Dynamics of Relativistic Particles and EM Fields
October 7, 20081
1
J.D.Jackson, ”Classical Electrodynamics”, 3rd Edition, Chapter 12
Dynamics of Relativistic Particles and EM Fields
Lagrangian Hamiltonian for a Relativistic Charged Particle
The equations of motion
~u ~
d~p
~
=e E + ×B
dt
c
(1)
dE
~
= e~u · E
(2)
dt
~ and B
~ can be written in
for a particle with charge e in external fields E
the covariant form (??)
e αβ
dU α
=
F Uβ
(3)
dτ
mc
where m is the mass, τ is the proper time, and U α = (γc, γ~u ) = p α /m is
the 4-velocity of the particle.
• The Lagrangian treatment of mechanics is based on the principle of of
least action or Hamilton’s principle.
• The system is described by generalized coordinates qi (t) & velocities
q̇i (t).
• The Lagrangian L is a functional of qi (t) and q̇u (t) and perhaps the
time.
Dynamics of Relativistic Particles and EM Fields
• The action A is the time integral of L along a possible path of the
system.
• The principle of least action states that the motion of a mechanical
system is such that in going from a configuration a at time t1 to a
configuration b in time t2 the action
Z t2
A=
L [qi (t), q̇i (t), t] dt
(4)
t1
is an extremum.
By considering small variations of coordinates and velocities away from
the actual path and requiring δA = 0 one obtains the Euler-Lagrange
equations of motion
∂L
d
∂L
−
=0
(5)
dt ∂ q̇i
∂qi
Dynamics of Relativistic Particles and EM Fields
Relativistic Lagrangian (Elementary)
From the 1st postulate of STR the action integral must be a Lorentz
scalar, because the equations of motion described by the extremum
condition δA = 0. Then if we set in (4) dt = γdτ we get
Z τ2
A=
γLdτ
(6)
τ1
since the proper time is Lorentz invariant the condition that A is
invariant requires that γL is also Lorentz invariant.
The Lagrangian for a free particle can be a function of the velocity (the
only invariant function of the velocity is ηαβ U α U β = Uα U α = c 2 ) and
the mass of the particle but not its position.
r
u2
2
(7)
1− 2
Lfree = −mc
c
and through (5) the free-particle equation of motion (remember that
~p = γm~u )
d
(γm~u ) = 0
(8)
dt
Dynamics of Relativistic Particles and EM Fields
Since the non-relativistic Lagrangian is T − V and V = eΦ the
interaction part of the relativistic Lagrangian must reduce in the
non-relativistic limit to
Lint → LNR
int = −eΦ
(9)
A wise guess is:
e
Uα Aα
(10)
γc
which may comes from the demand that γLint is :
• linear in the charge of the particle
• linear in the EM potentials
• translationally invariant
• a function no higher than the 1st time derivative of the particle
coordinates.
Another writing is:
e ~
(11)
Lint = −eΦ + ~u · A
c
and the combination of (7) and (11) yields the complete Lagrangian
r
e ~
u2
2
L = −mc
1 − 2 + ~u · A
− eΦ
(12)
c
c
(Verify that leads to the Lorentz force equation)
Lint = −
Dynamics of Relativistic Particles and EM Fields
~ conjugate to the position coordinate ~x is
The canonical momentum P
obtained by the definition
Pi ≡
e
∂L
= γmui + Ai
∂ui
c
(13)
Thus the conjugate momentum is
~ = ~p + e A
~
P
c
(14)
where ~p = γm~u is the ordinary kinetic momentum.
The Hamiltonian H is a function of the coordinate ~x and its conjugate
~ and is a constant of motion if the Lagrangian is not an
momentum P
explicit function of time, in terms of the Lagrangian is :
~ · ~u − L
H=P
(15)
~ and ~x we find (HOW?) that
by eliminating ~u in favor of P
~u = r
~ − eA
~
cP
2
~ − e ~A + m2 c 2
P
c
(16)
Dynamics of Relativistic Particles and EM Fields
This equation together with (12)
r
2
~ − eA
~ + m2 c 4 + eΦ
H=
cP
(17)
(Verify that from this Lagrangian you can get the Lorentz equation)
Equation (17) is an expression for the total energy W of the particle.
Actually, it differs by the potential energy term eΦ and by the
~ − (e/c)A].
~
replacement ~p → [P
These two modifications are actually a conseqency of considering
4-vectors. Notice that
2
2
~ − eA
~ = mc 2 2
(18)
(W − eΦ) − c P
is just the 4-vector scalar product
pα p α = (mc)
where
pα ≡
E
, ~p
c
=
2
1
~ − eA
~
(W − eΦ) , P
c
c
(19)
(20)
Thus the total energy W /c acts as the time component of a canonically
~ is given by (14).
conjugate 4-momentum P α of which P
Dynamics of Relativistic Particles and EM Fields
Relativistic Lagrangian (Covariant Treatment)
If ones wants to use of proper covariant description, has to abandon the
vectorial writing ~x , ~u and to replace them with the 4-vectors x α , U α .
Then the free particle Lagrangian (7) will be written as
Lfree = −
mc p
Uα U α
γ
remember that Uα U α = c 2 . The action integral will be:
Z τ2 p
A = −mc
Uα U α dτ
(21)
(22)
τ1
This invariant form can be the starting point for a variational calculation
leading to the equation of motion dU α /dτ = 0. One can further make
use of the constraint
Uα U α = c 2
(23)
or the equivalent one:
Uα
dU α
=0
dτ
(24)
Dynamics of Relativistic Particles and EM Fields
The integrant in (22) is:
r
q
dxα dx α
dτ = g αβ dxα dxβ
dτ dτ
i.e. the infinitesimal length element in 4-space. Thus the action integral
can be written as
Z s2 r
dxα dxβ
A = −mc
g αβ
ds
(25)
ds ds
s1
p
Uα
U α dτ
=
where the 4-vector coordinate of the particle is x α (s), where s is a
parameter monotonically increasing with τ , but otherwise arbitrary.
• The action integral is an integral along the world line of the particle
• The principle of least action is a statement that the actual path is the
longest path, the geodesic. We should keep in mind that
r
dxα dxβ
g αβ
ds = cdτ
(26)
ds ds
and then a straightforward variational calculation with (25) leads to


mc
d  dx α /ds 

=0
ds dxβ dx β 1/2
ds
(27)
ds
Dynamics of Relativistic Particles and EM Fields
or
d 2x α
=0
(28)
dτ 2
as expected for a free particle motion.
For a charged particle in an external field the form of the Lagrangian (11)
suggests that the manifesltly covariant form of the action integral is
#
Z s2 " Z s2 r
dx
e
dx
dx
β
α
α
α
g αβ
+
A (x) ds
(29)
A=−
mc
ds ds
c ds
s1
s1
m
Hamilton’s principle yields the Euler-Lagrange equations:
"
#
d
∂ L̃
− ∂ α L̃ = 0
ds ∂(dxα /ds)
where the Lagrangian is:
" r
L̃ = − mc
#
dx
dx
e
dx
β
α
α
+
Aα (x)
g αβ
ds ds
c ds
(30)
(31)
Then (30) becomes
m
d 2x α
e dAα (x) e dxβ α β
+
−
∂ A (x) = 0
dτ 2
c dτ
c dτ
Dynamics of Relativistic Particles and EM Fields
Since dAα /dτ = (dxβ /dτ )∂ β Aα this equation can be written as
m
dxβ
d 2x α
e α β
∂ A − ∂ β Aα
=
dτ 2
c
dτ
(32)
which is the covariant equation of motion (3).
Dynamics of Relativistic Particles and EM Fields
Relativistic Hamiltonian (Covariant Treatment)
The transition to conjugate momenta and a Hamiltonian is simple
enough. The conjugate 4-momentum is defined by
Pα = −
e
∂ L̃
= mU α + Aα
∂(dxα /ds)
c
(33)
which is in agreement with (4). A Hamiltonian can be determined
H̃ = Pα U α + L̃
(34)
the by eliminating U α by means of (33) leads to the expression
s
1
eAα
eAα
eAα
eAα
α
H̃ =
Pα −
P −
−c
Pα −
Pα −
m
c
c
c
c
(35)
Then by using the constraint
eAα
eAα
Pα −
= m2 c 2
Pα −
c
c
we get Hamilton’s equations:
Dynamics of Relativistic Particles and EM Fields
dx α
=
dτ
and
∂ H̃
1
=
∂Pα
m
α
dP
dτ
= −
eAα
P −
c
α
(36)
∂ H̃
e
=
∂xα
mc
Pβ −
eAβ
c
∂ α Aβ
These two equation can be shown to be equivalent to the Euler-Lagrange
equation (32).
Dynamics of Relativistic Particles and EM Fields
Motion in a Uniform, Static Magnetic Field
We consider the motion of charged particles in a uniform and static
magnetic field. The equations (1) and (2) are
d~p
e
~,
= ~v × B
dt
c
dE
=0
dt
(37)
where ~v is the particle’s velocity. Since the energy is constant in time,
the magnitude of the velocity is constant and so is γ.
Then the first equation can be written
d~v
= ~v × ω
~B
dt
where
ω
~B =
~
~
eB
ec B
=
γmc
E
(38)
(39)
is the gyration or precession frequency.
~ and a uniform
The motion is a circular motion perpendicular to B
~
translation parallel to B.
Dynamics of Relativistic Particles and EM Fields
The solution for the velocity is (HOW?)
~v (t) = vk~3 + ωB a(~1 − i~2 )e −iωB t
(40)
~3 : is a unit vector parallel to the field
~1 and ~2 : are the other two orthogonal unit vectors
vk : is the velocity component along the field, and
a : the gyration radius
One can see that (40) represents a counterclockwise rotation for positive
charge e
Further integration leads to the displacement of the particles
~ 0 + vk t~3 + ia(~1 − i~2 )e −iωB t
~x (t) = X
(41)
The path is a helix of radius a and pitch angle α = tan−1 (vk /ωB a).
The magnitude of the gyration radius a depends on the magnetic
~ and the transverse momentum ~p⊥ of the particle
induction B
cp⊥ = eBa
This relation allows for the determination of particle momenta. For
particle with charge equal to electron charge the momentum can be
written numerically as
p⊥ (MeV/c) = 3 × 10−4 Ba(Gauss-cm) = 300Ba(tesla-m)
(42)
Dynamics of Relativistic Particles and EM Fields
Figure: This three basic motions of charged particles in a magnetic field:
gyro, bounce between mirror points, and drift. The pitch angle α between
~ and the electron velocity ~v .
the directions of the magnetic field B
The angle between the direction of the magnetic field and a particle’s
spiral trajectory is referred to as the ”pitch angle”, which in a
non-uniform magnetic field changes as the ratio between the
perpendicular and parallel components of the particle velocity changes.
Pitch angle is important because it is a key factor in determining whether
a charged particle will be lost to the Earth’s atmosphere or not.
Dynamics of Relativistic Particles and EM Fields
Motion in Combined, Uniform, Static E- and B- Field
We will consider a charged particle moving in a combination of electric
~ and B,
~ both uniform and static, and for this study
and magnetic fields E
they will be considered perpendicular.
From the energy equation (2) we notice that the particle’s energy is not
constant in time. Consequently we can obtain a simple equation for the
velocity, as was done for a static magnetic field.
An appropriate Lorentz transformation can simplify the equations of
motion, here we consider a coordinate frame K 0 moving with velocity ~u
with respect to original frame K . Then the Lorentz force equation for a
particle in K 0 is
!
0
~0
~
v
×
B
d~p 0
0
~ +
=e E
dt 0
c
~ 0 and B
~ 0 can be estimated from relations of the previous
The fields E
chapter.
Dynamics of Relativistic Particles and EM Fields
Motion in Combined, Uniform, Static E- and B- Field
~ | < |B|
~
CASE: |E
~ 0 i.e.
~ 0 and B
If we chose ~u to be perpendicular to the orthogonal vectors E
~u = c
~ ×B
~
E
2
B
(43)
we find that the fields in K 0 (HOW?)
~0 = 0,
~0 = γ E
~ + ~u × B
~ =0
E
E
⊥
k
c
(44)
~0 = 0,
B
k
~ 0 = 1B
~ =
B
⊥
c
2
B −E
B2
2
1/2
~
B
~ 0 which
In the frame K 0 the only field acting is a static magnetic field B
~ but is weaker by a factor 1/γ . Thus
points in the same direction as B
0
the motion in K is the same as in the previous section, namely spiraling
around the lines of force.
Dynamics of Relativistic Particles and EM Fields
As viewed from the original frame, this gyration is accompanied by a
~ and B.
~
uniform “drift” ~u perpendicular to E
The direction of the drift is independent of the sign of the charge of the
particle.
~ ×B
~ drift of charged particles in perpendicular fields
Figure: E
Dynamics of Relativistic Particles and EM Fields
~ | > |B|
~
CASE: |E
The electric field is so strong that the particle is continually accelerated in
~ and its average energy continues to increase with time.
the direction of E
If we consider a Lorentz transformation to a system K 00 moving relative
to the first with velocity
~ ×B
~
E
~u 0 = c
(45)
2
E
we get in the K 00
~ 00 = 0 ,
E
k
~ =
~ 00 = 1 E
E
⊥
γ
E 2 − B2
E2
1/2
~
E
(46)
~ 00 = 0 ,
B
k
~ 00 = γ 0
E
⊥
~0
~
~ − u ×E
B
c
!
=0
Thus the particle, in the system K 00 , is acted on by a purely electrostatic
field which causes hyperbolic motion with ever-increasing velocity.
Dynamics of Relativistic Particles and EM Fields
Lowest Order Relativistic Corrections to the Lagrangian...
The interaction Lagrangian was given by (11). In its simpler form the
non-relativistic Lagrangian for two charged particles
q1 q2
(47)
LNR
int =
r
including lowest order relativistic effects is
q1 q2
1
(~v1 · ~r )(~v2 · ~r )
Lint =
−1 + 2 ~v1 · ~v2 +
(48)
r
2c
r2
For a system of interacting charged particles the complete Darwin
Lagrangian correct to order 1/c 2 can be written by expanding the
free-particle Lagrangian (7) for each particle and summing up all the
interaction terms of the form (48)
LDarwin
=
∼
1 X
1X
1 X qi qj
mi vi2 + 2
mi vi4 −
2
8c
2
rij
i
+
1
4c 2
i
∼
X
i,j
i,j
i
qi qj h
~vi · ~vj + (~vi · ~ˆrij )(~vj · ~ˆrij )
rij
(49)
Dynamics of Relativistic Particles and EM Fields
where rij = |~xi − ~xj |, and ~ˆrij is the unit vector in the direction ~xi − ~xj and
the “tilde” (∼) in the summation indicates omission of the self-energy
terms i = j.
Dynamics of Relativistic Particles and EM Fields
Lagrangian for the Electromagnetic Field
We now examine a Lagrangian description of the EM field in interaction
with specified external sources of charge and current.
The Lagrangian approach to to continuous fields is similar to the
approach used for discrete point particles.
The finite number of coordinates qi (t) and q̇i (t) are replaced by an
infinite number of degrees of freedom.
The coordinate qi is replaced by a continuous field φk (x) with a discrete
index (k = 1, 2, . . . , n) and a continuous index (x α ), i.e.
i → x α , k , qi → φk (x) q̇i → ∂ α φk (x)
Z
X
L=
Li (qi , q̇1 ) →
L(φk , ∂ α φk )d 3 x
(50)
i
d
dt
∂L
∂ q̇i
=
∂L
∂qi
→ ∂β
∂L
∂L
=
∂(∂ β φk )
∂φk
where L is the Lagrangian density, corresponding to a definite point in
space-time and equivalent to the individual terms in a discrete particle
Lagrangian like (49). For the EM-field the “coordinates” are Aα and the
“velocities” ∂ β Aα .
Dynamics of Relativistic Particles and EM Fields
The action integral take sthe form
Z Z
Z
3
A=
L d x dt = L d 4 x
(51)
and it will be Lorentz-invariant if the Lagrangian density L is a Lorentz
scalar. In analogy with the situation with discrete particles, we expect the
free-field Lagrangian at least to be quadratic in the velocities, that is,
∂ β Aα or F αβ . The only Lorentz invariant quadratic forms are Fαβ F αβ
and Fαβ F αβ (the last term is pseudoscalar under inversion).
Thus the Lfree will be a multipole of Fαβ F αβ and the Lint according to
(10) will be a multipole of Jα Aα . Thus the EM Lagrangian density is:
L=−
1
1
Fαβ F αβ − Jα Aα
16π
c
(52)
If we want to use it for the Euler-Lagrange equations given in (50) we get
L=−
1
1
gλµ gνσ (∂ µ Aσ − ∂ σ Aµ ) ∂ λ Aν − ∂ ν Aλ − Jα Aα
16π
c
(53)
Dynamics of Relativistic Particles and EM Fields
The term
∂L
∂(∂ β Aα )
in the Euler-Lagrange equations becomes (how?)
∂L
1
1
= − Fβα =
Fαβ
∂(∂ β Aα )
4π
4π
(54)
The other part of the Euler-Lagrange equations is
∂L
1
= − Jα
∂Aα
c
(55)
Thus the equations of motion for the EM field are
1 β
1
∂ Fβα = Jα
4π
c
(56)
which is a covariant form of the inhomogeneous Maxwell equations (??).
The conservation of the source current density cen be obtained from (56)
1 α β
1
∂ ∂ Fβα = ∂ α Jα
4π
c
The left hand side has a differential operator which is symmetric in α and
β, while Fαβ is antisymmetric. Again the contraction vanishes (why?)
and we have:
∂ α Jα = 0 .
(57)
Dynamics of Relativistic Particles and EM Fields
Proca Lagrangian; Photon Mass Effect
If we assume that the photon is not massless then the Lagrangian (52)
has to be modified by the addition of a “ mass” term, this is called Proca
Lagrangian
LProca = −
1
µ2
1
Fαβ F αβ +
Aα Aα − Jα Aα
16π
8π
c
(58)
The parameter µ has dimensions of inverse length and is the reciprocal
Compton wavelength of the photon (µ = mγ c/~). Instead of (56) the
Proca equations of motion are
∂ β Fβα + µ2 Aα =
4π
Jα
c
(59)
with the same homogeneous equations ∂α F αβ = 0 as in Maxwell theory.
In contrast to the Maxwell equations the potentials have real physical
(observable) significance through the mass term. In the Lorentz gauge
(59) can be written
4π
Aα + µ2 Aα =
Jα
(60)
c
Dynamics of Relativistic Particles and EM Fields
In the static limit takes the form
4π
Jα
(61)
c
If the source is a point charge q at rest in the origin then the only
non-vanishing component is A0 = Φ. the solution will be the spherically
symmetric Yukawa potential
q
Φ(x) = e −µr
(62)
r
i.e. we observe an exponential falloff of the static potentials and fields,
with 1/e distance equal to 1/µ.
Notice that the exponential factor alters the character of the Earth’s (and
other planets) magnetic fields sufficiently to permit us to set quite
stringent limits on the photon mass from geomagnetic data.
h
i
2 2 e −µr
µ2 r 2 e −µr
ˆ
ˆ
~
~ 3
~)−m
~
B(~x ) = 3~r (~r · m
1 + µr +
−
µ m
3
r3
3
r
∇2 Aα − µ2 Aα = −
The result shows that the Earth’s magnetic field will appear as a dipole
angular distribution plus an added constant magnetic field antiparallel to
~ . Measurements show that this “ external” field is less than 0.004 times
m
the dipole field at the magnetic equator which leads to
µ < 4 × 10−10 cm−1 or mγ < 8 × 10−49 g.
Dynamics of Relativistic Particles and EM Fields
Conservation Laws : Canonical Stress Tensor
In particle mechanics the transition to the Hamilton formulation and
conservation of energy is made by first defining the canonical momentum
variables
∂L
pi =
∂ q̇i
and then introducing the Hamiltonian
X
H=
pi q̇i − L
(63)
i
where dH/dt = 0 if ∂L/∂t = 0.
For fields the Hamiltonian is the volume integral over the 3-D space of
Hamiltonian density H i.e.
Z
H = Hd 3 x .
It is necessary that the Hamiltonian density H transform as the tt
component of a 2nd-rank tensor.
Dynamics of Relativistic Particles and EM Fields
If the Lagrangian density for some fields is a function of the field
variables φk (x) and ∂ α φk (x) with k = 1, 2, . . . , n the Hamiltonian
density is defined in analogy to (63) as
X
∂φk
∂L
H=
−L
∂(∂φk /∂t) ∂t
(64)
k
The first factor in the sum is the field momentum canonically conjugate
to φk (x) and ∂ α φk (x) is equivalent to the velocity q̇i .
The Lorentz transformation properties of H suggest that the covariant
generalization of the Hamiltonian density is the canonical stress tensor:
X ∂L
∂ β φk − g αβ L
(65)
T αβ =
∂(∂α φk )
k
For the free EM field Lagrangian:
Lem = −
1
Fµν F µν
16π
the canonical stress tensor is
T αβ =
∂Lem β λ
∂ A − g αβ Lem
∂(∂α Aλ )
Dynamics of Relativistic Particles and EM Fields
By using equation (54) we find
1
T αβ = − g αµ Fµλ ∂ β Aλ − g αβ Lem
4π
2
2
~
~
With L = E − B /8π and (??) we find (how?)
1 ~ 2 ~ 2 1 ~ ~
T 00 =
E +B +
∇ · ΦE
8π
4π
1 ~ ~ 1 ~
~)
T 0i =
E ×B +
∇ · (Ai E
4π
4π i
1 ~ ~ 1 ~
∂
i0
~
T
=
E ×B +
∇ × ΦB −
(ΦEi )
4π
4π
∂x0
i
i
(66)
(67)
If the fields are localized in some finite region of space the integrals over
all 3-space at fixed time in some inertial frame of the components T 00
and T 0i can be interpreted as the total energy and c times the total
momentum of the EM fields in that frame:
Z
Z 1
~2 +B
~ 2 d 3 x = Efield
T 00 d 3 x =
E
8π
(68)
Z
Z 1
~ ×B
~ d 3 x = cPfield
T i0 d 3 x =
E
4π
i
Dynamics of Relativistic Particles and EM Fields
The previous definitions of field energy and momentum densities suggests
that there should be a covariant generalization of the differential
conservation law (??) of Poynting’s theorem. That is:
∂α T αβ = 0
(69)
Consider
∂α T αβ
=
X
∂α
k
X
=
∂α
k
∂L
∂ β φk − ∂ β L
∂(∂α φk )
∂L
∂L
∂ β φk +
∂α ∂ β φk − ∂ β L
∂(∂α φk )
∂(∂α φk )
but because of the equation of motion (50) the first term can be
transformed so that
X ∂L
∂L
αβ
β
β
∂α T =
∂ φk +
∂ (∂α φk ) − ∂ β L
∂φk
∂(∂α φk )
k
α
since L = L(φk , ∂ φk ) the term in the square bracket is an implicit
differentiation, hence
∂α T αβ = ∂ β L(φk , ∂ α φk ) − ∂ β L = 0
Dynamics of Relativistic Particles and EM Fields
The conservation law (or continuity equation) (69) yields the
conservation of total energy and momentum upon integration over all of
3-space at fixed time
Z
Z
Z
0 = ∂α T αβ d 3 x = ∂0 T 0β d 3 x + ∂i T iβ d 3 x
If the fields are localized the 2nd integral (divergence) gives no
contribution. Then with the identification (68) we get
d
Efield = 0 ,
dt
d ~
Pfield = 0
dt
(70)
The results are valid for an observer at rest in the frame in which the
fields are specified.
Dynamics of Relativistic Particles and EM Fields
Conservation Laws : Symmetric Stress Tensor
The canonical stress tensor T αβ has a number of deficiencies for example
lack of symmetry ! ( see T 0i and T i0 ). This affects the consideration of
the angular momentum of the field
Z
~ × B)d
~ 3x
~Lfield = 1
~x × (E
4πc
The angular momentum density is expressed in terms of a 3rd-rank tensor
M αβγ = T αβ x γ − T αγ x β
(71)
Then as (69) implies conservation of energy and momentum (70) we
expect the vanishing of the 4-divergence
∂α M αβγ = 0
(72)
to imply conservation of the total angular momentum of the field.
Direct evaluation gives
0 = (∂α T αβ )x γ + T γβ − (∂α T αγ )x β − T βγ
then by using (69) one can see that the conservation of angular
momentum requires that T αβ be symmetric. Finally, the involvement of
the potentials in (66) makes it not gauge invariant.
Dynamics of Relativistic Particles and EM Fields
We will construct a symmetric, traceless, gauge-invariant stress tensor
Θαβ from the canonical stress tensor T αβ of (66).
We substitute ∂ β Aλ = −F λβ + ∂ λ Aβ and obtain
1
1 αβ
1 αµ
αβ
αµ
λβ
µν
T =
g Fµλ F + g Fµν F
−
g Fµλ ∂ λ Aβ
(73)
4π
4
4π
The first term in (73) is symmetric and gauge invariant while the last
term of (73), with the help of the source-free Maxwell equations, can be
written (how?)
1 λα
1 αµ
g Fµλ ∂ λ Aβ =
F ∂λ Aβ
4π
4π
1
1
F λα ∂λ Aβ + Aβ ∂λ F λα =
∂λ F λα Aβ
=
4π
4π
with the following properties: (why?)
Z
αβ
(i) ∂α TD = 0 , (ii) TD0β d 3 x = 0
TDαβ
≡
−
If we define the symmetric stress tensor Θαβ
1 αβ
1
αβ
αβ
αβ
αµ
λβ
µν
Θ = T − TD =
g Fµλ F + g Fµν F
4π
4
(74)
(75)
then the differential conservation law (69) will hold if it holds for T αβ .
Dynamics of Relativistic Particles and EM Fields
The explicit components of Θαβ are:
Θ00
=
Θ0i
=
Θij
=
1
8π
1
4π
E 2 + B2
i
~ ×B
~
E
1
1
−
E i E j + B i B j − δ ij E 2 + B 2
4π
2
The tensor Θαβ can be written in the schematic matrix form
!
u
c~
g
Θαβ =
(M)
c~g −Tij
(76)
(77)
where the time-time component is the energy density (??) the
time-space component is the momentum density (??) while the
space-space components are the negative of the Maxwell stress tensor
(??).
Dynamics of Relativistic Particles and EM Fields
The various covariant or mixed forms of the stress tensor are
!
!
u
−c~g
u −c~g
α
Θαβ =
Θ β=
(M)
(M)
−c~g −Tij
c~g Tij
!
c~g
u
β
Θα =
(M)
−c~g Tij
The differential conservation law
∂α Θαβ = 0
(78)
embodies Poynting’s theorem and conservation of momentum for free
fields. For example, for β = 0 we have
1 ∂u ~ ~
α0
0 = ∂α Θ =
+∇·S
c ∂t
where ~S = c 2~g is the Poynting vector, this is the source-free form of (??).
Similarly for β = i
3
0 = ∂α Θαi =
∂gi X ∂ (M)
−
T
∂t
∂xj ij
j=1
which is equivalent to (??).
Dynamics of Relativistic Particles and EM Fields
The conservation of the field angular momentum defined by
Θαβγ = Θαβ x γ − Θαγ x β
(79)
is assured by (78) and the symmetry of Θαβ .
Dynamics of Relativistic Particles and EM Fields
Conservation Laws for EM fields interacting with Charged
Particles
When external forces are present the Lagrangian for the Maxwell
equations is (52), the symmetric stress tensor for the EM field retains its
form (75) but the coupling to the external force makes its divergence
non-vanishing.
1
1
∂ µ Fµλ F λβ + ∂ β Fµλ F µλ
∂α Θαβ =
4π
4
1
1
=
(∂ µ Fµλ )F λβ + Fµλ ∂ µ F λβ + Fµλ ∂ β F µλ
4π
2
the 1st term can be transformed via (56) and we get
1
1
∂α Θαβ + F βλ Jλ =
Fµλ ∂ µ F λβ + ∂ µ F λβ + ∂ β F µλ
c
8π
the last two terms (in blue) can be transformed via the homogeneous
Maxwell equation ∂ µ F λβ + ∂ β F µλ + ∂ λ F βµ = 0 by −∂ λ F βµ = ∂ λ F µβ
1
1
∂α Θαβ + F βλ Jλ =
Fµλ ∂ µ F λβ + ∂ λ F µβ
c
8π
Dynamics of Relativistic Particles and EM Fields
The right-hand side is zero (why?) and thus the divergence of the stress
tensor is
1
∂α Θαβ = − F βλ Jλ
(80)
c
The time and space components of this equation are
1 ~
1 ∂u ~ ~
+ ∇ · S = − ~J · E
(81)
c ∂t
c
and
3
∂gi X ∂ (M)
1 ~ ~ −
Tij = − ρEi +
J ×B
∂t
∂xj
c
i
(82)
j=1
These are the conservation of energy and momentum equations for EM
fields interacting with sources described by J α = (cρ, ~J).
The negative of the right hand side term in (80) is called the Lorentz
force density,
1 βλ
1~ ~
1~ ~
β
~
f ≡ F Jλ =
J · E , ρE + J × B
(83)
c
c
c
Dynamics of Relativistic Particles and EM Fields
If the sources are a number of charged particles then the volume integral
of f β leads through the Lorents force equation (1) to the time rate of
change of the sum of the energies or momenta of all particles:
Z
f β d 3x =
β
dPparticles
dt
The conservation of the 4-momentum for the combined system of
particle and fields:
Z
d β
β
d 3 x ∂α Θαβ + f β =
Pfield + Pparticles
=0
dt
(84)
Dynamics of Relativistic Particles and EM Fields