path: root/Modern_Physics/Chapter2_2.ipynb

{
 "metadata": {
  "name": "MP-2"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": "The Special Theory of Relativity"
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.1 Page 22"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of varible\nv1=60.0; v2=40.0      #Velocities of cars wrt to observer in km/hr\n\n#calculation\nvr=v1-v2;         #relative velocity\n\n#result\nprint\"The value of relative velocity in km/h. is\",round(vr,3);\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The value of relative velocity in km/h. is 20.0\n"
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.2 Page 22"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import atan, pi\nimport numpy as np\nVa_w=[320.0,0.0]; Vw_g=[0.0, 65.0];            #Vp/q=[X Y]=>velocity of object p wrt q along X(east) and Y(north) directions.\n\n#calculation\nVa_g=Va_w + Vw_g;                    #net velocity\nk=np.linalg.norm(Va_g)                       #magnitude\ns=atan(Va_g[3]/Va_g[0])*180.0/pi;    #angle in rad*180/pi for conversion to degrees\n\n#result\nprint \"the velocity in x direction in Km/h is\", Va_w[0],\"in y direction in km/h is\",Vw_g[1]\nprint\"The magnitude of velocity Va/g(airplane wrt ground) in Km/h is\",round(k,3),\" at \",round(s,3),\" degrees north of east.\" ",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The magnitude of velocity Va/g(airplane wrt ground) in Km/h is 326.535  at  11.482  degrees north of east.\n"
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.4 Page 28"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\nLo=100.0*(10**3);c=3.0*(10**8); #Given values//all the quantities are converted to SI units \nd=2.2*(10**-6);      #time between its birth and decay\n\n#calculation\nt=Lo/c             #where Lo is the distance from top of atmosphere to the Earth. c is the velocity of light. t is the time taken\nu=sqrt(1-((d/t)**2));   # using time dilaion fromula for finding u where u is the minimum velocity in terms of c;\n\n#result\nprint\"Hence the minimum speed required in c is\",round(u,6);\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "Hence the minimum speed required in c is 0.999978\n"
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.5 Page 30"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#intiation of variable\nfrom math import sqrt\nLo=100.0*(10**3);  #Lo is converted to Km\nu=0.999978;      #//u/c is taken as u since u is represented in terms of c. \n\n#calculation\nL=Lo*(sqrt(1-u**2)); # from the length contraction formula\n\n#result\nprint\"Hence the apparent thickness of the Earth's surface in metres. is\",round(L,3)\nprint\"answer is slightly different in the book\"",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "Hence the apparent thickness of the Earth's surface in metres. is 663.321\nanswer is slightly different in the book\n"
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.6 Page 32"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\nL=65.0; c=3*10**8;u=0.8*c;      \n\n#calculation\nt=L/u ;          #The value of time taken as measured by the observer\n\n#result\nprint\"The time for rocket to pass a point as measured by O in musec is \",round(t*10**6,3);         #The value of time taken as measured by the observer\n\n#partb\nDo=65.0;                             #given length\nLo= L/sqrt(1-(u/c)**2);                 #contracted length of rocket\n\n#result\nprint\"Actual length according to O is \",round(Lo,3);\n\n#partc\nD=Do*(sqrt(1-(u/c)**2));       #contracted length of platform.\n\n#result\nprint\"Contracted length according to O'' is\",round(D,3);\n\n#partd\nt1=Lo/u;                    #time needed to pass according to O'.\nprint \"Time taken according to O is \",t1\n\n#part 3\nt2=(Lo-D)/u;                   #time intervals between the two instancs\nprint\"Time taken according to O'' is \",t2;\nprint'The value of t1 and t2 did not match';",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The time for rocket to pass a point as measured by O in musec is  0.271\nActual length according to O is  108.333\nContracted length according to O'' is 39.0\nTime taken according to O is  4.51388888889e-07\nTime taken according to O'' is  2.88888888889e-07\nThe value of t1 and t2 did not match\n"
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.7 Page 35"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nv1=0.6; u=0.8; c=1.0; # all the values are measured in terms of c hence c=1\n\n#calculation\nv= (v1+u)/(1+(v1*u/c**2));\n\n#result\nprint \"The speed of missile as measured by an observer on earth in c is\",round(v,3);",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The speed of missile as measured by an observer on earth in c is 0.946\n"
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.8 Page 37"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nw1=600.0;w2=434.0;  # w1=recorded wavelength;w2=actual wavelength\n               # c/w1 = c/w2 *(sqrt(1-u/c)/(1+u/c))\n    \n#calcualtion\nk=w2/w1;\nx=(1-k**2)/(1+k**2);         #solving for u/c\n\n#result\nprint\"The speed of galaxy wrt earth in c is\",round(x,3);\n",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The speed of galaxy wrt earth in c is 0.313\n"
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.9 Page 39"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\nv1x=0.6;v1y=0.0;v2x=0.0;v2y=.8;c=1.0;     # all the velocities are taken wrt c\nv21x=(v2x-v1x)/(1-(v1x*v2x/c**2));   #using lorentz velocity transformation\nv21y=(v2y*(sqrt(1-(v1x*c)**2)/c**2))/(1-v1y*v2y/c**2) \n\n#result\nprint\"The velocity of rocket 2 wrt rocket 1 along x and y directions is\",round(v21x,3),\" c &\", round(v21y,3),\"c respectively\"",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The velocity of rocket 2 wrt rocket 1 along x and y directions is -0.6  c & 0.64 c respectively\n"
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.10 Page 40"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\nu=0.8*c;L=65.0;c=3.0*10**8;            #all values are in terms of c\nt=u*L/(c**2*(sqrt(1-((u/c)**2))));  #from the equation 2.31 \n\n#result\nprint\"The time interval between the events is\",t, \"sec which equals\",round(t*10**6,3),\"musec.\"",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The time interval between the events is 2.88888888889e-07 sec which equals 0.289 musec.\n"
      }
     ],
     "prompt_number": 30
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.11 Page 41"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\nm=1.67*10**-27;c= 3*10**8;v=0.86*c;      #all the given values and constants\n\n#calculation\np=m*v/(sqrt(1-((v/c)**2)));            # in terms of Kgm/sec\n\n#result\nprint\"The value of momentum was found out to be in Kg-m/sec.\\n\",p;\n\n#part 2\nc=938.0;v=0.86*c;mc2=938.0                 # all the energies in MeV where mc2= value of m*c^2\npc=(mc2*(v/c))/(sqrt(1-((v/c)**2)));    #expressing in terms of Mev\n\n#result\nprint\"The value of momentum was found out to be in Mev.\",round(pc,3);",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The value of momentum was found out to be in Kg-m/sec.\n8.44336739668e-19\nThe value of momentum was found out to be in Mev. 1580.814\n"
      }
     ],
     "prompt_number": 34
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.12 Page 47"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\npc=1580.0; mc2=938.0;E0=938.0;    # all the energies in MeV mc2=m*c^2 and pc=p*c\n\n#result\nE=sqrt(pc**2+mc2**2); \nK=E-E0;                      #value of possible kinetic energy\n\n#result\nprint\"The relativistic total energy in MeV. is\",round(E,3);         #value of Energy E\nprint\"The kinetic energy of the proton in MeV.\",round(K,3);",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The relativistic total energy in MeV. is 1837.456\nThe kinetic energy of the proton in MeV. 899.456\n"
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.13 Page 47"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\nE=10.51; mc2=0.511; #all the values are in MeV\n\n#calculation\np=sqrt(E**2-mc2**2);   #momentum of the electron\nv=sqrt(1-(mc2/E)**2);  #velocity in terms of c\n\n#result\nprint\"The momentum of electron in MeV/c is\",round(p,3); \nprint\"The velocity of electron in c is\",round(v,5);",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The momentum of electron in MeV/c is 10.498\nThe velocity of electron in c is 0.99882\n"
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.14 Page 47"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\nk=50;mc2=0.511*10**-3;c=3.0*10**8; # all the values of energy are in GeV and c is in SI units\n\n#calculation\nv=sqrt(1-(1/(1+(k/mc2))**2));   #speed of the electron in terms of c\nk=c-(v*c);                       #difference in velocities\n\n#result\nprint\"Speed of the electron as a  fraction of c*10^-12 is.\",round(v*10**12,3); # v=(v*10^12)*10^-12; so as to obtain desired accuracy in the result\nprint\"The difference in velocities in cm/s.\",round(k*10**2,3);",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "Speed of the electron as a  fraction of c*10^-12 is. 9.99999999948e+11\nThe difference in velocities in cm/s. 1.567\n"
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.15 Page 48"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt, pi\nr=1.5*10**11; I=1.4*10**3;     #radius and intensity of sun\n\n#calculation\ns=4*pi*r**2                 #surface area of the sun\nPr=s*I                     # Power radiated in J/sec\nc=3.0*10**8;                #velocity of light\nm=Pr/c**2                  #rate od decrease of mass\nm=round(m,2)\n\n#result\nprint\"The rate of decrease in mass of the sun in kg/sec. is %.1e\" %m;",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The rate of decrease in mass of the sun in kg/sec. is 4.4e+09\n"
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.16 Page 48"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import pi, sqrt\nK=325; mkc2=498;  #kinetic energy and rest mass energy of kaons\nmpic=140.0; #given value\n\n#calculation\nEk=K+mkc2; \npkc=sqrt(Ek**2-mkc2**2); \n#consider the law of conservation of energy which yields Ek=sqrt(p1c^2+mpic^2)+sqrt(p2c^2+mpic^2)\n#The above equations (4th degree,hence no direct methods)can be solved by assuming the value of p2c=0.\np1c=sqrt(Ek**2-(2*mpic*Ek));\n#consider the law of conservation of momentum. which gives p1c+p2c=pkc implies\np2c=pkc-p1c;\nk1=(sqrt(p1c**2+(mpic**2))-mpic); #corresponding kinetic energies\nk2=(sqrt((p2c**2)+(mpic**2))-mpic);\n\n#result\nprint\"The corresponding kinetic energies of the pions are\", k1,\" MeV and\",round(k2,3),\" MeV.\"",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The corresponding kinetic energies of the pions are 543.0  MeV and 0.627  MeV.\n"
      }
     ],
     "prompt_number": 49
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": "Example 2.17 Page 49"
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": "#initiation of variable\nfrom math import sqrt\nmpc2=938.0;c=3.0*10**8;  #mpc2=mp*c^2,mp=mass of proton\n\n#calculation\nEt=4*mpc2;          #final total energy\nE1=Et/2;E2=E1;       #applying conservation of momentum and energy\nv2=c*sqrt(1-(mpc2/E1)**2);    #lorentz transformation\nu=v2;v=(v2+u)/(1+(u*v2/c**2)); \nE=mpc2/(sqrt(1-(v/c)**2));\nK=E-mpc2;\n\n#result\nprint\"The threshold kinetic energy in Gev\",round(K/10**3,3);",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "The threshold kinetic energy in Gev 5.628\n"
      }
     ],
     "prompt_number": 50
    }
   ],
   "metadata": {}
  }
 ]
}