Download Electromagnetism and Relativity

Chapter 10
Electromagnetism and Relativity
10.1
Introduction
“If I am moving at a large velocity along a light wave, what propagation velocity should I measure?”
This was a question young Einstein asked himself and in 1905, he published a monumental paper
on special relativity which formulated how to transform coordinates, velocity and electromagnetic
…elds between two inertial frames. Einstein postulated that:
1. all physical laws remain intact in any inertial frames, and
2. the light velocity c is invariant against inertial coordinate transformation.
Postulate 1 means that, for example, the Maxwell’s equations in a moving frame remain formally
identical to those in the laboratory frame, provided the spatial coordinates, time, and electromagnetic …elds are appropriately transformed. Electromagnetic …elds appear and disappear as we
change observing frame. For example, if a charge is moving in the laboratory frame, we observe
both electric and magnetic …elds. In the frame of the moving charge, the current and magnetic …eld
evidently disappear. However, the electromagnetic …elds (primed quantities) in the charge frame
are subject to the transformation
E0k = Ek ;
E0? =
(E? + V
B? ) ;
(10.1)
B0k = Bk ;
B0? =
(B?
E? ) ;
(10.2)
V
where k and ? are relative to the direction of the velocity V: Since in this example, Bk = 0
and B? = V E? in the laboratory frame, the magmatic …eld in the frame of the moving charge
indeed vanishes consistent with our intuition. The pertinent static Maxwell’s equations are satis…ed
in both frames
(10.3)
laboratory frame: r E = ; r B = 0 J
"0
in the frame of moving charge: r0 E0 =
1
0
"0
; r0
B0 = 0; (J0 = 0):
(10.4)
Here the primed operators and quantities are those in the moving frame which are subject to the
Lorentz transformation.
As shown in Chapter 8, electromagnetic …elds due to a charged particle moving at an arbitrary
velocity can be correctly formulated by the Lienard-Wiechert potentials which had been discovered
prior to the theory of relativity. Electromagnetic disturbances propagate at the velocity c regardless
of the velocity of the charge, just as sound waves emitted by a moving source propagate at a
sound velocity independent of the source velocity. A major di¤erence between sound waves and
electromagnetic waves occurs for stationary source and moving observer. For sound waves, if an
observer is approaching a source at a velocity VO ; the apparent sound velocity becomes cs + VO
because both cs and VO are well de…ned with respect to the medium of sound waves, namely, air. In
electromagnetic waves that can propagate in vacuum, there is no preferred inertial frame to de…ne
velocities and a moving observer will measure the same propagation velocity c regardless of the
relative velocity between two inertial frames. Of course, the frequency and wavelength are Doppler
shifted but the product
= 0 0 = c or the ratio !=k = ! 0 =k 0 = c remains invariant.
10.2
CGS-ESU System
In this Chapter, the CGS-ESU (Electro-Static Unit) unit system is used so that electromagnetic
…elds E (statvolt/cm ' 300 volt/cm = 3 104 volt/m) and B (gauss = 10 4 T) have the same
dimensions. In CGS-ESU, the Coulomb’s law is adopted to connect the mechanical world and
electromagnetic world. (Recall that in SI, the magnetic force is adopted to de…ne 1 ampere current
which in CGS-USU is 3 109 stat-ampere.) If two equal charges separated by 1 cm exert a force
of 1 dyne (= erg/cm = 10 7 J/10 2 m = 10 5 N) on each other, the charge is de…ned to be 1 ESU
' 3 109 C. The Coulomb’s law in CGS-ESU system is
Coulomb’s law: F =
q1 q2
(dyne).
r2
(10.5)
The electronic charge is e = 4:8 10 10 ESU (= 1:6 10 19 C). The potentials
and A also
have common dimensions (statvolt) in CGS-ESU. This is particularly convenient in theoretical
electrodynamics because the …elds E (polar vector) and B (axial or pseudo vector) are in fact
components of a uni…ed 4 4 …eld tensor and the potentials , A form a four vector ( ; A):The
Maxwell’s equations in this unit system are:
r E=4
r B = 0; r
; r
E=
B=
1 @B
;
c @t
4
1 @E
J+
;
c
c @t
(10.6)
(10.7)
and the relationships between the …elds and potentials are
E=
r
1 @A
; B=r
c @t
2
A:
(10.8)
The wave equations for the potentials are modi…ed as
1 @2
=
c2 @t2
1 @2
A =
c2 @t2
r2
r2
4
;
(10.9)
4
J;
c
(10.10)
subject to the Lorentz gauge,
r A+
1@
= 0:
c @t
(10.11)
Electromagnetic force in the CGS-ESU system is
1
F (dyne) = e E + v
c
the Poynting ‡ux is
S=
1
B ; f (dyne/cm3 ) = E + J
c
c
E
4
B; (erg cm
1
E
4
B; (dyne cm
E
B; (dyne sec cm
the ‡ux of momentum is
2
1
sec
2
);
);
B;
(10.12)
(10.13)
(10.14)
and the momentum density is
1
4 c
3
):
(10.15)
):
(10.16)
Electromagnetic energy density is
1
8
E 2 + B 2 ; (erg cm
3
The vacuum impedance for a plane wave is unity (dimensionless),
B=
c
k
!
E; jBj = jEj :
(10.17)
In CGS-ESU system, macroscopic proportional constants inevitably have unfamiliar units. For
example, the capacitance has dimensions of length (cm) as seen from its de…nition,
[C] =
[q]
= length.
[ ]
(10.18)
The conductivity relates the electric …eld and current density, J = E; and has dimensions of
frequency, sec 1 ; since
[J]
[q] cm 2 sec 1
[ ]=
=
= sec 1 :
[E]
[q] cm 2
3
10.3
Lorentz Transformation
The null result of Michelson-Morley’s extensive interference experiments to detect the ether velocity
was explained by Lorentz who assumed that a moving object contracts in the direction of its velocity
by a factor ;
q
1
0
2
L = L0 = 1
L0 :
(10.19)
This was followed by the …nding by Lorentz and Poincaré that if the spatial coordinates, time and
electromagnetic …elds are all transformed according to what is known as Lorentz transformation,
the Maxwell’s equations remain intact. If a relative velocity V is assumed in the x direction,
the laboratory coordinates (ct; x; y; z) and coordinates in the moving frame (ct0 ; x0 ; y 0 ; z 0 ) may be
assumed to be related through a linear transformation,
t0 =
0
x
y
=
0
0 (t
aV x)
1 (x
V t)
0
= y; z = z;
provided that the two coordinate systems coincide at t = 0: The invariance of the coordinates
perpendicular to the relative velocity, y = y 0 and z = z 0 ; follows from isotropy of space which is
implicitly assumed. For light pulse emitted at t = t0 = 0 and x = x0 = 0 in the positive x direction
(same direction as V ),
x0 = ct0 ; x = ct;
which yield
c=
c
1
01
V
;
acV
(10.20)
while for light pulse emitted in the negative x direction
x0 =
ct; x =
or
ct;
c+V
0 1 + acV
1
c=
(10.21)
From Eqs. (10.20) and (10.21), we …nd
0
=
1
=
and a =
1
:
c2
(10.22)
To determine ; consider a light pulse emitted along the y 0 axis (x0 = 0; that is, x = V t) in the
moving frame,
1
V2
V
x
=
c
1
t:
y 0 = ct0 = c t
c2
c2
4
In the laboratory frame, light propagation is tilted due to the relative motion between the two
coordinates, (ct)2 = y 2 + (V t)2 or
p
y = c2 V 2 t:
Since y 0 = y; we …nd
=r
1
V2
c2
1
:
(10.23)
Desired transformation between (x; t) and (x0 ; t0 ) is
x0 =
(x
V t);
V
t
x ;
c2
t0 =
and the four dimensional coordinates in the laboratory frame (ct; x; y; z) and those in the moving
frame (ct0 ; x0 ; y 0 ; z 0 ) are related through
2
6
6
6
6
4
ct0
x0
y0
z0
3
2
7 6
7 6
7=6
7 6
5 4
V
c
0
0
1
0
V
c
0
0
0
0
0
0
0
1
32
76
76
76
76
54
Its inverse transformation can be found by replacing V with
2
6
6
6
6
4
ct
x
y
z
3
2
7 6 V
7 6 c
7=6
7 6 0
5 4
0
V
c
0
0
0
0
1
0
0
0
0
1
3
ct
x
y
z
7
7
7
7
5
(10.24)
V;
32
76
76
76
76
54
ct0
x0
y0
z0
3
7
7
7
7
5
(10.25)
Graphically, Lorentz transformation may be visualized as contraction in the (ct; x) plane with a
quasi angle de…ned by
cosh = ; sinh = ; tanh = ;
as illustrated in Fig. (10-1). Note that the coordinates (ct0 ; x0 ) are not orthonormal if those in the
laboratory frame are so chosen. Since
tanh(
1
+
2)
=
tanh 1 + tanh
1 + tanh 1 tanh
2
2
=
1
1+
+
2
;
1 2
sum of two velocities cannot exceed c:
Time dilation and length contraction can be visualized as follows. A clock stationary at x = 0
in the laboratory frame moves along the vertical ct axis (x = 0) as time elapses. 1 second in the
laboratory frame appears as second in the moving frame. If seen from the moving frame, the
clock is moving and a moving clock ticks slower. If a clock is stationanry at x0 = 0 in the moving
5
Figure 10-1: Graphical representation of Lorentz transformation for the case = 0:5: Hyperbolic
curves show (ct)2 x2 = 1 (interval of time-like events) and (ct)2 x2 = 1 (length of space-like
object).
frame, it “travels”along the t0 axis (x0 = 0). In the laboratory frame, 1 second in the moving frame
appears as second. In both cases, a moving clock ticks slower.
Likewise, a stick one meter long in the moving frame is contarcted by a factor if seen from
the laboratory frame. Note that length measurements should be done for a common time in both
frames.
It is convenient to introduce a metric tensor de…ned by
2
6
6
gij = g = 6
6
4
ij
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
3
7
7
7
7
5
(10.26)
A contravariant vector (vector in ordinary sense) xi = (x0 ; x1 ; x2 ; x3 ) = (ct; x; y; z) can be converted
to a covariant vector through
xi = gij xj = (x0 ; x1 ; x2 ; x3 )
(10.27)
s2 = xi xi :
(10.28)
so that
6
Note that we follow Einstein’s convention, that is, repeated subscripts and superscripts mean
summation is to be taken,
3
X
gij xj =
gij xj ;
(10.29)
j=0
xi xi = x0 x0 + x1 x1 + x2 x3 + x4 x4 :
(10.30)
s2 can be either positive or negative. The Lorentz transformation for the coordinates
2
6
6
6
6
4
ct0
x0
y0
z0
3
2
7 6
7 6
7=6
7 6
5 4
0
0
0
0
1
0
0
0
0
0
0
1
32
76
76
76
76
54
ct
x
y
z
3
7
7
7
7
5
can be written in the form
x0i = Lij xj ;
(10.31)
where Lij is the Lorentz transformation mixed tensor,
Lij
2
6
6
=6
6
4
0
0
Its inverse tensor is
Lij
1
0
0
2
6
6
=6
6 0
4
0
0
0
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
1
3
(10.32)
3
(10.33)
7
7
7
7
5
7
7
7
7
5
A four dimensional vector that transforms according to the Lorentz transformation law is called
a four-vector. The “position” vector xi evidently forms a four vector. For an object moving at a
velocity v in the laboratory frame, the velocity four vector is
vi =
dxi
=
d
where
d =
1
dt =
(c; vx vy ; vz ) ;
(10.34)
q
(10.35)
1
2
dt;
is the proper time as measured in the moving frame of the object. This is the celebrated time
dilation e¤ect. The magnitude of the four velocity is constant,
v i vi = c2 :
7
(10.36)
The momentum four vector is de…ned by
E
;p
c
pi =
E
; px ; py ; pz ;
c
=
(10.37)
where
E = mc2 ;
(10.38)
is the energy and p is the momentum of mass of a particle having a mass m;
p = mv:
(10.39)
The magnitude of this four vector is also invariant,
i
p pi =
E
c
2
p2 =
2
2
(1
)(mc)2 = (mc)2 :
(10.40)
Example 1 How are the velocity and acceleration transformed?
For a relative velocity V between two inertial frames, spatial coordinates are transformed as
r0k = (rk
Vt); r0? = r? ;
(10.41)
where k and ? indicate components parallel and perpendicular to the velocity V: Since the time is
transformed as
v V
dt;
(10.42)
dt0 =
1
c2
the velocity in the direction of the relative velocity is transformed as
vk0
and the normal component as
0
v?
=
=
dr0k
dt0
=
dr0?
=
dt0
vk V
vV
c2
1
1
v?
;
vV
c2
(10.43)
:
(10.44)
For example, if two particles are approaching each with a velocity v relative to the laboratory frame,
the relative velocity in the frame of either particle is
2v
;
1 + (v=c)2
which cannot exceed c:Note that 2v itself can exceed c:However, a relative velocity between two
objects is meaningful only if it is measured in rest frame of either object. 2v pertains to an
observer in the laboratory frame in which both objects are moving. Therefore, 2v itself is not a
very meaningful velocity. As an example, consider head on collision of two protons. We assume
each proton has a velocity v = 0:9c relative to the laboratory frame. In the rest frame of either
8
proton, the two protons approach with a relative velocity
V =
2v
= 0:9945c;
1 + (v=c)2
and the kinetic energy available for nuclear interaction is
1)mc2 = 8 GeV.
(
To …nd transformation of acceleration, we note the acceleration parallel to the relative velocity
is transformed as
dvk0
ak
0
ak = 0 =
;
(10.45)
vV 3
3 1
dt
2
c
and the normal component as
a0? =
0
dv?
=
dt0
2
a?
vV 2
c2
1
+
2
v?
vV 3
c2
1
a V
:
c2
(10.46)
Note that transformation of the normal acceleration involves parallel acceleration as well.
The inverse transformation for the acceleration is
a0k
dvk
ak =
=
dt
a? =
2
3
a0?
1+
v0 V 2
c2
1+
(10.47)
v0 V 3
c2
2
0
v?
1+
a0 V
3 c2 :
v0 V
c2
(10.48)
In an instantaneously rest frame of a particle, v0 = V; and v? = 0: Then
a0k =
1
3
1
a0? =
The current density J and charge density
3 ak
V2
c2
2
=
3
ak ;
a? :
(10.49)
(10.50)
form the following four vector
J i = (c ; J) = (c ; Jx ; Jy ; Jz );
where
n0
(v) = en(v) = e q
1
J = en(v)v;
9
v 2
c
;
(10.51)
(10.52)
(10.53)
n0
n(v) = q
1
v 2
c
;
(10.54)
is the density of charged particles corrected for length contraction in the direction of the velocity v
and n0 is the charge density in the rest frame, v = 0: The magnitude of the current four vector is
J i Ji = (c 0 )2 = const.
(10.55)
where 0 is the proper charge density in the rest frame of the charge.
In Lorentz gauge, the potentials satisfy the decoupled wave equations,
r2
1 @2
c2 @t2
=
r2
1 @2
c2 @t2
A=
4
;
(10.56)
4
J:
c
(10.57)
Therefore, a resultant four vector potential is
Ai = ( ; A) = ( ; Ax ; Ay ; Az );
which satis…es the wave equation
1 @2
c2 @t2
r2 Ai =
4
Ji :
c
Noting
1 @2
c2 @t2
r2 =
@ @
= g ii @i @i = @ i @i ;
@xi @xi
the wave equation can readily be Lorentz transformed as
@ 0j @j0 A0i =
4 0
J
c i
since @ i @i is Lorentz invariant. The electromagnetic wave equation is thus Lorentz invariant which
guarantees the constancy of the wave propagation velocity c.
10.4
Transformation of Electromagnetic Fields
The electric and magnetic …elds, E and B; do not form four vectors. This is due to di¤erent
vectorial nature of the respective …elds. The electric …eld is a polar vector (or true vector) because
it changes the sign if coordinates are reversed, r ! r: In contrast, the magnetic …eld
1
B(r) =
c
Z
(r
r0 ) J(r0 ) 0
dV ;
jr r0 j3
10
remains unchanged against coordinate inversion since both r r0 and J(r) change sign. The magnetic …eld is an axial vector (or pseudo vector). Rather they are components of an antisymmetric
pseudo tensor,
2
3
0
Bz By
6
7
B ij = 4 Bz
0
Bx 5
By Bx
0
The conventional magnetic Lorentz force
1
fm = J
c
now takes a form
B
1
i
fm
= Jj B ij ; (i; j = 1; 2; 3)
c
where
Jj = ( Jx ; Jy ; Jz )
is the covariant current density. Combining the electric …eld and magnetic …eld into a single …eld
tensor
2
3
0
Ex
Ey
Ez
6
7
6 Ex
0
Bz By 7
ij
6
7;
F =6
0
Bx 7
4 Ey Bz
5
Ez
By Bx
0
the conventional electromagnetic force
1
f = E+ J
c
can be rewritten as
B
1
f i = F ij Jj ; i = 1; 2; 3; j = 0; 1; 2; 3:
c
The component f 0
1
f0 = E J
c
indicates the work done by the electromagnetic …eld. A resultant force four vector is
fj =
1
E J; f
c
:
Using the …eld tensor in Eq. (), we now reformulate Maxwell’s equations as follows. Since
E=
r
1 @A
=
c @t
rA0
1 @A
; B=r
c @t
the …eld tensor can be written as
F ij = @ i Aj
11
@ j Ai
A;
where @ i is the contravariant derivative
@i =
1@
;
c @t
@
=
@xi
@
;
@x
@
;
@y
@
@z
:
For example,
F ii = @ i Ai
@ i Ai = 0; (no summation here)
F 01 = @ 0 A1
F 12 = @ 1 A2
and so on. Di¤erentiating F ij = @ i Aj
1 @Ax @
+
= Ex ;
c @t
@x
@Ay
@Ax
+
= Bz ;
@ 2 A1 =
@x
@y
@ 1 A0 =
@ j Ai with respect to xi covariantly, we obtain
@i F ij = @i @ i Aj
@i @ j Ai = @i @ i Aj
@ j @i Ai :
The …rst term in the RHS is
@i @ i Aj =
1 @2
c2 @t2
r 2 Aj =
4 j
J ;
c
while the second term vanishes because of our choice of Lorentz gauge,
@i Ai =
1@
+ r A = 0:
c @t
Thus
@i F ij =
4 j
J :
c
For j = 0; noting J 0 = c ; we recover Gauss’law,
@i F ij = r E = 4
:
For j = 1; 2; 3; we also recover j-th component of generalized Ampere’s law,
r
B=
4
1 @E
J+
:
c
c @t
Another identity satis…ed by F ij is
@ i F jk + @ j F ki + @ k F ij = 0;
as can be readily checked by substituting F ij = @ i Aj @ j Ai : When i = 0; j = 1; k = 2; Eq. ()
yields
@ 0 F 12 + @ 1 F 20 + @ 2 F 01 = 0;
12
or
1@
( Bz )
c @t
@
Ey
@x
@
( Ex ) = 0;
@y
E)z =
1 @Bz
:
c @t
that is, we recover Faraday’s law
(r
Furthermore, for i = 1; j = 2; k = 3; we recover
r B = 0:
In order to …nd how the …eld tensor F ij is Lorentz transformed, let us consider an arbitrary
contravariant vector B j de…ned by
B j = F ij Ai :
After Lorentz transformation, this becomes
B 0j = F 0ij A0i ;
where
B 0j = Ljk B k ;
A0i = Lik Ak :
Then,
Ljk B k = F 0ij Lim Am
Ljk F lk Al = F 0ij Lil Al
Since Al is arbitrary, we obtain
F 0ij Lim = Ljn F mn :
Multiplying both sides by Lrm and noting
Lim Lrm =
r
i;
we …nd
F 0ij = Lim Ljn F mn = Lim F mn Lnj :
For example,
F 001 = L0m L1n F mn
= L00 L11 F 01 + L01 L10 F 10
=
=
2
Ex +
2 2
Ex = F 01 ;
13
Ex
that is, the electric …eld parallel to the relative velocity V is invariant. For F 002 ; we …nd
F 002 = L00 L22 F 02 + L01 L22 F 12
=
and so on. The overall result is
2
0
6
6
Ex
F 0ij = 6
6 (E
Bz )
y
4
(Ez + By )
(Ey
Ex
0
Bz );
(Ey
(Bz
(Bz
Ey )
(By + Ez )
Bz )
Ey )
0
Bx
(Ez + By )
(By + Ez )
Bx
0
3
7
7
7
7
5
For a relative velocity in an arbitrary direction, the electromagnetic …elds are transformed according
to
E0k = Ek ;
B0k
E0? = (E? +
B0? =
= Bk ;
(B?
B? );
E? );
where k and ? indicate components parallel and perpendicular to the relative velocity V:
The invariance of Bx ; the magnetic …eld parallel to V; can be seen from the following observation.
Bx appears only in F 23 = F 32 : Since the component F ij transforms similar to the coordinates
xi and xj ; Bx trarnsforms as y and z which are invariant. Therefore, Bx does not change through
Lorentz transformation. Likewise, F 02 = Ey transforms as x0 = ct and x2 = y;
Ey0 = (Ey
F 03 =
Bz );
Ez as
Ez0 = (Ez + By );
and so on. F 01 = Ex = F 01 is invariant since the Lorentz transformation corresponds to
rotation in the (ct; x) plane and F 00 = F 11 = 0; F 01 = F 10 form an antisymmetric tensor.
Example 2 A charge e is moving at a velocity V along the x axis. Find the electric …eld and
compare it with the …eld expected from the Lienard-Wirchert potentials.
In the frame of the moving charge, the scalar potential and electric …eld are
0
where
r0 =
=
e
er0
0
;
E
=
;
r0
r03
p
x02 + y 02 + z 02 :
The vector potential in the moving frame is zero, A0 = 0: The x component of the electric …eld is
14
invariant,
Ex = Ex0
ex0
=
r03
= e
(x
2 (x
[
2
= e(1
)
V t)
V t)2 + y 2 + z 2 ]3=2
x Vt
2
V t)2 + (1
(x
Similarly,
Ey = Ey0 = e(1
2
Ez = Ez0 = e(1
2
)(y 2 + z 2 )
3=2
:
y
)
V t)2 + (1
(x
2
)(y 2 + z 2 )
3=2
2
)(y 2 + z 2 )
3=2
z
)
V t)2 + (1
(x
;
:
Therefore, the electric …eld in the laboratory frame is
2
E = e(1
(x
)
V t)ex + y + z
2
V t)2 + (1
(x
)(y 2 + z 2 )
3=2
:
Equivalence of this expression to the Coulomb …eld emerging from the Lienard-Wiechert potentials,
2
E = e(1
n
)
3 R2
;
ret
can be readily proven by noting
n
1
n
(x
=
(x
ret
V t)ex + y + z
2
V t)2 + (1
)(y 2 + z 2 )
1=2
;
where “ret” means the retarded time. Denoting the angle between the x axis and the position
vector R = (x V t)ex + y + z by ; we …nd
E=
eR
R3 (1
2
1
2
sin2 )3=2
:
The Coulomb …eld is “radial” with respect to the present location of the charge. At
Ek =
and at
e
(1
R2
2
= =2;
E? =
)=
= 0;
e 1
;
R2 2
e
:
R2
In highly relativistic case, the …eld is dominated by components perpendicular to the velocity.
15
Angular dependence of the electric …eld for the case
2
1
2
(1
= 0:9 is shown below in polar plot.
sin2 )3=2
3
2
1
-2
-1
1
2
-1
-2
-3
Example 3 Show that the quantities E 2
B 2 and E B are invariant in Lorentz transformation.
Since
2
;
E 2 = Ek2 + E?
and
E0k = Ek ;
E0? = (E? +
B? );
we …nd
E 02 = Ek2 +
2
(E? +
B? )2 :
B 02 = Bk2 +
2
(B?
E? )2 :
Similarly,
Then
E 02
B 02 = Ek2
= Ek2
= E2
Bk2 +
Bk2 +
B2:
2
2
(E? +
(1
2
B? )2
2
)(E?
2
2
B?
)+2
16
E? )2
(B?
2
[E? (
B ? ) + B? (
E? )]
For E B;
E0 B0 = Ek0 Bk0 + E0? B0?
= Ek Bk +
= Ek Bk +
2
2
(E? +
(1
2
B? ) (B?
E? )
)E? B?
= E B:
Note that
(A
B) (C
D) = (A C)(B D)
(A D)(B C):
The invariance of E B means that if E and B are normal to each other in one reference frame,
they remain so in any other frames. If E or B is zero in one reference frame, in other frames they
are normal to each other. Furthermore, if the …elds in the laboratory frame are E and B; there
exists a frame moving at a velocity
1
E B
V
=c 2
;
(V =c)2
E + B2
wherein electric and magnetic …elds are parallel to each other.
10.5
Energy and Momentum Tensor
As shown in Chapter 1 and 3, the electromagnetic force
1
f = E+ J
c
B;
can be expressed as the divergence of the Maxwell’s stress tensor T ij ;
fi =
@j T ij =
1
@j
4
1 2
(E + B 2 ) ij
2
Ei Ej
Bi Bj ; (i; j = 1; 2; 3):
The time component of the force four vector was
1
f0 = J E:
c
Since
J E=
1 @
(E 2 + B 2 )
8 @t
where S is the Poynting vector
S=
c
E
4
r S;
B;
f 0 can be written as
f0 =
1 @
(E 2 + B 2 )
8 c @t
17
1
r (E
4
B) ;
and a resultant four dimensional Maxwell’s stress tensor is
2
1
(E 2 + B 2 ) 41 (E B)x 41 (E B)y
6 81
6 4 (E B)x
T 11
T 12
T =6
6 1 (E B)y T 21 (= T 12 )
T 22
4 4
1
B)z T 31 (= T 13 ) T 32 = (T 23 )
4 (E
1
4
(E B)z
T 13
T 23
T 33
3
7
7
7
7
5
The quaintly
1
1
S=
E
c
4
B;
is the momentum ‡ux density and thus
1
4 c
E
B;
is the momentum density of electromagnetic …elds. The angular momentum density is accordingly
given by
1
r (E B);
4 c
and the total electromagnetic angular momentum is
Z
1
r (E B)dV:
4 c
In four dimensional form, the angular momentum tensor can thus be de…ned by
Z
1
ij
K =
(xi T ik xj T ik )d k ;
c
where d k is the “area” element in the four dimensional (hyper) space having the dimensions of
volume (cm3 ):
10.6
Relativistic Mechanics
In terms of the velocity four vector
v i = (c; v) = (c; vx ; vy ; vz );
and the …eld tensor
F
ij
2
6
6
=6
6
4
0
Ex
Ey
Ez
Ex
0
Bz
By
18
Ey
Bz
0
Bx
Ez
By
Bx
0
3
7
7
7
7
5
the force four vector to act on a charged particle having a charge e can be formulated as
e
F i = F ij vj ;
c
where
vj = (c; v) = (c; vx ; vy ; vz );
is the covariant four velocity. F i reduces to
F i = (e
E; F);
where
F = e(E +
B);
is the space component of the electromagnetic force. Note that
F=e
E;
re‡ecting the fact that the magnetic …eld does not do any work on charged particles.
The magnitude of the velocity four vector is constant,
v i vi =
2
2
(1
)c2 = c2 :
The corresponding momentum four vector
E
;p ;
c
pi =
also has a constant magnitude,
E
c
pi p i =
2
p2 = (mc)2 ;
where
E 2 = (cp)2 + (mc2 )2 ;
is the energy of the particle.
The Lagrangian in nonrelativistic mechanics,
1
L = mv 2 + e(
2
);
A
can be readily generalized as
L =
=
2
mc
mc2
q
1
2
+ e(
1
2
+ v (P
q
19
A
)
mv)
e ;
where
P=
e
e
@L
= mv + A = p+ A;
@v
c
c
is the canonical momentum. The equation of motion can be derived from Lagrange equation
@L
@v
d
dt
= rL:
Noting
r(v A) = v rA + A rv + v
= v rA + v
r
r
A+A
r
v
A;
since the velocity should be …xed in carrying out spatial di¤erentiation, we …nd
d
e
e
(p + A) = (v rA + v
dt
c
c
r
A) er :
Noting
dA
@A
=
+ v rA; r
dt
@t
A = B;
E=
r
1 @A
;
c @t
we recover the familiar equation of motion for a charged particle,
dp
d
1
= ( mv) = e E + v
dt
dt
c
B :
Since the momentum p and energy E are related through
p=
v
E;
c2
the acceleration a = dv=dt can be readily found,
a=
e
[E +
m
B
(
E)] :
(10.58)
Example 4 Analyze the motion of an electron in the Coulomb …eld of a heavy ion carrying a
charge Ze:
Since the electric …eld is static, the energy of electron is conserved,
p
E = c p2 + (mc)2
Ze2
= E0 (const.)
r
(10.59)
By suitable choice of coordinates, the problem can be made two dimensional so that electron motion
is con…ned in the plane (r; ): The momentum can be decomposed into
p2 = p2r +
20
L
r
2
;
(10.60)
where L = rp ; the angular momentum, is also conserved. Then,
s
c p2r +
In nonrelativistic limit v
2
L
r
+ (mc)2
Ze2
= E0 .
r
(10.61)
c, we have
1
2m
s
2
L
r
p2r +
Ze2
= const.
r
(10.62)
In the limit r ! 0; the LHS diverges. (Note that in nonrelativistic limit, the momentum is bounded.)
This means that in nonrelativistic limit, the electron cannot approach the ion inde…nitely. In
relativistic case, however, the LHS of Eq. () remains …nite when r ! 0 provided pr ! 1:
The electron trajectory can be found by di¤erentiating Eq. (10.61) with respect to time,
L2 1
Ze2
+
= 0:
m r3
r2
d
(m r)
_
dt
(10.63)
L = const. means
m r2 _ = L = const.
Then time derivative can be converted into angular derivative through
L
d
d
=
2
dt
m r d
(10.64)
and Eq. () reduces to
d2 1
+ 1
d 2r
2
where
=
When
Ze2
1
= 2 2 E0 ;
r
L c
Ze2
:
cL
(10.65)
(10.66)
< 1; the solution for r( ) is quasi-oscillatory,
r( ) =
b
p
1 + a cos( 1
2
)
;
(10.67)
where a is an integration constant determined by the initial condition and
b=
(cL )2 (Ze2 )2
:
Ze2 E0
(10.68)
p
2 is in general an irrational number, the oscillation is quasi-periodic without closed
Since 1
p
2 ) in
orbits. The electron cannot approach the ion inde…nitely. If > 1; or Ze2 > L ; cos( 1
p
2
Eq. (10.67) becomes cosh(
1 ) and in this case the electron can collapse on the ion.
21
Example 5 Find the power radiated by a charge explicitly in terms of the external electric and
magnetic …elds.
In nonrelativistic limit, the radiation power is given by the Larmor’s formula,
P =
2 (ea)2
:
3 c3
(10.69)
This is still applicable in an instantaneous rest frame of the charged particle,
P =
2 (ea0 )2
;
3 c3
(10.70)
where a0 is the acceleration in that frame. According to Example 1, the acceleration in the laboratory frame a and that in the instantaneous rest frame of the charge a0 are related through
a0 =
3
ak +
2
a? ;
(10.71)
as can be found by choosing v = V: Since
ak =
e
e
E ; a? =
(E? +
m 3 k
m
B);
(10.72)
we …nd for the radiation power
P
=
=
=
2 (ea0 )2
3 c3
2 e4 h 2
Ek + 2 (E? +
3 m2 c3
2 e4 2
(E +
B)2
3 m2 c3
B)2
(
i
E)2 :
(10.73)
If the electric …eld is parallel to the velocity and there is no magnetic …eld as in linear accelerators,
the power becomes independent of the particle energy,
P =
2 e4
E2:
3 m2 c3 k
In the absence of the electric …eld, we obtain
P =
2 e4
3 m2 c3
22
2
(
B)2 :
10.7
Radiation Damping
Consider an electron subject to an acceleration a for a duration T: In nonrelativistic limit, the
electron acquires a kinetic energy of order
1
Ekin = m(aT )2 ;
2
while the amount of energy radiated is
Erad =
2 (ea)2
T:
3 c3
It is reasonable to conjecture that the radiation energy should not exceed the kinetic energy, Erad <
Ekin for otherwise, one must wonder where the radiation energy comes from. This imposes a limit
on T;
e2
T &
= ' 10 23 sec,
mc3
where
is approximately the transit time of light over the classical electron radius,
re =
e2
= 2:85
mc2
10
13
cm.
(To realize such a time scale in cyclotron motion in a magnetic …eld, we need an unrealistically
large magnetic …eld, B ' 1012 T.) In this time scale, radiation reaction on dynamics of charged
particle is expected to become signi…cant. However, the uncertainty principle imposes a more strict
constraint,
E t & ~:
If we choose the self energy of the electron for E = mc2 ; we …nd
t&
where
~c e2
~
=
= 137 ;
mc2
e2 mc3
e2
1
=
;
~c
137
is the …ne structure constant. Therefore, in nonrelativistic cases, quantum mechanical e¤ects
become important well before recoil force due to radiation need to be considered.
However, in radiation from a charge undergoing harmonic motion, radiation reaction is well
de…ned and has been observed experimentally. Recoil force exerted by radiation may be estimated
from energy balance,
Z
T
F vdt =
0
=
Z
2 e2 T 2
a dt
3 c3 0
Z
2 e2 T d2 v
v 2 dt:
3 c3 0
dt
23
Then
F =
2 e2 d2 v
;
3 c3 dt2
and the equation of motion of harmonic oscillator is modi…ed as
x
•
...
x + ! 20 x =
eE
e
m
i!t
:
Solution for x(t) is
x(t) =
e i!t
eE
m !2 + i !3
! 20
;
which remains bounded at the resonance ! ' ! 0 due to radiation damping. The shift in the
resonance frequency is given by
5 3 2
!'
! :
8 0
24