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 the time being. We need clear 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 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
the 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.
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. 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 doesn't 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 appropriate for systems much larger than one wavelength (a good approximation in astronomy), 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 $dE_\nu d\nu$ 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
$$dE_\nu d\nu = I_\nu \cos\theta d\sigma d\Omega dt d\nu
$$
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 (e.g., specific intensity is conserved along a ray
path in empty space) are true for all
narrow frequency ranges, 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 (e.g., 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).
(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 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 is 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.
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 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 telescope observing it, its flux density can be
measured but its spectral brightness cannot. 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:
$$1{\rm ~Jansky} = 1{\rm ~Jy} \equiv 10^{-26} {\rm ~W~m}^{-2}{\rm
~Hz}^{-1}$$
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. Since
$\int_{\rm source}d \Omega \propto 1/r^2$,
$$S_\nu \propto r^{-2}$$
(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
$$\bbox[border:3px blue solid,7pt]{L_\nu = 4 \pi r^2 S_\nu}\rlap{\quad
\rm {(2A4)}}$$
where the distance $r$ between the source and the observer is much
larger than the dimensions of the source itself. The bolometric luminosity
$L$ 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?
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
$$\theta_\odot = \sin\biggl({R_\odot \over{r_{\odot}}}\biggr) \approx
4.7 \times 10^{-3} {\rm ~rad}$$
$$S_\nu = \int_{\rm Sun} I
_\nu \cos\theta d\Omega$$
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:
$$L_\nu = 4 \pi r^2 S_\nu = 4 \pi (1.5
\times 10^{13}{\rm ~cm})^2
\times 1.24 \times 10^{-17}{\rm ~erg~s}^{-1}{\rm ~cm}^{-2}{\rm
~Hz}^{-1}$$
$$L_\nu = 3.5 \times 10^{10} {\rm ~erg~s}^{-1}{\rm ~Hz}^{-1}$$