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**Mass energy equivalence**

Relation between momentum and energy

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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

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**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

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Classical expression

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**Examples for proving equivalence between energy and mass**

Nuclear fission Nuclear fusion Nuclear reaction processes Phenomenon of pair production

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Nuclear fission

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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

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Example 2 A nucleus of mass m emits a gamma ray of frequency Show that the loss of internal energy by the nucleus is not but is Solution The momentum of gamma ray photon is According to the law of conversation of momentum , the nucleus having mass m will recoil with the momentum in the back ground direction . Therefore, the loss of energy recoiling is

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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 (mpc2=140 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

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**Solution The initial relativistic total energy**

Ek= K+ mkc2 =325 MeV MeV =823 MeV Total initial momentum Total energy for final system consisting of two pions is i

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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

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**Relation between momentum and energy**

Relativistic momentum of a particle moving with a velocity v is given by P=mv (1) Where (2) M0 being the rest mass of the particle , from relativity we have E= mc (3) From 1 and 2 we have

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**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

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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 Where v is the velocity of the frame S` with respect to the frame S in which the velocity of the particle is U, hence U=c we have Clearly the particle has the same speed c and zero rest mass for all observers in inertial frames.

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Chapter 29 Relativity.

Chapter 29 Relativity.

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