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Massimo Capaccioli » 23.Fundamentals - Part II

Tutorial on galaxy photometry

Let us follow the series of processes through which the image signal of a galaxy arrives to us, is collected by the telescope, is registered by our CCD [what is it?] detector, and is finally read as a file to be eventually processed. The clear understanding of this pipeline of happenings allows us to carry on a proper data reduction and a correct error analysis.

Let $G(x,y)$ be the surface intensity distribution map of the galaxy as a function of $x,y$ coordinated in the plane of the sky. We put the origin at the galaxy center [What does this mean? Geometrical center or brightness peak?].

Let $S(x,y)$ be instead the surface intensity distribution map of the sky brightness: the light from the unresolved cosmic background, the atmospheric airglow, and the human pollution. In the ideal case $S$ is a constant, but this never happens. For simplicity we neglect here the contributions to $S$ by sources other than our target galaxy, such as Milky Way objects and resolved extragalactic objects, which though exist and must be accounted for. We also ignore defects, cosmic rays, and noise.

Seeing convolution

The combined signal:

$J(x,y) = G(x,y) + S(x,y)$

is then blurred while passing through the Earth atmosphere (which takes place just a few tens of meters above the telescope and is named “atmospheric seeing”) and through the instrument (which has vibrations and optical elements with a finite resolution).

The result is a convolved signal:

$J_c(x,y) = \int\!\!\!\!\int J\left(x^{\prime}, y^{\prime}\right)PSF\left(x-x^{\prime}, y-y^{\prime}\right) dx^{\prime}dy^{\prime} =G(x,y)\otimes PSF(x,y) + S(x,y),$

where $\otimes$ indicates a convolution product and $PSF(x,y)$ is the “point-spread function” of the overall “seeing” and represents the convolution of point-like source. Oversimplifying, the $PSF$ is classically characterized by one 2-D symmetric Gaussian function with variance $\sigma_{PSF}$ (< 1 arcsec in fair/good ground-based astronomical sites [does the Hubble Space Telescope have a finite PSF?]). A way to quantify the generic PSF is to give the full-width at half-maximum, FWHM, which is the width of the function, whatever its shape is, at half the distance from the peak [show that, for a Gaussian PSF, $FWHM=2.3\sigma$ ]. You noted that above the convolution by the PSF was not applied to the night sky signal S. The reason is that, having decided to ignore fore-background sources, the characteristic frequencies of S are orders of magnitude lower that those of the point-spread function: in conclusion, S is not altered by the convolution.

Components of an observed image

Now, $J_c(x,y)$ reaches the detector where it is integrated and sampled in picture elements (pixels) of a given size (e.g. $20\mu m\times 20\mu m)$ . This is translated into an angular size by the knowledge of the equivalent focal length of the instrument [how?]. Because of the sampling, $J_c(x,y)$ turns into the average value within the pixel of indices i,j:

$J_c(x,y) \rightarrow \langle(x,y)\rangle + {\rm noise} = J_a(i,j)$ .

Moreover, since each pixel has its own response to the signal $F$ (gain and zero point), and since electronic detectors require a sort of start-up signal (bias $B$ ) and produce thermal signals (dark current $D$ ) kept under control by lowering the temperature [why?], what we read is:

$J_r(i,j) = J_a(i,j)\times F_{CCD}(i,j) + B_{CCD}(i,j) + D_{CCD}(i,j)$ ,

or: $J_r(i,j) = \langle J_c(i,j)\rangle\times F_{CCD}(i,j) + B_{CCD}(i,j) + D_{CCD}(i,j) =$

$\langle G(x,y)\otimes PSF(x,y) + S(x,y)\rangle\times F_{CCD}(i,j) + B_{CCD}(i,j) + D_{CCD}(i,j)$ ,

from where we need to extract $G(i,j)$ ; not $G(x,y)$ , since what is lost by the pixel averaging ( $x\rightarrow i$ , $y\rightarrow j$ ) cannot be recovered but marginally. In order to reach our goal, first of all we have to set up a proper observational strategy to recover all the information that we will later need.

Tutorial on galaxy photometry: reduction steps

The steps are as follows.

If the detector field is small than our target galaxy, there is no direct way to estimate the background directly on the science exposure. There are ways to overcome the problem [can you guess one?]. Note that the background must be accurately defined to avoid increasingly large errors in the galaxy outskirts. The risk is a stretched galaxy for a too faint estimate of S, a truncated object for a too bright estimate. [Try to evaluate this effect on a toy model].

The PSF is usually mapped using the non-saturated images of bright galaxies present in the science exposure. The way this PSF is used for deconvolution, if the case, will be briefly discussed later.

At the beginning of the observing night, and at the end, several exposures of a uniform artificial source (or of the sky at sunset/dawn) are secured to obtain the flat-field map F, and other exposures for bias and dark current.

In this brief account we neglected further complications such as non-linearity (as in photographic emulsions); tiling of CDD mosaics, which requires multiple exposures slightly shifted one from the other (dithered exposures) to remove the inter-tile gaps and causes grave unevenness in the background; fringes (same as Newton rings), due to reflecting surfaces in the detector (particularly in the IR); contamination by neighboring celestial sources; cosmic rays and defects removal; photometric calibration, which requires the observation of standard stars, and the extinction correction.

Tutorial on galaxy photometry

The result of all these passages and corrections are galaxy surface brightness maps which, at best, are progressively unreliable

while approaching the peaks (nucleus, point-like sources) depending of the original resolution and on the deconvolution treatment (if any), and
at increasing distances from the center, where the decreasing surface brightness competes with the night sky: typically $\mu_B(sky)=22.3\ mag/arcsec^2$ ).

[Compute the surface brightness at which the signal is 1/100 that of the sky. Find a way to estimate, for a given instrumental setup, what the exposure time should be to achieve a signal-to-noise ration of 3.]

The figure shows a star imaged by a CCD. It is clear the seeing blurring by an otherwise point-like source. Individual pixels are discernible (gray scale for the discretized intensity). The detector is read by successively shifting the rows to the top register where the pixels of the front row are sent sequentially to the preamplifier and the A/D converter. [Meditate on which can be the map of the read-out noise.]

Fourier Transform (1)

The Fourier transform is a linear operator mapping a function into another. In practice, it decomposes a function into the continuous spectrum of its own frequencies. The inverse transform, instead, synthesizes the function from the frequency spectrum. A Lebesgue integrable function of complex variable, $f(t)$ , transforms and anti-transforms as:

$F(\omega)=\displaystyle{ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-\infty}f(t)e^{-i\omega t}dt}$ ;

$f(t)=\displaystyle{\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}F(\omega)e^{i\omega t}d\omega }$ .

Fourier Transform (2)

If $t$ is time and $\omega$ an angular frequency (in radiants), by writing: $F(\omega)=A(\omega)e^{i\phi(\omega)}$ , then each component of: $f(t)=\displaystyle{\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}A(\omega)e^{i\left[\omega t+\phi(\omega)\right]}d\omega}$ ,

is a sinusoid of amplitude $A(\omega)$ , frequency $\omega$ , and phase $\phi(\omega)$ at $t = 0$ .

Prove that $f(t)$ is the antitransform of its transform: $f(t)=\displaystyle{\frac{1}{2\pi}\int_{-\infty}^{+\infty}\left[\int_{-\infty}^{+\infty}f(t)e^{-i\omega t}dt\right]e^{i\omega t} d\omega}$ .

Fourier transform (3)

Any function $f(t)$ can be written as the sum of an even and an odd function. A function is even if

$f(t) = f(-t)$ ;

it is odd if

$f(-t) = -f(t)$ .

Therefore, the function:

$e(t) = \frac{1}{2}\left(f(t) + f(-t)\right)$

is the even component of f, while:

$o(t) = \frac{1}{2}\left(f(t) - f(-t)\right)$

is the odd one, and we can write:

$f(t) = e(t) + o(t).$

It follows that:

$F(\omega)=\sqrt{\frac{2}{\pi}}\displaystyle{\int_{-\infty}^{+\infty}e(t)\cos(\omega t) dt-i\sqrt{\frac{2}{\pi}}\int_{-\infty}^{+\infty} o(t)\sin(\omega t)dt}$ .

Jean Baptiste Joseph Fourier (1768-1830).

Properties of the Fourier Transform

Properties of the Fourier transform are:

$\mbox{ Linearity: }\hspace{5pt} af_1(t)+bf_2(t)= aF_1(\omega)+bF_2(\omega) ;\\$

$\mbox{ Time reversal: }\hspace{5pt} f(-t)\Leftrightarrow F(-\omega) ;$

$\mbox{ Time shift: }\hspace{5pt} f(t-t_0)\Leftrightarrow F(\omega)e^{it_0\omega} ;\\$

$\mbox{ Modulation: }\hspace{5pt} f(t)e^{it\omega_0} \Leftrightarrow F(\omega-\omega_0) ;\\$

$\mbox{ Scaling: }\hspace{5pt} f(at)\Leftrightarrow \displaystyle{\frac{F(\omega/a)}{a}}\hspace{20pt} a\neq0 ;\\$

$\mbox{ Integration: }\hspace{5pt} \int_{-\infty}^t f(s)ds \Leftrightarrow\frac{1}{i\omega}F(\omega)+\pi F(0)\delta(\omega) ;\\$

$\mbox{ Convolution theorem: }\hspace{5pt} \mbox{if \ } g(t)=f(t)\otimes h(t)=\int_{-\infty}^{+\infty}f(s)h(t-s)ds = \int_{-\infty}^{+\infty} f(t-s)h(s)ds, \ \mbox{then\ } G(t)=F(t)\cdot H(t);$

$\mbox{ Parceval theorem: }\hspace{5pt} \int_{-\infty}^{+\infty} \mid f(t)\mid^2dt=\int_{-\infty}^{+\infty} \mid F(\omega)\mid^2d\omega,$

i.e. the power is the same in the space of the function and in that of its transform.

Gaussian convolution

It is easy to show that a Gaussian function:

$g(t)= g_0 e^{-\frac{\left(t-t_0\right)^2}{2\sigma^2}}$ ,

is an eigenfunction of the Fourier transform. You simply compute the transform:

$F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}g(t)e^{-i\omega t}dt= \frac{g_0}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-\frac{\left(t-t_0\right)^2}{2\sigma^2}} e^{-i\omega t}dt=$ $\\[5pt] =\frac{g_0}{\sqrt{2\pi}}\left(\int_{-\infty}^{+\infty}e^{-\frac{\left(t-t_0\right)^2}{2\sigma^2}} \cos(\omega t) dt + i\int_{-\infty}^{+\infty}e^{-\frac{\left(t-t_0\right)^2}{2\sigma^2}} \sin(\omega t)dt \right).$

The second integral is null since the integrand is symmetric. Note that, if $\sigma_1\ge\sigma_2$ , then $1\le \sigma/\sigma_1\le 1.41$ ; i.e., the variance of the broader Gaussian may vary by not more than 41% because of a convolution with a thinner Gaussian. The first can be solved analytically to give:

$G(\omega)=\displaystyle{\frac{g_0}{2\sigma}e^{-2\sigma^2\omega^2},$

which is a Gaussian with variance equal to $1/(2\sigma)}$ .

Also, the convolution of two Gaussian functions with variances $\sigma_1$ and and $\sigma_2$ and centers at $t_1$ and $t_2$ is again a Gaussian with center at $(t_1+t_2)$ and variance: $\sigma=\sqrt{\sigma_1^2+\sigma_2^2}$ .

Variance of the Gaussian resulting from the convolution of two Gaussians.

Seeing convolution: toy model

Let us first ask ourselves why the mathematical equivalent of seeing is a (two-dimensional) convolution. The answer is that an exposure is the sum (time integration) of a same image slightly shifted in two (independent) directions by the atmosphere and by the telescope according to a sort of weighting function, the $PSF$ , which tells us for how long (relative to the total exposure time) the image has been in one relative position. Remember that the $PSF$ is the blurred image of a point source, and that the timing of the process is given by the frequency of seeing blurring, $\sim 30\ Hz$ .

Seeing convolution: toy model

The effects of seeing on the image of a galaxy are easily understood using toy models for the light distribution. For sake of simplicity we will discuss a light profile, $I(R)$ , rather than the full galaxy image, and a one-dimensional Gaussian $PSF$ :

$G(R) = \exp\left(-R^2/2\sigma_*^2\right)$ .

The light profile is the cross-section of the surface intensity map through the center of the object. It is unique for circularly symmetric galaxies.

Seeing convolution: toy model

Let us now assume that the light profile can be approximated by the sum of two Gaussians, a narrower one for the core (c) and broader one for the envelope (e):

$I(R) = \exp\left(-R^2/2\sigma_c^2\right) + \exp\left(-R^2/2\sigma_e^2\right)$ .

For nearby large galaxies and a reasonable seeing, it is $\sigma_\star \ll \sigma_e$ . This implies that the galaxy envelope will be unaffected by atmospheric blurring. [No longer true, if the galaxy distance increases. Why?] The figure shows the effects of convolution on a spherical r^1/4galaxy: the center is depressed and the energy redistributed all around it.

The core instead can be narrow enough, $\sigma_*\simeq \sigma_c$ , and even narrower that the seeing (then we say that it is unresolved). In this case the convolution will be effective: the result will be a less peaked profile. Since the process is conservative, though, the energy lost by the peak will be redistribute at larger radii. In summary, a seeing convolved profile will be fainter that the unconvolved one near the center, than brighter and farther out unmodified.

From Capaccioli and de Vaucouleurs, Ap.J.S., 52, 465, 1983.

Sampling theorem and deconvolution

Can we recover what seeing has washed out? In order to answer this question we need to recall a basic theorem of the information theory: the Nyquist-Shannon sampling theorem.

A signal $f(t)$ is band-limited if it has no spectral components with frequency $v>\nu_B,$ , i.e. if its Fourier transform:

$F(\omega) =0$ for $|\omega|>2\pi v_B$

The sampling theorem states that $f(t)$ can be perfectly reconstructed from a discrete sampling at a rate $R>2v_B$ of samples per unit of $t$ . The smallest sampling frequency, named after the American mathematician Harry Nyquist, is then: $\nu_N = 2\nu_B$ , and the Nyquist sampling interval $T=2 /\nu_B$ .

Let us analyze the consequence of this for an image taken with a seeing with $p=(\alpha/2)(1/s)$ , where $s$ is scale of the focal plane of the telescope (remember that $s = 206265/L$ , with $L =$ effective focal length of the instrument). According to the sampling theorem, a larger pixel would imply a loss of resolution. A smaller pixel would also be ill-chosen. It would provide no better resolution and a net loss if the signal-to-noise ratio. Note that:

$S/N=\displaystyle{\frac{Ip^2}{\sqrt{Np^2+R}}}$ ,

where $I$ is the surface density of the signal, $N$ that of the noise, and $R$ the detector noise, independent on the pixel size.

Sampling theorem and Deconvolution

We can now answer the question posed at the beginning of the previous page. One may indeed deconvolve seeing-blurred images up to about the scale of the seeing itself. And how can that be done? In principle it is straightforward. We know that, if:

$h(x,y) = f(x,y)\otimes g(x,y)$ ,

then the Fourier transformed functions obey the simple relation:

$H(x,y) = F(x,y)\cdot G(x,y)$ .

If $h$ is a galaxy image obtained by convolving $f$ with $g$ (PSF), then the anti-transform of:

$F(x,y) = H(x,y)/G(x,y)$ ,

where $H$ and $G$ are transforms of observables, is $f$ , i.e. precisely what we need. Unfortunately, this simple procedure [but is it really simple to perform a FT on an image?] hits the wall of noise. In fact, it can be shown that it amplifies dramatically the noise [explain heuristically]. Thus, any deconvolution procedure must be accompanied by a proper noise control. This latter is just a full chapter of the information theory!

Atmospheric extinction

How do we correct our photometric measurements for the extinction operated by the Earth atmosphere? In order to derive a simplified but working expression, let’s assume that the atmosphere (at small enough zenithal distances $z$ ; see figure) may be sketched by parallel layers (plane-parallel approximation). Note that the zenithal distance $z$ for a source at equatorial coordinates $(\alpha, \delta)$ observed at the sidereal time $t_s$ from a place on Earth at latitude $\phi$ is given by the equation: $\sin(90-z)=\sin\phi\sin\delta+\cos\phi\cos\delta\cos(t_s-\alpha)$ . The relative loss of flux $I_\lambda$ for a variation of altitude $dh$ is proportional to the length of the true step $dh\sec z$ and to the property of the layer, i.e. to the absorption coefficient $k_\lambda(h)=k_\lambda(x)$ , where we have explicitly indicated the dependence on $\lambda$ . The term $\sec z$ is named airmass. It is: $\displaystyle{\frac{dI_\lambda}{I_\lambda}=-k_\lambda(h)\sec z\ dh=-\sec z\ d\tau_\lambda},$ where $\tau_\lambda$ is the optical depth. In passing: the atmospheric extinction is mostly due to Reyleigh scattering, which has a cross section depending on $\lambda^{-4}$ . [Knowing this, as yourself why the sky is blue while the Sun is yellow.]

Plane parallel approximation for the atmosphere about the Zenith.

Atmospheric extinction

Integration from the observer level $h_0$ to infinity:

$\displaystyle{\int_{h_0}^\infty \frac{dI_\lambda}{I_\lambda}= -sec z \int_{h_0}^\infty d\tau_\lambda}$ ,

gives:

$\displaystyle{I_\lambda(h_0)=I_\lambda(\infty)e^{-sec z\ \tau_\lambda(\infty)}}$ ,

where:

$\displaystyle{\tau(\infty)=\int_{h_0}^\infty d\tau_\lambda= \int_{h_0}^\infty k_\lambda(h) dh}$ .

In magnitudes:

$-2.5\log\left[I_\lambda(h_0)\right]=-2.5\log\left[I_\lambda(\infty)\right]+2.5\tau_\lambda\log_e\sec z,$

from where the magnitude outside the atmosphere, $m_{out}$ , is:

$m_{obs}-m_{out}=-k_\lambda\sec z=-k_\lambda X(z),$

with $\sec z=0$ (see figure) gives the wanted correction:

$\Delta m_\lambda=m_{obs}-m_{out}.$

Pickering (2002) gave a better formula for the generalize airmass: $X=1/\sin\left[h+244/(165+47h^{1.1}) \right].$

Derivation of m_Î» by extrapolating the observations of the same star at different angular distances on the same night.

Interstellar Medium (ISM)

Outer space, in between Milky Way (MW) stars and among galaxies, is not empty. It is just a very low density environment, far more empty than the best vacuums in terrestrial labs (1 atom/cm³, and a mean free path of $\sim 1\ lyr$ ).

Literarily, the interstellar medium (ISM) is the gas (99% of the total mass) and dust (1%) filling the interstellar space among stars in galaxies, and bleeding in/from the intergalactic space. It is a turbulent ensemble of ions, atoms, molecules, larger dust grains, cosmic rays, and (galactic) magnetic fields, which emits, absorbs, and reflects radiation. As a consequence of the primordial nucleosynthesis, by number of nuclei (not mass!), $\sim 89\%$ of the gas is H, $9\%$ He, and $2\%$ heavier elements (“metals” in astronomical jargon).

Dust is made visible by the obscuration of foreground sources (as the Coal Sack) or revealed by the emission of associated molecules. Unawareness of the effects of the Galactic dust on the photometry of stars has led to major mistakes in the 20-th century.

Extinction has two effects: it reduces the intensity and reddens the radiation, as scattering [Rayleigh but not Thomson? Which one?] is more efficient in the blue than in the red.

Gas may either re-shine energy of nearby sources (e.g. Orion Nebula), emit its own photons in form of emission line spectra (e.g. HI extensions of spiral galaxies or the Orion Nebula) and non-thermal continua (e.g. emission from extragalactic radio sources), or capture photons causing absorption lines in other sources’ continuum spectra (see, e.g., the Lyman-alpha forests in the spectra of distant quasars). Some intracluster stars have also been identified in the space within clusters of galaxies: all over they are possibly up to 10% in light of the baryons.

ISM

Let’s us list the properties of the various ingredients and structures according to their temperature (called the phase of the interstellar medium):

cold ( $\sim 10\ K$ ),
warm ( $100\div 1000\ K$ )
hot ( $> 10^6\ K$ ).

If the emission is thermal, the colder the gas cloud is, the longer are the wavelengths of its emission (Wien’s law). So, cold nebulae (atoms and molecules) shine in the microwave and radio domain, hot regions of interstellar gas (ions) emit in the optical, and very hot fluid envelops of clusters (plasmas) in the X-ray region.

Carina Nebula with priming dark molecular clouds. Credit: NASA, ESA, Hubble Heritage team.

Orion Nebula. Credit: ESO.

Galactic extinction (1)

How do we correct our photometric measurements for the extinction operated by the dust in the Galaxy? First of all note that the observed color:

$C_{\lambda_1\lambda_2}=m_{\lambda_1} - m_{\lambda_2}$ , differs from the intrinsic color as $m_\lambda=m_\lambda^0 + A_\lambda$ , where the suffix indicated a de-reddened magnitude and $A_\lambda$ is the absorption term. It is:

$C_{\lambda_1\lambda_2}=m_{\lambda_1}-m_{\lambda_2}=m_{\lambda_1}^0-m_{\lambda_2}^0+(A_{\lambda_1}-A_{\lambda_2})$ .

Thus:

$C_{\lambda_1\lambda_2}=m_{\lambda_1}^0-m_{\lambda_2}^0+A_{\lambda_1}-A_{\lambda_2}= C_{\lambda_1\lambda_2}^0+E_{\lambda_1\lambda_2}$ ,

$E_{\lambda_1\lambda_2}=C_{\lambda_1\lambda_2}-C_{\lambda_1\lambda_2}^0=A_{\lambda_1}-A_{\lambda_2}$ ,

and:

$E_{\lambda_1\lambda_2}=A_{\lambda_1}-A_{\lambda_2}=A_{\lambda_1}\left(1-\displaystyle{\frac{A_{\lambda_2}}{A_{\lambda_1}}} \right)=A_{\lambda_1} Q$ ,

and finally:

$A_{\lambda_1}=(1/Q) E_{\lambda_1\lambda_2}$ .

Galactic extinction (2)

The color excess is known by guessing the value of the intrinsic color. More serious is the problem of the coefficient 1/Q, which depends on the nature of the scattering material and may vary from 1 to 5. A simplified formula, applicable to galaxies, has been proposed by Holmberg (1958):

$A_{pg}=e_\lambda\cos\vert B\vert$ ,

where $B$ is the Galactic latitude [what is it?].

More accurate extinction maps, which account for the patchy distribution of the dust in the Galaxy, have been produced by Burstein and Heiles (Ap.J., 225, 40, 1978) on the assumption of the proportionality between neutral hydrogen and dust column density. HI column densities are provided by the sky mapping at 21 cm.

ISM ingredients (1)

Neutral Hydrogen (HI): most abundant in interstellar medium ( $\rightarrow$ high lifetime ( $\rightarrow$ forbidden line). In the MW it is concentrated in the spiral arms.
Interstellar Molecules: made mostly of H, He, C, N, O, these fragile ensembles, producing spectral bands by rotation and vibration, are only found in the dark centers of dense gas and dust nebulae, usually sites of protostar/star cluster formation.
Interstellar Dust: produced in the envelopes about red supergiants, and ejected into ISM by stellar winds and the planetary nebula (PN) phase, dust particles are mostly a few μm C and Si grains. They completely absorb any photon striking them and are collisionally heated by surrounding gas (processes which may destroy the grains), but re-radiate the energy in the far-infrared (FIR). Dust grains serve as sites for the formation of molecules and organic compounds.

ISM ingredients (2)

Dark Nebulae: named “holes in the heaven” by William Herschel, they were then identified as obscuring clouds by American astronomer E.E. Barnard. Dense and cold cores of $5\div 50\ M_\odot$ , spanning $1\ lyr$ and containing molecular H (H₂), CO. He, and silicate dust, named Bok globules, are sites of star formation.
Molecular Clouds: dense concentrations (up to 10⁹ atoms/cm³) of low temperature ( $T\sim20\div200\ K$ ) gas and dust, up to $50\ pc$ wide and containing up to $10^6 M_\odot$ (the largest are said Large Molecular Clouds), favor the formation of molecules at the surface of dust grains. The condition may favor the collapse, leading to protostars [see the Jean mass concept].
Reflection Nebulae: simple scattering of radiation of nearby luminous massive stars. The nebula is bluer on average than the illuminating star, as the scattering is Rayleigh’s: particles smaller than wavelength of incident radiation and cross section going as: $\left(1+\cos^2\theta\right)/\lambda^4$ , where $\theta$ is the scattering angle.

ISM ingredients (3)

Masers: often star formation is associated with powerful sources of microwave radiation, called OH or H2 masers. The radiation from newborn stars pumps molecules into excited meta-stable energy states, which transition downward when stimulated by a passing photon (as in lasers). This generates amplification, directionality and coherence.
HII regions: large regions of hydrogen photo-ionized by young bright stars, surrounded by neutral H. The UV flux determines the size of the HII region [Strömgren sphere]. HII regions are associated to new stars, then to molecular clouds, and to type II supernovae. Found in spiral arms.
Planetary Nebulae (PNe): the ejected envelopes of AGB stars, hotter and more compact than HII regions. The green color is from emission lines of oxygen.

ISM ingredients (4)

Supernova Remnants (SNRs): what is left over by a SN explosion are expanding shells of gas moving away from dead star at supersonic velocities (1% c). They compress the ISM (shock waves), heat it (up to 10⁷ K) and make it glow (X-rays). Structure and asymmetries in SNRs are due to the lumpy density distribution of ISM: expansion runs faster where the density is lower. SNRs slow down with time and merge with ISM, returned material enriched in metals. A supernova remnant is a major source of energy for ISM.
Cosmic Rays: protons (90%) , α-particles (about 10%), and rare heavier nuclei, accelerated to relativistic velocities (from 100 MeV up to 10²⁰ eV), producing showers of particles when hitting Earth’s atmosphere. Being electrically charged (positive but very few α-particles), they are deviated by Galactic and terrestrial magnetic fields, so it is hard to trace back their trajectories. They originate from Galactic sources (SN explosions) as well as extragalactic.

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J. Binney & S. Tremaine, Galactic Dynamics, Capter 6, Cambridge Univ. Press, 1987.

J. Binney & S. Tremaine, Galactic Astronomy, Capter 10, Cambridge Univ. Press, 1998.

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J. Binney & S. Tremaine, Galactic Dynamics, Capter 6, Cambridge Univ. Press, 1987.

J. Binney & S. Tremaine, Galactic Astronomy, Capter 10, Cambridge Univ. Press, 1998.

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A. Toomre & J. Toomre, Ap.J., 178, 623, 1972.

W. Freedman, "The Measure of Cosmological Parameters", stro-ph/0202006.

R. Sanders, "Observational Cosmology" astro-ph/0402065.

J. Jensen, J. Tonry and J. Blakeslee, "The Extragalactic Distance Scale", astro-ph/0304427.

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