Brightness and Flux Density

Astronomers learn about an astronomical source by measuring the strength of its radiation as a function of direction on the sky (by mapping or imaging) and frequency (spectroscopy), plus other quantities (time, polarization) that we ignore for now. We need precise and quantitative definitions to describe the strength of radiation and how it varies with distance between the source and the observer. The concepts of brightness and flux density are deceptively simple, but they regularly trip up experienced astronomers.  It is very important to understand them clearly because they are so fundamental.

We start with the simplest possible case of radiation traveling from a source through empty space (so there is no absorption, scattering, or emission along the way) to an observer.  In the ray-optics approximation, radiated energy flows in straight lines. This approximation is valid only for systems much larger than the wavelength $\lambda$ of the radiation, a criterion easily met by astronomical sources.   You may find it helpful to visualize electromagnetic radiation as a stream of light particles (photons), essentially bullets that travel in straight lines at the speed of light.  To motivate the following mathematical definitions, imagine you are looking at the Sun.  The "brightness" of the Sun appears to be about the same over most of the Sun's surface, which looks like a nearly uniform disk even though it is a sphere. This means, for example, that a photograph of the Sun would expose the film equally across the Sun's disk. It also turns out that the exposure would not change if photographs were made at different distances from the Sun, from points near Mars, the Earth, and Venus, for example.


Sun seen from MarsSun seen from EarthSun seen from Venus

The Sun in three imaginary photos taken from a long distance (left), medium distance (center), and short distance (right) would have a constant brightness but increasing angular size.



Only the angular size of the Sun changes with the distance between the Sun and the observer. The photo taken from near Venus would not be overexposed, and the one from near Mars would not be underexposed. The number of photons falling on the film per unit area per unit time per unit solid angle does not depend on the distance between the source and the observer. The total number of photons falling on the film per unit area per unit time (or the total energy absorbed per unit area per unit time) does decrease with increasing distance. Thus we distinguish between the brightness of the Sun, which does not depend on distance, and the apparent flux, which does.

Note also that the number of photons per unit area hitting the film is proportional to $\cos\theta$ if the normal to the film is tilted by an angle $\theta$ from the ray direction.  This is just the same projection effect that reduces the amount of water collected by a tilted rain gauge by $\cos\theta$. Likewise at the source, such as the spherical Sun, the projected area perpendicular to the line of sight scales as $\cos\theta$. 

Using the ray-optics approximation, we can define the specific intensity (sometimes called spectral intensity or spectral brightness, spectral radiance, or loosely, just brightness) $I_\nu$ in terms of


Specific intensity measured by a detector whose normal is an angle $\theta$ from the line of sight.


The surface containing $d\sigma$ can be any surface, real or imaginary; that is, it could be the physical surface of the detector, the source, or an imaginary surface anywhere along the ray. If energy $dE_\nu$ flows through $d\sigma$ in time $dt$ in the frequency range $\nu$ to $\nu + d\nu$ within the solid angle $d\Omega$ on a ray which points an angle $\theta$ away from the surface normal, then
$$dE_\nu = I_\nu \cos\theta d\sigma d\Omega dt d\nu $$
Since power is defined as energy per unit time, the power $dP$ received in the solid angle $d \Omega$ and in the frequency range $\nu$ to $\nu + d\nu$ is
$$\qquad dP \quad= \quad I_\nu  \cos\theta d\sigma d\Omega d\nu$$

$${\rm (Watts)~~~~~~~~~~~~~~~~~~~~~(m}^2~{\rm sr}~{\rm ~Hz)}$$

Thus the quantitative definition of specific intensity or spectral brightness is
$$\bbox[border:3px blue solid,7pt]{I_\nu \equiv {dP \over  \cos\theta d\sigma \, d\nu \, d\Omega}}\rlap{\quad \rm {(2A1)}}$$

and the mks units of $I_\nu$ are W m$^{-2}$ Hz$^{-1}$ sr$^{-1}$. 

The reasons for specifying an infinitesimal frequency range $d \nu$ are: (1) The detailed spectra of sources carry astrophysically important information, (2) source properties (e.g., opacity) may vary with frequency, and (3) most general theorems about radiation are true for all narrow frequency ranges (e.g., specific intensity is conserved along a ray path in empty space), so they are also true for broadband radiation (e.g., the total intensity integrated over all frequencies is also conserved along a ray path in empty space).


Theorem: Specific intensity is conserved (is constant) along any ray in empty space.

This follows directly from geometry.  Consider two surfaces, $d\sigma_1$ and $d\sigma_2$ along a ray of length $r$.


Let $d\Omega_1$ be the solid angle subtended by $d\sigma_2$ as seen from the center of the surface $d\sigma_1$.
Let $d\Omega_2$ be the solid angle subtended by $d\sigma_1$ as seen from the center of the surface $d\sigma_2$.
Then
$$d\Omega_1 = {\cos\theta_2 d\sigma_2 \over r^2}$$
$$d\Omega_2 = {\cos\theta_1 d\sigma_1 \over r^2}$$

The energy $dW_1$ flowing through the area $d\sigma_1$ in solid angle $d\Omega_1$ is
$$dW_1 = (I_\nu)_1 \cos \theta_1 d\Omega_1 d\sigma_1 d\nu$$
$$dW_1 = (I_\nu)_1 \cos \theta_1 \biggl({\cos\theta_2 d\sigma_2 \over r^2}\biggr) d\sigma_1 d\nu$$
Likewise
$$dW_2 = (I_{\nu})_2 \cos \theta_2 \biggl({\cos\theta_1 d\sigma_1 \over r^2}\biggr) d\sigma_2 d\nu$$
Since the radiation energy is conserved in free space (where there is no absorption or emission), $dW_1 = dW_2$, so $$(I_\nu)_1 = (I_\nu)_2 \quad{\rm QED}$$



This conservation theorem has two important consequences:

(1) Brightness is independent of distance.  Thus the camera setting for a good exposure of the Sun would be the same, regardless of whether the photograph was taken close to the Sun (from near Venus, for example) or far away from the Sun (from near Mars, for example), so long as the Sun is resolved in the photograph. 

(2) Brightness is the same at the source and at the detector. Thus you can think of brightness in terms of energy flowing out of the source or as energy flowing into the detector.


Two ways of looking at brightness.



The total intensity, the specific intensity integrated over all frequencies,
$$I \equiv \int_0^\infty I_\nu d\nu$$
is also conserved.  The conservation of brightness also applies to any lossless optical system, a system of lenses and mirrors for example, that can change the direction of a ray.  No passive optical system can increase the specific intensity or total intensity of radiation.  If you look at the Moon through a large telescope, the Moon will appear bigger (in angular size) but not brighter. Many people are disappointed when they see a large, nearby galaxy (e.g., Andromeda) through a telescope because it looks so dim; they expected to see a brilliantly glowing disk of stars, as in the photograph below.  The difference is not in the telescope; it is in the detector—the photograph appears brighter only because the photograph has summed the light over a long exposure time.


photo of Andromeda galaxy

No passive optical system (e.g., a telescope) can increase the specific intensity of an extended source.  The Andromeda galaxy (M31) appears much brighter in this photograph than it does to the eye, either with or without the aid of a telescope, only because a long exposure accumulates more light. Image credit


If a source is discrete, meaning that it subtends a well-defined solid angle, the spectral power received by a detector of unit projected area is called the source flux density $S_\nu$.  Equation (2A1) implies
$${dP \over d\sigma d\nu} = I_\nu \cos\theta d\Omega$$
so integrating over the solid angle subtended by the source yields
$$\bbox[border:3px blue solid,7pt]{S_\nu  \equiv  \int_{\rm source}I _\nu(\theta,\phi) \cos\theta d\Omega}\rlap{\quad \rm {(2A2)}}$$
If the source angular size is $ \ll 1{\rm ~rad}$, $\cos\theta \approx 1$ and the expression for flux density is much simpler:
$$\bbox[border:3px blue solid,7pt]{S_\nu\approx \int_{\rm source}I_\nu(\theta,\phi) d\Omega }\rlap{\quad \rm {(2A3)}} $$
This is usually the case for astronomical sources, and astronomers almost never use flux densities to describe sources so extended that the $\cos\theta$ factor must be retained (e.g., the emission from our Galaxy).

In practice, when do we use spectral brightness and when do we use flux density to describe a source?  If a source is unresolved, meaning that it is much smaller in angular size than the point-source response of the eye or telescope observing it, its flux density can be measured but its spectral brightness cannot.  Calling the red giant star Betelgeuse a "bright star" is misleading.  The intensity of this relatively cool star is lower than the intensity of a hotter star that is scarcely visible to the eye.  Betelgeuse appears "brighter" than most other stars because it subtends a much larger solid angle and therefore its flux is higher.  If a source is much larger than the point-source response, its spectral brightness at any position on the source can be measured directly, but its flux density must be calculated by integrating the observed spectral brightnesses over the source solid angle.  Consequently, flux densities are normally used to describe only relatively compact sources.


This figure illustrates the definition of flux density.



The mks units of flux density, W m$^{-2}$ Hz$^{-1}$, are much too big for practical astronomical use, so we define smaller ones:
$$1{\rm ~Jansky} = 1{\rm ~Jy} \equiv 10^{-26} {\rm ~W~m}^{-2}{\rm ~Hz}^{-1} \equiv 10^{-23} {\rm ~erg~s}^{-1}~{\rm cm}^{-2}{\rm ~Hz}^{-1}$$
and 1 milliJanksy = 1 mJy $\equiv 10^{-3}$ Jy, 1 microJansky = 1 $\mu$Jy $\equiv 10^{-6}$ Jy.  Unlike source brightness, flux density is not independent of source distance $d$. Since $\int_{\rm source}d \Omega \propto 1/d^2$,
$$S_\nu \propto d^{-2}$$
(the inverse-square law). The specific intensity or brightness is an intrinsic property of a source, while the flux density of a source also depends on the distance between the source and the observer. 

The spectral luminosity $L_\nu$ of a source is defined as the total power per unit bandwidth radiated by the source at frequency $\nu$; its mks units are W Hz$^{-1}$.  The area of a sphere of radius $r$ is $4 \pi r^2$, so the inverse-square law relation between spectral luminosity and flux density in free space is
$$\bbox[border:3px blue solid,7pt]{L_\nu = 4 \pi d^2 S_\nu}\rlap{\quad \rm {(2A4)}}$$
where the distance $d$ between the source and the observer is much larger than the dimensions of the source itself.  Note that spectral luminosity is an intrinsic property of the source.  Spectral luminosity does not depend on the distance $d$ between the source and the observer because the $d^2$ in Eq. 2A4 cancels the $d^{-2}$ dependence of $S_\nu$.  The bolometric luminosity $L_{\rm bol}$ of a source is defined as the integral over all frequencies of the spectral luminosity:
$$L_{\rm bol} \equiv \int_0^\infty L_\nu d \nu$$


Example of spectral brightness, flux density, and spectral luminosity:

What is the specific intensity of the Sun at $\nu = 10$ GHz if the Sun is a nearly ideal blackbody with temperature $T \approx 5800$ K?

Since
$${h\nu \over k T} = {6.63 \times 10^{-27}{\rm erg~s~} \times 10^{10}{\rm ~Hz} \over 1.38 \times 10^{-16}{\rm ~erg~K~}^{-1} \times 5800{\rm ~K}} = 8 \times 10^{- 5} \ll 1$$
so we can use the Rayleigh-Jeans approximation
$$I_\nu = B_\nu \approx {2 k T \nu^2 \over c^2}$$
for the spectral brightness of a black body. 
$$I_\nu \approx {2 \times 1.38 \times 10^{-16}{\rm ~erg~K}^{-1}~ 5800{\rm ~K~} (10^{10}{\rm ~s}^{- 1})^2 \over (3 \times 10^{10}{\rm ~cm~s}^{-1})^2}$$
$$I_\nu \approx 1.78 \times 10^{-13} {{\rm erg} \over {\rm cm}^2{\rm ~(sr)}}\biggl({{\rm s}^{-1} \over {\rm Hz}}\biggr)$$
Note that Hz = s$^{-1}$ and that both angles (e.g., rad) and solid angles (e.g., sr) are dimensionless, so the units in parentheses above can be dropped.  Note also that most astrophysical theory is done in cgs units but radio observations are usually reported in mks units since engineers use mks. Recall that
$$ 1{\rm ~W} = 1{\rm ~J~s}^{-1} = 10^7 {\rm ~erg~s}^{-1}$$
$$1{\rm ~m} = 10^2 {\rm ~cm}$$
so
$$I_\nu \approx 1.78 \times 10^{-13} \biggl({ 10^{-7} {\rm ~W} \over ( 10^{-2}\rm{ ~m})^2 {\rm ~sr~Hz}}\biggr)$$
$$I_\nu \approx 1.78 \times 10^{-16} { {\rm W} \over {\rm m}^2{\rm ~sr~Hz}}$$
at $\nu = 10$ GHz.  This result is a property of the Sun and does not depend on the observer's distance.


What is the flux density of the Sun at $\nu = 10$ GHz, measured at the Earth? The Sun has a radius $R_\odot = 7.0 \times 10^{10}$ cm, and its distance from the Earth $\equiv$ 1 astronomical unit (AU)  is $r_\odot \approx 1.49598 \times 10^{13}$ cm.

The angular radius of the Sun viewed from the Earth is
$$\theta_\odot = \sin^{-1}\biggl({R_\odot \over{r_{\odot}}}\biggr) \approx 4.7 \times 10^{-3} {\rm ~rad}$$
$$S_\nu = \int_{\rm Sun} I _\nu \cos\theta d\Omega$$
Remembering that in spherical coordinates an infinitesimal solid angle is $d\Omega = \sin\theta d\theta d\phi$, we have
$$S_\nu = I_\nu \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^{\theta_\odot} \cos\theta (\sin\theta d\theta d\phi)$$ Defining $x \equiv \sin\theta$ so $dx = \cos\theta d\theta$ we get
$$S_\nu = 2 \pi I_\nu \int_0^{\sin\theta_\odot} x dx$$
$$S_\nu = \pi I_\nu \sin^2 \theta_\odot$$

Since $\theta_\odot \ll1$ rad, $\sin\theta_\odot \approx \theta_\odot $ (and $\cos\theta \approx 1$ above can be ignored), leading to the much simpler
$$S_\nu \approx \int_{\rm Sun} I_\nu d\Omega$$
$$S_\nu \approx I_\nu \Omega_\odot \approx \pi I_\nu \theta^2_\odot$$
$$S_\nu \approx 1.78 \times 10^{-16} {{\rm ~W} \over {\rm m}^2{\rm ~Hz~sr}} \times \pi (4.7 \times 10^{-3} {\rm ~rad})^2$$
$$S_\nu \approx 1.24 \times 10^{-20} {{\rm ~W} \over {\rm m}^2{\rm ~Hz}} = 1.24 \times 10^6{\rm ~Jy}$$
at $\nu = 10$ GHz.   This result is not a property of the Sun alone because it varies with the distance of the observer from the Sun.

What is the spectral luminosity of the Sun at $\nu = 10$ GHz? 
Convert the flux density to cgs units ($S_\nu = 1.24 \times 10^{-17}$ erg s$^{-1}$ cm$^{-2}$ Hz$^{-1}$) and use the inverse-square law:

$$L_\nu = 4 \pi r^2_\odot S_\nu = 4 \pi (1.5 \times 10^{13}{\rm ~cm})^2
\times 1.24 \times 10^{-17}{\rm ~erg~s}^{-1}{\rm ~cm}^{-2}{\rm ~Hz}^{-1}$$ $$L_\nu = 3.5 \times 10^{10} {\rm ~erg~s}^{-1}{\rm ~Hz}^{-1}$$
This spectral luminosity is an intrinsic property of the Sun; it does not depend on the distance to the Sun.