Gas In A Box

Thomas-Fermi approximation for the degeneracy of states

$p=\backslash frac\{h\}\{2L\}\backslash sqrt\{n\_x^2+n\_y^2+n\_z^2\}~~~~~~~~~n\_i=1,2,3,\backslash ldots$

$n=\backslash sqrt\{n\_x^2+n\_y^2+n\_z^2\}=\backslash frac\{2Lp\}\{h\}$

  = \frac{4\pi f}{3} \left(\frac{Lp}{h}\right)^3 

   N_i = \frac{g_i}{\Phi} 

/dd>
math>\Phi=e^{\beta(\epsilon_i-\mu)}	for particles obeying Maxwell-Boltzmann statistics
math>\Phi=e^{\beta(\epsilon_i-\mu)}-1	for particles obeying Bose-Einstein statistics
math>\Phi=e^{\beta(\epsilon_i-\mu)}+1	for particles obeying Fermi-Dirac statistics

with β = 1/kT  with k  being Boltzmann's constant, T  being temperature, and μ being the chemical potential. Using the continuum approximation, the number of particles dN  with energy between E  and E+dE  is now written:

$dN=\; \backslash frac\{dg\}\{\backslash Phi\}$

The energy distribution function

We are now in a position to determine some distribution functions for the "gas in a box". The distribution function for any variable A is P_AdA and is equal to the fraction of particles which have values for A  between A  and A+dA

$P\_A~dA\; =\; \backslash frac\{dN\}\{N\}\; =\; \backslash frac\{dg\}\{N\backslash Phi\}$

It follows that:

$\backslash int\_A\; P\_A~dA\; =\; 1$

The distribution function for the absolute value of the momentum is:

$P\_p~dp\; =\; \backslash frac\{Vf\}\{N\}~\backslash frac\{4\backslash pi\}\{h^3\backslash Phi\}~p^2dp$

and the distribution function for the energy E  is:

$$

P_E~dE = P_p\frac{dp}{dE}~dE For a particle in a box, the relationship between energy E  and momentum p  is different for massive and massless particles. For massive particles, we have

$E=\backslash frac\{p^2\}\{2m\}$

while for massless particles:

$E=pc\backslash ,$

where m  is the mass of the particle and c  is the speed of light. Using these relationships we have:

• For massive particles

  $$

  dg = \left(\frac{Vf}{\Lambda^3}\right) \frac{2}{\sqrt{\pi}}~\beta^{3/2}E^{1/2}~dE

  $$

  P_E~dE = \frac{1}{N}\left(\frac{Vf}{\Lambda^3}\right) \frac{2}{\sqrt{\pi}}~\frac{\beta^{3/2}E^{1/2}}{\Phi}~dE where Λ is the thermal wavelength of the gas.

  $$

  \Lambda =\sqrt{\frac{h^2 \beta }{2\pi m}} This is an important quantity, since when Λ is on the order of the interparticle distance (V/N)^1/3 , quantum effects begin to dominate and the gas can no longer be considered to be a Maxwell-Boltzmann gas.
• For massless particles

  $dg\; =\; \backslash left(\backslash frac\{Vf\}\{\backslash Lambda^3\}\backslash right)\backslash frac\{1\}\{2\}~\backslash beta^3E^2~dE$

  $$

  P_E~dE = \frac{1}{N}\left(\frac{Vf}{\Lambda^3}\right) \frac{1}{2}~\frac{\beta^3E^2}{\Phi}~dE where Λ is now the thermal wavelength for massless particles.

  $\backslash Lambda\; =\; \backslash frac\{ch\backslash beta\}\{2\backslash ,\backslash pi^\{1/3\}\}$

  Specific Examples

  The following sections give an example of results for some specific cases.

  Massive Maxwell-Boltzmann particles

  For this case:

  $\backslash Phi=e^\{\backslash beta(E-\backslash mu)\}\backslash ,$

  Integrating the energy distribution function and solving for N gives

  $N\; =\; \backslash left(\backslash frac\{Vf\}\{\backslash Lambda^3\}\backslash right)\backslash ,\backslash ,e^\{\backslash beta\backslash mu\}$

  Substituting into the original energy distribution function gives

  $P\_E~dE\; =\; 2\; \backslash sqrt\{\backslash frac\{\backslash beta^3\; E\}\{\backslash pi\}\}~e^\{-\backslash beta\; E\}~dE$

  which are the same results obtained classically for the Maxwell-Boltzmann distribution. Further results can be found in the article on the classical ideal gas.

  Massive Bose-Einstein particles

  For this case:

  $\backslash Phi=e^\{\backslash beta\; \backslash epsilon\}/z-1\backslash ,$

  where z is defined as

  $z=e^\{\backslash beta\backslash mu\}\backslash ,$

  Integrating the energy distribution function and solving for N gives the particle number

  $N\; =\; \backslash left(\backslash frac\{Vf\}\{\backslash Lambda^3\}\backslash right)\backslash textrm\{Li\}\_\{3/2\}(z)$

  Where Li_s(z) is the polylogarithm function and Λ is the thermal wavelength. The polylogarithm term must always be positive and real, which means its value will go from 0 to ζ(3/2) as z  goes from 0 to 1. As the temperature drops towards zero, Λ will become larger and larger, until finally Λ will reach a critical value Λ_c where z=1 and

  $N\; =\; \backslash left(\backslash frac\{Vf\}\{\backslash Lambda\_c^3\}\backslash right)\backslash zeta(3/2)$

  The temperature at which Λ=Λ_c is the critical temperature. For temperatures below this critical temperature, the above equation for the particle number has no solution. The critical temperature is the temperature at which a Bose-Einstein condensate begins to form. The problem is, as mentioned above, that the ground state has been ignored in the continuum approximation. It turns out, however, that the above equation for particle number expresses the number of bosons in excited states rather well, and so we may write:

  $$

  N=\frac{g_0 z}{1-z}+\left(\frac{Vf}{\Lambda^3}\right)\textrm{Li}_{3/2}(z) where the added term is the number of particles in the ground state. (The ground state energy has been ignored.) This equation will hold down to zero temperature. Further results can be found in the article on the ideal Bose gas.

  Massless Bose-Einstein particles (e.g. black body radiation)

  The most common massless bose gas is a gas of photons in a black body. Taking the "box" to be a black body cavity, the photons are continually being absorbed and re-emitted by the walls. The number of photons is not a conserved quantity, since we are actually in a relativistic regime. In the derivation of Bose-Einstein statistics, when the restraint on the number of particles is removed, this is effectively the same as setting the chemical potential (μ) to zero. Therefore:

  $$

     \Phi=e^{\beta E}-1\, 

  The spectral energy density (energy per unit volume per unit frequency) is

  $U\_\backslash nu~d\backslash nu\; =\; \backslash frac\{NE\}\{V\}\; P\_E\; \backslash frac\{dE\}\{d\backslash nu\}~d\backslash nu$

  = \frac{4\pi f h\nu^3 }{c^3}~\frac{1}{e^{h\nu/kT}-1}~d\nu where E=hν has been used, and since the radiation is the same in all directions, and propagates at the speed of light (c), the spectral intensity (energy/time/area/solid angle/frequency) is

  $I\_\backslash nu\; =\; \backslash frac\{U\_\backslash nu\backslash ,c\}\{4\backslash pi\}$

  which yields

  $$

     I_\nu~d\nu = \frac{f h\nu^3 }{c^2}~\frac{1}{e^{h\nu/kT}-1}~d\nu  

  Since photons have two spin states, we have f=2 which yields Planck's law of black body radiation. Note that if we had carried out this procedure for massless Maxwell-Boltzmann particles, we would recover Wien's distribution which approximates a Planck's distribution for high temperature or low density.

  Massive Fermi-Dirac particles (e.g. electrons in a metal)

  For this case:

  $\backslash Phi=e^\{\backslash beta(E-\backslash mu)\}+1\backslash ,$

  Integrating the energy distribution function gives

  $N=\backslash left(\backslash frac\{Vf\}\{\backslash Lambda^3\}\backslash right)\backslash left-\backslash textrm\{Li\}\_\{3/2\}(-z)\backslash right$

  Where again, Lis(z) is the polylogarithm function and Λ is the thermal de Broglie wavelength. Further results can be found in the article on the ideal Fermi gas.

  References

  • Huang, Kerson, "Statistical Mechanics", John Wiley & Sons, New York, 1967.
  • A. Isihara, "Statistical Physics", Academic Press, New York, 1971.
  • L. D. Landau and E. M. Lifshitz, "Statistical Physics, 3rd Edition Part 1", Butterworth-Heinemann, Oxford, 1996.
  • Zijun Yan, "General thermal wavelength and its applications", Eur. J. Phys. 21 (2000) 625-631. http://www.iop.org/EJ/article/0143-0807/21/6/314/ej0614.pdf

   

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