# How is the visual luminosity of a star calculated?

My idea is a planet that orbits a red dwarf at a distance in which the apparent magnitude of this seen from the planet is the same as that of the sun seen from Earth. Does this mean that the daylight on the planet will be as intense as that of Earth?

I know that the red dwarfs emit most of their energy in the infrared spectrum, invisible to humans, therefore, a planet located at a distance from it in which it receives the same amount of energy as that received by the Earth of the sun, it will be darker, despite having the same equilibrium temperature. My question is, is the apparent magnitude measured considering the entire electromagnetic spectrum or just the visible spectrum?

If it is not enough for the star to have the same apparent magnitude as the sun so that the daylight of the planet is the same as that of the Earth, how is its visual magnitude calculated?

• This sounds like it might be a question for the astrologers but maybe also see this: standard reference values for apparent magnitude. Aug 31, 2019 at 20:45
• Your title asks about luminosity, in the question you ask about 'apparent magnitude', but luminosity and magnitude are different concepts for a star, magnitude is the inverse logarithm of energy/time at some distance (a fixed reference distance for absolute magnitude, viewer's distance for apparent mag.). For both luminosity and magnitude, it's possible to define it in terms of energy per unit time over all frequencies, or relative to some frequency filters. If someone talks about "visual" magnitude or luminosity they are presumably talking only about energy in the visual frequency range. Aug 31, 2019 at 21:20

The visual luminosity is the luminosity in the visual spectrum. If you start with the blackbody temperature of the star, you can calculate the blackbody spectrum, and then derive the visual luminosity by integrating the spectrum from the lower end of the visual range to the upper.

Since a larger fraction of the spectrum of a red dwarf lies outside the human visual range, a planet at a distance where it will receive the same visual illumination as the Earth will be receiving considerably more infrared, and will thus be hotter. The red sun will also be larger in the sky, and so have lower surface brightness. Daylight will still not be exactly the same, as the distribution of wavelengths will be different, so colors will be slightly off (equivalent to illumination by an incandescent bulb vs. sunlight), and the differing sensitivity of the human eye across different wavelengths means the red-heavy spectrum will actually still look darker even if it has exactly the same amount of power in the visual spectrum as Earth's sunlight. Figuring out how much closer the world would have to be to make up for the lowered sensitivity of the human eye and match perceptual brightness is tricky, but references for the apparent brightness of various kinds of incandescent lamps might be useful there.

Well, in order to achieve same temperature as the Earth, total absorbed radiation(across the entire spectrum) must be ~ same as Earth.

According to wikpedia, luminosity of red dwarf is 3% - 0.01% of sun, while habitable zone is in distance of 1/4 to 1/30 AU[averages for differently sized red dwarves], making total intensity of light 3.4.4 = 48%(almost an "orange dwarf") to just 9%(very small red dwarf).

However, problems with red dwarf habitability is more than just lack of light. Tidal effects on such planet would likely cause it to quickly reach tidal locking (planet would only show one side to its sun) and apart from that, there's also problem with uneven output of red dwarves.

Thus if you really want to go for that setting, it's better to go for the largest possible red dwarf size, or settle for "orange dwarf" as a star.

For more details, check out this wiki link: https://en.wikipedia.org/wiki/Habitability_of_red_dwarf_systems

L is the luminosity of the star; R is the star's radius; T is the star's temperature, measured in Kelvins; L☉ is the luminosity of the Sun, equal to 3.828 * 10²⁶ W; R☉ is the Sun's radius, equal to 695700 km; T☉ is the temperature of the Sun, equal to 5778 K.

Equation for star brightness calculation; P = σ * A * T⁴

• This seems like it'd have weird units. Is there a constant missing somewhere? Feb 29, 2020 at 1:42
• @jdunlop $\sigma$ there is the Stefan-Boltzmann constant which does indeed have weird units, making $P$ end up as just watts. Radiated power isn't very useful by itself for computing visual luminosity, so this seems like half an answer. Feb 29, 2020 at 13:08