Star Cooler Than Venus Found

crossconnects writes to mention that Discovery is reporting that astronomers have found a nearby star with a mild surface temperature of 660 degrees fahrenheit. "The spectacularly unspectacular object is of special interest because it falls right smack in the middle of the final frontier that divides mega-planets from the puniest stars. Stars in that realm theoretically qualify as an entirely new stellar type -- what's called a Y class dwarf."

Comments

Publication at arXiv.org

[Jugalator]

Here's the actual publication on the discovery:
http://arxiv.org/abs/0802.4387

... or straight to the PDF:
http://arxiv.org/pdf/0802.4387v2

Re:Nuclear fusion?

[Tenebrousedge]

Stars above about 13 jupiter masses fuse deuterium and above 65 jupiter masses also fuse lithium, according to Wikipedia. Below 13 jupiter masses, well, it's hard to say...

Re:Fahrenheit

[servognome]

Even with an American audience at the temperatures discussed Fahrenheit has no real meaning.
The usefulness of Fahrenheit is how the range of 0 - 100 reflects weather temperatures people have experienced.
Temperatures beyond common experience are better expressed in Celsius.

Re:Nuclear fusion?

[Sique]

If you don't have much fusion, and not very much convection within the star, then the heat gets to the surface very slowly (it can take up to billions of years!), and distributed along the while surface the energy stream is low.

Re:Sigh, Bad English / Hmm - Biosphere?

[IndustrialComplex]

That is a good point, I forgot that it is mostly infrared radiation. However, just because there is no visible light, does that mean it is generating no radiation above the visible spectrum? I'm assuming that this star is fusing Deuterium and Tritium, which I believe does produce Gamma Rays.

Brown dwarfs have been observed to produce X-rays and Gamma rays. So just because this one produces no visible light does not mean it isn't producing a large amount of high energy radiation.

Re:Fahrenheit

[ultranova]

The "temperature" of the Big Bang is the theoretical hottest you can ever get, since at that point all mass was in the form of energy, and therefore you had the maximum energy at the maximum density. Nothing can ever exceed that.

Since the volume of space at t=0 was zero, but the energy content was not, the temperature at t=0 is infinite. That isn't useful for determining a scale. Alternatively, if the energy content was zero, then the temperature is lim(x->0) x/x, which is 0. If energy is zero but volume is nonzero, then temperature is 0/x (x>0), which is also 0. And if both volume and energy are nonzero, then you get a very large but finite amount, which is exceeded by classical black holes, where singularity has zero volume but nonzero energy content.

This is, of course, all assuming that neither volume nor energy can be (were not) negative, since if they are, you get all kinds of extra nastiness.

Thus, if you knew what that was, you could assign it a fixed value as your upper end of the scale.

How do you define this numerical value ? How are you going to make such a scale any less arbitrary than Kelvin scale ?

I propose an alternative: since temperature is determined by the average kinetic energy of a particle, use that as a gauge: at the temperature of 1 base unit, the average kinetic energy per particle is 1 Joule.

Alternatively, use the blackbody radiation: the base unit corresponds to the temperature of a blackbody object who's peak of radiation has wavelength of 1 meter.