The self-consistency problem in EM radiation
Charged particles are the sources for EM ¤elds, but the particles respond
to the ¤elds through the Lorentz force. Accelerated particles radiate, and the
energy radiated must come from the particlezs energy, and so tha back-reaction
of the ¤eld son the particles cannot be ignored. Letzs use a simple example to
see when this e¤ect is likely to be important.
A particle with charge q and acceleration a radiates power (by the Larmor
formula)
2 q 2 a2
P =
3 c3
Thus the energy lost in time t is
¢E = P t =
2 q 2 a2
t
3 c3
If the particle starts from rest, it gains energy in time t equal to
1
1
K = mv 2 = m (at)2
2
2
Thus the energy lost by radiation is negligible if
¢E¿ 4
2 q 2 a2
t
3 c3
or
tÀ
¿
K
1
ma2 t2
2
(1)
4 q2
´ 2¿
3 mc3
where we have de¤ned
2 q2
2 r0
=
(2)
3 mc3
3 c
where the last equality gives the result for an electron, and r0 is the classical
electron radius. Thus this e¤ect is important for very short time-scale phenomena, or in the very ¤rst instants of a particlezs acceleration. Since ¿ / 1=m;
the e¤ect is greatest for the particles with smallest mass. For an electron
¿ =
¿ (electron) =
¡
¢2
4:8 £ 10 ¡10 esu
2
= 6 £ 10 ¡24 s
3 (9 £ 10 ¡28 gm) (3 £ 10 10 cm)3
(3)
This is a very short time!
Another example. For a particle undergoing oscillatory motion with amplitude d and frequency ! 0 , its energy is
E ' m! 20 d2
1
and its acceleration is
a ' ! 20 d
With t equal to the period, equation (1) for this system takes the form
¡
¢2
2 q2 ! 20 d 1
3
c3
!0
!0 ¿
¿
m! 20 d2
¿
1
So again the e¤ect is important for systems with very short periods.
Radiation reaction
Neglecting radiation, the equation of motion for a particle acted upon by an
external force is:
d~v
m
= F~e x t
dt
To incorporate the energy loss due to radiation, we assume there is an e¤ective
force F~ rad that also acts on the particle, so that the true equation of motion is:
m
d~v
= F~ e xt + F~ rad
dt
The known properties of this force are:
² F~ra d must ! 0 if the acceleration a ! 0; because then no power is radiated.
² F~ra d should be proportional to q 2 because P is. (The sign of the charge
should not appear)
² F~ra d most likely involves the timescale ¿ :
Now we invoke energy conservation. The work done by this force over a time
interval must equal the energy radiated. Thus
Z t2
Z t2
Z t2
2 q 2 d~v d~v
F~ rad :~v dt = ¡
P (t) dt = ¡
¢
dt
3
dt
t1
t1
t1 3 c dt
Integrate by parts:
Z
t2
t1
¯t
Z t2
2 q 2 d~v ¯¯ 2
2 q2 d2~v
F~ra d :~v dt = ¡
¢
~
v
+
¢ ~v dt
¯
3
2
3 c3 dt
t1 3 c dt
t1
The integrated term is zero if:
² We have uniform circular motion so that
d~
v
dt
¢ ~v = 0 at all times
² The motion is periodic and the time interval t2 ¡ t1 is a whole number of
periods
² The acceleration lasts for a ¤nite time less than, and included within,
t2 ¡ t1
2
In any of these cases, we have:
Z t2
Z
~ rad :~v dt =
F
t1
or
Z
t2
t1
µ
t2
t1
2 q 2 d2~v
¢ ~v dt
3 c3 dt2
2 q 2 d2~v
F~ rad ¡
3 c3 dt2
¶
:~v dt = 0
Thus in a time-averaged sense, we may say that
2 q2 d2 ~v
d2 ~v
F~ rad =
= m¿ 2
3
2
3 c dt
dt
(4)
This result satis¤es the preoperties we previously enumerated for F~ rad : The
equation of motion for the particle now becomes:
µ
¶
d~v
d2~v
m
¡ ¿ 2 = F~ ex t
(5)
dt
dt
This is the Abraham-Lorentz Equation of motion.
Implications of the Abraham-Lorentz equation
To understand the implications of this, we solve the equation using an integrating factor: Let
d~v
= ~u et=¿
dt
Then
d2 ~v
d~u t=¿
1
=
e + ~u et=¿
dt2
d¿
¿
and inserting this result into equation (5), we have
¡m¿
d~u t=¿
e
= F~ ex t
d¿
We may now integrate directly to get:
~u =
d~v ¡ t=¿
¡1
e
=
dt
m¿
Z
F~ e xt (t0 ) e ¡t
0
=¿
dt0
To avoid an unknow integration constant, wezd like to inetgrate over a ¤xed
interval, with the variable t at one end. But we know the e¤ects of radiation
reaction are negligible at large times, so
d~v
F~e xt
!
as t ! 1
dt
m
Thus we use t ! 1 as the other limit. Then
Z
0
d~v
et=¿ 1 ~
=
Fe x t (t0) e¡t =¿ dt0
dt
m¿ t
3
(6)
An integral over the future! Letting s = (t ¡ t0 ) =¿ ; this becomes
d~v
dt
=
=
Z
¡1 ¡ 1 ~
F ex t (t ¡ s¿ ) es ds
m 0
Z 0
1
F~ ex t (t ¡ s¿ ) es ds
m ¡1
To verify that this makes sense, letzs expand our function
#
Z 0 "
d~v
1
dF~ e xt (t)
~
=
F ex t (t) + s¿
+ ¢ ¢ ¢ es ds
dt
m ¡1
dt
F~e xt (t) s 0
¿ dF~e x t (t)
0
e j¡1 +
(ses ¡ es )j¡ 1 + ¢ ¢ ¢
m
m
dt
F~e xt (t)
¿ dF~e x t (t)
¡
m
m
dt
=
=
The nth term involves the integral
In
=
=
Z
0
0
¡1
sn esds = nsn es j¡1 ¡ n
Z
0
sn¡1 es ds
¡1
n
¡nIn¡1 = n (n ¡ 1) In¡2 = (¡1) n!
Thus
d~v
1 X
dn F~ e xt
=
(¡1)n ¿ n
dt
m
dtn
n=0
where the ¤rst term is the Newtonian result, while the remaining terms are the
corrections for radiation reaction, and are increasingly negligible as n increases.
We expect
dn F~ e xt
F~ e xt
¿n
» ¿ n n ¿ F~ e xt
n
dt
T
The integral over the future extends over a time roughly or order ¿ because
the exponential reduces the integrand appreciable for t0 >; and so this rather
bizarre result doe not violate "macroscopic causality: The uncertainty principle
restricts our knowledge:
¢E ¢t
¢t
& ~
&
~
¢E
Then if the change in energy is or order mc2 ; the particlezs rest energy, the
restirtcion on observable time intervals is
¢t &
~
~c e 2
3
=
= 137 £ ¿ » 200¿
mc2
e2 mc3
2
4
So the interval in the future that determines the radiation reaction e¤ects is
within the quantum uncertainty.
Now letzs apply this to an electron in an atom. The electron is bound by a
restoring force F~re s = ¡m!20 ~x and so the equation of motion, including radiation
reaction, is
d2
~x
dt2
d2
m 2 ~x
dt
m
=
F~r es + F~ rad + F~ ex t
=
¡m! 20~x ¡ m¿
d2~v
~ e xt
+F
dt2
Letzs begin by considering the case where F~ ex t = 0: The zeroth order solution,
ignoring radiatin reaction, is
~x = ~x0 cos ! 0 t
so letzs look for a solution
~x = ~x 0 e¡i!t
Then stu¢ng in, we get
3
¡m! 2 ~x + m!20 ~x + m¿ (¡i!) ~x = 0
and we expect ! = ! 0 + ° where ° ¿ ! 0 : Thus
¡ (! 0 + °)2 + ! 20 + i¿ (! 0 + °)3 = 0
and expanding to ¤rst order, we have
¡2! 0 ° + i¿ ! 30
=
°
=
0
(! 0 ¿ )
!2 ¿
¡
i
!0 = i 0 ´ i
2
2
2
Now the solution is
~x (t) = ~x 0 cos !0 te¡ ¡t=2
and its Fourier transform is
Z 1
1
~x (!) = p
~x 0 cos ! 0 te¡¡t=2 ei!t dt
2¼ 0
· ¡¡t=2 i(!+! 0) t
¸1
1
e
e
e¡¡t=2 ei(!¡ ! 0)t
p
=
~x 0
+
¡¡=2 + i (! + ! 0 )
¡¡=2 + i (! ¡ ! 0 ) 0
2 2¼
·
¸
1
1
1
p ~x 0
=
+
¡=2 ¡ i (! + ! 0 ) ¡=2 ¡ i (! ¡ ! 0 )
2 2¼
"
#
1
¡=2 ¡ i (! + ! 0 ) + ¡=2 ¡ i (! ¡ ! 0 )
p ~x 0
=
2
2 2¼
(¡=2) ¡ i (! + ! 0 ) ¡=2 ¡ i (! ¡ ! 0 ) ¡=2 ¡ (! 2 ¡ ! 20 )
"
#
1
¡=2 ¡ i!
= p ~x 0
2¼
(¡=2)2 ¡ i!¡ ¡ (!2 ¡ !20 )
5
Then we can compute the power radiated:
P (t) =
W (t) =
2 e2
2
a (t)
3 c3
Z
Z
2 e2 + 1
2 e 2 +1
2
2
a (t) dt =
ja (!)j d!
3 c3 ¡1
3 c3 ¡1
where
"
#
1
¡=2 ¡ i!
~a (!) = ¡! ~x (!) = ¡! p ~x 0
2
2¼
(¡=2) ¡ i!¡ ¡ (! 2 ¡ ! 20 )
2
2
and thus
dW
d!
2 e2
ja (!)j2
3 c3
2
2 e 2 m! 4 x20
(¡=2) + !2
3 mc3 2¼ ! 2 ¡2 + (!2 ¡ !20 ) 2
=
=
Result is very peaked around ! = ! 0 so adding positive and negative frequency
contributions, and noting that ¡ ¿ ! 0 , the power spectrum for positive (physical) frequencies is
dW
d!
=
=
'
2
m!20 x 20
! 20 ¿
2¼ ¡2 + !20 (1 ¡ ! 2 =! 20 )2
1
2
¡
m! 20 x20 £
2
¼ ¡ 2 + ! 20 (1 ¡ ! 2 =! 20 )2
E0 Á (!)
where E 0 is the initial energy of the oscillator and
2
¡
¼ ¡ 2 + ! 20 (1 ¡ !2 =! 20 )2
Á (!) =
is the line shape function.
Z 1
Z 1
2
¡
2¡
d!
d! =
2
2
2
2 2
2
2
2
2
¼
¼!
0
¡ + ! 0 (1 ¡ ! =!0 )
¡ =! 0 + (1 ¡ ! 2 =! 20 )
0 0
Now let
¡
1 ¡ ! 2 =! 20
¡
¢
2!d!
!20
=
=
¡
tan µ
!0
¡
2d!
sec2 µdµ ' ¡
!0
!0
6
As ! ! 1; µ ! ¡¼=2
and as ! ! 0;
tan µ !
so µ » ¼=2: Thus
Z
1
Á (!) d!
!0
À1
¡
Z
¼=2
=
¡
¼! 20
=
1
¼=1
¼
0
¡¼ =2
¡ sec2 µdµ
¡2 =! 20 sec2 µ
Thus the total energy radiated is the initial energy of the oscillator, as expected.
Now letzs look at a driven oscillator. This is a model for an atom encountering an incoming electromagnetic wave.
m
d2x
d3 x
= ¡m! 20 x + m¿ 3 ¡ eE 0 e¡i!t
2
dt
dt
We expect the response to be at the driving frequency: x = x 0 e¡i-t: Then:
3
¡! 2 x + ! 20 x ¡ (¡i!) ¿ x = ¡
and thus,
x (t) =
e
E0
m
e
1
E0 2
e¡ i!t
m
(! ¡ ! 20 + i! 3 ¿ )
where as usual the real part is assumed. Thus
a (t) =
e
¡!2
E0 2
e¡i!t
m (! ¡ !20 + i! 3 ¿ )
and and the time-averaged radiated power is:
P =
1 2 e2
1 e2 4 e2 2
1
2
ja
(t)j
=
!
E
2 3 c3
3 c3 m 2 0 (! 2 ¡ ! 2 )2 + (!3 ¿ )2
0
and the scattering cross section is:
¾
=
P
8¼ e4
!4
=
2
cE 0 =8¼
3 c4 m 2 (! 2 ¡ ! 20 )2 + (!3 ¿ )2
=
¾T
!4
2
(! 2 ¡ ! 20 ) + (! 3 ¿ )2
(7)
(a) For ! À !0 ; we have ¾ ¼ ¾ T : The electron scatters radiation as if it
were free.
(b)For ! ¿ ! 0 ;
µ ¶4
!
¾ = ¾T
!0
7
and we have Rayleigh scattering.
(c)For ! » ! 0 ;
¾
=
=
=
! 40
¾T
2
! 20 )
2
!20
= ¾T
2
(! 2 ¡
+ (! 30 ¿ )
! 20 (1 ¡ !2 =! 20 ) + ¡2
¾T ¼ 2
¡
¿ 2 ¼ !20 (1 ¡ !2 =! 20 )2 + ¡2
¾T ¼
Á (!)
¿ 2
and we have an absorption line with the Lorentz pro¤le. In the plot below, we
used ¡=!0 = 1=5 in equation (7).
25
20
15
y
10
5
0
0.5
1
1.5
x
2
2.5
3
In a log-log plot it looks like:
.1e2
.1
x
.5
5.
.1e2
1.
.1
.1e-1
y
.1e-2
.1e-3
The total scattering cross section (frequency inegrated) for the line is:
Z
¾T ¼ 1
¾ =
Á (!) d! =
¿ 2 0
¾T
=
¼
2¿
8
where we used our previous result for the integral over the line pro¤le . Thus
¾
8¼ e4
1
¼
e2
3 c4 m 2 2 23 mc
3
=
= 2¼ 2
e2
¼e2
= (2¼)
mc
mc
which is independent of ! 0 . Thus we may write the result in terms of frequency
º as:
¼e2
¾ (º ) =
f ' (º)
mc
where
Z
Z
Z
Á (!) d! =
Á (º ) 2¼dº = ' (º ) d!
' (º ) =
'
and
2
¡
¡=2¼
=4
2
2
¼ ¡2 + ! 20 (1 ¡ ! 2 =!20 )2
¡ + ! 0 (1 ¡ !=!0 ) 2 (1 + !=! 0 )2
¡=2¼
¡=2¼
4
2 2
2
2 =
2
2
2
2
¡ + (2¼) º 0 (1 ¡ º =º 0 ) (2)
(2¼) (º ¡ º 0 ) + (¡=2)
Z
1
Á (º) dº = 1
0
The fudge factor f is called the oscillator strength. It may be measured or
calculated (for simple atoms) using quantum mechanics. It takes into account
deviations from classical theory. Values of f are tabulated in reference works
such as Allenzs Astrophysical Quantities.
9