RELATION BETWEEN MOMENTUM AND ENERGY

Mass energy equivalence

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= FdxBut force being rate of change of linear

momentum of the particle , is given by

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NUCLEAR FISSIONNUCLEAR FUSIONNUCLEAR REACTION PROCESSES PHENOMENON OF PAIR PRODUCTION

Examples for proving equivalence between energy and mass

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

SolutionRate of energy radiated

yearpertonsyearpertons

c

Eismassinlossannual

yearperradiatedenergy

uteperradiatedenergyTotal

cmerg

1414

20

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2

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27

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109

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102.42101504

min

min//102.42

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

SolutionThe 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

Solution

The initial relativistic total energy Ek= K+ mkc2 =325 MeV + 498 MeV =823

MeVTotal initial momentum

Total energy for final system consisting of two pions is i

MeVcmEcP kkk 655498823 22222

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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 iiWe 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= mc2 (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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