energy-momentum relation e2 p2c2 mc2 (that famous equation…..) · 2016-08-25 · energy-momentum...
TRANSCRIPT
Energy-momentum relation(That Famous Equation…..)
E = !
umc
2
p = !
umu
E2 ! p
2c
2= m
2c
4" 2
u1!
u2
c2
#
$%&
'(= m
2c
4" 2
u
1
" 2
u
= m2c
4
E
2= p
2c
2+ mc
2( )2
E = mc2
Energy:
Momentum:
Squaring and subtracting the second equation from the first:
Energy-momentum relation:
If the particle is at rest, p=0. Thus the rest energy is
Energy-momentum relation
E
2= p
2c
2+ mc
2( )2
E
2! p
2c
2= mc
2( )2
The rest mass of a particle mc2 is invariant in all inertial frames.Thus the quantity
is also invariant in all inertial frames.
Note: The total rest mass of a composite system is not equal to thesum of the rest masses of the individual particles.
Particles with zero mass (example: photons) have total energy:
E = pc
Energy-momentum relation
E
2= p
2c
2+ mc
2( )2
Energy is often expressed in electron-volts (eV):
Some Rest Mass Values: Photon = 0 MeV, Electron = 0.511 MeV,Proton = 938.28 MeV
It is also convenient to express mass m and momentum p in energyunits mc2 and pc.
1eV = 1.60 !10"19 J , 1MeV = 10
6 eV
1J = 1kgm2
s2
#
$%&
'(
Energy-momentum relation
If u approaches c (relativistic limit)
E ! pc
If u/c << 1 (classical limit)
E ! mc2
pc
pc
mc2
mc2
E
E
Two identical particles with rest mass 12 kgapproach each other with equal but oppositevelocities u1 = –u2 = 0.6c. Find the total restmass of this system.
( )
( )( )
2 2 2
2 2 21 2
22 2 2
2 2
1 11.25
3
1 1 0.6
2 2 1.25 12 30 ; 0
3000
tot tot
tot tottot
u c
E E E mc kg c kg c p
E p kg c
cc cm kg
!
!
= = =
" "
= + = = = # =
$ %$ % #$ %= " = " =& '& ' & ' & '( )( ) ( )
x
y u1 = 0.6c u2 = !0.6c
m1 = 12 kg m2 = 12 kg
Solving for E, m, K and p
Find the kinetic energy, rest energy and total energy of a 1g particlewith speed of 0.8c.
Solving for E, m, K and p
!u=
1
1" u2
c2
=1
1" 0.80( )2
=5
3#!
u"1=
5
3"1=
2
3
E 0 = mc2= 1 .0 $ 10
"3 kg( ) 3 .0 $ 10
8 m / s( )
2
= 9.0 $ 1013
J
The rest energy is E0 = mc2 = 9.0 $ 1013 J.
The kinetic energy is K = ! p "1( )E 0
# K =2
3"1
%&'
()*
9.0 $1013
J( ) = 3.0 $ 1013
J
The total energy is E = E0 + K = 9.0 $ 1013 J + 3.0 $ 1013 J = 1.2 $ 1014 J.
At what speed is a particle’s kinetic energy twice its rest energy?
Solving for E, m, K and p
E = umc2 = E0 + K. For K = 2E0,
umc2 = E0 + 2E0 = 3mc2
⇒γ u =
1
1− u2 c2= 3⇒1− u2
c2 =19⇒ u =
83
c = 0.943c
What is the speed and momentum pc (MeV) of an electron (rest mass 0.511 MeV) with a total energy of 12 MeV?
!u=
E
mc2=
12 MeV
0.511 MeV= 23.483
u
c= 1"
1
!u
2= 1"
1
23.48( )2= 0.9991
pc = E2" mc
2( )2
= 12 MeV( )2" 0.511 MeV( )
2= 11.99MeV
Solving for E, m, K and p
Conservation of mass-energy
The sum of the relativistic total energies of a system of particles beforeinteraction is equal to the sum of the relativistic total energies of thesystem of particles after the interaction. The total relativistic energy of theith particle in the system is given by
Ei= !
ui
mic
2=
mic
2
1" ui
2c
2
Conservation of mass-energy
Consider an inelastic collision of two particles of equal mass approaching eachother with speed v in the lab frame S.
Applying the conservation of mass-energy:
Before collision After collision
v v1 2 1 2V=0
!
u1
mc2+ !
u2
mc2= !
uf
Mc2
M =2m
1! u2
c2
= 2m +2K
c2
Note that total mass is not conserved. Kinetic energy has been convertedto mass.
Conservation of mass-energy
Nuclear Fission: The decay of a heavy nucleus into several lighter nuclei.Example:
By conservation of mass-energy:
Nuclear fission is the basis of nuclear reactors as well as nuclear bombs.
236
U !144
Ba +89
Kr + 3n
mU
c2=
mBa
c2
1! uBa
2c
2
+m
Krc
2
1! uKr
2c
2
+
mn
1
c2
1! un
1
2c
2
+
mn
2
c2
1! un
2
2c
2
+
mn
3
c2
1! un
3
2c
2
The rest mass of the uranium nucleus is greater than the sum of the masses of thedaughter products. The extra mass is converted to kinetic energy causing theproducts of the fission to be ejected at extremely high velocities.
Fusion: Collision of nucleii in which the kinetic energy is used to overcome therepulsion of protons to form a heavier nucleus. In this case the total incomingenergy must equal the rest energy + the binding energy:
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )
2
2
2 22 2
2 22 2
2 2 222
2
4702 MeV 940 MeV 4795 MeV
169 MeV 140 MeV 219 MeV
4795 MeV + 219 MeV = 5014 MeV
5014 MeV 4871 MeV 1189 MeV
where 4702 MeV +169 MeV = 4871 MeV
n n n
n
n
E pc mc
E pc mc
E E E
mc E pc
pc pc pc
K E mc
! !!
!
!
"
" ""
"
" ""
= + = + =
= + = + =
= + =
= # = # =
= + =
= # 5014 MeV 1189 MeV = 3825 MeV= #
A Σ particle decays into a neutron (pc = 4702 MeV) and pion (pc = 169MeV). Find the total rest mass and kinetic energy of the Σ particle.
Conservation of mass-energy
The reaction that powers the sun consists of the fusion of four protons intoa helium nucleus. How much energy is released in each fusion? (Mass of ahelium nucleus = 6.64 x 10-27kg).
Conservation of mass-energy
4mp – mHe = 4(1.67 × 10 27 kg) – 6.64 × 10 27 kg = 0.04 × 10 27 kg
Mass lost in the fusion process:
Energy released:
E = mc2 = (0.04 × 10 27 kg) (3.0 × 108 m/s)2 = 3.6 × 10 12 J
General Relativity
Postulates:The laws of physics are the same in all frames (including acceleratingframes).
Principle of equivalence: A gravitational field in a region around a pointis equivalent to an accelerating reference frame.
Consequences:• Curvature of spacetime• Explanation of gravity• Gravitational time dilation, gravitational red shift