34
$\begingroup$

Apparent magnitude is a rather complex way to determine the brightness of a star. Quoting the introduction text from the linked to Wikipedia page:

The apparent magnitude (m) of a celestial body is a measure of its brightness as seen by an observer on Earth, adjusted to the value it would have in the absence of the atmosphere. The brighter the object appears, the lower the value of its magnitude. Generally the visible spectrum (vmag) is used as a basis for the apparent magnitude, but other regions of the spectrum, such as the near-infrared J-band, are also used. In the visible spectrum Sirius is the brightest star in the night sky, whereas in the near-infrared J-band, Betelgeuse is the brightest.

While it is of course a useful measure also for the slightly more casual, non-scientific observer to help determine the observed star from its neighboring cluster, or identification in general, I always wondered if there is a way to measure apparent magnitude with enthusiast-class equipment, and what would these procedures be?

                              apparent magnitude

                              Apparent magnitude scale and observational limits (Source: ESA Science)

Also interesting would be a description of what level of precision and enthusiast can get with such equipment while measuring apparent magnitude of a distant star. If you need specifics to answer the question, such as precise available equipment or subject of observations, please feel free to choose at will any that would broadly match the capabilities of enthusiast-grade equipment.

$\endgroup$
2
  • 1
    $\begingroup$ You may already have a DSLR, but if you're ready to buy it, a CCD would cost as much. Do you want an answer with CCDs? $\endgroup$
    – Cheeku
    Sep 28, 2013 at 2:17
  • $\begingroup$ @Cheeku - Sure, any method that's accessible to enthusiasts will do, thereof my last sentence. If you know of more ways, then the better. There ought not be many ways anyway, so I thought to ask it like that, have answerers demonstrate their ingenuity a little LOL $\endgroup$
    – TildalWave
    Sep 28, 2013 at 4:21

2 Answers 2

20
$\begingroup$

The easiest way to determine the magnitude of a given star is probably to use the Pogson relation. The idea is to determine the magnitude of a star knowing the magnitude of a reference star; it is thus quite easy, using a well-known reference as Vega or Sirius.

The Pogson relation is given by:

$$m_1-m_2=-2.5\ log\ \left({\frac{E_1}{E_2}}\right)$$

where $m_1$ and $m_2$ are the magnitudes of star 1 and star 2 (your reference star), and $E_1$ and $E_2$ brightness (that could be in arbitrary units, which is a good news for you).

With a CCD detector the brightness is easy to determine since it's just the flux of a pixel. You don't have to convert it in physical units (as Watt per square meter) since you have a ratio $\left({\frac{E_1}{E_2}}\right)$, so the raw intensity is enough to determine the magnitude.

As for any other measurement with a CCD camera, you have of course to determine the bias, dark and flat-field and to reduce your data before estimating any magnitude. The precision will depend on the number of observations performed and, as always, it is better to perform several short exposure images than a single long exposure one.

$\endgroup$
4
$\begingroup$

There is a difficulty with measuring magnitude: the atmosphere! It is sometimes clear sometimes hazy and it can be very hard to know just how much light is being lost to a high layer of thin cirrostratus, or a passing puff of mist.

Amateurs have measured star magnitudes by the comparison technique. This is literally comparing the star with other nearby stars of known magnitude and estimating "is it brighter or less bright"

If there are convenient nearby stars of similar magnitude, a good estimate of 0.1 magnitudes is possible by a skilled observer. There are long term sequence of the magnitude of variable stars that have been produced by this method.

It is harder if there is no nearby star of similar magnitude. It is easier to compare two stars in the same field of view, rather than have to move you eyes from one star to another.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .