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PHY452H1S Basic Statistical Mechanics. Lecture 8: Midterm review, thermodynamics. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on February 7, 2013

[Click here for a PDF of this post with nicer formatting]

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Exam review

  • Product of IID random vars. What can we say about this with the CLT

    \begin{aligned}X = x_1 x_2 \cdots x_N\end{aligned} \hspace{\stretch{1}}(1.2.1)

    We can’t apply CLT directly, but we can apply it to \ln X

    \begin{aligned}Y = \ln X = \sum_i \ln x_i = \sum_i y_i\end{aligned} \hspace{\stretch{1}}(1.2.2)

    So

    \begin{aligned}\tilde{P}(Y) = \frac{1}{{\sqrt{ 2 \pi N \sigma_y^2}}} \exp\left( {-\frac{(Y - N \mu_y)^2}{N \sigma_y^2}} \right)\end{aligned} \hspace{\stretch{1}}(1.2.3)

    Of interest here is how probabilities change under change of variables. If Y = \ln X, this works like

    \begin{aligned}P(X) dX = \tilde{P}(Y) dY\end{aligned} \hspace{\stretch{1}}(1.2.4)

    \begin{aligned}P(X) = \tilde{P}(Y = \ln X) \frac{dY}{dX}.\end{aligned} \hspace{\stretch{1}}(1.2.5)

    This is illustrated roughly in (Fig 1)

    Fig 1: Change of variables for a probability distribution

     

  • Random walkThis was a unit stepping problem as illustrated in (Fig2).

    Fig2: Unit one dimensional random walk

     

    \begin{aligned}\left\langle{{X}}\right\rangle = \sum_i \left\langle{{x_i}}\right\rangle\left\langle{{X^2}}\right\rangle = \left\langle{{\left( {\sum_{i = 1}^N x_i} \right)\left( {\sum_{j = 1}^N x_i} \right)}}\right\rangle= \left\langle{{\left( {\sum_{i = 1}^N x_i^2} \right)\left( {\sum_{i \ne j = 1}^N x_i x_j} \right)}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.2.6)

  • State and prove Liouville’s theorem

    \begin{aligned}\frac{d\rho}{dt} = \frac{\partial {\rho}}{\partial {t}} + \sum_{i = 1}^{3N} \frac{\partial {\left( {\dot{x}_i \rho} \right)}}{\partial {x_i}}+\frac{\partial {\left( {\dot{x}_i \rho} \right)}}{\partial {x_i}}\end{aligned} \hspace{\stretch{1}}(1.2.7)

  • Collisions in 1DThis was a one direction collision problem for N particles, a portion of which is illustrated in (Fig 3).

    Fig3: One dimensional collision of particles

    The statement here that the collision was in 1D was a hint that we can actually calculate the result.

    Start with a pair of collisions and work out that the velocities are exchanged.

    \begin{aligned}(v_i, v_{i+1}) \rightarrow (v_{i+1}, v_{i})\end{aligned} \hspace{\stretch{1}}(1.2.8)

    We have only exchange velocities. This won’t result in all the phase space being explored, and was meant to show that things are extremely restrictive in 1D.

  • Harmonic oscillator in 1D.This problem ends up essentially requiring the evaluation of the area in phase space of an ellipse in phase space as in (Fig 4).

    Fig4: 1D classical SHO phase space

    The result is \pi a b \times \text{area}.

Gibbs paradox

We’d been discussing Gibbs paradox, considering the doubled volume as in (Fig 5).

Fig5: Gibbs paradox two volumes allowed to mix

and defined a \underlineAndIndex{statistical entropy} that requires division of \Omega by N!. We have

\begin{aligned}S_statistical = k_B \ln \underbrace{\Omega}_{\text{Number of states}}\end{aligned} \hspace{\stretch{1}}(1.3.9)

Thermodynamics – phenomenology

Let’s move on to discuss some concepts of thermodynamics

First law of thermodynamics

This is the law of \underlineAndIndex{Energy conservation}! \footnote{This requires some corrections if relativisitic systems are studied.}

\begin{aligned}\underbrace{dE}_{\text{Change in energy}}=\underbrace{d W}_{\text{work done on the system}}+\underbrace{d Q}_{\text{Heat supplied to the system}}\end{aligned} \hspace{\stretch{1}}(1.4.10)

Here the d differentials indicate that the change depends is path dependent. For example the number of times a piston is moved back and forth can be significant, even given a specific change in the total differential change in the piston position. The total change of energy in such a change is not path dependent.

Consider a piston setup compressing some gas taking the position of the pistion from X \rightarrow X + dX as in (Fig 6).

Fig6: Piston and gas

\begin{aligned}d W = \sum_i \underbrace{f_i}_{\text{Generated force}}\underbrace{dX_i}_{\text{Controllable macroscopic coordinate}}\end{aligned} \hspace{\stretch{1}}(1.4.11)

What is d Q? These are the microscopic, and uncontrollable changes in energy.

Processes

  1. Change X_i \rightarrow X_i + dX_i
  2. Change d Q

Adiabatic process

These are those that are thermally isolated, or

\begin{aligned}d Q = 0.\end{aligned} \hspace{\stretch{1}}(1.4.12)

The only way to change energy is to change the macroscopic X_i, so the change in energy is given by

\begin{aligned}dE = d W = \sum_i f_i dX_i\end{aligned} \hspace{\stretch{1}}(1.4.13)

These are what we are used to thinking about as conservative systems. Imagine, for example, a force applied to a spring, slowly so that there is no heat loss, we can figure out the total change in energy by summing differential changes to the spring.

We can imagine that there is some way to change the position that results in a change of energy. This is of course just the force as in (Fig 7).

\begin{aligned}-\frac{dE}{dx_i} \sim \text{force}.\end{aligned} \hspace{\stretch{1}}(1.4.14)

Fig 7: A possible energy position relationship

We can imagine that we started with an initially thermally isolated system, with the macroscopic parameter unchanged, but somebody sneaks in and heats the system, before re-thermally isolating it. We’d then have a system that perhaps looks like (Fig 8).

Fig 8: Level curves for adiabatic and non-adiabatic changes to the system

The jumps from one curve from another is the product of the sneaky heat suppliers.

Now, if we think about this in a higher dimensional space, we’d instead have level surfaces.

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