{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%pylab inline"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Populating the interactive namespace from numpy and matplotlib\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from sympy import *\n",
      "from pylab import *"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 16
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import matplotlib.pyplot as plt"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 3
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from sympy.interactive import printing\n",
      "printing.init_printing(use_latex=True)\n",
      "from IPython.display import display\n",
      "a,x,Z = symbols('a,x,Z')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "## Problem 1"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "### 1D"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The partition function for 1D system is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$Z = \\sum_i e^{-\\beta E_i}$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The probability on a state $E_i$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$P(E_i) = \\frac{e^{-\\beta E_i}}{Z}$$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "In our case, there are only two possible energy states, $E_1 = -\\mu B$, $E_2 = \\mu B$. So the magnetization is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ M =  \\mu N \\left( e^{\\beta \\mu B} - e^{-\\beta \\mu B} \\right)/Z$$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Write magnetization as a function of $T$ and $T$."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$M(T,B) = \\mu N \\frac{ \\left( e^{\\mu B /(k_B T)} - e^{- \\mu B/(k_B T)} \\right)  }  {  \\left(  e^{\\mu B /(k_B T)} + e^{- \\mu B/(k_B T)}  \\right)  } = \\mu N \\tanh(\\mu B /(k_B T))$$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "###### $M$ vs $T$ Plots"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "To plot $M~T$, we need to make a quantity with dimension of $T$ . $\\bar T = \\mu B/k_B$ will do the work."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$\\frac{M(T)}{\\bar M} = e^{\\bar T/T} - e^{-\\bar T/T} $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "in which, $\\bar M = \\mu N$ ."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "At $T \\rightarrow \\infty $ limit, we have the magnetization $M$ is 0. This is because high temperature distroys the alignment of dipoles."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "tanh(0)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$0.0$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "png": "iVBORw0KGgoAAAANSUhEUgAAABoAAAAOBAMAAADDIxFwAAAAMFBMVEX///8AAAAAAAAAAAAAAAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAEJmJdjLNVN0iZu+7\nq0QgoRR7AAAAdklEQVQIHWNgEDJRZQABCJ3OIDEBxAPTXBsYmBOAHAjNc4GB/SOQB6H5LzDwfQfy\nILS8AgPfFyAPQs83YGD9DORBaFSevAFUJZjmV2BgB5sCpjkOMHCBbIDQLBsYmBKAPCjdziAdwPiN\ngQFEMzAIu1xlYOiF0AD+DiarJQxxrwAAAABJRU5ErkJggg==\n",
       "prompt_number": 5,
       "text": [
        "0.0"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "At $T\\rightarrow \\infty$, we have $\\frac{M}{\\bar M}$ 1."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "tanh(inf)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "latex": [
        "$$1.0$$"
       ],
       "metadata": {},
       "output_type": "pyout",
       "png": "iVBORw0KGgoAAAANSUhEUgAAABkAAAAPBAMAAADjSHnWAAAAMFBMVEX///8AAAAAAAAAAAAAAAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAzRAiu5mrdu/dZoky\nVEQKohj3AAAAeklEQVQIHWNgYBBiAAGm6NIDQMrkE4jDwHiBaTIDg0oYhLeCgWEjUJAdwnNjYLiv\nAOd9Y2B4fwHGY/oK5BXAeMy/GRjOb8DOYwLKIVQyAPXdh5vCMIWBYT3ChgoGhkCo7foCDJwXmHoY\nGFgzfmYxcCYwMG0vOwCUQwYADrco9JpgipkAAAAASUVORK5CYII=\n",
       "prompt_number": 6,
       "text": [
        "1.0"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "At very low temperature, the dipoles have no random motion so all dipoles aligned together thus $M$ is 1."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Now plot out this result."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "t = linspace(0, 100, 1000)\n",
      "mt = tanh(1/t)\n",
      "\n",
      "fig11 = plt.figure()\n",
      "\n",
      "axes = fig11.add_axes([0.1, 0.1, 0.8, 0.8]) # left, bottom, width, height (range 0 to 1)\n",
      "\n",
      "axes.plot(t, mt, 'r')\n",
      "\n",
      "axes.set_xlabel('T bar')\n",
      "axes.set_ylabel('M bar')\n",
      "axes.set_title('M ~ T');"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stderr",
       "text": [
        "-c:2: RuntimeWarning: divide by zero encountered in divide\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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       "text": [
        "<matplotlib.figure.Figure at 0xa6fcbac>"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "This plot agrees with our results of the limits."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "###### $M$ vs $B$ Plots"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Define a unit magnetic induction $\\bar B = k_B T/\\mu$. Write the magnetization as a function of $B$."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ \\frac{M(B)}{\\bar M} = \\tanh(B/\\bar B) $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "At the limit of $B\\rightarrow 0$, we have $M\\rightarrow 0$, which is because no magnetic field is there to align the dipoles. (No interactions between dipoles.)"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "At the limit of $B\\rightarrow \\infty$, we have $M\\rightarrow 1$, which is because infinite magnetic field can align all the dipoles to one direction no matter how strong the thermal random motion is."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Plot the result."
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "b = linspace(0, 5, 100)\n",
      "mb = tanh(b)\n",
      "\n",
      "fig12 = plt.figure()\n",
      "\n",
      "axes = fig12.add_axes([0.1, 0.1, 0.8, 0.8]) # left, bottom, width, height (range 0 to 1)\n",
      "\n",
      "axes.plot(b, mb, 'r')\n",
      "\n",
      "axes.set_xlabel('B/Bbar')\n",
      "axes.set_ylabel('M/Mbar')\n",
      "axes.set_title('M ~ B');"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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evRAYCLGxZlciIuJzPCcsqhfiadW2iEir87ywEBGRVucZU2ePHTO6n0pKoE0b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\n",
       "text": [
        "<matplotlib.figure.Figure at 0xa75c5ac>"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "###### Heat Capacity"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Energy of the system is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ E = - \\mu B N \\tanh(\\mu B /(k_B T)) $$"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "t13e = linspace(0, 5, 100)\n",
      "e1 = -tanh(t13e)\n",
      "\n",
      "fig13e = plt.figure()\n",
      "\n",
      "axes = fig13e.add_axes([0.1, 0.1, 0.8, 0.8]) # left, bottom, width, height (range 0 to 1)\n",
      "\n",
      "axes.plot(t13e, e1, 'r')\n",
      "\n",
      "axes.set_xlabel('T/Tbar')\n",
      "axes.set_ylabel('E/Ebar')\n",
      "axes.set_title('E ~ T');"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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Tf3v16kV2dnbt/ezsbHpp2KZxffoYw1obN5qdRETEIVo19kBMTAwAZ86cISwsjDvvvBOL\nxcLevXsZPHiwwwK6pJkz4fe/h7g4s5OIiNhdo0NbqampRoMGujYWi4X77rvP7uGuxSmHtgAqKqBH\nD9i7F3r2NDuNiEizNPUzttEeSUBAAGFhYQCcO3eOtm3b1j72t7/9zQYR3Vi7dsbVEz/4ABISzE4j\nImJXjfZIIiIiSE9PByAyMpIDBw40+JiZnLZHAnDiBNxzjzHpfkURFhFxFTabbAec98PamfXuDRER\n4ASbW4qI2FOTCok006xZ4ATrbURE7KnROZLc3FzmzJlDdXU1eXl5tbcB8vLyHBbQpY0bB88+C19+\nCQMGmJ1GRMQuGp0jWb58eb3xsZr7FouF+Ph4h4VsjFPPkdR4/XXIyTEm3kVEXEiLV7YvWLCAsWPH\nEhERYfNwtuIShaSgAMLCICsLfH3NTiMi0mQtnmy/5ZZbWLx4MQMGDCA+Pp5Vq1ZRXFxs05AeITAQ\nxoyBZcvMTiIiYhdN2v03PT2d5ORkUlJSuHjxIqNGjWLs2LHceeedjsrZIJfokQCkpcFjjxmnBHt7\nm51GRKRJbLZp4w+VlpaSkpLCZ599xpIlS5od0BZcppAA3HUXvPSStk0REZfR4qGtt956q/b2mjVr\nam/7+vqyf/9+04uIy5k7F37zG7NTiIjYXKOFZMWKFbW3FyxYUOex5ORk+yVyVxMmwMmTxqnAIiJu\nRAsSHcXHB/7jP2DxYrOTiIjYlAqJIz31FKxdC4WFZicREbGZRifbvb29ad++PQAVFRW0a9eu9rGK\nigouXrzomIRX4VKT7TWeegq6d4dXXjE7iYjIVdntrC1n4pKF5OhRGDkSMjON7eZFRJyUTXf/FRsK\nD4c774TERLOTiIjYhMMLSVFREVarldDQUEaPHk1JSUmD7UpKSpg4cSJhYWGEh4eTlpbm4KR29OKL\nsGgRVFWZnUREpMUcXkgSEhKwWq1kZGQQHR1NQiNXEHzuued44IEHOHbsGF999VXt1Rrdwj33gL8/\nfPKJ2UlERFrM4XMkffr0Yfv27QQEBFBQUEBUVBTHjx+v06a0tJSIiAhOnTp11edyyTmSGuvWwX/9\nF+zbBxaL2WlEROpp8TXb7aWwsJCAgADAuC58YQOnwmZmZtKtWzeeeOIJDh48yMCBA1m8eHHtWWRX\nmjdvXu3tqKgooqKi7BXdtmJijC1Ttm0zJt9FREyWmppKamrqdf+cXXokVquVgoKCet+fP38+8fHx\ndXYR9vPzo6ioqE67L774grvvvpvdu3czePBg5s6dS6dOnXjttdfqhnflHgnA//4vrF4N2ilARJyQ\nqT2SlJSURh+rGdIKDAzk9OnT+Pv712sTFBREUFAQgwcPBmDixImNzqW4tKlT4Ze/hPR04/ruIiIu\nyOGT7bGxsSR+f+prYmIicQ3shhsYGEhwcDAZGRkAbNmyhb59+zo0p0O0aQM//SnMn292EhGRZnP4\nZHtRURGTJk0iOzubkJAQVq9eTefOncnPz2fmzJls3LgRgIMHDzJjxgwqKyu59dZbWbZsGb4/uMKg\nyw9tAZSXwy23wJYtcPvtZqcREamlle2uZOFC2L8fVq40O4mISC0VEldy9izceiukphrXdxcRcQLa\nIsWVdOhgXPhKcyUi4oLUI3EWZWVGr2TXLggNNTuNiIh6JC6nUyeYM0e9EhFxOeqROJPSUujd25gr\nCQ83O42IeDj1SFyRry/87Gfwq1+ZnUREpMnUI3E25eVGr2TdOhg0yOw0IuLB1CNxVe3bwy9+YRwi\nIi5AhcQZTZ8OGRmwfbvZSURErkmFxBm1bg2vvQY//zm429CdiLgdFRJnNWWKsbYkKcnsJCIiV6XJ\ndmeWnAzPPQeHD4OPj9lpRMTDaLLdHYwZAz17wvvvm51ERKRR6pE4u8OHjUvx/v3v0KWL2WlExINo\n91938vTTxsaOb79tdhIR8SAqJO6ksBD69oW0NOjVy+w0IuIhNEfiTgIC4IUXjO1TREScjAqJq3j+\neTh0yDiTS0TEiaiQuIq2beG//xtmz4Zz58xOIyJSS4XElYwdC7ffDosWmZ1ERKSWJttdTVaWsSvw\nF19ASIjZaUTEjWmy3V2FhBjzJXPnmp1ERARQIXFNP/0pHD1qXLNERMRkGtpyVdu3w9SpcOSIcWVF\nEREb04JET/DMM3DpEnzwgdlJRMQNqZB4grIyY8X7hx/CiBFmpxERN6PJdk/QqRP87ncwc6ZxrXcR\nEROoR+IOHnsMbr5Z60tExKY0tOVJvvkG7rgD/vxnuO8+s9OIiJvQ0JYnufFGWLIE4uOhtNTsNCLi\nYdQjcSezZsHZs8bku4hIC6lH4okWLoQ9e2DNGrOTiIgHUY/E3ezdCzExsH8/BAWZnUZEXJh6JJ7q\nzjvhuedg8mS4cMHsNCLiAdQjcUeXLsG4cdC/P7z5ptlpRMRF6fRfT/fNNxAZaSxYHD/e7DQi4oJU\nSAR27YKHHzbmTX70I7PTiIiLcdo5kqKiIqxWK6GhoYwePZqSkpIG273xxhv07duXfv368dhjj3H+\n/HkHJ3UD99wDP/sZTJwIFRVmpxERN+XwQpKQkIDVaiUjI4Po6GgSEhLqtcnKymLJkiUcOHCAQ4cO\nUVVVxcqVKx0d1T288AL06gVPPQXqvYmIHTi8kCQlJREfHw9AfHw8a9eurdemU6dO+Pj4UF5ezsWL\nFykvL6d79+6OjuoeLBZYutS4bsnbb5udRkTcUCtHv2BhYSEBAQEABAQEUFhYWK+Nn58fL7zwAj16\n9KBdu3bcf//9jBo1qsHnmzdvXu3tqKgooqKi7BHbtbVvD2vXwl13we23w5gxZicSESeUmppKamrq\ndf+cXSbbrVYrBQUF9b4/f/584uPjKS4urv2en58fRUVFddp9/fXXxMTEsGPHDnx9fXnkkUeYOHEi\nU6dOrRtek+3XZ+dOY/I9NRXCw81OIyJOrqmfsXbpkaSkpDT6WEBAAAUFBQQGBnL69Gn8/f3rtfni\niy8YOnQoXbt2BeDhhx9m9+7d9QqJXKdhw4yt5h94AHbvNraeFxFpIYfPkcTGxpKYmAhAYmIicXFx\n9dr06dOHtLQ0KioqqK6uZsuWLYTrf9C2MW2acSGsBx4wrrAoItJCDl9HUlRUxKRJk8jOziYkJITV\nq1fTuXNn8vPzmTlzJhs3bgTgrbfeIjExES8vLyIjI/nDH/6Aj49P3fAa2mqe6mrjeu9ffw0bN0Lr\n1mYnEhEnpAWJcnUXLxrzJR07GtvOe3ubnUhEnIzTLkgUJ9GqFaxcCfn58OMfa42JiDSbCokna98e\nkpKMNSZz56qYiEizqJB4uo4dYdMm49Tgl19WMRGR6+bwBYnihDp3hs8/hxEjjEKSkGCsiBcRaQL1\nSMTQtSts2wZbt8Ls2cY1TUREmkCFRC7r2hW2bIGDB2H6dKiqMjuRiLgAFRKpy9cXkpMhNxcefRTO\nnTM7kYg4ORUSqe+GG2DDBvDxgVGj4NtvzU4kIk5MhUQa1qYN/OlPxv5cQ4caq+BFRBqgQiKN8/Iy\nzuB67jmjoOzcaXYiEXFCKiRybbNmwbJlxpYq77+vtSYiUof22pKmO3kS4uLg7rvh3XeN4S8RcVva\na0tsr1cv+NvfoKgIhg+HU6fMTiQiTkCFRK5Px47w8ccwdSoMGQKrVpmdSERMpqEtab79+2HyZIiK\ngnfegQ4dzE4kIjakoS2xv4ED4cABuHAB+veH7dvNTiQiJlCPRGxj/XrjuiYTJ8KCBcaiRhFxaeqR\niGPFxMChQ8Yq+H79jEv4iohHUI9EbO/zz421J3fcAb/5DQQHm51IRJpBPRIxz+jRcPiw0TOJiIDX\nX4fycrNTiYidqJCIfbRtC/Pmwd698NVXcNtt8NFHus6JiBvS0JY4xu7d8JOfGNvSv/aaMaeiqzCK\nOLWmfsaqkIjjVFdDUhL86lfQurVRUMaMUUERcVIqJOK8Ll2CTz4xhr5at4YXXzROG27VyuxkInIF\nFRJxfpcuwaZN8OabkJ8Pc+dCfDx06mR2MhFBZ22JK/DygvHjYccOYyJ+xw4ICYFnn4Xjx81OJyJN\npEIizmHoUFi92ljU6Odn7N91772wfDl8953Z6UTkKjS0Jc7pwgVjdfzSpcaVGR9+GKZMgREjwNvb\n7HQiHkFzJOI+8vNh5UpYsQJyc+GRR4zJ+XvuUVERsSMVEnFPGRlGUfn0U8jLg9hYePBBGDlSG0WK\n2JgKibi/zEyjoGzYAPv2GfMsY8eC1Qrh4VqfItJCKiTiWcrKYMsW2LwZ/vIXqKgweikjRhiXBQ4N\nVWERuU4qJOLZTp0yCsr27cZpxefOwbBhcNddxiWCBw7UUJjINaiQiFwpOxt27YK0NNizxzjN+NZb\nITLSKCoREcZuxb6+ZicVcRoqJCJXc/68sdX9gQOXjyNH4MYbjYLSty+EhV0+OnY0O7GIw6mQiFyv\nS5eMIbFDh4yicuyYcfz978a2Lb17G3Mtt94Kt9xiHD17Qteumn8Rt6RC4qFSU1OJiooyO4ZTafF7\ncumSsZYlI8M4Tp26fGRmGosne/QwjuBgCAoyju7d4aabjOPGG40tYZyE/k7q03tSX1M/Yx2+3eqa\nNWuYN28ex48fZ9++fURGRjbYLjk5mblz51JVVcWMGTP4z//8TwcndU36x1Bfi98TL6/LxWHkyPqP\nl5UZczD/+IexYDI315iPyc2F06eNo6wMunUDf38ICDC+dut2+eja1Tj8/IyvXbpAmzbNz3wN+jup\nT+9J8zm8kPTr149PP/2Up59+utE2VVVVPPvss2zZsoXu3bszePBgYmNjCQsLc2BSkSbq1Aluv904\nGnP+PPzzn8ZRWGh8/de/4JtvjF7Ot98aR1GR8bW42NhWv3Pny4evr3F06mQcHTsaR4cOl7/ecEPd\no31742u7dtoFQOzG4YWkT58+12yzd+9eevXqRUhICACTJ09m3bp1KiTiutq0MYa9goOb1r662rjO\nfXExlJZePkpK4MwZo4dTVmb0es6cMY6zZ40NLr/7zrhdXl738PExCkq7dsbp0KtXG7fbtjWONm0u\nf23d2vhac7vmvo+PcdvH5/Jx5f1Wrep/9fY2vtYcNfe9vevfrjm8vBr/6uWlOSkn45RXEsrLyyP4\nin9wQUFB7Nmzp8G2Fv1B1fPqq6+aHcHp6D0BKiuNo7QUgFdLSkwO5Hz0d9I8dikkVquVgoKCet9f\nsGABMTEx1/z5phYHTbSLiJjPLoUkJSWlRT/fvXt3cnJyau/n5OQQFBTU0lgiImIHpp6P2FiPYtCg\nQZw4cYKsrCwqKytZtWoVsbGxDk4nIiJN4fBC8umnnxIcHExaWhrjxo1j7NixAOTn5zNu3DgAWrVq\nxbvvvsv9999PeHg4jz76qCbaRUSclMsuSNQ6k7qefPJJNm7ciL+/P4cOHTI7jlPIyclh2rRp/POf\n/8RisfDUU08xZ84cs2OZ6ty5c9x3332cP3+eyspKHnzwQd544w2zYzmFqqoqBg0aRFBQEOvXrzc7\njlMICQmhU6dOeHt74+Pjw969exts55KFpKqqittuu63OOpMVK1Z4dK9lx44ddOjQgWnTpqmQfK+g\noICCggIGDBjA2bNnGThwIGvXrvXovxOA8vJy2rdvz8WLFxk2bBiLFi1i2LBhZscy3a9//Wv279/P\nmTNnSEpKMjuOU+jZsyf79+/Hz8/vqu2cZ8+G63DlOhMfH5/adSaebPjw4XTp0sXsGE4lMDCQAQMG\nANChQwfCwsLIz883OZX52rdvD0BlZSVVVVXX/JDwBLm5uWzatIkZM2bobNAfaMr74ZKFpKF1Jnl5\neSYmEmeXlZVFeno6Q4YMMTuK6S5dusSAAQMICAhgxIgRhIeHmx3JdM8//zwLFy7Ey4n2Q3MGFouF\nUaNGMWjQIJYsWdJoO5d817QIUa7H2bNnmThxIosXL6ZDhw5mxzGdl5cXX375Jbm5ufz1r38lNTXV\n7Eim2rBhA/7+/kRERKg38gO7du0iPT2dzZs3895777Fjx44G27lkIdE6E2mqCxcuMGHCBB5//HHi\n4uLMjuNUfH19GTduHF988YXZUUy1e/dukpKS6NmzJ1OmTGHr1q1MmzbN7FhO4aabbgKgW7duPPTQ\nQ41Otru9MGtTAAADCElEQVRkIdE6E2mK6upqpk+fTnh4OHPnzjU7jlP45ptvKPl+a5SKigpSUlKI\niIgwOZW5FixYQE5ODpmZmaxcuZKRI0fy4Ycfmh3LdOXl5Zw5cwaA7777js8//5x+/fo12NYlC4nW\nmdQ3ZcoUhg4dSkZGBsHBwSxbtszsSKbbtWsXf/zjH9m2bRsRERFERESQnJxsdixTnT59mpEjRzJg\nwACGDBlCTEwM0dHRZsdyKho6NxQWFjJ8+PDav5Xx48czevToBtu65Om/IiLiPFyyRyIiIs5DhURE\nRFpEhURERFpEhURERFrEKa+QKOKsvv32W0aNGgUYe3l5e3vTrVs3LBYLe/bsYfbs2WRkZFBcXExl\nZSWZmZncdtttAPziF7/g3XffZdGiRQwcONDMX0PEplRIRK5D165dSU9PB4zLsnbs2JGf/OQntY/v\n2bOHAwcOYLFY+Mc//sH48eNr2wO8++67zX7tixcv0qqV/smK89HQlkgLXHn2/LFjxwgNDa1dh9DY\nmfUfffQRERER9OvXj3379gHGRqRDhw4lMjKSe+65h4yMDACWL19ObGws0dHRWK1WO/82Is2j/96I\n2MjmzZtrL9R2NRUVFaSnp7Njxw6efPJJDh06RFhYGDt27MDb25stW7bw85//nI8//hiA9PR0Dh06\nROfOne39K4g0iwqJiI18/vnnLF++/JrtpkyZAhhb/5eVlVFWVkZpaSnTpk3j5MmTWCwWLl68WNt+\n9OjRKiLi1DS0JWID5eXllJSUEBgY2Kyf/+Uvf0l0dDSHDh1i/fr1VFRU1D5Wc/0QEWelQiJiA9u2\nbWPkyJHXbFddXc2qVasA2LlzJ507d6ZTp06UlZVx8803A2ifNHE5KiQiLVAzsb5582bGjBnT6ONX\n3m/bti2RkZHMmjWLpUuXAvDiiy/y8ssvExkZSVVVVe3PWSwWbSIoTk+bNorYwMCBA9m7dy/e3t5m\nRxFxOBUSERFpEQ1tiYhIi6iQiIhIi6iQiIhIi6iQiIhIi6iQiIhIi6iQiIhIi/w/PBsECefjDVkA\nAAAASUVORK5CYII=\n",
       "text": [
        "<matplotlib.figure.Figure at 0xa81956c>"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Heat capacity is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ C = \\frac{\\partial}{\\partial T} E = \\frac{\\mu^2 B^2 N}{k_B} \\frac{ (1 - \\tanh^2(\\mu B/(k_B T)) }{T^2} $$"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "t13 = linspace(0, 5, 100)\n",
      "c1 = tanh(t13)\n",
      "\n",
      "fig13 = plt.figure()\n",
      "\n",
      "axes = fig13.add_axes([0.1, 0.1, 0.8, 0.8]) # left, bottom, width, height (range 0 to 1)\n",
      "\n",
      "axes.plot(t13, c1, 'r')\n",
      "\n",
      "axes.set_xlabel('T/Tbar')\n",
      "axes.set_ylabel('C/Cbar')\n",
      "axes.set_title('C ~ T');"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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udUkiIj7Fd9pQa9YYFzlSUACQl5dndQkeR59JY/pMGtNn0jqmhYXdbicmJoao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\nhYWIiLRIYSEiIi1SWIiISIv+H6Fl2wXvsqUxAAAAAElFTkSuQmCC\n",
       "text": [
        "<matplotlib.figure.Figure at 0xa76188c>"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "### 2D"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The energy of a dipole can be written as $E(\\theta) = -\\mu B \\cos\\theta$, in which $\\theta$ is the angle between the dipole and magnetic field."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Partition function is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ Z = \\int _ \\theta 2 e^{-\\beta E(\\theta)} \\mathrm d \\theta  = 2\\pi \\mathrm {BesselI}(0, \\mu B \\beta) $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "where $\\mathrm{BesselI}$ is the modified Bessel function of the first kind."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Then we can calculate magnetization, which is given by"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ M = \\frac{1}{Z}\\int _ \\theta 2N\\mu \\cos\\theta e^{-\\beta E(\\theta)} d \\theta  $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The result given by Mathematica is (so bad that I can't make sympy to deal with such integrations.)"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ M = \\mu N \\frac{ \\mathrm{BesselI}(1, B \\beta \\mu)  }{ \\mathrm{BesselI}(0, B \\beta \\mu) }$$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Energy should be"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ E = - \\mu B N \\frac{ \\mathrm{BesselI}(1, B \\beta \\mu)  }{ \\mathrm{BesselI}(0, B \\beta \\mu) }$$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Heat capacity is the derivitive of energy over temperature."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ C = B^2 \\mu^2 N  \\frac{-2 \\mathrm{BesselI}(0, B \\beta \\mu)^2 +  \\mathrm{BesselI}(0,\\mu\\beta B)(  \\mathrm{BesselI}(0,\\mu\\beta B) + \\mathrm{BesselI}(2,\\mu\\beta B)   ) }{2 k T^2 \\mathrm{BesselI}(0, B \\beta \\mu)^2} $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Plot out these results using Mathematica."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "![](https://raw.github.com/emptymalei/StatisticalPhysics/master/equilibrium/homework/images/magb2.png)"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "![](https://raw.github.com/emptymalei/StatisticalPhysics/master/equilibrium/homework/images/magt2.png)"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "![](https://raw.github.com/emptymalei/StatisticalPhysics/master/equilibrium/homework/images/capt2.png)"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "### 3D"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Energy of a dipole is $E(\\theta) = -\\mu B \\cos\\theta$."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Then we can calculate the partition function, which is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ Z = \\int_\\theta \\int _ \\phi N e^{-\\beta E(\\theta)}\\sin\\theta d\\theta d \\phi = 4\\pi N\\frac{ \\sinh(\\mu B \\beta)}{\\mu B \\beta} $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Magnetization is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ M =  -\\frac{\\mu N}{\\mu B \\beta} + \\mu N \\coth(\\mu B \\beta) = \\mu N \\left(  - \\frac{\\bar B}{B} + \\coth \\left(\\frac{B}{\\bar B}\\right) \\right) $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "where $\\bar B = \\frac{1}{\\mu \\beta}$. Define $\\bar M = \\mu N$ .Then we can write down the dimensionless equation, which is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ \\frac{M}{\\bar M} =  - \\frac{\\bar B}{B} + \\coth \\left(\\frac{B}{\\bar B}\\right) $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Total energy is"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ E = \\frac{1}{Z} \\int _ \\phi \\int _ \\theta (-\\mu B \\cos\\theta N) e^{-\\mu E(\\theta)}\\sin\\theta d \\theta d\\phi = - N \\left( \\frac{1}{\\beta} - B \\mu \\coth(\\mu \\beta B) \\right)  $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Take the derivative with respect to temperature, we have the heat capacity."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ C = N \\left( 1 - \\left(\\frac{\\bar T}{T} \\right)^2 \\mathrm{csch}^2 \\left( \\frac{\\bar T}{T} \\right) \\right) $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Plot out the results."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "![](https://raw.github.com/emptymalei/StatisticalPhysics/master/equilibrium/homework/images/magb3.png)"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "![](https://raw.github.com/emptymalei/StatisticalPhysics/master/equilibrium/homework/images/magt3.png)"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "![](https://raw.github.com/emptymalei/StatisticalPhysics/master/equilibrium/homework/images/capt3.png)"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "## Problem 2"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "#### First law of thermodynamics"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Energy can be transfered from one form to another, or from one system to another, but can never be disapear. The total energy of a closed system is conserved. For any infinitesimal process, the change of internal energy is equivalent to the sum of work done to other systems and heat obsorbed from other systems, which is to say, in the language of mathematiccs,"
     ]
    },
    {
     "cell_type": "heading",
     "level": 4,
     "metadata": {},
     "source": [
      "$$ \\mathrm d U = \\mathrm d Q + \\mathrm d W $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "#### Entropy"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Basicly, at high temperature and in large systems, entropy is the dominant of all marco properties of a system. Entropy indicates the reversibility of a process, defines Kelvin and restricts us from reaching absolute zero."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "The most important property of entropy is that it is a function that only depends on state of the system not the path. This property can lead us to many interesting results."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "#### Free energy"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "As I have learned in undergrad stat mech, free energy is a quantity defined to make our life easier when dealing with systems which has fixed temperature and fixed volume. In that case, free energy can never grow larger. Mathematically,"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ \\mathrm d F \\leq 0 $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Suppose we have a gas system going through a reversible process, free energy"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "$$ \\mathrm d F = - S\\mathrm d T - p \\mathrm d V $$"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "For fixed $T$ and $V$, we have $\\mathrm d F = 0$. For irreversible process, it is less than 0."
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "#### Chemical potential"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Chemical potential is the increased Gibbs free energy when one mole particle is added to the system."
     ]
    }
   ],
   "metadata": {}
  }
 ]
}