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Massimo Capaccioli » 24.Fundamentals - Part III

Maxwell-Boltzmann distribution (1)

In the following we recap the formulae which leading the statistics in the processes of excitation and ionization of atoms. The Maxwell-Boltzmann distribution accounts for the velocities of non-relativistic gas [what is gas in this case] particles subject to short range interactions only [gravity too?] and sufficiently far from quantum effects. It applies to systems in thermodynamic equilibrium (i.e. at thermal, mechanical, radiative, and chemical equilibrium). The number of particles with velocity between $v$ and $v+ dv$ is:

$\displaystyle{ n_vdv=n\left(\frac{m}{2\pi kT}\right)^{3/2} \exp\left(\frac{-mv^2}{2kT}\right)4\pi v^2dv} ,$

where $n$ is the total number of particles [show that the distribution is normalized. Hint: integrate $n_vdv$ over all velocities], each of mass $m$ [Note the dependence on the mass. What does it happen in a gas if for some reason each particle splits into two?], $k=1.38\times 10^{-16} erg\ K^{-1} ,$ is the Boltzmann constant, and $T$ the temperature in Kelvin.

Maxwell-Boltzmann distribution (2)

In terms of probability:

$\displaystyle{ P(v) dv= \frac{n_v}{n} dv=\left(\frac{m}{2\pi kT}\right)^{3/2} \exp\left(\frac{-mv^2}{2kT}\right)4\pi v^2dv} ,$

from where $\displaystyle{ \frac{dP(v)}{dv}=0}$ gives the most probable velocity: $\displaystyle{v_{mp}=\sqrt{\frac{2kT}{m}} }$ , and $\displaystyle{ \frac{\int_0^{\infty} vP(v)dv}{\int_0^{\infty} P(v)dv} }$ is the mean velocity $\displaystyle{ \overline{v}=\sqrt{\frac{3kT}{m}}} .$

For a population of atoms in equilibrium, the statistics of the excitation states is given by the Boltzmann equation: $\displaystyle{ \frac{N_b}{N_a} = \frac{P(E_b)}{P(E_a)}= \frac{g_b e^{-E_b/kT} }{g_a e^{-E_a/kT}} =\frac{g_b}{g_a} e^{-\left(E_b-E_a\right)/kT}} ,$

providing the number of atoms in the state “b” relative to that in the state “a”, or the ratio of the relative probabilities. The statistical weight $g$ is the measure of the degeneracy of the corresponding state (“a” or “b”), i.e. the number of combinations of the atomic quantum numbers of the level [which are they?] giving the same energy $E .$ The term $e^{-E_i/kT}$ is the Boltzmann factor for the j-th level.

Saha equation (1)

The relative populations of two ionized states separated by the energy $\chi_i$ are accounted for by an equation written by the Indian physicist Meghnad Saha in 1920:

$\displaystyle{ \frac{N_{i+1}}{N_i}= \frac{2}{n_e}\frac{Z_{i+1}}{ Z_i}\left(\frac{2\pi m_e KT}{h^2}\right)^{3/2} e^{-\chi_i/kT}} .$

Here $Z$ is the partition function:

$\displaystyle{Z=\sum_{j=1}^\infty g_j e^{-(E_j-E_1)/kT}} ,$

which accounts for the number of way the atom may arrange its electrons for the same level of energy; $n_e$ is the number density of electrons of mass $m_e = 9.11\times10^{-31} kg (P_e=n_e kT)$ , $h$ is the Plank constant, and $\chi_i$ the ionization potential. It is useful to note, similarly to Lotka-Volterra predator-prey model, excitation and ionization work one against the other: the more you ionize your gas, the less is left over for excitation.

Saha equation (2)

This concept is well exemplified by the case of the hydrogen atom and its degree of ionization in a stellar atmosphere. The Saha equation will be used to calculate the fraction of atoms that are ionized,

$N_{II}/N_{\mathrm{total}} = N_{II}/(N_{I}+N_{II})$

as the temperature $T$ varies between 5000 K and 25000 K. A hydrogen ion is just a proton, thus: $Z_{II}=1$ .

The energy of the first excited state of hydrogen is $E_2 - E_1 = 10.2\ eV$ above the ground state energy. $10.2\ eV >> kT$ for the temperature regime under consideration, thus the Boltzmann factor $e^{-(E_2 - E_1)/kT}<<1$ and so nearly all of the HI atoms are therefore in the ground state. The partition function then simplifies to $Z_I = 2$ . Inserting these values into the Saha equation with $\chi_I = 13.6\ eV$ gives the ratio of ionized to neutral hydrogen, $N_{II} / N_I$ . This ratio is then used to find the fraction of ionized hydrogen, $N_{II}/N_{\mathrm{total}}$ , by writing:

$\frac{N_{II}}{N_{\mathrm{total}}} = \frac{N_{II}}{N_I + N_{II}} = \frac{N_{II}/N_I}{1 + N_{II}/N_I}$

Saha equation (3)

This figure shows that when $T=5000\ K$ , essentially none of the hydrogen atoms are ionized.

At about $8300\ K$ , $5 \%$ of the atoms have become ionized. Half of the hydrogen is ionized at a temperature of $9600\ K$ .

When $T$ has risen to $11300\ K$ , all but $5\%$ of the hydrogen is in the form of HII.

Thus the ionization of hydrogen takes place within a temperature interval of approximately $3000\ K$ .

NII/Ntotal for hydrogen from the Saha equation when Pe=20Nm^(-2). Fifty percent ionization occurs at T≈9600K

Saha equation (4)

Now we can see why the Balmer lines are observed to attain their maximum intensity at a temperature of $9520\ K$ , instead of at the much higher characteristics temperatures [Which order of magnitude?] required to excite electrons to the $n=2$ energy level of hydrogen.

The strenght of the Balmer lines depends on $N_2 / N_{\mathrm{total}}$ , i.e. the fraction of all hydrogen atoms that are in the first excited state. This is found by combining the results of the Boltzmann and Saha equations.

Because virtually all of the neutral hydrogen atoms are in either the ground state or the first excited state, we can employ the approximation $N_1 + N_2 \simeq N_I$ and write:

$\frac{N_2}{N_{\mathrm{total}}}= \left( \frac{N_2}{N_1 + N_2} \right) \left( \frac{N_I}{N_{\mathrm{total}}} \right) = \left( \frac{N_2 / N_1}{1 + N_2 / N_1} \right) \left( \frac{1}{1 + N_{II}/N_I} \right)$

As we can see from the figure on the right, the hydrogen gas would produce the most intense Balmer lines at a temperature of $9900\ K$ , in good agreement with the observations. The diminishing strenght of the Balmer lines at higher temperatures is due to the rapid ionization of hydrogen above $10000\ K$

N2/Ntotal for hydrogen from the Boltzmann and Saha equations, assuming Pe=20Nm-2. The peak occurs at approximately 9900 K.

Non-thermal processes

Non-thermal radiation processes are:

Bremsstrahlung (free-free as compared to free-bound and bound-free processes): acceleration on (high speed, low mass) charged particles. The spectrum increases with frequency to a maximum value, then it is cut-off.
Compton scattering: exchange of energy between high energy photons with low energy electrons (direct effect) and by cold photons with relativistic electrons (inverse). The low energy tail of the Compton scattering is Thomson scattering, where the particles do not exchange energy.
Synchrotron: acceleration of ultra-relativistic charged particles by a magnetic field (spectrum peaks at a critical frequency an rapidly decreases redward).

Electromagnetic jargon

The 21-cm line of neutral hydrogen

Neutral hydrogen (HI) consists of a single proton orbited by a single electron. The spin of the electron and proton can be either parallel or antiparallel (hyperfine structure):

$\displaystyle{S_z=\pm\frac{1}{2}\hslash}.$

Because of magnetic interactions between the particles, the parallel spin configuration has slightly more energy than other one. Therefore, the transition from the parallel to the antiparallel spin configuration of an hydrogen atom at the ground state (lowest energy state) generates the emission of a low energy photon at: $\nu = 1420.40575\ MHz,$ or $\lambda\simeq 21\ cm,$ i.e. in the radio domain. This transition is highly forbidden with an extremely small probability of $2.9\times10^{-15} s^{-1}.$ This means that an isolated atom of neutral hydrogen undergoes this transition in $\sim 10^7\ yr$ (possible only at very low densities). The very long lifetime allows the line to have a very narrow width [why?]; most of the broadening comes from the Doppler shift. The lifetime gets shorter by collisions with other atoms and by interaction with the cosmic microwave background radiation. Jan Oort and his student Hendrik van de Hulst were first to guess that, in spite of the inefficiency of the process, the huge amount of HI in the interstellar space should secure a detectable signal. Actually, since World War II, the 21 cm emission line of HI has become a powerful way to gauge the kinematics of galaxies.

Electron Volt

How good is your measure: signal and noise

The signal $S$ is the number of photons detected from a source. But not all of the incoming photons are actually detected and converted into electrons [modify the concept for photographic emulsions and all non linear detectors].

The ratio electron/photon is the “quantum efficiency”, typically of the order of $60\div 90\%$ in modern detectors (Charge Coupled Devices = CCDs). So, in the following the signal will refer to the number of electrons (to use photons you need to consider the gain):

$S=s\times t\ [e^-],$

where $s\ [e^-/t]$ is the number of electrons per second, and $t\ [s]$ the total exposure time. The latter may be split in $k_{dit}$ sub-exposures (dithers) of fixed length $t_{dit}$ (for reasons which include saturation, cosmic rays removal and, in mosaics, filling of image gaps), so that:

$t=k_{dit}\times t_{dit},$

and:

$S=s\times k_{dit} t_{dit}.$

The error on $S$ is the noise $N,$ also measured in $e^-,$ which comes from the combination of random contributions to the signal from various sources. We want to calculate the so called signal-to-noise ration, $S/N,$ disentangling the various contributions.

S/N: intrinsic sources of noise

The noise associated to $n$ incoming electrons, $N=\sqrt{n},$ has the following sources.

Object’s noise: $N_{obj}=\sqrt{n_{obj}},$ is associated to the $n_{obj}$ electrons arising from the signal of the source, so:

$S_{obj}=\sqrt{S}=\sqrt{s\times t}=\sqrt{s\times k_{dit}\times t_{dit}} .$

Sky noise:

$N_{sky} = \sqrt{s_{sky}}$

is associated to the signal $s_{sky}$ coming from each of all the unresolved sources within the area (measured in number of pixels $n_{pix}$ ) over where the source is measured, so:

$N_{sky} \sqrt{n_{pix}\times s\times t} = \sqrt{n_{pix}\times s_{sky}\times k_{dit}\times t_{dit}}.$

Dark noise: is the shot noise applied to the dark current. It is due to $n_{dk}$ thermal electrons (per pixel and per hour) produced by the detector (in spite of the cooling devices): $N_{dk} \sqrt{n_{pix}\times n_{dk} \times t}= \sqrt{n_{pix}\times n_{dk} \times k_{dit}\times t_{dit}} .$

S/N: read-out noise

It is another source of noise associated to the reading of the detector, which depends on the detector and on the mode it is read. The $n_{RON}$ is the read-out noise, in $e^-,$ for every reading (you add noise at each reading) and per pixel:

$N_{RON}=\sqrt{n_{pix}\times n_{RON}^2}= \left(\sqrt{n_{pix}}\right) \times n_{RON}.$

Reading the detector $k_{dit}$ times:

$N_{RON}= \left(\sqrt{n_{pix}\times k_{dit}}\right) \times n_{RON}.$

S/N: cumulative effect: signal-to-noise

Summing up all sources:

$N_{total}=\sqrt{n_{obj}^2+n_{sky}^2+n_{dk}^2+n_{sky}^2},$ and:

$\displaystyle{\frac{S}{N}=\frac{s_{obj}\times t}{\sqrt{s_{obj}\times t + n_{pix}\times s_{sky}\times t + n_{pix}\times n_{dark}\times t + n_{pix}\times k_{dit}\times n_{RON}^2}}} .$

S/N: observation of a bright source

Consider a bright enough source so that the dominant noise is its Poisson noise, the others being negligible. So:

$\displaystyle{\frac{S}{N}=\displaystyle{\frac{s_{obj}\times t}{\sqrt{s_{obj}\times t}}} = \sqrt{s_{obj} \times t }\propto \sqrt{t}} .$

Application: bright stars in imaging and spectra. Note that in these cases the problem is usually not that of reaching a good S/N, but rather not to saturate the detector [what happens then?].

S/N: sky-noise dominated observation

Let’s consider the case of a faint star with a bright sky background: $s_{sky}\gg s_{obj}.$ The signal-to-noise equations writes:

$S/N=\displaystyle{\frac{s\times t}{\sqrt{n_{pix}\times s_{sky}\times t}}}\propto \displaystyle{\frac{\sqrt{t}}{\sqrt{n_{pix}}}} .$

The consequences are that: $S/N$ increases as $n_{pix}$ decreases (but not as fast). Since, for point-like sources, $n_{pix} \simeq FWHM^2,$ one sees that the better the seeing, the better the signal-to-noise.

The signal-to-noise ratio increases as:

$\sqrt{t_{tot}} = \sqrt{k_{dit}\times t_{dit}},$

showing that in this one case the splitting of the exposures into some shorter exposures does not affect the total $S/N.$

S/N: example of sky-noise dominated observation

Let’s ask the following question: given a sky-noise dominated observation of a bright star, which is the aperture size through which the $S/N$ is maximum for a given seeing? To this purpose we assume that, due to seeing, the stellar image is Gaussian:
$I(r)=I_0e^{-r^2/2\sigma^2},$
where $\sigma=\mbox{\it FWHM}/2.355$ measures the seeing.
Then, within $n_{pix}=\pi R^2$ pixels, the signal is:

$s=\int_0^R I(r)2\pi r dr .$

Given the seeing $\sigma$ and the exposure time $t,$ the signal-to-noise ratio is:

$\displaystyle{ \frac{S}{N} = \displaystyle{\frac{s\times t}{\sqrt{n_{pix}\times s_{sky}\times t}}} \propto \frac{\int_0^R I(r)2\pi r dr }{R}\propto \frac{\int_0^{R/\sigma} e^{-x^2/2} x dx}{x}}.$

Computed numerically (see figure) the maximum of $S/N$ is for $x=R/\sigma=1.585,$ that is the most favorable aperture has a radius $R=1.585\sigma=0.67 \ FWHM.$

S/N: read-out-noise dominated observation

If sky and dark are very small, and the RON is left to dominate, then:

$\dispalystyle{\frac{S}{N}= \frac{s\times t}{\sqrt{n_{pix}\times k_{dit}\times n_{RON}^2}} \propto \frac{1}{\sqrt{n_{pix}}}\frac{1}{\sqrt{k_{dit}}}}.$

First term: $1/\sqrt{n_{pix}}$ By binning the data (reading the combination of $m\times m$ pixels as a single value) you gain in signal-to-noise ratio a factor $m,$ but loose the same in resolution.

Second term: $1/\sqrt{k_{di}}.$ You may try to reduce the number of dithers to the minimum, making the sub-exposures as long as possible. The cost is more cosmic rays and, in case of mosaic detectors, a higher unevenness of the weights of the single pixels [why?].