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The hotter something is glowing the more white/blue it appears. A dying medium sized star expands, cools and becomes a red giant for a while, but eventually it is going to gravitationally collapse (once enough Iron (Fe) is accumulated in the core). Then it blows the outer layers away and what is left collapses into a white dwarf.

What makes the dwarf shine? and why is it white?

Does the luminosity decreases as the object cools down, or is there some other reaction that keeps it glowing for a long time?

Can a white dwarf turn brown or black never to be seen again?

Do all white dwarfs turn into Neutron stars eventually?

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3 Answers

up vote 3 down vote accepted

White dwarves used to be the interior of a star, which was the hottest part of the star. They shine white because they are still very hot from this past part of their history. As they age, they will cool, and as they cool, they will lose temperature, and their blackbody profile will shift to redder and redder colors, and eventually into the infared and radio ranges where they won't seem to shine at all to the naked eye.

And yes, a white dwarf state is a stable final state of a star, so long as it does not interact further with any matter. If that happens, it is possible to have a White Dwarf supernova

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So a white dwarf will not necessarily be "white", but follow a typical hot body radiation distribution and subsequent redding as it cools. And it cools due to radiative heat transfer only? So temperature should drop kind of exponentially. –  ja72 Feb 7 '11 at 16:20
    
@ja72: see Omega Centauri's answer below. They are very hot and have a relatively small surface area, and radiation is not a particularly effective way to lose heat. White holes stay hot for a long time. –  Jerry Schirmer Feb 7 '11 at 16:22
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""So temperature should drop kind of exponentially."" No! Stefan-Bolzmann says that power is ~ to T^4 –  Georg Feb 7 '11 at 16:28
    
As a coda to what Georg said, if you model the white dwarf as an ideal gas, this gives you $T \;\alpha \;\frac{1}{\sqrt[3]{C + \frac{16\sigma t}{27N^{4}k^{4}}}}$, which is quite a slow fall-off indeed. –  Jerry Schirmer Feb 7 '11 at 16:42
    
@ja72 right - what everyone is getting at above is that while conduction drops off exponentially, it's not how white dwarfs cool. Cooling by pure radiation is very slow. –  kharybdis Feb 7 '11 at 17:44
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To amplify a bit bit on Jerry's answer. Because of its small surface area, and large thermal mass (typically about a half the mass of the sun) the cooling time of white dwarves is billions of years. As he says they do cool, however the universe isn't old enough to have created condensed red dwarves. The stars currently called red dwarfs, are main sequence (hydrogen burning) low mass stars with lifetimes on the order of a trillion years.

I guess, we've all simply forgotten to answer his other question "do all white dwarves become neutron stars/". The answer is no. The white dwarfs pressure is maintained by electron degeneracy pressure, they do not contract appreciably as they cool down, and dense as they are they are orders of magnitude less dense than nuclear matter. A white dwarf has to exceed the Chadreshekar mass for core collapse in order to become a neutron star. And if enough mass is added to a white dwarf to exceed that limit, you get a thermonuclear runaway reaction leading to a type 1a supernova instead.

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Last week I wrote an answer to:

Please clarify how entropy increases when matter gravitationally coalesces

I illustrate how a collapsing body will increase its entropy. This also corresponds to an increased temperature. The imploded white dwarf has its internal energy compressed into a very small volume. Hence they initially shine at a very high temperature.

[further]

The interior of a white dwarf has Fermi-Dirac statistics, but the gas on the outer layers is probably more ordinary. Further, that is what we detect. Consider a stellar core imploding from about $1.0R_{sol}$, probably more like less than this, which we will write at $5\times 10^{5}$, to about $5\times 10^3$. A rough estimate would be the natural gas law, in fact in the form Boyle wrote it in $$ T~=~\frac{V}{V_0}T_0~\simeq~10^6\times T_0 $$ The initial temperature of the core is about $10^7$K, which gives a pretty outstanding temperature of about $10^{13}$K

This is not realistic, for as the white dwarf forms it radiates most of its energy to eject off the outer layers into a nebular cloud. So the outer layer of the core collapses inwards more at a temperature comparable to a stellar surface. So that knocks this estimate down three or four orders of magnitude. In fact a newly minted white dwarf as a temperature of about $10^7$ K So direct simple calculations are not very accurate.

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I believe this is a significant part of the physics which the OP did not appreciate: the initially very high temperature due to gravitational collapse. Wikipedia (en.wikipedia.org/wiki/White_dwarf#Radiation_and_cooling) gives temperatures on the order of 10^4 to 10^7 K. If you edit the comment to give a calculation which demonstrates the right order of magnitude for the problem, I'll +1 it :-) –  genneth Feb 7 '11 at 17:07
    
Well, I guess that deserves an upvote anyway :-) –  genneth Feb 7 '11 at 20:47
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