Relativistic Energy and Momentum in Particle Physics
Recapitulate
New definition of Momentum
New concept of Total Relativistic Energy
Use of Momentum Energy Conservation
Zero rest mass particle, having finite
energy and momentum but moving with
speed
c
.
New Definitions
Useful relationships
Example 9: Inelastic Collision
A particle of rest mass
m
o
and kinetic
energy
6m
o
c
2
strikes and sticks to an
identical particle at rest. what is the rest
mass and speed of the resultant particle?
Initial Momentum
Initial Energy
After the collision
There exists a single particle with
E
=
8m
o
c
2
and
p
=
. We, therefore, get
Find speed
Alternately
Momentum Energy Transformation
Example 10: Decay into photons
The speed of a particle of rest mass
135
MeV / c
2
is found to be
0.8
c in a frame
S’
.
The frame
S’
is observed to move with a
speed of
0.6
c in another frame
S
. All the
motions are collinear and are taken to be
along
x
-axis. In
S’
, it is found that the
particle decays into two photons. Each of
the photon makes an angle
’
with the initial
direction of the particle.
Find the angle
’
and the frequency of the
photon in this frame. Find the angle
and
frequency in
S
frame.
θ
’
In
S’
Initial particle momentum and energy
If
E’
p1
and
E’
p2
are the energies of the two
photons,
This gives
In
S
: Method1
(particle energy transformation)
Now we can apply similar conservation laws
in
S
and get the following results.
Method2
(photon energy
transformation)
For Photon 1 in
S’
frame.
Transformation to
S
frame
C
-Frame and
L
-Frame
Centre of Mass Frame (
C
-Frame) is
defined as a frame in which the sum of
momenta of all the particles are zero.
In
STR
, problems can be simplified in
C
-
Frame
Many
HEP
experiments are planned in
C
-
Frame so that energy is not consumed in
accelerating c.m.
Example
11:
Decay of a single particle
A particle of rest mass M
o
and total
energy 3M
o
c
2
, decays into two identical
particles of rest mass m
o
. Find the
velocities of the two particles, given that
the decay products move long the
direction of motion of the parent
particle.
In
L
Frame
L
Frame Conservation Laws
Conservation of Energy and Momentum
In addition
It is not easy to solve these equations.
Go to
C
Frame
As there only a single particle initially,
we shall have following in
C
frame.
Or transform to a frame, with
Conservation Laws in
C
Frame
Conservation of energy and momentum
The two particles will move in opposite
direction with the following speed.
These speeds can now be transformed back
to
L
frame.
Example12:
Electron Collision
An electron of total energy
1.4
MeV
collides with another electron, which is at
rest in the laboratory (
L
) frame. After
collision the target electron scatters at an
angle of
45
o
in the center of mass (
C
)
frame.
Find the energy and the momentum
components of the target electron after
the scatter in the center of mass and
laboratory frames. (Rest mass energy of
electron
0.51
MeV)
Issues
Energies known in
L
frame but angles
given in
C
frame.
Find Energy in
C
frame of the two
particles.
Let
C
frame travel in
L
frame with a
speed
v
.
Find momentum and Energy in
C
frame.
Energy and Momentum in
L
Frame
Total initial energy of the two electrons
The momentum of the first electron
Method 1
For C frame
This gives
Now evaluate
and then Energy.
Method 2:
The length of four vector
In
C
Frame
The length of four vector is same but the
total momentum is zero. Hence we can
evaluate the total energy.
In
C
Frame
The energy of the two electrons and the
magnitude of the momenta are same. We
therefore get the following.
The magnitude of x and y component of the
momenta are given by
Relative speed of
C
frame in
L
frame
First Electron in
L
frame
Second Electron in
L
frame
Check Conservation in
L
Frame
After the collision the sum of energy and
momenta components
This matches with the initial values within the
truncation errors.
Energy considerations
In which frame it takes lesser energy to
perform such an experiment?
In
L-
Frame the energy is 1.91 MeV
In
C-
Frame the energy is 1.396 MeV
Example13:
Minimum Energy for a process
A proton (
p
) with kinetic energy
K
is
incident on another proton at rest in
L
frame. After the interaction four
particles are found, three proton and
one anti proton. Find the least value of
K
needed for this process to happen.
C
-Frame
The least energy required would be
when all the four particles are created
at rest. But that is possible only when
initial momentum is zero, which is in
C
-
frame. The energy in
C
-frame is given
by following.
The square of length of four vector in this
frame is , therefore
Length of the four vector in
L
-Frame
The square of length of four vector in
L
frame is given by the following
Equating the two
This is the least K needed in L frame.
C
-Frame
This length is same in all frames. For least
value of
K
the absolute value of length
should be least. But in C-frame we need
at least an energy of
4m
o
c
2
to create
four particles of rest mass
m
o
. Any
additional kinetic energy given to these
particles will lead an increase in the
absolute value of length. Remember sum
of momenta is zero in C-frame.
Discuss the new definition of momentum and the concept of total relativistic energy. Explore the use of momentum-energy conservation in scenarios involving zero rest mass particles. An example of an inelastic collision is provided to demonstrate the application of these principles.
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Presentation Transcript
Recapitulate New definition of Momentum New concept of Total Relativistic Energy Use of Momentum Energy Conservation Zero rest mass particle, having finite energy and momentum but moving with speed c.
New Definitions = = p E mu mc dp dt 2 = F m u = m o 2 2 1 / c
Useful relationships 2 2 2 2 4 E = = = p c + m c 0 2 2 K mc m c o + 2 ( 1) m c u o 2 2 2 2 p c K 2 m c K o
Example 9: Inelastic Collision A particle of rest mass mo and kinetic energy 6moc2 strikes and sticks to an identical particle at rest. what is the rest mass and speed of the resultant particle?
Initial Energy = + + = 2 2 2 2 E m c 6 m c m c 8 m c o o o o Initial Momentum ( ) 2 = = 2 2 p 7 m c m c 48 m c o o o = + 2 2 2 2 4 E p c m c o
After the collision There exists a single particle with E=8moc2 and p= . We, therefore, get 48 o m c = = + 2 4 2 4 2 4 64 m c 48 4 m c m M c o o o M o o = + 2 2 2 2 4 E p c m c o
Find speed = = = = = 2 2 2 E M c 8 m c 2 M c u o o o 2 u p M u u o = 48 m c 8 m u o o 3 = u c 2
Alternately 1 = 2 2 u c 1 2 2 u c 1 4 = 1 2 3 = u c 2
Momentum Energy Transformation x p p p E c p p p x 0 1 0 0 0 0 i y y 0 0 i 0 0 = 1 0 z z E c i i vE c x y z = = = p p ; p p p ; p x y z 2 ( ) = E E vp x
Example 10: Decay into photons The speed of a particle of rest mass 135 MeV / c2 is found to be 0.8 c in a frame S . The frame S is observed to move with a speed of 0.6 c in another frame S. All the motions are collinear and are taken to be along x-axis. In S , it is found that the particle decays into two photons. Each of the photon makes an angle with the initial direction of the particle.
Find the angle and the frequency of the photon in this frame. Find the angle and frequency in S frame. E p1 S S y y E x O O E p2 x z z
In S Initial particle momentum and energy 1 5 3 = = u 1 (0.8) 2 5 3 = = = 2 E m c 135 225 MeV u o 5 3 135 c MeV c = = = p m u 0.8 c 180 u o 2
If Ep1and Ep2 are the energies of the two photons, p p + = E E E 225 E c E c MeV 1 2 p MeV c p 1 2 + = cos cos 180 c p p E 1 2 = sin sin c
This gives p p = = E E 112.5 MeV 1 2 180 225 o = cos = 36.87
In S: Method1(particle energy transformation) vE c = + p p 2 MeV c ( ) = + 0.6 225 1.25 180 MeV c = 393.75
( ) = + E E vp ( ) = = + 0.6 180 1.25 416.25 225 MeV MeV Now we can apply similar conservation laws in S and get the following results. 416.25 2 393.75 cos ; 416.25 = = = E E MeV 208.125 MeV p 1 p 2 = = o 18.92
Method2 (photon energy transformation) For Photon 1 in S frame. p = E 112.5 MeV 1
p E p x 1 = p cos 1 c 180 225 MeV c MeV c = = 112.5 90 p E p y 1 = p sin 1 c MeV c MeV c = = 112.5 0.6 67.5
Transformation to S frame p vE c p x 1 = + p p p x 1 1 2 MeV c ( ) = + 0.6 112.5 1.25 90 MeV c = 196.875
MeV c p y = = p p 67.5 p y 1 1 ( ) 1 = + E E vp 1 1.25 208.125 1 112.5 MeV x ( ) = = + 0.6 90 MeV 67.5 196.875 18.92 = tan = o
C-Frame and L-Frame Centre of Mass Frame (C-Frame) is defined as a frame in which the sum of momenta of all the particles are zero. In STR, problems can be simplified in C- Frame Many HEP experiments are planned in C- Frame so that energy is not consumed in accelerating c.m.
Example 11: Decay of a single particle A particle of rest mass Mo and total energy 3Moc2, decays into two identical particles of rest mass mo. Find the velocities of the two particles, given that the decay products move long the direction of motion of the parent particle.
In L Frame = = 2 E 3 3 M c o 2 1 = L 2 8 = 3 8 = = p 3 M c 8 M c L o o 3
L Frame Conservation Laws Conservation of Energy and Momentum 2 3 o M c M c In addition 2 1 2 2 L E = = + E E 1 L 2 L = + 8 p p o 1 L 2 L = + 2 2 2 4 E p c m c + L 1 L o 2 2 2 4 p c m c 2 L o It is not easy to solve these equations.
Go to C Frame As there only a single particle initially, we shall have following in C frame. = = 2 p 0; E M c x c o
Or transform to a frame, with 8 = = 3, v c 3 2 3 M c c 8 = = p 3 8 M c c 0 o x o 2 3 8 = = 2 2 E 3 3 M c c 8 M c M c c o o o 3
Conservation Laws in C Frame Conservation of energy and momentum = 2 2 M c 2 m c o C o M m = o C 2 o
The two particles will move in opposite direction with the following speed. 2 4 M m 1 = = v 1 c 1 c o 2 2 C o These speeds can now be transformed back to L frame.
Example12: Electron Collision An electron of total energy 1.4 MeV collides with another electron, which is at rest in the laboratory (L) frame. After collision the target electron scatters at an angle of 45o in the center of mass (C) frame.
Find the energy and the momentum components of the target electron after the scatter in the center of mass and laboratory frames. (Rest mass energy of electron 0.51 MeV)
Issues Energies known in L frame but angles given in C frame. Find Energy in C frame of the two particles. Let C frame travel in L frame with a speed v. Find momentum and Energy in C frame.
Energy and Momentum in L Frame Total initial energy of the two electrons = 1.4 .51 1.91 + = E MeV L The momentum of the first electron MeV c = = 2 2 p (1.4) (0.51) 1.304 L = + 2 2 2 2 4 E p c m c o
Method 1 MeV c v 1.4 c MeV 1 = p 1.304 x 2 v 0.51 c MeV 2 = p 0 x 2
For C frame 1 2 + = p p 0 x x This gives MeV c v v 1.4 c MeV 1.304 2 0.51 c MeV = + = 0 0 2 v .683 c Now evaluate and then Energy.
Method 2: The length of four vector ( ) 2 E c ( ) 2 Li p Li 2 = 2 2 (1.304) (1.91) 2 MeV c = 1.9482 MeV c = 1.396 i
In C Frame The length of four vector is same but the total momentum is zero. Hence we can evaluate the total energy.
( ) 2 E c Ci 2 MeV c = 1.396 i ( ) 2 2 E c MeV c Ci = 1.949 2
In C Frame The energy of the two electrons and the magnitude of the momenta are same. We therefore get the following. 1.396 2 = = E 0.698 MeV C MeV c = = 2 2 p (0.698) (0.51) 0.477 C
The magnitude of x and y component of the momenta are given by 0.477 MeV c = = = p p 0.337 Cx Cy 2
Relative speed of C frame in L frame 0.698 0.51 = = 1.369 2 1 1.873 1 1.873 = = = 2 0.466 2 = v 0.683 c
First Electron in L frame = + 0.683 0.698) p 1.369 (0.337 MeV c MeV 1 Lx = 1.114 = p 0.337 1 Ly c = + 0.683 0.337) E = 1.369 (0.698 MeV 1 1.27 L
Second Electron in L frame = + 0.683 0.698) p 1.369 ( 0.337 MeV c MeV = 2 Lx = 0.191 p 0.337 2 Ly c = 0.683 0.337) E = 1.369 (0.698 MeV 2 0.64 L
Check Conservation in L Frame After the collision the sum of energy and momenta components = + = E 1.27 0.64 1.91 MeV L MeV c = + 0.191 1.305 = p 1.114 Lx = = p 0.337 0.337 0 Ly This matches with the initial values within the truncation errors.
Energy considerations In which frame it takes lesser energy to perform such an experiment? In L-Frame the energy is 1.91 MeV In C-Frame the energy is 1.396 MeV
Example13: Minimum Energy for a process A proton (p) with kinetic energy K is incident on another proton at rest in L frame. After the interaction four particles are found, three proton and one anti proton. Find the least value of K needed for this process to happen.
C-Frame The least energy required would be when all the four particles are created at rest. But that is possible only when initial momentum is zero, which is in C- frame. The energy in C-frame is given by following.
= 2 E 4 m c C o The square of length of four vector in this frame is , therefore 2 2 16 m c o
Length of the four vector in L-Frame The square of length of four vector in L frame is given by the following E p c 2 2 2 ( ) 2 + 2 K 2 c m c 2 K c o = + 2 m K o 2 2 = 2 2 2 m K 4 m c o o
Equating the two = 2 2 2 2 16 = m c 2 m K 4 m c o m c o o 2 K 6 o This is the least K needed in L frame.
