Free-free Radio Emission from an HII Region

Why should an HII region emit radio radiation at all? Because charged particles are being accelerated electrostatically, and free accelerated charges radiate according to Larmor's formula. Recall that only the component of the electric field perpendicular to the line-of-sight contributes to the electromagnetic radiation at large distances $r$:
$$E_\bot = {q \dot{v} \sin \theta \over r c^2} ,$$
where $q$ is the charge, $\theta$ is the angle between the acceleration $\dot{v}$ and the line-of-sight to the observer, and $c$ is the speed of light.

All manner of electrostatic interactions between various charged particles take place in an HII region, but most do not emit significant amounts of radiation. The magnitude of the acceleration is equal to the Coulomb force
$$ F = { q_1 q_2 \over r^2_{12}}$$
divided by the particle mass $m$. The mass of even the lightest ion (the mass of a hydrogen ion is the proton mass $m_{\rm p} \approx 1.66 \times 10^{-24}$ g) is about two thousand times the electron mass $m_{\rm e} \approx 9.11 \times 10^{-28}$ g. In an electron-ion collision, the electron will therefore radiate at least $({m_{\rm p} / m_{\rm e}})^2 \approx 4 \times 10^6$ as much power as the ion, so we may ignore as negligible the radiation from all of the ions. Interactions between identical particles also do not radiate significantly because the accelerations of the two particles are equal in magnitude but opposite in direction: $\dot{v_1} = -\dot{v_2}$. Thus their radiated electric fields are equal in magnitude but opposite in sign; the net $E_\bot$ approaches zero at large distances where $r \approx r_1 \approx r_2$. We can therefore ignore the radiation from electron-electron collisions. Only electron-ion collisions are important, and only the electrons radiate.

Radio Radiation From a Single Electron-ion Interaction

Consider the radiation from an electron passing by a much more massive ion of charge $Ze$, where $Z = 1$ for a singly ionized atom like hydrogen. Each electron-ion interaction will generate a single pulse of radiation. We will calculate the total energy emitted and the (approximate) frequency spectrum of the pulse. The exact spectrum of an individual pulse will not be needed since there is a broad distribution of electron energies and impact parameters to smear out such detail in the total spectrum of  an HII region.

Radio photons are produced by weak interactions, meaning the change $\Delta E_{\rm e}$ of electron kinetic energy is much smaller than the intial kinetic energy $E_{\rm e}$. The reason is that the energy $E_\gamma = h \nu$ of a radio photon is much smaller than the average kinetic energy of an electron in an HII region. This numerical comparison is an example of how astrophysical "intuition" leads to a simplification of the electron-ion scattering problem.

The mean electron energy in a plasma of temperature $T$ is
$$E_{\rm e} = {3 k T \over 2}$$
For an HII region with $T \sim 10^4$ K, this is
$$E_{\rm e} \approx {3 \times 1.38 \times 10^{-16} {\rm ~erg~K}^{-1} \times 10^4 {\rm ~K} \over 2} \approx 2 \times 10^{-12} {\rm ~erg~} \approx 1 {\rm ~eV} .$$ 

[This is another useful conversion factor to remember: 1 eV is the typical energy associated with the temperature $T \approx 10^4$ K.] The energy of a photon is $E_\gamma = h \nu$. For example, a radio photon of frequency $\nu = 10$ GHz has energy
$$E_\gamma \approx 6.63 \times 10^{-27} {\rm ~erg~s~} \times 10^{10} {\rm ~Hz~} \approx 6.63 \times 10^{-17} {\rm ~erg~} \approx 4 \times 10^{-5} {\rm ~eV} .$$
The great inequality $E_\gamma \ll E_{\rm e}$ means that most radio photons are produced by weak interactions which cause the trajectory of the electron to deflect by only a small angle ($\ll 1$ radian). We may make the approximation that the electron path is nearly straight as it interacts with an ion to produce radio radiation.

During the interaction, the electron will be accelerated electrostatically both parallel to and perpendicular to its nearly straight path:
$$F_\parallel = m_{\rm e} \dot{v}_\parallel = {Z e^2 \over l^2} \sin \psi = {Z e^2 \sin \psi \cos^2 \psi \over b^2}$$
$$F_\bot = m_{\rm e} \dot{v}_\perp = {Z e^2 \over l^2} \cos \psi = {Z e^2 \cos^3 \psi \over b^2} ,$$
where $b$ is the impact parameter of the interaction, the minimum value of the distance $l$ between the electron and the ion.

For a given impact parameter $b$, the equations above can be solved to show that the maximum of $\dot{v}_\parallel$ is actually about 0.38 $\times$ the maximum of $\dot{v}_\perp$. Even so, the radio radiation arising from $\dot{v}_\parallel$ is negligible. Plotting the variation with time of $\dot{v}_\parallel$ and $\dot{v}_\perp$ during the interaction shows pulses with quite different shapes:

The pulse duration is $\tau \approx b /v$. The $\dot{v}_\parallel$ pulse is roughly a sine wave of frequency $\omega \sim \tau^{-1} \approx v / b$, which we will soon show is much higher than radio frequencies for all relevant impact parameters $b$. That is, the parallel acceleration produces infrared radiation but not radio radiation. The $\dot{v}_\perp$ pulse is a single peak whose frequency spectrum extends from zero up to $\sim v /b$ (recall that Fourier transform of a Gaussian is a Gaussian), so it is rich in radio frequencies.

Considering only the acceleration perpendicular to the electron velocity, we get the instantaneous power emitted from Larmor's formula:
$$P = {2 \over 3} {e^2 \dot{v}_\perp^2 \over c^3} = {2 e^2 \over 3 c^3} {Z^2 e^4 \over m_{\rm e}^2} \biggl( { \cos^3 \psi \over b^2} \biggr)^2$$
The total energy $W$ emitted by the pulse is
$$W = \int_{-\infty}^\infty P dt .$$
Since $(\Delta E_{\rm e}) / E_{\rm e} = E_\gamma / E_{\rm e} \ll 1$, we can make the approximation that the electron velocity is nearly constant. From the interaction diagram we see that
$$ v = {d x \over d t} {\rm \quad~and~\quad} \tan \psi = {x \over b}$$
so
$$v = {b d \tan \psi \over d t} = {b \sec^2 \psi d \psi \over dt} = {b  d \psi \over \cos^2 \psi  dt}$$
and
$$d t = { b \over v} { d \psi \over \cos^2 \psi}~.$$
The total pulse energy becomes
$$W = \int_{-\infty}^\infty P d t = {2 \over 3} {Z^2 e^6 \over c^3 m_{\rm e}^2 b^4} \int_{-\infty}^\infty \cos^6 \psi  d t~.$$
By changing the variable of integration from $t$ to $\psi$ and then invoking symmetry, we get
$$W = {2 \over 3} {Z^2 e^6 \over c^3 m_{\rm e}^2 b^4} \int_{-\pi / 2}^{\pi /2} { b \over v} {\cos^6 \psi \over \cos^2 \psi}  d \psi = {4 \over 3} {Z^2 e^6 \over c^3 m_{\rm e}^2 b^3 v} \int_0^{\pi /2} \cos^4 \psi  d \psi$$

Evaluating the integral yields
$$\int_0^{\pi/2} \cos^4 \psi d \psi = {3 \pi \over 16}$$
so the energy radiated by a single electron-ion interaction characterized by impact parameter $b$ and velocity $v$ is:
$$W = {\pi Z^2 e^6 \over 4 c^3 m_{\rm e}^2} \biggl( { 1 \over b^3 v} \biggr)$$

This energy is emitted in a single pulse of duration $\tau \approx b /v $, so the pulse power spectrum (the spectrum of the squared Fourier components) is nearly flat over all frequencies $\nu < \nu_{\rm max} \approx (2 \pi \tau)^{-1} \approx v /(2 \pi b)$ and falls rapidly at higher frequences. (Recall that the Fourier transform of a Gaussian is a Gaussian.) In principle we could calculate the actual Fourier transform of the pulse shape, but that would only introduce an unnecessary complication into an already complicated calculation. The ranges of velocities $v$ and impact parameters $b$ characterizing electron-ion interactions in an HII region are so wide that averaging over all collision parameters will wash out any fine detail in the spectrum associated with a particular $v$ and $b$.

The cutoff frequency $\nu_{\rm max}$ is much higher than any radio frequency; it corresponds to infrared radiation. As we will soon calculate, the typical electron speed in a $T \sim 10^4$ K HII region is $v \approx 7 \times 10^7$ cm s$^{-1}$ and the typical impact parameter is $b \approx 10^{-7}$ cm, so $\nu_{\rm max} \approx 10^{14}$ Hz, corresponding to the near-infrared wavelength $\lambda_{\rm min} \approx 3\mu$m.

We make the approximation that the power spectrum shown above is flat out to $\nu = \nu_{\rm max}$ and zero at higher frequencies.  Then the average energy per unit frequency emitted during a single interaction is approximately
$$W_\nu \approx {W \over \nu_{\rm max}} =\biggl( {\pi Z^2 e^6 \over 4 c^3 m_{\rm e}^2 b^3 v} \biggr) \biggl( { 2 \pi b \over v} \biggr)$$ $$W_\nu \approx {\pi^2 \over 2} {Z^2 e^6 \over c^3 m_{\rm e}^2 } \biggl( { 1 \over b^2 v^2} \biggr) {\rm ,~~} \nu < \nu_{\rm max} \approx {v \over 2 \pi b} \approx 10^{14} {\rm ~Hz}$$

Radio Radiation From an HII Region

Next we need to find the distributions of $v$ and $b$ to evaluate the radio emission from an HII region. The distribution of $v$ depends on the electron temperature $T$, and the distribution of $b$ depends on the electron number density $N_{\rm e}$ (cm$^{-3}$) and the ion number density $N_{\rm I}$  (cm$^{-3}$).

In LTE, the average kinetic energies of electrons and ions are equal. Since the electrons are much less massive, their speeds are much higher, and we can make the approximation that the ions are nearly stationary during an interaction.

Then the number of electrons passing any ion per unit time with impact parameter $b$ to $b + db$ and speed range $v$ to $v + dv$ is
$$N_{\rm e} (2 \pi b  d b) v  f(v)  dv~.$$
Here $f(v)$ is the normalized ($\int f(v) dv = 1$) speed distribution of the electrons. The number $N(v,b)$ of such encounters per unit volume per unit time is given by $$N(v,b) dv db = (2 \pi b  db) [v f(v)  dv] N_{\rm e} N_{\rm i} .$$ The spectral power at frequency $\nu$ emitted isotropically per unit volume will be $4 \pi \epsilon_\nu$, where $\epsilon_\nu$ is the familiar emission coefficient from radiative transfer. Thus
$$4 \pi \epsilon_\nu = \int_{b = 0}^\infty \int_{v = 0}^\infty W_\nu (v, b) N(v, b) dv db$$
Substituting the results above gives
$$4 \pi \epsilon_\nu = \int_{b = 0}^\infty \int_{v = 0}^\infty \biggl( { \pi^2 Z^2 e^6 \over 2 c^3 m_{\rm e}^2 b^2 v^2} \biggr) 2 \pi b  db  N_{\rm e} N_{\rm i} v f(v)  dv$$
$$4 \pi \epsilon_\nu = {\pi^3 Z^2 e^6 N_{\rm e} N_{\rm i} \over c^3 m_{\rm e}^2} \int_{v = 0}^\infty {f(v) \over v}  dv \int_{b = 0}^\infty {d b \over b}$$
You can see right away that we have run into a problem: the integral $$\int_{b = 0}^\infty {db \over b}$$
diverges logarithmically. There must be some physical limits on the range of the impact parameter $b$ that prevent this divergence. For the time being we will just call those limits $b_{\rm min}$ and $b_{\rm max}$; later we will identify their physical meanings and evaluate them numerically.  Meanwhile, $$4 \pi \epsilon_\nu = {\pi^3 Z^2 e^6 N_{\rm e} N_{\rm i} \over c^3 m_{\rm e}^2} \int_{v = 0}^\infty {f(v) \over v}  dv \int_{b_{\rm min}}^{b_{\rm max}} {d b \over b}$$

The distribution $f(v)$ of electron speeds in LTE is the nonrelativistic Maxwellian distribution
$$\bbox[border:3px blue solid,7pt]{f(v) = {4 v^2 \over \sqrt{\pi}} \biggl( { m \over 2 k T} \biggr)^{3/2} \exp \biggl( - {m v^2 \over 2 k T} \biggr)}\rlap{\quad \rm {(4B1)}}$$
(see the derivation here or in any thermodynamics textbook such as Kittel's Elementary Statistical Physics).
Having $f(v)$ we can now evaluate the integral over the electron speeds:
$$ \int_{v = 0}^\infty { f(v) \over v}  dv = {4 \over \sqrt{\pi}} \biggl( { m_{\rm e} \over 2 k T} \biggr)^{3/2} \int_{v = 0}^\infty v \exp \biggl( - { m_{\rm e} v^2 \over 2 k T} \biggr)  dv .$$
Let $u \equiv m_{\rm e}v^2 / (2 k T)$ so $du = mv dv / (kT)$:
$$ \int_{v = 0}^\infty { f(v) \over v}  dv = {4 \over \sqrt{\pi}} \biggl( {m_{\rm e} \over 2 k T} \biggr)^{3/2} \int_{u = 0}^\infty {k T \over m_{\rm e}} e^{-u}  du = {4 \over \sqrt{\pi} 2} \biggl( { m_{\rm e} \over 2 k T} \biggr)^{1/2} \int_{u = 0}^\infty e^{-u}  du$$
$$\int_{v=0}^\infty {f(v) \over v}  dv = \biggl( {2 m_{\rm e} \over \pi k T} \biggr)^{1/2}$$

In conclusion, we have the following equation for the free-free emission coefficient:
$$\bbox[border:3px blue solid,7pt]{\epsilon_\nu = {\pi^2 Z^2 e^6 N_{\rm e} N_{\rm i} \over 4 c^3 m_{\rm e}^2 } \biggl( {2 m_{\rm e} \over \pi k T} \biggr)^{1/2} \ln \biggl( {b_{\rm max} \over b_{\rm min}} \biggr)}\rlap{\quad \rm {(4B2)}}$$

The next problem is to estimate the minimum and maximum impact parameters $b_{\rm min}$ and $b_{\rm max}$.  Since only the logarithms of both parameters are needed, these estimates don't have to be precise.

To estimate the minimum impact parameter $b_{\rm min}$, we note that the net momentum impulse
$$\Delta P = \int_{-\infty}^{\infty} f dt = \int_{-\infty}^{\infty}e E dt$$
comes from $E_\perp = E \cos \psi$ only because the contribution from $E_\parallel$ is antisymmetric about $t = 0$.  Thus
$$\Delta P = \int_{-\infty}^{\infty} e {Z e \cos \psi \over l^2} dt
= Z e^2 \int_{-\infty}^{\infty} {\cos^3 \psi \over b^2} dt$$
Changing the variable of integration from $t$ to $\psi$ using
$$ dt = {b \over v} d \tan \psi = {b \over v} \sec^2 \psi d \psi $$
gives
$$\Delta P = {Z e^2 \over b v} \int_{-\pi /2}^{\pi /2} \cos \psi d \psi =
{2 Z e^2 \over b v}$$
The maximum possible momentum transfer during the interaction is twice the initial momentum $m_{\rm e} v$ of the electron, so
$$\bbox[border:3px blue solid,7pt]{b_{\rm min} \approx {Z e^2 \over m_{\rm e} v^2}}\rlap{\quad \rm {(4B3)}}$$

This result comes from a purely classical treatment of the interaction (See also Jackson's Classical Electrodynamics, section 13.1 and problem 13.1, for a more detailed discussion). The uncertainty principle does imply a quantum-mechanical limit
$$b_{\rm min} = {\hbar \over m_{\rm e} v}~,$$
but this lower limit is generally smaller than the classical limit in HII regions and hence may be ignored.  Following Jackson, section 13.3, we define $\eta$ is the ratio of the classical to quantum limits
$$ \eta = {Z e^2 \over 3 k T} / {\hbar \over m_{\rm e} v} =
{Z e^2 \over \hbar v} $$
For an HII region with $T \approx 10^4$, $ v  \approx (3 k T / m_{\rm e})^{1/2}$ and $\eta \approx 3$, so the classical limit applies. Only for unusually cold plasmas, $T \ll 1000$ K, would the quantum-mechanical limit be needed.

There are two ways to estimate the relevant upper limit $b_{\rm max}$ to the impact parameter. Because electrostatic forces always dominate gravity on small scales, electrons in the vicinity of a nearly stationary ion are free to rearrange themselves to neutralize, or shield, the ionic charge. The characteristic scale length of this shielding is called the Debye length. From Jackson's Classical Electrodynamics, the Debye length is roughly
$$D \approx \biggl( { k T \over 4 \pi N_{\rm e} e^2 } \biggr)^{1/2}$$
The Debye length is quite large in the low-density plasma of a typical HII region. For example, if $T \approx 10^4$ K and $N_{\rm e} \approx 10^3$ cm$^{-3}$,
$$D \approx \biggl[ { 1.38 \times 10^{-16} {\rm ~erg~K}^{-1} \times 10^4 {\rm ~K} \over 4 \pi \times 10^3 {\rm ~cm}^{-3} \times (4.8 \times 10^{-10} {\rm ~statcoulomb})^2 } \biggr]^{1/2} \approx 22 {\rm ~cm}.$$

An independent upper limit to the impact parameter is the largest value of $b$ that can generate a significant amount of power at some relevant radio frequency $\nu$. Recall that the pulse power per unit bandwidth is small above angular frequencies
$$\omega \approx {v \over b}$$
so
$$ b_{\rm max} \approx { v \over \omega} = {v \over 2 \pi \nu},$$
where $\nu$ is the radio frequency being considered.

Clearly, the relevant upper limit $b_{\rm max}$ in any particular situation is the smaller of these two upper limits. To see which is smaller, we calculate the limits for a typical HII region and a typical radio frequency.

Example: Estimate $b_{\rm min}$ and $b_{\rm max}$ for a pure HII region ($Z = 1$) with $T \approx 10^4$ K observed at $\nu = 10$ GHz $= 10^{10}$ Hz.

$$b_{\rm min} \approx {Z e^2 \over m_{\rm e} v^2} \approx {Z e^2 \over 3 k T }$$
$$b_{\rm min} \approx { (4.8 \times 10^{-10} {\rm ~statcoulomb})^2 \over 3 \times 1.38 \times 10^{-16} {\rm ~erg~K}^{-1} \times 10^4 {\rm ~K}} \approx 5.6 \times 10^{-8} {\rm ~cm}$$

$$b_{\rm max} \approx {v \over 2 \pi \nu} \approx \biggl( { 3 k T \over m_{\rm e} } \biggr)^{1/2} / (2 \pi \nu)$$
so the largest impact parameter giving significant power at $\nu = 10^{10}$ Hz is
$$ b_{\rm max} \approx \biggl( { 3 \times 1.38 \times 10^{-16} {\rm ~erg~K }^{-1} \times 10^4 {\rm ~K} \over 9.1 \times 10^{-28} {\rm ~g} } \biggr)^{1/2} / ( 2 \pi \times 10^{10} {\rm s}^{-1} ) \approx 1.1 \times 10^{-3} {\rm ~cm}.$$
The maximum impact parameter capable of generating power at radio frequencies is much smaller than the Debye length in an HII region, so the Debye length is irrelevant. It takes so long for an electron to move a $D \approx 22$ cm Debye length that it emits only at unobservably low frequencies ($\nu < 1 $ MHz). The Debye length becomes relevant only in very dense plasmas such as the solar chromosphere ($N_{\rm e} \approx 10^{12}$ cm$^{-3}$).

Our simple estimate of the ratio
$${b_{\rm max} \over b_{\rm min} } \approx \biggl( { 3 k T \over m_{\rm e} } \biggr)^{1/2} (2 \pi \nu)^{-1} \biggl( { 3 k T \over Z e^2} \biggr) \approx \biggl( { 3 k T \over m_{\rm e} } \biggr)^{3/2} { m _{\rm e} \over 2 \pi Z e^2 \nu}$$
is very close to the result of Oster's (1961, Rev. Modern Physics, 33, 525) very detailed derivation and presented as Equation 9.29 in the textbook. Next we note that the ratio $(b_{\rm max} / b_{\rm min})$ is roughly $10^4$, which is much greater than the fractional velocity range $\sigma_v / v \approx 1$ in the Maxwellian velocity distribution. Also note that
$$ \ln \biggl({b_{\rm max} \over b_{\rm min}} \biggr) \sim 10$$
varies slowly with changes in either $b_{\rm max}$ or $b_{\rm min}$, so small uncertainties in these limits have little effect on the calculated emission coefficient of an HII region.

Since the HII region is in local thermodynamic equilibrium (LTE) at some temperature $T$, we can use Kirchoff's law to calculate the absorption coefficient from the emission coefficient and the blackbody brightness law: $$\kappa_\nu = { \epsilon_\nu \over B_\nu(T)}~.$$
In the Rayleigh-Jeans limit
$$\kappa _\nu = {\epsilon_\nu c^2 \over 2 k T \nu^2}$$
and
$$\bbox[border:3px blue solid,7pt]{\kappa_\nu = {1 \over \nu^2 T^{3/2} } \biggl[ {Z^2 e^6 \over c} N_{\rm e} N_{\rm i} {1 \over \sqrt { 2 \pi (m_{\rm e} k)^3 } } \biggr] {\pi^2 \over 4} \ln \biggl( { b_{\rm max} \over b_{\rm min} } \biggr)}\rlap{\quad \rm {(4B4)}}$$

Because the $b_{\rm max}$ varies slowly with frequency, the absorption coefficient is not exactly proportional to $\nu^{-2}$. A good numerical approximation is $\kappa _\nu \propto \nu^{-2.1}$.

The total opacity $\tau_\nu$ of an HII region is the integral of $ -\kappa_\nu$ along the line of sight.

$$ \tau_\nu = \int_{\rm los} -\kappa_\nu  ds \propto \int { N_{\rm e} N_{\rm i} \over \nu^{2.1} T^{3/2} } ds \approx \int { N_{\rm e}^2 \over \nu^{2.1} T^{3/2} } ds$$

At frequencies low enough that $\tau_\nu \gg 1$, the HII region becomes opaque, its spectrum approaches that of a black body with temperature $T \sim 10^4$ K, and the flux density varies as $S \propto \nu^2$. At very high frequencies, $\tau_\nu \ll 1$, the HII region is nearly transparent, and
$$S_\nu \propto {2 k T \nu^2 \over c^2} \tau_\nu \propto \nu^{-0.1}$$
On a log-log plot, the overall spectrum of a uniform HII region looks like this, with the spectral break corresponding to the frequency at which $\tau_\nu \approx 1$:

The spectral slope on a log-log plot is often called the spectral index and denoted by $\alpha$. Thus
$$\alpha \equiv \pm {d \log S \over d \log \nu}$$
Beware the $\pm$ sign! Unfortunately both conventions are found in the literature, and you have to look carefully at each paper to find out which one is being used. With the $+$ sign convention, the low-frequency spectral index of a uniform HII region would be $\alpha = +2$. The $-$ convention was introduced in the early days of radio astronomy because most sources discovered at low frequencies are stronger at low frequencies than at high frequencies. Thus $\alpha = +0.7$ might mean
$${d \log S \over d \log \nu} = -0.7$$
The spectrum of an inhomogeneous HII region will be similar well above the break frequency, but the break will be more gradual and the low-frequency slope will be somewhat less than +2. For example, ionized winds from stars are quite inhomogeneous. Mass conservation in a steady, constant-velocity, isothermal spherical wind implies that the electron density is inversely proportional to the square of the distance from the star: $N_{\rm e} \propto r^{-2}$. The low-frequency spectral spectral slope of free-free emission by such a wind is closer to $+0.6$ than $+2$.

In the astronomical literature you will encounter the term emission measure (EM) defined by the integral of $N_{\rm e}^2$ along the line-of-sight and usually expressed in astronomically convenient units:
$$\bbox[border:3px blue solid,7pt]{{ EM \over {\rm pc~cm}^{-6} } \equiv \int_{\rm los} \biggl( { N_e \over {\rm cm} ^{-3} } \biggr)^2 d \biggl( { s \over {\rm pc}} \biggr) }\rlap{\quad \rm {(4B5)}}$$
Since $\kappa_\nu$ is proportional to $N_{\rm e} N_{\rm i} \approx N_{\rm e}^2$, the optical depth $ \tau$ is proportional to the emission measure. The emission measure is commonly used to parameterize $\tau$ in astronomically convenient units:
$$\tau_\nu \approx 3.014 \times 10^{-2} \biggl( { T_{\rm e} \over {\rm K} } \biggr)^{-3/2} \biggl( { \nu \over {\rm GHz}} \biggr)^{-2} \biggl( { EM \over {\rm pc~cm}^{-6}} \biggr) \langle g_{\rm ff} \rangle~,$$
where the free-free Gaunt factor $\langle g _{\rm ff} \rangle$ absorbs the slow frequency dependence associated with the logarithmic term in $\kappa_\nu$:
$$\langle g_{\rm ff} \rangle \approx \ln \biggl[ 4.955 \times 10^{-2} \biggl( {\nu \over {\rm GHz}} \biggr)^{-1} \biggr] + 1.5 \ln \biggl( { T_{\rm e} \over {\rm K}} \biggr)$$
We end up with a very good approximation for free-free opacity that is easy to evaluate numerically:
$$\bbox[border:3px blue solid,7pt]{\tau_\nu \approx 3.28 \times 10^{-7} \biggl( {T_{\rm e} \over 10^4 {\rm ~K} } \biggr)^{-1.35} \biggl( {\nu \over {\rm GHz} } \biggr)^{-2.1} \biggl( { EM \over {\rm pc~cm}^{-6} } \biggr)}\rlap{\quad \rm {(4B6)}}$$
From the optical depth $\tau$ and the electron temperature $T_{\rm e}$ we can calculate the brightness temperature $T_{\rm b}$:
$$T_{\rm b} = T_{\rm e} (1 - e^{-\tau} )~.$$
We don't know the structure of the HII region along the line-of-sight, so it is common to approximate the geometry of an HII region by a circular cylinder whose axis lies along the line-of-sight, and whose axis length equals its diameter. We also suppose that the temperature and density are constant throughout this volume. Then it is very easy to estimate physical parameters of the HII region (e.g., electron density, temperature, emission measure, production rate $N_{\rm Ly}$ of ionizing photons) from the observed radio spectrum, once the distance to the HII region is known (see problem set 5).

A useful approximation relating the production rate of ionizing photons to the free-free spectral luminosity $L_\nu$ at high frequencies (where $\tau \ll 1$) of an HII region in ionization equilibrium is
$$\bbox[border:3px blue solid,7pt]{\biggl( { N_{\rm Ly} \over {\rm s}^{-1} } \biggr) \approx 6.3 \times 10^{52} \biggl( { T_{\rm e} \over 10^4 {\rm ~K} } \biggr)^{-0.45} \biggl( { \nu \over {\rm GHz} } \biggr)^{0.1} \biggl( { L_\nu \over 10^{20} {\rm ~W~Hz}^{-1} } \biggr)}\rlap{\quad \rm {(4B7)}}$$

Example: The interstellar medium of our Galaxy contains a diffuse ionized component, some of which is "warm" ($T_{\rm e} \approx 10^4$ K) and some is "hot" ($T_{\rm e} \approx 10^6$ K). These two phases are roughly in pressure equilibrium so the hot medium is less dense by a factor of $\sim10^2$. The combination of high $T_{\rm e}$ and low $N_{\rm e}$ of the hot phase means that only the warm component contributes significantly to the free-free opacity of the ISM. There must be some frequency $\nu$ below which the disk becomes opaque and we cannot see out of the Galaxy, even in the direction perpendicular to the disk.

From the observed brightness spectrum in the direction perpendicular to the disk, Cane (1979, MNRAS, 189, 465) found that $\tau \approx 1$ at $\nu \approx 3$ MHz. We can use this result to estimate the typical electron density in the warm ISM.
$$ 1 \approx 3.28 \times 10^{-7} \times (1)^{-1.35} \times {0.003}^{-2.1} \times \langle N_{\rm e}^2 \rangle \times 1000 {\rm ~pc}$$
$$\langle N_{\rm e}^2 \rangle^{1/2} \approx \biggl( { 0.003^{2.1} \over 3.28 \times 10^{-4} } \biggr)^{ 1/2} \approx 0.1 {\rm ~cm}^{-3}$$

Free-free emission accounts for about 10% of the 1 GHz continuum luminosity in most spiral galaxies. It is the strongest component in the frequency range from $\nu \approx 30$ GHz to $\nu \approx 200$ GHz, above which thermal emission from cool dust grains dominates. Free-free absorption flattens the low-frequency spectra of spiral galaxies, and the frequency at which $\tau \approx 1$ is higher in galaxies with high star-formation rates, especially if the star formation is confined to a compact region near the nucleus.