Post on 10-Feb-2016
description
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 ClerkMaxwell
(1831-1879)
Edward Williams Morley
(1838 – 1923)
Albert Abraham Michelson
(1852 – 1931)
Postulates of the special theory
• 1) The laws of physics are the same to all inertial observers
• 2) The speed of light is the same to all inertial observers
• Formulation of physics that explicitly incorporates these two postulates is called covariant
• The space and time comprise a single entity: spacetieme
• 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
• Scalar is a tensor of rank 0• Vector is a tensor of rank 1• Matrix is a tensor of rank 2• Etc.
• Tensor product of two tensors of ranks m and n is a tensor of rank (m + n)
• 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)
7.5
n
ijA ..
AiAijA
mnmn
ijij CBA
........
2
........
mnmn
ikj
jijk DBA
Tensors
• Tensor product of two vectors is a matrix
• 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
• 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)
• 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)
7.5
ijji CBA
DBAi
ii
ij
jij DBA
ikj
jkij DBA
Metrics, covariant and contravariant vectors
• Vectors, which describe physical quantities, are called contravariant vectors and are marked with superscripts instead of a subscripts
• 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)
• 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)
• Magnitude: square root of the scalar product of a contravariant vector and its covariant counterpart
7.47.5
iA
ijg
ij
jij AAg
3D Euclidian Cartesian coordinates
• Contravariant infinitesimal coordinate vector:
• Metric
• Covariant infinitesimal coordinate vector:
• Magnitude:
7.47.5
dzdydx
dr i
3
1j
jiji drgdr
100010001
g
dzdydx
dzdydx
100010001
ii drdr
3
1i
iidrdr 222 )()()( dzdydx
3D Euclidian spherical coordinates
• Contravariant infinitesimal coordinate vector:
• Metric
• Covariant infinitesimal coordinate vector:
• Magnitude:
7.47.5
dddr
dr i
3
1j
jiji drgdr
22
2
sin0000001
rrg
drdrdr
dddr
rr
22
2
22
2
sinsin0000001
ii drdr
3
1i
iidrdr 222222 sin drdrdr
Hilbert space of quantum-mechanical wavefunctions
• Contravariant vector (ket):
• Covariant vector (bra):
• Magnitude:
• Metric:
7.47.5
David Hilbert(1862 – 1943)
4D spacetime
• Contravariant infinitesimal coordinate 4-vector:
• Metric
• Covariant infinitesimal coordinate vector:
7.47.5
dzdydxcdt
dx
3
0
dxgdx
1000010000100001
g
dzdydx
cdt
dzdydxcdt
1000010000100001
4D spacetime
• Magnitude:
• This magnitude is called differential interval
• Interval (magnitude of a 4-vector connecting two events in spacetime):
• Interval should be the same in all inertial reference frames
• 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
7.47.5
3
0
dxdx 22222 dzdydxdtc
ds
22222 zyxtcs
Lorentz transformations
• We consider two inertial reference frames S and S’; relative velocity as measured in S is v :
• Then Lorentz transformations are:
• Lorentz transformations can be written in a matrix form
7.2
ctrrrrctct
2
)1()(');('
Hendrik AntoonLorentz
(1853 – 1928)
211;
cv
3
0
'
xLx
Lorentz transformations7.2
3
0
'
xLx
2
22
2
2
2
22
2
)1(1)1()1(
)1()1(1)1(
)1()1()1(1
vv
vvv
vvv
vvv
vv
vvv
vvv
vvv
vv
zzyzxz
zyyyxy
zxyxxx
zyx
L
Lorentz transformations
• If the reference frame S‘ moves parallel to the x axis of the reference frame S:
• If two events happen at the same location in S:
• Time dilation
7.2
100001000000
L
)0,0,(
zzyy
ctxxxctct
''
)(')('
)('
)('
222
111
xc
tt
xc
tt
))(('' 121212 xxc
tttt
12 xx
)('' 1212 tttt
vtxc
vctxctxx
ttcv
'
;';1;0
Lorentz transformations
• If the reference frame S‘ moves parallel to the x axis of the reference frame S:
• If two events happen at the same time in S:
• Length contraction
7.2
100001000000
L
)0,0,(
zzyy
ctxxxctct
''
)(')('
)(')('
222
111
ctxxctxx
))(('' 121212 ttcxxxx
12 tt
)('' 1212 xxxx
Velocity addition
• If the reference frame S‘ moves parallel to the x axis of the reference frame S:
• If the reference frame S‘‘ moves parallel to the x axis of the reference frame S‘:
7.3
1000010000'''00'''
'''
SSL
100001000000
'
SSL
Velocity addition
• The Lorentz transformation from the reference frame S to the reference frame S‘‘:
• On the other hand:
7.3
1000010000)'1(')'('00)'(')'1('
''''''
SSSSSS LLL
1000010000''''''00''''''
''
SSL
)'(''''')'1('''
'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 system, we define a contravariant vector of four-velocity:
7.4
ddxu
dctdu )(0
dcd )(
c
ddxu 1
ddx
)(
ddx
dtdx xv
zy vuvu 32 ;
z
y
x
vvvc
Four-velocity
• Magnitude of four-velocity
7.4
z
y
x
z
y
x
vvvc
vvvc
1000010000100001
3
0
uu
3
0
3
0
uug 2222 )()()()( zyx vvvc
cv
;1
12
2
222 )()()(1c
vvvczyx
2
2
1cvc
1c c
23
0
cuu
HermannMinkowski
(1864 - 1909)
Minkowski spacetime
• Lorentz transformations for parallel axes:
• How do x’ and t’ axes look in the x and t axes?
• t’ axis:
• x’ axis:
7.1
0'x
zzyyctxxxctct
';')(');('
0 ctx
vx
cxt
x’
t’t
x0't
0 xct 2cvx
cxt
Minkowski spacetime
• When
• How do x’ and t’ axes look in the x and t axes?
• t’ axis:
• x’ axis:
7.1
0'x 0 ctx
cxt
t
x0't
0 xct
1cv
ctx
xct cxt
Minkowski spacetime
• Let us synchronize the clocks of the S and S’ frames at the origin
• Let us consider an event
• In the S frame, the event is to the right of the origin
• In the S‘ frame, the event is to the left of the origin
7.1
x’
t’t
x
Minkowski spacetime
• Let us synchronize the clocks of the S and S’ frames at the origin
• Let us consider an event
• In the S frame, the event is after the synchronization
• In the S‘ frame, the event is before the synchronization
7.1
x’
t’t
x
Minkowski spacetime7.1
0)( 2 s
0)( 2 s
223
0
cmpp
Four-momentum
• For an object moving relative to a laboratory system, we define a contravariant vector of four-momentum:
• Magnitude of four-momentum
7.4
mup 00 mup cm
11 mup ;xvm zy vmpvmp 32 ;
z
y
x
vmvmvmcm
3
0
pp
3
0
3
0
ppg 2222 )()()()( zyx vmvmvmcm
2
2
1cvmc mc
Four-momentum
• Rest-mass: mass measured in the system where the object is at rest
• For a moving object:
• The equation has units of energy squared
• If the object is at rest
223
0
cmpp
7.4
m
2222 )()()()( zyx vmvmvmcm
mm~222
3
0
)()~()~( mcvmcmpp
222 )~()()~( vmmccm
222222 )~()()~( cvmmccm
2222 )~
()( cpmc
2~cmE
22222 )~
()( cpmcE
0~p
2220 )(mcE
20 mcE
Four-momentum7.4
20 mcE
z
y
x
vmvmvmcE
~~~
Four-momentum
• Rest-mass energy: energy of a free object at rest – an essentially relativistic result
• For slow objects:
• For free relativistic objects, we introduce therefore the kinetic energy as
7.42
0 mcE
22~ cmcmE 2/122 )1( mc0
2/122 )1( mcE
21
22 mc
2
22
21
cvmc
2
2
0mvE
0
2
2EEmv
0EET
2mcET 22 mcmc 22222 )~
()( mccpmc
Non-covariant Lagrangian formulation of relativistic mechanics
• 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
3,2,1
ixV
dtdp
i
i
7.9
i
i
dxdVmv
dtd
21 iii
i
xL
xV
vLmv
;1 2
22
21
1
mcv
mvi
iVmcL 22 1
VTL
Non-covariant Lagrangian formulation of relativistic mechanics
• For an electromagnetic potential, the Lagrangian is similar
• The equations of motion should look like
• Recall our derivations in “Lagrangian Formalism”:
3,2,1
ixL
dtdp
i
i
7.9
3
121 i
i
ii
ii
i
xAv
xqqAmv
dtd
)()(1 2
BvEqAvqtAqmv
dtd i
3
1
22 1i
iivAqqmcL
ii
i qAmvp
21
Non-covariant Lagrangian formulation of relativistic mechanics
• Example: 1D relativistic motion in a linear potential
• The equations of motion:
• Acceleration is hyperbolic, not parabolic
7.9
maxc
xmcdtd
22
constaxmaxcmcL ;22
cat
xc
x
22
22 )(
atc
catcx
22220 )( catc
acxx
Useful results
21
1
2
2)(1
1
cv
2
22
)(1
1
cv
2222 ))(( cvc
2
222)(ccv
22222 )( ccv 222 )( cvc
Non-covariant Hamiltonian formulation of relativistic mechanics
• We start with a non-covariant Lagrangian:
• Applying a standard procedure
• Hamiltonian equals the total energy of the object
LvpHi
ii
3
1
7.98.4
VmcL 22 1
Vmcvvmi
ii
223
1
1 Vmcvm
2
2)(
Vmcccm
2
2
22 Vcm 2
22 mcmcT
VmcT 2 VET 0
VETH 0
2
222)(ccv
Non-covariant Hamiltonian formulation of relativistic mechanics
• We have to express the Hamiltonian as a function of momenta and coordinates:
7.98.4
VcmH 2
222 )( cvc
Vcvmc 222 )(
Vcmvmc 4222 )(
z
y
x
vmvmvmcm
p
VEpc 20
22 )(
VEpcH 20
22 )(
More on symmetries
• Full time derivative of a Lagrangian:
• Form the Euler-Lagrange equations:
• If
dtdL
M
m
M
mm
mm
m
qqLq
qL
tL
1 1
M
m
M
mm
mm
m
qqLq
qL
dtd
tL
1 1
M
mm
m
qqL
dtd
tL
1
LqqL
dtd
tL M
mm
m1
dt
dH
0tL constLq
qLH
M
mm
m
1
Non-covariant Hamiltonian formulation of relativistic mechanics
• Example: 1D relativistic harmonic oscillator
• The Lagrangian is not an explicit function of time
• The quadrature involves elliptic integrals
7.98.4
constEHtL
tot 0
2
2
22
32 kx
xc
mcVcmH
totEkx
xc
mc
2
2
22
3
2
222 kxxcmcL
2
2
322
22
kxE
mccxtot
2322
2
0)2()2(
)2(
mckcxcE
dxkxEtttot
tot
Covariant Lagrangian formulation of relativistic mechanics: plan A
• 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
7.10
3,2,1,0;'
ddxx
2
1
),',(
dxxI
Covariant Lagrangian formulation of relativistic mechanics: plan A
• Equations of motion
• We need to find Lagrangians producing equations of motion for the observable behavior
• First approach: use previously found Lagrangians and replace time and velocities according to the rule:
7.10
3,2,1
idtd
ddx
dtdx ii
2
1
),',(
dxxI
xxdd
'
cxt
0
ddtddxi
cx
dd
x i
0
'
0''
xxc
i
Covariant Lagrangian formulation of relativistic mechanics: plan A
• Then
• So, we can assume that
• Attention: regardless of the functional dependence, the new Lagrangian is a homogeneous function of the generalized velocities in the first degree:
7.10
3,2,1;''0 i
xxc
dtdx ii
2
1
2
1
0
0
0
'',,),,(
cxd
xxc
cxxLdtxtxLI
ii
t
t
ii
cxt
0
2
1
0
0
0 1'',,
dddx
cxxc
cxxL
ii
2
1
0
00
'',,'
dxxc
cxxL
cx i
i
0
00
'',,')',(
xxc
cxxL
cxxx
ii
)',()',( xxaaxx
Covariant Lagrangian formulation of relativistic mechanics: plan A
• From Euler’s theorem on homogeneous functions it follows that
• Let us consider the following sum
7.10
3
0 ''
xx
)',()',( xxaaxx
3
0
'
xx
3
0
3
0 '''
xx
xx
3
0, '''
xx
xx
3
0,
2
'''
xxxx 0
3
0,
23
0,
2
''"'
'''
xxxx
xxxx
Covariant Lagrangian formulation of relativistic mechanics: plan A
• If three out of four equations of motion are satisfied, the fourth one is satisfied automatically
7.10
3
0,
23
0,
23
0 '''"
''''
xxxx
xxxx
xx
3
0,
22
'''"
'''
xxxx
xxxx
3
0
3
0
"''
''
'
xxx
xxx
x
3
0 ''
xddx
0''
3
0
xxxd
d xxd
d
'
Example: a free particle
• We start with a non-covariant Lagrangian
7.10
22 1 mcL222
2 1
cz
cy
cxmc
0
00
'',,')',(
xxc
cxxL
cxxx
ii
2
0
32
0
22
0
1
2 ''
''
''
1
cxxc
cxxc
cxxc
mc
3
1
2
02
''1
i
i
xxmc
3,2,1''0
ixxc
dtdx ii
3
1
2
0
02
0
00
''1'
'',,'
i
iii
xx
cxmc
xxc
cxxL
cx
3
1
220 ''i
ixxmc
Example: a free particle
• Equations of motion
7.10
3
1
2200
00
'''',,'
i
ii
i xxmcxxc
cxxL
cx
3
0
''
xxmc
3
0
''
xxmc
xxdd
'
0'''
3
0
xxmcxd
d0
''
'21
3
0
xx
mcxdd
dd
ddx
ddxx '
ddu
dxud'
Example: a free particle
• Equations of motion of a free relativistic particle
7.10
02
'3
0
ddu
ddu
ddmcu
dd
ux 0
3
0
uu
mcudd
cuu
3
0
0
cmcu
dd
0)(
dmud 0
ddp
0)()()(
dvmd
dvmd
dvmd zyx
0d
dE
Covariant Lagrangian formulation of relativistic mechanics: plan B
• Instead of an arbitrary invariant parameter, we can use proper time
• However
• 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
7.10
2
3
0
3
0
cddx
ddx
uu
Covariant Lagrangian formulation of relativistic mechanics: plan B
• In this case, the equations of motion will look like
• 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
7.10
xudd
3
0 2
umu
0)(
dmud
Covariant Lagrangian formulation of relativistic mechanics: plan B
• 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
7.10
Covariant Lagrangian formulation of relativistic mechanics: plan B
• Example: 1D relativistic motion in a linear potential
• In a specific inertial frame, the non-covariant Lagrangian was earlier shown to be
• The covariant form of this problem is
• In a specific inertial frame, the interaction vector will be reduced to
7.10
constaxmaxcmcL ;22
3
0
3
0 2
xGumu
1 maG
Example: relativistic particle in an electromagnetic field
• For an electromagnetic field, the covariant Lagrangian has the following form:
• The corresponding equations of motion:
7.107.6
3
0
32103
0
),,,(2
uxxxxqAumu
3
2
1
/
AAA
c
A
3
0
)()(
xuAA
qdmud
ABtAE
3
0
uFq
00
00
1
xyz
xzy
yzx
zyx
cBcBEcBcBE
cBcBEEEE
cF
3
01
1)(
uFq
dmud
Example: relativistic particle in an electromagnetic field
• Maxwell's equations follow from this covariant formulation (check with your E&M class)
7.107.6
00
00
1
xyz
xzy
yzx
zyx
cBcBEcBcBE
cBcBEEEE
cF
3
01
1)(
uFq
dmud
)()( zyyzxx
vcBvcBcEcq
dtvmd
)( zyyzxx
vBvBEcq
dtdp
xBvEcq )(
Covariant Lagrangian formulation of relativistic mechanics: plan B
• 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
7.10
Covariant Hamiltonian formulation of relativistic mechanics: plan A
• In ‘Plan A’, Lagrangians are homogeneous functions of the generalized velocities in the first degree
• Let us try to construct the Hamiltonians using canonical approach (Legendre transformation)
• ‘Plan A’: a bad idea !!!
8.4
)',()',( xxaaxx
3
0 ''
xx
3
0
'
x
3
0
''
x
x 'x
0
Covariant Hamiltonian formulation of relativistic mechanics: plan B
• In ‘Plan B’: instead of an arbitrary invariant parameter, we use proper time
• 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:
8.4
3
0
up
3
0
u
u up
23
0
cuu
Covariant Hamiltonian formulation of relativistic mechanics: plan B
• For a free particle:
8.4
3
0
up
3
0 2
umu
up
3
0 2
umuu
3
0 2
uumgu
umg mu
mpu /
3
0 2
mpp
3
0
3
0
3
0 22
mpp
mpp
mpp
3
0 2
mpp
Covariant Hamiltonian formulation of relativistic mechanics: plan B
• For a particle in an electromagnetic field:
8.4
3
0
up
3
0
3
0 2
uqAumu
up
3
0
3
0 2
uqAumu
u qAmu
mqApu
3
0
3
0
3
0
)(2
))(()(
qApqA
mqApqAp
mqApp
3
0 2))((
mqApqAp
Relativistic angular momentum
• For a single particle, the relativistic angular momentum is defined as an antisymmetric tensor of rank 2 in Minkowski space:
• 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
pxpxm
0)()(/)(0)(/)()(0/
///0
zyzxz
yzyxy
xzxyx
zyx
yvzvmxvzvmmvctczEzvyvmxvyvmmvctcyEzvxvmyvxvmmvctcxE
czEmvctcyEmvctcxEmvct
m
Relativistic angular momentum
• Evolution of the relativistic angular momentum is determined by:
• For open systems, we have to define generalized relativistic torques in a covariant form
7.8
)(
pxpx
dd
ddm
ddpx
ddxp
ddpx
ddxp
ddpx
ddpxumuumu
From the equations of motion
ddpx
ddpx
ddm
N
ddm
Relativistic kinematics of collisions
• The subject of relativistic collisions is of considerable interest in experimental high-energy physics
• Les us assume that the colliding particle 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
• Conservation of a four-momentum includes conservation of a linear momentum and conservation of energy
7.7
0
ddp
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)
• Conservation of the four-momentum of the system:
• 1) Rearrange the equation (supposed we are not interested in particle D)
• 2) Magnitude squared of each side of the equation:
7.7
)()()()( DCBA pppp
)()()()( CBAD pppp
3
0
)()()()()()(
CBACBA pppppp
3
0
)()(
DD pp
Relativistic kinematics of collisions7.7
3
0
)()()()()()(
CBACBA pppppp
3
0
)()(
DD pp
3
0
2222
)()()()()()(2
)()()()(
CBCABA
CBAD
pppppp
cmcmcmcm
3
0
3
0,
3
0
)()()()()()(
jijiji pppgppp
3
0
3
0
3
0
)()(2)()()()(
jijiji pppppp
ji
23
0
)()()( cmpp iii
Example: electron-positron pair annihilation
• Annihilation of an electron and a positron produces two photons
• Conservation of the four-momentum of the system:
• Let us assume that the positron is initially at rest:
• 1) Rearrange the equation
21 ee
)()()()(21
pppp
2;0 mcEp
)()()()(
12pppp
Example: electron-positron pair annihilation
• 2) Magnitude squared of each side of the equation:
3
0
)()()()()()(11
pppppp
3
0
)()(22
pp
3
0
2222
)()()()()()(2
)()()()(
11
12
pppppp
cmmcmccm 021 mm
0)()()()()()()(3
0
211
ppppppmc
)()()()(
12pppp
Example: electron-positron pair annihilation
0)()()()()()()(3
0
211
ppppppmc
3
0
)()(
pp
pp
cE
cE
0;2 pmcE
mE
3
0
)()(1
pp
1
1
ppc
EcE
mE
1
3
0
)()(1
pp
1
1
ppc
EcE
12 cos1
1 pp
cEE
Example: electron-positron pair annihilation
• The photon energy will be at a maximum when emitted in the forward direction, and at a minimum when emitted in the backward direction
0cos)( 122
1
1
1
pp
cEE
mEmEmc
11 pcE
0cos)( 122 11
1
pc
EcEE
mEmEmc
mEmcc
pcEmE
21
2 )(cos
1
12
22
cos)()(
1
pcEmcEmcmcE