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# Posts Tagged ‘specific heat’

## 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.

September 05, 2013 Large volume fermi gas density

April 30, 2013 Ultra relativistic spin zero condensation temperature

April 24, 2013 Low temperature Fermi gas chemical potential

## Bose gas specific heat above condensation temperature

Posted by peeterjoot on May 9, 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: Bose gas specific heat above condensation temperature ([1] section 7.1.37)

Equation 7.1.33 provides a relation for specific heat

\begin{aligned}\frac{C_{\mathrm{V}}}{N k_{\mathrm{B}}} = \left(\frac{\partial {}}{\partial {T}}\left( \frac{3}{2} T \frac{ g_{5/2}(z) } { g_{3/2}(z) } \right)\right)_v.\end{aligned} \hspace{\stretch{1}}(1.0.1)

Fill in the details showing how this can be used to find

\begin{aligned}\frac{C_{\mathrm{V}}}{N k_{\mathrm{B}}} = \frac{15}{4} \frac{ g_{5/2}(z) }{ g_{3/2}(z) }-\frac{9}{4} \frac{ g_{3/2}(z) }{ g_{1/2}(z) }.\end{aligned} \hspace{\stretch{1}}(1.0.2)

With

\begin{aligned}g_{{3/2}}(z) = \frac{\lambda^3}{v} = \frac{h^3}{\left( 2 \pi m k_{\mathrm{B}} T \right)^{3/2}}\end{aligned} \hspace{\stretch{1}}(1.0.3)

we have for constant $v$

\begin{aligned}\left({\partial {g_{3/2}}}/{\partial {T}}\right)_{{v}}= -\frac{3}{2}\frac{h^3}{\left( 2 \pi m k_{\mathrm{B}} \right)^{3/2} T^{5/2}}= -\frac{3}{2 T} g_{{3/2}}(z).\end{aligned} \hspace{\stretch{1}}(1.0.3)

From the series expansion

\begin{aligned}g_{{\nu}}(z) = \sum_{k = 1}^\infty \frac{z^k}{k^\nu},\end{aligned} \hspace{\stretch{1}}(1.0.5)

we have

\begin{aligned}z \frac{\partial {}}{\partial {z}} g_{{\nu}}(z) = z\sum_{k = 1}^\infty k \frac{z^{k-1}}{k^\nu}=\sum_{k = 1}^\infty \frac{z^{k}}{k^{\nu-1}}= g_{{\nu-1}}(z).\end{aligned} \hspace{\stretch{1}}(1.0.5)

Taken together we have

\begin{aligned}-\frac{3}{2 T} g_{{3/2}}(z) &=\left({\partial {g_{3/2}}}/{\partial {T}}\right)_{{v}} \\ &=\left({\partial {z}}/{\partial {T}}\right)_{{v}}\frac{\partial {}}{\partial {z}} g_{{3/2}}(z) \\ &=\frac{1}{{z}} \left({\partial {z}}/{\partial {T}}\right)_{{v}}z \frac{\partial {}}{\partial {z}} g_{{3/2}}(z) \\ &=\frac{1}{{z}} \left({\partial {z}}/{\partial {T}}\right)_{{v}}g_{{1/2}}(z),\end{aligned} \hspace{\stretch{1}}(1.0.5)

or

\begin{aligned}\frac{1}{{z}} \left({\partial {z}}/{\partial {T}}\right)_{{v}} = -\frac{3}{2 T} \frac{g_{{3/2}}(z)}{g_{{1/2}}(z)}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

We are now ready to evaluate the derivative and find the specific heat

\begin{aligned}\frac{C_{\mathrm{V}}}{N k_{\mathrm{B}}} &= \left(\frac{\partial {}}{\partial {T}}\left( \frac{3}{2} T \frac{ g_{5/2}(z) } { g_{3/2}(z) } \right)\right)_v \\ &=\frac{3}{2} \frac{ g_{5/2}(z) }{ g_{3/2}(z) }+\frac{3 T}{2} \left({\partial {z}}/{\partial {T}}\right)_{{v}}\frac{\partial {}}{\partial {z}}\left( \frac{ g_{5/2}(z) } { g_{3/2}(z) } \right) \\ &=\frac{3}{2} \frac{ g_{5/2}(z) }{ g_{3/2}(z) }-\frac{9 T}{4} \frac{g_{{3/2}}(z)}{g_{{1/2}}(z)}z\frac{\partial {}}{\partial {z}}\left( \frac{ g_{5/2}(z) } { g_{3/2}(z) } \right) \\ &=\frac{3}{2} \frac{ g_{5/2}(z) }{ g_{3/2}(z) }-\frac{9 }{4} \frac{g_{{3/2}}(z)}{g_{{1/2}}(z)}\not{{\frac{ g_{3/2}(z) }{ g_{3/2}(z) }}}+\frac{9 }{4} \frac{\not{{g_{{3/2}}(z)}}}{\not{{g_{{1/2}}(z)}}}\frac{ g_{5/2}(z) \not{{g_{1/2}(z)}}}{ \left( g_{3/2}(z) \right)^{\not{{2}}} } \\ &=\frac{3}{2} \frac{ g_{5/2}(z) }{ g_{3/2}(z) }-\frac{9 }{4} \frac{g_{{3/2}}(z)}{g_{{1/2}}(z)}+\frac{9 }{4} \frac{ g_{5/2}(z) }{ g_{3/2}(z) } \\ &=\frac{15}{4} \frac{ g_{5/2}(z) }{ g_{3/2}(z) }-\frac{9 }{4} \frac{g_{{3/2}}(z)}{g_{{1/2}}(z)}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

This is the desired result.

# References

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

## 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.

## PHY452H1S Basic Statistical Mechanics. Problem Set 7: BEC and phonons

Posted by peeterjoot on April 10, 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

## Question: Bose-Einstein condensation (BEC) in one and two dimensions

Obtain the density of states $N(\epsilon)$ in one and two dimensions for a particle with an energy-momentum relation

\begin{aligned}E_\mathbf{k} = \frac{\hbar^2 \mathbf{k}^2}{2 m}.\end{aligned} \hspace{\stretch{1}}(1.1)

Using this, show that for particles whose number is conserved the BEC transition temperature vanishes in these cases – so we can always pick a chemical potential $\mu < 0$ which preserves a constant density at any temperature.

We’d like to evaluate

\begin{aligned}N_d(\epsilon) \equiv\sum_\mathbf{k}\delta(\epsilon - \epsilon_\mathbf{k})\approx\frac{L^d}{(2 \pi)^d} \int d^d \mathbf{k} \delta\left( \epsilon - \frac{\hbar^2 k^2}{2 m} \right),\end{aligned} \hspace{\stretch{1}}(1.2)

We’ll use

\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}}(1.3)

where the roots of $g(x)$ are $x_0$. With

\begin{aligned}g(k) = \epsilon - \frac{\hbar^2 k^2}{2 m},\end{aligned} \hspace{\stretch{1}}(1.4)

the roots $k^{*}$ of $g(k) = 0$ are

\begin{aligned}k^{*} = \pm \sqrt{\frac{2 m \epsilon }{\hbar^2}}.\end{aligned} \hspace{\stretch{1}}(1.5)

The derivative of $g(k)$ evaluated at these roots are

\begin{aligned}g'(k^{*}) &= -\frac{\hbar^2 k^{*}}{m} \\ &= \mp \frac{\hbar^2}{m}\frac{\sqrt{2 m \epsilon}}{ \hbar } \\ &= \mp \frac{\hbar \sqrt{2 m \epsilon} }{m}.\end{aligned} \hspace{\stretch{1}}(1.6)

In 2D, we can evaluate over a shell in $k$ space

\begin{aligned}N_2(\epsilon) &= \frac{A}{(2 \pi)^2} \int_0^\infty 2 \pi k dk\left( \delta \left( k - k^{*} \right) + \delta \left( k + k^{*} \right) \right)\frac{m}{\hbar \sqrt{2 m \epsilon} } \\ &= \frac{A}{2 \pi} \not{{k^{*}}}\frac{m}{\hbar^2 \not{{k^{*}}} }\end{aligned} \hspace{\stretch{1}}(1.7)

or

\begin{aligned}\boxed{N_2(\epsilon) = \frac{2 \pi A m}{h^2}.}\end{aligned} \hspace{\stretch{1}}(1.8)

In 1D we have

\begin{aligned}N_1(\epsilon) &= \frac{L}{2 \pi} \int_{-\infty}^\infty dk\left( \delta \left( k - k^{*} \right) + \delta \left( k + k^{*} \right) \right)\frac{m}{\hbar \sqrt{2 m \epsilon} } \\ &= \frac{2 L}{2 \pi} \frac{m}{\hbar \sqrt{2 m \epsilon} }.\end{aligned} \hspace{\stretch{1}}(1.9)

Observe that this time for 1D, unlike in 2D when we used a radial shell in $k$ space, we have contributions from both the delta function roots. Our end result is

\begin{aligned}\boxed{N_1(\epsilon) =\frac{2 L}{h} \sqrt{\frac{m}{2 \epsilon}}.}\end{aligned} \hspace{\stretch{1}}(1.10)

To consider the question of the BEC temperature, we’ll need to calculate the density. For the 2D case we have

\begin{aligned}\rho = \frac{N}{A} &= \frac{1}{A} A \int \frac{d^2 \mathbf{k}}{(2 \pi)^2} f(e_\mathbf{k}) \\ &= \frac{1}{A} \frac{2 \pi A m}{h^2}\int_0^\infty d\epsilon \frac{1}{{ z^{-1} e^{\beta \epsilon} -1 }} \\ &= \frac{2 \pi m}{h^2 \beta}\int_0^\infty dx \frac{1}{{ z^{-1} e^{x} -1 }} \\ &= -\frac{2 \pi m k_{\mathrm{B}} T}{h^2} \ln (1 - z) \\ &= -\frac{1}{{\lambda^2}} \ln (1 - z).\end{aligned} \hspace{\stretch{1}}(1.11)

Recall for the 3D case that we had an upper bound as $z \rightarrow 1$. We don’t have that for this 2D density, so for any value of $k_{\mathrm{B}} T > 0$, a corresponding value of $z$ can be found. That is

\begin{aligned}z &= 1 - e^{-\rho \lambda^2} \\ &= 1 - e^{-\rho h^4/(2 \pi m k_{\mathrm{B}} T)^2}.\end{aligned} \hspace{\stretch{1}}(1.1.12)

For the 1D case we have

\begin{aligned}\rho &= \frac{N}{L} \\ &= \frac{1}{L} L \int \frac{dk}{2 \pi} f(e_\mathbf{k}) \\ &= \frac{1}{L} \frac{2 L}{h} \sqrt{\frac{m}{2}}\int_0^\infty d\epsilon \frac{1}{{\sqrt{\epsilon}}}\frac{1}{{ z^{-1} e^{\beta \epsilon} -1 }} \\ &= \frac{1}{{h}} \sqrt{\frac{2 m}{\beta}} \int_0^\infty \frac{x^{1/2 - 1}}{z^{-1} e^x - 1} \\ &= \frac{1}{{h}} \sqrt{\frac{2 m}{\beta}} \Gamma(1/2) f^-_{1/2}(z),\end{aligned} \hspace{\stretch{1}}(1.1.12)

or

\begin{aligned}\rho= \frac{1}{{\lambda}} f^-_{1/2}(z).\end{aligned} \hspace{\stretch{1}}(1.1.12)

See fig. 1.1 for plots of $f^-_\nu(z)$ for $\nu \in \{1/2, 1, 3/2\}$, the respective results for the 1D, 2D and 3D densities respectively.

Fig 1.1: Density integrals for 1D, 2D and 3D cases

We’ve found that $f^-_{1/2}(z)$ is also unbounded as $z \rightarrow 1$, so while we cannot invert this easily as in the 2D case, we can at least say that there will be some $z$ for any value of $k_{\mathrm{B}} T > 0$ that allows the density (and thus the number of particles) to remain fixed.

## Question: Estimating the BEC transition temperature

Find data for the atomic mass of liquid ${}^4$ He and its density at ambient atmospheric pressure and hence estimate its BEC temperature assuming interactions are unimportant (even though this assumption is a very bad one!).

For dilute atomic gases of the sort used in Professor
Thywissen’s lab
, one typically has a cloud of $10^6$ atoms confined to an approximate cubic region with linear dimension 1 $\mu\,m$. Find the density – it is pretty low, so interactions can be assumed to be extremely weak. Assuming these are ${}^{87}$ Rb atoms, estimate the BEC transition temperature.

With an atomic weight of 4.0026, the mass in grams for one atom of Helium is

\begin{aligned}4.0026 \,\text{amu} \times \frac{\text{g}}{6.022 \times 10^{23} \text{amu}} &= 6.64 \times 10^{-24} \text{g} \\ &= 6.64 \times 10^{-27} \text{kg}.\end{aligned} \hspace{\stretch{1}}(1.15)

With the density of liquid He-4, at 5.2K (boiling point): 125 grams per liter, the number density is

\begin{aligned}\rho &= \frac{\text{mass}}{\text{volume}} \times \frac{1}{{\text{mass of one He atom}}} \\ &= \frac{125 \text{g}}{10^{-3} m^3} \times \frac{1}{{6.64 \times 10^{-24} g}} \\ &= \frac{125 \text{g}}{10^{-3} m^3} \times \frac{1}{{6.64 \times 10^{-24} g}} \\ &= 1.88 \times 10^{28} m^{-3}\end{aligned} \hspace{\stretch{1}}(1.16)

In class the $T_{\mathrm{BEC}}$ was found to be

\begin{aligned}T_{\mathrm{BEC}} &= \frac{1}{k_{\mathrm{B}}} \left( \frac{\rho}{\zeta(3/2)} \right)^{2/3} \frac{ 2 \pi \hbar^2}{M} \\ &= \frac{1}{{1.3806488 \times 10^{-23} m^2 kg/s^2/K}} \left( \frac{\rho}{ 2.61238 } \right)^{2/3} \frac{ 2 \pi (1.05457173 \times 10^{-34} m^2 kg / s)^2}{M} \\ &= 2.66824 \times 10^{-45} \frac{\rho^{2/3}}{M} K.\end{aligned} \hspace{\stretch{1}}(1.17)

So for liquid helium we have

\begin{aligned}T_{\mathrm{BEC}} &= 2.66824 \times 10^{-45} \left( 1.88 \times 10^{28} \right)^{2/3} \frac{1}{{ 6.64 \times 10^{-27} }} K \\ &= 2.84 K.\end{aligned} \hspace{\stretch{1}}(1.18)

The number density for the gas in Thywissen’s lab is

\begin{aligned}\rho &= \frac{10^6}{(10^{-6} \text{m})^3} \\ &= 10^{24} m^{-3}.\end{aligned} \hspace{\stretch{1}}(1.1.19)

The mass of an atom of ${}^{87}$ Rb is

\begin{aligned}86.90 \,\text{amu} \times \frac{10^{-3} \text{kg}}{6.022 \times 10^{23} \text{amu}} = 1.443 \times 10^{-25} \text{kg},\end{aligned} \hspace{\stretch{1}}(1.1.19)

which gives us

\begin{aligned}T_{\mathrm{BEC}} &= 2.66824 \times 10^{-45} \left( 10^{24} \right)^{2/3} \frac{1}{{ 1.443 \times 10^{-25} }} K \\ &= 1.85 \times 10^{-4} K.\end{aligned} \hspace{\stretch{1}}(1.1.19)

## Question: Phonons in two dimensions

Consider phonons (quanta of lattice vibrations) which obey a dispersion relation

\begin{aligned}E_\mathbf{k} = \hbar v \left\lvert {\mathbf{k}} \right\rvert\end{aligned} \hspace{\stretch{1}}(1.1.22)

for small momenta $\left\lvert {\mathbf{k}} \right\rvert$, where $v$ is the speed of sound. Assuming a two-dimensional crystal, phonons only propagate along the plane containing the atoms. Find the specific heat of this crystal due to phonons at low temperature. Recall that phonons are not conserved, so there is no chemical potential associated with maintaining a fixed phonon density.

The energy density of the system is

\begin{aligned}\frac{E}{V} &= \int \frac{d^2 \mathbf{k}}{(2 \pi)^2} \frac{\epsilon}{ e^{\beta \epsilon} - 1 } \\ &= \int d\epsilon \frac{N(\epsilon)}{V} \frac{\epsilon}{ e^{\beta \epsilon} - 1 }.\end{aligned} \hspace{\stretch{1}}(1.23)

For the density of states we have

\begin{aligned}\frac{N(\epsilon) }{V} &= \int \frac{d^2 \mathbf{k}}{(2 \pi)^2} \delta( \epsilon - \epsilon_\mathbf{k} ) \\ &= \frac{1}{{(2 \pi)^2}} 2 \pi \int_0^\infty k dk \delta( \epsilon - \hbar v k ) \\ &= \frac{1}{{2 \pi}} \int_0^\infty k dk \delta \left( k - \frac{\epsilon}{\hbar v} \right) \frac{1}{{\hbar v}} \\ &= \frac{1}{{2 \pi}} \frac{\epsilon}{(\hbar v)^2}.\end{aligned} \hspace{\stretch{1}}(1.24)

Plugging back into the energy density we have

\begin{aligned}\frac{E}{V} &= \frac{2 \pi}{(\hbar v)^2}\int_0^\infty d\epsilon \frac{\epsilon^2}{ e^{\beta \epsilon} - 1 } \\ &= \frac{\pi \left( k_{\mathrm{B}} T \right)^3 }{(\hbar v)^2}\zeta(3),\end{aligned} \hspace{\stretch{1}}(1.25)

where $\zeta(3) \approx 2.40411$. Taking derivatives we have

\begin{aligned}\boxed{C_V = \frac{dE}{dT} = V\frac{3 \pi k_{\mathrm{B}}^3 T^2 }{(\hbar v)^2}\zeta(3).}\end{aligned} \hspace{\stretch{1}}(1.1.26)

## Unresolved question about energy distribution around mean energy

Posted by peeterjoot on April 10, 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.)]

In [1] is an expansion of

\begin{aligned}P(E) \propto e^{-\beta E} g(E),\end{aligned} \hspace{\stretch{1}}(1.0.1)

around the mean energy $E^{*} = U$. The first derivative part of the expansion is simple enough

\begin{aligned}\frac{\partial }{\partial E} \left( e^{-\beta E} g(E) \right) &= \left( -\beta g(E) + g'(E) \right)e^{-\beta E} \\ &= g(E) e^{-\beta E}\left( -\beta + (\ln g(E))' \right)\end{aligned} \hspace{\stretch{1}}(1.0.1)

The peak energy $E^{*}$ will be where this derivative equals zero. That is

\begin{aligned}0 = g(E^{*}) e^{-\beta E^{*}}\left( -\beta + {\left.{{(\ln g(E))'}}\right\vert}_{{E = E^{*}}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.3)

or

\begin{aligned}{\left.{{\frac{\partial }{\partial E}\left( \ln g(E) \right)}}\right\vert}_{{E = E^{*}}} = \beta\end{aligned} \hspace{\stretch{1}}(1.0.4)

With

\begin{aligned}S = k_{\mathrm{B}} \ln g\end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}\frac{1}{{k_{\mathrm{B}}}} \left( \frac{\partial S}{\partial E} \right)_{E = U} &= \frac{1}{{k_{\mathrm{B}} T}} \\ &= \beta\end{aligned} \hspace{\stretch{1}}(1.0.5b)

We have

\begin{aligned}\left( \frac{\partial \ln g(E) }{\partial E} \right)_{E = U} = \beta\end{aligned} \hspace{\stretch{1}}(1.0.6)

so that

\begin{aligned}E^{*} = U.\end{aligned} \hspace{\stretch{1}}(1.0.7)

So far so good. Reading the text, the expansion of the logarithm of $P(E)$ around $E = E^{*} = U$ wasn’t clear. Let’s write that out in full. To two terms that is

\begin{aligned}\ln e^{-\beta E} g(E)= \underbrace{\ln e^{-\beta U} g(U)}_{- \beta U + \frac{1}{{k_{\mathrm{B}}}} S}+ {\left.{{\frac{\partial }{\partial E} \left( \ln e^{-\beta E} g(E) \right)}}\right\vert}_{{E = U}}+ \frac{1}{2}{\left.{{\frac{\partial^2 }{\partial E^2} \left( \ln e^{-\beta E} g(E) \right)}}\right\vert}_{{E = U}}(E - U)^2.\end{aligned} \hspace{\stretch{1}}(1.0.7)

The first order term has the derivative of the logarithm of $e^{-\beta E}g(E)$. Since the logarithm is monotonic and the derivative of $e^{-\beta E}g(E)$ has been shown to be zero at $E = U$, this must be zero. We can also see this explicitly by computation

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{-\beta e^{-\beta E} g(E) + e^{-\beta E} g'(E)}{e^{-\beta E} g(E)}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{-\beta g + g'}{g}}}\right\vert}_{{E = U}} \\ &= -\beta+{\left.{{(\ln g)'}}\right\vert}_{{E = U}} \\ &= -\beta + \frac{1}{{k_{\mathrm{B}}}} {\left.{{\frac{\partial S}{\partial E}}}\right\vert}_{{E = U}} \\ &= -\beta + \frac{1}{{k_{\mathrm{B}} T}} \\ &= -\beta + \beta \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.7)

For the second derivative we have

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{\partial }{\partial E} \left( -\beta + (\ln g)' \right)}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{\partial }{\partial E} \frac{g'}{g}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{g''}{g} - \frac{(g')^2}{g^2}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{g''}{g}}}\right\vert}_{{E = U}} - ((\ln g)')^2 \\ &= {\left.{{\frac{g''}{g}}}\right\vert}_{{E = U}} - \beta^2.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Somehow this is supposed to come out to $k_{\mathrm{B}} T^2 C_{\mathrm{V}}$? Backing up, we have

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{\partial^2 }{\partial E^2} \ln g}}\right\vert}_{{E = U}} \\ &= \frac{1}{{k_{\mathrm{B}}}}{\left.{{\frac{\partial^2 S}{\partial E^2}}}\right\vert}_{{E = U}}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

I still don’t see how to get $C_{\mathrm{V}} = {\partial U}/{\partial T}$ out of this? $C_{\mathrm{V}}$ is a derivative with respect to temperature, but here we have derivatives with respect to energy (keeping $\beta = 1/k_{\mathrm{B}} T$ fixed)?

# References

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

## 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.

## 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.

## 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)