७ जुलै, २०१४

Less than 3,000° Celsius is considered "extremely, extremely cold" for a star.

UW-Milwaukee astronomer David Kaplan has detected this invisible star, a "white dwarf" or dying star, and "they realized, the star was so cold that the carbon making it up would have crystallized, effectively making it into an Earth-sized diamond."

२६ टिप्पण्या:

Nonapod म्हणाले...

I assume it's still irradiates enough heat to not be considered a black dwarf, and also since black dwarfs are hypothetical objects that shouldn't yet exist in our Universe.

SteveR म्हणाले...

"I'm just an old dwarf star but I'll be a diamond some day"

gerry म्हणाले...

black dwarfs are hypothetical objects that shouldn't yet exist in our Universe.

Where's Crack when you need him?

MadisonMan म्हणाले...

If it's 3200 K, or thereabouts, it's emitting radiation with a peak around 1 micron, by Wien's Law.

So it would certainly be visible if you were next to it, because it's emitting in the visible part of the electromagnetic spectrum, too. And if you happen to be looking at around 1 micron, you'd see it. That it's 900 light years away makes detection difficult from Earth, but not because it's "invisible".

Terrible science writing by the Journal Sentinel writer.

madAsHell म्हणाले...

effectively making it into an Earth-sized diamond

shhhh.....don't tell my wife.

Curious George म्हणाले...

MadisonMan said...
If it's 3200 K, or thereabouts, it's emitting radiation with a peak around 1 micron, by Wien's Law.

So it would certainly be visible if you were next to it, because it's emitting in the visible part of the electromagnetic spectrum, too. And if you happen to be looking at around 1 micron, you'd see it. That it's 900 light years away makes detection difficult from Earth, but not because it's "invisible".

Terrible science writing by the Journal Sentinel writer."

From the link: "Scientists know the star must be very cold because it is invisible to their telescopes, implying that it is radiating little light or heat."

Terrible reading by MadisonMan ;-)

Original Mike म्हणाले...
ही टिप्पणी लेखकाना हलविली आहे.
Original Mike म्हणाले...

There are five things, off the top of my head, that will determine signal power:

1). Star temperature
2). Star diameter
3). Star distance
4). Telescope aperture
5). Spectral sensitivity of the detector
6). Transmission of the atmosphere at the frequencies in question

I'm not sure what your beef is, C.G.

YoungHegelian म्हणाले...

I don't understand why the scientists would think that white dwarf material would be crystalline carbon.

That there would be some highly compressed carbon, yes, but I would think it would be mostly iron & a core of heavier elements & then a lot of that God-only-knows-what stuff that remains when the core of a large star collapses under gravity into a dwarf star, blowing off the lighter outer layers.

SJ म्हणाले...

It kind of hangs on what a star is, and what is "cold" in the range of temperatures that stars exist in.

For comparison, a match or a candle typically burn at temperatures near 1800 K.*

Stars need to be very large (to sustain nuclear fusion in their core). I think the Sun has a surface temperatures above 5000 K. I strongly suspect that brighter, hotter stars exist. The Sun's core temperature is thought to be in the range of 15000000 K. With those numbers in mind, a temperature of ~3300K is kind of cold.

The odd carbon-crystal structure, which might be a giant diamond, is a possibility. However, observation is limited. This conclusion is a combination of theory and observation.




*To convert temperature in degree-C to Kelvin, subtract 273 degrees. Scientists tend to use Kelvin for stars and other really hot stuff, because...well, they do. The Kelvin scale is an "absolute" scale, with no negative values. It provides a good reference for addressing how far from absolute zero a temperature is.

David म्हणाले...

Where are Liz Taylor and Richard Burton in all this?

Original Mike म्हणाले...

"That there would be some highly compressed carbon, yes, but I would think it would be mostly iron & a core of heavier elements & then a lot of that God-only-knows-what stuff that remains when the core of a large star collapses under gravity into a dwarf star, blowing off the lighter outer layers."

Very little iron nor God-only-knows-what produced by a star like this. It mainly consists of the end product of whatever fusion reaction(s) its mass supported. Carbon is a common result for stars of the right mass.

traditionalguy म्हणाले...

So the universe is slowly turning into carbon diamonds...or something.

Science is fun.

The Godfather म्हणाले...

Do you think Jane Fonda still has her Barbarella costume? 'Cause this could make a great movie!

El Camino Real म्हणाले...

Finally! A diamond big enough for Queen Michelle's tranny finger.

Mark म्हणाले...

YH, any star big enough to fuse to iron will be big enough to go supernova. Stars the size of the sun stop at carbon, hence diamond. (A minor point is that the carbon will concentrate to the core, surrounded by lighter fusion products. I'll let an astrophysicist weigh in on the proportions and how they'd be layered.)

Mark म्हणाले...

By the way, I bet De Beers already owns that star.

traditionalguy म्हणाले...

Nobody tell Obama's EPA Czaress. She will use this story to issue Regs banning stars as carbon fusion pollutants in order to save asthma suffering white dwarfs.

And what's more, starlight causes skin cancer to the little white guys.

Revenant म्हणाले...

That there would be some highly compressed carbon, yes, but I would think it would be mostly iron & a core of heavier elements

Stars small enough to eventually become white dwarfs aren't large enough to produce iron in the first place, as I understand it.

tim in vermont म्हणाले...

De Beers right now is patrolling the planet buying up and burning plans for warp drives.

Guildofcannonballs म्हणाले...

These little punk ass bitch scientists don't know This is a link to Sir Michael Starr.

Gabriel म्हणाले...

@SJ:The Kelvin scale is an "absolute" scale, with no negative values.

Actually, there are negative Kelvin values, but they are not colder than absolute zero; rather they are hotter than any finite temperature.

If the system is in a state such that adding energy reduces entropy instead of increasing it, that makes the temperature (dU/dS) negative. Because a loss of energy will increase the entropy, such a system will give up energy to any at finite temperature which it is in thermal contact with, no matter how hot. making the negative temperature hotter than any finite positive temperature.

Vile Pliskin म्हणाले...

I call dibs.

Original Mike म्हणाले...

"I call dibs."

Damn.

AustinRoth म्हणाले...

Futurerama predicted this.

MikeB म्हणाले...

< 3000° Celsius

Stellar sweater weather.