1
$\begingroup$

We know that the absolute magnitude of a single star can be calculated with the formulae: $$M_{\rm bol} = 4.75 -2.5 \log_{10} \left(\frac{L_*}{L_{\odot}}\right)\ ,$$ $$M = -2.5 \log_{10}(L_* / L_0),$$ where $L_*$ is the luminosity of the given star, and $L_0$ being the luminosity of a magnitude 0 star at a distance of 10 pc ($79L_\odot)$.

However, this only works for a single star, not multiple star systems. Is the formula for the absolute magnitude of a multiple star system defined as $L_*=L_1+L_2+...+L_n$ (the sum of all luminosities in the system), where $L_n$ is the luminosity of a star in the system, or is it something else?

$\endgroup$
1
  • $\begingroup$ I do not think you can talk of the luminosity of a star at 10pc. The luminosity is the electromagnetic energy emitted per second and is not dependent on the distance to the object.You would probably need the Irradiance instead, which is the amount of energy received per surface area. The Irradiance changes with distance to the object. $\endgroup$
    – Dieudonné
    Commented Nov 3, 2020 at 8:09

1 Answer 1

4
$\begingroup$

Yes, you are correct. The luminosities can be added. Luminosity is the amount of electromagnetic energy emitted per unit of time (measured in $ \textrm{J} \cdot \textrm{s}^{-1} $ or $\textrm{W}$). So if you have a multiple star system, the total amount of energy emitted (i.e. the total luminosity) is simply the sum of the energy emitted by each of its components.

This does not take into account any absorption of the radiation of one of the components by one of the others.

In the book by Jean Meeus [1], the formula for adding stellar magnitudes can be found in Chapter 56:

$$ m = -2.5 \log_{10} \sum_{i=1}^{n} 10^{-0.4 m_i} $$

This formula is for the apparent magnitude, but would also apply for the absolute magnitude, which, after all, is only the apparent magnitude at $10\,\textrm{pc}$ distance.

$$ M = -2.5 \log_{10} \sum_{i=1}^{n} 10^{-0.4 M_i} $$

[1] Jean Meeus, 1998, Chapter 56: Stellar Magnitudes, from Astronomical Algorithms, Willmann-Bell Inc. Second Edition.

$\endgroup$
2
  • $\begingroup$ So what is the formula for absolute magnitude? $\endgroup$
    – WarpPrime
    Commented Nov 2, 2020 at 22:30
  • $\begingroup$ The same, but with the visible magnitude m replaced by the absolute magnitude M. I've added it to the answer for completeness sake. $\endgroup$
    – Dieudonné
    Commented Nov 3, 2020 at 8:02

You must log in to answer this question.

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