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Posts Tagged ‘Fermi gas’

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

May 30, 2013 Bernoulli polynomials and numbers and Euler-MacLauren summation

May 09, 2013 Bose gas specific heat above condensation temperature

May 09, 2013 A dumb expansion of the Fermi-Dirac grand partition function

April 30, 2013 Ultra relativistic spin zero condensation temperature

April 30, 2013 Summary of statistical mechanics relations and helpful formulas

April 24, 2013 Low temperature Fermi gas chemical potential

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Large volume Fermi gas density

Posted by peeterjoot on September 5, 2013

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

Here’s part of a problem from our final exam. I’d intended to redo the whole exam over the summer, but focused my summer study on world events instead. Perhaps I’ll end up eventually doing this, but for now I’ll just post this first part.

Question: Large volume Fermi gas density (2013 final exam pr 1)

Write down the expression for the grand canonical partition function Z_{\mathrm{G}} of an ideal three-dimensional Fermi gas with atoms having mass m at a temperature T and a chemical potential \mu (or equivalently a fugacity z = e^{\beta \mu}). Consider the high temperature “classical limit” of this ideal gas, where z \ll 1 and one gets an effective Boltzmann distribution, and obtain the equation for the density of the particles

\begin{aligned}n = \frac{1}{{V \beta}} \frac{\partial {\ln Z_{\mathrm{G}}}}{\partial {\mu}}\end{aligned} \hspace{\stretch{1}}(1.1)

by converting momentum sums into integrals. Invert this relationship to find the chemical potential \mu as a function of the density n.

Hint: In the limit of a large volume V:

\begin{aligned}\sum_\mathbf{k} \rightarrow V \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}.\end{aligned} \hspace{\stretch{1}}(1.2)

Answer

Since it was specified incorrectly in the original problem, let’s start off by verifing the expression for the number of particles (and hence the number density)

\begin{aligned}\frac{1}{{\beta}} \frac{\partial {}}{\partial {\mu}} \ln Z_{\mathrm{G}} &= \frac{1}{{\beta}} \frac{\partial {}}{\partial {\mu}} \ln \sum_N z^N e^{-\beta E_N} \\ &= \frac{1}{{\beta}} \frac{1}{{\Omega}}\frac{\partial {}}{\partial {\mu}} \sum_N z^N e^{-\beta E_N} \\ &= \frac{1}{{\beta}} \frac{1}{{\Omega}}\sum_N e^{-\beta E_N} \frac{\partial {}}{\partial {\mu}} e^{\mu \beta N} \\ &= \frac{1}{{\Omega}}\sum_N N z^N e^{-\beta E_N} \\ &= \left\langle{{N}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.3)

Moving on to the problem, we’ve seen that the Fermion grand canonical partition function can be written

\begin{aligned}Z_{\mathrm{G}} = \prod_\epsilon \left( { 1 + z e^{-\beta \epsilon}} \right),\end{aligned} \hspace{\stretch{1}}(1.4)

so that our density is

\begin{aligned}n &= \frac{N}{V} \\ &= \frac{1}{{ V \beta }}\frac{\partial {}}{\partial {\mu}} \ln \prod_\epsilon \left( { 1 + z e^{-\beta \epsilon}} \right) \\ &= \frac{1}{{ V \beta }}\sum_\epsilon \frac{\partial {}}{\partial {\mu}} \ln \left( { 1 + z e^{-\beta \epsilon}} \right) \\ &= \frac{1}{{ V \beta }}\sum_\epsilon \frac{\partial {z}}{\partial {\mu}}\frac{e^{-\beta \epsilon}}{1 + z e^{-\beta \epsilon}} \\ &= \frac{1}{{ V }}\sum_\epsilon z\frac{e^{-\beta \epsilon}}{1 + z e^{-\beta \epsilon}}.\end{aligned} \hspace{\stretch{1}}(1.5)

In the high temperature classical limit, where z \ll 1 we have

\begin{aligned}n \\ &\approx \frac{1}{ V } \sum_\epsilon z e^{-\beta \epsilon} \\ &\approx z \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}e^{-\beta \epsilon} \\ &= \frac{2 z}{(2 \pi)^2} \int_0^\infty k^2 dk e^{-\beta \frac{\hbar^2 k^2}{2m} } \\ &= \frac{z}{2 \pi^2} \left( { \frac{\beta \hbar^2}{2m} } \right)^{-3/2} \int_0^\infty x^2 e^{-x^2} dx \\ &= \frac{z}{2 \pi^2} \left( { \frac{2m}{\beta \hbar^2} } \right)^{3/2} \frac{\sqrt{\pi}}{4} \\ &= \frac{z}{8 \pi^{3/2}} \frac{\left( {8 \pi^2 m k_{\mathrm{B}} T } \right)^{3/2}}{h^3} \\ &= z \frac{\left( { 2 \pi m k_{\mathrm{B}} T } \right)^{3/2}}{h^3}.\end{aligned} \hspace{\stretch{1}}(1.6)

This is

\begin{aligned}\boxed{n = \frac{z}{\lambda^3},}\end{aligned} \hspace{\stretch{1}}(1.7)

where

\begin{aligned}\lambda = \frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}}.\end{aligned} \hspace{\stretch{1}}(1.8)

Inverting for \mu we have

\begin{aligned}\mu = \frac{1}{\beta} \ln z\end{aligned} \hspace{\stretch{1}}(1.10)

or

\begin{aligned}\boxed{\mu = k_{\mathrm{B}} T \ln \left( { n \lambda^3 } \right).}\end{aligned} \hspace{\stretch{1}}(1.10)

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Low temperature Fermi gas chemical potential

Posted by peeterjoot on April 24, 2013

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Question: Low temperature Fermi gas chemical potential

[1] section 8.1 equation (33) provides an implicit function for \mu \equiv k_{\mathrm{B}} T \ln z

\begin{aligned}n = \frac{4 \pi g}{3} \left( \frac{2m}{h^2}  \right)^{3/2}\mu^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{ (k_{\mathrm{B}} T)^2 }{ \mu^2 } + \cdots  \right),\end{aligned} \hspace{\stretch{1}}(1.0.1)

or

\begin{aligned}E_{\mathrm{F}}^{3/2} = \mu^{3/2} \left( 1 + \frac{\pi^2}{8} \frac{ (k_{\mathrm{B}} T)^2 }{ \mu^2 } + \cdots \right).\end{aligned} \hspace{\stretch{1}}(1.0.2)

In class, we assumed that \mu was quadratic in k_{\mathrm{B}} T as a mechanism to invert this non-linear equation. Without making this quadratic assumption find the lowest order, non-constant approximation for \mu(T).

Answer

To determine an approximate inversion, let’s start by multiplying eq. 1.0.2 by \mu^{1/2}/E_{\mathrm{F}}^2 to non-dimensionalize things

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}}  \right)^{1/2} = \left( \frac{\mu}{E_{\mathrm{F}}} \right)^2 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.3)

or

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}}  \right)^{1/2} =\frac{1}{{ 1 - \left( \frac{\mu}{E_{\mathrm{F}}} \right)^{3/2} }}\frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2.\end{aligned} \hspace{\stretch{1}}(1.0.4)

If we are looking for an approximation in the neighborhood of \mu = E_{\mathrm{F}}, then the LHS factor is approximately one, whereas the fractional difference term is large (with a corresponding requirement for k_{\mathrm{B}} T/E_{\mathrm{F}} to be small. We must then have

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}} \right)^{3/2} \approx 1 - \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.4)

or

\begin{aligned}\mu\approx E_{\mathrm{F}}\left(1 - \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right)^{2/3}\approx E_{\mathrm{F}}\left(1 - \frac{2}{3} \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right).\end{aligned} \hspace{\stretch{1}}(1.0.4)

This gives us the desired result

\begin{aligned}\boxed{\mu \approx E_{\mathrm{F}}\left(1 - \frac{\pi^2}{12} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right).}\end{aligned} \hspace{\stretch{1}}(1.0.7)

References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

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A final pre-exam update of my notes compilation for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

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

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

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

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Relativisitic Fermi gas

Posted by peeterjoot on April 20, 2013

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Question: Relativisitic Fermi gas ([1], pr 9.3)

Consider a relativisitic gas of N particles of spin 1/2 obeying Fermi statistics, enclosed in volume V, at absolute zero. The energy-momentum relation is

\begin{aligned}\epsilon = \sqrt{(p c)^2 + \epsilon_0^2},\end{aligned} \hspace{\stretch{1}}(1.1)

where \epsilon_0 = m c^2, and m is the rest mass.

Find the Fermi energy at density n.

With the pressure P defined as the average force per unit area exerted on a perfectly-reflecting wall of the container.

Set up expressions for this in the form of an integral.

Define the internal energy U as the average \epsilon - \epsilon_0.
Set up expressions for this in the form of an integral.

Show that P V = 2 U/3 at low densities, and P V = U/3 at high densities. State the criteria for low and high densities.

There may exist a gas of neutrinos (and/or antineutrinos) in the cosmos. (Neutrinos are massless Fermions of spin 1/2.) Calculate the Fermi energy (in eV) of such a gas, assuming a density of one particle per \text{cm}^3.

Attempt exact evaluation of the various integrals.

Answer

We’ve found [3] that the density of states associated with a 3D relativisitic system is

\begin{aligned}\mathcal{D}(\epsilon) = \frac{4 \pi V}{(c h)^3} \epsilon \sqrt{\epsilon^2 -\epsilon_0^2},\end{aligned} \hspace{\stretch{1}}(1.0.2)

For a given density n, we can find the Fermi energy in the same way as we did for the non-relativisitic energies, with the exception that we have to integrate from a lowest energy of \epsilon_0 instead of 0 (the energy at \mathbf{p} = 0). That is

\begin{aligned}n &= \frac{N}{V} \\ &= \left( 2 \frac{1}{{2}} + 1 \right)\frac{4 \pi}{(c h)^3} \int_{\epsilon_0}^{\epsilon_{\mathrm{F}}}d\epsilon \epsilon \sqrt{ \epsilon^2 -\epsilon_0^2} \\ &= \frac{8 \pi}{(c h)^3}\frac{1}{{3}} {\left.{{\left( x^2 - \epsilon_0^2 \right)^{3/2}}}\right\vert}_{{\epsilon_0}}^{{\epsilon_{\mathrm{F}}}} \\ &= \frac{8 \pi}{3 (c h)^3}\left( \epsilon_{\mathrm{F}}^2 - \epsilon_0^2 \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Solving for \epsilon_{\mathrm{F}}/\epsilon_0 we have

\begin{aligned}\frac{\epsilon_{\mathrm{F}}}{\epsilon_0} =\sqrt{\left( \frac{3 (c h)^3 n}{8 \pi \epsilon_0^3} \right)^{2/3}+ 1}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

We’ll see the constant factor above a number of times below and designate it

\begin{aligned}n_0 = \frac{8 \pi}{3} \left( \frac{\epsilon_0}{c h} \right)^3,\end{aligned} \hspace{\stretch{1}}(1.0.2)

so that the Fermi energy is

\begin{aligned}\frac{\epsilon_{\mathrm{F}}}{\epsilon_0} =\sqrt{\left( \frac{n}{n_0} \right)^{2/3}+ 1}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

For the pressure calculation, let’s suppose that we have a configuration with a plane in the x,y orientation as in fig. 1.1.

Fig 1.1: Pressure against x,y oriented plane

 

It’s argued in [4] section 6.4 that the pressure for such a configuration is

\begin{aligned}P = n \int p_z u_z f(\mathbf{u}) d^3 \mathbf{u},\end{aligned} \hspace{\stretch{1}}(1.7)

where n is the number density and f(\mathbf{u}) is a normalized distribution function for the velocities. The velocity and momentum components are related by the Hamiltonian equations. From the Hamiltonian eq. 1.1 we find \footnote{ Observe that by squaring and summing one can show that this is equivalent to the standard relativisitic momentum p_x = \frac{m v_x}{\sqrt{ 1 - \mathbf{u}^2/c^2}}.} (for the x-component which is representative)

\begin{aligned}u_x \\ &= \frac{\partial {\epsilon}}{\partial {p_x}} \\ &= \frac{\partial {}}{\partial {p_x}}\sqrt{(p c)^2 +\epsilon_0^2} \\ &= \frac{ p_x c^2 }{\sqrt{(p c)^2 +\epsilon_0^2}}.\end{aligned} \hspace{\stretch{1}}(1.8)

For \alpha \in \{1, 2, 3\} we can summarize these velocity-momentum relationships as

\begin{aligned}\frac{u_\alpha}{c} = \frac{ c p_\alpha }{ \epsilon }.\end{aligned} \hspace{\stretch{1}}(1.9)

Should we attempt to calculate the pressure with this parameterization of the velocity space we end up with convergence problems, and can’t express the results in terms of f^+_\nu(z). Let’s try instead with a distribution over momentum space

\begin{aligned}P=n \int \frac{(c p_z)^2}{\epsilon} f(c \mathbf{p}) d^3 (c \mathbf{p}).\end{aligned} \hspace{\stretch{1}}(1.10)

Here the momenta have been scaled to have units of energy since we want to express this integral in terms of energy in the end. Our normalized distribution function is

\begin{aligned}f(c \mathbf{p})\propto \frac{\frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }}}{\int \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }} d^3 (c \mathbf{p})},\end{aligned} \hspace{\stretch{1}}(1.11)

but before evaluating anything, we first want to change our integration variable from momentum to energy. In spherical coordinates our volume element takes the form

\begin{aligned}d^3 (c \mathbf{p}) &= 2 \pi (c p)^2 d (c p) \sin\theta d\theta \\ &= 2 \pi (c p)^2 \frac{d (c p)}{d \epsilon} d \epsilon \sin\theta d\theta.\end{aligned} \hspace{\stretch{1}}(1.12)

Implicit derivatives of

\begin{aligned}c^2 p^2 = \epsilon^2 - \epsilon_0^2,\end{aligned} \hspace{\stretch{1}}(1.13)

gives us

\begin{aligned}\frac{d (c p)}{d\epsilon}= \frac{\epsilon}{c p}=\frac{\epsilon}{\sqrt{\epsilon^2 -\epsilon_0^2}}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Our momentum volume element becomes

\begin{aligned}d^3 (c \mathbf{p}) \\ &= 2 \pi (c p)^2 \frac{\epsilon}{\sqrt{\epsilon^2 - \epsilon_0^2 }}d \epsilon \sin\theta d\theta \\ &= 2 \pi \left( \epsilon^2 - \epsilon_0^2 \right)\frac{\epsilon}{\sqrt{\epsilon^2 - \epsilon_0^2 }}d \epsilon \sin\theta d\theta \\ &= 2 \pi \epsilon \sqrt{ \epsilon^2 - \epsilon_0^2} d \epsilon \sin\theta d\theta.\end{aligned} \hspace{\stretch{1}}(1.0.14)

For our distribution function, we can now write

\begin{aligned}f(c \mathbf{p}) d^3 (c \mathbf{p})= C \frac{\epsilon \sqrt{ \epsilon^2 - \epsilon_0^2} d \epsilon }{ z^{-1} e^{\beta \epsilon} + 1 }\frac{ 2 \pi \sin\theta d\theta }{ 4 \pi \epsilon_0^3 },\end{aligned} \hspace{\stretch{1}}(1.0.14)

where C is determined by the requirement \int f(c \mathbf{p}) d^3 (c \mathbf{p}) = 1

\begin{aligned}C^{-1} = \int_{0}^\infty \frac{(y + 1)\sqrt{ (y + 1)^2 - 1} dy }{ z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 }.\end{aligned} \hspace{\stretch{1}}(1.0.14)

The z component of our momentum can be written in spherical coordinates as

\begin{aligned}(c p_z)^2= (c p)^2 \cos^2\theta= \left( \epsilon^2 - \epsilon_0^2 \right)\cos^2\theta,\end{aligned} \hspace{\stretch{1}}(1.0.18)

Noting that

\begin{aligned}\int_0^\pi \cos^2\theta \sin\theta d\theta =-\int_0^\pi \cos^2\theta d(\cos\theta)= \frac{2}{3},\end{aligned} \hspace{\stretch{1}}(1.0.19)

all the bits come together as

\begin{aligned}P &= \frac{C n}{3 \epsilon_0^3 } \int_{\epsilon_0}^\infty\left( \epsilon^2 - \epsilon_0^2 \right)^{3/2} \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }} d \epsilon \\ &= \frac{n \epsilon_0}{3} \int_{1}^\infty\left( x^2 - 1 \right)^{3/2} \frac{1}{{ z^{-1} e^{\beta \epsilon_0 x} + 1 }} dx.\end{aligned} \hspace{\stretch{1}}(1.0.19)

Letting y = x - 1, this is

\begin{aligned}P= \frac{C n \epsilon_0}{3} \int_{0}^\infty \frac{ \left( (y + 1)^2 - 1 \right)^{3/2} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 } dy.\end{aligned} \hspace{\stretch{1}}(1.0.19)

We could conceivable expand the numerators of each of these integrals in power series, which could then be evaluated as a sum of f^+_\nu(z e^{-\beta \epsilon_0}) terms.

Note that above the Fermi energy n also has an integral representation

\begin{aligned}n &= \left(2\left( \frac{1}{{2}} \right) + 1\right)\int_{\epsilon_0}^\infty d\epsilon \mathcal{D}(\epsilon) \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1}} \\ &= \frac{8 \pi}{(c h)^3} \int_{\epsilon_0}^\infty d\epsilon\frac{\epsilon \sqrt{\epsilon^2 - \epsilon_0^2} }{ z^{-1} e^{\beta \epsilon} + 1} \\ &= \frac{8 \pi \epsilon_0^3}{(c h)^3} \int_{0}^\infty dy\frac{(y + 1)\sqrt{(y + 1)^2 - 1} }{ z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1},\end{aligned} \hspace{\stretch{1}}(1.0.19)

or

\begin{aligned}\boxed{n = \frac{3 n_0}{C}.}\end{aligned} \hspace{\stretch{1}}(1.0.23)

Observe that we can use this result to remove the dependence of pressure on this constant C

\begin{aligned}\boxed{\frac{P}{n_0 \epsilon_0}= \int_{0}^\infty dy \frac{ \left( (y + 1)^2 - 1 \right)^{3/2} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 }.}\end{aligned} \hspace{\stretch{1}}(1.0.24)

Now for the average energy difference from the rest energy \epsilon_0

\begin{aligned}U &= \left\langle{{\epsilon - \epsilon_0}}\right\rangle \\ &= \int_{\epsilon_0}^\infty d\epsilon \mathcal{D}(\epsilon) f(\epsilon) (\epsilon - \epsilon_0) \\ &= \frac{8 \pi V}{(c h)^3}\int_{\epsilon_0}^\infty d\epsilon \frac{ \epsilon(\epsilon - \epsilon_0) \sqrt{ \epsilon^2 - \epsilon_0 } }{ z^{-1} e^{\beta \epsilon} + 1} \\ &= \frac{8 \pi V \epsilon_0^4}{(c h)^3}\int_{0}^\infty dy\frac{ y ( y - 1 ) \sqrt{ (y + 1)^2 - 1 }}{ z^{-1} e^{\beta \epsilon} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.24)

So the average energy density difference from the rest energy, relative to the rest energy, is

\begin{aligned}\boxed{\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0} =3 n_0 \int_{0}^\infty dy \frac { y (y + 1)\sqrt{(y + 1)^2 - 1} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 }.}\end{aligned} \hspace{\stretch{1}}(1.0.26)

From eq. 1.0.24 and eq. 1.0.26 we have

\begin{aligned}\frac{1}{{n_0}} &= 3 \frac{V \epsilon_0} {\left\langle{{\epsilon - \epsilon_0}}\right\rangle} \int_{0}^\infty \frac { y (y + 1)\sqrt{(y + 1)^2 - 1} dy } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 } \\ &= \frac{\epsilon_0}{P} \int_{0}^\infty \frac{ \left( (y + 1)^2 - 1 \right)^{3/2} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 } dy,\end{aligned} \hspace{\stretch{1}}(1.0.26)

or

\begin{aligned}P V =\frac{U}{3}\frac{ \int_{0}^\infty \frac{ \left( (y + 1)^2 - 1 \right)^{3/2} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 } dy}{ \int_{0}^\infty \frac { y (y + 1)\sqrt{(y + 1)^2 - 1} dy } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 }}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

This ratio of integrals is supposed to resolve to 1 and 2 in the low and high density limits. To consider this let’s perform one final non-dimensionalization, writing

\begin{aligned}\begin{aligned} \\ x &= \beta \epsilon_0 y \\ \theta &= \frac{1}{{\beta \epsilon_0}} = \frac{k_{\mathrm{B}} T}{\epsilon_0} \\ \bar{\mu} &= \mu - \epsilon_0 \\ \bar{z} &= e^{\beta \bar{\mu}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.29)

The density, pressure, and energy take the form

\begin{aligned}\frac{n}{n_0}= 3 \theta\int_{0}^\infty dx\frac{(\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} }{ \bar{z}^{-1} e^{x} + 1}\end{aligned} \hspace{\stretch{1}}(1.0.30a)

\begin{aligned}\frac{P}{n_0 \epsilon_0}= \theta \int_{0}^\infty dx \frac{ \left( (\theta x + 1)^2 - 1 \right)^{3/2} } { \bar{z}^{-1} e^{x} + 1 }\end{aligned} \hspace{\stretch{1}}(1.0.30b)

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} =3 \theta^2 \int_{0}^\infty dx \frac { x (\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} } { \bar{z}^{-1} e^{x} + 1 }.\end{aligned} \hspace{\stretch{1}}(1.0.30c)

We can rewrite the square roots in the number density and energy density expressions by expanding out the completion of the square

\begin{aligned}(1 + \theta x) \sqrt{ (1 + \theta x)^2 - 1}=(1 + \theta x) \sqrt{ 1 + \theta x + 1 }\sqrt{ 1 + \theta x - 1 }= \sqrt{2 \theta} x^{1/2} (1 + \theta x) \sqrt{ 1 + \frac{\theta x}{2}},\end{aligned} \hspace{\stretch{1}}(1.0.30c)

Expanding the distribution about \bar{z} e^{-x} = 0, we have

\begin{aligned}\frac{1}{ \bar{z}^{-1} e^{x} + 1}=\frac{\bar{z} e^{-x}}{ 1 + \bar{z} e^{-x}}=z e^{-x} \sum_{s = 0}^\infty (-1)^s \left( \bar{z} e^{-x} \right)^s,\end{aligned} \hspace{\stretch{1}}(1.0.32)

allowing us to write, in the low density limit with respect to \bar{z}

\begin{aligned}\frac{n}{n_0}= 3 \sqrt{2}\theta^{3/2} \sum_{s=0}^\infty(-1)^s\bar{z}^{s + 1}\int_{0}^\infty dx x^{1/2}(1 + \theta x) \sqrt{ 1 + \frac{\theta x}{2}} e^{-x(1 + s)} \end{aligned} \hspace{\stretch{1}}(1.0.33a)

\begin{aligned}\frac{P}{n_0 \epsilon_0}= \theta\sum_{s=0}^\infty(-1)^s\bar{z}^{s + 1} \int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2} e^{-x(1 + s)} \end{aligned} \hspace{\stretch{1}}(1.0.33b)

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} =3 \sqrt{2} \theta^{5/2} \sum_{s=0}^\infty(-1)^s\bar{z}^{s + 1} \int_{0}^\infty dx x^{3/2} (1 + \theta x) \sqrt{ 1 + \frac{\theta x}{2}} e^{-x(1 + s)} .\end{aligned} \hspace{\stretch{1}}(1.0.33c)

Low density result

An exact integration of the various integrals above is possible in terms of special functions. However, that attempt (included below) introduced an erroneous extra factor of \theta. Given that this end result was obtained by tossing all but the lowest order terms in \theta and \bar{z}, let’s try that right from the get go.

For the pressure we have an integrand containing a factor

\begin{aligned}\left( (\theta x + 1)^2 -1 \right)^{3/2}&= \left( \theta x + 1 - 1 \right)^{3/2}\left( \theta x + 1 + 1 \right)^{3/2} \\ &= \theta^{3/2} x^{3/2} 2^{3/2} \left( 1 + \frac{\theta x}{2} \right)^{3/2} \\ &= 2 \sqrt{2} \theta^{3/2} x^{3/2} \left( 1 + \frac{\theta x}{2} \right)^{3/2}\approx2 \sqrt{2} \theta^{3/2} x^{3/2} \end{aligned} \hspace{\stretch{1}}(1.0.33c)

Our pressure, to lowest order in \theta and \bar{z} is then

\begin{aligned}\frac{P}{\epsilon_0 n_0} = 2 \sqrt{2} \theta^{5/2} \bar{z} \int_0^\infty x^{3/2} e^{-x} dx= 2 \sqrt{2} \theta^{5/2} \bar{z} \Gamma(5/2).\end{aligned} \hspace{\stretch{1}}(1.0.33c)

Our energy density to lowest order in \theta and \bar{z} from eq. 1.0.33c is

\begin{aligned}\frac{U}{V \epsilon_0 n_0} &= 3 \sqrt{2} \theta^{5/2} \bar{z} \int_{0}^\infty dx x^{3/2} e^{-x} \\ &= 3 \sqrt{2} \theta^{5/2} \bar{z} \Gamma(5/2).\end{aligned} \hspace{\stretch{1}}(1.0.33c)

Comparing these, we have

\begin{aligned}\frac{1}{{\epsilon_0 n_0\sqrt{2} \theta^{5/2} \bar{z} \Gamma(5/2)}} &= 3 \frac{V}{U} \\ &= \frac{2}{P},\end{aligned} \hspace{\stretch{1}}(1.0.37)

or in this low density limit

\begin{aligned}\boxed{P V = \frac{2}{3} U.}\end{aligned} \hspace{\stretch{1}}(1.0.38)

High density limit

For the high density limit write \bar{z} = e^y, so that the distribution takes the form

\begin{aligned}f(\bar{z}) &= \frac{1}{ \bar{z}^{-1} e^{x} + 1} \\ &= \frac{1}{ e^{x - y} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.39)

This can be approximated by a step function, so that

\begin{aligned}\frac{P}{n_0 \epsilon_0}\approx \int_{0}^y \theta dx\left( (\theta x + 1)^2 - 1 \right)^{3/2} \end{aligned} \hspace{\stretch{1}}(1.0.40a)

\begin{aligned}\frac{U}{V \epsilon_0 n_0} \approx3 \int_{0}^\infty \theta dx \theta x (\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} \end{aligned} \hspace{\stretch{1}}(1.0.40b)

With a change of variables u = \theta x + 1, we have

\begin{aligned}\begin{aligned}\frac{P}{n_0 \epsilon_0} &\approx \int_{1}^{\theta y + 1x} du\left( u^2 - 1 \right)^{3/2} \\ &=\frac{1}{8} \left((2 \theta y (\theta y+2)-3) \sqrt{\theta y (\theta y+2)} (\theta y+1)+3 \ln \left(\theta y+\sqrt{\theta y (\theta y+2)}+1\right)\right) \\ &\approx\frac{1}{4} \left( \theta \ln \bar{z} \right)^4\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.41a)

\begin{aligned}\begin{aligned}\frac{U}{V \epsilon_0 n_0} &\approx3 \int_{1}^{\theta y + 1x} (u^2 - u)\sqrt{u^2 - 1} \\ &=\frac{3}{24} \left(\sqrt{\theta y (\theta y+2)} (\theta y (2 \theta y (3 \theta y+5)-1)+3)-3 \left(\ln \left(\theta y+\sqrt{\theta y (\theta y+2)}+1\right)\right)\right) \\ &\approx\frac{3}{4} \left( \theta \ln \bar{z} \right)^4\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.41b)

Comparing both we have

\begin{aligned}\frac{4}{\epsilon_0 n_0 \left( \theta \ln \bar{z} \right) } = \frac{1}{{P}} = \frac{3 V}{U},\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}\boxed{P V = \frac{1}{{3}} U.}\end{aligned} \hspace{\stretch{1}}(1.0.43)

\begin{aligned}{\left.{{\epsilon_{\mathrm{F}}}}\right\vert}_{{n = 1/(0.01)^3}} = 6.12402 \times 10^{-35} \text{J} \times 6.24150934 \times 10^{18} \frac{\text{eV}}{\text{J}} = 3.82231 \times 10^{-16} \text{eV}\end{aligned} \hspace{\stretch{1}}(1.0.43)

Wow. That’s pretty low!

Pressure integral

Of these the pressure integral is yields directly to Mathematica

\begin{aligned}\begin{aligned} \int_{0}^\infty & dx\left( (\theta x + 1)^2 - 1 \right)^{3/2} e^{-x(1 + s)} \\ &=\frac{3 \theta e^{(s+1)/\theta}}{(s + 1)^2} K_2\left( \frac{s+1}{\theta } \right) \\ &=\frac{3 \sqrt{\frac{\pi }{2}} \theta ^{3/2}}{(s+1)^{5/2}}+\frac{45 \sqrt{\frac{\pi }{2}} \theta ^{5/2}}{8 (s+1)^{7/2}}+\frac{315 \sqrt{\frac{\pi }{2}} \theta ^{7/2}}{128 (s+1)^{9/2}}-\frac{945 \sqrt{\frac{\pi }{2}} \theta ^{9/2}}{1024 (s+1)^{11/2}}+\frac{31185 \sqrt{\frac{\pi }{2}} \theta ^{11/2}}{32768 (s+1)^{13/2}} + \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.45)

where K_2(z) is a modified Bessel function [5] of the second kind as plotted in fig. 1.2.

Fig 1.2: Modified Bessel function of the second kind

 

Plugging this into the series for the pressure, we have

\begin{aligned}\frac{P}{n_0 \epsilon_0}= 3 \left( \frac{k_{\mathrm{B}} T}{\epsilon_0} \right)^2\sum_{s=0}^\infty(-1)^s\frac{\left( \bar{z} e^{\epsilon_0/k_{\mathrm{B}} T} \right)^{s + 1}}{(s + 1)^2}K_2\left( (s+1) \epsilon_0/k_{\mathrm{B}} T \right).\end{aligned} \hspace{\stretch{1}}(1.0.46)

Plotting the summands 3 (-1)^s \frac{\theta^2}{(s + 1)^2} \left( \bar{z} e^{ 1/\theta} \right)^{s + 1} K_2\left((s+1)/\theta\right) for \bar{z} = 1 in fig. 1.4 shows that this mix of exponential Bessel and quadratic terms decreases with s.

Plotting this sum in fig. 1.3 numerically to 10 terms, shows that we have a function that appears roughly polynomial in \bar{z} and \theta.

Fig 1.3: Pressure to ten terms in z and theta

 

Fig 1.4: Pressure summands

 

For small \bar{z} it can be seen graphically that there is very little contribution from anything but the s = 0 term of this sum. An expansion in series for a few terms in \bar{z} and \theta gives us

\begin{aligned}\begin{aligned}\frac{P}{\epsilon_0 n_0}&=\sqrt{\pi} \theta^{5/2} \left(\frac{3 \bar{z}}{\sqrt{2}}-\frac{3 \bar{z}^2}{8}+\frac{\bar{z}^3}{3 \sqrt{6}}-\frac{3 \bar{z}^4}{32 \sqrt{2}}+\frac{3 \bar{z}^5}{25 \sqrt{10}}\right) \\ &+\sqrt{\pi} \theta^{7/2} \left(\frac{45 \bar{z}}{8 \sqrt{2}}\right) -\frac{45 \bar{z}^2}{128}+\frac{5 \bar{z}^3}{24 \sqrt{6}}-\frac{45 \bar{z}^4}{1024 \sqrt{2}}+\frac{9 \bar{z}^5}{200 \sqrt{10}}\\ &+\sqrt{\pi} \theta^{9/2} \left(\frac{315 \bar{z}}{128 \sqrt{2}}-\frac{315 \bar{z}^2}{4096}+\frac{35 \bar{z}^3}{1152 \sqrt{6}}-\frac{315 \bar{z}^4}{65536 \sqrt{2}}+\frac{63 \bar{z}^5}{16000 \sqrt{10}}\right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.47)

This allows a k_{\mathrm{B}} T \ll m c^2 and \bar{z} \ll 1 approximation of the pressure

\begin{aligned}\frac{P}{\epsilon_0 n_0} = \frac{3}{2} \sqrt{2 \pi} \bar{z} \theta^{5/2}.\end{aligned} \hspace{\stretch{1}}(1.0.48)

Number density integral

For the number density, it appears that we can evaluate the integral using integration from parts applied to eq. 1.0.30.30

\begin{aligned}\frac{n}{n_0}= \theta\int_{0}^\infty dx\frac{3 (\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} }{ \bar{z}^{-1} e^{x} + 1}=\theta\int_{0}^\infty dx\left( \frac{d}{dx} \left( (\theta x + 1)^2 - 1 \right) ^{3/2} \right)\frac{1}{ \bar{z}^{-1} e^{x} + 1}={\left.{{\theta\left( (\theta x + 1)^2 - 1 \right)^{3/2}\frac{1}{ \bar{z}^{-1} e^{x} + 1}}}\right\vert}_{{0}}^{{\infty}}-\theta\int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2}\frac{ -\bar{z}^{-1} e^{x} }{ \left( \bar{z}^{-1} e^{x} + 1 \right)^2}=\theta\int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2}\frac{ \bar{z} e^{-x} }{ \left( 1 + \bar{z} e^{-x} \right)^2}.\end{aligned} \hspace{\stretch{1}}(1.0.48)

Expanding in series, gives us

\begin{aligned}\frac{n}{n_0}=\theta\sum_{s = 0}^\infty\binom{-2}{s}\bar{z}^{s + 1} \int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2} e^{-x(s + 1)}=3 \theta^2\sum_{s = 0}^\infty\binom{-2}{s}\frac{\left( \bar{z} e^{1/\theta} \right)^{s + 1}}{(s + 1)^2}K_2\left( \frac{s+1}{\theta } \right).\end{aligned} \hspace{\stretch{1}}(1.0.48)

Here the binomial coefficient has the meaning given in the definitions of \statmechchapcite{nonIntegralBinomialSeries}, where for negative integral values of b we have

\begin{aligned}\binom{b}{s}\equiv(-1)^s \frac{-b}{-b + s} \binom{-b+s}{-b}.\end{aligned} \hspace{\stretch{1}}(1.0.51)

Expanding in series to a couple of orders in \theta and \bar{z} we have

\begin{aligned}\frac{n}{n_0} = \frac{\sqrt{2 \pi}}{36} \theta^{1/2} \left(\left(2 \sqrt{3} \bar{z} - 9/\sqrt{2} \right) \bar{z} +18 \right) \bar{z}+\frac{5 \sqrt{ 2 \pi}}{576} \theta^{3/2} \left(\left(4 \sqrt{3} \bar{z} - 27/\sqrt{2}\right) \bar{z} +108 \right) \bar{z}+ \cdots\end{aligned} \hspace{\stretch{1}}(1.0.52)

To first order in \theta and \bar{z} this is

\begin{aligned}\frac{n}{n_0} = \frac{1}{{2}} \sqrt{ 2 \pi } \bar{z} \theta^{1/2},\end{aligned} \hspace{\stretch{1}}(1.0.53)

which allows a relation to pressure

\begin{aligned}P V = 3 N (k_{\mathrm{B}} T)^2 /\epsilon_0.\end{aligned} \hspace{\stretch{1}}(1.0.54)

It’s kind of odd seeming that this is quadratic in temperature. Is there an error?

Energy integral

Starting from eq. 1.0.30c and integrating by parts we have

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} &= 3 \theta^2 \int_{0}^\infty dx \frac { x (\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} } { \bar{z}^{-1} e^{x} + 1 } \\ &= -\theta^2 \int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2}\frac{d}{dx} \left( \frac{x} { \bar{z}^{-1} e^{x} + 1 } \right) \\ &= -\theta^2 \int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2}\left( \frac{1} { \bar{z}^{-1} e^{x} + 1 } - \frac{x \bar{z}^{-1} e^{x} } { \left( \bar{z}^{-1} e^{x} + 1 \right)^2 } \right) \\ &= \theta^2 \int_{0}^\infty dx \left( (\theta x + 1)^2 - 1 \right)^{3/2} \frac{ (x - 1)\bar{z}^{-1} e^{x} - 1} { \left( \bar{z}^{-1} e^{x} + 1 \right)^2 } \\ &= \theta^2 \int_{0}^\infty dx \left( (\theta x + 1)^2 - 1 \right)^{3/2} \frac{ (x - 1)\bar{z} e^{-x} - \bar{z}^2 e^{-2 x}} { \left( 1 + \bar{z} e^{-x} \right)^2 } \\ &= \theta^2\sum_{s=0}^\infty \binom{-2}{s} \int_{0}^\infty dx \left( (\theta x + 1)^2 - 1 \right)^{3/2} \left( (x - 1)\bar{z} e^{-x} - \bar{z}^2 e^{-2 x} \right) (\bar{z} e^{-x})^s \\ &= \theta^2\sum_{s=0}^\infty \binom{-2}{s} \bar{z}^{s + 1} \int_{0}^\infty dx \left( (\theta x + 1)^2 - 1 \right)^{3/2} \left( (x - 1) e^{-x(s + 1)} - \bar{z} e^{-x(s + 2)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.54)

The integral with the factor of x doesn’t have a nice closed form as before (if you consider the K_2 a nice closed form), but instead evaluates to a confluent hypergeometric function [6]. That integral is

\begin{aligned}\int_0^{\infty } x \left((\theta x+1)^2-1\right)^{3/2} e^{-x (1+s)} dx = \frac{15 \sqrt{\pi } \theta^3 U\left(-\frac{3}{2},-4,\frac{2 (s+1)}{\theta }\right)}{8 (s+1)^5},\end{aligned} \hspace{\stretch{1}}(1.0.54)

and looks like fig. 1.5. Series expansion shows that this hypergeometricU function has a \theta^{3/2} singularity at the origin

Fig 1.5: Plot of HypergeometricU, and with theta^5 scaling

 

\begin{aligned}U\left(-\frac{3}{2},-4,\frac{2 (s+1)}{\theta }\right)=\frac{2 \sqrt{2} \sqrt{s+1} s+2 \sqrt{2} \sqrt{s+1}}{\theta^{3/2}}+\frac{21 \sqrt{s+1}}{2 \sqrt{2} \sqrt{\theta }}+ \cdots\end{aligned} \hspace{\stretch{1}}(1.57)

so our multiplication by \theta^5 brings us to zero as seen in the plot. Evaluating the complete integral yields the unholy mess

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} &= \sum_{s=0}^\infty \theta^2 (-1)^s (s+1) \bar{z}^{s+1} \Bigl( \\ &\frac{105 \sqrt{\pi } \theta^3 U\left(-\frac{1}{2},-4,\frac{2 (s+1)}{\theta }\right)}{16 (s+1)^5} \\ &- \frac{3 \sqrt{\pi } \theta^2 U\left(-\frac{1}{2},-2,\frac{2 (s+1)}{\theta }\right)}{2 (s+1)^3} \\ &- \frac{3 \sqrt{\pi } \theta^2 \bar{z} U\left(-\frac{1}{2},-2,\frac{2 (s+2)}{\theta }\right)}{2 (s+2)^3} \\ &+\frac{(\theta -2) (-3 \theta +2 s+2) e^{\frac{s+1}{\theta }} K_2\left(\frac{s+1}{\theta }\right)}{\theta (s+1)^2} \\ &-\frac{2 (\theta -2) e^{\frac{s+1}{\theta }} K_1\left(\frac{s+1}{\theta }\right)}{\theta (s+1)} \\ &+\frac{\bar{z} (-3 \theta +2 s+4) e^{\frac{s+2}{\theta }} K_2\left(\frac{s+2}{\theta }\right)}{(s+2)^2} \\ &-\frac{2 \bar{z} e^{\frac{s+2}{\theta }} K_1\left(\frac{s+2}{\theta }\right)}{s+2} \Bigr),\end{aligned} \hspace{\stretch{1}}(1.58)

to first order in \bar{z} and \theta this is

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} =\frac{9}{4} \sqrt{2 \pi} \bar{z} \theta^{7/2}.\end{aligned} \hspace{\stretch{1}}(1.59)

Comparing pressure and energy we have for low densities (where \bar{z} \approx 0)

\begin{aligned}\frac{1}{{\epsilon_0 n_0 \sqrt{2 \pi} \bar{z} \theta^{5/2}}} = \frac{3}{2} \frac{1}{{P}} = \frac{9}{4} \theta \frac{V}{U},\end{aligned} \hspace{\stretch{1}}(1.0.60)

or

\begin{aligned}\theta P V = \frac{2}{3} U.\end{aligned} \hspace{\stretch{1}}(1.0.61)

It appears that I’ve picked up an extra factor of \theta somewhere, but at least I’ve got the 2/3 low density expression. Given that I’ve Taylor expanded everything anyways around \bar{z} and \theta this could likely have been done right from the get go, instead of dragging along the messy geometric integrals. Reworking this part of this problem like that was done above.

References

[1] Kerson Huang. Introduction to statistical physics. CRC Press, 2001.

[2] Peeter Joot. Basic statistical mechanics., chapter {Non integral binomial coefficient}. \natexlab{a}. URL http://sites.google.com/site/peeterjoot2/math2013/phy452.pdf.

[3] Peeter Joot. Basic statistical mechanics., chapter {Relativisitic density of states}. \natexlab{b}. URL http://sites.google.com/site/peeterjoot2/math2013/phy452.pdf.

[4] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

[5] Wolfram. BesselK, \natexlab{a}. URL http://reference.wolfram.com/mathematica/ref/BesselK.html. [Online; accessed 11-April-2013].

[6] Wolfram. HyperGeometricU, \natexlab{b}. URL http://reference.wolfram.com/mathematica/ref/HypergeometricU.html. [Online; accessed 17-April-2013].

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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.

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | 1 Comment »

PHY452H1S Basic Statistical Mechanics. Lecture 18: Fermi gas thermodynamics. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 26, 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.)]

Disclaimer

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

Review

Last time we found that the low temperature behaviour or the chemical potential was quadratic as in fig. 1.1.

\begin{aligned}\mu =\mu(0) - a \frac{T^2}{T_{\mathrm{F}}}\end{aligned} \hspace{\stretch{1}}(1.1.1)

Fig 1.1: Fermi gas chemical potential

 

Specific heat

\begin{aligned}E = \sum_\mathbf{k} n_{\mathrm{F}}(\epsilon_\mathbf{k}, T) \epsilon_\mathbf{k}\end{aligned} \hspace{\stretch{1}}(1.1.2)

\begin{aligned}\frac{E}{V} &= \frac{1}{{(2\pi)^3}} \int d^3 \mathbf{k} n_{\mathrm{F}}(\epsilon_\mathbf{k}, T) \epsilon_\mathbf{k} \\ &= \int d\epsilon N(\epsilon) n_{\mathrm{F}}(\epsilon, T) \epsilon,\end{aligned} \hspace{\stretch{1}}(1.1.3)

where

\begin{aligned}N(\epsilon) = \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)^{3/2}\sqrt{\epsilon}.\end{aligned} \hspace{\stretch{1}}(1.1.4)

Low temperature C_{\mathrm{V}}

\begin{aligned}\frac{\Delta E(T)}{V}=\int_0^\infty d\epsilon N(\epsilon)\left( n_{\mathrm{F}}(\epsilon, T) - n_{\mathrm{F}}(\epsilon, 0) \right)\end{aligned} \hspace{\stretch{1}}(1.1.5)

The only change in the distribution fig. 1.2, that is of interest is over the step portion of the distribution, and over this range of interest N(\epsilon) is approximately constant as in fig. 1.3.

Fig 1.2: Fermi distribution

Fig 1.3: Fermi gas density of states

\begin{aligned}N(\epsilon) \approx  N(\mu)\end{aligned} \hspace{\stretch{1}}(1.0.6a)

\begin{aligned}\mu \approx  \epsilon_{\mathrm{F}},\end{aligned} \hspace{\stretch{1}}(1.0.6b)

so that

\begin{aligned}\Delta e \equiv\frac{\Delta E(T)}{V}\approx N(\epsilon_{\mathrm{F}})\int_0^\infty d\epsilon\left( n_{\mathrm{F}}(\epsilon, T) - n_{\mathrm{F}}(\epsilon, 0) \right)=N(\epsilon_{\mathrm{F}})\int_{-\epsilon_{\mathrm{F}}}^\infty d x (\epsilon_{\mathrm{F}} + x)\left( n_{\mathrm{F}}(\epsilon + x, T) - n_{\mathrm{F}}(\epsilon_{\mathrm{F}} + x, 0) \right).\end{aligned} \hspace{\stretch{1}}(1.0.7)

Here we’ve made a change of variables \epsilon = \epsilon_{\mathrm{F}} + x, so that we have near cancelation of the \epsilon_{\mathrm{F}} factor

\begin{aligned}\Delta e &= N(\epsilon_{\mathrm{F}})\epsilon_{\mathrm{F}}\int_{-\epsilon_{\mathrm{F}}}^\infty d x \underbrace{\left( n_{\mathrm{F}}(\epsilon + x, T) - n_{\mathrm{F}}(\epsilon_{\mathrm{F}} + x, 0) \right)}_{\text{almost equal everywhere}}+N(\epsilon_{\mathrm{F}})\int_{-\epsilon_{\mathrm{F}}}^\infty d x x\left( n_{\mathrm{F}}(\epsilon + x, T) - n_{\mathrm{F}}(\epsilon_{\mathrm{F}} + x, 0) \right) \\ &\approx N(\epsilon_{\mathrm{F}})\int_{-\infty}^\infty d x x\left( \frac{1}{{ e^{\beta x} +1 }} - {\left.{{\frac{1}{{ e^{\beta x} +1 }}}}\right\vert}_{{T \rightarrow 0}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.8)

Here we’ve extended the integration range to -\infty since this doesn’t change much. FIXME: justify this to myself? Taking derivatives with respect to temperature we have

\begin{aligned}\frac{\delta e}{T} &= -N(\epsilon_{\mathrm{F}})\int_{-\infty}^\infty d x x\frac{1}{{(e^{\beta x} + 1)^2}}\frac{d}{dT} e^{\beta x} \\ &= N(\epsilon_{\mathrm{F}})\int_{-\infty}^\infty d x x\frac{1}{{(e^{\beta x} + 1)^2}}e^{\beta x}\frac{x}{k_{\mathrm{B}} T^2}\end{aligned} \hspace{\stretch{1}}(1.0.9)

With \beta x = y, we have for T \ll T_{\mathrm{F}}

\begin{aligned}\frac{C}{V} &= N(\epsilon_{\mathrm{F}})\int_{-\infty}^\infty \frac{ dy y^2 e^y }{ (e^y + 1)^2 k_{\mathrm{B}} T^2} (k_{\mathrm{B}} T)^3 \\ &= N(\epsilon_{\mathrm{F}}) k_{\mathrm{B}}^2 T\underbrace{\int_{-\infty}^\infty \frac{ dy y^2 e^y }{ (e^y + 1)^2 } }_{\pi^2/3} \\ &= \frac{\pi^2}{3} N(\epsilon_{\mathrm{F}}) k_{\mathrm{B}} (k_{\mathrm{B}} T).\end{aligned} \hspace{\stretch{1}}(1.0.10)

Using eq. 1.1.4 at the Fermi energy and

\begin{aligned}\frac{N}{V} = \rho\end{aligned} \hspace{\stretch{1}}(1.0.11a)

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2 m}\end{aligned} \hspace{\stretch{1}}(1.0.11b)

\begin{aligned}k_{\mathrm{F}} = \left( 6 \pi^2 \rho \right)^{1/3},\end{aligned} \hspace{\stretch{1}}(1.0.11c)

we have

\begin{aligned}N(\epsilon_{\mathrm{F}}) &= \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)^{3/2}\sqrt{\epsilon_{\mathrm{F}}} \\ &= \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)^{3/2}\frac{\hbar k_{\mathrm{F}}}{\sqrt{2m}} \\ &= \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)^{3/2}\frac{\hbar }{\sqrt{2m}} \left( 6 \pi^2 \rho \right)^{1/3} \\ &= \frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)\left( 6 \pi^2 \frac{N}{V} \right)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.0.12)

Giving

\begin{aligned}\frac{C}{N} &= \frac{\pi^2}{3} \frac{V}{N}\frac{1}{{4 \pi^2}}\left( \frac{2m}{\hbar^2} \right)\left( 6 \pi^2 \frac{N}{V} \right)^{1/3}k_{\mathrm{B}} (k_{\mathrm{B}} T) \\ &= \left( \frac{m}{6 \hbar^2} \right)\left( \frac{V}{N} \right)^{2/3}\left( 6 \pi^2 \right)^{1/3}k_{\mathrm{B}} (k_{\mathrm{B}} T) \\ &= \left( \frac{ \pi^2 m}{3 \hbar^2} \right)\left( \frac{V}{\pi^2 N} \right)^{2/3}k_{\mathrm{B}} (k_{\mathrm{B}} T) \\ &= \left( \frac{ \pi^2 m}{\hbar^2} \right)\frac{\hbar^2}{2 m \epsilon_{\mathrm{F}}}k_{\mathrm{B}} (k_{\mathrm{B}} T),\end{aligned} \hspace{\stretch{1}}(1.0.13)

or

\begin{aligned}\boxed{\frac{C}{N} = \frac{\pi^2}{2} k_{\mathrm{B}} \frac{ k_{\mathrm{B}} T}{\epsilon_{\mathrm{F}}}.}\end{aligned} \hspace{\stretch{1}}(1.0.14)

This is illustrated in fig. 1.4.

Fig 1.4: Specific heat per Fermion

 

Relativisitic gas

  1. Relativisitic gas

    \begin{aligned}\epsilon_\mathbf{k} = \pm \hbar v \left\lvert {\mathbf{k}} \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.0.15)

    \begin{aligned}\epsilon = \sqrt{(m_0 c^2)^2 + c^2 (\hbar \mathbf{k})^2}\end{aligned} \hspace{\stretch{1}}(1.0.16)

  2. graphene
  3. massless Dirac Fermion

    Fig 1.5: Relativisitic gas energy distribution

     

    We can think of this state distribution in a condensed matter view, where we can have a hole to electron state transition by supplying energy to the system (i.e. shining light on the substrate). This can also be thought of in a relativisitic particle view where the same state transition can be thought of as a positron electron pair transition. A round trip transition will have to supply energy like 2 m_0 c^2 as illustrated in fig. 1.6.

    Fig 1.6: Hole to electron round trip transition energy requirement

     

Graphene

Consider graphene, a 2D system. We want to determine the density of states N(\epsilon),

\begin{aligned}\int \frac{d^2 \mathbf{k}}{(2 \pi)^2} \rightarrow \int_{-\infty}^\infty d\epsilon N(\epsilon),\end{aligned} \hspace{\stretch{1}}(1.0.17)

We’ll find a density of states distribution like fig. 1.7.

Fig 1.7: Density of states for 2D linear energy momentum distribution

 

\begin{aligned}N(\epsilon) = \text{constant factor} \frac{\left\lvert {\epsilon} \right\rvert}{v},\end{aligned} \hspace{\stretch{1}}(1.0.18)

\begin{aligned}C \sim \frac{d}{dT} \int N(\epsilon) n_{\mathrm{F}}(\epsilon) \epsilon d\epsilon,\end{aligned} \hspace{\stretch{1}}(1.0.19)

\begin{aligned}\Delta E \sim \underbrace{T}_{\text{window}}\times\underbrace{T}_{\text{energy}}\times\underbrace{T}_{\text{number of states}}\sim T^3\end{aligned} \hspace{\stretch{1}}(1.0.20)

so that

\begin{aligned}C_{\mathrm{Dimensionless}} \sim T^2\end{aligned} \hspace{\stretch{1}}(1.0.21)

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PHY452H1S Basic Statistical Mechanics. Lecture 17: Fermi gas thermodynamics. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 26, 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.)]

Disclaimer

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

Fermi gas thermodynamics

  • Energy was found to be

    \begin{aligned}\frac{E}{N} = \frac{3}{5} \epsilon_{\mathrm{F}}\qquad \text{where} \quad T = 0.\end{aligned} \hspace{\stretch{1}}(1.2.1)

  • Pressure was found to have the form fig. 1.1

    Fig 1.1: Pressure in Fermi gas

  • The chemical potential was found to have the form fig. 1.2.

    \begin{aligned}e^{\beta \mu} = \rho \lambda_{\mathrm{T}}^3\end{aligned} \hspace{\stretch{1}}(1.0.2a)

    \begin{aligned}\lambda_{\mathrm{T}} = \frac{h}{\sqrt{ 2 \pi m k_{\mathrm{B}} T}},\end{aligned} \hspace{\stretch{1}}(1.0.2b)

    so that the zero crossing is approximately when

    \begin{aligned}e^{\beta \times 0} = 1 = \rho \lambda_{\mathrm{T}}^3.\end{aligned} \hspace{\stretch{1}}(1.0.3)

    That last identification provides the relation T \sim  T_{\mathrm{F}}. FIXME: that bit wasn’t clear to me.

    Fig 1.2: Chemical potential in Fermi gas

How about at other temperatures?

  • \mu(T) = ?
  • E(T) = ?
  • C_{\mathrm{V}}(T) = ?

We had

\begin{aligned}N = \sum_k \frac{1}{{e^{\beta (\epsilon_k - \mu)} + 1}} = \sum_{\mathbf{k}} n_{\mathrm{F}}(\epsilon_\mathbf{k})\end{aligned} \hspace{\stretch{1}}(1.0.4)

\begin{aligned}E(T) =\sum_k \epsilon_\mathbf{k} n_{\mathrm{F}}(\epsilon_\mathbf{k}).\end{aligned} \hspace{\stretch{1}}(1.0.5)

FIXME: references to earlier sections where these were derived.

We can define a density of states

\begin{aligned}\sum_\mathbf{k} &= \sum_\mathbf{k} \int_{-\infty}^\infty d\epsilon  \delta(\epsilon  - \epsilon_\mathbf{k}) \\ &= \int_{-\infty}^\infty d\epsilon \sum_\mathbf{k}\delta(\epsilon  - \epsilon_\mathbf{k}),\end{aligned} \hspace{\stretch{1}}(1.0.6)

where the liberty to informally switch the order of differentiation and integration has been used. This construction allows us to write a more general sum

\begin{aligned}\sum_\mathbf{k} f(\epsilon_\mathbf{k}) &= \sum_\mathbf{k} \int_{-\infty}^\infty d\epsilon  \delta(\epsilon  - \epsilon_\mathbf{k}) f(\epsilon_\mathbf{k}) \\ &= \sum_\mathbf{k}\int_{-\infty}^\infty d\epsilon \delta(\epsilon  - \epsilon_\mathbf{k})f(\epsilon_\mathbf{k}) \\ &=\int_{-\infty}^\infty d\epsilon  f(\epsilon_\mathbf{k})\left( \sum_\mathbf{k} \delta(\epsilon  - \epsilon_\mathbf{k}) \right).\end{aligned} \hspace{\stretch{1}}(1.0.7)

This sum, evaluated using a continuum approximation, is

\begin{aligned}N(\epsilon ) &\equiv \sum_\mathbf{k}\delta(\epsilon  - \epsilon_\mathbf{k}) \\ &= \frac{V}{(2 \pi)^3} \int d^3 \mathbf{k} \delta\left( \epsilon  - \frac{\hbar^2 k^2}{2 m} \right) \\ &= \frac{V}{(2 \pi)^3} 4 \pi \int_0^\infty k^2 dk \delta\left( \epsilon  - \frac{\hbar^2 k^2}{2 m} \right)\end{aligned} \hspace{\stretch{1}}(1.0.8)

Using

\begin{aligned}\delta(g(x)) = \sum_{x_0} \frac{\delta(x - x_0)}{\left\lvert {g'(x_0)} \right\rvert},\end{aligned} \hspace{\stretch{1}}(1.0.9)

where the roots of g(x) are x_0, we have

\begin{aligned}N(\epsilon ) &= \frac{V}{(2 \pi)^3} 4 \pi \int_0^\infty k^2 dk \delta\left( k - \frac{\sqrt{2 m \epsilon }}{\hbar} \right)\frac{m \hbar }{ \hbar^2 \sqrt{2 m \epsilon }} \\ &= \frac{V}{(2 \pi)^3} 2 \pi \frac{2 m \epsilon }{\hbar^2}\frac{2 m \hbar }{ \hbar^2 \sqrt{2 m \epsilon }} \\ &= V \left( \frac{2 m}{\hbar^2} \right)^{3/2} \frac{1}{{4 \pi^2}} \sqrt{\epsilon }.\end{aligned} \hspace{\stretch{1}}(1.0.10)

In 2D this would be

\begin{aligned}N(\epsilon ) \sim  V \int dk k \delta \left( \epsilon  - \frac{\hbar^2 k^2}{2m} \right) = V \frac{\sqrt{2 m \epsilon }}{\hbar} \frac{m \hbar}{\hbar^2 \sqrt{ 2 m \epsilon }} \sim  V\end{aligned} \hspace{\stretch{1}}(1.0.11)

and in 1D

\begin{aligned}N(\epsilon ) &\sim  V \int dk \delta \left( \epsilon  - \frac{\hbar^2 k^2}{2m} \right) \\ &= V \frac{m \hbar}{\hbar^2 \sqrt{ 2 m \epsilon }} \\ &\sim  \frac{1}{{\sqrt{\epsilon }}}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

What happens when we have linear energy momentum relationships?

Suppose that we have a linear energy momentum relationship like

\begin{aligned}\epsilon_\mathbf{k} = v \left\lvert {\mathbf{k}} \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.0.13)

An example of such a relationship is the high velocity relation between the energy and momentum of a particle

\begin{aligned}\epsilon_\mathbf{k} = \sqrt{ m_0^2 c^4 + p^2 c^2 } \sim  \left\lvert {\mathbf{p}} \right\rvert c.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Another example is graphene, a carbon structure of the form fig. 1.3. The energy and momentum for such a structure is related in roughly as shown in fig. 1.4, where

Fig 1.3: Graphene bond structure

 

Fig 1.4: Graphene energy momentum dependence

 

\begin{aligned}\epsilon_\mathbf{k} = \pm v_{\mathrm{F}} \left\lvert {\mathbf{k}} \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.0.15)

Continuing with the 3D case we have

FIXME: Is this (or how is this) related to the linear energy momentum relationships for Graphene like substances?

\begin{aligned}N = V \int_0^\infty\underbrace{n_{\mathrm{F}}(\epsilon )}_{1/(e^{\beta (\epsilon  - \mu)} + 1)}\underbrace{N(\epsilon )}_{\epsilon ^{1/2}}\end{aligned} \hspace{\stretch{1}}(1.0.16)

\begin{aligned}\rho &= \frac{N}{V} \\ &= \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\int_0^\infty d\epsilon  \frac{\epsilon ^{1/2}}{z^{-1} e^{\beta \epsilon } + 1} \\ &= \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\left( k_{\mathrm{B}} T \right)^{3/2}\int_0^\infty dx \frac{x^{1/2}}{z^{-1} e^{x} + 1}\end{aligned} \hspace{\stretch{1}}(1.0.17)

where z = e^{\beta \mu} as usual, and we write x = \beta \epsilon . For the low temperature asymptotic behavior see [1] appendix section E. For z large it can be shown that this is

\begin{aligned}\int_0^\infty dx \frac{x^{1/2}}{z^{-1} e^{x} + 1}\approx \frac{2}{3}\left( \ln z \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{1}{{(\ln z)^2}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.18)

so that

\begin{aligned}\rho &\approx  \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\left( k_{\mathrm{B}} T \right)^{3/2}\frac{2}{3}\left( \ln z \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{1}{{(\ln z)^2}} \right) \\ &= \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\frac{2}{3}\mu^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{1}{{(\beta \mu)^2}} \right) \\ &= \left( \frac{2m}{\hbar^2 } \right)^{3/2} \frac{1}{{ 4 \pi^2}}\frac{2}{3}\mu^{3/2}\left( 1 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{\mu} \right)^2 \right) \\ &= \rho_{T = 0}\left( \frac{\mu}{ \epsilon_{\mathrm{F}} } \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{\mu} \right)^2 \right)\end{aligned} \hspace{\stretch{1}}(1.0.19)

Assuming a quadratic form for the chemical potential at low temperature as in fig. 1.5, we have

Fig 1.5: Assumed quadratic form for low temperature chemical potential

 

\begin{aligned}1 &= \left( \frac{\mu}{ \epsilon_{\mathrm{F}} } \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{\mu} \right)^2 \right) \\ &= \left( \frac{\epsilon_{\mathrm{F}} - a T^2}{ \epsilon_{\mathrm{F}} } \right)^{3/2}\left( 1 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{\epsilon_{\mathrm{F}} - a T^2} \right)^2 \right) \\ &\approx  \left( 1 - \frac{3}{2} a \frac{T^2}{\epsilon_{\mathrm{F}}} \right)\left( 1 + \frac{\pi^2}{8} \frac{(k_{\mathrm{B}} T)^2}{\epsilon_{\mathrm{F}}^2} \right) \\ &\approx  1 - \frac{3}{2} a \frac{T^2}{\epsilon_{\mathrm{F}}} + \frac{\pi^2}{8} \frac{(k_{\mathrm{B}} T)^2}{\epsilon_{\mathrm{F}}^2},\end{aligned} \hspace{\stretch{1}}(1.0.20)

or

\begin{aligned}a = \frac{\pi^2}{12} \frac{k_{\mathrm{B}}^2}{\epsilon_{\mathrm{F}}},\end{aligned} \hspace{\stretch{1}}(1.0.21)

We have used a Taylor expansion (1 + x)^n \approx  1 + n x for small x, for an end result of

\begin{aligned}\mu = \epsilon_{\mathrm{F}} - \frac{\pi^2}{12} \frac{(k_{\mathrm{B}} T)^2}{\epsilon_{\mathrm{F}}}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

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