cases where fluctuations are important
The result above was found by using an equation of state and is typical. It shows the same dependence as in the energy fluctuation of a canonical ensemble. Consider two cases where fluctuations become important. They are important in small systems and around critical points. The former case is not considered in this course, but look at some examples of the latter.
h2. Single-component systems
Consider the transition between a liquid and a gas in a single-component system. A plot of the Gibbs free energy as a function of volume is superimposed over a phase diagram. The letters <math>S</math> and <math>U</math> denote on the free energy plots stable and unstable atates. At the transition between liquid and gas, a barrier between the states disappears. There is an absence of a restoring force and there are large fluctuations. See Chapter 10 of Callen. The spontaneous volume fluctuations are large at <math>T_c</math>
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!Phase_diagram_with_gibbs_energy_plot.PNG!
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<math>\mathcal{H} = - \frac{1}{V} \frac{\Delta V}{\partial P}</math>
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</center>
There is a phenomena called "critical opulesence" wherein droplets of liquid form and strongly scatter light. This is shown with <math>NaCl</math>. Below <math>T_c</math>, the material is transparent. Connect the system to a steam generator and keep under high pressure. The substance become milky-white. There are droplets large enough to scatter light. There are spontaneous fluctuation between the two phases. Increasing the temperature causes the system to turn into a gas.
h2. Multi-component system
Consider order to disorder transitions in <math>Cu-Zn</math>. There is a second-order transition. At the transition, the property <math>C_v</math> diverges as a result of large energy. There can be wild fluctuations and local islands and orderings. Similarly consider ferromagnetic to paramagnetic transition. The divergence of <math>C_v</math> is a sign of phase transitions and is used to calculate where they occur. Fluctuations are related to linear response coefficient.
h2. Summary
h1. Determine equivalence of ensembles
When there are small fluctuations in the canonical, microcanonical, grand canonical, and isobaric-isothermal ensembles, the ensembles are the same. The ensembles are useful under different circumstances and mathematically convenient under different conditions. In the thermodynamic limit where the number of particles approaches infinity, the fluctuations are exceedingly small away from the critical points. The results obtained in one ensemble are the same as in any other.
h2. Adsorbed atoms
Consider the adsorption of oxygen on platinum and the use of the grand canonical ensemble. First, calculate <math>N(\mu)</math> in the grand canonical ensemble through the known framework.
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<math>Q(\mu, V, T)</math>
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<math>N(\mu) = kT \left ( \frac{\partial \ln \theta}{\partial \mu} \right )_{V, T}</math>
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Second, calculate <math>\mu (N) </math> in the canonical ensemble.
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<math>Q(N, V, T)</math>
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<math>\mu(N) = -kT \left ( \frac{\partial \ln Q}{\partial N} \right )_{V, T}</math>
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The same curve is found but shifted. Choose the ensemble to work in based on mathematical convenience.
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!Mu_versus_N.PNG!
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h2. Canonical versus microcanonical
Consider the partition function of a canonical system. Assume that the distribution is a delta spike. The probability, <math>P(E)</math>, is a delta function around <math>E^*</math>
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<math> Q(N, V, T) = \sum_E \Omega (N, V, E) e^{- \beta E}</math>
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<math> Q(N, V, T) = \begin{cases} E=E^* \\ \mbox{all other probability} = 0 \end{cases} </math>
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<math> Q(N, V, T) = \Omega (N, V, E) e^{- \beta E^*}</math>
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<math>F = -k_B T \ln \left ( \Omega e^{- \beta E^*} \right )</math>
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<math>F = -k_B T \ln \Omega+k_B T \beta E^*</math>
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<math>F = -ST+E^*</math>
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If variation is very small, the canonical ensemble is equivalent to the microcanonical ensemble.
h2. Summary
Looked at variance around average of mechanical variables and related to linear response function. Any ensemble can be used and the same result be obtained.
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