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Transcript
Mass energy equivalence RELATION BETWEEN MOMENTUM AND ENERGY In quantum mechanics , we considered that kinetic energy could be increased only increasing by its velocity But now dealing with relativistic mechanics we take mass variation into account Relationship between mass and energy If force F acting on a particle ,produces a displacement dx then the work done by the force = Fdx the work done must be equal to the gain in the kinetic energy dE of the particle dE= Fdx But force being rate of change of linear momentum of the particle , is given by d dv dm F mv m v dt dt dt dv dm dE m dx v dx dt dt dx 2 dE mvdv v dm v dt 2 m0 m0 2 2 2 2 2 2 m m m c m v m0 c 2 2 1 v c 1 v 2 c2 2 differentiating c 2mdm m 2vdv v 2mdm 0 2 2 2 c 2 dm mvdv v 2 dm 0 dE c 2 dm m E c 2 dm c 2 m m0 E mc2 m0c 2 m0 putting m0 m 1 v 2 c2 v c, we for 1 v E 2 c 2 1 2 m0c 2 1 v 2 c2 m0c 2 have 1 v 2 2c 2 3 v 4 8c 4 .... 4 1 3 v E m0c 2 1 v 2 2c 2 3 v 4 8c 4 .... 1 m0v 2 2 8 c2 Examples for proving equivalence between energy and mass NUCLEAR FISSION NUCLEAR FUSION NUCLEAR REACTION PROCESSES PHENOMENON OF PAIR PRODUCTION Nuclear fission Example 1 What is the annual loss in the mass of the sun , if the earth receives heat energy approximately 2 cal/cm2/min , The earth sun distance is about 150x106 km Solution Rate of energy radiated 2 4.2 107 erg / cm 2 / min Total energy radiated per min ute 4 150 10 2 4.2 10 11 2 7 energy radiated per year 4 150 10 2 4.2 10 11 2 7 5.3 105 E annual loss in mass is 2 c 4 150 10 2 4.2 10 11 2 7 5.3 105 9 1020 1.4 1014 tons per year 1.4 1014 tons per year Example 2 A nucleus of mass m emits a gamma ray of frequency 0 .Show that the loss of internal energy h by the nucleus is not h0 but is h0 1 02 2mc Solution The momentum of gamma ray photon is p h 0 c According to the law of conversation of momentum , the nucleus having mass m will recoil with the momentum h in the back ground direction . c Therefore, the loss of energy recoiling is 0 1 2 m 2v 2 p 2 h0 E mv where photon energy h0 2 2 2m 2m 2mc h0 2 h0 the total loss h0 h 0 1 2 2m 2mc 2 Example 3 A certain accelerator produces a beam of neutral K- mesons or kaons mkc2=498 MeV . Consider a kaon that decays in flight into two pions (mc2140 MeV) Show that the kinetic energy of each pion in the special case in which the pions travel parallel or anti parallel to the direction of the kaon beam or 543 MeV and 0.6 Mev Solution The initial relativistic total energy Ek= K+ mkc2 =325 MeV + 498 MeV =823 MeV Total initial momentum Pk c E mk c 2 k 2 2 8232 4982 655MeV Total energy for final system consisting of two pions is E E1 E2 i p1c 2 m c 2 2 p2c 2 m c 2 2 823 MeV Applying conservation of momentum , the final momentum of the two pions system along the beam direction is P1 + P2 and setting this equal to the initial momentum Pk , one obtains P1c +P2c =Pkc = 655 MeV ii We have now two equations in the two unknown P1 and P2 , solving we find P1c= 668 MeV or -13 MeV iii K Pc2 m0c 2 2 m0c 2 K1 6682 1402 140 543 MeV K2 132 1402 140 0.6 MeV Relation between momentum and energy Relativistic momentum of a particle moving with a velocity v is given by P=mv Where m0 m (1) (2) M0 being the rest mass of the particle , from relativity we have E= mc2 (3) From 1 and 2 we have 2 4 2 2 2 m0 c c m0 v 2 2 2 2 4 2 2 2 2 4 E c P m c c m v m c 2 2 2 2 0 1 v c 1 v c 1 v2 c2 or E 2 c 2 P 2 m 02c 4 ( 4) Particles with zero rest mass Photon and Graviton are the familiar examples of particles with zero rest mass . a particle with zero rest mass always moves with the speed of light in vacuum . According to 4 , if m0 =0 , we have E= Pc E 2 p mv v 2 as E mc c E Pc pv 2 v 2 c c or v c i.e., a particle with zero mass ( rest mass ) always moves with the speed of light in vacuum . The velcity of the particle observed in some other inertial frame S` is U U v 2 1 Uv c Where v is the velocity of the frame S` with respect to the frame S in which the velocity of the particle is U, cv U c hence U=c we have 1 cv c 2 Clearly the particle has the same speed c and zero rest mass for all observers in inertial frames.