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There is a mass-luminosity relationship of the form $L \propto M^a$, where $a = 3.5$, that is is applicable to main-sequence stars. Is there a similar relationship that holds for brown dwarfs?

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  • $\begingroup$ See also astronomy.stackexchange.com/questions/32771/… $\endgroup$
    – ProfRob
    Jan 10, 2023 at 12:03
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    $\begingroup$ Thanks, @ProfRob. I've checked out the link and I found the formula for mass-luminosity-time relation given by you, however this is for Main-sequence only and not for brown dwarfs, right? $\endgroup$
    – user47732
    Jan 10, 2023 at 15:30

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The linked question applies specifically to brown dwarfs during their deuterium-burning phase, which lasts from 3-30 million years, depending on the brown dwarf mass. The conclusion there, is that there is a luminosity plateau, akin to the main-sequence, where luminosity is closely-related to mass. However, after that, the luminosity depends on both mass and age so there is no unique relationship between luminosity and mass and no simple relationship that can be written as a formula.

Details:

It is worth showing this diagram again (from Burrows et al. 1997) which shows the luminosity evolution of stars, brown dwarfs (and planets) as a function of age for different masses. The regions where there is a good relationship between luminosity and mass will be horizontal lines on this plot. The luminosity-mass relationship works in main sequence stars (especially low-mass main sequence stars) because their luminosity does not change for long periods after they have initiated nuclear burning.

Thes situation for brown dwarfs is far more complex (the green lines). During deuterium-burning, the green curves are more-or-less flat, implying a close relationship between mass and luminosity. Thereafter you see that brown dwarfs continuously cool and therefore the luminosity is a function of both mass and age. By looking at the wobbling of the cooling curves and the fact that some of the curves meet or even cross at certain ages, you can see that there is no simple mathematical formula that can be written down to represent this relationship - one must use the theoretical curves (and different models can have significant disagreements in detail). There is probably a luminosity-mass relationship at a given age, but in general, you wouldn't know the age of a brown dwarf.

So, for example, if you know the luminosity is $10^{-4}$ solar luminosities, the plots below show that if you don't know the age, the object could have a mass anywhere between 0.004 - 0.07 solar masses.

Luminosity evolution from Burrows (1997)

Incidentally, it is worth noting that there isn't a single power-law formula relating luminosity and mass across the whole main sequence and that for stars of about 0.8 solar masses or higher, one should also take account of (or correct for) the age of the star. For example, the Sun is about a factor of 1.5 more luminous than when it started on the main sequence, but its mass is almost the same (slightly less actually).

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  • $\begingroup$ Thanks a lot!. So there isn't a equation for it, Right? and What if the mass given from the Mass- luminosity relation equation can vary a bit, So that I don't have to supply the age of the star? $\endgroup$
    – user47732
    Jan 10, 2023 at 12:01
  • $\begingroup$ @ScienceAJ I don;t understand your comments. You asked if there is a mass-luminosity relationship. No, there isn't. There is some sort of mass-luminosity-age relationship. If you ignore the age, then you can see that at a given luminosity the brown dwarf could have almost any mass within the brown dwarf (or planetary) regime. $\endgroup$
    – ProfRob
    Jan 10, 2023 at 12:49
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    $\begingroup$ Thanks a lot!. Moreover what do you mean by a simple formula, by that do you mean that there is not formulated formula at all or there is some complex formula? $\endgroup$
    – user47732
    Jan 11, 2023 at 13:16
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    $\begingroup$ Thanks a lot!. So can the Main-sequence formula be interchanged with Brown dwarf equation ? $\endgroup$
    – user47732
    Jan 11, 2023 at 14:30
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    $\begingroup$ Thanks. Alright, @ProfRob $\endgroup$
    – user47732
    Jan 11, 2023 at 14:33

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