- Gas in a box
In

quantum mechanics , the results of the quantumparticle in a box can be used to look at the equilibrium situation for a quantum ideal**gas in a box**which is a box containing a large number of molecules which do not interact with each other except for instantaneous thermalizing collisions. This simple model can be used to describe the classicalideal gas as well as the various quantum ideal gases such as the ideal massiveFermi gas , the ideal massiveBose gas as well asblack body radiation which may be treated as a massless Bose gas.Using the results from either

Maxwell-Boltzmann statistics ,Bose-Einstein statistics orFermi-Dirac statistics , and considering the limit of a very large box, theThomas-Fermi approximation is used to express the degeneracy of the energy states as a differential, and summations over states as integrals. This enables thermodynamic properties of the gas to be calculated with the use of thepartition function or thegrand partition function . These results will be applied to both massive and massless particles. More complete calculations will be left to separate articles, but some simple examples will be given in this article.**Thomas-Fermi approximation for the degeneracy of states**For both massive and massless

particles in a box , the states of a particle areenumerated by a set of quantum numbers ["n"_{"x"}, "n"_{"y"}, "n"_{"z"}] . The absolute value of the momentum is given by::$p=frac\{h\}\{2L\}sqrt\{n\_x^2+n\_y^2+n\_z^2\}~~~~~~~~~n\_i=1,2,3,ldots\; ~~~~~~\; for\; ~\; i=x,\; y,\; ~or\; ~\; z$

where "h " is

Planck's constant and "L " is the length of a side of the box.Each possible state of a particle can be thought of as a point on a 3-dimensionalgrid of positive integers. The distance from the origin to any point will be:$n=sqrt\{n\_x^2+n\_y^2+n\_z^2\}=frac\{2Lp\}\{h\}$

Suppose each set of quantum numbers specify "f " states where "f " isthe number of internal degrees of freedom of the particle that can be altered bycollision. For example, a spin 1/2 particle would have "f=2", one for each spinstate. For large values of "n ", the number of states with absolute value of momentum less than or equal to "p " from the above equation is approximately

:$g=left(frac\{f\}\{8\}\; ight)\; frac\{4\}\{3\}pi\; n^3\; =\; frac\{4pi\; f\}\{3\}\; left(frac\{Lp\}\{h\}\; ight)^3$

which is just "f " times the volume of a sphere of radius "n " divided by eightsince only the octant with positive "n

_{i }" is considered. Using a continuum approximation, the number ofstates with absolute value of momentum between "p " and "p+dp " istherefore:$dg=frac\{pi\}\{2\}~f\; n^2,dn\; =\; frac\{4pi\; fV\}\{h^3\}~\; p^2,dp$

where "V=L

^{3}" is the volume of the box. Notice that in using thiscontinuum approximation, the ability to characterize the low-energystates is lost, including the ground state where "n_{i }=1". For most cases thiswill not be a problem, but when considering Bose-Einstein condensation, in which alarge portion of the gas is in or near the ground state, theability to deal with low energy states becomes important.Without using the continuum approximation, the number of particles withenergy ε

_{i }is given by:$N\_i\; =\; frac\{g\_i\}\{Phi\}$

where:

::with β = "1/kT ",

Boltzmann's constant "k",temperature "T", andchemical potential "μ". Using the continuum approximation, the number of particles "dN " with energy between"E " and "E+dE " is now written as::$dN=\; frac\{dg\}\{Phi\}$

**The energy distribution function**Using the results derived from the previous sections of this article, some distribution functions for the "gas ina box" can now be determined. The distribution function for any variable A is P

_{A}dA andis equal to the fraction of particles which have values for "A " between"A " and "A+dA":$P\_A~dA\; =\; frac\{dN\}\{N\_T\}\; =\; frac\{dg\}\{N\_TPhi\}$

where $N\_T$ is total number of particles.

It follows that:

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

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

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

and the distribution function for the energy "E " is:

:$P\_E~dE\; =\; P\_pfrac\{dp\}\{dE\}~dE$

For a particle in a box (and for a free particle as well), the relationship between energy "E " and momentum "p " is different for massive and massless particles. For massive particles,

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

while for massless particles,

:$E=pc,$

where "m " is the mass of the particle and "c " is the speed of light.Using these relationships,

- For massive particles
:$dg\; =\; left(frac\{Vf\}\{Lambda^3\}\; ight)frac\{2\}\{sqrt\{pi~eta^\{3/2\}E^\{1/2\}~dE$

:$P\_E~dE\; =\; frac\{1\}\{N\_T\}left(frac\{Vf\}\{Lambda^3\}\; ight)frac\{2\}\{sqrt\{pi~frac\{eta^\{3/2\}E^\{1/2\{Phi\}~dE$

where Λ is the

thermal wavelength orthermal de Broglie wavelength of the gas.:$Lambda\; =sqrt\{frac\{h^2\; eta\; \}\{2pi\; m$

This is an important quantity, since when Λ is on the order of theinter-particle distance $(V/N\_T)^\{1/3\}$, quantum effects begin todominate and the gas can no longer be considered to be a Maxwell-Boltzmann gas.See

configuration integralfor a detailed derivation of the expression for$displaystyle\; Lambda$. - For massless particles
:$dg\; =\; left(frac\{Vf\}\{Lambda^3\}\; ight)frac\{1\}\{2\}~eta^3E^2~dE$

:$P\_E~dE\; =\; frac\{1\}\{N\_T\}left(frac\{Vf\}\{Lambda^3\}\; ight)frac\{1\}\{2\}~frac\{eta^3E^2\}\{Phi\}~dE$

where Λ is now the

thermal wavelength for massless particles.:$Lambda\; =\; frac\{cheta\}\{2,pi^\{1/3$

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

**Massive Maxwell-Boltzmann particles**For this case:

:$Phi=e^\{eta(E-mu)\},$

Integrating the energy distribution function and solving for $N\_T$ gives

:$N\_T=\; left(frac\{Vf\}\{Lambda^3\}\; ight),,e^\{etamu\}$

Substituting into the original energy distribution function gives

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

which are the same results obtained classically for the

Maxwell-Boltzmann distribution . Further results can be found in the article on theclassical ideal gas .**Massive Bose-Einstein particles**For this case:

:$Phi=e^\{eta\; epsilon\}/z-1,$

where "z" is defined as

:$z=e^\{etamu\},$

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

:$N\_T=\; left(frac\{Vf\}\{Lambda^3\}\; ight)\; extrm\{Li\}\_\{3/2\}(z)$

where Li

_{s}(z) is thepolylogarithm function and Λ is thethermal wavelength . The polylogarithm term must always be positiveand real, which means its value will go from 0 to ζ(3/2) as "z " goes from0 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:$N\_T\; =\; left(frac\{Vf\}\{Lambda\_c^3\}\; ight)zeta(3/2)$

The temperature at which Λ=Λ

_{c}is the critical temperature. Fortemperatures below this critical temperature, the above equation for the particle numberhas no solution. The critical temperature is the temperature at which a Bose-Einsteincondensate begins to form. The problem is, as mentionedabove, that the ground state has been ignored in the continuum approximation. It turnsout, however, that the above equation for particle number expresses the number of bosons in excited statesrather well, and thus::$N\_T=frac\{g\_0z\}\{1-z\}+left(frac\{Vf\}\{Lambda^3\}\; ight)\; extrm\{Li\}\_\{3/2\}(z)$

where the added term is the number of particles in the ground state. (The groundstate 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)**For the case of massless particles, the massless energy distribution function must be used. It is convenient to convert this function to a frequency distribution function:

:$P\_\; u~d\; u\; =\; frac\{h^2\}\{N\_T\}left(frac\{Vf\}\{Lambda^3\}\; ight)frac\{1\}\{2\}~frac\{eta^3\; u^2\}\{e^\{(h\; u-mu)/kT\}-1\}~d\; u$

where Λ is the thermal wavelength for massless particles. The spectral energy density (energy per unit volume per unit frequency) is then

:$U\_\; u~d\; u\; =\; left(frac\{N\_T,h\; u\}\{V\}\; ight)\; P\_\; u~d\; u\; =\; frac\{4pi\; f\; h\; u^3\; \}\{c^3\}~frac\{1\}\{e^\{(h\; u-mu)/kT\}-1\}~d\; u$

Other thermodynamic parameters may be derived analogously to the case for massive particles. For example, integrating the frequency distribution function and solving for $N\_T$ gives the number of particles:

:$N\_T=frac\{16,pi\; V\}\{c^3h^3eta^3\},mathrm\{Li\}\_3left(e^\{mu/kT\}\; ight)$

The most common massless Bose gas is a

photon gas in ablack body . Taking the "box" to be a black body cavity, the photons are continually being absorbed and re-emitted by the walls. When this is the case, the number of photons is not conserved. In the derivation ofBose-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. Furthermore, since photons have two spin states, the value of "f" is 2. The spectral energy density is then:$U\_\; u~d\; u\; =\; frac\{8pi\; h\; u^3\; \}\{c^3\}~frac\{1\}\{e^\{h\; u/kT\}-1\}~d\; u$

which is just the spectral energy density for

Planck's law of black body radiation . Note that the Wien distribution is recovered if this procedure is carried out for massless Maxwell-Boltzmann particles, which approximates a Planck's distribution for high temperatures or low densities.In certain situations, the reactions involving photons will result in the conservation of the number of photons (e.g.

light-emitting diode s, "white" cavities). In these cases, the photon distribution function will involve a non-zero chemical potential. (Hermann 2005)Another massless Bose gas is given by the

Debye model for heat capacity. This considers a gas ofphonons in a box and differs from the development for photons in that the speed of the phonons is less than light speed, and there is a maximum allowed wavelength for each axis of the box. This means that the integration over phase space cannot be carried out to infinity, and instead of results being expressed inpolylogarithm s, they are expressed in the relatedDebye function s.**Massive Fermi-Dirac particles (e.g. electrons in a metal)**For this case:

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

Integrating the energy distribution function gives

:$N\_T=left(frac\{Vf\}\{Lambda^3\}\; ight)left\; [-\; extrm\{Li\}\_\{3/2\}(-z)\; ight]$

Where again, Li

_{s}(z) is thepolylogarithm function and Λ is thethermal de Broglie wavelength . Further results can be found in the article on theidealFermi gas .**References*** cite journal| last = Herrmann| first = F.| coauthors = Würfel, P.

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