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 and is:
where is the total number of particles [show that the distribution is normalized. Hint: integrate over all velocities], each of mass [Note the dependence on the mass. What does it happen in a gas if for some reason each particle splits into two?], is the Boltzmann constant, and the temperature in Kelvin.
In terms of probability:
from where gives the most probable velocity: , and is the mean velocity
For a population of atoms in equilibrium, the statistics of the excitation states is given by the Boltzmann equation:
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 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 The term is the Boltzmann factor for the j-th level.
The relative populations of two ionized states separated by the energy are accounted for by an equation written by the Indian physicist Meghnad Saha in 1920:
Here is the partition function:
which accounts for the number of way the atom may arrange its electrons for the same level of energy; is the number density of electrons of mass , is the Plank constant, and 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.
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,
as the temperature varies between 5000 K and 25000 K. A hydrogen ion is just a proton, thus: .
The energy of the first excited state of hydrogen is above the ground state energy. for the temperature regime under consideration, thus the Boltzmann factor and so nearly all of the HI atoms are therefore in the ground state. The partition function then simplifies to . Inserting these values into the Saha equation with gives the ratio of ionized to neutral hydrogen, . This ratio is then used to find the fraction of ionized hydrogen, , by writing:
This figure shows that when , essentially none of the hydrogen atoms are ionized.
At about , of the atoms have become ionized. Half of the hydrogen is ionized at a temperature of .
When has risen to , all but of the hydrogen is in the form of HII.
Thus the ionization of hydrogen takes place within a temperature interval of approximately .
Now we can see why the Balmer lines are observed to attain their maximum intensity at a temperature of , instead of at the much higher characteristics temperatures [Which order of magnitude?] required to excite electrons to the energy level of hydrogen.
The strenght of the Balmer lines depends on , 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 and write:
As we can see from the figure on the right, the hydrogen gas would produce the most intense Balmer lines at a temperature of , 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
Non-thermal radiation processes are:
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):
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: or i.e. in the radio domain. This transition is highly forbidden with an extremely small probability of This means that an isolated atom of neutral hydrogen undergoes this transition in (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.
The signal 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 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):
where is the number of electrons per second, and the total exposure time. The latter may be split in sub-exposures (dithers) of fixed length (for reasons which include saturation, cosmic rays removal and, in mosaics, filling of image gaps), so that:
The error on is the noise also measured in 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, disentangling the various contributions.
The noise associated to incoming electrons, has the following sources.
Object’s noise: is associated to the electrons arising from the signal of the source, so:
is associated to the signal coming from each of all the unresolved sources within the area (measured in number of pixels ) over where the source is measured, so:
Dark noise: is the shot noise applied to the dark current. It is due to thermal electrons (per pixel and per hour) produced by the detector (in spite of the cooling devices):
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 is the read-out noise, in for every reading (you add noise at each reading) and per pixel:
Reading the detector times:
Summing up all sources:
Consider a bright enough source so that the dominant noise is its Poisson noise, the others being negligible. So:
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?].
Let’s consider the case of a faint star with a bright sky background: The signal-to-noise equations writes:
The consequences are that: increases as decreases (but not as fast). Since, for point-like sources, one sees that the better the seeing, the better the signal-to-noise.
The signal-to-noise ratio increases as:
showing that in this one case the splitting of the exposures into some shorter exposures does not affect the total
Let’s ask the following question: given a sky-noise dominated observation of a bright star, which is the aperture size through which the is maximum for a given seeing? To this purpose we assume that, due to seeing, the stellar image is Gaussian:
where measures the seeing.
Then, within pixels, the signal is:
Given the seeing and the exposure time the signal-to-noise ratio is:
Computed numerically (see figure) the maximum of is for that is the most favorable aperture has a radius
If sky and dark are very small, and the RON is left to dominate, then:
First term: By binning the data (reading the combination of pixels as a single value) you gain in signal-to-noise ratio a factor but loose the same in resolution.
Second term: 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?].
8. Spiral arms
10. Scale relations
14. Galaxy dynamics
19. Galaxy clusters
James Binney and Michael Merrifield, Galactic Astronomy, Princeton Univ. Press, 1998.