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PHY452H1S Basic Statistical Mechanics. Problem Set 6: Max entropy, fugacity, and Fermi gas

Posted by peeterjoot on April 3, 2013

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Disclaimer

This is an ungraded set of answers to the problems posed.

Question: Maximum entropy principle

Consider the “Gibbs entropy”

\begin{aligned}S = - k_{\mathrm{B}} \sum_i p_i \ln p_i\end{aligned} \hspace{\stretch{1}}(1.1)

where p_i is the equilibrium probability of occurrence of a microstate i in the ensemble.

For a microcanonical ensemble with \Omega configurations (each having the same energy), assigning an equal probability p_i= 1/\Omega to each microstate leads to S = k_{\mathrm{B}} \ln \Omega. Show that this result follows from maximizing the Gibbs entropy with respect to the parameters p_i subject to the constraint of

\begin{aligned}\sum_i p_i = 1\end{aligned} \hspace{\stretch{1}}(1.2)

(for p_i to be meaningful as probabilities). In order to do the minimization with this constraint, use the method of Lagrange multipliers – first, do an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i,\end{aligned} \hspace{\stretch{1}}(1.3)

then fix \alpha by demanding that the constraint be satisfied.

For a canonical ensemble (no constraint on total energy, but all microstates having the same number of particles N), maximize the Gibbs entropy with respect to the parameters p_i subject to the constraint of

\begin{aligned}\sum_i p_i = 1,\end{aligned} \hspace{\stretch{1}}(1.4)

(for p_i to be meaningful as probabilities) and with a given fixed average energy

\begin{aligned}\left\langle{{E}}\right\rangle = \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.5)

where E_i is the energy of microstate i. Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i - \beta \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.6)

then fix \alpha, \beta by demanding that the constraint be satisfied. What is the resulting p_i?

For a grand canonical ensemble (no constraint on total energy, or the number of particles), maximize the Gibbs entropy with respect to the parameters p_i subject to the constraint of

\begin{aligned}\sum_i p_i = 1,\end{aligned} \hspace{\stretch{1}}(1.7)

(for p_i to be meaningful as probabilities) and with a given fixed average energy

\begin{aligned}\left\langle{{E}}\right\rangle = \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.8)

and a given fixed average particle number

\begin{aligned}\left\langle{{N}}\right\rangle = \sum_i N_i p_i.\end{aligned} \hspace{\stretch{1}}(1.9)

Here E_i, N_i represent the energy and number of particles in microstate i. Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i - \beta \sum_i E_i p_i - \gamma \sum_i N_i p_i,\end{aligned} \hspace{\stretch{1}}(1.10)

then fix \alpha, \beta, \gamma by demanding that the constrains be satisfied. What is the resulting p_i?

Answer

Writing

\begin{aligned}f = S - \alpha \sum_{j = 1}^\Omega p_j,= -\sum_{j = 1}^\Omega p_j \left( k_{\mathrm{B}} \ln p_j + \alpha  \right),\end{aligned} \hspace{\stretch{1}}(1.11)

our unconstrained minimization requires

\begin{aligned}0 = \frac{\partial {f}}{\partial {p_i}}= -\left( k_{\mathrm{B}} \left( \ln p_i + 1  \right) + \alpha  \right).\end{aligned} \hspace{\stretch{1}}(1.12)

Solving for p_i we have

\begin{aligned}p_i = e^{-\alpha/k_{\mathrm{B}} - 1}.\end{aligned} \hspace{\stretch{1}}(1.13)

The probabilities for each state are constant. To fix that constant we employ our constraint

\begin{aligned}1 = \sum_{j = 1}^\Omega p_j= \sum_{j = 1}^\Omega e^{-\alpha/k_{\mathrm{B}} - 1}= \Omega e^{-\alpha/k_{\mathrm{B}} - 1},\end{aligned} \hspace{\stretch{1}}(1.14)

or

\begin{aligned}\alpha/k_{\mathrm{B}} + 1 = \ln \Omega.\end{aligned} \hspace{\stretch{1}}(1.15)

Inserting eq. 1.15 fixes the probability, giving us the first of the expected results

\begin{aligned}\boxed{p_i = e^{-\ln \Omega} = \frac{1}{{\Omega}}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

Using this we our Gibbs entropy can be summed easily

\begin{aligned}S &= -k_{\mathrm{B}} \sum_{j = 1}^\Omega p_j \ln p_j \\ &= -k_{\mathrm{B}} \sum_{j = 1}^\Omega \frac{1}{{\Omega}} \ln \frac{1}{{\Omega}} \\ &= -k_{\mathrm{B}} \frac{\Omega}{\Omega} \left( -\ln \Omega  \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{S = k_{\mathrm{B}} \ln \Omega.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

For the “action” like quantity that we want to minimize, let’s write

\begin{aligned}f = S - \alpha \sum_j p_j - \beta \sum_j E_j p_j,\end{aligned} \hspace{\stretch{1}}(1.0.16)

for which we seek \alpha, \beta such that

\begin{aligned}0 &= \frac{\partial {f}}{\partial {p_i}} \\ &= -\frac{\partial {}}{\partial {p_i}}\sum_j p_j\left( k_{\mathrm{B}} \ln p_j + \alpha + \beta E_j  \right) \\ &= -k_{\mathrm{B}} (\ln p_i + 1) - \alpha - \beta E_i,\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}p_i = \exp\left( - \left( \alpha - \beta E_i \right) /k_{\mathrm{B}} - 1  \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Our probability constraint is

\begin{aligned}1 &= \sum_j \exp\left( - \left( \alpha - \beta E_j \right) /k_{\mathrm{B}} - 1  \right) \\ &= \exp\left( - \alpha/k_{\mathrm{B}} - 1  \right)\sum_j \exp\left( - \beta E_j/k_{\mathrm{B}}  \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\exp\left( \alpha/k_{\mathrm{B}} + 1  \right)=\sum_j \exp\left( - \beta E_j/k_{\mathrm{B}}  \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Taking logs we have

\begin{aligned}\alpha/k_{\mathrm{B}} + 1 = \ln \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}}  \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

We could continue to solve for \alpha explicitly but don’t care any more than this. Plugging back into the probability eq. 1.0.16 obtained from the unconstrained minimization we have

\begin{aligned}p_i = \exp\left( -\ln \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}}  \right)  \right)\exp\left( - \beta E_i/k_{\mathrm{B}}  \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{p_i = \frac{   \exp\left( - \beta E_i/k_{\mathrm{B}}  \right)}{   \sum_j    \exp\left( - \beta E_j/k_{\mathrm{B}}  \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

To determine \beta we must look implicitly to the energy constraint, which is

\begin{aligned}\left\langle{{E}}\right\rangle &= \sum_i E_i p_i \\ &= \sum_iE_i\left( \frac{ \exp\left( - \beta E_i/k_{\mathrm{B}}  \right) } { \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}}  \right) }  \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{\left\langle{{E}}\right\rangle = \frac{   \sum_i E_i \exp\left( - \beta E_i/k_{\mathrm{B}}  \right)}{   \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}}  \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

The constraint \beta (=1/T) is given implicitly by this energy constraint.

Again write

\begin{aligned}f = S - \alpha \sum_j p_j - \beta \sum_j E_j p_j - \gamma \sum_j N_j p_j.\end{aligned} \hspace{\stretch{1}}(1.0.16)

The unconstrained minimization requires

\begin{aligned}0 &= \frac{\partial {f}}{\partial {p_i}} \\ &= -\frac{\partial {}}{\partial {p_i}}\left( k_{\mathrm{B}} (\ln p_i + 1) + \alpha + \beta E_i + \gamma N_i  \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}p_i = \exp\left( -\alpha/k_{\mathrm{B}} - 1  \right) \exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}}  \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

The unit probability constraint requires

\begin{aligned}1 &= \sum_j p_j \\ &= \exp\left( -\alpha/k_{\mathrm{B}} - 1  \right) \sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}}  \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\exp\left( -\alpha/k_{\mathrm{B}} - 1  \right) =\frac{1}{{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}}  \right)}}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

Our probability is then

\begin{aligned}\boxed{p_i = \frac{\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}}  \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}}  \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

The average energy \left\langle{{E}}\right\rangle = \sum_j p_j E_j and average number of particles \left\langle{{N}}\right\rangle = \sum_j p_j N_j are given by

\begin{aligned}\left\langle{{E}}\right\rangle = \frac{E_i\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}}  \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}}  \right)}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

\begin{aligned}\left\langle{{N}}\right\rangle = \frac{N_i\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}}  \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}}  \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

The values \beta and \gamma are fixed implicitly by requiring simultaneous solutions of these equations.

Question: Fugacity expansion ([3] Pathria, Appendix D, E)

The theory of the ideal Fermi or Bose gases often involves integrals of the form

\begin{aligned}f_\nu^\pm(z) = \frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{x^{\nu - 1}}{z^{-1} e^x \pm 1}\end{aligned} \hspace{\stretch{1}}(1.36)

where

\begin{aligned}\Gamma(\nu) = \int_0^\infty dy y^{\nu-1} e^{-y}\end{aligned} \hspace{\stretch{1}}(1.37)

denotes the gamma function.

Obtain the behavior of f_\nu^\pm(z) for z \rightarrow 0 keeping the two leading terms in the expansion.

For Fermions, obtain the behavior of f_\nu^\pm(z) for z \rightarrow \infty again keeping the two leading terms.

For Bosons, we must have z \le 1 (why?), obtain the leading term of f_\nu^-(z) for z \rightarrow 1.

Answer

For z \rightarrow 0 we can rewrite the integrand in a form that allows for series expansion

\begin{aligned}\frac{x^{\nu - 1}}{z^{-1} e^x \pm 1} &= \frac{z e^{-x} x^{\nu - 1}}{1 \pm z e^{-x}} \\ &= z e^{-x} x^{\nu - 1}\left( 1 \mp z e^{-x} + (z e^{-x})^2 \mp (z e^{-x})^3 + \cdots  \right)\end{aligned} \hspace{\stretch{1}}(1.38)

For the kth power of z e^{-x} in this series our integral is

\begin{aligned}\int_0^\infty dx z e^{-x} x^{\nu - 1} (z e^{-x})^k &= z^{k+1}\int_0^\infty dx x^{\nu - 1} e^{-(k + 1) x} \\ &= \frac{z^{k+1}}{(k+1)^\nu}\int_0^\infty du u^{\nu - 1} e^{- u} \\ &= \frac{z^{k+1}}{(k+1)^\nu} \Gamma(\nu)\end{aligned} \hspace{\stretch{1}}(1.39)

Putting everything back together we have for small z

\begin{aligned}\boxed{f_\nu^\pm(z) =z\mp\frac{z^{2}}{2^\nu}+\frac{z^{3}}{3^\nu}\mp\frac{z^{4}}{4^\nu}+ \cdots}\end{aligned} \hspace{\stretch{1}}(1.40)

We’ll expand \Gamma(\nu) f_\nu^+(e^y) about z = e^y, writing

\begin{aligned}F_\nu(e^y) &= \Gamma(\nu) f_\nu^+(e^y) \\ &= \int_0^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1} \\ &= \int_0^y dx \frac{x^{\nu - 1}}{e^{x - y} + 1}+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1}.\end{aligned} \hspace{\stretch{1}}(1.41)

The integral has been split into two since the behavior of the exponential in the denominator is quite different in the x  y ranges. Observe that in the first integral we have

\begin{aligned}\frac{1}{{2}} \le \frac{1}{e^{x - y} + 1} \le \frac{1}{{1 + e^{-y}}}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Since this term is of order 1, let’s consider the difference of this from 1, writing

\begin{aligned}\frac{1}{e^{x - y} + 1} = 1 + u,\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}u = \frac{1}{e^{x - y} + 1} - 1 &= \frac{1 -(e^{x - y} + 1)}{e^{x - y} + 1} \\ &= \frac{-e^{x - y} }{e^{x - y} + 1} \\ &= -\frac{1}{1 + e^{y - x}}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

This gives us

\begin{aligned}F_\nu(e^y) &= \int_0^y dx x^{\nu - 1} \left( 1 - \frac{ 1 } { 1 + e^{y - x} }  \right)+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1} \\ &= \frac{y^\nu}{\nu}-\int_0^y dx \frac{ x^{\nu - 1}  } { 1 + e^{y - x} }+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Now let’s make a change of variables a = y - x in the first integral and b = x - y in the second. This gives

\begin{aligned}F_\nu(e^y) = \frac{y^\nu}{\nu}-\int_0^\infty da \frac{ (y - a)^{\nu - 1}  } { 1 + e^{a} }+\int_0^\infty db \frac{(y + b)^{\nu - 1}}{e^{b} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

As a gets large in the first integral the integrand is approximately e^{-a} (y-a)^{\nu -1}. The exponential dominates this integrand. Since we are considering large y, we can approximate the upper bound of the integral by extending it to \infty. Also expanding in series we have

\begin{aligned}F_\nu(e^y)  &\approx \frac{y^\nu}{\nu}+\int_0^\infty da \frac{ (y + a)^{\nu - 1} -(y - a)^{\nu - 1}  } { 1 + e^{a} } \\ &= \frac{y^\nu}{\nu}+\int_0^\infty da \frac{1}{{e^a + 1}}\left( \left( \frac{1}{{0!}} y^{\nu-1} a^0 + \frac{1}{{1!}} (\nu-1) y^{\nu-2} a^1 + \frac{1}{{2!}} (\nu-1) (\nu-2) y^{\nu-3} a^2 + \cdots  \right) - \left( \frac{1}{{0!}} y^{\nu-1} (-a)^0 + \frac{1}{{1!}} (\nu-1) y^{\nu-2} (-a)^1 + \frac{1}{{2!}} (\nu-1) (\nu-2) y^{\nu-3} (-a)^2 + \cdots  \right)  \right) \\ &= \frac{y^\nu}{\nu}+ 2\int_0^\infty da \frac{1}{{e^a + 1}}   \left( \frac{1}{{1!}} (\nu-1) y^{\nu-2} a^1 + \frac{1}{{3!}} (\nu-1) (\nu-2) (\nu - 3)y^{\nu-4} a^3 + \cdots  \right) \\ &= \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} \frac{y^{\nu - 1 - j}}{j!} \left( \prod_{k = 1}^j (\nu-k)  \right)\int_0^\infty da \frac{a^j}{e^a + 1} \\ &= \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} \frac{y^{\nu - 1 - j}}{j!} \frac{ \Gamma(\nu) } {\Gamma(\nu - j)}\int_0^\infty da \frac{a^j}{e^a + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

For the remaining integral, Mathematica gives

\begin{aligned}\int_0^\infty da \frac{a^j}{e^a + 1}=\left( 1 - 2^{-j} \right) j! \zeta (j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

where for s > 1

\begin{aligned}\zeta(s) = \sum_{k=1}^{\infty} k^{-s}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

This gives

\begin{aligned}F_\nu(e^y)  \approx \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} y^{\nu - 1 - j}\frac{ \Gamma(\nu) } {\Gamma(\nu - j)}\left( 1 - 2^{-j} \right) \zeta(j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}f_\nu^+(e^y)  &\approx y^\nu\left( \frac{1}{\nu \Gamma(\nu)} + 2 \sum_{j = 1, 3, 5, \cdots} \frac{ 1 } {\Gamma(\nu - j) y^{j + 1} } \left( 1 - 2^{-j} \right) \zeta(j+1)  \right) \\ &= \frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + 2 \sum_{j = 1, 3, 5, \cdots} \frac{ \Gamma(\nu + 1) } {\Gamma(\nu - j) } \left( 1 - 2^{-j} \right) \frac{\zeta(j+1)}{ y^{j + 1} }  \right),\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}\boxed{f_\nu^+(e^y)  \approx \frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + 2 \nu \sum_{j = 1, 3, 5, \cdots} (\nu-1) \cdots(\nu - j) \left( 1 - 2^{-j} \right) \frac{\zeta(j+1)}{ y^{j + 1} }  \right).}\end{aligned} \hspace{\stretch{1}}(1.0.42)

Evaluating the numerical portions explicitly, with

\begin{aligned}c(j) = 2 \left(1-2^{-j}\right) \zeta (j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

\begin{aligned}\begin{aligned}c(1) &= \frac{\pi^2}{6} \\ c(3) &= \frac{7 \pi^4}{360} \\ c(5) &= \frac{31 \pi^6}{15120} \\ c(7) &= \frac{127 \pi^8}{604800},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.42)

so to two terms (j = 1, 3), we have

\begin{aligned}\boxed{f_\nu^+(e^y)  \approx \frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + \nu(\nu-1) \frac{\pi^2}{6 y^{2}} + \nu(\nu-1)(\nu-2)(\nu -3) \frac{7 \pi^4}{360 y^4}  \right).}\end{aligned} \hspace{\stretch{1}}(1.0.42)

In order for the Boson occupation numbers to be non-singular we require \mu less than all \epsilon. If that lowest energy level is set to zero, this is equivalent to z < 1. Given this restriction, a z = e^{-\alpha} substitution is convenient for investigation of the z \rightarrow 1 case. Following the text, we'll write

\begin{aligned}G_\nu(e^{-\alpha})=\Gamma(\nu)f_\nu^-(e^{-\alpha}) = \int_0^\infty dx \frac{x^{\nu - 1}}{e^{x + \alpha} - 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

For \nu = 1, this is integrable

\begin{aligned}\frac{d}{dx} \ln\left( 1 - e^{-x - \alpha}  \right) &= \frac{e^{-x - \alpha}}{ 1 - e^{-x - \alpha} } \\ &= \frac{1}{{ e^{x + \alpha} - 1}},\end{aligned} \hspace{\stretch{1}}(1.0.42)

so that

\begin{aligned}G_1(e^{-\alpha}) &= \int_0^\infty dx \frac{1}{e^{x + \alpha} - 1} \\ &= {\ln \left( 1 - e^{-x - \alpha}  \right)}_{0}^{\infty} \\ &= \ln 1 - \ln \left( 1 - e^{- \alpha}  \right) \\ &= -\ln \left( 1 - e^{- \alpha}  \right).\end{aligned} \hspace{\stretch{1}}(1.0.42)

Taylor expanding 1 - e^{-\alpha} we have

\begin{aligned}1 - e^{-\alpha} = 1 - \left( 1 - \alpha + \alpha^2/2 - \cdots \right).\end{aligned} \hspace{\stretch{1}}(1.0.42)

Noting that \Gamma(1) = 1, we have for the limit

\begin{aligned}\lim_{\alpha \rightarrow 0} G_1(e^{-\alpha}) \rightarrow - \ln \alpha,\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}\lim_{z\rightarrow 1} f_\nu^-(z)= -\ln (-\ln z).\end{aligned} \hspace{\stretch{1}}(1.0.42)

For values of \nu \ne 1, the denominator is

\begin{aligned}e^{\alpha + x} - 1 = (\alpha + x) + (\alpha + x)^2/2 + \cdots\end{aligned} \hspace{\stretch{1}}(1.0.42)

To first order this gives us

\begin{aligned}f_\nu^-( e^{-\alpha} ) \approx \frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{1}{x + \alpha}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Of this integral Mathematica says it can be evaluated for 0 < \nu < 1, and has the value

\begin{aligned}\frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{1}{x + \alpha}=\frac{\pi}{\sin(\pi\nu)} \frac{1}{\alpha^{1 - \nu} \Gamma (\nu )}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

From [1] 6.1.17 we find

\begin{aligned}\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)},\end{aligned} \hspace{\stretch{1}}(1.0.42)

with which we can write

\begin{aligned}\boxed{f_\nu^-( e^{-\alpha} ) \approx \frac{ \Gamma(1 - \nu)}{ \alpha^{1 - \nu} }.}\end{aligned} \hspace{\stretch{1}}(1.0.42)

Question: Nuclear matter ([2], prob 9.2)

Consider a heavy nucleus of mass number A. i.e., having A total nucleons including neutrons and protons. Assume that the number of neutrons and protons is equal, and recall that each of them has spin-1/2 (so possessing two spin states). Treating these nucleons as a free ideal Fermi gas of uniform density contained in a radius R = r_0 A^{1/3}, where r_0 = 1.4 \times 10^{-13} \text{cm}, calculate the Fermi energy and the average energy per nucleon in MeV.

Answer

Our nucleon particle density is

\begin{aligned}\rho &= \frac{N}{V} \\ &= \frac{A}{\frac{4 \pi}{3} R^3} \\ &= \frac{3 A}{4 \pi r_0^3 A} \\ &= \frac{3}{4 \pi r_0^3} \\ &= \frac{3}{4 \pi (1.4 \times 10^{-13} \text{cm})^3} \\ &= 8.7 \times 10^{37} (\text{cm})^{-3} \\ &= 8.7 \times 10^{43} (\text{m})^{-3}\end{aligned} \hspace{\stretch{1}}(1.67)

With m for the mass of either the proton or the neutron, and \rho_m = \rho_p = \rho/2, the Fermi energy for these particles is

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2m} \left( \frac{6 \pi (\rho/2)}{2 S + 1} \right)^{2/3},\end{aligned} \hspace{\stretch{1}}(1.68)

With S = 1/2, and 2 S + 1 = 2(1/2) + 1 = 2 for either the proton or the neutron, this is

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2 m} \left( \frac{3 \pi^2 \rho}{2} \right)^{2/3}.\end{aligned} \hspace{\stretch{1}}(1.69)

\begin{aligned}\begin{aligned}\hbar &= 1.05 \times 10^{-34} \,\text{m^2 kg s^{-1}} \\ m &= 1.67 \times 10^{-27} \,\text{kg}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.70)

This gives us

\begin{aligned}\epsilon_{\mathrm{F}} &= \frac{(1.05 \times 10^{-34})^2}{2 \times 1.67 \times 10^{-27}} \left( \frac{3 \pi^2 }{2} \frac{8.7 \times 10^{43} }{2}  \right)^{2/3}\text{m}^4 \frac{\text{kg}^2}{s^2} \frac{1}{{\text{kg}}} \frac{1}{{\text{m}^2}} \\ &= 3.9 \times 10^{-12} \,\text{J} \times \left( 6.241509 \times 10^{12} \frac{\text{MeV}}{J} \right) \approx 24 \text{MeV}\end{aligned} \hspace{\stretch{1}}(1.71)

In lecture 16

we found that the total average energy for a Fermion gas of N particles was

\begin{aligned}E = \frac{3}{5} N \epsilon_{\mathrm{F}},\end{aligned} \hspace{\stretch{1}}(1.72)

so the average energy per nucleon is approximately

\begin{aligned}\frac{3}{5} \epsilon_{\mathrm{F}}  \approx  15 \,\text{MeV}.\end{aligned} \hspace{\stretch{1}}(1.73)

Question: Neutron star ([2], prob 9.5)

Model a neutron star as an ideal Fermi gas of neutrons at T = 0 moving in the gravitational field of a heavy point mass M at the center. Show that the pressure P obeys the equation

\begin{aligned}\frac{dP}{dr} = - \gamma M \frac{\rho(r)}{r^2},\end{aligned} \hspace{\stretch{1}}(1.74)

where \gamma is the gravitational constant, r is the distance from the center, and \rho(r) is the density which only depends on distance from the center.

Answer

In the grand canonical scheme the pressure for a Fermion system is given by

\begin{aligned}\beta P V &= \ln Z_{\mathrm{G}} \\ &= \ln \prod_\epsilon \sum_{n = 0}^1 \left( z e^{-\beta \epsilon}  \right)^n \\ &= \sum_\epsilon \ln \left( 1 + z e^{-\beta \epsilon}  \right).\end{aligned} \hspace{\stretch{1}}(1.75)

The kinetic energy of the particle is adjusted by the gravitational potential

\begin{aligned}\epsilon &= \epsilon_\mathbf{k}- \frac{\gamma m M}{r} \\ &= \frac{\hbar^2 \mathbf{k}^2}{2m}- \frac{\gamma m M}{r}.\end{aligned} \hspace{\stretch{1}}(1.76)

Differentiating eq. 1.75 with respect to the radius, we have

\begin{aligned}\beta V \frac{\partial {P}}{\partial {r}} &= -\beta \frac{\partial {\epsilon}}{\partial {r}}\sum_\epsilon \frac{z e^{-\beta \epsilon}}{ 1 + z e^{-\beta \epsilon} } \\ &= -\beta\left( \frac{\gamma m M}{r^2}  \right)\sum_\epsilon \frac{1}{ z^{-1} e^{\beta \epsilon} + 1} \\ &= -\beta\left( \frac{\gamma m M}{r^2}  \right)\left\langle{{N}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.77)

Noting that \left\langle{{N}}\right\rangle m/V is the average density of the particles, presumed radial, we have

\begin{aligned}\boxed{\frac{\partial P}{\partial r} &= -\frac{\gamma M}{r^2} \frac{m \left\langle N \right\rangle}{V} \\ &= -\frac{\gamma M}{r^2} \rho(r).}\end{aligned} \hspace{\stretch{1}}(1.0.78)

References

[1] M. Abramowitz and I.A. Stegun. \emph{Handbook of mathematical functions with formulas, graphs, and mathematical tables}, volume 55. Dover publications, 1964.

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

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

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