Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I learned that the sun from space appears white. So a white dwarf with the same outer temperature will appear the same color. – ja72 Oct 18 '14 at 14:16
up vote 4 down vote accepted

White dwarf stars 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

share|cite|improve this answer
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
""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. – spencer nelson Feb 7 '11 at 17:44

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 dwarfs is billions of years. As he says they do cool, however the universe isn't old enough to have created condensed red dwarfs. 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 dwarfs become neutron stars/". The answer is no. The white dwarf's 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 Chandrasekhar 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.

share|cite|improve this answer

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.


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.

share|cite|improve this answer
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 ( 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
The core that forms the white dwarf is already degenerate. The fact that it is degenerate tells you that it can't be hotter than about $10^{8}$ K. The back-of-the-envelope calculation you are attempting simply does not work. The core is cooling from the moment it is formed. The surface temperature is determined by radiative diffusion between the degenerate interior and the non-degenerate outer layers and has nothing to do with the history or mass of the envelope. – Rob Jeffries Mar 12 at 15:30

The first paragraph is incorrect. We do not expect white dwarf progenitors to have iron cores. The majority will have cores of degenerate carbon and oxygen.

White dwarfs are "born" with interior temperatures of a few $10^{7}$ K (and if hotter, then neutrino losses ensure they cool to this temperature on timescales of a million years). Their interiors are approximately isothermal (degenerate electrons have a long mean free path) and contain thermal energy in the non-degenerate carbon and oxygen nuclei (the degenerate electrons have essentially zero heat capacity). If the nuclei behaved as an ideal gas (with heat capacity $3k/2$ per ion, we can see that a $\sim 1 M_{\odot}$ white dwarf contains of order $10^{41}$ J of thermal energy. It is this that makes them "shine".

To get out of the interior, this energy must be transferred across a thin, non-degenerate layer at their surfaces. This acts like an insulating blanket and restricts the energy flow, that is mainly due to radiative diffusion. Calculations and models suggest that the effective surface temperature of a white dwarf is a factor of $\sim 100$ less than the interior as a result.

When they are "born" white dwarfs have surface temperatures of $\sim 100,000$ K. This radiates across the whole visible spectrum and our eyes would perceive this as a blue-ish-white. Even after a billion years, the surface temperature would be of order 10,000 K and the white dwarf would still appear white.

The cooling rate of a white dwarf slows down drastically with age, because it turns out that the luminosity is proportional to the interior temperature to the power of 3.5. This means it takes billions of years to cool to temperatures where the white dwarf surface might appear yellow or orange. In addition, the interior of a white dwarf crystallises as it cools, and this (i) increases the heat capacity to $3k$ per ion (ii) releases latent heat of crystallisation; both of which significantly delay white dwarf cooling further.

The coolest (and hence oldest) isolated white dwarfs in our Galaxy have luminosities of $10^{-5} L_{\odot}$ and radii of about 5000 km. A quick calculation suggests surface temperatures of 3800 K - which might appear vaguely reddish to the eye.

In fact there maybe some cooler, massive white dwarfs in the Galaxy. Theoretically, there may be isolated white dwarfs formed up to about $1.1M_{\odot}$ and white dwarfs in binary systems may accrete mass and grow to about $1.38 M_{\odot}$ (before exploding as type Ia supernovae or collapsing - see below). These can cool more quickly - their high interior densities mean that the temperature can fall below the Debye temperature of their extremely dense ($\geq 10^{12}$ kg/m$^3$) crystalline interiors. Their heat capacity then falls rapidly with temperature and they can quickly fade out of sight. An example that may only have a surface temperature of 3000K has recently been found in a binary system with a pulsar.

Some representative cooling curves are shown below (from Althaus et al. 2010) that illustrate these arguments. It seems quite likely that there could be white dwarfs out there that are essentially invisible in the visible spectrum, but these would likely have to be very massive ($1.2<M/M_{\odot}<1.38$) companions to other stars.

White dwarf cooling curves

It is highly unlikely that most white dwarfs will turn into neutron stars. They will just cool at almost constant radius, since the electron degeneracy pressure that supports them is independent of temperature. However, if they were to accrete matter, such that they eventually exceed their Chandrasekhar masses, they will become unstable (actually this takes place before the "classic Chandrasekhar mass" because of GR and inverse beta decay) and they could collapse. The outcome could be a type Ia supernova that blows the white dwarf apart, but it might be possible in some circumstances to have an accretion induced collapse to a neutron star.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.