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# absolute magnitude

absolute magnitude: see magnitude.

In astronomy, absolute magnitude (also known as absolute visual magnitude) is the apparent magnitude an object would have if it were at a standard luminosity distance (10 parsecs, 1 AU, or 100 km depending on object type) away from the observer, in the absence of astronomical extinction. It allows the overall brightnesses of objects to be compared without regard to distance.

The absolute magnitude uses the same convention as the visual magnitude, with a factor of about 2.512 difference in brightness between steps in magnitude. The Milky Way, for example, has an absolute magnitude of about −20.5. So a quasar at an absolute magnitude of −25.5 is 100 times brighter than our galaxy (because 2.5125 is approximately equal to 100). If this particular quasar and our galaxy could be seen side by side at the same distance, the quasar would be 5 magnitudes (or 100 times) brighter than our galaxy.

## Stars and galaxies (M)

In stellar and galactic astronomy, the standard distance is 10 parsecs (about 32.616 light years, or 3 × 1014 kilometres). A star at ten parsecs has a parallax of 0.1" (100 milli arc seconds).

In defining absolute magnitude it is necessary to specify the type of electromagnetic radiation being measured. When referring to total energy output, the proper term is bolometric magnitude. The bolometric magnitude can be computed from the visual magnitude plus a bolometric correction, $M_\left\{bol\right\}=M_V+BC$. This correction is needed because very hot stars radiate mostly ultraviolet radiation, while very cool stars radiate mostly infrared radiation (see Planck's law). The dimmer an object (at a distance of 10 parsecs) would appear, the higher its absolute magnitude. The lower an object's absolute magnitude, the higher its luminosity. A mathematical equation relates apparent magnitude with absolute magnitude, via parallax.

Many stars visible to the naked eye have an absolute magnitude which is capable of casting shadows from a distance of 10 parsecs; Rigel (−7.0), Deneb (−7.2), Naos (−6.0), and Betelgeuse (−5.6). For comparison, Sirius has an absolute magnitude of 1.4 and the Sun has an absolute visual magnitude of 4.83 (it actually serves as a reference point). The Sun's absolute bolometric magnitude is 4.75.

Absolute magnitudes for stars generally range from −10 to +17. The absolute magnitude for galaxies can be much lower (brighter). For example, the giant elliptical galaxy M87 has an absolute magnitude of −22.

### Computation

One can compute the absolute magnitude $M!,$ of an object given its apparent magnitude $m!,$ and luminosity distance $D_L!,$:
$M = m - 5 \left(\left(log_\left\{10\right\}\left\{D_L\right\}\right) - 1\right)!,$

where $D_L!,$ is the star's luminosity distance in parsecs, which are approximately 3.2616 light-years.

For nearby astronomical objects (such as stars in our galaxy) the luminosity distance DL is almost identical to the real distance to the object, because spacetime within our galaxy is almost Euclidean. For much more distant objects the Euclidean approximation is not valid, and General Relativity must be taken into account when calculating the luminosity distance of an object.

In the Euclidean approximation for nearby objects, the absolute magnitude $M!,$ of a star can be calculated from its apparent magnitude and parallax:

$M = m + 5 \left(log_\left\{10\right\}\left\{p\right\} + 1\right)!,$

where p is the star's parallax in arcseconds.

You can also compute the absolute magnitude $M!,$ of an object given its apparent magnitude $m!,$ and distance modulus $mu!,$:

$M = m - \left\{mu\right\}!,$

#### Examples

Rigel has a visual magnitude of $m_V = 0.18$ and distance about 773 light-years
$M_V = 0.18 - 5 cdot \left(log_\left\{10\right\} frac\left\{773\right\}\left\{3.2616\right\} - 1\right) = -6.7$

Vega has a parallax of 0.129", and an apparent magnitude of +0.03

$M_V = 0.03 + 5 cdot \left(1 +log_\left\{10\right\}\left\{0.129\right\}\right) = +0.6$

Alpha Centauri A has a parallax of 0.742" and an apparent magnitude of −0.01

$M_V = -0.01 + 5 cdot \left(1 +log_\left\{10\right\}\left\{0.742\right\}\right) = +4.3$

The Black Eye Galaxy has a visual magnitude of mV=+9.36 and a distance modulus of 31.06.

$M_V = 9.36 - 31.06 = -21.7$

### Apparent magnitude

Given the absolute magnitude $M!,$, for objects within our galaxy you can also calculate the apparent magnitude $m!,$ from any distance $d!,$:

$m = M + 5 \left(log_\left\{10\right\}\left\{d\right\} - 1\right)!,$

For objects at very great distances (outside our galaxy) the luminosity distance DL must be used instead of d.

Given the absolute magnitude $M!,$, you can also compute apparent magnitude $m!,$ from its parallax $p!,$:

$m = M - 5 \left(log_\left\{10\right\}p + 1\right)!,$

Also calculating absolute magnitude $M!,$ from distance modulus $mu!,$:

$m = M + \left\{mu\right\}!,$

## Planets (H)

For planets, comets and asteroids a different definition of absolute magnitude is used which is more meaningful for nonstellar objects.

In this case, the absolute magnitude is defined as the apparent magnitude that the object would have if it were one astronomical unit (au) from both the Sun and the observer and at a phase angle of zero degrees. This is a physical impossibility, as it requires the observer to be located at the centre of the Sun, but it is convenient for purposes of calculation.

To convert a stellar or galactic absolute magnitude into a planetary one, subtract 31.57.

### Apparent magnitude

The absolute magnitude can be used to help calculate the apparent magnitude of a body under different conditions.

$m = H + 2.5 log_\left\{10\right\}\left\{\left(frac\left\{d_\left\{BS\right\}^2 d_\left\{BO\right\}^2\right\}\left\{p\left(chi\right) d_0^4\right\}\right)\right\}!,$

where

$d_0!,$ is 1 au, $chi!,$ is the phase angle, the angle between the Sun-Body and Body-Observer lines; by the law of cosines, we have:

$cos\left\{chi\right\} = frac\left\{ d_\left\{BO\right\}^2 + d_\left\{BS\right\}^2 - d_\left\{OS\right\}^2 \right\} \left\{2 d_\left\{BO\right\} d_\left\{BS\right\}\right\}!,$

$p\left(chi\right)!,$ is the phase integral (integration of reflected light; a number in the 0 to 1 range)

Example: (An ideal diffuse reflecting sphere) - A reasonable first approximation for planetary bodies

$p\left(chi\right) = frac\left\{2\right\}\left\{3\right\} \left(\left(1 - frac\left\{chi\right\}\left\{pi\right\}\right) cos\left\{chi\right\} + \left(1/pi\right) sin\left\{chi\right\} \right)!,$

A full-phase diffuse sphere reflects ⅔ as much light as a diffuse disc of the same diameter
Distances:
$d_\left\{BO\right\}!,$ is the distance between the observer and the body
$d_\left\{BS\right\}!,$ is the distance between the Sun and the body
$d_\left\{OS\right\}!,$ is the distance between the observer and the Sun

#### Example

Moon
$H_\left\{Moon\right\}!,$ = +0.25
$d_\left\{OS\right\}!,$ = $d_\left\{BS\right\}!,$ = 1 au
$d_\left\{BO\right\}!,$ = 384.5 Mm = 2.57 mau

How bright is the Moon from Earth?
Full Moon: $chi!,$ = 0, ($p\left(chi\right)!,$ ≈ 2/3)
$m_\left\{Moon\right\} = 0.25 + 2.5 log_\left\{10\right\}\left\{\left(frac\left\{3\right\}\left\{2\right\} 0.00257^2\right)\right\} = -12.26!,$
(Actual -12.7) A full Moon reflects 30% more light at full phase than a perfect diffuse reflector predicts.
Quarter Moon: $chi!,$ = 90°, $p\left(chi\right) approx frac\left\{2\right\}\left\{3pi\right\}!,$ (if diffuse reflector)
$m_\left\{Moon\right\} = 0.25 + 2.5 log_\left\{10\right\}\left\{\left(frac\left\{3pi\right\}\left\{2\right\} 0.00257^2\right)\right\} = -11.02!,$
(Actual approximately -11.0) The diffuse reflector formula does better for smaller phases.

## Meteors

For a meteor, the standard distance is a height of 100 km at the observer's zenith.

## External links

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