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There are a large fraction of M dwarfs which are claimed to be brown dwarfs.

Why do we still use M-type and not create a new stellar type like L, T and Y? The main signature of M is TiO absorption bands. They appear in M-type brown dwarfs?

Is a spectrum enough to judge whether a celestial object is a brown dwarf or not? Or do we need to know its mass?

The earliest type is M3.5. It means M-type objects older than M3.5 are necessarily brown dwarfs? Or is there a safe value, say, M9, is necessarily a brown dwarf? A brown dwarf is necessarily colder than a red dwarf?

There should be more brown dwarfs or red dwarfs?

The reason we do not need to create a new stellar type is the spectra of brown dwarfs are more like red dwarf's spectra?

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  • $\begingroup$ en.wikipedia.org/wiki/…. Says the class type tells you what is in the spectrum, but you cannot deduce the mass cleanly. $\endgroup$ – eshaya Nov 10 '16 at 17:55
  • $\begingroup$ Please stop changing the question and adding new questions. $\endgroup$ – Rob Jeffries Nov 12 '16 at 9:11
  • $\begingroup$ @Rob All questions fit the title. e.g. whether M-type brown dwarfs have TiO absorption bands or not. I appreciate you typed a lot, but the questions have not been answered. $\endgroup$ – questionhang Nov 12 '16 at 9:21
  • $\begingroup$ M-type objects older than M3.5 are not necessarily brown dwarfs? A M3.5 brown dwarf is hotter than a M9 red dwarf? I do not know whether there is a M9 red dwarf. $\endgroup$ – questionhang Nov 12 '16 at 9:28
  • $\begingroup$ What do you mean by older than M3.5? Do you mean cooler? $\endgroup$ – Rob Jeffries Nov 12 '16 at 9:31
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The spectral type of an object is almost entirely determined by the temperature of its photosphere. ie Saying something is type M3.5 is just a measure of its surface temperature. An M3.5 brown dwarf is at a very similar temperature to an M3.5 star.

Brown dwarfs begin their lives as hot balls of gas and gradually cool with time. They start off as M-type objects and then cool to become L-type and then eventually, T-type objects. Stars on the other hand, start their lives by cooling, but stabilise their temperatures at a mass-dependent value that corresponds to an M-type classification or even as cool as type L2 for the very lowest mass stars.

Given that stars and brown dwarfs appear to have been born throughout Galactic history, then a range of masses are possible for any spectral type. In order to use the surface temperature (or spectral type) as an indicator of whether something is a brown dwarf, then you also need to know its age. In general that is not known, unless the object can be demonstrated to be part of a cluster of stars (of known age).

In order to be a brown dwarf the mass should be below about $0.075 M_{\odot}$, but unless the object is in a binary system, this cannot be measured. So what people do is trust the models(!) and compare the luminosity (or estimated temperature) with model evolutionary tracks at a given age. This gives an age- and model-dependent mass that can be used to claim that an object is a brown dwarf.

The fact that brown dwarfs can have an M-type spectral class when they are young, but then move through L- to T-dwarfs as they get older is not a problem, it is a confirmation that the evolutionary models are (roughly) correct!

A plot might help (from the work of Burrows et al. (1997)). This shows the evolution of temperature with time. Each track represents a brown dwarf of a given mass (the tracks from top to bottom) are masses of 0.2 - 0.08$M_{\odot}$, representing stars (in blue). Then come masses of 0.07 to 0.015$M_{\odot}$ representing brown dwarfs (in green, that will never ignite hydrogen in their cores to any extent). Then below this are the "planetary mass objects" from 14 Jupiter masses to 0.3 Jupiter masses at the very bottom (in blue - the definition here is that these objects don't even ignite deuterium in their cores).

The horizontal lines mark the boundaries of the spectral type classifications (M at the top, L in the middle, T at the bottom). Notice how the stars level-off in temperature (when they begin H-burning) but can have M-type or even L-type classifications, but the brown dwarfs cool throughout their lives. Even objects as low as $0.01M_{\odot}$ begin their lives as "M-type" objects.

To answer some of your more specific queries: The warmest spectral type below which you can almost guarantee that an object is a brown dwarf is $\sim$ L2. Any object warmer than this could be a star.

A brown dwarf is always cooler than a low-mass star of the same age.

Current observational estimates are that there is about 1 brown dwarf for every 4 low-mass stars - the weighted mean ratio for stars between 0.08 and 1 solar mass to brown dwarfs from observations of several clusters is $4.3 \pm 1.9$, according to Andersen et al. (2008); the commonly adopted/assumed "Chabrier Initial Mass Function" has 4.9 stars for every brown dwarf.

The spectrum of a brown dwarf and an M-dwarf star at the same temperature are very difficult to distinguish. The M-dwarf is only slightly bigger and this might lead to small gravity-dependent differences. Lithium is not burned in brown dwarfs less than $0.06 M_{\odot}$, whereas it is burned completely in low-mass stars on a mass-dependent timescale. Thus an older (than say 1 billion years) M-dwarf star would certainly have burned its lithum, whereas a younger M-type brown dwarf of lower mass would still have lithium.

Evolution of brown dwarf temperatures

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    $\begingroup$ @HDE@ Rob Both of your answers help, but I still.. itch. I clarified my question. $\endgroup$ – questionhang Nov 11 '16 at 8:59
  • $\begingroup$ You replot fig.7 of Burrows1997? How do you get the data? It helps much, but please take a look at the bold face. $\endgroup$ – questionhang Nov 11 '16 at 11:17
  • $\begingroup$ @questionhang In case you weren't notified, he updated the question. $\endgroup$ – Sir Cumference Nov 11 '16 at 17:54
  • $\begingroup$ @Sir yes, I know. the bold face, I specify it again. $\endgroup$ – questionhang Nov 12 '16 at 2:27
  • $\begingroup$ ..1 brown dwarf for every 4 low-mass stars...reference? fig9.21 of press.princeton.edu/titles/8963.html gives a different value, but it is limited in 8pc. $\endgroup$ – questionhang Nov 12 '16 at 10:10
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There are a couple different tests you could use to determine whether the object is a brown dwarf of a red dwarf.

You could try to use the lithium test to determine whether the object in question is a brown dwarf or a low-mass red dwarf. Brown dwarfs generally don't reach temperatures high enough for lithium burning to occur at any significant rate, and so in low-mass M-type red dwarfs, there should be no evidence of a lithium line at 670.8 nm. This was originally proposed from models by Rebolo et al. (1992). The test isn't perfect, as some objects will burn some lithium at different points in their lives, but it can work, and it's certainly a good starting point.

Lithium isn't the only atmospheric component whose presence is temperature-dependent - titanium oxide, vanadium oxide, carbon monoxide, nitrogen are other examples - so by looking for spectral lines, it might be possible to determine the temperature of the atmosphere and compare the spectra to models of brown dwarf and red dwarf spectra (see this for more). However, some of these only disappear at temperatures much lower than values for which red dwarfs and brown dwarfs could be confused.

Brown dwarfs also have a very small range of radii. $R\propto M$ for low-mass stars, whereas $R\propto M^{-1/3}$ for brown dwarfs supported by degeneracy pressure (see Oppenheimer et al. (1998)). If you can measure the effective surface temperature of the object, $T_{\text{eff}}$, you should be able to determine its radius by assuming a black body-like temperature-luminosity relation. $T_{\text{eff}}$ could be determined in a few different ways, including using atmospheric models.

I honestly don't see the need for an additional spectral class that would likely overlap the M class. It would create more confusion than already exists, and Morgan-Keenan system already in use is honestly complicated enough, and has been simplified from more complicated and problematic schemes. See the first table here, taken from Table 5 of Kirkpatrick et al. (1999).

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