Transcript Slide 1
Relativistic Classical Mechanics
XIX century crisis in physics: some facts
•
Maxwell : equations of electromagnetism are not invariant under Galilean transformations
•
Michelson and Morley : the speed of light is the same in all inertial systems
James Clerk Maxwell (1831-1879) Albert Abraham Michelson (1852 – 1931) Edward Williams Morley (1838 – 1923)
Postulates of the special theory
•
1) The laws of physics observers are the same to all inertial
•
2) The speed of light observers is the same to all inertial
•
Formulation of physics that explicitly incorporates these two postulates is called covariant
•
The space and time comprise a single entity: spacetime
•
A point in spacetime is called event
•
Metric of spacetime is non-Euclidean
7.1
Tensors
•
Tensor of rank n is a collection of elements grouped through a set of n indices
A n
..
• • • •
Scalar Vector Matrix Etc.
is a tensor of rank 0 is a tensor of rank 1 is a tensor of rank 2
A A A i ij
• •
Tensor product of two tensors of ranks m and n is a tensor of rank (m + n)
A n B
m
C
n
m
..
Sum over a coincidental index in a tensor product of two tensors of ranks m and n is a tensor of rank (m +
+ n
– 2)
j A
n B
m
D ik n
..
m
2 ..
7.5
Tensors
•
Tensor product of two vectors is a matrix
A i B j
C ij
7.5
•
Sum over a coincidental index in a tensor product of two tensors of ranks 1 and 1 (two vectors) is a tensor
•
of rank 1 + 1 – 2 = 0 (scalar): scalar product of two vectors
A i B i
D i
Sum over a coincidental index in a tensor product of
•
two tensors of ranks 2 and 1 (a matrix and a vector) is a tensor of rank 2 + 1 – 2 = 1 (vector)
j A ij B j
D i
Sum over a coincidental index in a tensor product of two tensors of ranks 2 and 2 (two matrices) is a tensor of rank 2 + 2 – 2 = 2 (matrix)
j A ij B jk
D ik
Metrics, covariant and contravariant vectors
7.4
7.5
•
Vectors, which describe physical quantities, are called contravariant vectors superscripts and are marked with instead of a subscripts
A i
•
For a given space of dimension N, we introduce a concept of a metric – N x N matrix uniquely defining the symmetry of the space (marked with subscripts )
g ij
•
Sum over a coincidental index in a product of a
•
metric and a contravariant vecor is a covariant vector or a 1-form (marked with subscripts )
j g ij A j
A i
Magnitude : square root of the scalar product of a contravariant vector and its covariant counterpart
3D Euclidian Cartesian coordinates
7.4
7.5
•
Contravariant
•
Metric
g
1 0 0
infinitesimal coordinate vector:
0 0
dr i
1 0 0 1
dx dy dz
•
Covariant
1 0 0 0 1 0
infinitesimal coordinate vector:
0 0 1
dx dy dz
dx dy dz
dr i
dr i dr i
j
3 1
g ij dr j
•
Magnitude :
i
3 1
dr i dr i
(
dx
) 2 (
dy
) 2 (
dz
) 2
1 0 0
3D Euclidian spherical coordinates
7.4
7.5
•
Contravariant
•
Metric
g
1 0 0
infinitesimal coordinate vector:
r
0 0 2
r
2 0 0 sin 2
dr i
dr d
d
•
r
0 0 2
Covariant
r
2 0 0 sin 2
infinitesimal coordinate vector:
d dr d
r
2
r
2 sin
dr d
2
d
dr i
dr dr i i
j
3 1
g ij dr j
•
Magnitude :
i
3 1
dr i dr i
dr
2
r
2
d
2
r
2 sin 2
d
2
Hilbert space of quantum-mechanical wavefunctions
7.4
7.5
•
Contravariant vector (ket):
•
Covariant vector (bra):
•
Magnitude :
David Hilbert (1862 – 1943) •
Metric :
• •
Contravariant Metric
g
1 0 0 0
4D spacetime
7.4
7.5
infinitesimal coordinate 4-vector :
0 0
dx
0 0 1 0 0 1 0 0 0 1
cdt
dx dy dz
•
Covariant infinitesimal coordinate vector:
dx
3 0
g
dx
1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1
cdt
dx dy dz
cdt
dx dy dz
•
Magnitude : 4D spacetime
3 0
dx
dx
c
2
dt
2
dx
2
dy
2
dz
2 7.4
7.5
•
This magnitude is called differential interval
ds
•
Interval (magnitude of a 4-vector connecting two events in spacetime):
s
c
2
t
2
x
2
y
2
z
2 •
Interval should be the same frames in all inertial reference
•
The simplest set of transformations that preserve the invariance of the interval relative to a transition from one inertial reference frame to another: Lorentz transformations
7.2
Lorentz transformations
•
We consider two inertial reference frames S and S ’; relative velocity as measured in S is v :
•
Then Lorentz transformations are:
ct
' (
ct
r
);
r v
c
; '
r
( 1 1
r
) 2 2 ( 1 )
ct
•
Lorentz transformations can be written in a matrix form
x
' 3 0
L
x
Hendrik Antoon Lorentz (1853 – 1928)
7.2
Lorentz transformations
x
' 3 0
L
x
L
x y z
1 (
x
1 )
v x v
2 ( 1 )
v x v y v
2 ( 1 )
v x v z v
2
y
( 1 ( 1 )
v x v y v
2 1 )
v y v
2 ( 1 )
v y v z v
2 ( ( 1 ( 1 ) 1 )
v
1 )
z v v x v v
2
v y v z
2
v z z
2
7.2
Lorentz transformations L
•
If the reference frame S of the reference frame S:
( , 0 , 0 ) 0 0 • 0 0 0 0 1 0 0 0 0 1
ct
'
x
'
y
'
z
'
y
(
ct
(
x
z t
' 2
t
' 1
x ct
) ) (
t
2
t
' 1
t
' 2
t
1
c
(
t
1 (
t
2
c
c
(
x
2
If two events happen at the same location in S:
x
2
x
1 )
x
2 )
x
1 ))
x
1
t
' 2
t
' 1 (
t
2
t
1 ) •
Time dilation
v c
0 ; 1 ;
t
'
t
;
x
'
x
ct
x
vct c
x
vt
7.2
Lorentz transformations L
•
If the reference frame S of the reference frame S:
( , 0 , 0 ) 0 0 • 0 0 0 0 1 0 0 0 0 1
ct
'
x
'
y
'
z
'
y
(
ct
(
x
x
)
ct
)
z x
' 2
x
' 1 (
x
2
x
' 1
x
' 2 (
x
1 (
x
2
ct
1 )
ct
2 )
x
1
If two events happen at the same time in S:
c
(
t t
2 2
t
1 ))
t
1
x
' 2
x
' 1 (
x
2
x
1 ) •
Length contraction
Velocity addition
•
If the reference frame S ‘ moves parallel to the x axis of the reference frame S: L
S
S
' 0 0 0 0 0 0 1 0 0 0 0 1 •
If the reference frame S ‘‘ moves parallel to the x axis of the reference frame S ‘: L
S
'
S
'' 0 0 ' ' ' 0 0 ' ' ' 0 0 1 0 0 0 0 1 7.3
7.3
Velocity addition L
S
• •
frame S to the reference frame S ‘‘:
S
''
L
S
'
S
''
L
S
S
' ' ( ' 1 ( 0 0 ' ) ' )
L
S
S
''
The Lorentz transformation from the reference On the other hand:
0 0 '' '' '' 0 0 '' '' '' 0 0 1 0 0 0 0 1 ' ( 1 ' ' ' ( 1 ' ) ' ' ' ' ' ( ' ) ' ( 0 0 ' ' 1 ' ) ' ) ' ' 0 0 1 0 0 0 0 1
Four-velocity
•
Proper time is time measured in the system where the clock is at rest
•
For an object moving relative to a laboratory
u
0
u
1
system, we define a contravariant vector of four velocity :
d
(
ct
)
d
dx d
u
2
v dx d
d u
(
c d
)
dx
d
y
;
u
3
d v dx
(
z
c
)
dx dt
v x
v v
v c x y z
7.4
7.4
Four-velocity
•
Magnitude of four-velocity
3 0
u
u
g
u
u
(
c
) 2 (
v x
) 2 (
v y
) 2 (
v z
) 2
c
1 (
v x
) 2 (
v y
) 2 (
v z
) 2
c
2
c
1
v
2
c
2
c
1
c
1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
v v
v c x y z
c
v v v x y z
3 0
u
u
c
2 1 1 2 ;
v c
7.1
Minkowski spacetime
ct
' •
Lorentz transformations for parallel axes:
(
ct
x
);
x
' (
x
ct
)
t
y
'
y
;
z
'
z
•
How do x ’ and t’ axes look in the x and t axes?
t’
Hermann Minkowski (1864 - 1909)
x’
ct
•
t
’ axis:
t
x
c
v x
' 0
x
•
x
’ axis:
t
' 0
x
0
x
t
x c
vx c
2
ct
0
x
Minkowski spacetime
•
When
v
c
1
t
•
How do x ’ and t’ axes look in the x and t axes?
•
t
’ axis:
x
ct t
x
' 0
x c
•
x
’ axis:
ct
x ct
x
0
t
' 0
t
x c x
ct
0
x
7.1
7.1
Minkowski spacetime
• •
Let us synchronize the clocks of the S and S ’ frames at the origin
t t’
Let us consider an event
•
In the S frame, the event is to the right of the origin
x’ x
•
In the S ‘ frame, the event is to the left of the origin
7.1
Minkowski spacetime
• •
Let us synchronize the clocks of the S and S ’ frames at the origin
t t’
Let us consider an event
•
In the S frame, the event is after the synchronization
x’ x
•
In the S ‘ frame, the event is before the synchronization
Minkowski spacetime
(
s
) 2 0 7.1
(
s
) 2 0
7.4
Four-momentum
p
p
0 •
For an object moving relative to a laboratory system, we define a contravariant vector of four-
momentum :
mu
p
1
mu
1
m
v x
;
mu
0
m
c p
2
m
v y
; •
Magnitude of four-momentum
3 0
p
3
m
v z p
p
m m m
v m
v v c x y z
g
p
p
(
m
c
) 2 (
m
v x
) 2 (
m
v y
) 2 (
m
v z
) 2
mc
1
v
2
c
2
mc
3 0
p
p
m
2
c
2
3 0
p
p
m
2
c
2
Four-momentum
(
m
c
) 2 (
m
v x
) 2 (
m
v y
) 2 (
m
v z
) 2 •
Rest-mass : mass measured in the system where the object is at rest
m
7.4
( 3 0
p
•
p
For a moving object:
(
c
) 2 (
v
) 2 (
mc
) 2
m
(
c
) 2 (
mc
) 2
c
2 ) 2 • • (
mc
2 ) 2 (
v
) 2
c
2 (
mc
2 ) 2 (
p
) 2
c
2
The equation has units of energy squared
E
2 (
mc
2 ) 2 (
p
) 2
c
2
If the object is at rest
p
0
E
0 2 (
mc
2 ) 2
E
0
E
mc
2 ( ~
c
2
v
) 2
Four-momentum
~
c v x
~ ~
E v v y z
E
0
mc
2 7.4
E
0
mc
2
Four-momentum
•
Rest-mass energy : energy of a free object at rest – an essentially relativistic result
E E
c
2
m
c
2
mc
2 ( 1 2 ) 1 / 2 •
mc
2 •
For slow objects:
( 1 2 ) 1 / 2
mc mv
2 2 2 1
E
0 2 2
E
0
mc
2 1
v
2
c
2 2
E
0
mv
For free relativistic objects, we introduce therefore
2 2
the kinetic energy as
T
E
E
0
T
E
mc
2
mc
2
mc
2 (
mc
2 ) 2 ( ~
p
) 2
c
2
mc
2 7.4
Non-covariant Lagrangian formulation of relativistic mechanics
7.9
d dt
•
As a starting point, we will try to find a non covariant Lagrangian formulation (the time variable is still separate)
•
The equations of motion should look like
mv i
1 2
mv i
1 2
dV dx i
v i mv i
mc
2 1 1 2 2
L
v i L
T
V
;
V
x i
L
x i L
mc
2
i i dp
dt
1 , 2 , 3
V
x i
1 2
V
Non-covariant Lagrangian formulation of relativistic mechanics
7.9
p i
d dt
• •
For an electromagnetic potential, the Lagrangian is similar
L
mc
2 1 2
q
q i
3 1
A i v i
•
The equations of motion should look like
mv i
1 2
qA i d dt
mv i
1 2
qA i
q
x i i
dp i dt
i
3 1
L
x i
1 , 2 , 3
v i
A i
x i
mv i
1 2
q
A
t
q v
(
A
)
q
(
E
v
B
)
Non-covariant Lagrangian formulation of relativistic mechanics
7.9
d dt
•
Example: 1D relativistic motion in a linear potential
L
mc c
2 2
xma
;
a
const
•
The equations of motion:
mc x
c
2
x
2
ma x
x
0
c a
x
c
2
x
2
at
c c
2 (
at
) 2
x
c
2 2
atc
c c
2 (
at
) 2 •
Acceleration is hyperbolic , not parabolic
Useful results
1 1 2 1 1 (
v c
2 ) 2 (
c
2 (
v
) 2 ) 2
c
2 2 1 1 (
v
) 2
c
2 (
v
) 2
c
2
c
2 2 2 (
v
) 2 2
c
2
c
2
c
2 (
v
) 2
c
2
7.9
Non-covariant Hamiltonian formulation
8.4
of relativistic mechanics
•
We start with a non-covariant Lagrangian:
L
mc
2 1 2
V
•
m
c
2
Applying a standard procedure
H
3
c i
2 2 1
m
v
i mc
v i
2
mc
2
V
H
1
m
c
2
T
2
V E
0
V
T
V
m
i
3 ( 1
v
)
p
2
i v i
mc
2
V
L mc
2
T
V E
0
V
(
v
) 2 •
Hamiltonian equals the total energy of the object
c
2
c
2 2
T
mc
2
mc
2
7.9
Non-covariant Hamiltonian formulation
8.4
of relativistic mechanics
•
We have to express the Hamiltonian as a function of momenta and coordinates:
H
m
c
2
V
mc
2 (
v
) 2
c
2
V
c
2 (
m v
) 2
m
2
c
4
V
c
2 (
p
) 2
E
0 2
V
c
2 (
v
) 2
c
2
H
c
2 (
p
) 2
E
0 2
V p
m m m
v m
v v c x y z
More on symmetries
•
Full time derivative of a Lagrangian:
dL dt
L t
M m
1
q L m q
m
M m
1
q
L m
•
Form the Euler-Lagrange equations:
q
m
L
t
M
m
1
d dt
L
q
m
q
m
m M
1
L
q
m
q
m
L
t
d dt m M
1
L
m q
m
L
t
d dt
m M
1
L
q
m
•
If
L
t q
m
L
dH dt
0
H
M m
1
L
q
m q
m
L
const
7.9
Non-covariant Hamiltonian formulation
8.4
of relativistic mechanics
L
•
Example: 1D relativistic harmonic oscillator
mc c
2
x
2
kx
2 2
H
m
c
2
V
c mc
3 2
x
2
L
t
•
The Lagrangian is not an explicit function of time
0
H
E tot
const c mc
3 2
x
2
kx
2 2
E tot
kx
2 2
x
2
c
2 2
E
2
mc
3
tot
kx
2 2
t
t
0 ( 2
cE
( 2
E tot tot
kx
2
kcx
2 ) 2 )
dx
( 2
mc
3 ) 2 •
The quadrature involves elliptic integrals
Covariant Lagrangian formulation of relativistic mechanics: plan A
7.10
•
So far, our canonical formulations were not Lorentz invariant – all the relationships were derived in a specific inertial reference frame
•
We have to incorporate the time variable as one of the coordinates of the spacetime
•
We need to introduce an invariant parameter ,
•
describing the progress of the system in configuration space:
Then
x
'
dx
d
; 0 , 1 , 2 , 3
I
1 2 (
x
,
x
' , )
d
7.10
Covariant Lagrangian formulation of
•
I
relativistic mechanics: plan A
1 2 (
x
,
x
' , )
d
Equations of motion
d d
x
'
x
•
We need to find Lagrangians producing equations of motion for the observable behavior
t
•
First approach: use previously found Lagrangians
x
and replace time and velocities according to the rule:
dx i c
0
i dx dt
i
1 , 2 ,
dx d
3
i d
dt
d d
dt
d d
x
i
'
x c
0
c x
'
i x
' 0
7.10
Covariant Lagrangian formulation of
dx i dt
• •
c
x
'
i x
' 0
Then
2 1 ;
relativistic mechanics: plan A
i
1 , 2 , 3
L
I
x i
, 2
t t
1
L
(
x
0
c
,
x i c
,
t
,
x
'
i x
' 0
x
i
)
dt
1
c
So, we can assume that
dx
0
d
( 1 2
L
d
x
,
x i x
' 1 , 2 )
x
0
c x
' 0
c
,
c L
x
' 0
c x
'
i x
' 0
x i L
,
d x
0
c x i
,
x
0
c
,
c x
0
c t
x
'
i x
' 0 ,
c
x
0
d
x
'
i x
' 0
c
•
Attention: regardless of the functional dependence, the new Lagrangian is a homogeneous function of the generalized velocities in the first degree:
(
x
,
ax
' )
a
(
x
,
x
' )
Covariant Lagrangian formulation of relativistic mechanics: plan A
(
x
,
ax
' )
a
(
x
,
x
' ) 7.10
• •
From Euler’s theorem on homogeneous functions it follows that
3 0
x
'
x
'
Let us consider the following sum
3 0
x
'
x
3 0
x
'
x
3 0
x
'
x
' , 3 0
x
'
x
x
'
x
'
x
'
x
' 2
x
x
' 0 , 3 0
x
'
x
' 2
x
x
' , 3 0
x
'
x
" 2
x
'
x
'
7.10
Covariant Lagrangian formulation of
, 3 0 3 0
x
'
x
'
relativistic mechanics: plan A
x
, 3 0
x
'
x
'
x
2
x
' , 3 0
x
"
x
'
x
' 2
x
'
x
' 2
x
x
'
x
"
x
' 2
x
'
x
' 3 0
x
' 3 0
x
x
'
x
'
x
'
x
'
x
" 3 0
x
'
d d
x
' 3 0
d d
x
'
x
x
' 0
d d
x
'
x
•
If three out of four equations of motion are satisfied, the fourth one is satisfied automatically
7.10
Example: a free particle
L
•
We start with a non-covariant Lagrangian
2 2
mc
2 1 2
mc
2 1
x c
c
z
c
2
mc
2
x
' 1 1
c x c
' 0 2
x
' 2
c x c
' 0 2
x
' 3
c
x c
' 0 2
mc
2
dx i
c i dt
1 , 2 , 3
x
'
x
' 0
i
1
i
3 1
x x
' 0
i
' 2
x c
' 0
L
x i
, (
x
,
x
' )
x
0
c
,
c x
' 0
c x x
'
i
' 0
L
x i
,
x
0
c
mc
2 ,
c x
'
i x
' 0
x
' 0
c
1
i
3 1
x x
' '
i
0 2
mc
2
i
3 1 2
Example: a free particle
x
' 0
c L
x i
,
x
0
c x
'
i
,
c x
' 0
mc
2
i
3 1 2
mc
3 0
x
'
x
' 7.10
d d
mc
3 0
x
'
x
' •
x
'
Equations of motion
mc
3 0
x
'
x
'
d d
0
x
'
x
'
dx
d
dx
d
d
d
u
d
d
x
1 2
d d
d
u
x
'
d
mcx
' 3 0
x
'
x
' 0
7.10
Example: a free particle
x
' 2
u
d d
3 0
d
mcu
u
d
d
d
u
d
d
0
d d
mcu
c
0
d
(
mu
d
) 0
d d
mcu
3 0
u
u
0
dp d
0 •
Equations of motion of a free relativistic particle
d
(
m
v x d
)
d
(
m
v d
y
)
d
(
m d
v z
) 0
dE d
0 3 0
u
u
c
Covariant Lagrangian formulation of relativistic mechanics: plan B
7.10
• •
Instead of an arbitrary invariant parameter, use proper time
However
3 0
u
u
3 0
dx
d
dx
d
c
2
we can
•
Thus, components of the four-velocity are not independent: they belong to three-dimensional manifold ( hypersphere ) in a 4D space
•
Therefore, such Lagrangian formulation has an inherent constraint
•
We will impose this constraint only after obtaining the equations of motion
Covariant Lagrangian formulation of relativistic mechanics: plan B
7.10
•
In this case, the equations of motion will look like
d d
u
x
•
But now the Lagrangian does not have to be a homogeneous function to the first degree
•
Thus, we obtain freedom of choosing Lagrangians from a much broader class of functions that produce Lorentz-invariant equations of motion
•
E.g., for a free particle we could choose
d
(
mu
d
) 0 3 0
mu
u
2
Covariant Lagrangian formulation of relativistic mechanics: plan B
7.10
•
If the particle is not free, then interaction terms have to be added to the Lagrangian – these terms must generate Lorentz-invariant equations of motion
•
In general, these additional terms will represent interaction of a particle with some external field
•
The specific form of the interaction will depend on the covariant formulation of the field theory
•
Such program has been carried out for the following fields: electromagnetic, strong/weak nuclear, and a weak gravitational
Covariant Lagrangian formulation of relativistic mechanics: plan B
7.10
•
Example: 1D relativistic motion in a linear potential
•
In a specific inertial frame, the non-covariant Lagrangian was earlier shown to be
L
mc c
2
x
2
xma
;
a
const
•
The covariant form of this problem is
3 0
mu
u
2 3 0
G
x
•
In a specific inertial frame, the interaction vector will be reduced to
G
ma
1
Example: relativistic particle in an electromagnetic field
7.10
7.6
• •
For an electromagnetic field, the covariant Lagrangian has the following form:
d
(
mu
d
) 3 0
q
mu
2
The corresponding equations of motion:
A
u
3 0 3 0
qA
( (
A
x
u
x
0 , )
x
1 ,
x q
2 , 3 0
x
3 )
u
F
u
A
E
B
A
1
A
/
A
3 2
c
A
t
A F
1
c
0
E E E x y z E x
0
cB z
cB y E y
cB z
0
cB x
E cB cB x
0
z y
d
(
mu
1
d
)
q
3 0
F
1
u
Example: relativistic particle in an electromagnetic field
7.10
7.6
d
(
mu
1
d
)
q
3 0
F
1
u
d
(
m
v x
)
q
(
E x
c
cB z
v y
dt c dp dt
•
x
q c
(
E x
B z v y
B y v z
)
q c
(
E
v
B
)
x
Maxwell's equations follow from this covariant
cB y
v z
)
formulation (check with your E&M class)
F
1
c
0
E E E x y z E x
0
cB z
cB y E y
cB z
0
cB x
E cB cB x
0
z y
Covariant Lagrangian formulation of relativistic mechanics: plan B
7.10
•
What if we have many interacting particles ?
•
Complication #1 : How to find an invariant parameter describing the evolution? (If proper time, then of what object?)
•
Complication #2 : How to describe covariantly the interaction between the particles? (Information cannot propagate faster than a speed of light – action-at-a-distance is outlawed)
•
Currently, those are the areas of vigorous research
Covariant Hamiltonian formulation of relativistic mechanics: plan A
8.4
•
In ‘Plan A’, Lagrangians are homogeneous functions
of the generalized velocities in the first degree
(
x
,
ax
)
a
(
x
,
x
' ) 3 0
x
'
x
' •
Let us try to construct the Hamiltonians using canonical approach (Legendre transformation)
x
' 3 0
x
' 3 0
x
'
x
' 0 •
‘Plan A’: a bad idea !!!
Covariant Hamiltonian formulation of relativistic mechanics: plan B
8.4
p
•
In ‘Plan B’: instead of an arbitrary invariant parameter, we use proper time
u
3 0
p
u
3 0
u
u
•
We have to express four-velocities in terms of conjugate momenta and substitute these expressions into the Hamiltonian to make it a function of four coordinates and four-momenta
•
Don’t forget about the constraint:
3 0
u
u
c
2
8.4
Covariant Hamiltonian formulation of relativistic mechanics: plan B
•
For a free particle:
3 0
mu
u
2
p
u
3 0
u
u
3 0
p
u
p
mu
u
2 /
m
3 0 3
u
3 0
mg
u
u
2
p
m
p
3 0
p
3 0 2
m p
p
2
m p
3 0
mg
u
p
p
2
m
0
p
p
2
m
mu
Covariant Hamiltonian formulation of relativistic mechanics: plan B
8.4
p
u
3 0 •
u
p
p
(
For a particle in an electromagnetic field:
p m
qA
m
u
qA
) 3 0 3 3
mu
0 3 0 0
mu
( ( 2
u
2
p
p
u
qA
3 3 0 3 0
qA
p
qA
)(
p
u u
2
m
)( 0 2
m qA
p
u
qA
qA
)
mu
) 3 0
qA
qA
(
p
qA
)
7.8
Relativistic angular momentum
•
For a single particle, the relativistic angular momentum is defined as an antisymmetric tensor of rank 2 in Minkowski space:
m
x
p
x
p
m
xE yE zE
/ /
c c
/
c
0
ct
mv x ct
mv y ct
mv z ct
mv x
m
(
yv x
0
xv y
)
m
(
zv x
xE
/
c
xv z
)
ct
mv y
m
(
xv y
m
(
zv y
0
yE
/
c yv x
)
yv z
)
ct
mv m
(
m
(
z xv z yv z
0
zE zv x zv y
/ ) )
c
•
This tensor has 6 independent elements; 3 of them coincide with the components of a regular angular momentum vector in non-relativistic limit
7.8
Relativistic angular momentum
dm
d
•
Evolution of the relativistic angular momentum is determined by:
mu
u
d d
(
x
p
mu
u
x
x
p
)
dp
d
p
x
dx
d
dp
d
x
dp d
From the equations of motion
p
dm
d
x
dx
d
dp
d
x
x
dp
d
dp
d
•
For open systems, we have to define generalized relativistic torques in a covariant form
dm
d
N
Relativistic kinematics of collisions
•
The subject of relativistic collisions is of considerable interest in experimental high-energy physics
•
Let us assume that the colliding particles do not interact outside of the collision region, and are not affected by any external potentials and fields
•
We choose to work in a certain inertial reference
•
frame; in the absence of external fields, the four momentum of the system is conserved
dp
d
Conservation of a four-momentum includes conservation of a linear momentum and conservation of energy
0 7.7
Relativistic kinematics of collisions
•
Usually we know the four-momenta of the colliding particles and need to find the four-momenta of the collision products
•
There is a neat trick to deal with such problems:
•
1) Rearrange the equation for the conservation of the four-momentum of the system so that the four momentum for the particle we are not interested in stands alone on one side of the equation
•
2) Write the magnitude squared of each side of the equation using the result that the magnitude squared of a four-momentum is an invariant
7.7
Relativistic kinematics of collisions
•
Let us assume that we have two particles before the collision (A and B) and two particles after the collision (C and D)
7.7
•
Conservation of the four-momentum of the system:
(
p A
) (
p B
) (
p C
) (
p D
) •
1) Rearrange the equation (supposed we are not interested in particle D)
(
p D
) (
p A
) (
p B
) (
p C
) •
2) Magnitude squared of each side of the equation:
3 3 0 (
p A
) (
p B
) 0 ( (
p D p C
) ) (
p
(
D p
)
A
) (
p B
) (
p C
)
Relativistic kinematics of collisions
3 0 (
p A
) 3 0 (
p i
) (
p j
) (
p B
3 0 ) ( (
p D
)
p C
) ( , 3 0 (
p i
)
g
(
p D
) (
p j
)
p A
) (
p B
) 3 0 (
p i
) (
p C
) (
p j
)
i
7.7
j
3 0 (
p i
) (
p j
) 3 0 ( 3 0 (
p i
) (
p i
) (
p j
)
p i
) (
m i c
) 2 2 3 0 (
p i
) (
p j
) (
m D c
) 2 (
m A c
) 2 (
m B c
) 2 (
m C c
) 2 2 3 0 (
p A
) (
p B
) (
p A
) (
p C
) (
p B
) (
p C
)
Example: electron-positron pair annihilation
•
Annihilation of an electron and a positron produces two photons
e
e
1 2 •
Conservation of the four-momentum of the system:
(
p
) (
p
) (
p
1 ) (
p
2 ) •
Let us assume that the positron is initially at rest :
p
0 ;
E
mc
2 •
1) Rearrange the equation
(
p
2 ) (
p
) (
p
) (
p
1 )
Example: electron-positron pair
(
p
2 )
annihilation
(
p
) (
p
) (
p
1 ) •
2) Magnitude squared of each side of the equation:
3 0 (
p
) (
p
3 0 ) ( (
p
2
p
1 ) ) (
p
2 ( )
p
) (
p
) (
p
1 ) (
m
2
c
) 2 (
mc
) 2 (
mc
) 2 (
m
1
c
) 2
m
1
m
2 0 2 3 0 (
p
) (
p
) (
mc
) 2 (
p
) (
p
1 ) 3 0 (
p
) (
p
) (
p
) (
p
1 ) (
p
) (
p
1 ) (
p
) (
p
1 ) 0
Example: electron-positron pair
(
mc
) 2 3 0 (
p
) (
p
)
annihilation
(
p
) (
p
1 ) (
p
) (
p
1 ) 0 3 0 (
p
) (
p
)
E
c E
E
p
c
mc
2 ;
p
p
0
E
m
3 0 (
p
) (
p
1 )
E
c E
1
c
p
p
1
E
1
m
3 0 (
p
) (
p
1 )
E
c E
1
c
p
p
1
E
E
1
c
2
p
p
1 cos 1
Example: electron-positron pair
(
mc
) 2
E
m
E
1
m
annihilation
E
E
1
c
2
p
p
1 cos 1 0
E
1
c
p
1 (
mc
) 2
E
1
m
E
m
E
1
m
E
E
1
c
2
E c
2
p
cos
c
1
E
1
c
p
cos 1 0 (
mc
)
E
2 1
E
(
m mc
2
mc
2 (
E
mc
) 2
c
p E
) cos 1 •
The photon energy will be at a maximum when emitted in the forward direction, and at a minimum when emitted in the backward direction