1
Synchrotron radiation
1.1
Qualitative discussion
Consider a particle moving along a circular path. The radiation is beamed into
a cone of opening angle » 1=°; as shown above. Thus an observer at P sees
radiation while the particle is pointed within an angle » 1=° of P; and thus
while the particle travels an arc of length s » °2 r; where r is the radius of the
circle. The particle travels this distance in a time
¢t =
2r
°v
Radiation emitted at A at time t1 reaches P , distance d away, at time
T 1 = dc + t1:
Radiation emitted at B at time t2 = t1 + ¢t = t1 + 2r=°v reaches P at time
2r
T 2 = dc ¡ °c
+ t1 + 2r
: The length of the observed pulse is
°v
2r
¢T = T 2 ¡ T 1 =
°
Now °2 =
1
1¡¯ 2
and so ¯ 2 = 1 ¡
1
=
¯
Thus:
2r
¢T =
°c
µ
µ
1
°2
1¡
µ
1
1
¡
v
c
¶
and then:
1
°2
1
¡1
¯
¶ ¡1=2
¶
2r
=
°c
' 1+
µ
1
2° 2
¶
1
r
¡1 = 3
¯
° c
But for an extremely relativistic particle, ! 0 = v=r ' c=r; so ¢T = 1=°3 ! 0 :
Now a pulse of width ¢T has Fourier components up to at least 1=¢T ; and so
the observed frequencies extend to at least °3 ! 0 ; where !0 = eB=°mc = ! B =°:
An exact calculation (See eg Jackson Ch 14) shows that the spectrum actually peaks around this high frequency º ¼ ° 2 º B : We can ¤nd a typical value for
1
ºB :
ºB =
eB
4:8 £ 10 ¡10 esu £ 1 £ 10¡ 6 G
=
B¹G
2¼mc
2¼ (511keV/c)
Units: In this unit system
1G =
1 esu
2
(1 cm)
and
1 esu 2
1 cm
1 erg =
So
ºB
=
4:8 £ 10 ¡10 esu £ 1 £ 10 ¡6 G
3 £ 10 10 cm/s B¹G
2¼ (511 £ 1:6 £ 10 ¡9 erg)
4:8 £ 10 ¡10 esu £ 1 £ 10 ¡6 esu/cm2
3 £ 10 10 cm/s B¹G
2¼ (511 £ 1:6 £ 10 ¡9 erg)
= 2:8 Hz B¹G
=
(1)
This is a useful number to remember.
For cyclotron radiation, the power radiated is
µ
¶2
2 e2 2
2 e2 evB?
P =
a =
3 c3
3 c3
mc
=
2 e4 v 2 2
B
3 m 2 c5 ?
For relativistic particles, v ' c; we can obtain the correct result by writing the
formula in covariant form. Since P » E=t and both E and t are 0¡components
of 4-vectors, we may write
2 e2 # #
P =¡ 3a¢ a
3c
where
µ
¶
#
dv
d° d
#
a=
= ° c ; (°v)
d¿
dt dt
and
d°
d
1
~a
=
p
= ° 3 ~¯ ¢
dt
dt 1 ¡ ~v ¢ ~v =c2
c
Thus
#
#
¡a ¢ a
#
¶2
d°
2 2
= ¡°
¡v
¡° a
dt
"
#
µ
¶2
¡
¢
a
2
2 2
2
3~ ~
2
= ° ° a ¡c ° ¯¢
1¡¯
c
·
³
´2 ¸
4
2
2 ~
= ° a ¡ ° ¯ ¢ ~a
2
"µ
d°
c
dt
¶2
2
2
µ
For synchrotron radiation with ~a?~v ;
#
#
¡ a ¢ a = ° 4 a2
and thus
P
=
=
µ
¶2
2 e2 4 evB?
°
3 c3
°mc
4
2 e
2
° 2 B?
3 m2 c3
If ~v makes an angle ® with the magnetic ¤eld, then B? = B sin ® and
P =
2 e4 2 2
° B sin 2 ®
3 m 2 c3
For an isotropic distribution of particle velocities
Z
¢
1 +1 ¡
2
< sin 2 ® >=
1 ¡ ¹2 d¹ =
2 ¡1
3
Thus
4 8¼e4 2 B 2
4
°
= ¾ T c° 2 uB
2
3
9m c
8¼
3
This should be compared with the result for Compton scattering (eqn 1 in those
notes)
4
P = ¾ T c°2 u p h
3
Synchrotron radiation may be viewed as scattering of virtual photons associated
with the magnetic ¤eld.
P »
3
1.2
Spectrum of emitted radiation
1.3
For a power law distribution of electrons, we may compute the radiated spectrum as we did for compton
scattering. We approximate by assuming that all the
emission is at the peak frequency, and using a delta
function.
Z
Z
¡
¢
1
1
4
jº =
Pº (°) n (°) d° =
¾T c° 2uB ± º ¡ ° 2 ºB K° ¡pd°
4¼
4¼ ´3
³
p
Z
± ° ¡ º=º B
¾T c
= K
uB ° 2
° ¡p d°
3¼
2°º B
µr ¶¡p+1
¾T c
1
º
= K
uB
3¼
2º B
ºB
¾ T c º (p¡3)=2
uB B
º ¡(p¡1)=2
3¼
2
µ
¶
8¼e4 c B2
eB (p¡3)=2 ¡(p¡1)=2
= K
º
3m2c3 6¼ 8¼ 2¼mc
³ e ´(p¡3)=2
e4
= K
B (p+1)=2º ¡(p¡1)=2
2
2
18¼m c 2¼mc
Thus the emiited spectrum is also a power law whose
slope re¢ects the slope of the underlying electron spectrum.
p¡1
jº / º¡® with ® =
2
The emission is also proportional to the electron density (through K ) and to the magnetic ¤eld strength
= K
jº / B (p+1)=2 = B ®+1
Astrophysical sources of synchrotron radiation
1.3.1
Supernova remnants
Some supernova remnants have a shell structure- wezll talk about these more
when we get to ¢uids and shock waves. But some look "¤lled" on the sky.
These sources are called plerions. The prototypical source is the Crab Nebula.
These sources typically exhibit power-law spectra in the radio with spectral
index ® » 0:8: The pulsar (remnant of the exploded star) accelerates electrons
to high energy, and these particles produce synchroton radiation. The crab
radiates all the way out to gamma rays.
4
If all the particles are accelerated in the initrial explosion, then as the remnant expands the density of particles (and also the magnetic ¤eld) decreases.
This leads to a dimming of the source. The resulting relation between synchrotron surface brightness and size was used to determine distances. However
there are pitfalls because there are di¤erent phases in SNR evolution, as wezll
see.
An electron of energy °mc2 loses energy at a rate P and so has a lifetime
¡
¢3
°mc2
°mc2
9 m 3 c5
9 mc2
¿ =
= 4 e4
=
=
2 2
P
4° e4 B 2
4° ce 4B 2
9 m2c 3 ° B
19
(511 keV) 3
2
° 4 (3 £ 10 10 cm/s) (4:8 £ 10¡ 10 esu)4 (10 ¡6 G)2 B¹G
=
So
¿
=
19
°4
¢3
1:6 £ 10 ¡9 erg/keV
³
´2
4
2
cm/s) (4:8 £ 10 ¡10 esu) 10 ¡6 esu/cm
B2¹G
3
(511 keV)
(3 £ 10 10
¡
=
3
1
20 erg
cm3 ¢ s
2 7:721 8 £ 10
°B¹G
esu 6
=
1 7:7 £ 10 20 s
1
13
=
y
2
2 2:5 £ 10
7
°B¹G ¼ £ 10 s/y
°B¹G
(2)
The value of ° needed to produce a photon of energy E is given by
°2
So
°
=
=
=
=
eB
E
=º=
2¼mc
h
r
r
E 2¼mc
E 2¼mc2
=
h eB
hc eB
s
v
u
Eke V u
2¼ (511 keV) (1 keV)
³
´
t
B¹G (1:24 keV ¢ nm) (4:8 £ 10¡10 esu) 10¡ 6 esu/cm2
s
v
Eke V u
2¼ (511) (1:6 £ 10¡ 9 erg)
u
³
´
t
B¹G (1:24 £ 10 ¡7 cm) (4:8 £ 10¡ 10 esu) 10¡ 6 esu/cm2
s
Eke V
2:9 £ 108
B¹G
(3)
Thus the lifetime of the particle in terms of the energy of the emitted photon is
r
1 2: 5 £ 10 13 y B¹G
¿ =
B2¹G 2:9 £ 108
Eke V
=
8: 6 £ 104 y
q
3
B¹G
Ek eV
5
So if we observe °¡rays with energy 10 MeV, the lifetime is
8: 6 £ 104 y
1
¿ » p q
= 860 y q
104 B3¹G
B3¹G
Since the crab was formed in 1054, 950 years ago, we conclude that energetic
particles must be continually supplied to the remnant.
1.3.2
Extragalactic radio sources
These are most often associated with giant elliptical galaxies, and they are
double-lobed. Each lobe is fed by a jet that runs from the central engine at
the nucleus out to the lobe. The specturm is a power law with ® » 1: Letting
E = °mc2 be the electron energy, the total luminosity is
Z
L =
P (°) N (°) d°
Z °m a x
4
=
¾ T c° 2 uB K °¡ p d°
3
°m i n
¸°
3¡ p
4
° 3¡p m a x
4
° 3¡ p ¡ ° min
=
¾ T cuB K
= ¾ T cuB K max
(4)
3
3 ¡ p °m i n
3
3¡p
We can obtain °max from the maximum frequency in the observed spectrum.
Then (equation1)
r
r
2¼º m ax mc
ºm a x
°m ax =
=
eB
3 (Hz/¹G) B¹G
Then since ® = (p ¡ 1) =2; p ¡ 1 = 2® and 3 ¡ p = 2 (1 ¡ ®) :
L
=
=
=
µ
¶ (3¡p)=2 (3¡p)=2
4
B2
2¼mc
º max
¡ º (3¡p)=2
min
¾T c K
3
8¼
eB
3¡ p
µ
¶ 1¡®
·
1¡® ¸
1
2¼mc
K 1+® º 1¡®
max ¡ º min
¾T c
B
3
e
2¼
2 (1 ¡ ®)
· 1¡®
³
´
1¡ ® ¸
1
mc 1¡ ® K
1+ ® º max ¡ º min
¾T c
®B
3
e
(2¼)
2 (1 ¡ ®)
The total energy in electrons is:
Ee
=
=
=
Z
°mc2 N (°) d°
¯
2¡p ¯° max
2 °
¯
Kmc
2 ¡ p ¯° min
Kmc2
6
2¡p
°2¡p
max ¡ ° min
2¡ p
(5)
We can replace the unobservable parameter K with the observable Luminosity
L; (equation 5) and 2 ¡ p = 1 ¡ 2®
2¡ p
Ee
= mc2
=
=
p
° 2¡
max ¡ °min
2¡p
L
° 3¡p ¡° 3¡p
4
¾ cuB max3¡ p min
3 T
2¡ p
3 mc ° 2¡p
max ¡ ° min L 3 ¡ p
8¼
3¡ p
2
4 ¾ T ° 3¡p
max ¡ ° min B 2 ¡ p
2¡p
mc 1 ¡ (° min=°max )
L 2 (1 ¡ ®)
6¼
¾ T ° max 1 ¡ (° min=°max )3¡p B 2 1 ¡ 2®
The energy stored in the magnetic ¤eld in the source volume V is
UB =
B2
V
8¼
Thus the total energy is
1¡2®
Ee + UB = 6¼
mc
1 ¡ (° min=°max )
L 2 (1 ¡ ®) B 2
+
V
2(1¡
®)
¾ T ° max 1 ¡ (° =°
B2 1 ¡ 2®
8¼
)
min
max
Writing the result in terms of the observed frequencies, we have
E e + U B = 6¼
r
eB
mc 1 ¡ (º min=º max )(1¡2®)=2 L 2 (1 ¡ ®) B 2
+
V
2¼º m ax mc ¾ T 1 ¡ (º min=º max )(1¡®) B 2 1 ¡ 2®
8¼
The ¤rst term is (roughly) inversely proportional to B3=2 while the second
is directly proportional to B2 :
E tot =
with
A=
3
¾T
r
A
V
+ B2
B 3=2
8¼
(1¡ 2®)=2
2¼emc 1 ¡ (º min=º max )
2 (1 ¡ ®)
L
º m ax 1 ¡ (º min=º max )(1¡ ®) 1 ¡ 2®
(6)
Thus we can minimize the energy required to produce the observed luminosity
by adjusting B :
dE to t
3 A
BV
=¡
+
=0 )B =
dB
2 B5=2
4¼
µ
6¼A
V
With this value we ¤nd
UB =
µ
6¼A
V
¶ 4=7
4
V
6 7 V 3=7 4=7
=
A
8¼
8 ¼ 3=7
7
¶2=7
(7)
while
Ee = A
µ
6¼A
V
¶ ¡3=7
3
= 67
V 3=7 4=7
4
A
= UB
¼ 3=7
3
Thus the least possible total energy is required if the energy is almost equally
divided between the electrons and the magnetic ¤eld. This condition is referred
to as "equipartition". Then the total energy is
E tot
=
=
7 V 3=7 4=7
A
4 (6¼)3=7
" r
# 4=7
(1¡ 2®)=2
7 V 3=7L 4=7 3
2¼emc 1 ¡ (º min=º max )
2 (1 ¡ ®)
4 (6¼)3=7 ¾ T
º m ax 1 ¡ (º min=º max )(1¡ ®) 1 ¡ 2®
Example: Cygnus A
® = 0:75; Flux = 1:3 £ 104 Jy @ 100 MHz. Angular size of each lobe is 38",
distance is 220 Mpc. º min » 1 MHz and º max » 1500 MHz. Then
µ
¶
4 3
V =2
¼r
3
with
r
=
=
Thus
V =
and
L
=
=
=
=
=
Z
º max
º min
µ
F0
(µ
1
38
¼
(220 Mpc) £
£
2
60 £ 60
180
2:0 £ 10 ¡2 Mpc = 20 kpc
¢3
8 ¡
¼ 20 £ 3 £ 10 21 cm = 1:8 £ 10 69 cm3
3
º
º0
¶¡ ®
dº £ 4¼d2
¶ 1¡® µ
¶ 1¡ ®)
F0 º 0
º max
º min
¡
£ 4¼d2
1¡ ®
º0
º0
¡
¢¡
¢(
µ ¶ 0:25 )
1:3 £ 104 Jy 100 £ 10 6 Hz
1
:25
15 ¡
£ 4¼ (220 Mpc)2
0:25
10
³
´¡
¢
2
1:3 £ 104 £ 10 ¡26 W/m ¢ Hz 100 £ 106 Hz
¡
¢2
f1:4g £ 4¼ 220 £ 3 £ 1022 m
0:25
3:985 £ 10 37 W = 4 £ 1044 erg/s ' 10 11 L ¯
8
Then
Be q
=
=
µ
µ
6¼A
V
¶ 2=7
6¼
1:8 £ 1069 cm3
¶ 2=7 "
3
¾T
r
(1¡ 2®)=2
2¼emc 1 ¡ (º min=º max )
2 (1 ¡ ®)
L
º m ax 1 ¡ (º min =º max )(1¡ ®) 1 ¡ 2®
# 2=7
# 2=7 µ
¶ 2=7 "
¶ 1=7
¡1=4
6¼
3
1 ¡ (1=1500)
0:5
2¼emc
44
=
4 £ 10 erg/s
1:8 £ 1069 cm3
0:66 £ 10 ¡24 cm2 1 ¡ (1=1500)1=4 ¡0:5
º m ax
µ
¶
µ
¶
1=7
2=7
2¼ £ 4:8 £ 10 ¡10 esu 9 £ 10 ¡28 g 3 £ 10 10 cm/s
erg/s
= 3: 912 6
1500 £ 10 6 Hz
cm5
µ
¶2=7
esu2
= 5 £ 10¡ 5
(esu ¢ g ¢ cm)1=7
cm6 ¢ s
µ
Check the units:
µ
¶ 2=7
esu2
1=7
(esu ¢ g ¢ cm)
cm6 ¢ s
=
=
µ
µ
esu5 ¢ g ¢ cm
cm 12 ¢ s 2
¶ 1=7
esu5 ¢ esu2 =cm
cm14
=
¶ 1=7
µ
=
esu5 ¢ erg/cm
cm12
¶ 1=7
esu
=G
cm2
Whew! Thus the equipartition ¤eld is about 50 ¹G. The minimum energy is
then
E tot
=
=
=
7
7 Be2q
UB =
V
3
3 8¼
¡
¢2
¢
7 5 £ 10 ¡5 G ¡
1:8 £ 10 69 cm3
3
8¼
4: £ 10 59 erg
and the ° we need to get a 1500 MHz photon is
r
r
º ma x
1500 £ 106
°max =
=
= 3: £ 103
3 (Hz/¹G) B ¹G
3 £ 50
and the lifetime of this electron is (equation2):
¿
=
1
1
13
13
y=
y
2 2:5 £ 10
2 2:5 £ 10
°B¹G
3000 (50)
=
3: £ 10 6 y
Thus sources like Cygnus A are either very short-lived (by astronomical standards) or the relativistic particles must be resupplied by reacceleration in situ
or transport from a central engine.
9
The jets visible in this radio image may be the source of resupply, but, if so, it
is an ine¢cient process. (We can see the jets, therefore they radiate away some
of the transported energy.)
When observations of B can be made, it appears that B is less than the
equipartition value. This increases the energy requirements but also increases
the source lifetime.
10