Thermal Bremsstrahlung''Radiation due to the acceleration of a charge in the Coulomb feld of another charge
is called bremsstrahlung or free-free emission”
A full understanding of the process requires a quantum treatment, since photons ofenergies [may be] are comparable to that of the emitting particle can be produced
However, a classical treatment is justifed in some (most of ) regimes, and the formulas
so obtained have the correct functional dependence for most of the physical parameters.
Therefore, we frst give a classical treatment and then state the quantum results ascorrections (Gaunt factors) to the classical formulas''George B. Rybicki & Alan P. Lightman in ''Radiative processes in astrophysics'' p. 155

Processi di radiazione e MHD –THERMAL BRESSTRAHLUNG– Daniele Dallacasa

(Thermal) Bremsstrahlungon textbooks

G. Ghisellini: “Radiative Processes in High Energy Astrophysics” § 2
M. Longair: “High Energy Astrophysics” § 6.3 – 6.6
G. B. Rybicki & A.P. Lightman: “Radiative Processes in Astrophyiscs” § 5
C. Fanti & R. Fanti: “Lezioni di radioastronomia” § 5
T. Padmanabhan: “Theoretical Astrophysics” § 6.9

Radiative processes: foreword
Continuum from radiative processes from free-free transitions
Classical physics holds when λ dB =√2h 1− v / c

2smaller than d =the typical scale of the interaction

mv→ Quantum mechanics(Δ x Δ p≥ h / 2 π)can be avoided
hFor non relativistic velocities≪ dmv
h =Planck ' s constantm = mass of the(lightest)particlev = velocity of the(lightest)particle
d = characteristic scale of the interactionor λ of the electromagnetic wave if it is involved)
λ dB = de Broglie wavelength

Implication → Particles act like points when h ν ≪ Eparticle

Luis (7th duke of) de Broglie 1892 -1987(Nobel 1929)

Bremsstrahlung

Foreword

Free electrons are accelerated (decelerated) in the Coulomb feld of atomic (ionized)
nuclei (free-free) and radiate energy
If Electrons and nuclei are in thermal equilibrium at the temperature T, their emission
Is termed “thermal bremsstrahlung”
Electrons move at velocity v wrt the ions and during ''collisions'' are defected. The
electrostatic interaction decelerates the electron which emits a photon bringing
away part of its kinetic energy
N.B. e and p have also homologous interactions (e with e; p with p) but they do not emit radiation
since the electric dipole is zero.

Astrophysical examples: 1. HII regions (104 oK)
2. Galactic bulges & hot-coronae (107 oK)
3. Intergalactic gas in clusters/groups (107-8 oK)
Bremsstrahlung is the main cooling process for high T (> 107 K) plasmabBremsstrahlung

single event

Radiated power (single particle) is given by the Larmor's formula:v
dt=2e3c2e_2a3acceleration is from Coulomb's law
F = ma ≃ −2ZeZe+→a =more appropriately x ∼ x(t )then2Zexm x2
where x= actual distance between electron (e-) and ion (Ze+)

P(t ) =

22e23 c32b

P = −dEv'a (t) =2 62Z e1 1

3 c3

m2 [ x(t )]4x(t)e_Bremsstrahlung

single event

Radiated power (single particle) is given by the Larmor's formula:v

P(t) =

3c23a( t) =Remarks:

2 613m [x(t)]2Z e3c1241.

a ∼ m−1→

2.

−2P ∼ m−2a ∼ x→P ∼ xe_b2ev')t(xZe+−4Radiated power is relevant
1. for electrons (positrons) only i.e. light particles
2. when the two charges are at the closest distance: (x ∼ b)
The interaction lasts very shortly: Δ t ∼2bv

where v = velocity

e_Bremsstrahlung

Pulses and spectrum

The emitted energy in a single “collision” (consider x ~ b)
PΔt =( )22 e2 Ze23c3mx22bv=4 Z2 e 6 132x ≃b33c m b v

P(t)Radiation comes in pulses Δ t long
Asymmetric profle ( decrease of the e– velocity)t

Fourier analysis of the pulse → Spectral distribution of the radiated energy
Flat up to a frequency

νmax ≃1≈

2Δ t
Related to the kinetic energy of the electrons(exponential decay beyond νmax )

v

P(ν)

4b

νmaxν

P(ν)

t

νmaxν

P(t)

Bremsstrahlung

radiated energy

The radiated energy per unit frequency (assuming fat spectrum) from asingle event can be written asPΔtΔν≈PΔtνmax= 2 P(Δ t) ≃216 Z 2 e6

113 c me b vb v

≈3 2 2 22 2P(ν)νmax

Exercise:
Determine the range of distances (order of mag)in which the released energy is signifcant−10e = 4.8⋅10

Remember.... Bohr's radius
Typical ISM densities−28g−13cm

me = 9⋅10statC

ao ∼ 5⋅10−3nISM ∼ 1 cmv = 100 km / sνBremsstrahlung(un)real plasma

Unphysical case: a cloud with ion number density
nZ and free electron number density neIn a unit time, each electron experiences a numberof collisions with an impact parameter between b and b + db is
2 π nZ v b db
The total number of collisions is

2 π ne nZ v bdb

Total emissivity from a cloud emitting via bremsstrahlung (single velocity!)Jbr (v , ν) ==32 π e32edEd νdV dt

63c m v

ne nZ Z2= 2 π ne nZ v
bmax

∫bmindbb=bmax∫b33c m12e

2 23c m b v

min

612ev

32 π e32 616 Z e

ne nZ Z ln2bdb =( )bmax
bminBremsstrahlungimpact parameter

Evaluation of bmax :at a given ν , we must consider only those interactions where electrons have impact
parameters corresponding to v max > v and thenbmax ≤v

is a function of frequency

4ν

Evaluation of bmin :a) classical physics requires that Δ v ≤ v
Δv ≃ aΔt =2Ze2

me b2bv≤ v→bmin−c ≥22 Ze

me v2b) quantum physics and the Heisemberg principle Δ p Δ x ≥ h / 2 πhhh
Δ p= me Δ v ≈ me v ≥≈→ bmin −q≥2πΔx2 π bmin
2 π me v
we must consider which bmin has to be used

Thermal bremsstrahlungelectron velocities / total emissivity

A set of particles at thermal equilibrium follow Maxwell-Boltzmann distribution:f (v)dv = 4 πThen( )
me

2πk T

3/ 2−me v2 / 2 kT 2ev dv

ne (v) = ne f (v)dv is the number density of electrons whose velocity is

between v and v + dv and ne (v) replaces ne in Jbr (ν)

When the plasma is at equilibrium, the process is termed thermal bremsstrahlung
Total emissivity:
must integrate over velocity (energy) distribution (0 ≤ v < ∞ ?)dEJbr (ν , T) ==dt dV d ν∞∫v J (ν , v)f (v)dv =min

br= 6.8⋅10−38 T−1 / 2 e−h ν / kT ne nZ Z2 gff (ν , T)
T ~ particle energy−1−3−1erg s cm Hz

high energy cut-off (T)Plasma density (ne2)Bremsstrahlungbmax ≃v

evaluating bmin , bmax
frequency dependent!....velocity dependent!

4ν
3

10e.g. if ~ v = 10 km / s and ν =10 Hz then bmax ≃6−1

10 m s−5≈
2.5×10m
10 −1

4 ×10 sThermal distribution → range of velocities
31
Order of magnitude
kT = m〈 v 2 〉22〈 v〉 ≃ √ 3kT =√−233 ×1.38× 102 −2 o −1

kgm s
−31

9.11× 10K Tkg= 0.68×104 √ T m / s

4

For a minimum temperature of 10 K we get a typical velocity of 680 km / s.
10At 10 Hz we get:bmin(q) ≥bmin(c) ≥h2 π me v
2Ze2me v

2....

bmax ≃≈v
4ν

56.8 × 10 m / s≃10 −14× 10 s

−346.626× 102 −1kgm s−312 π 9.11× 10≃ 1.7 ×10−5 m
5−10≈
0.17×10m−1kg × 6.8 ×10 m sFYI,Computatcions
never askedBremsstrahlungclassical/quantum limit

which [classical/quantum] limit prevails:bmin−qbmin−cbut≈h / (2 π me v)
222 Z e / (me v )v = √ 3 k T / me=

h4 π ZethenExamples:
Hot IGM cluster gas (T~108)c2>1 when v≥0.01cvc

≈

137 v
Z c
bmin− qbmin− c≈√

137 3kZc

meT

1/2HII region (T~104)

Bremsstrahlung( )Gaunt factorbmax
3√g (ν , T) =lnffπbmin

The Gaunt Factor ranges from about 1 to about 10 in the radio domain

Values around 10-15 in the radio domain, slightly higher than 1 at higher energies

bremsstrahlung – interlude: photon frequency .vs. temperature
Consider an electron moving at a speed of v =1000 km s-1.
In case it would radiate all its kinetic energy in a single interaction
1262m v = 0.5 ×9.1 ×10 −31 kg ×10 m s−1  = 4.55 ×10 −19 J22
E4.55 ×10 −19 kg m s−2
14
the frequency of the emitted radiation is  ===6.87×10Hz−342 −1h
6.626 ×10kg m s
E = h =in case this electron belongs to a population of particles at thermal equilibrium for whichv = 1000 km s −1 is the typical speed, determine which is the temperature of the plasma
E
T =k=

4.55 ×10 −19 kg m 2 s−2

1.38 ×10 −23 kg m s− 2 K
24=
3.30×10K−1= 6.626 ×10−34 J s−1− 16−1− 23−1
Boltzmann's constant: 1.381× 10 ergK = 1.381× 10 JK
me= 9.10938188× 10−31 kg
Planck's constant:−276.626 ×10−1erg sbremsstrahlung – interlude (2): the cut-off frequency
The cut-off frequency in the thermal bremsstrahlung spectrum depends on the temperature
only, and is conventionally set when the exponential equals 1/e.
h
kT=1k = Th=

1.38 ×10 −23 kg m 2 s− 2 K −1
6.626 × 10− 34

10×T=
2.08×10×T2 −1kg m s

It is interesting to determine the cut-off frequency for a warm plasma (HIIregion, T ~ 10 4 K) and for a hot plasma (IGM in clusters of galaxies,T ~ 10 8 K).
⇨ In observations, the measure of the cut-off frequency is a way todetermine the plasma temperature

Hz 

Thermal bremsstrahlung

total emissivity(2)Total emissivity does not depend on frequency(except the cut – off) and the spectrum is fat
Jbr (ν , T)= 6.8⋅10−38 T−1 / 2 e−h ν / kT ne nZ Z2 gff (ν , T)erg s−1 cm−3 Hz−1

Total emitted energy per unit volume: must integrate
over velocity (energy) distribution(0≤ v <∞ ?) at agiven frequency ν , the velocity of the particle v must be as such
122mv ≥ hν→vmin ≥and integrating over the whole spectrum:
Jbr (T) =dE√2hνme≃ 2.4⋅10−27 T 1 / 2 ne nZ Z2 ḡff (T)dt dVḡff (T) average gaunt factor−1−3erg s cm

Thermal bremsstrahlung

cooling timeLet's consider the total thermal energy of a plasma and the radiative losses:
the cooling time istot

tbr =≃E thJbr (T)=3 / 2(ne +np )kT
Jbr (T)

3ne kT−27 1 / 2 2

2.4⋅10T

ne ḡff

111.8⋅10=

ne = np

we can assume
Tne ḡff

1/2

sec

6⋅10=→

3

ne ḡff

T1 /2

yr

N.B. The “average” Gaunt Factor is about a factor of a few
Astrophysical examples:
2−33HIIregions:ne ∼ 10 − 10 cm

Galaxy clusters

ne ∼ 10−3−3cm;;

4T ∼ 10 K
78T ∼ 10 − 10 K(thermal) bremsstrahlung is the main cooling process at temperatures above T ∼ 107 K
N.B.-2- all galaxy clusters have bremsstrahlung emission

Bremsstrahlung

practical applications(1)A galactic HII region: M42 (Orion nebula)
Distance: 1.6 kly ~ 0.5 kpc angular size ~ 66 ' → 1.1
T =104 oo

K , S1.4GHz = 410 Jy

Let's get some astrophysical information
ν cutoff ∼S =tbr =kh

T = 2.08⋅1010Jbr (ν , T)⋅V( )ToK

Hz

24πD6⋅103

ne ḡff

T1/ 2yr−27 1 / 2 2jbr (T) = 2.4⋅10L br = jbr⋅VPay attention to the UNITS

Tne ḡff−1−3erg s cm−1erg s

courtesy of NRAO/AUIBremsstrahlung

practical applications(2)A cluster of galaxies:

Coma (z=0.0232), size ~ 1 Mpc

Typical values of IC hot
gas have radiative coolingtime exceeding 10 Gyr
All galaxy clusters are
X–ray emitters

Bremsstrahlung

practical applications(3)Cluster of galaxies: a laboratory for plasma physics

Bremsstrahlung

self-absorption (1)Thermal free-free absorption: energy gained by a free moving electron.
For a Maxwellian velocity (energy) distribution, i.e at thermal equilibrium, Kirchhoff's law holds
S(ν) =j(ν , T)μ(ν , T)hν

= B(ν , T) = 2

j(ν , T) = μ(ν , T)B(ν , T)

c321eh ν / kT

−1

[Kirchhoff ' slaw ]For an isotropically emitting plasma cloud via thermal bremsstrahlungdE
dt dV d ν

= 4 π Jbr (ν , T)

Therefore we can get
μ br (ν , T) =Jbr ( ν , T )4 π B(ν , T)=−1 / 21
4π

−38 −1 / 2

6.8⋅10−3⇒μ br (ν , T) ≈ T⇒strong dependency on frequencyT−h ν / kTne nz ν (1 −e)

2ne nZ Z ḡff (ν , T)c22h ν−h ν / kT(1− e3)Bremsstrahlung

self-absorption (2)– at high frequencies μ(ν , T) is negligible(ν–3and e– hv / kT)– at low frequencies (h ν ≪ kT)
μbr (ν , T) ≈ T−1 / 2ne nz ν (1− e)−1 / 2−3hν≈Tμbr (ν , T) =ḡff (T)≈104e−3ne nz ν63 me h c−h ν / kT

(1− 1+( )2π

3 k me1 /2kT2)

ne nZ Z T

= ne nz T − 3 / 2 ν−2− 3 / 2 −2ν ḡff (T)therefore

μ br (ν , T) ≈ 0.018 T−3 / 2 −22

ν ne nZ Z gff (T) ≃ 0.2 T−3 / 2 −2

ν ne nZ Z2

Bremsstrahlung

brightness/spectrum (1)The total brightness from a free-free emitting cloud:let's consider emission + absorption
B(ν , T) =1

jbr (ν , T)−τ(ν , T)(1−e4 π μ br (ν , T)−τ (ν , T))=BBB (ν , T)(1 −e)according to radiative transfer
Therefore:B(ν , T) ≈TT−1 / 22ne nZ Z ḡff (ν , T)

−3 / 2 −22

ν ne nZ Z ḡff (ν , T)

− τ(ν , T)

(1− e2− τ(ν , T)) ≈ T ν (1−eThe opacity [ τ(ν , T) = μ(ν , T)l ; where l = mean free path] determines
the spectral shape and the regime:
1. optically thick
2. optically thin

τ (ν , T) ≫ 1
τ (ν , T) ≪ 1)

Bremsstrahlung

brightness/spectrum(fgures from Ghisellini)

T=10 7K, Sphere with R = 10 15cm

Sphere with R = 10 15cm
np = ne =10 10 cm – 3

ne =10 10 – 10 14 cm – 3− τ(ν , T)

2B(ν , T) ≈ T ν (1− e()1τ ≫ 1 → (1− e ) = 1− − τe−τ)≈ 1→ B( ν , T) ≈ T ν2

τ ≪ 1 → (1− e− τ ) = ( 1− 1+ τ − o(τ )) ≈ τ = μ lo
2but μ(ν) ∼ ν−22→ B(ν , T) ≈ T ν τ ≈ T ν0

Bremsstrahlung

brightness/spectrum

T=10 7K, Sphere with R = 10 15cm

Sphere with R = 10 15cm
np = ne =10 10 cm – 3

Fixed size and density↑Emissivity decreases with temperature [ Jbr (ν , T ) ≈ T−1 / 2(fgures from Ghisellini)

ne =10 10 – 10 14 cm – 3

Fixed size and temperature]

The cut-off frequency increases linearly with T ...andSelf-absorption becomes less effective [μ(ν , T) ≈ T − 3 / 2 ]↑Emissivity increases with density [ Jbr (ν , T) ≈ ne2 ]The cut-off frequency stays constant2Self-absorption reaches higher ν [μ(ν , T) ≈ ne ]Bremsstrahlung :real X-ray spectra (3)Coma cluster: Conlon + 2016

The cut-off in these X-ray spectra is ameasure of the plasma temperature
If distance is known, from the fux,
we can derive the density (ne)

Relativistic Bremsstrahlung
Minor remark: the Gaunt Factor is slightly different wrt the case of ''thermal br.''
The emissivity for relativistic bremsstrahlung as a function of v for a given velocity v is:jrel , br (v , ν) =32 π e

62 3e

13 mc v

ne nZ Z ln2( )
183
Z

1/3–δ

E

The typical velocities of the particles are now relativistic and theenergy (velocity) distribution is described by a power law:

Log N(E)

N.B. T is not a relevant concept any more, v must be usedne (E) ≈ ne , o E−δ
Integrating over velocities (energies)∞∞

jrel , br (ν) ≈ ∫h ν ne (E)nZ Z 2 dE ≈ ∫h ν E−δ dE ≃The emitted spectrum is a power – law !

E−δ+ 11− δLog E≈ ν−δ+1

Bremsstrahlung
solar wind:ne ~ 4 – 7

cm-3

vswind~ 300 – 900 km s-1
T ~ 150 000 K(not consistent with v!)

fares:
relativistic particles
with power–law energy
distribution

the sunFinal remarks on bremsstrahlung emission:a photon is emitted when a moving electron deeply penetrates the Coulomb
field of positive ion and then is decelerated
the energy of the photon can never exceed the kinetic energy of the electron
generally the ''classical'' description of the phenomenon is adequate, thequantum mechanics requires an appropriate Gaunt Factor only
the individual interaction produces polarized radiation (fixed direction of
E and B fields of the e-m wave wrt the acceleration)
all the interactions produced in a hot plasma originate unpolarized emission(random orientation of the plane of interaction)
we have either ''thermal'' or ''relativistic'' bremsstrahlung, depending on thepopulation (velocity/energy distribution) of the electrons.
Annoying question.... Given that only electrons radiate, they can cool, while
positive ions do not... therefore, what happens to the thermal equilibrium?

10 Nov 2017

Total emissivity (per unit volume)Jbr (ν , T)= 6.8⋅10−38 T−1 / 2 e−h ν /kT ne nZ Z2 gff (ν , T)
Total emitted energy per unit volumedWJbr (T) =≃ 2.4⋅10−27 T1 / 2 ne nZ Z2 ḡff (T)dt dV−1−1tbr =11ne ḡff

T1 /2

sec→=

6⋅103ne ḡff

T 1/ 2−3erg s cmCooling Time
1.8⋅10−3erg s cm Hz

yr−1ḡff (T)average gaunt factor

18 Sep 2018

NGC2264 – Cone Nebula

Total emissivity (per unit volume)Jbr (ν , T)= 6.8⋅10−38 T−1 / 2 e−h ν /kT ne nZ Z2 gff (ν , T)

−1−3erg s cm Hz−1