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Massimo Capaccioli » 22.Fundamentals

Apparent magnitudes

Astronomers call “flux” $\phi_\lambda$ the energy per unit wavelength coming from a source S and crossing the unit area perpendicularly (outside the atmosphere) in the time unit. It does not matter how extended the source is, provided your photometer sees it all, and how far it is. According to a formula due to the American astronomer Robert Norman Pogsom, who used the notion that the eye is a logarithmic detector [expand the concept], the quantity: $m_\lambda=m_0-2.5 \log\phi_\lambda,$ is the apparent magnitude of S at $\lambda,$ where $m_0$ is the zero point constant (arbitrarily fixed in any photometric system; see examples later). The same concept applies to a finite band centered at $\lambda_0$ and characterized by the transfer function $T(\lambda):$ $\displaystyle{ \lambda_0=\frac{\int \lambda'T(\lambda')d\lambda'}{\int T(\lambda')d\lambda'}}.$ The magnitude is called bolometric if all photons (of any frequency) are transmitted and recorded. Observed apparent magnitudes are affected by various sources of systematic errors, most of which tend to make them fainter.

Absolute magnitudes

If $D$ is the distance of the source S, measured in parsec, then: $m_\lambda=m_0-2.5\log \left(\phi_0\frac{10^2}{D^2}\right)=M_\lambda+5\log D-5,$ with $\phi_0$ being the flux at the standard distance of 10 pc and $M_\lambda$ the corresponding magnitude, thus named absolute magnitude. If $m_\lambda$ is the apparent magnitude as observed, i.e. not yet corrected for reddening and for the other causes of dimming, then the quantity: $(m_\lambda)_{obs}-M_\lambda,$ is called apparent distance modulus. It becomes a true distance modulus after all corrections are applied. Since the reddening tends to vanish as $\lambda \rightarrow +\infty$ it is clear that (at least for this cause of dimming) the apparent distance modulus tends to the true value as the observing band moves redwards. As a rule of the thumb, remember that a distance of 1 Mpc is equivalent to a true distance modulus of 25 mag, and that the increase of one order of magnitude in D implies a 5 magnitude fainter distance modulus.

Color index and surface brightness

The color index: $C_{\lambda 1, \lambda 2}=\left(m\left(\lambda_1\right)-m\left(\lambda_2\right)\right)$ is the magnitude difference for two bands (colors) of the same source at two wavelengths $\lambda_1>\lambda_2$ ; for instance, $(B-V)$ is the blue minus visual color index, where B and V are two bands of the Morgan-Johnson photometric system [see later]. Apart from absorption and other effects on the apparent magnitudes, colors are distance independent [think about]. If S is extended, one may consider only the flux $\varphi_\lambda$ from a unit angular element (a unit solid angle) and define the “surface brightness” accordingly: $\mu_x=\mu_0-2.5\log \varphi_\lambda,$ where $\varphi_\lambda$ is distance independent [be sure that you understand why and which are the assumptions needed to make the statement true. Hints: extinction and non uniformity of S]. Obviously the integration over all the area of S turns the surface brightness into a distance dependent total magnitude: $m_\lambda=m_0-2.5\log\left(\int_s 10^{0.4 (\mu_0-\mu_\lambda)}dS\right).$ Measurements of magnitude and surface brightness may by attenuated by extinction (atmospheric and interstellar) and altered by radial velocities (K-term) [see later], but only magnitudes are affected by distance. We shall see these various effects in the followings. We shall also remember, in comparing objects at different distances, that they will be at different cosmic time, and thus at a different evolutionary stage.

A sample of photometric systems

Characteristics of some well-known and widely used photometric systems.

A sample of photometric systems

Johnson filters against the characteristics of two CCDs of the CFH12K camera installed at the CFHT on Mauna Kea. The average night sky spectrum at the Observatory is also shown; see the night sky brightness. Credit: CFHT.

Black body

Some recapitulation of the basic notions of the physics of radiation are in order. A black body is an idealized object absorbing all the electromagnetic radiation falling on it. No radiation passes through it and none is reflected [try to remember why it so important in physics].

The specific intensity $B(\lambda, T)$ of a blackbody ([ $J~m^{-2}~sr^{-1}~s^{-1} m^{-1}$ ], power per unit area per unit wavelenght per unit solid angle) is represented by the so called Planck law [which require quantization, i.e. a new physics]:

$B(\lambda,T) d \lambda=\frac{2hc^2}{\lambda^5}\frac 1 {\exp (\frac {hc}{k\lambda T})-1}d\lambda$

where

$2hc^2=1.19\times10^{-16} ~J~m^2~s^{-1},$

and $hc/k=1.439 \times 10^{-2}~m~K.$

Note that $B$ depends on $T$ only! As a function of frequency it writes as: $B(v,T)dv=\frac{2hv^3}{c^2}\frac 1{\exp (\frac{hv}{kT})-1}dv.$

Approximations of Planck’s law

At long wavelengths (low frequencies), $(hc/k\lambda T)\ll 1,$ the Planck formula writes as: $B(\lambda, T)d \lambda=\frac{2 ck T}{\lambda^4}d\lambda.$

A short wavelenghts (high frequency), $(hc/k\lambda T)\gg 1,$ it writes as: $B(\lambda, T)d \lambda=\frac{2ck}{\lambda^5}\exp \left(-\frac{hc} {k\lambda T}\right)d\lambda.$

The two regimes are named after Wien, $(hc/k\lambda T)\gg 1$ , and after Rayleigh-Jeans, $(hc/k\lambda T)\ll1$ ;

the second is unphysical as it diverges “towards” the UV (the so-called ultraviolet catastrophe).

The separation between the two regimes occurs at

$hc/(k\lambda_{break }T)=(1.439\times 10^{-2})/(\lambda_{break }T)=1$

at a wavelength that depends on the temperature:

$\lambda_{break}[m]=1.439\times 10^{-2}T^{-1}[K]$ (see insert with sampled values)

Peak wavelength vs. temperature for a black body.

Stafan-Boltzmann law

Let’s integrate the specific intensity for the unit surface in any direction (semi-sphere) and for all frequencies. We obtain the Stefan-Boltzmann law:

$L=\int_0^{2\pi} d\phi\int_0^{\pi/2} \sin\theta \cos\theta d\theta\int_0^{+\infty} B_\nu(T)d\nu=\pi \int_0^{+\infty} B_\nu(T)d\nu=\sigma T^4\ [W m^{-2}]$

where $L$ is the bolometric luminosity per unit area and:

$\sigma=\frac{2\pi^5k^4}{15h^3c^2}=5.669\times10^{-8}[Wm^{-2}K^{-4}].$

This is a fundamental law.

Stefan-Boltzmann via thermodynamics

Note that the Stefan-Boltzmann law my be derived by thermodynamical principles alone, without introducing quantization. Assume a box of volume $V$ containing energy with density $u.$ The total internal energy is: $U=uV=3PV,$ since the pressure is:

$P=\frac u 3.$

As: $dU=TdS-PdV,$ where $S$ is the entropy, then

$d(3PV)=TdS-PdV=3PdV+3VdP,$

or:

$dS=4\frac P T dV+3\frac V T dP=\left(\frac{\partial S}{\partial V}\right)dV+\left(\frac{\partial S}{\partial P\right) dP,}$

which are called Maxwell’s relations.

It is: $\left(\frac{\partial S}{\partial V}\right)=4\frac{ P}{ T}$

and $\left( \frac {\partial S} {\partial P} \right)=3\frac V T.$

Stefan-Boltzmann via thermodynamics

From the symmetry of second derivatives it follows:

$\frac \partial {\partial P}\left(\frac{\partial S}{\partial V}\right)_V=\frac \partial {\partial V}\left(\frac {\partial S}{\partial P}\right)_P.$

Developing the derivatives and noting that in the LHS member $V~~\text{as}~~P$ is constant:

$\frac 1 P \frac{dP}{dT}=\frac 4 T,$

from where:

$P=\frac u 3 \propto T^4.$

Wien’s law

The Wien’s law establishes a relation between the wavelength (or the frequency) where the emission of the black body reach the maximum and $T$ of the black body.

It is: $\lambda_{max}T=2.898\times 10^{-3} [m~ K],$

or $\frac{v_{max}} T=5.879 \times 10^{10} [Hz~ K^{-1}]$

[Be careful in deriving the second relation from the first one using the fact that $\lambda \nu=c.$ What you have to do correctly is to set to zero the derivate of $B$ relative to either $\lambda$ or $v$ .]

It is useful to remember that a black body obeys Lambert’s law: intensity is independent of the direction of observation $\theta$ or $dP_v=B_vd\Sigma\cos\theta d\Omega d v=B_vd\sigma d\Omega dv.$

Colors of stars

Since normal stars, of which galaxies are made up, are typically at a quasi-thermodynamical equilibrium (LTE), their colors are assigned by the peaks of their energy distribution, $F(\lambda).$

Under this condition, the latter has a quasi-Plank behavior:

$F(\lambda, T)=\frac{2hc^2}{\lambda^5}\frac 1 {e^{hc/\lambda kT}-1},$

and thus $F(\lambda)$ obeys Wien’s law:

$\lambda_{max}T=2.898\times 10^7 ~[\mathring {A} ~K].$

Then stellar colors are related to spectral types and effective temperatures (those resulting from the forced application of the Stefan-Boltzmann law to non-black body spectra).The largest departures occur in the atmospheres of cool stars, due to blanketing (bands redistributing blue energy into the red).

Correspondences of spectral types of stars with effective temperature, color, and average (B-V) color index.

The HR diagram

In stars there is tight correlation between color (effective temperature) and luminosity. This correlation is shown by the famous Hertzsprung-Russell (HR) or color-magnitude (CM) diagram, where we identify [recap of a previous lecture]:

the Main Sequence, where stars of all masses burn hydrogen, more quickly if more massive according to the relation: $L\propto M^4$ [show that, under simplified hypoteses, the lifetime $t_{MS}\propto M^{-3}$ ],
the Supergiant Branch, where bright stars evolve quickly,
the Giant Branch, where stars evolve after central hydrogen is exhausted, and
the White Dwarf cloud, where low mass stars drift to death.

A young population possesses some massive stars; they are blue (high temperature) and luminous enough to dominate the color of the population. An old population has no blue stars; its brightest sources are the low mass stars that have evolved into red giants; so it looks red.

The Hertzsprung-Russell Diagram. Credit: ESO.

Colors of galaxies

Optical colors in galaxies are caused by a mix of various dyes:

blue from young massive stars;
red by old and low mass stars;
blends of red to blue from hot nebulae (HII for red, [OIII] for yellow, [OII] for blue];
contributions from thermal/non thermals sources in bands other than optical (from gamma to radio).

Colors are further affected by dust reddening and by redshift.

Spectroscopic redshifts

Colors have become of paramount importance in the large surveys of medium-distant galaxies. Being gross samplings of the spectral energy distribution (SED), they provide a way to determine the (photometric) redshifts of distant objects, essential to derive absolute (distance-free) photometric parameters. The efficiency of the method relies in the identification of the spectral breaks [see later], which have not been washed out by the integration (smoothing) of the SED through the transmission functions of the filters. Clearly the estimate is worse than plain spectroscopy; its precision depends on the filter set (number and distribution of the color bands available to the survey), on the photometric accuracy, and on the procedure actually adopted to apply the basic idea of photometric redshift: that of finding a redshift and a template spectrum (out of a data base of standard spectra) matching, after shifting and integrating, the multiband photometry of an object. Note that the problem is not only that of finding a solution; this must also be unique. To test it, usually researchers compare direct spectroscopic redshifts with the photometric ones in a subset of their sample.

Spectroscopic redshifts

The figure gives an example of photometric redshifts. Photometric redshift of galaxies in the Hubble Deep Field (HDF) South, a small area, $\sim 1$ arcmin squared, of the Southern sky where very deep observations in the IR bands J, H and K (1.1, 1.6, and 2.2 μm) have been obtained by the Hubble Space Telescope (HST). Black dots are photometric measurements made with the ESO VLT in five bands (UBVRI) and by HST (H). They are superimposed to best-fitting spectra taken from a library of more than 400,000 synthetic spectra of galaxies at various redshifts.

Credit: ESO.

Malmquist bias

Photometric samples of celestial objects as stars or galaxies suffer many problems having to do with the effects of distance and redshift. In two papers (1922, Lund Medd., Ser. I, 100, 1; 1936, Stockholm Obs. Medd., 26), the Swedish astronomer Gunnar Malmquist (1893-1982) analyzed a very subtle selection effect affecting flux-limited samples of objects, which appear brighter and brighter on average as the distance increases. In order to derive the expression to correct for the so-called Malmquist bias, lets assume that the following conditions hold:

no interstellar absorption;
homogeneity, i.e. the luminosity function of the sample objects, $\phi (M),$ independent of the distance $r: d\phi/dr =0$ ;
isotropy, i.e. spatial star density, dependent on $r$ only: $\rho=\rho(r)$ ;
completeness, i.e. the sample contains all the objects brighter than an apparent limiting magnitude $m_{lim}$ and no one fainter;
the sample objects share the same spectral type, intrinsic mean magnitude $M_0$ and dispersion $\sigma$ about it.

Let $N(m,r)$ be the number of objects with magnitude between $m$ and $m+dm$ in a spherical layer within the solid angle $\omega$ bounded by $r$ and $r+dr$ .

Malmquist bias

It follows: $N(m,r)dmdr=\omega\phi(M)\rho r^2dmdr,$ where, as always, $M=m+5-5\log ~r$ (as we assumed no absorption). The average over the entire sample gives a different value than a subsample. The average number of all objects within $m$ and $m + dm$ is given by a basic equation of stellar statistics: $a(m)=\int_0^\infty N(m,r)dr=\omega\int_0^\infty \phi(M) \rho r^2 dr.$ If we count the objects up to $m_{lim}$ at a given distance $r$ we get: $\psi(m_{lim, r})=\int_0^\infty\phi (M)dm=\int_{-\infty}^{M_{lim}}\phi (M)dM,$ where: $M_{lim}=m_{lim}+5-5\log~r}$ at any $r.$ Then the total number of objects brighter than $m_{lim}$ is:

$A(m_{lim})=\int_{-\infty}^{m_{lim}}\int_0^\infty N(m,r)drdm=\omega\int_{-\infty}^{m_{lim}}\int_0^\infty\phi (M)\rho r^2 drdm \left\{\begin{array}{ll} \int_{-\infty}^{m_{lim}}a(m)dm \\ \\ \omega \int_0^\infty\psi (M_{lim})\rho r^2 dr\end{array} \right$

Let’s compute, using Lagrange’s Theorem of the Mean, the average values of $M$ for a given apparent magnitude: $\overline {M}_ma(m)=\omega \int_0^\infty M\phi (M) \rho r^2 dr.$

Malmquist bias

We now adopt a Gaussian luminosity function:

$\phi (M)=\frac 1 {\sigma \sqrt{2\pi}}\exp \left[-\frac{(M-M_0)^2}{2\sigma^2}\right].$

Deriving this function:

$\phi ' (M)=\phi(M)\left[-\frac{(M-M_0)}{\sigma^2}\right],$

then: $M\phi (M)=M_0\phi (M)-\sigma^2\phi '(M),$

and:

$\overline M_m a(m)=\omega \int_0^\infty M\phi (M)\rho r^2 dr=\omega \int_0^\infty (M_0 \phi(M) - \sigma^{2}\phi'(M))\rho r^2dr=\omega M_0\int_0^\infty \phi (M)\rho r^2 dr - \omega \sigma^2\int_0^\infty\phi' (M)\rho r^2 dr=$ $=M_0a(m)-\sigma^2\omega \int_0^\infty \phi^{'} (M)\rho r^2 dr.$

Malmquist bias

Note that the RHS is the derivative:

$da(m)/dm=\omega \int_0^\infty \phi^{'}(M)\rho r^2 dr.$

It follows:

$[\overline M_m-M_0]a(m)=-\sigma^2 \omega \int_0^\infty \phi^{'}(M)\rho r^2 dr.$

Malmquist bias

In conclusion:

$[\overline M_m-M_0]a(m)=-\sigma^2 \frac{da(m)}{dm},$

$[\overline M_m-M_0]=-\sigma^2 \frac 1 {a(m)}\frac{da(m)}{dm}, [\overline M_m-M_0]=-\sigma^2 \frac{d\log a(m)}{dm}, \overline M_m=M_0-\sigma^2 \frac {d\log a (m)}{dm}.$

This expression tells us that, by looking at objects of our sample with apparent magnitude $m,$ their mean total magnitude $\overline M_m$ is brighter than the true total magnitude $M_0$ by the quantity: $-\sigma^2 \frac {d\log a(m)}{dm}$ which is null only if $\sigma=0,$ that is the luminosity function is a $\delta$ function, or if the logarithmic gradient of $a(m)$ is zero. This is called Malmquist bias. The reason for this effect is that, in a magnitude limited sample, the “depth” of the explored volume is larger for brighter objects.

Completeness correction

Objects all at the same distance do not suffer any Malmquist bias. Still the variety of luminosities, from bright to faint will determine another problem. Let’s consider, for instance, a cluster of objects (stars or galaxies) for which we want to built the luminosity function (LF) $m$ by the following Monte Carlo experiment. Using the proper (noisy) PSF, sum up to a star of assigned magnitude $m$ to the image of the cluster, placing it in a randomly defined position, than perform a blind analysis of the image (i.e. run your photometric extractor) and verify if the fake stellar image has been recovered. The number of successes over a large number of trials gives you the completeness lat the magnitude level $m$ . You may ask yourself what happened to the star when you did not recover it. Likely it went on top of a brighter star, loosing its identity and contributing to make the bright star brighter. For all that it is apparent that the incompleteness in a crowded field causes an excess at the bright end of the LF and an increasing deficiency at the faint end.

K-term

Let’s write again the relation of the distance , $M$ holding for a universe which is transparent, Euclidean, and at rest:

$m=M -5+5\log D,$

where:

$10^{-0.4 m}=\phi=\int_0^{+\infty}f(\lambda)T(\lambda)d\lambda,$

is the flux density in the photometric band delimited by the (finite) transfer function $T(\lambda)$ and relative to the energy distribution $f(\lambda).$ If the universe is expanding, and thus sources are exposed to an increasing redshift, the equation at the top of the page changes. We shall add [why add?] a term which accounts for the fact that the spectrum of the source is shifted and stretched by redshift while the observing window does not change.

K-term

In fact, for a source at redshift $z=\Delta \lambda/\lambda$ it is:

$\lambda_{obs}=\lambda (1+z),$

and the observed flux density is: $\phi_z=\frac 1 {1+z}\int_0^{+\infty}f(\lambda_{obs})T(\lambda)d\lambda =\frac 1 {1+z}\int_0^{+\infty}f(\lambda(1+z))T(\lambda)d\lambda.$

The coefficient $1/(1+z)$ represents the energy-loss of photons, while the variable $\lamba(1+z)$ reminds us that, while increases, the energy distribution stretches towards the red (all happens as if the it were the band to narrow and move towards the blue). In other words, if you for instance measure the V magnitude of a redshifted galaxy, you actually observe a bluer region than for a galaxy at $z=0$ and through a band which is $(1+z)$ narrower.

K-Term

In conclusion, the observed magnitude is:

$m_z=-2.5 \log \phi_z=-2.5\log \left( f\frac{f_z}f\right)=$ $=m_0-2.5\log\left(\frac 1 {1+z}\frac{\int_0^{+\infty}f(\lambda(1+z))T(\lambda)d\lambda}{\int_0^{+\infty}f(\lambda)T(\lambda)d\lambda} \right)=$ $=m_0+2.5\log (1+z)-2.5 \log \frac{\int_0^{+\infty}f(\lambda(1+z))T(\lambda)d\lambda}{\int_0^{+\infty}f(\lambda)T(\lambda)d\lambda}=m_0+K.$