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# Posts Tagged ‘average’

## Final version of my phy452.pdf notes posted

Posted by peeterjoot on September 5, 2013

I’d intended to rework the exam problems over the summer and make that the last update to my stat mech notes. However, I ended up studying world events and some other non-mainstream ideas intensively over the summer, and never got around to that final update.

Since I’m starting a new course (condensed matter) soon, I’ll end up having to focus on that, and have now posted a final version of my notes as is.

Since the last update the following additions were made

September 05, 2013 Large volume fermi gas density

April 30, 2013 Ultra relativistic spin zero condensation temperature

April 24, 2013 Low temperature Fermi gas chemical potential

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

## open system variance of N

Posted by peeterjoot on March 16, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

## Question: Variance of $N$ in open system ([1] pr 3.14)

Show that for an open system

\begin{aligned}\text{var}(N) = \frac{1}{{\beta}} \left({\partial {\bar{N}}}/{\partial {\mu}}\right)_{{V, T}}.\end{aligned} \hspace{\stretch{1}}(1.0.1)

## Answer

In terms of the grand partition function, we find the (scaled) average number of particles

\begin{aligned}\frac{\partial {}}{\partial {\mu}} \ln Z_{\mathrm{G}} &= \frac{\partial {}}{\partial {\mu}} \ln \sum_{r,s} e^{\beta \mu N_r - \beta E_s} \\ &= \frac{1}{{Z_{\mathrm{G}}}} \sum_{r,s} \beta N_r e^{\beta \mu N_r - \beta E_s} \\ &= \beta \bar{N}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Our second derivative provides us a scaled variance

\begin{aligned}\frac{\partial^2 {{}}}{\partial {{\mu}}^2} \ln Z_{\mathrm{G}} &= \frac{\partial {}}{\partial {\mu}} \left( \frac{1}{{Z_{\mathrm{G}}}} \sum_{r,s} \beta N_r e^{\beta \mu N_r - \beta E_s} \right) \\ &= \frac{1}{{Z_{\mathrm{G}}}} \sum_{r,s} (\beta N_r)^2 e^{\beta \mu N_r - \beta E_s}-\frac{1}{{Z_{\mathrm{G}}^2}} \left( \sum_{r,s} \beta N_r e^{\beta \mu N_r - \beta E_s} \right)^2 \\ &= \beta^2 \left( \bar{N^2} - {\bar{N}}^2 \right)\end{aligned} \hspace{\stretch{1}}(1.0.3)

Together this gives us the desired result

\begin{aligned}\text{var}(N) &= \frac{1}{{\beta^2}}\frac{\partial {}}{\partial {\mu}} \left( \beta \bar{N} \right) \\ &= \frac{1}{{\beta}}\frac{\partial {\bar{N}}}{\partial {\mu}}.\end{aligned} \hspace{\stretch{1}}(1.0.4)

# References

[1] E.A. Jackson. Equilibrium statistical mechanics. Dover Pubns, 2000.

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

## One dimensional random walk

Posted by peeterjoot on February 3, 2013

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

## Question: One dimensional random walk

Random walk in 1D by unit steps. With the probability to go right of $p$ and a probability to go left of $1 -p$ what are the first two moments of the final position of the particle?

## Answer

This was a problem from the first midterm. I ran out of time and didn’t take the answer as far as I figured I should have. Here’s a more casual bash at the problem.

First we need the probabilities.

One step: $N = 1$

Our distance (from the origin) can only be $X = \pm 1$.

\begin{aligned}P_{X = -1} = p^{0} (1 -p)^1\end{aligned} \hspace{\stretch{1}}(1.0.1)

\begin{aligned}P_{X = 1} = p^1 (1-p)^{1 - 1}\end{aligned} \hspace{\stretch{1}}(1.0.2)

Two steps: $N = 2$

We now have three possibilities

\begin{aligned}P_{X = -2} = p^{0} (1 -p)^{2 - 0}\end{aligned} \hspace{\stretch{1}}(1.0.3)

\begin{aligned}P_{X = 0} = 2 p^1 (1-p)^{2 - 1}\end{aligned} \hspace{\stretch{1}}(1.0.4)

\begin{aligned}P_{X = 2} = p^2 (1-p)^{2 - 2}\end{aligned} \hspace{\stretch{1}}(1.0.5)

Three steps: $N = 3$

We now have three possibilities

\begin{aligned}P_{X = -3} = p^{0} (1 - p)^{3 - 0}\end{aligned} \hspace{\stretch{1}}(1.0.6)

\begin{aligned}P_{X = -1} = 3 p^1 (1 - p)^{3 - 1}\end{aligned} \hspace{\stretch{1}}(1.0.7)

\begin{aligned}P_{X = 1} = 3 p^2 (1 - p)^{3 - 2}\end{aligned} \hspace{\stretch{1}}(1.0.8)

\begin{aligned}P_{X = 3} = p^3 (1-p)^{3 - 3}\end{aligned} \hspace{\stretch{1}}(1.0.9)

General case

The pattern is pretty clear, but we need a mapping from the binomial index to the the final distance. With an index $k$, and a guess

\begin{aligned}D(k) = a k + b,\end{aligned} \hspace{\stretch{1}}(1.0.10)

where

\begin{aligned}D(0) = -N = b\end{aligned} \hspace{\stretch{1}}(1.0.11)

\begin{aligned}D(N) = a N + b = (a - 1)N = N.\end{aligned} \hspace{\stretch{1}}(1.0.12)

So

\begin{aligned}D(k) = 2 k - N,\end{aligned} \hspace{\stretch{1}}(1.0.13)

and

\begin{aligned}k = \frac{D + N}{2}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Our probabilities are therefore

\begin{aligned}\boxed{P_{X = D} = \binom{N}{(N + D)/2} p^{(N + D)/2}(1 - p)^{(N - D)/2}.}\end{aligned} \hspace{\stretch{1}}(1.0.15)

First moment

For the expectations let’s work with $k$ instead of $D$, so that the expectation is

\begin{aligned}\left\langle{{X}}\right\rangle &= \sum_{k = 0}^N (2 k - N) \binom{N}{k} p^k (1 - p)^{N - k} \\ &= 2 \sum_{k = 0}^N k \frac{N!}{(N-k)!k!} p^k (1 - p)^{N - k} - N \\ &= 2 N p \sum_{k = 1}^N \frac{(N-1)!}{(N - 1 - (k - 1))!(k-1)!} p^{k-1} (1 - p)^{N - 1 - (k - 1)} - N \\ &= 2 N p \sum_{s = 0}^{N-1} \binom{N-1}{s} p^{s} (1 - p)^{N -1 - s} - N.\end{aligned} \hspace{\stretch{1}}(1.0.16)

This gives us

\begin{aligned}\boxed{\left\langle{{X}}\right\rangle = N( 2 p - 1 ).}\end{aligned} \hspace{\stretch{1}}(1.0.17)

Second moment

\begin{aligned}\left\langle{{X^2}}\right\rangle &= \sum_{k = 0}^N (2 k - N)^2 \binom{N}{k} p^k (1 - p)^{N - k} \\ &= 4 \sum_{k = 0}^N k^2 \binom{N}{k} p^k (1 - p)^{N - k}- 4 N^2 p+ N^2 \\ &= 4 N p \sum_{k = 1}^N k \frac{(N-1)!}{(N - 1 - (k - 1))! (k-1)!} p^{k-1} (1 - p)^{N - k} + N^2(1 - 4 p) \\ &= 4 N p \sum_{s = 0}^N (s + 1) \frac{(N-1)!}{(N - 1 - s)! s!} p^s (1 - p)^{N - 1 - s} + N^2(1 - 4 p) \\ &= 4 N p ((N-1) p + 1) + N^2(1 - 4 p) \\ &= N^2 ( 1 - 4 p + 4 p^2 ) + 4 N p ( 1 - p ).\end{aligned} \hspace{\stretch{1}}(1.0.18)

So the second moment is

\begin{aligned}\boxed{\left\langle{{X^2}}\right\rangle = N^2 ( 1 - 2 p )^2 + 4 N p ( 1 - p )}\end{aligned} \hspace{\stretch{1}}(1.0.19)

From this we see that the variance is just this second term

\begin{aligned}\sigma^2 = \left\langle{{X^2}}\right\rangle - \left\langle{{X}}\right\rangle^2 = 4 N p ( 1 - p ).\end{aligned} \hspace{\stretch{1}}(1.0.20)