[AAVSO-DIS] Zero Point

[Michael Newberry]

Comments below.

Michael Newberry

----- Original Message -----
From: "Steve Robinson" <srobinso at mindspring.com>
To: "'Mike'" <mnewberry at axres.com>
Sent: Monday, August 27, 2001 1:12 PM
Subject: RE: [AAVSO-DIS] Zero Point



Thanks in advance for helping me understand this.  I have annotated your
kind email below.

SR
-----Original Message-----
From: Mike [SMTP:mnewberry at axres.com]
Sent: Monday, August 27, 2001 2:32 PM
To: aavso-discussion at informer2.cis.McMaster.CA
Subject: Re: [AAVSO-DIS] Zero Point

>The photometric zero point is a numerical offset which shifts your
>instrumental magnitude onto the standard photometric system. The zero point
>may also be used to describe moving your raw instrumental magnitude onto a
>"cooked" instrumental magnitude which has yet to account for small
>corrections due to atmospheric extinction and filter bandpass effects.
>
>A raw instrumental magnitude may be calculated as -2.5log(f) where f is the
>"flux", or electrons per second from the object. Flux is derived from
>"Counts", C, (a.k.a. "DN", or Digital Number), CCD gain, g, and exposure
>
>[Steve Robinson] I understand Count to be the Z value in Mira, but this is
for only one pixel.  It would seem that count should apply >not to a point,
but to all of the light for the star.

    The "count" applies to the amount of light you measure, whether that is
from 1 pixel or "n" pixels. The count is summed over the pixels covered by
the measuring aperture. This gives the net count above the sky background,
which then becomes a magnitude, and finally a standard system-like value
after application of the zero point. The fraction of total light collected
inside the selected aperture size affects the value of the zero point---If
the aperture does not contain 100% of the starlight, then the zero point
will be smaller than if the aperture contained 100% of the light. This is
because an aperture encircling < 100% of the light will underestimate the
star count, thus requiring a smaller zero point to shift it to the same
positive magnitude value. That is OK, provided you apply the same aperture
size and zero point to all stars measured in the same image, and that all
stars on the same image have the same point spread function ("PSF")---that
is, they have the same FWHM, or "size". If in doubt about such changes
within an image, or the stars change size over the frame, then use an
aperture large enough to contain 100% of the light. You can measure the FWHM
in MIRA using the FWHM tool or the Radial Profile plot. Look at where the
star merges into the background noise. That is not quite the radius
containing all the light, but is getting close. There is an entire topic in
itself about choosing an optimal aperture size and this is where things also
get much more complicated. Just pick an aperture size about 3x the FWHM and
use that.

>time t, using f = C * g / t. Doing this calculation for some number of
>measured counts, you will get a large negative number, e.g., -12.42. But

    The minus sign in "-2.5log" turns the flux into a negative raw
magnitude. Don't blame me for defining the magnitude scale backward---blame
the Greeks. Then blame Pogson who, a couple of centuries ago, had the chance
to rectify things but decided to simply put the Greek system on a better
mathematical footing. Hopefully there are no Pogson descendants on this
list!.

>[Steve Robinson]  I don't understand how this can result in a negative
number.  C is positive, as is gain.  Time is positive as well, >ergo f
should be positive.??
>actually, the object may have a system magnitude of 9.58. The zero point is
>the numeric shift required to move the raw mag onto the system mag scale.
>Therefore we write the magnitude equation as  m = Z - 2.5log(f), using a
>zero point value, Z. In this example, the zero point is Z = 9.58 - (-12.42)
>= +22.0. So your magnitude equation for this instrument and filter would be
>m = 22.0 - 2.5log(f). Please excuse the fact that I have swept things like
>airmass and extinction and color terms under the rug in this definition.
>Either they will have been already included in your calculation of the flux
>f, or you will correct for them after the fact. If correcting after the
>fact, then you will need another small correction to the magnitude "m" to
>account for extinction and filter bandpass ("color term") effects before
you
>get a true magnitude measurement on the standard photometric system. In the
>latter case, with corrections yet to be applied, the zero point has been
>used as an offset to give a magnitude close to the photometric system
value.
>
>Another way to look at the zero point is that it is the magnitude of an
>object yielding 1 e- (or count) above the background. This is because m =
>Z - 2.5log(0) = Z.
>
>In general, the larger the telescope, the larger will be the zero point. A
>larger telescope will record more flux in a given exposure time, leading to
>a larger number for the raw magnitude, hence requiring a larger zero point
>value to shift it to the same standard system magnitude. The value 22.0 in
>my example is about what one would expect for a telescope in the 24" class.
>
>Michael Newberry

----- Original Message -----
From: "Steve Robinson" <srobinso at mindspring.com>
To: <aavso-discussion at informer2.cis.McMaster.CA>
Sent: Monday, August 27, 2001 10:29 AM
Subject: [AAVSO-DIS] Zero Point


> Can anyone give me a good definition of the term zero point?
>
> Thanks
>
> Steve Robinson
>
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> aavso-discussion mailing list
> aavso-discussion at mailman.McMaster.CA
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