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

dtdmv

dtdvmmv

dtdF

0222

11

222

220

222222

202

220

2

mdmvvdvmmdmc

atingdifferenti

cmvmcmcv

mmcv

mm

vdtdxdmvmvdvdE

dxdtdmvdx

dtdvmdE

2

42

044222

0

44222122

2022

20

220

20

20

22

2

22

83

211....8321

....83211

,

11

0

0

cvvmcvcvcmE

cvcvcv

havewecvfor

cmcv

cmEcv

mm

putting

cmmcEmmcdmcE

dmcdE

dmvmvdvdmc

m

m

Classi

cal

expressi

on

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

cEismassinlossannual

yearperradiatedenergy

uteperradiatedenergyTotal

cmerg

1414

20

57211

2

57211

7211

27

104.1104.1

109103.5102.42101504

103.5102.42101504

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

00h

2

00 21

mchh

chp 0

ch 0

mhh

mchhlosstotalthe

henergyphotonwheremch

mp

mvmmvE

21

2

22221

002

20

0

02

20

2222

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

MeVTotal initial momentum

Total energy for final system consisting of two pions is i

MeVcmEcP kkk 655498823 22222

MeVcmcpcmcpEEE 823222

2222

121

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

MeVK

MeVK

cmcmPcK

6.014014013

543140140668

222

221

20

220

2

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

220

1 cvmm

)4(

11

42222

4222

220

2

22

42022242222

0

0

cmPcEor

cmcvvmc

cvcmvmccmPcE

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

cvor

cPcv

cEvp

mcEascEvmvp

22

22

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.

21 cUvvUU

cccvvcU

21