where \(d^3\vec{v} = dv_x\ dv_y\ dv_z\), or a volume element in the phase space. The product of a velocity distribution funciton with a phase space volume plays a role analogous to the probability density \(\Psi^*\Psi\) in quantum theory.
Maxwell proved that the probability distribution funciton is proportional to \(e^{-mv^2/(2kT)}\), where \(m\) is the molecular mass, \(v\) is the molecular speed, \(k\) is Boltzmann’s constant, and \(T\) is the absolute temperature. Then, we may write
-where \(C\) is aproportionality factor and \(\beta \equiv (kT)^{-1}\); not to be confused with \(\beta = v/c\) from relativity. We can expand the above expression as
-and use the properties of exponents to rewrite as three factors that each contain one of the three velocity components. They are defined as
-Recalling that \(a=\beta m/2\), we have
-and
-With this distribution we can calculate \(\bar{v}_x\) (mean value of \(v_x\)) as
-because \(v_x\) is an odd function. The result makes sense because in a random distribution of velocities one woul expect the velocity components to be evenly distributed about \(v_x = 0\). The mean value of \(v_x^2\) is
-Alternatively, one could use an integral table with exponential functions. The final result via substitution is
-There’s nothing special about the \(x\)-direction, so the results for the \(x\)-, \(y\)-, and \(z\)-velocity components are identical. The three components can be summed to find the mean translational kinetic energy \(\overline{K}\) of a molecule:
-This is one of the principal results of kinetic theory.
@@ -853,8 +853,8 @@The space between \(r\) and \(r+dr\) is a spherical shell, and thus, we must integrate over that volume to transform to the radial distribution. These two distributions are related by the volume of spherical shell, or \(4\pi r^2\). We may write
-Returning to the problem of the speed distribution \(F(v)\) from the velocity distribution \(f(\vec{v})\). We need to integrate over a spherical shell in the velocity phase space, where we replace \(r\rightarrow v\) from the previous example. The desired speed distribution is
@@ -862,18 +862,16 @@Maxwell speed distribution
-The Maxwell speed distribution as derived from purely classical considerations, where it gives a nonzero probability of finding a particle with a speed greater than \(c\). Therefore, it is only valid in the classical limit.
The assymetry of the distribution curve leads to an interesting result:
the most probable speed \(v_{\rm mp}\), the mean speed \(\bar{v}\), and the root-mean-square speed \(v_{\rm rms}\) are all slightly different from each other.
To find the most probable speed, we simply find maximum speed in the probabilty curve, or
-Most probable speed \(v_{\rm mp}\)
-See a list of integrals containing exponential functions and using \((k=2,\ n=4,\ a=\beta m/2)\) to find
-Root-mean-square speed \(v_{\rm rms}\)
-A particle moving with the mean squared speed has a kinetic energy
@@ -931,8 +929,8 @@in keeping with our basic law of kinetic theory.
The standard deviation of the molecular speeds \(\sigma_v\) is
-which can be applied to each type of gas. The only difference between results will be due to the molar mass \(m\). The molar mass of \(\rm H_2\) is \(2.01568\ {\rm u}\), which is found by multiplying the mass of hydrogen by two. The conversion from atomic mass units \({\rm u}\) to \(\rm kg\) is \(1.660539 \times 10^{-27}\ {\rm kg/u}\). Let’s re-write the mean molecular speed so that all the physical constants are combined and so that we can use atomic mass units directly. This gives
-Then, we compute the mean molecular speed for molecular hydrogen \(\rm H_2\) as
@@ -1010,8 +1008,8 @@What fraction of the molecules in an ideal gas in equilibrium has speeds with \(\pm1\%\) of the most probable speed \(v_{\rm mp}\)?
The Maxwell speed distribution function provides the probability of finding a particle within an interval of speeds. The fraction of molecules within the a given speed interval is equal to the integrated probability over the interval. Mathematically, this is expressed by the number of molecules at a particular speed \(N(v)\):
-The indefinite integral introduces the error function, which is beyond the scope of this course. However, we can obtain an approximate solution by calculating \(F(v_{\rm mp})\) and multiplying by \(dv \approx \Delta v = 0.02v_{\rm mp}\). This solution works for a small window, where wider intervals are easily evaluated using numerical methods (e.g., Simpson’s rule).
@@ -1085,11 +1083,221 @@The Maxwell speed distribution is only valid in the classical limit, which prohibits its use at exceptionally high temperatures (i.e., gas particle velocities near \(c\)). An ideal gas is dilute, which means that the gas molecules rarely interact or collide with one another due to their large (relative) separations. When collisions occur, they can be treated as totally elastic, and have no effect on the distributions and mean values.
+When the density of matter is higher (i.e., solid or liquids), the assumption of noninteracting particles may no longer be valid. If molecules, atoms, or subatomic particles are closely packed together, the Pauli exclusion principle prevents two particles in identical quantum states from sharing the same s=pace. This limits the allowed energy states of any particle (fermion), which affects the distribution of energies for a system of particles.
Because energy levels are fundamental in quantum theory, we rewrite the Maxwell speed distribution in terms of energy rather than velcocity. For a monatomic gas, the energy is all translational kinetic energy. We can use the following relations
+Similar to our derivation of the Maxwell speed distribution (see Eq.(8.13)), we can transform \(F(v) \rightarrow F(E)\),
+ +to get the Maxwell-Boltzmann energy distribution. The factor \(e^{-\beta E}\) is important because Boltzmann showed that this factor is a characteristic of any classical system, regardless of how quantities other than molecular speeds may affect the energy of a given state. We define the Maxwell-Boltzmann factor for classical systems as
+ +where \(A\) is a normalization constant. The energy distribution for a classical system will then have the form
+where \(n(E)\) is a distribution that represents the number of particles within a bin of energy from \(E\) to \(E+dE\).
+The function \(g(E)\) is known as the density of states, or the number of states available per unit energy range. The density of states is an essential element in al distributions. The factor \(F_{\rm MB}\) is the relative probability that an energy state is occupied at a given temperature.
In quantum theory, particles are described by wave functions. Indentical particles cannot be distinguished from one another if there is a significant overlap of their wave functions. It is this characteristic that makes quantum statistics different from classical stastics.
+Suppose that we have a system of just two particles. Each particle has an equal probability (0.5) of existing in either of two energy states. If the particles are distinguishable (e.g., labeled A and B), then the possible configurations are
+Both in state 1,
Either A or B in each state, but not together, or
Both in state 2.
These configurations can be illustrated using a probability table:
+State 1 |
+State 2 |
+
|---|---|
AB |
++ |
A |
+B |
+
B |
+A |
+
| + | AB |
+
Each of the four configurations are equally likely, where the probability of each is one-fourth (0.25). If the two particles are indistinguishable, then our probability table changes:
+State 1 |
+State 2 |
+
|---|---|
XX |
++ |
X |
+X |
+
| + | XX |
+
Now there are only three equally likely configurations, where each have a probability of one-third (\({\sim}0.33\)).
+Two kinds of quantum distributions are needed because some particles obey the Pauli exclusion principle and others do not.
+Particles that obey the Pauli exclusion princple have half-integers spins and are called fermions. Protons, neutrons, and electrons are examples of fermions.
Particles with zero or integer spins do not obey the Pauli exclusion principle and are known as bosons. Photons and pions are examples of bosons.
Note
+Atoms and molecules consisting of an even number of fermions must be bosons when considered as a whole, because their total spin will be zero or an integer. Conversely, atoms and molecules with an odd number of fermions are fermions.
+The probability distribution of fermions are given by the Fermi-Dirac distribution:
+ +Similarly the Bose-Einstein distribution is valid for bosons and is
+ +In each case \(B_i\) (\(B_{\rm FD}\) or \(B_{\rm BE}\)) represents a normalization factor, and \(g(E)\) is the density of states appropriate for a particular situation.
+Note
+The Fermi-Dirac and Bose-Einstein distributions look very similar, where they differ only by the normalization constant and by the sign attached to the \(1\) in the denominator.
+Both the Fermi-Dirac and Bose-Einstein distributions reduce to the classical Maxwell-Boltzmann distribution when \(B_ie^{\beta E} \gg 1\) \((\text{recall }x\pm 1 \approx x\) for \(x\gg 1)\) and \(A = 1/B_i\). This means that the Maxwell-Boltzmann factor is much less than 1 (i.e., the probability that a particular energy state will be occupied is much less than 1).
+Distributions |
+Properties |
+Examples |
+Function |
+
|---|---|---|---|
Maxwell-Boltzmann |
+Identical, |
+Ideal gases |
+\(F_{\rm MB} = Ae^{-\beta E}\) |
+
Bose-Einstein |
+Identical, indistinguishable |
+Liquid \(\rm ^4He\), |
+\(F_{\rm BE} = \frac{1}{B_{\rm BE}\ e^{\beta E}-1}\) |
+
Fermi-Dirac |
+Identical, indistinguishable |
+Electron gas |
+\(F_{\rm FD} = \frac{1}{B_{\rm FD}\ e^{\beta E}+1}\) |
+
Exercise 8.5
+Assume that the Maxwell-Boltmann distribution is valid in a gas of atomic hydrogen. What is the relative number of atoms in the ground state and first excited state at: \(293\ {\rm K}\), \(5000\ {\rm K}\), and \(10^6\ {\rm K}\)?
+The relative number of atoms between the ground and first excited states can be determined by using the ratio of \(n(E_2)\) to \(n(E_1)\). Mathematically this is given as
+where the density of states \(g(E_1) = 2\) for the ground state (spin up or down) and \(g(E_2) = 8\) for the first excited state (see Electron shells). For atomic hydrogen, \(E_2-E_1 = 10.2\ {\rm eV}\). Putting this together we can write
+for a given temperature \(T\). Using the Boltzmann constant in \(\rm eV/K\), we can easily calculate the relative numbers as
+At very high temperatures, the exponential factor approaches 1, so the result simply approaches the ratio of the density of states.
+import numpy as np
+from scipy.constants import physical_constants
+
+##For atomic hydrogen only
+
+def energy(n):
+ #n = energy level
+ return -E_o/n**2 #energy in eV
+
+def relative_states(n1,n2,T):
+ #n2 = energy level 2; n2 > n1
+ #n1 = energy level 1
+ #T = temperature in K
+ g2, g1 = 2*n2**2, 2*n1**2
+ E2,E1 = energy(n2),energy(n1)
+ return (g2/g1)*np.exp(-(E2-E1)/(k_eV*T))
+
+k_eV = physical_constants['Boltzmann constant in eV/K'][0]
+E_o = physical_constants['Rydberg constant times hc in eV'][0]
+
+print("The relative number of atoms at 293 K is %1.2e." % relative_states(1,2,293))
+print("The relative number of atoms at 293 K is %1.2e." % relative_states(1,2,5000))
+print("The relative number of atoms at 293 K is %1.2f." % relative_states(1,2,1e6))
+The relative number of atoms at 293 K is 1.21e-175.
+The relative number of atoms at 293 K is 2.07e-10.
+The relative number of atoms at 293 K is 3.55.
+