<p>Before protostars become &quot;stars&quot; and begin nuclear fusion, they are just balls of gas. Proto- brown dwarfs are also just balls of gas, but their cores <em>never</em> become hot enough for significant hydrogen ignition.</p>
<p>Roughly speaking, for a ball of gas, its contraction timescale is given by the amount of gravitational potential energy it has divided by the rate at which it loses energy from its surface.</p>
<p>The amount of potential energy energy depends on mass squared. The luminosity at the surface also depends indirectly on mass, but with a steeper dependence than the potential energy. This means that the contraction timescale is always longer for lower mass objects - roughly as <span class="math-container">$M^{-1}$</span> for low-mass objects.</p>

CC BY-SA 4.0
2024-09-28T09:33:47.963Z
2024-09-28T12:04:38.283Z
2024-09-28T12:04:38.283Z
ProfRob
<p>I know that larger mass stars form quicker, and lower mass stars form slower. Does this work for brown dwarfs? Would brown dwarfs take a very long time to form? If so, how long? I couldn't find anything online.</p>

CC BY-SA 4.0
2024-05-13T05:11:46.090Z
There is a difficulty in definition.  The time it takes a star to form is the time from when the gas cloud start to collapse, until it reaches "Zero age" on the main sequence, when hydrogen fusion begins.  But a brown dwarf never achieves hydrogen fusion, so in this sense it is always "forming".   You could change your definition of "zero age", but then you are not comparing like with like.
James K
CC BY-SA 4.0
2024-05-13T07:43:33.950Z
Agree with JamesK, you need to define a point in the evolutionary sequence that is shared by both stars and brown dwarfs. Before nuclear fusion, both are just balls of gas and the Kelvin Helmholtz contraction timescale is longer for brown dwarfs.
ProfRob
CC BY-SA 4.0
2024-05-13T23:48:20.663Z
Ahh, you're right, I hadn't thought of that. I could use the point when the star stops accreting, but that's just the PMS phase in real stars. Thanks for pointing that out
Astrovis
<brown-dwarf><formation>
How long does it take for brown dwarfs to form?
Astrovis
# How long does it take for brown dwarfs to form?

(Asked by: username_0 on 2024-05-13)

I know that larger mass stars form quicker, and lower mass stars form slower. Does this work for brown dwarfs? Would brown dwarfs take a very long time to form? If so, how long? I couldn't find anything online.

**Tags:** brown-dwarf, formation
> **Comments:**
> **username_1**: There is a difficulty in definition.  The time it takes a star to form is the time from when the gas cloud start to collapse, until it reaches "Zero age" on the main sequence, when hydrogen fusion begins.  But a brown dwarf never achieves hydrogen fusion, so in this sense it is always "forming".   You could change your definition of "zero age", but then you are not comparing like with like.  
> **username_2**: Agree with JamesK, you need to define a point in the evolutionary sequence that is shared by both stars and brown dwarfs. Before nuclear fusion, both are just balls of gas and the Kelvin Helmholtz contraction timescale is longer for brown dwarfs.  
> **username_0**: Ahh, you're right, I hadn't thought of that. I could use the point when the star stops accreting, but that's just the PMS phase in real stars. Thanks for pointing that out  

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## 1 answers

### Answer by username_2 on 2024-09-28. Score: 3 | Accepted answer

Before protostars become "stars" and begin nuclear fusion, they are just balls of gas. Proto- brown dwarfs are also just balls of gas, but their cores *never* become hot enough for significant hydrogen ignition.

Roughly speaking, for a ball of gas, its contraction timescale is given by the amount of gravitational potential energy it has divided by the rate at which it loses energy from its surface.

The amount of potential energy energy depends on mass squared. The luminosity at the surface also depends indirectly on mass, but with a steeper dependence than the potential energy. This means that the contraction timescale is always longer for lower mass objects - roughly as $M^{-1}$ for low-mass objects.


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