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Giovanni Covone » 5.Equilibrium configurations the ISM

Introduction

Topic of this lecture: equilibrium configuration in the ISM systems.

Thermodynamical properties of the ISM: temperature, pressure, density, energy content.

Important concept: thermal phase.

The ISM is very far from thermal equilibrium. This fact makes the study of the physics of interstellar systems a challenging work.

However: we can find other types of equilibria in the ISM.

These are: Kinetic Equilibrium, Excitation Equilibrium, Ionization Equilibrium, and Pressure Equilibrium.

Let us start from considering the kinetic equilibrium.

Kinetic equilibrium

The Maxwell distribution of velocities reads:

$f({\bf v}) \, d^3 v \, = \, \left(\frac{\mu}{2 \pi k T} \right)^{3/2} \, {\rm exp} \left( \frac{\mu v^2}{2 k T} \right) \, d v_x \, d v_y \, d v_z \, ,$
where $\mu$ is the reduced mass of the particles.

Maxwellian distribution of velocities is typical of an ideal gas at equilibrium, with kinetic temperature $T$ .
In an ideal gas, all collisions are elastics.

As in the ISM most collisions in the ISM are elastic, we expect that the particle velocities will follow a Maxwellian distribution.

Therefore, a system in the ISM with a Maxwellian distribution of the velocities is said to be at the kinetic equilibrium.

Maxwell distribution of velocities

The Maxwell distribution of velocities in a system at kinetic equilibrium.

Kinetic equilibrium in the ISM

How long does it take for a system in the ISM to reach such a configuration?

Timescale for elastic collision between electrons:

$t_{\rm ee} \sim \, 10^4 \left( \frac{E_{\rm e}}{1 {\rm eV}} \right) ^{3/2} \, n_{\rm e}^{-1} {\rm s}$

Time scale for electron-hydrogen collision:

$t_{\rm eH} \sim \, 2 \times 10^7 \left( \frac{E_{\rm e}}{1 {\rm eV}} \right)^{3/2} \, n_{\rm e}^{-1} {\rm s}$

Which timescale determines the rapidity in reaching the kinetic equilibrium? How they compare with other time scale?

Result: ISM systems very quickly thermalize.

Excitation equilibrium

Consider a gas locally at the thermodynamical equilibrium (LTE).

$n_{\rm i}^*$ is the relative populations within an atom or molecule; $n_{\rm i}$ the true level populations.

Boltzmann equation:
$\frac{n_u^*}{n_l^*} \, = \, \frac{g_u}{g_i} \, e^{- \Delta E_{ul} / k T} \,$
where $g_{\rm i}$ is the statistical weight of the given energetic level, and $T$ is the kinetic temperature of the system.

The population of a level is referred with respect to the ground state:

$\frac{n_u^*}{n_0^*} \, = \, \frac{g_u}{g_0} \, e^{- \Delta E_{u0} / k T} \,$

The sum over all the states is related to the Partition Function of the system, $f (T)$ :
$n^* \, \equiv \, \Sigma_{i=0}^{\infty} \, n_i^*\, = \, \frac{n_0^*}{g_0} \, \Sigma_{i=0}^{\infty} g_i \, e^{-E_i / kT} \,= \, \frac{n_0^*}{g_0} \, f(T) \, .$

Excitation equilibrium (cont.)

This allows us to determine the fractional level in a given level $j$ :

$\frac{n_j^*}{n^*} \, = \, \frac{g_j}{f(T)} \, e^{-E_i / kT} \, \, .$

However, we already know very well that systems are far away from LTE.
So, how does this apply to the ISM?

How can we determine the absolute values of the level populations, $n_i$ ?

Two possible formalisms: Departure coefficients and Excitation Temperature.

Departure coefficients

First it is possible to determine the so-called “departure coefficients”.

The departure coefficients are defined as:

$b_i = n_i / n^*_i \, .$

These parameters measure the “departure” of the true level population from the local thermodynamical equilibrium predicted by the Boltzamn equation.

So, they can be used to determine the ratio of absolute populations:
$\frac{n_u}{n_l} \, = \, \frac{b_u}{b_l} \, \frac{n^*_u}{n_l^*} \, .$

Excitation temperature

A second method: the Excitation Temperature, $T_{\rm exc}$ .

The ratio of the true populations in two levels is given by:

$\frac{n_u}{n_l} \, = \, \frac{g_u}{g_i} \, e^{- \Delta E_{\rm ul} / k T_{\rm exc}} \, .$

Relation between the two different definition of temperature.

So, what is the meaning of the excitation temperature?

It is not the actual, physical temperature of any physical system.
Note: do not confuse the excitation temperature with the physical kinetic temperature of the gas in the ISM.

Ionization equilibrium

Source for ionization: the radiation field from nearby hot stars, the general interstellar radiation field.

In very hot plasma (where $T_{\rm e} > 10^6 {\rm K}$ ) collisional ionization is dominant.

At equilibrium: average number density of ionized elements does not change in time:

$X^r + h \nu \, \leftrightarrow \, X^{(r+1)} + e^- \, .$

At ionization equilibrium:

$N (X^r) \, N_{\rm photons} \, \sigma_{\rm ioniz} \, c = \, N (X^{r+1}) \, N_e \, \sigma_{\rm recomb} \, \Delta {\mathbf v} \, ,$

where $\Delta {\mathbf v}$ is the relative velocity between the ion and the electron.

Recombination cross-section is proportional to $v^{-2}$ . Hence:
$\frac{n (X^{r+1})}{n(X^r)} \, = \, \frac{1}{n_e} \, \left( \frac{\Gamma}{\alpha(T)} \right) \, ,$

with the recombination rate $\alpha(T)$ and the radiation field term $\Gamma$ .

Where do the electrons come from?

Sources of electrons

Sources of electron in HII and HI regions.

In HII regions: dominant source of electrons is ionized hydrogen.
Hydrogen is ionized by photons more energetic than 13.6 eV. A smaller contribution comes from Helium photoionization.

In HI regions: the spectrum of the radiation field is almost null above 13.6.
Only species with ionization potential below 13.6 eV can be photo-ionized.

The primary sources of electron: surface of dust grains.

Other mechanisms in the neutral regions are provided by cosmic rays and diffuse X-rays from the hot components of the ISM (that is , $T>10^5 {\rm K}$ ): they can ionize hydrogen deep inside the HI region, where UV photons cannot penetrate.

Sources of electrons (cont.)

Sources of electrons: dust versus gas.

The ratio of dust to gas is approximatively 1:100. However, dust is indeed the dominant source of electrons in the ISM, by even a factor 100.

Let us consider a system with no dust. The lower limit on the fractional ionization $x$ in the neutral regions HI is given by the relative abundances of C and H:

$x \equiv \frac{n_e}{n} > \frac{n_C}{n_H} \simeq \, 4 \times 10^{-4} \, .$
The mean interstellar electron density is measured from pulsar signals dispersion.
Typical value: $n_e \sim 0.03 {\rm cm}^{-3}$ .

H density: $\sim 1 {\rm cm}^{-3}$ .
Hence, ionization fraction about $x \sim 0.01 - 0.04$ .

Most electrons come from photo-ejection from dust grains.
Dust grains play an important role in almost all phases of the ISM.

Pressure equilibrium

A simple model of the ISM: cold an dense interstellar clouds embedded in a warm, ionized, low density medium.

How this coexistence could be possible?
A description in terms of pressure balance between the two components.

When thermal pressure is dominant:

$n_H \, T_{\rm cool} \simeq (n_e + n_p) \, T_{\rm warm} \, .$

Other sources of pressure need to be considered.

Magnetic pressure, $B^2 / 8 \pi$ : interstellar magnetic fields.

Cosmic ray pressure: protons with energies of a few MeV.

Hydrodynamic pressure: stellar winds, or supernovae explosions.

Radiation pressure: from starlight and background radiation.

Pressure Equilibrium (cont.)

The thermal pressures of the cold clouds and the warm ionized medium are very close.
Indeed, typical values for the cold clouds are:
$n_H \geq 10 {\rm cm}^{-3}$ and $T \leq 300 {\rm K}$

for the warm ionized medium $n_H \leq 0.3 {\rm cm}^{-3}$ and $T \leq 10^4 {\rm K}$ . Hence:
$\frac{P}{k} = k T \simeq 3000 {\rm cm}^{-3} \, {\rm K}$

Comparison with pressure from typical magnetic fields in the ISM. Typical values of magnetic fields are in the range 2 – 8 micro Gauss. For a magnetic field $B_0 \sim 3 \mu G$

$\left( P / k_B \right) \, \simeq \, \frac{B_0^2}{8 \pi k} \simeq 2600 {\rm cm}^{-3} \, ,$

With ordered magnetic field: magnetic pressure even larger than the thermal pressure.
Disordered magnetic fields produce pressure about one third smaller.