updated Ch 8

Chapter_8/statistical-physics.ipynb

@@ -463,11 +463,12 @@
     "**Mean speed $\\bar{v}$**\n",
     "```\n",
     "\n",
-    "\\begin{align*}\n",
+    "```{math}\n",
+    ":label: mean_speed_stat\n",
     "\\bar{v} &= 4\\pi C \\left(\\frac{1}{2(\\beta m/2)^2}\\right) = 8\\pi \\left(\\frac{\\beta m}{2\\pi}\\right)^{3/2} \\left(\\frac{1}{(\\beta m)^2}\\right), \\\\\n",
     " &= \\frac{4}{\\sqrt{2\\pi}} \\left(\\frac{1}{\\sqrt{\\beta m}}\\right), \\\\ &= \\frac{2}{\\sqrt{\\pi}} \\sqrt{\\frac{2kT}{m}}, \\\\\n",
     " &= \\frac{2}{\\sqrt{\\pi}} v_{\\rm mp}.\n",
-    "\\end{align*}\n",
+    "```\n",
     "\n",
     "We define the **root-mean-square** $v_{\\rm rms}$ to be $ v_{\\rm rms} \\equiv \\left(\\overline{v^2}\\right)^{1/2}$.  We cannot simply use the result for the mean speed $\\bar{v}$ directly.  Instead, we need to find the mean using the probability function $F(v)$, or\n",
     "\n",
@@ -920,10 +921,283 @@
     "## Fermi-Dirac Statistics\n",
     "\n",
     "### Introduction to Fermi-Dirac Theory\n",
+    "The Fermi-Dirac distribution helps us understand how collections of fermions behave, one of its most useful applications is for the problem of electrical conducion.\n",
+    "\n",
+    "First, we consider the role of the factor $B_{\\rm FD}$.  Like any other normalization factor, one could simply integrate over all allowed values of $F_{\\rm FD}$.  The factor $B_{\\rm FD}$ is temperature dependent because of the parameter $\\beta\\ (=1/kT)$.  We express this temperature dependence as\n",
+    "\n",
+    "\\begin{align}\n",
+    "B_{\\rm FD} = e^{-\\beta E_{\\rm F}},\n",
+    "\\end{align}\n",
+    "\n",
+    "where $E_{\\rm F}$ is the **Fermi energy**.  Then we can rewrite the Fermi-Dirac distribution as\n",
+    "\n",
+    "\\begin{align}\n",
+    "F_{\\rm FD} = \\frac{1}{e^{\\beta(E-E_{\\rm F})}+1}.\n",
+    "\\end{align}\n",
+    "\n",
+    "Notice what happens when $E=E_{\\rm F}$, $F_{\\rm FD} = 1/2$ (exactly).  As a result, it is common to define the Fermi energy as the energy at which $F_{\\rm FD} = 1/2$.\n",
+    "\n",
+    "Consider the temperature dependence of $F_{\\rm FD}$, where in the limit as $T \\rightarrow 0$:\n",
+    "\n",
+    "\\begin{align}\n",
+    "F_{\\rm FD} &= \\begin{cases} 1, & \\text{for } E < E_{\\rm F}, \\\\ 0,  & \\text{for } E > E_{\\rm F}. \\end{cases}\n",
+    "\\end{align}\n",
+    "\n",
+    "- At $T=0$, fermions occupy the lowest energy levels available to them.  They cannot all be in the lowest level because that would violate the Pauli exclusion principle.  Rather, fermions will fall all available energy levels up to a particular energy (i.e., the Fermi energy).\n",
+    "- Near $T=0$, there is little chance thermal agitation will kick a fermion to an energy above $E_{\\rm F}$.\n",
+    "- As $T$ increases from zero, more fermions jump to higher energy levels, resultying in a smooth transition rather than a sharp step.\n",
+    "\n",
+    "It is sometimes useful to consider a **Fermi temperature** ($T_{\\rm F}\\equiv E_{\\rm F}/k)$.\n",
+    "\n",
+    "- When $T\\ll T_{\\rm F}$, the step function approximation is reasonably accurate. \n",
+    "- When $T \\gg T_{\\rm F}$, $F_{\\rm FD}$ approaches a simple decaying exponential.  Recall that at sufficiently high temperatures, we expect Maxwell-Boltzmann statistics to be reasonably accurate.\n",
     "\n",
     "### Classical Theory of Electrical Conduction\n",
+    "In 1900 [Paul Drude](https://en.wikipedia.org/wiki/Paul_Drude) developed a theory of electrical conduction to explain the observed conductivity of metals.  His model assumed that the electrons existed as a gas of free particles.  This is a fair assumption because (in a conductor) the outermost electron is weakly bound to the atom and easily stripped away, even by a very weak electric field.\n",
+    "\n",
+    "In the Drude model, the mteal is treated as a lattice of positive ions with a electron gas that is free to flow through it.  Just as in the case of ideal gases, electrons have a thermal kinetic energy proportional to temperature.  The mean speed of an electron at room temperature is ${\\sim}10^5\\ {\\rm m/s}$.  But, the particle velocities in a gas are directed randomly.  Therefore, ther will be no net flow of electrons unless an electric field is applied to the conductor.\n",
+    "\n",
+    "When an electric field is applied, the negatively charged electrons flow in the direction *opposite* that of the field.  According to Drude, their flow is severely restricted by collisions with the lattice ions.  Using some basic assumptions from classical mechanics, Drude showed that the current in a conductor should be linearly proportional to the applied electric field.  This is consistent with Ohm's law.\n",
+    "\n",
+    "Unfortunately, the numerical predictions of the theory were not as successful.  One important prediction was that the electrical conductivity $\\sigma$ could be expressed as\n",
+    "\n",
+    "\\begin{align}\n",
+    "\\sigma &= \\frac{n}{m}e^2\\tau,\n",
+    "\\end{align}\n",
+    "\n",
+    "where $n$ is the number density of conduction electrons, $e$ is the electron charge, $\\tau$ is the average time between electron-ion collisions, and $m$ is the electron mass.  It is possible to measure $n$ through the [Hall effect](https://en.wikipedia.org/wiki/Hall_effect).\n",
+    "\n",
+    "The parameter $\\tau$ is not as easy to measure, but it can be estimated using [transport theory](https://en.wikipedia.org/wiki/Charge_transport_mechanisms).  The best estimates of $\\tau$ produced a value of $\\sigma$ that is about one order of magnitude too small for most conductors.  The Drude theory is incorrect in this prediction.\n",
+    "\n",
+    "```{margin}\n",
+    "**Mean free path**\n",
+    "```\n",
+    "\n",
+    "The mean time between collisions $\\tau$ should be simply the mean distance $\\ell$ traveled by an electron between collisions (i.e., the **mean free path**) divided by the mean speed $\\bar{v}$ of the electrons, $\\tau = \\ell/\\bar{v}$.\n",
+    "\n",
+    "Then the electrical conductivity is expressed as\n",
+    "\n",
+    "\\begin{align}\n",
+    "\\sigma &= \\frac{n\\ell}{m\\bar{v}}e^2.\n",
+    "\\end{align}\n",
+    "\n",
+    "Equation {eq}`mean_speed_stat` shows that the mean speed is proportional to the square root of the absolute temperature.  According to the Drude model, the conductivity is proportional to $T^{-1/2}$.  For most conductors the conductivity is nearly proportional to $T^{-1}$, except at low temperatures.  Drude's classical model fails again.\n",
+    "\n",
+    "There is the problem of calculating the electronic contribution to the heat capacity of a solid conductor.  The heat capacity of a solid can be almost completely accounted for by considering the six degrees of freedom from lattice vibrations.  This gives a molar heat capacity of $c_{\\rm V} = 3R$.  According to the equipartion theorem, we need to add another $3(\\frac{1}{2}R)$ for the heat capacity of the electron gas, giving a total of $\\frac{9}{2}R$.\n",
+    "\n",
+    "This is not consistent with experimental results.  The electronic contribution to the heat capacity depends on temperature and is only ${\\sim}0.02R$ per mole at room temperature.  \n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "### Quantum Theory of Electrical Conduction\n",
+    "To obtain a better solution, we turn to quantum theory where it is necesary to understand how the electron energies are distributed in a conductor.  We can use the Fermi-Dirac distribution, but we have a problem to find $g(E)$, or the number of allowed states per unit energy.  \n",
+    "\n",
+    ">What energy values should we use?\n",
+    "\n",
+    "Let's retain the \"free electron\" assumption of the Drude model and use the results of the 3-D infinite square-well potential because it corresponds physically to a cubic lattice of ions.  The allowed energies are\n",
+    "\n",
+    "\\begin{align*}\n",
+    "E = \\frac{h^2}{8mL^2}\\left(n_1^2 + n_2^2 + n_3^2 \\right),\n",
+    "\\end{align*}\n",
+    "\n",
+    "where $L$ is the side-length of a cube and $n_i$ are the integer quantum numbers.  \n",
+    "\n",
+    "Let's first solve the problem at $T=0$.  The distribution at room temperature is very much like $T=0$ because $T_{\\rm F}$ is high for conductive metals (e.g, $T_{\\rm F} = 80,000\\ {\\rm K}$ for $\\rm Cu$).  \n",
+    "\n",
+    "The above allowed energies can be rewritten as\n",
+    "\n",
+    "\\begin{align*}\n",
+    "E = r^2 E_c,\n",
+    "\\end{align*}\n",
+    "\n",
+    "where $r^2\\ (= n_1^2 + n_2^2 + n_3^2)$ represents the \"radius\" of a sphere in phase space (also dimensionless) and $E_c$ is just a constant, not the ground state energy.\n",
+    "\n",
+    "We have defined $r$ in this way to construct a geometric solution to the problem of deteriming $g(E)$.  Think of the $n_i$ as the \"coordianates\" of a 3-D number space.  The number of allowed states up to \"radius\" $r$ (or up to energy $E$) is directly related to the spherical \"volume\" $\\frac{4}{3}\\pi r^3$.  The exact number of states up to radius $r$ is\n",
+    "\n",
+    "\\begin{align}\n",
+    "N_r &= (2)\\left(\\frac{1}{8}\\right) \\left(\\frac{4}{3}\\pi r^3 \\right).\n",
+    "\\end{align}\n",
+    "\n",
+    "The extra factor of $2$ comes from spin degeneracy (one spin up and spin down electron).  The factor of $1/8$ is necessary because we are restricted to positive quantum numbers (i.e., one octant of a 3-D number space).\n",
+    "\n",
+    "The number of states $N_r$ can be expressed as a function of $E$:\n",
+    "\n",
+    "\\begin{align}\n",
+    "N_r &= \\frac{1}{3} \\pi r^3 = \\frac{1}{3} \\pi \\left(\\sqrt{\\frac{E}{E_c}}\\right)^3, \\\\\n",
+    "&= \\frac{1}{3} \\pi \\left(\\frac{E}{E_c}\\right)^{3/2}. \n",
+    "\\end{align}\n",
+    "\n",
+    "At $T=0$, the Fermi energy is the energy of the highest occupied energy level.  If there are a total of $N$ electrons, then\n",
+    "\n",
+    "$$ N = \\frac{1}{3} \\pi \\left(\\sqrt{\\frac{E_{\\rm F}}{E_c}}\\right)^3. $$\n",
+    "\n",
+    "We solve for $E_{\\rm F}$ to obtain\n",
+    "\n",
+    "```{math}\n",
+    ":label: Fermi_energy_cube\n",
+    "E_{\\rm F} &= \\left(\\frac{3N}{\\pi}\\right)^{2/3}E_c, \\\\\n",
+    "&= \\frac{h^2}{8m}\\left(\\frac{3N}{\\pi L^3}\\right)^{2/3}.\n",
+    "```\n",
+    "\n",
+    "Equation {eq}`Fermi_energy_cube` is useful because the ratio $N/L^3$ is well knonw for most condutors, where it is the number density of conduction electrons at $T=0$."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "```{exercise}\n",
+    ":class: orange \n",
+    "\n",
+    "**Calculate the Fermi energy and temperature for copper.**\n",
+    "\n",
+    "Equation {eq}`Fermi_energy_cube` can be used to compute the Fermi energy, provided the number density of conduction electrons is known.  For copper, $N/V = 8.47 \\times 10^{28}\\ {\\rm m^{-3}}.$  We use $V = L^3$ for a cube.  This allows us to compute the Fermi energy as\n",
+    "\n",
+    "\\begin{align*}\n",
+    "E_{\\rm F} &= \\frac{\\left(6.626 \\times 10^{-34}\\ {\\rm J\\cdot s}\\right)^2}{8\\left(9.11 \\times 10^{-31}\\ {\\rm kg}\\right)} \\left[ \\frac{3\\left( 8.47 \\times 10^{28}\\ {\\rm m^{-3}} \\right)}{\\pi}\\right]^{2/3}, \\\\\n",
+    "&= 1.13 \\times 10^{-18}\\ {\\rm J} = 7.03\\ {\\rm eV}.\n",
+    "\\end{align*}\n",
+    "\n",
+    "The Fermi temperature is\n",
+    "\n",
+    "\\begin{align*}\n",
+    "T_{\\rm F} = \\frac{E_{\\rm F}}{k} = \\frac{1.13 \\times 10^{-18}\\ {\\rm J}}{1.381 \\times 10^{-23}\\ {\\rm J/K}} = 8.16 \\times 10^4\\ {\\rm K}.\n",
+    "\\end{align*}\n",
+    "```"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The Fermi energy for copper is 1.1e-18 J or 7.03 eV.\n",
+      "The Fermi temperature for copper is 8.16e+04 K.\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy as np \n",
+    "from scipy.constants import h, k, m_e, eV\n",
+    "\n",
+    "def Fermi_energy(NV):\n",
+    "    #NV = number density of conduction electrons N/V\n",
+    "    return h**2/(8*m_e)*(3*NV/np.pi)**(2./3.)\n",
+    "\n",
+    "def Fermi_temp(E_F):\n",
+    "    #E_F = Fermi energy\n",
+    "    return E_F/k\n",
+    "\n",
+    "NV_Cu = 8.47e28 #number density of conduction electrons for Copper \n",
+    "EF_Cu = Fermi_energy(NV_Cu)\n",
+    "TF_Cu = Fermi_temp(EF_Cu)\n",
+    "\n",
+    "print(\"The Fermi energy for copper is %1.1e J or %1.2f eV.\" % (EF_Cu,EF_Cu/eV))\n",
+    "print(\"The Fermi temperature for copper is %1.2e K.\" % TF_Cu)\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "The density of states can be calculated by differentiating $N_r$ with respect to energy:\n",
+    "\n",
+    "$$ g(E) = \\frac{N_r}{dE} = \\frac{\\pi}{2}E_c^{-3/2}E^{1/2}. $$\n",
+    "\n",
+    "This result can be expressed in terms of $E_{\\rm F}$ as\n",
+    "\n",
+    "\\begin{align}\n",
+    "g(E) &= \\frac{\\pi}{2} \\left( E_{\\rm F}^{-3/2} \\frac{3N}{\\pi}\\right) E^{1/2}, \\\\\n",
+    "&= \\frac{3}{2}N E_{\\rm F}^{-3/2} E^{1/2}.\n",
+    "\\end{align}\n",
+    "\n",
+    "Because we are considering $T=0$, it is possible to use the step function form of the Fermi-Dirac factor.  We have\n",
+    "\n",
+    "\\begin{align}\n",
+    "n(E) &= \\begin{cases} g(E), &\\text{for }E< E_{\\rm F}, \\\\\n",
+    "0, & \\text{for } E>E_{\\rm F}.\n",
+    "\\end{cases} \\end{align}\n",
+    "\n",
+    "With the distribution function $n(E)$, the mean electron energy can be calculated as\n",
+    "\n",
+    "\\begin{align}\n",
+    "\\overline{E} &= \\frac{1}{N} \\int_0^\\infty n(E) E\\ dE = \\frac{1}{N} \\int_0^{E_{\\rm F}} g(E)E\\ dE, \\\\\n",
+    "&= \\frac{3}{2} \\int_0^{E_{\\rm F}} \\left(\\frac{E}{E_{\\rm F}}\\right)^{3/2} \\ dE, \\\\\n",
+    "&= \\frac{3}{2 E_{\\rm F}^{3/2}} \\int_0^{E_{\\rm F}} E^{3/2}\\ dE = \\frac{3}{5} E_{\\rm F}.\n",
+    "\\end{align}"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "We can now find the electronic contribution to the heat capacity of a conductor.  Recall that the heat capacity at constant volue is $C_{\\rm V} = \\partial U/\\partial T$, where $U$ is the internal energy.  Using our result for $\\overline{E}$, we have\n",
+    "\n",
+    "$$ U = N\\overline{E} = \\frac{3}{5}N E_{\\rm F}. $$\n",
+    "\n",
+    "The important question for determining the heat capacity is how $U$ increases with temperature because the energy levels are filled up to $E_{\\rm F}$.  We expect that only those electrons within about $kT$ of $E_{\\rm F}$ will absorb thermal energy and jumpt to a higher state.  The fraction of electrons capable of engaging in this thermal process is on the order of $kT/E_{\\rm F}$.  \n",
+    "\n",
+    "The exact number of electrons depends on the temperature because the shape of the curve for $n(E)$ changes with temperature.  In general, we can say \n",
+    "\n",
+    "\\begin{align}\n",
+    "U = \\frac{3}{5}N E_{\\rm F} + \\alpha N \\left(\\frac{kT}{E_{\\rm F}}\\right) kT,\n",
+    "\\end{align}\n",
+    "\n",
+    "where $\\alpha$ is a constant $>1$ due to the shape of the distribution curve.  Therefore the electronic contribution to the heat capacity is\n",
+    "\n",
+    "\\begin{align*}\n",
+    "C_{\\rm V} = \\frac{\\partial U}{\\partial T} = 2\\alpha N k^2 \\frac{T}{E_{\\rm F}}.\n",
+    "\\end{align*}\n",
+    "\n",
+    "The molar heat capacity is\n",
+    "\n",
+    "\\begin{align}\n",
+    "c_{\\rm V} = 2\\alpha R \\frac{T}{T_{\\rm F}},\n",
+    "\\end{align}\n",
+    "\n",
+    "where $R=N_{\\rm A}k$ and $E_{\\rm F} = kT_{\\rm F}$.\n",
+    "\n",
+    "[Arnold Sommerfeld](https://en.wikipedia.org/wiki/Arnold_Sommerfeld) used the correct distribution $n(E)$ at room temperature and found $\\alpha = \\pi^2/4$.  With $T_{\\rm F} = 80,000\\ {\\rm K}$ for copper, we obtain $c_{\\rm V} = 0.02R$, which is also seen experimentally.\n",
+    "\n",
+    "```{margin}\n",
+    "**Fermi speed**\n",
+    "```\n",
+    "\n",
+    "To address the problem of electrical conductivity, we can replace the mean speed $\\bar{v}$ with what is called a **Fermi speed** $u_{\\rm F}$, defined from $E_{\\rm f} = \\frac{1}{2}mu_{\\rm F}^2$.  We can justify this change because conduction electrons are the most loosely bound to their atoms.  Therefore, these electrons must be at the highest energy level.\n",
+    "\n",
+    "At room temperature, the highest energy level is close to the Fermi energy.  This means that we should use\n",
+    "\n",
+    "\\begin{align*}\n",
+    "u_{\\rm F} &= \\sqrt{\\frac{2E_{\\rm F}}{m}} \\approx 1.6 \\times 10^6\\ {\\rm m/s},\n",
+    "\\end{align*}\n",
+    "\n",
+    "for copper.\n",
+    "\n",
+    "Unfortunately, this is an even higher speed than $\\bar{v}$.  It turns out that there is a problem with the classical value Drude used for the mean free path $\\ell$.  He assumed that the ions were so large and occupied so much space in a solid, the mean free path could be no more than ${\\sim}0.1\\ {\\rm nm}$.\n",
+    "\n",
+    "In quantum theory the ions are not hard spheres, and the electrons can be thought of as waves.  Accounting for the wavelike property of electrons, the mean free path can be longer than Drude estimated.  Einstein calculated the value of $\\ell$ to be ${\\sim}40\\ {\\rm nm}$ for copper at room temperature.  This gives a conductivity of\n",
+    "\n",
+    "$$ \\sigma = \\frac{n\\ell}{mu_{\\rm F}} e^2 \\approx 6 \\times 10^7\\ {\\rm \\Omega^{-1}\\cdot m^{-1}}, $$\n",
+    "\n",
+    "which is just right.\n",
+    "\n",
+    "Finally, let's consider the temperature dependence of the electrical conductivity.  As the temperature of a conductor is increased, ionic vibrations become more severe.  Thus electron-ion collsions will become more frequent, the mean free path will become smaller, and the conductivity will be reduced.\n",
+    "\n",
+    "According to elementary transport theory, the mean free path is inversely proportional to the cross-sectional area of these ionic scatterers.  Let's assume that the ions are harmonic oscillators.  The energy of a harmonic oscillator is proportional to the square of its vibration amplitude.\n",
+    "\n",
+    "Therefore, the mean free path is inversely proportional to vibration energy (and temperature). We summarize this by\n",
+    "\n",
+    "$$ \\sigma \\propto \\ell \\propto r^{-2} \\propto U^{-1} \\propto T^{-1}. $$\n",
     "\n",
-    "### Quantum Theory of Electrical Conduction"
+    "The electrical conductivity varies inversely with temperature.  This is another success for quantum theory, where electrical conductivity is observed to vary inversely with temperature for most pure metals."
    ]
   },
   {