Proxima centauri main sequence duration

  • #1
erm just a quick question, as i read something that didnt quite make sense to me anyway, our sun will stay in its current form for about another 5 billion years before inflating into a red giant, now i was just browsing wikipedia reading up about various things, and was reading the article about proxima centauri which is the nearest star to us besides our sun. it stated that proxima centauri is one seventh the diameter or our own star, which understandably it would be in the main sequence for a longer period of time, however the time scale just seems a hell of a lot larger than can be right

"The mixing of the fuel at Proxima Centauri's core through convection and the star's relatively low energy production rate suggest that it will be a main-sequence star for another four trillion years, or nearly 300 times the current age of the universe.
surely that cant be right, in essence it will be in its current form for 4,000,000,000,000 years...
 

Answers and Replies

  • #2
Nabeshin
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The relationship between time on the main sequence and mass is not linear! I recall that the relationship between rates of nuclear fusion and mass scales like M^20 or something absurdly high like that (don't recall the exact power, but I remember it is very large!), so four trillion years is certainly not out of the question.
 
  • #3
Vanadium 50
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The Eddington Relationship is L ~ M3.5. The time on the main sequence is M/L, so therefore M-2.5 or so. A star that is 1/10 the mass of the sun would therefore live 300 times as long: ~3 trillion years.

It's actually more than that, because Eddington is an approximation, and the exponent gets bigger for smaller stars.

We can do a more accurate estimate with Proxima. If it has 1/7th the radius and half the temperature, it will have 0.0013 the luminosity. It's mass is 0.13 solar masses, so that means it will burn 100x as long: a trillion years or so.

In fact, red dwarfs live longer than that, because they burn their fuel more efficiently than the sun does, but a factor of 5-10. So several trillion years is the right ballpark.
 
  • #4
interesting, thank you both for your replies
 
  • #5
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Nabeshin, that's temperature-dependence, not mass-dependence. The power-law dependence is an approximation; it's actually more like
[itex]\exp\left(-\left(T_b/T\right)^{1/3}\right)[/itex]
where Tb is related to the electrostatic repulsion of the fusing nuclei.
 
  • #6
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The Sun will only fuse about 10% of its hydrogen while on the Main Sequence because its core, where fusion happens, essentially doesn't mix with the rest of the Sun. A red dwarf, like Proxima, is fully convective, so ~90% of its hydrogen will be fused while on the Main Sequence. Then it's core will contract somewhat and fuse hydrogen in an outer shell as helium builds up in the core. Eventually almost all the hydrogen will be fused into helium-4, leaving a helium white dwarf. Total lifespan will be about 5 trillion years. A good source is a paper by Greg Laughlin, Fred Adams & Peter Bodenheimer, "The End of the Main Sequence", which presents the results of detailed modeling of low mass red dwarfs.
 
  • #7
Vanadium 50
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Convection was the reason I said "they (red dwarfs) burn their fuel more efficiently than the sun does" However, the larger factor is just that they burn slower.
 
  • #8
Nabeshin
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Nabeshin, that's temperature-dependence, not mass-dependence. The power-law dependence is an approximation; it's actually more like
[itex]\exp\left(-\left(T_b/T\right)^{1/3}\right)[/itex]
where Tb is related to the electrostatic repulsion of the fusing nuclei.
Ah, thanks for the correction :)
 
  • #9
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Convection was the reason I said "they (red dwarfs) burn their fuel more efficiently than the sun does" However, the larger factor is just that they burn slower.
Hi Vanadium
You also said the exponent gets larger for lower mass stars. Not so. Using Iben's stellar evolution figures it peaks at 4.75 for near solar mass stars and decreases as the mass goes down. The lowest mass red-dwarfs, and the highest mass stars oddly enough, have a mass-luminosity exponent of roughly ~3, though the exact temperature requires detailed modelling of the star's atmosphere.
 
  • #10
Vanadium 50
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Are you sure - I thought it grew and then flattened as you went down. I could be wrong - I'm doing this from memory.
 
  • #11
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Are you sure - I thought it grew and then flattened as you went down. I could be wrong - I'm doing this from memory.
All my best work is from memory ;-)

Iben's papers are available from NASA's ADS if you want to check. Also there's the series of papers from Chabrier, Baraffe & others which models low mass stars, also available from the ADS.
 
  • #12
Drakkith
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The Sun will only fuse about 10% of its hydrogen while on the Main Sequence because its core, where fusion happens, essentially doesn't mix with the rest of the Sun. A red dwarf, like Proxima, is fully convective, so ~90% of its hydrogen will be fused while on the Main Sequence. Then it's core will contract somewhat and fuse hydrogen in an outer shell as helium builds up in the core. Eventually almost all the hydrogen will be fused into helium-4, leaving a helium white dwarf. Total lifespan will be about 5 trillion years. A good source is a paper by Greg Laughlin, Fred Adams & Peter Bodenheimer, "The End of the Main Sequence", which presents the results of detailed modeling of low mass red dwarfs.
Why doesn't the suns core mix with the rest of the sun, or at least a part of it?
 
  • #13
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Why doesn't the suns core mix with the rest of the sun, or at least a part of it?
The Sun's core loses its generated heat by radiation sufficiently efficiently for convection to be surpressed. The radiative core is surrounded by a convective region, and above that is the outer layers where the heat escapes to space. In low mass stars the heat production is a lot lower and the star much denser. For example, Proxima Centauri masses 0.123 Solar, but its radius is 0.145 solar i.e. it's 40 times denser than the Sun. As a result the convective region is much deeper. In the lowest mass red-dwarfs the convective zone goes all the way to the centre of the star. Thus very efficient use of the available fusion fuel.

In the Sun, once the core's fuel is depleted, the core continues to release energy via gravitational contraction and fusion burning begins in a shell around the depleted core. As the core contracts it heats up and because of the temperature sensitivity of the reactions the energy production rate rises. The outer layers bloat and the Sun inexorably becomes a Red Giant - basically the same picture for all stars between about 0.25 and 8 Solar masses.
 
  • #14
Drakkith
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So the suns core, where the fusion reactions are taking place, contains enough fuel to power itself for about 10 billion years before it becomes a red giant? I'm assuming that the core region doesn't burn hydrogen very fast at all compared to the amount of fuel availble in the core. Is that about right? Do we know the size/density of the suns core? Is it possible to calculate about how much fuel is available compared to the amount used per second in reactions?
 
  • #15
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So the suns core, where the fusion reactions are taking place, contains enough fuel to power itself for about 10 billion years before it becomes a red giant?
The luminosity rises from 0.7 to 1.84 times the present levels, but yes there's enough for 10 Gyr.

I'm assuming that the core region doesn't burn hydrogen very fast at all compared to the amount of fuel availble in the core. Is that about right?
It is. The fusion reaction rate for proton-proton reactions that make deuterium - the reactions of which actually provide most of the power - is very slow. The half-life for a proton in the Sun's core is about 8 billion years. Deuterium OTOH fuses in milliseconds so it never builds up. Proton-proton reactions are SLOW...

Do we know the size/density of the suns core?
We've know it roughly since the 19th Century when the Lane-Embden equation was first solved. Since then our modelling has improved out of sight, while we can do helioseismology to peak inside, as well as catch neutrinoes to "sample" the fusion reactions happening there.

Is it possible to calculate about how much fuel is available compared to the amount used per second in reactions?
Yep. About 600 million tonnes per second is fused. At that rate the total available, since it has been fusing for 4.5 billion years already, is about 8 billion years worth. Remember the Sun's luminosity is rising, so that will be used up in 5.5 billion years according to the latest models.
 
  • #16
Drakkith
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Know a good information source for the sun? Mass of the core, density, dimensions, ETC.
Thanks. =)
 
  • #17
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Hi Drakkith
Search the ADS and you'll find plenty. Look up "Astrophysical Database Server" to find it.
 
  • #18
Drakkith
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Hi Drakkith
Search the ADS and you'll find plenty. Look up "Astrophysical Database Server" to find it.
Awesome, thanks!
 
  • #19
Vanadium 50
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Graal pointed out an important idea: convection.

The ideas of stellar modeling are not very hard to understand. One of the most important is that every erg of energy produced in the core has to be radiated away at the surface. There are two ways to do this: radiation, and convection. The denser the medium, the less efficient radiation is: the more "stuff" that's in the way. On the other hand, mass transfer (convection) becomes more efficient. So dense stars (which means red drwarfs) are convective.
 
  • #20
Drakkith
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I'm assuming the math is pretty complicated to figure out the burn rate of a particular star, given its mass, temp, ETC.
 
  • #21
Vanadium 50
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Not really. It's a long problem, not a hard one. Take a star, and dice it up into many small volume elements.

You know quite a few things about these elements: The heat that leaves this element is equal to the heat that enters plus the heat that is produced. The heat that is produced is a function of the temperature, pressure and composition of that volume element. The force on that element is the vector sum of the force due to gravity and the force due to pressure. And so on.

Now, do this for every element and you have a model of a star. It's not so much a hard problem as a long one.
 

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