Scattering
S.M.Lea
1 Polarization
~ in terms of two orthogonal polarization vectors e^1 and
Letzs describe the electric ®eld E
^e2 : Then:
~ =E
~ ¢ ^e1 + E
~ ¢ ^e2
E
and
¯ ¯2 ³
´2 ³
´2
¯ ~¯
~ ¢ e^1 + E
~ ¢ ^e2
¯ E¯ = E
¯ ¯2
¯~ ¯
Thus the power radiated, which is proportional to ¯E
¯ ; is the sum of the powers radiated
into the two orthogonal polarizations.
~ / ^n £ (^
Now in the non-relativistic case, or with ~a parallel to ~¯; E
n£~
a) and so
~
E ¢ ^e1 / (^
n £ (^
n £ ~a)) ¢ e^1
=
=
(^
n (^
n ¢ ~a) ¡ ~a) ¢ e^1
(^
n ¢ ~a) n
^ ¢ e^1 ¡ ~a ¢ e^1
But since n
^ is perpendicular to ^e1 ; then
~ ¢ ^e1 / ~a ¢ e^1
E
Generally this is the easiest way to compute the power.
2 Thomson scattering
Let the incident wave be described by:
³ h
i´
~ (~x ; t) = ^"0 E0 exp i ~k ¢ ~x ¡!t
E
where ^"0 is a vector describing the polarization. The electric ®eld is incident on an electron
and gives it an acceleration:
³ h
i´
¡e
~a =
^"0 E 0 exp i ~k ¢ ~x¡!t
m
1
and then the power radiated into polarization i is:
dP i
d-
=
=
and the time average is
e2
2
j~a ¢ ^e1 j
4¼c3
´2 ³
³ h
i´
´2
e2 ³ e
~k ¢ ~x ¡ !t ¢ e^¤1
E
"
^
exp
i
0
0
4¼c3 m
dP i
e4
E 20
>=
(^"0 ¢ ^e¤i )2
d4¼m 2 c3 2
Note: by using the complex conjugate of the polarization vector to compute the absolute
value, we can include circular polarizations in our results.
The differential scattering cross section is de®ned as:
d¾
energy radiated/unit time/unit solid angle
´
dincident -ux
and thus
<
d¾ i
d-
=
=
=
2
e4
2
"0 ¢ ^e¤i )
8¼ m2 c3 E 0 (^
cE 20 =8¼
e4
(^"0 ¢ ^e¤i )2
2 4
m c
r20 (^"0 ¢ ^
e¤i )2
where
r0 =
e2
mc2
is the classical electron radius.
Letzs assume the incident wave is incident along the z¡axis and is linearly polarized
along the x^ ¡axis: ^" 0 = x^ : An outgoing wave scattered at angle µ may be described in terms
of the polarization vectors
e1 = cos µ (^
^
x cos Á + y
^ sin Á) ¡ ^z sin µ
and
as shown in the diagram:
e2 = ¡^
^
x sin Á + ^
y cos Á
2
(1)
Then
"0 ¢ ^e1 = cos µ cos Á
^
and
"0 ¢ ^e2 = ¡ sin Á
^
Thus:
d¾
d-
d¾ 1
d¾ 2
+
d-¡
d¢
= r02 cos2 µ cos 2 Á + sin2 Á
=
For the perpendicular incident polarization ^
"0 = ^
y; we have:
¡
¢
d¾
= r20 cos2 µ sin 2 Á + cos 2 Á
dand thus the differential scattering cross section for unpolarized incident radiation is:
¢
d¾
r2 ¡
= 0 1 + cos 2 µ
d2
and thus the total scattering cross section is:
µ
¶¯ +1
Z
Z
¢
d¾
r2 +1 ¡
¹3 ¯¯
¾ =
d- = 2¼ 0
1 + ¹2 d¹ = ¼r02 ¹ +
d2 ¡1
3 ¯ ¡1
8¼ 2
=
r = 0:665 £ 10 ¡24 cm2
3 0
3
This result is valid for hº ¿ mc2 = 511 keV, and an electron at rest or moving with v ¿ c:
3 Compton scattering
For high-energy photons it is important to include the photon momentum and regard the
scattering process as a particle collision. Then we have:
Energy conservation:
mc2 + hº = °mc2 + hº 0
(2)
and momentum conservation: x¡component
hº
hº 0
= °mv cos Á +
cos µ
c
c
and y¡component
(3)
hº 0
0 = °mv sin Á ¡
sin µ
(4)
c
Now square equations (3) and (4) and add:
µ 0
¶2 µ 0
¶2
hº
hº
hº
2
2
2
(°mv cos Á) + (°mv sin Á)
= (°mv) =
sin µ +
cos µ ¡
c
c
c
µ 0 ¶2 µ ¶2
2
0
hº
hº
2h ºº
=
+
¡
cos µ
c
c
c2
But
1
¯ 2 = 1 ¡ 2 ) ° 2 ¯ 2 = °2 ¡ 1
°
So
µ
¶
µ 0 ¶2
¡ 2
¢ mc2 2
º
º0
° ¡1
=1+
¡ 2 cos µ
(5)
hº
º
º
Then squaring equation (2) gives:
µ 2 ¶2 µ
¶2
µ 0 ¶2
mc
º0
º
º0
2
(° ¡ 1)
= 1¡
= 1+
¡2
(6)
hº
º
º
º
Subtracting equations (6) and (5), we get:
µ 2 ¶2
mc
º0
2 (° ¡ 1)
= 2 (1 ¡ cos µ)
hº
º
Then using equation (2) again gives:
µ 2 ¶2
µ
¶
º0
mc
hº
º0
(1 ¡ cos µ) =
1
¡
º
hº
mc2
º
µ
¶
0
2
2
º
mc
mc
1 ¡ cos µ +
=
º
hº
hº
4
and so
º
0
= º
=
µ
mc2
hº
hº
1 + mc
2
¶
1
1 ¡ cos µ +
º
(1 ¡ cos µ)
¡ mc2 ¢
hº
The differential cross section is modi®ed by the frequency shift:
¯
µ 0 ¶2
d¾ ¯¯
º
2
¤ 2
=
r
(^
"
¢
e
^
)
0
0
i
d- ¯ QM
º
5