nathennguyen_hw2.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 52,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.61740530956282e-27\n",
      "1.854822648511579e+17\n"
     ]
    }
   ],
   "source": [
    "\"problem 1\" \n",
    "import math \n",
    "numberdensity = 10e9 #1/mpc^-3\n",
    "rstar = 2.269e-14 #mpc or 7e8 m\n",
    "\n",
    "print (math.pi*(rstar**2)) # gives the area of each star to the observer\n",
    "\n",
    "#numberdensity*(4/3 *math.pi *(r**2)) the number of stars as you expand out the sphere\n",
    "#multiply these two values and set it equal to the surface area of a sphere, \n",
    "#basically saying when the entire area of the sphere is filled by the areas of the stars\n",
    "#numberdensity*(4/3 *math.pi *(r**2)) * math.pi*(rstar**2) = 4*math.pi*(r**2) we get\n",
    "r = (3/(numberdensity*(math.pi*(rstar**2))))\n",
    "     \n",
    "print (r)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "#comparing these two values, we see there is a massive distance. \n",
    "#1.85e+17 is a huge amount of space, but possible for an olberian universe since \n",
    "#t= inf and the universe itself is infinite.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1957394566.8990486\n",
      "1.9573945668990487\n"
     ]
    }
   ],
   "source": [
    "\"problem 2\"\n",
    "#current H0 = 70 +/- 7 km*s^-1*mpc^-1, while Hubble himself calculated 500km*s^-1*mpc^-1\n",
    "#becuase z = (H0/c)*r, or written v = H0 * r, the hubble consatnt is calculated by v/r = H0\n",
    "#Hubble thought the r for the cephid stars was roughly 1/7 of what it actually is. \n",
    "#if the H0 was what Hubble originally calculated, with t= 1/H0\n",
    "H0 = 1.62e-17 # conversion from 500 km*s^-1*mpc^-1 to 1/s\n",
    "t = 1/H0\n",
    "#31536000 seconds in a year\n",
    "print (t/31536000)\n",
    "print (t/31536000/1e+9)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "#if Hubble was correct, then the Hubble time, the estimate for the time of the universe, would only be 2 gigayears\n",
    "#from above we know that v = H0 *r, the Hubble time gives us the age of the universe if relative velocities are constant.\n",
    "#that would mean in order to fit our current model, the universe would have to be accelerating. \n",
    "#comparitively this makes sense, if the universe had H0 = 500, but it was as big as we see now, then in order to reconcile\n",
    "#with the current age of the universe, the Hubble constant would need to start smaller, and then grow in value until it would\n",
    "#reach H0 = 500, a larger v means a larger H0, v = H0 *r.\n",
    "#In short, if the universe was accelearting, the Hubble constant would give a larger value, which means we OBSERVE/CALCULATE\n",
    "#A smaller hubble time than the initial start of the universe. The universe would appear younger than it actually is. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "''"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "\"problem 3\"\n",
    "#this is a 2d positively curving space,i.e. k=+1 which gives #Sk = R*sin^2(r/R), thus we have the relation\n",
    "#ds^2 = dr^2 + R*sin^2(r/R)dtheata^2\n",
    "#rearrange for dthea, we get\n",
    "\"dtheata = sqrt[(ds^2-dr^2)/(R*sin^2(r/R))]\"\n",
    "#for positvely curved spaces, Sk approaches 0 as r/R approaches pi. (r approaches pi*R) \n",
    "#however, for calculating dtheta, because we divive by Sk, as Sk approaches 0, 1/Sk approaches inf, thus\n",
    "\"r -> pi*R, dtheta -> infinity\"\n",
    "''"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "\"problem 4\"\n",
    "from scipy.integrate import quad\n",
    "import numpy as np\n",
    "import pandas as pd\n",
    "import matplotlib.pyplot as plt\n",
    "%matplotlib inline\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "def nth_root(val, n):\n",
    "    ret = val**(1./n)\n",
    "    return ret + 1 if (ret + 1) ** n == val else ret"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.4193336415231441\n"
     ]
    }
   ],
   "source": [
    "def integrand(x):\n",
    "    return (x**(3/2))\n",
    "#we will treat the age of the universe as 14.4 gigayeras, \n",
    "\n",
    "integral0, err = quad(integrand, 0, 1)\n",
    "\n",
    "ans0 = integral0 * 14.4\n",
    "print (nth_root(ans0,5))\n",
    "\n",
    "#because we get ln(timefinal/timeinitial), using 0 gets an error, ln(0) is not defined, so for this case I used 1.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "step = 0.01 \n",
    "values = [] \n",
    "a = np.arange(0,1,step)\n",
    "for i in a: \n",
    "    intergral0,err = quad(integrand,0,i)\n",
    "    \n",
    "    values.append(integral0)\n",
    "    \n",
    "a = np.asarray(a)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "100\n",
      "100\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "import pandas as pd\n",
    "import matplotlib.pyplot as plt\n",
    "%matplotlib inline\n",
    "t = np.linspace(0,14.4,100) \n",
    "t2 = np.linspace(0,14.4,a.size)\n",
    "eq2 = ((5/2)*(1/14.4)*t)**(2/5)\n",
    "\n",
    "#I algebraically derived a = (5/2*H0*t)**(2/5)\n",
    "\n",
    "print (t2.size)\n",
    "print (a.size)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 46,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[<matplotlib.lines.Line2D at 0x7fc979ce9e80>]"
      ]
     },
     "execution_count": 46,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
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IHD8eTjsNmjYNuhKR5PP5js+pM7QON190M8PbDKdMSfVfhllSLL370kvwyCPa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\nIgp1EZEQUaiLiITI/wcf7J30YCru8gAAAABJRU5ErkJggg==\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7fc979ce9f28>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "\n",
    "t = np.linspace(-14.4,14.4,100)   \n",
    "#plt.plot(t,(4*(1 - t/14.4))**(1/5))\n",
    "plt.plot(t,eq2)\n",
    "plt.plot(t2,a)\n",
    "#plt.hlines((0), -15, 15, color='m', linewidth=2) "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "# We see that our algebraically derived graph approaches our numerically derived value. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.4.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}