Jody Wilson

This report describes a
method of calibrating surface brightnesses in a telescopic or all-sky image
using standard stars.

Given:

-Flux of standard star at the Earth in **photons cm ^{-2} s^{-1} Å^{-1}**
for the wavelength observed (look up in standard star files)

-Total detected brightness (*S*) of standard star image in **Data
Numbers (DN)**

-Spectral width of the filter in **Angstroms** (approximately FWHM ´ peak transmission)

-Solid angle *θ*^{2}
of a pixel in **radians** (for
telescopic images only)

**Step 1:** Calculate airless star flux seen by instrument (F).

The
airless flux *F* is the spectral flux
(photons cm^{-2} s^{-1} Å^{-1} from table) multiplied by
the filter’s spectral width:

*F = (spectral flux) **´** (spectral width)* (1)

This
is the flux from the standard star that the instrument would see of there was
no attenuation of the starlight by the Earth’s atmosphere.

**Step 2:** Calculate the apparent airless omnidirectional
surface brightness of the standard star image (*I*).

Assume
that the standard star has an angular size of exactly one pixel, that it is
emitting isotropically, and that it is at some
arbitrary distance *R* from the
telescope.

The area *a*
of the “star pixel” is then

*a = *(*Rθ*)^{2}, (2)

where *θ* is the angular width of one pixel
in radians. We may now define an omnidirectional
surface brightness *I* (units of
photons cm^{-2} s^{-1}) of this star pixel such that its total
brightness (in photons s^{-1}) is *Ia*.*R* can be expressed as
the product of the star pixel’s total brightness divided by the area of a
sphere of radius *R*:

*F = Ia*/(4π*R*^{2}).
(3)

This is the same flux
calculated in step 1. Substituting for *a* from equation (2):

*F*
= *I θ*^{2}/ 4π. (4)

Solving for *I* (photons cm^{-2} s^{-1}):

*I*
= 4π*F* /*θ*^{2}

One Rayleigh
is 10^{6} photons cm^{-2} s^{-1}, so we finally have:

*I _{Rayleighs}* = [4π

**Step 3:** Calibrate the image.

*3A: Telescopic images*

Given
a standard star of measured brightness *S*
and apparent surface brightness *I* with
an exposure time *t _{1}*, and a
second image with exposure time

image_{Rayleighs}_{ }= image_{DN} · (t_{1}/t_{2}) · (I/S) (6)

All-sky
images have three characteristics which make them more difficult to calibrate than
telescopic images:

1. The angular size of a pixel depends on its location in
the image.

2. Some all-sky images cannot be flat-fielded, so any
center-to-edge changes in instrument sensitivity are unknown.

3. All airmasses from 1 to >10
are represented in a single image, so an “airless” calibration for one part of
an image does not apply to other parts.

So
calibration of all-sky images with standard stars requires calculating the
solid angle *ω _{pixel}*
of each pixel in the image. For telescopic images this is simply

Consider
a pixel in an all-sky image at a distance of r pixels from the zenith.

The true zenith distance of this pixel can be represented
as a function of its pixel distance *r*.
This function, *z*(*r*), is determined by measuring star locations and fitting a
function to the zenith distances at the time the image was taken.

Once *z*(*r*) is known, the latitudinal
size of a pixel on the meridian is easily calculated:

Δ*z* = |*z*(*r*) – *z*(*r*+1)| = *z*'(*r*). (7)

Consider an annulus of radius *r* pixels and thickness of one pixel. Its
total area in pixels is

*A*(*r*) = 2π*r*
(8)

The angular circumference
of this annulus is 360º sin(*z*), so the solid angle of the annulus is:

*Ω*(*r*) = 360º sin[*z*(*r*)]
*z*'(*r*). (9)

Thus the solid angle of
each pixel in the annulus is, from equations (8) and (9):

*ω _{pixel}*(

Now
substituting *ω _{pixel}*(