Bremmstrahlung
Spring 2004
An electron approaches a nucleus of charge Z: As a result of the encounter, the electron is accelerated and moves o¤ in a di¤erent direction.
In the absence of radiation, the electron would emerge from the encounter
at the same speed that it entered. But the radiated photon decreases the
electronzs KE.
We start by looking at encounters in which the electronzs direction of
motion changes very little. The electric force produces an impulse
Z
I = e E?dt
perpendicular to the electronzs path. In this approximation the impulse
along the path is zero. The electric ¤eld component we need is
Ze
Zeb
E? = 2 cos µ = ¡
¢3=2
r
b2 + (vt)2
Thus
2
I = Ze b
Z
+1
¡1
¡
1
1
b2 + (vt)2
¢3=2 dt
Let vt = b tan µ: Then
2
I = Ze b
Z
¼=2
(b=v) sec2 µdµ
b3 sec3 µ
¡¼=2
Ze2
2
bv
Thus the electronzs momentum changes by an amount
=
¢p = I
The length of the encounter is approximately 2b=v; since the electric ¤eld
decreases rapidly once vt ¸ b: (This approximation gets better if the electron
moves relativistically.) Thus
¢p
Ze2 v
Ze2
'
2 = 2
¢t
bv 2b
b
and the average acceleration is
a=
1 ¢p
Ze2
=
m ¢t
mb2
and it occurs over the interval ¡¢t=2 < t < ¢t=2: Thus the Fourier transform is
¯+b=v
Z +b=v
1
Ze2 i!t
1 Ze2 ei!t ¯¯
1 2Ze2 sin !b
v
p
p
a (!) = p
e
dt
=
=
2¼ ¡b=v mb2
2¼ mb2 i! ¯¡b=v
2¼ mb2 !
Thus, from the Larmor formula, the radiation spectrum is:
Ã
!2
¯
dP ¯¯
2 e2
2 e2
1 2Ze2 sin !b
2
v
=
a (!) = 3 p
d! ¯one electron 3 c3
3c
2¼ mb2 !
At low frequencies, ! ¿ b=v; the spectrum is ¢at:
¯
µ
¶2
dP ¯¯
2 e2
1 2Ze2
4 Z 2e6
'
p
=
d! ¯ one electron 3 c3
3¼ m2b2v2 c3
2¼ mbv
At higher frequencies, ! À v=b; the emission oscillates wildly and is close to
zero. Converting to frequency º; where ! = 2¼º;
dP
8 Z2 e6
2¼v
=
for º <
2
2
2
3
dº
3m b v c
b
2
Now we sum up over all the electrons. The total number of electrons passing
the ion between distances b and b + db per second is nv2¼bdb and thus
Z
Z
1
8 Z 2 e6
4 Z 2e 6
4 Z 2e6
jº =
nv2¼bdb
=
ndb
=
n ln b
4¼
3 m 2b 2v 2c 3
3 m2bvc3
3 m2vc3
The factor 1=4¼ appears because jº is the emission per steradian.
Now we need to ponder the limits of the integral. The logarithm diverges
at b = 0 and at b = 1; so there must be maximum and minimum values
of b. We already found bmax above: it is approximately 2¼v=º: Quantum
mechanical considerations determine bmin :
4 Z 2e6
bmax
jº =
n ln
2
3
3 m vc
bmin
Now this result still assumes only one ion interacting with all the electrons.
Of course, there are many ions. So accounting for interactions with all the
ions, we have:
4 Z 2e 6
bmax
jº =
n en i ln
2
3
3 m vc
bmin
The last step is to integrate over the electron speed distribution- a Maxwellian
in a thermal plasma. The lower limit is not zero, because the electron cannot
radiate more than its own energy. Thus
1 2
mv
= hº
2 min
Then:
jº =
=
=
=
µ
¶
Z
4 Z 2e6
bmax 1
1 ³ m ´3=2
mv 2
n n ln
exp ¡
4¼v2 dv
3 m2c3 e i bmin p 2hº=m v 2¼kT
2kT
Z
µ 2¶
16¼ Z 2e 6 ³ m ´3=2
bmax kT 1
v2
mv
¡ 12 m kT
ne ni ln
e
d
p
2
3
3 m c 2¼kT
bmin m
2kT
2hº =m
¯
16¼ Z2 e6 ³ m ´1=2
bmax
v2 ¯ 1
¡ 12 m kT
n
n
ln
¡e
¯p
e
i
bmin
2hº=m
3 (2¼)3=2 m2c3 kT
³
´
8 Z 2e6 m 1=2
bmax ¡ hº
p
n en i ln
e kT
2
bmin
3 2¼ m c kT
This result is an approximation. Compare the exact result
hº
16 ³ ¼ ´1=2 Z 2e 6 ³ m ´1=2
jº =
ne ni e¡ kT g
2
3 6
m c kT
3
where g is the Gaunt factor. We have captured the correct dependence on
all the physical parameters. The ratio of the two expressions is
r
approx
8 3 6
= p
= 0:28
exact
3 2¼ 16 ¼
It is also useful to look at the total energy radiated, as this leads to cooling
of the plasma. The electrons radiate, but collisions reestablish equilibrium
Z
Z 1
16 ³ ¼ ´1=2 Z 2e6 ³ m ´1=2
hº
P = 4¼ jº dº = 4¼
n e ni
e¡ kT gdº
2
3 6
m c kT
0
³
´
³
´
2
6
1=2
1=2
16 ¼
Z e
m
kT
= 4¼
ne ni g
2
3 6
m c pkT
h
³
´
2 6
1=2
16 ¼
Z e
kT
= 4¼
n en ig
3=2
3 6
m c h
The important features to notice are that P _ n 2T 1=2:
Putting in numbers, we get
nn
jº = 5:4 £ 10¡39Z 2 e1=2i e¡hº=kT gf f erg ¢ cm¡3 ¢ s¡1 ¢ Hz¡1 ¢ ster ¡1
T
Averaging over cosmic abundances, we ¤nd
n2e ¡hº=kT
j º = 6:2 £ 10
e
gf f erg ¢ cm¡3 ¢ s¡1 ¢ Hz¡1 ¢ ster¡1
1=2
T
And integrating over frequencies gives
¡39
P = 1:4 £ 10¡27 Z2 ne ni T 1=2 erg ¢ cm¡3 ¢ s¡1
Astrophysical sites where Bremmstrahlung is an important emission mechanism include HII regions (T » few £103 K, so kT =h » 1:4£10¡16 erg/K£few£103
K= 6:6 £10¡27 erg¢s = 6£ 1013 Hz, so the emission is in the IR and the radio)
and clusters of galaxies (T »few£107 ¡ 108 K, emission in the x-ray.)
. The cooling time is
(ne + ni ) kT
3 £ (1:4 £ 10¡16 erg/K) T (K)
=
P
1:4 £ 10¡27 Z2 ne T 1=2 erg ¢ cm¡3 ¢ s¡1
T 1=2
T 1=2
= 3:0 £ 1011
s = 104
s
n
n
If T = 108 K and n = 10¡2 cm¡3; as in clusters; then ¿ = 1010 yr, the age of
the universe.
¿ =
3
2
4