Peeter Joot's (OLD) Blog.

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Posts Tagged ‘chemical potential’

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

[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: 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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PHY452H1S Basic Statistical Mechanics. Lecture 20: Bosons. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 2, 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.

Bosons

In order to maintain a conservation of particles in a Bose condensate as we decrease temperature, we are forced to change the chemical potential to compensate. This is illustrated in fig. 1.1.

Fig 1.1: Chemical potential in Bose condensation region

 

Bose condensatation occurs for T < T_{\mathrm{BEC}}. At this point our number density becomes (except at \mathbf{k} = 0)

\begin{aligned}n(\mathbf{k}) = \frac{1}{{e^{\beta \epsilon_\mathbf{k}} - 1}}.\end{aligned} \hspace{\stretch{1}}(1.2.1)

Except for \mathbf{k} = 0, n(\mathbf{k}) is well defined, and not described by this distribution. We are forced to say that

\begin{aligned}N = N_0 + \sum_{\mathbf{k} \ne 0} n(\mathbf{k}) = N_0 + V\int \frac{d^3 \mathbf{k}}{(2 \pi)^3} \frac{1}{{ e^{\beta \epsilon_\mathbf{k}} - 1 }}.\end{aligned} \hspace{\stretch{1}}(1.2.1)

Introducing the density of states, our density is

\begin{aligned}\rho = \rho_0 + \int_0^\infty d\epsilon \frac{N(\epsilon)}{e^{\beta \epsilon} - 1 },\end{aligned} \hspace{\stretch{1}}(1.2.3)

where

\begin{aligned}N(\epsilon) = \frac{1}{{4 \pi^2}} \left( \frac{2m}{\hbar} \right)^{3/2} \epsilon^{1/2}.\end{aligned} \hspace{\stretch{1}}(1.2.4)

We worked out last time that

\begin{aligned}\rho = \rho_0 + \rho \left( \frac{T}{T_{\mathrm{BEC}}} \right)^{3/2},\end{aligned} \hspace{\stretch{1}}(1.2.4)

or

\begin{aligned}\rho_0 = \rho \left( 1 - \left( \frac{T}{T_{\mathrm{BEC}}} \right) ^{3/2} \right).\end{aligned} \hspace{\stretch{1}}(1.2.6)

This is plotted in fig. 1.2.

Fig 1.2: Density variation with temperature for Bosons

 

\begin{aligned}\rho_0 = \frac{N_{\mathbf{k} = 0}}{V}.\end{aligned} \hspace{\stretch{1}}(1.7)

For T \ge T_{\mathrm{BEC}}, we have \rho_0 = 0. This condensation temperature is

\begin{aligned}T_{\mathrm{BEC}} \propto \rho^{2/3}.\end{aligned} \hspace{\stretch{1}}(1.8)

This is plotted in fig. 1.3.

Fig 1.3: Temperature vs pressure demarkation by T_BEC curve

 

There is a line for each density that marks the boundary temperature for which we have or do not have this condensation phenomina where \mathbf{k} = 0 states start filling up.

Specific heat: T < T_{\mathrm{BEC}}

\begin{aligned}\frac{E}{V} &= \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} \frac{1}{{ e^{\beta \hbar^2 k^2/2m} - 1}}\frac{\hbar^2 k^2}{2m} \\ &= \int_0^\infty d\epsilon N(\epsilon) \frac{1}{{ e^{\beta \epsilon} - 1 }} \epsilon \\ &\propto \int_0^\infty d\epsilon \frac{\epsilon^{3/2}}{ e^{\beta \epsilon} - 1 } \\ &\propto \left( k_{\mathrm{B}} T \right)^{5/2},\end{aligned} \hspace{\stretch{1}}(1.9)

so that

\begin{aligned}\frac{C}{V} \propto \left( k_{\mathrm{B}} T \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.10)

Compare this to the classical and Fermionic specific heat as plotted in fig. 1.4.

Fig 1.4: Specific heat for Bosons, Fermions, and classical ideal gases

 

One can measure the specific heat in this Bose condensation phenomina for materials such as Helium-4 (spin 0). However, it turns out that Helium-4 is actually quite far from an ideal Bose gas.

Photon gas

A system that is much closer to an ideal Bose gas is that of a gas of photons. To a large extent, photons do not interact with each other. This allows us to calculate black body phenomina and the low temperature (cosmic) background radiation in the universe.

An important distinction between a photon sea and some of these other systems is that the photon number is actually not fixed.

Photon numbers are not “conserved”.

If a photon interacts with an atom, it can impart energy and disappear. An excited atom can emit a photon and change its energy level. In a thermodynamic system we can generally expect that introducing heat will generate more photons, whereas a cold sink will tend to generate fewer photons.

We have a few special details that distinguish photons that we’ll have to consider.

  1. spin 1.
  2. massless, moving at the speed of light.
  3. have two polarization states.

Because we do not have a constraint on the number of particles, we essentially have no chemical potential, even in the grand canonical scheme.

Writing

\begin{aligned}\lambda = \left\{\begin{array}{l l}+1 & \quad \mbox{Right circular polarization} \\ -1 & \quad \mbox{Left circular polarization}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.11)

Our number density, since we have no chemical potential, is of the form

\begin{aligned}n_{\mathbf{k}, \lambda}= \frac{1}{{e^{\beta \epsilon_{\mathbf{k}, \lambda}} - 1 }},\end{aligned} \hspace{\stretch{1}}(1.12)

Observe that the average number of photons in this system is temperature dependent. Because this chemical potential is not there, it can be quite easy to work out a number of the thermodynamic results.

Photon average energy density

We’ll now calculate the average energy density of the photons. The energy of a single photon is

\begin{aligned}\epsilon_{\mathbf{k}, \lambda} = \hbar c k = \hbar \omega,\end{aligned} \hspace{\stretch{1}}(1.2.13)

so that the average energy density is

\begin{aligned}\frac{E}{V} &= \sum_{\mathbf{k}, \lambda} \frac{1}{{ e^{ \beta \epsilon_\mathbf{k}} - 1}} \epsilon_\mathbf{k}\rightarrow\underbrace{2}_{\text{number of polarizations}}\int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\frac{ \hbar c k}{ e^{ \beta \epsilon_\mathbf{k}} - 1} \\ &= 2 \int_0^\infty d\epsilon \underbrace{\frac{1}{{(2 \pi)^3}} 4 \pi \frac{\epsilon^2}{(\hbar c)^3} }_{\text{Photon density of states}}\frac{\epsilon}{e^{\beta \epsilon} - 1} \\ &= \frac{1}{{\pi^2}} \frac{1}{{ (\hbar c)^3 }} \int_0^\infty d\epsilon \frac{\epsilon^3}{e^{\beta \epsilon} - 1}\end{aligned} \hspace{\stretch{1}}(1.2.13)

Mathematica tells us that this integral is

\begin{aligned}\int_0^\infty d\epsilon \frac{\epsilon^3}{e^{\beta \epsilon} - 1} =\frac{\pi ^4}{15 \beta ^4},\end{aligned} \hspace{\stretch{1}}(1.2.13)

for an end result of

\begin{aligned}\frac{E}{V} =\frac{\pi^2}{15} \frac{1}{{(\hbar c)^3}} \left( k_{\mathrm{B}} T \right)^4.\end{aligned} \hspace{\stretch{1}}(1.2.13)

Phonons and other systems

There is a very similar phenomina in matter. We can discuss lattice vibrations in a solid. These are called phonon modes, and will have the same distribution function where the only difference is that the speed of light is replaced by the speed of the sound wave in the solid. Once we understand the photon system, we are able to look at other Bose distributions such as these phonon systems. We’ll touch on this very briefly next time.

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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.

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PHY452H1S Basic Statistical Mechanics. Lecture 16: Fermi gas. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 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

Review

Continuing a discussion of [1] section 8.1 content.

We found

\begin{aligned}n_{\mathbf{k}} = \frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}}\end{aligned} \hspace{\stretch{1}}(1.2.1)

With no spin

\begin{aligned}\int n_\mathbf{k} \times \frac{d^3 k}{(2\pi)^3} = \rho\end{aligned} \hspace{\stretch{1}}(1.2.2)

Fig 1.1: Occupancy at low temperature limit

 

Fig 1.2: Volume integral over momentum up to Fermi energy limit

 

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m}\end{aligned} \hspace{\stretch{1}}(1.2.3)

gives

\begin{aligned}k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.4)

\begin{aligned}\sum_\mathbf{k} n_\mathbf{k} = N\end{aligned} \hspace{\stretch{1}}(1.2.5)

\begin{aligned}\mathbf{k} = \frac{2\pi}{L}(n_x, n_y, n_z)\end{aligned} \hspace{\stretch{1}}(1.2.6)

This is for periodic boundary conditions \footnote{I filled in details in the last lecture using a particle in a box, whereas this periodic condition was intended. We see that both achieve the same result}, where

\begin{aligned}\Psi(x + L) = \Psi(x)\end{aligned} \hspace{\stretch{1}}(1.2.7)

Moving on

\begin{aligned}\sum_{k_x} n(\mathbf{k}) = \sum_{p_x} \Delta p_x n(\mathbf{k})\end{aligned} \hspace{\stretch{1}}(1.2.8)

with

\begin{aligned}\Delta k_x = \frac{2 \pi}{L} \Delta p_x\end{aligned} \hspace{\stretch{1}}(1.2.9)

this gives

\begin{aligned}\sum_{k_x} n(\mathbf{k}) = \sum_{n_x} \frac{L}{2\pi} \Delta k_x \rightarrow \frac{L}{2\pi} \int d k_x\end{aligned} \hspace{\stretch{1}}(1.2.10)

Over all dimensions

\begin{aligned}\sum_{\mathbf{k}} n_\mathbf{k} = \left( \frac{L}{2\pi} \right)^3 \left( \int d^3 \mathbf{k} \right)n(\mathbf{k})=N\end{aligned} \hspace{\stretch{1}}(1.2.11)

so that

\begin{aligned}\rho = \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\end{aligned} \hspace{\stretch{1}}(1.2.12)

Again

\begin{aligned}k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.13)

Example: Spin considerations

{example:basicStatMechLecture16:1}{

\begin{aligned}\sum_{\mathbf{k}, m_s} = N\end{aligned} \hspace{\stretch{1}}(1.2.14)

\begin{aligned}\sum_{\mathbf{k}, m_s} \frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}} = (2 S + 1)\left( \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} n(\mathbf{k}) \right)L^3\end{aligned} \hspace{\stretch{1}}(1.2.15)

This gives us

\begin{aligned}k_{\mathrm{F}} = \left( \frac{ 6 \pi^2 \rho }{2 S + 1} \right)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.16)

and again

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m}\end{aligned} \hspace{\stretch{1}}(1.2.17)

}

High Temperatures

Now we want to look at the at higher temperature range, where the occupancy may look like fig. 1.3

Fig 1.3: Occupancy at higher temperatures

 

\begin{aligned}\mu(T = 0) = \epsilon_{\mathrm{F}}\end{aligned} \hspace{\stretch{1}}(1.2.18)

\begin{aligned}\mu(T \rightarrow \infty) \rightarrow - \infty\end{aligned} \hspace{\stretch{1}}(1.2.19)

so that for large T we have

\begin{aligned}\frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}} \rightarrow e^{-\beta(\epsilon_k - \mu)}\end{aligned} \hspace{\stretch{1}}(1.2.20)

\begin{aligned}\rho &= \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} e^{\beta \mu} e^{-\beta \epsilon_k} \\ &= e^{\beta \mu} \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} e^{-\beta \epsilon_k} \\ &= e^{\beta \mu} \int dk \frac{4 \pi k^2}{(2 \pi)^3} e^{-\beta \hbar^2 k^2/2m}.\end{aligned} \hspace{\stretch{1}}(1.2.21)

Mathematica (or integration by parts) tells us that

\begin{aligned}\frac{1}{{(2 \pi)^3}} \int 4 \pi^2 k^2 dk e^{-a k^2} = \frac{1}{{(4 \pi a )^{3/2}}},\end{aligned} \hspace{\stretch{1}}(1.2.22)

so we have

\begin{aligned}\rho &= e^{\beta \mu} \left( \frac{2m}{ 4 \pi \beta \hbar^2} \right)^{3/2} \\ &= e^{\beta \mu} \left( \frac{2 m k_{\mathrm{B}} T 4 \pi^2 }{ 4 \pi h^2} \right)^{3/2} \\ &= e^{\beta \mu} \left( \frac{2 m k_{\mathrm{B}} T \pi }{ h^2} \right)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.2.23)

Introducing \lambda for the thermal de Broglie wavelength, \lambda^3 \sim T^{-3/2}

\begin{aligned}\lambda \equiv \frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}},\end{aligned} \hspace{\stretch{1}}(1.2.24)

we have

\begin{aligned}\rho = e^{\beta \mu} \frac{1}{{\lambda^3}}.\end{aligned} \hspace{\stretch{1}}(1.2.25)

Does it make any sense to have density as a function of temperature? An inappropriately extended to low temperatures plot of the density is found in fig. 1.4 for a few arbitrarily chosen numerical values of the chemical potential \mu, where we see that it drops to zero with temperature. I suppose that makes sense if we are not holding volume constant.

Fig 1.4: Density as a function of temperature

 

We can write

\begin{aligned}\boxed{e^{\beta \mu} = \left( \rho \lambda^3 \right)}\end{aligned} \hspace{\stretch{1}}(1.2.26)

\begin{aligned}\frac{\mu}{k_{\mathrm{B}} T} = \ln \left( \rho \lambda^3 \right)\sim -\frac{3}{2} \ln T\end{aligned} \hspace{\stretch{1}}(1.2.27)

or (taking \rho (and/or volume?) as a constant) we have for large temperatures

\begin{aligned}\mu \propto -T \ln T\end{aligned} \hspace{\stretch{1}}(1.2.28)

The chemical potential is plotted in fig. 1.5, whereas this - k_{\mathrm{B}} T \ln k_{\mathrm{B}} T function is plotted in fig. 1.6. The contributions to \mu from the k_{\mathrm{B}} T \ln (\rho h^3 (2 \pi m)^{-3/2}) term are dropped for the high temperature approximation.

Fig 1.5: Chemical potential over degenerate to classical range

Fig 1.6: High temp approximation of chemical potential, extended back to T = 0

Pressure

\begin{aligned}P = - \frac{\partial {E}}{\partial {V}}\end{aligned} \hspace{\stretch{1}}(1.2.29)

For a classical ideal gas as in fig. 1.7 we have

Fig 1.7: Ideal gas pressure vs volume

 

\begin{aligned}P = \rho k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.2.30)

For a Fermi gas at T = 0 we have

\begin{aligned}E &= \sum_\mathbf{k} \epsilon_k n_k \\ &= \sum_\mathbf{k} \epsilon_k \Theta(\mu_0 - \epsilon_k) \\ &= \frac{V}{(2\pi)^3} \int_{\epsilon_k < \mu_0} \frac{\hbar^2 \mathbf{k}^2}{2 m} d^3 \mathbf{k} \\ &= \frac{V}{(2\pi)^3} \int_0^{k_{\mathrm{F}}} \frac{\hbar^2 \mathbf{k}^2}{2 m} d^3 \mathbf{k} \\ &= \frac{V}{(2\pi)^3} \frac{\hbar^2}{2 m} \int_0^{k_{\mathrm{F}}} k^2 4 \pi k^2 d k\propto k_{\mathrm{F}}^5\end{aligned} \hspace{\stretch{1}}(1.2.31)

Specifically,

\begin{aligned}E(T = 0) = V \times \frac{3}{5} \underbrace{\epsilon_{\mathrm{F}}}_{\sim k_{\mathrm{F}}^2}\underbrace{\rho}_{\sim k_{\mathrm{F}}^3}\end{aligned} \hspace{\stretch{1}}(1.2.32)

or

\begin{aligned}\frac{E}{N} = \frac{3}{5} \epsilon_{\mathrm{F}}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}E = \frac{3}{5} N \frac{\hbar^2}{2 m} \left( 6 \pi^2 \frac{N}{V} \right)^{2/3} = a V^{-2/3},\end{aligned} \hspace{\stretch{1}}(1.2.34)

so that

\begin{aligned}\frac{\partial {E}}{\partial {V}} = -\frac{2}{3} a V^{-5/3}.\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}P &= -\frac{\partial {E}}{\partial {V}}  \\ &= \frac{2}{3} \left( a V^{-2/3} \right)V^{-1} \\ &= \frac{2}{3} \frac{E}{V} \\ &= \frac{2}{3} \left( \frac{3}{5} \epsilon_{\mathrm{F}} \rho \right) \\ &= \frac{2}{5} \epsilon_{\mathrm{F}} \rho.\end{aligned} \hspace{\stretch{1}}(1.2.36)

We see that the pressure ends up deviating from the classical result at low temperatures, as sketched in fig. 1.8. This low temperature limit for the pressure 2 \epsilon_{\mathrm{F}} \rho/5 is called the degeneracy pressure.

Fig 1.8: Fermi degeneracy pressure

 

References

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

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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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open system variance of N

Posted by peeterjoot on March 16, 2013

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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.

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