Radiation damping, line pro®les and Rayleigh
scattering
S.M.Lea
1 Radiation damping.
An accelerating particle radiates energy at the rate
2 e2 2
a
3 c3
The energy radiated must come from the electronzs kinetic energy, so
P =
dE
d 1
mv 2
=P =
mv 2 »
dt
dt 2
tloss
And thus the time scale for the particle to lose an appreciable amount of energy is:
µ ¶
3 mc3 v 2
3 mc3 2
t0
tloss »
»
t0 = t0
2
2
2
2 e a
2 e
¿
where t0 = v=a is a characteristic time assocuated with the particlezs motion. (For
circular motion or oscillatory motion, t0 is approximately the period.) In this expression
the timescale
¡
¢2
4:0 £ 10 ¡10 esu
2 e2
2
¿ =
=
= 4 £ 10¡ 24 s
(1)
3 mc3
3 (9 £ 10 ¡28 g) (3 £ 10 10 cm/s)3
where the numerical value is for an electron. Heavier particles have smaller values of ¿ :
Thus the loss time is in general very long, and the radiation loss process is most important
for systems with small values of t0: (This result is surprisng, perhaps.)
To ®nd the force that acts on the electron to remove energy from it, look at the integrated
energy loss:
Z t2
4E =
P (t)dt
t1
t2
=
Z
t1
~ rad ¢~
¡F
v dt =
1
Z
t2
t1
2 e2 d
v ¢~
~
a dt
3 c3 dt
Integrate by parts:
=
Z
t2
t1
¯ t2 Z t2
¯
2 e2
2 e2 d
~
¯ ¡
¡ Fra d ¢~
vdt =
v
~
¢
a
~
a ¢v
~
~ dt
¯
3
3 c3
t1 3 c dt
t1
The integrated term is zero if the motion is periodic, and t2 ¡ t1 is a whole number of
periods, or for circular motion where v
~ is perpendicular to ~
a : Then we have:
¶
Z t2 µ
2
2
e
d
~ ra d ¡
F
a ¢~
~
v dt = 0
3 c3 dt
t1
Thus it is reasonable to suppose that in a time-averaged sense we may write:
2
d
~ ra d = 2 e d ~
F
a =m¿ ~
a
3 c3 dt
dt
(2)
2 Application to atomic systems
Model an atom as a classical, damped oscillator with natural frequency ! 0 :
including the radiation damping, the equation of motion for the system is:
m
d2 ~
x
d3
2
=
¡m!
x
~
+m¿
x
~
0
dt2
dt3
Then,
(3)
We expect a solution of the form:
x=~
¹
x0 (t) cos (! 0 t)
where the amplitude x
~0 (t) is a slowly decreasing function of time due to the damping.
Then from equation 3, we then have:
µ
¶
µ
¶
d3
d d2 ~
x
d
d3
2
x
~
=
=
¡!
x
~
+¿
x
~
0
dt3
dt dt2
dt
dt3
µ ¶
d¹
x
' ¡! 20
dt
to zeroth order in the small quantity ! 0 ¿ : We only need zeroth order since this derivative
is multiplied by another ¿ in equation 3. So thedifferential equation becomes:
µ ¶
d2 ~
x
d¹
x
2
2
= ¡!0 x
~¡¿ ! 0
2
dt
dt
Now look for a solution of the form:
x
¹=~
x0 e®t
We ®nd
®2 + !20 + ¿ ®! 20 = 0
which has the solutions:
®=
where
¡! 20 ¿ §
q
2
(! 20 ¿ ) ¡ 4! 20
2
= §i! 0
p
1 ¡ ¡2 =4 ¡
¡
2
¡ = ! 20 ¿
(4)
Thus the solution is a damped oscillation at a 2frequency very close to the natural oscillation
frequency.
The solution that satis®es the initial condition ~
x=x
~0
¡¡t=2
x (t) = ~
~
x0 e
at t = 0 is:
cos !0 t
and its Fourier transform is:
Z 1
1
x (!) = p
~
x0 e¡ ¡t=2 cos ! 0 te i!t dt
~
2¼ 0
µ
¶¯ 1
1
exp (¡¡t=2 + i! 0 t + i!t)
exp (¡¡t=2 ¡ i! 0 t + i!t) ¯¯
=
p ~
x0
¡
¯
¡¡=2 + i (! 0 + !)
¡¡=2 + i (¡!0 + !)
2 2¼
0
µ
¶¯
¯
1
1
1
¯
=
p ~
x0 ¡
+
¡¡=2 + i (! 0 + !)
¡¡=2 + i (¡! 0 + !) ¯
2 2¼
This function is strongly peaked around the two values of ! = §! 0 :
Now the energy radiated by the system per unit frequency is:
¯2
dE
2 e2
2 e2 ¯¯ 2
2
¯
=
j~
a
(!)j
=
!
x
~
(!)
d!
3 c3
3 c3
2 e2 x 20
!4
=
3 c3 8¼ (¡=2) 2 + (! ¡ ! 0 )2
for frequencies near ! 0 : There is an equal contribution for frequncies near ¡!0 ; which
corresponds to the same real frequency. Thus, including both contributions, and using
equation 1 we get:
dE
x2
!20
= 0 m¿ !20
2
d!
4¼
(¡=2) + (! ¡ ! 0 )2
where we set ! ' ! 0 except where they are subtracted. Simplifying, and using equation
4:
dE
x 20
¡=2
=
m!20
2
2
d!
2¼
(¡=2) + (! ¡ ! 0 )
1
=
m!20 x 20 Á (!)
2
where 12 m! 20 x 20 is the oscillatorzs energy at t = 0 and Á (!) is the Lorentz line pro®le
function that describes how energy is distributed in frequency.
Note that:
Z 1
Z
1 1
¡=2
Á (!) d! =
d!
¼ 0 (¡=2) 2 + (! ¡ ! 0 )2
0
Z
1 ¼=2 sec2 µ
=
dµ
¼ µmin 1 + tan2 µ
where we used the substitution ! ¡ ! 0 = (¡=2)tanµ;
µmin = tan
since ¡ ¿ ! 0 :
Thus:
Z
0
as required.
1
¡1
Á (!) d! =
and the lower limit is
(¡2! 0 =¡) ' ¡¼=2
1 ³ ¼ ³ ¼ ´´
¡ ¡
=1
¼ 2
2
3
3 Scattering by atoms
If the oscillator is now driven by an externally applied force due to an incoming EM
wave, The equation of motion becomes:
m
d2 ~
x
d3
2
~
=
¡m!
x
~
+m¿
x¡e E
~
0
dt2
dt3
or
d2 ~
x
d3
2
~ 0 cos !t = ¡ e Re E
~ 0 ei!t
+
!
x
~
¡
¿
x = ¡e E
~
0
2
3
dt
dt
m
¡
¢
The expected solution is an oscillation at the same frequency !: So letting ~
x = Re ~
x0 ei!t
we ®nd:
¡ 2
¢
e~
¡! + ! 20 + i¿ ! 3 ~
x0 = ¡ E
0
m
or
1
~0 e
x
~0 = ¡ E
m (¡! 2 + ! 20 + i¿ ! 3 )
and the radiated power at frequency ! is:
P (!) =
2 e2 2
2 e2 2 e2
!4
´ cos 2 (!t + Á)
a =
E0 2 ³
3
3
3c
3c
m (! 2 ¡ !2 ) 2 + (¿ ! 3 )2
0
where the phase shift Á is given by:
tan Á =
Now we take the time average to get:
< P (!) >=
¿ !3
! 2 ¡ !20
1 e4
!4
2
³
´
E
0
2
2
3 m 2 c3
(! 2 ¡ ! 2 ) + (¿ ! 3 )
0
and the scattering cross section is:
¾ (!) =
=
< P (!) >
8¼ e4
!4
³
´
=
2
2
4
cE 0 =8¼
3 m c (!2 ¡ !2 )2 + (¿ ! 3 )2
0
¾T ³
!4
2
2
(! 2 ¡ ! 20 ) + (¿ !3 )
Now letzs look at some limits:
For ! À !0
´
¾ (!) = ¾ T
At high frequencies the electron }looks like} a free electron, and the cross section is the
Thomson scattering cross section
For ! ¿ !0
!4
¾ (!) = ¾ T 4
!0
which is Rayleigh scattering.
4
(5)
Finally, near the resonance, ! ¼ ! 0 ; we have:
!4
¾ (!) =
¾T ³
(! ¡ ! 0 ) (! + ! 0 ) + (¿ !3 )
'
¾T ³
(! ¡ ! 0 )2 (2! 0 )2 + (¿ ! 30 )
=
=
=
=
¾T ³
2
2
!40
2
! 20
2
2
(! ¡ ! 0 ) 4 + (¿ ! 20 )
´
2
´
´
¾T
! 20 ¿ =2
³
´
2¿ (! ¡ ! 0 )2 + (¡=2)2
8¼ e4 3mc3
¡=2
³
´
2
4
2
2
2
3 m c 4e
(! ¡ ! 0 ) + (¡=2)
2¼
e2
¡=2
³
´
mc (! ¡ ! 0 )2 + (¡=2)2
And we regain the Lorentz line pro®le. Using our previous result for the inetgral, we get:
µ 2¶
Z
¼e
¾ (!) d! = 2¼
mc
or, equivalently:
Z
¼e2
¾ (º ) dº =
mc
The classical calculation should be modi®ed to allow for quantum mechanical effects by
including the oscilltor strength f : Then
¾ (º) = f
¼e2
Á (º)
mc
5
Lorentz pro®le with ¡ = 0:1
:1=2
1
¼ (:1=2)2 +(!¡5)2
6
5
4
3
2
1
0
4
4.2
4.4
4.6
4.8
5
6
5.2
5.4
5.6
5.8
6