Please wait for the animation to completely load.
In this animation, you can choose a specific idealized system by pushing the radio buttons. Restart. You can change the value of kBT with the slider and the value of the constant in the equation described below for the specific system (labeled as C). You can also plot the distribution function n(ε) and also n(ε)/g(ε), which is the occupancy, f(ε), and should match the shape of the distributions found in Section 15.4.
Different assemblies of particles are described by the relevant distribution function (number of particles with a given energy):
|Bose-Einstein||ni = gi/(exp(α + βεi) - 1)||n(ε)dε = g(ε)dε/(exp(α + βε) - 1)|
|Fermi-Dirac||ni=gi/(exp(α + βεi) + 1)||n(ε)dε = g(ε)dε/(exp(α + βε) + 1)|
|Maxwell-Boltzmann (Classical)||ni=gi/exp(α + βεi)||n(ε)dε = g(ε)dε/exp(α + βε)|
where ni is the number of particles in region in phase space (a given box, see Section 15.3), εi is the energy of that region (or box) and gi is the number of states associated with a given region in phase space (a given box), also called a density of states. The constants α and β must be set so that Σni = N (or ∫ dn = N) and Σniεi = E (or ∫ εn(ε)dε = E). From the definition of temperature, β = 1/kBT for all distributions (see Section 15.4). However, the value of α varies by distribution, so we will consider several specific applications.4
In order to determine which distribution to use, we must remember that the fundamental difference between quantum-mechanical distributions and the classical distribution is whether or not the particles are distinguishable. In both cases, the particles are identical, but, for the quantum distributions, they are countable, but indistinguishable. One way to determine whether we can consider particles distinguishable or not is whether or not the wave function for each particle can be considered separately from the other particle or not. For example, the wave function for the two electrons in a helium atom cannot be written as two individual wave functions that do not overlap significantly, but the wave function for the electrons in two neighboring hydrogen atoms (in a diffuse gas) can be considered independently of each other (because the wave function between the two essentially goes to zero and they no longer need to be considered entangled states).
Ideal gas is a model of hard spheres that collide into each other and do not otherwise interact with each other. For a diffuse gas, the molecules are considered distinguishable so we can use the Maxwell-Boltzmann (classical) distribution. Furthermore, the number of particles is generally quite large so that energy quantization is not noticeable (at normal temperatures). Therefore, we can use a continuous distribution.
It is easier to determine g(p)dp, the number of states with momentum between p and dp, and then using ε = p2/2m rewrite g(p)dp as g(ε)dε:
g(p) dp = ∫∫∫∫∫ dx dy dz dpx dpy dpz /h3 = V 4πp2dp/h3 ,
where V is the volume integral of dx dy dz and 4πp2dp is due to changing dpx dpy dpz from Cartesian to spherical coordinates and integrating. We find α by requiring that ∫ n(ε)dε = N (with limits of integration from 0 to infinity):
n(ε)dε = (4πV/h3)m3/2 ε1/2 exp(−α) exp(−ε/kBT) dε ,
which, upon integration, gives exp(−α) = (Nh3/V) (2πmkBT)−3/2 so that finally
n(ε)dε/N = 2π/(πkBT)3/2 ε1/2 exp(−ε/kBT) dε .
Click on the Ideal Gas button. You can change the value of kBT with the slider and the value of (Nh3/V) (2πm)−3/2 (labeled as C). How does the distribution change with C? Explain.
Blackbody Radiation (see also Section 4.1) is the radiation from a perfect emitter (absorber) and is due to photons, which are indistinguishable particles of spin 1 (and therefore are bosons). This means that we will use Bose-Einstein statistics. The energy levels of the photons are quantized, but there are so many different energy levels, we can use the continuous distribution function. The number of states, g(p)dp in a given region of phase space (with an energy between p and p + dp) is double that of the ideal gas because there are two polarization states. Using momentum as p = ε/c we can find g(ε)dε as below:
g(ε)dε = 8Vπε2dε/(hc)3 .
Furthermore, the total number of photons, N, does not have to remain constant. This means that we can set α = 0. This gives
n(ε)dε = (8Vπ/h3c3) ε2dε /(exp(ε/kBT) − 1) .
Click on the distribution for blackbody radiation (since α = 0, changing C in the animation does not change anything). Notice that the number of particles (area under the n(ε) curve) is not constant. Look at f(ε). What happens when the temperature goes to zero?
Free Electron Gas is used to describe the energy distribution of electrons in metals. Electrons in the conduction bands of metals that are essentially free to move throughout the material, so we describe this as a free electron gas. Since this is a gas of electrons, however, we must use Fermi-Dirac statistics. Again, although the energies are quantized, there are many energy levels in the energy band and so we can use a continuous distribution. As before, we need to find g(ε)dε and α. Again, it is easier to use g(p)dp and it is twice the value of the ideal gas expression because there are two spin states. Using ε = p2/2m (and thus, dp = (m/2ε)1/2dε), we have
g(ε)d(ε) = 8Vπ(2m3)1/2ε1/2dε/h3 .
Now, we need to find α. It turns out that to be useful to describe the temperature dependence of α as α = −εF/kBT, where εF is the Fermi energy of a metal. So,
n(ε)dε = (8Vπ/h3)(2m3)1/2ε1/2dε/(exp((−εF + ε)/kBT) + 1) .
Notice what happens when T = 0. When ε > εF , n(ε)dε = 0. When εF > ε, n(ε)dε = g(ε)dε. In words, then, the states below the Fermi energy have particles in them (at T = 0), but the states above the Fermi energy do not. Notice that for blackbody radiation above, when T = 0, n(ε)dε = 0 except for the state ε = 0. For a boson, at T = 0 all the particles are in the lowest energy region in phase space, but for a fermion many regions in phase space are occupied because the Pauli exclusion principle does not allow particles to share states. This limit allows for the calculation of the Fermi energy at T = 0: ∫0εF n(ε)dε = ∫0εF g(ε)dε = N and the limits of integration are from 0 to εF because n(ε)dε = 0 for ε > εF. Solving, then, we find
εF = h2/2m (3N/8πV)2/3 . (15.8)
Click on the Free Electron Gas distribution. For a metal with εF = 5 eV, find what happens to the distribution as T = 0? What about the occupancy? At room temperature, how does β compare with the Fermi level? This particular analysis of metals as a free electron gas is only valid at low temperatures (so electrons are not excited to higher energy bands) so we keep the value of kBT < 0.2 eV in the animation.
4Development follows A. Beiser, Modern Physics, 6th ed, McGraw-Hill, 2003, pp. 296-331 and J. R. Taylor, C. D. Zapfiratos and M. A. Dubson, Modern Physics, 2nd ed, Prentice Hall, 2004, pp. 502-526.
Section by Anne J. Cox and William F. Junkin III
© 2006 by Prentice-Hall, Inc. A Pearson Company