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