<p>The density of white dwarfs is not hypothetical, it can be measured. The short answer is that the density is so high that a stable star can <em>only</em> be supported by electron degeneracy.</p>

<p>Sirius B is an example. The radius can be estimated by combining the luminosity of the white dwarf with its temperature estimated from spectroscopy. The mass can then be determined from the binary motion (with Sirius A), or gravitational redshift or from estimating the surface gravity from the width of absorption lines in its spectrum. These tell us that Sirius B has a radius of $0.0084 R_{\odot}$ and a mass of about $1 M_{\odot}$ (e.g. <a href="https://arxiv.org/abs/1201.3822" rel="noreferrer">Holberg et al. 2012</a>). This leads to an estimate of the <em>average</em> density of $2.4 \times 10^{9}$ kg/m$^3$.</p>

<p>Now, we know that for such an object to find any kind of equilibrium that the pressure must be greatest in the middle. Since for any plausible equation of state, the pressure increases with density, then the white dwarf must be <em>even</em> denser than the average at its centre.</p>

<p>From there it is a standard piece of bookwork to show why the white dwarf <em>must</em> be supported by electron degeneracy pressure and I will repeat it here.</p>

<p>An electron gas can only escape degeneracy <em>if</em> $E_F - m_e c^2 &lt; k_{b}T$, where $E_F - m_e c^2$ is the kinetic energy associated with the <a href="https://en.wikipedia.org/wiki/Fermi_energy" rel="noreferrer">Fermi energy</a> of the electrons, $m_e$ is the electron mass and $T$ is the temperature.</p>

<p>The Fermi energy can be derived from the following formula (again, standard bookwork, this is not exotic physics, it is the same physics that describes how conduction takes place in metals).
$$ E_F = \left[ \left(\frac{3h^3n_e}{8\pi}\right)^{2/3}c^2 + m_e^2 c^4 \right]^{1/2},$$
where $n_e$ is the electron number density. Now it turns out that the temperatures we are going to need to avoid degeneracy (see below) are easily going to be sufficient to completely ionise the contents of the white dwarf. If we assume the white dwarf is made out of carbon or oxygen, then the relationship between mass density and $n_e$ is just $\rho = 2 n_e m_u$, where $m_u$ is an atomic mass unit (the 2 just comes from the fact that there are 2 mass units per free electron). We could mess about with the composition <em>a little</em> to change this relationship, but it doesn't matter very much.</p>

<p>Making the appropriate substitutions and using $\rho = 2.4 \times 10^{9}$ kg/m$^3$, we obtain $E_F = 0.75$ MeV (or $1.2\times10^{-13}$ J). If we now use the test
$$ T &gt; \frac{E_F - m_e ,c^2}{k_b},$$
we find that to avoid electron degeneracy then the temperature inside the white dwarf must exceed $2.75 \times 10^{9}$ K (and be much higher in the centre of the white dwarf where the density and Fermi energy are larger). If temperatures were this high inside a white dwarf and it was made of something like carbon or oxygen then nuclear fusion would take place (rapidly) and the luminosity of the white dwarf would not be as low as it is.</p>

<p>But what if it were made of iron? There would be no nuclear fusion, even at these high temperatures. OK, leaving aside that if the interior really were that hot, it could not be hidden very easily by an atmosphere, then this would not be a stable situation. The star would be losing copious heat via neutrino losses. If the gas were non-degenerate, the star would rapidly contract and become even hotter and even denser, and if you are not going to allow for degeneracy (i.e. you postulate that quantum mechanics does not exist), then rather quickly the white dwarf ends up being a black hole. </p>

<p>A back of the envelope estimate would be to divide the gravitational potential energy by the current luminosity. Taking Sirius B as an example, the timescale is
$$ \tau \sim \frac{GM^2}{RL},$$
where $L$ is the luminosity. Taking the current luminosity from photons at the surface this timescale would be a few $10^{10}$ years (plausibly stable). But if the interior were at $\sim 10^{10}$ K, the neutrino losses would be many orders of magnitude greater and the timescale would be just millions of years.</p>

<p>We know that this isn't happening to white dwarfs because we know they are stable and long-lived. We see them in clusters of stars of known age and associated with stars that are much older than millions of years. For instance we know that stars like Sirius A are hundreds of millions of years old.
We can calculate using the theory of electron degenerate structures that the faintest white dwarfs we see have been cooling for 10 billion years without collapsing - precisely the age expected for the galactic disk. The cooling ages of white dwarfs in clusters match the ages of those clusters very well.</p>

<p>Degenerate matter is very common on earth. The <a href="http://farside.ph.utexas.edu/teaching/sm1/lectures/node86.html" rel="noreferrer">conduction electrons</a> in a metal form a degenerate gas. If you mean the kind of degenerate matter in white dwarfs, then no, at least not in any stable state. The pressures required to constrain such energy densities are too large to create in the lab.</p>

CC BY-SA 3.0
2017-03-07T00:39:02.693Z
2017-03-07T16:25:07.027Z
2017-03-07T16:25:07.027Z
ProfRob
<p>How do we know degenerate matter exists on white dwarfs? Is it purely hypothetical or have we observed it before? Have we ever created a form of degenerate matter on earth?</p>

<white-dwarf><degenerate-matter>
White Dwarf and Degenerate Matter
Aidan Powers
# White Dwarf and Degenerate Matter

(Asked by: username_0 on 2017-03-06)

How do we know degenerate matter exists on white dwarfs? Is it purely hypothetical or have we observed it before? Have we ever created a form of degenerate matter on earth?

**Tags:** white-dwarf, degenerate-matter


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## 1 answers

### Answer by username_1 on 2017-03-07. Score: 10

The density of white dwarfs is not hypothetical, it can be measured. The short answer is that the density is so high that a stable star can *only* be supported by electron degeneracy.

Sirius B is an example. The radius can be estimated by combining the luminosity of the white dwarf with its temperature estimated from spectroscopy. The mass can then be determined from the binary motion (with Sirius A), or gravitational redshift or from estimating the surface gravity from the width of absorption lines in its spectrum. These tell us that Sirius B has a radius of $0.0084 R\_{\odot}$ and a mass of about $1 M\_{\odot}$ (e.g. [Holberg et al. 2012](https://arxiv.org/abs/1201.3822)). This leads to an estimate of the *average* density of $2.4 \times 10^{9}$ kg/m$^3$.

Now, we know that for such an object to find any kind of equilibrium that the pressure must be greatest in the middle. Since for any plausible equation of state, the pressure increases with density, then the white dwarf must be *even* denser than the average at its centre.

From there it is a standard piece of bookwork to show why the white dwarf *must* be supported by electron degeneracy pressure and I will repeat it here.

An electron gas can only escape degeneracy *if* $E\_F - m\_e c^2 < k\_{b}T$, where $E\_F - m\_e c^2$ is the kinetic energy associated with the [Fermi energy](https://en.wikipedia.org/wiki/Fermi_energy) of the electrons, $m\_e$ is the electron mass and $T$ is the temperature.

The Fermi energy can be derived from the following formula (again, standard bookwork, this is not exotic physics, it is the same physics that describes how conduction takes place in metals).
$$ E\_F = \left[ \left(\frac{3h^3n\_e}{8\pi}\right)^{2/3}c^2 + m\_e^2 c^4 \right]^{1/2},$$
where $n\_e$ is the electron number density. Now it turns out that the temperatures we are going to need to avoid degeneracy (see below) are easily going to be sufficient to completely ionise the contents of the white dwarf. If we assume the white dwarf is made out of carbon or oxygen, then the relationship between mass density and $n\_e$ is just $\rho = 2 n\_e m\_u$, where $m\_u$ is an atomic mass unit (the 2 just comes from the fact that there are 2 mass units per free electron). We could mess about with the composition *a little* to change this relationship, but it doesn't matter very much.

Making the appropriate substitutions and using $\rho = 2.4 \times 10^{9}$ kg/m$^3$, we obtain $E\_F = 0.75$ MeV (or $1.2\times10^{-13}$ J). If we now use the test
$$ T > \frac{E\_F - m\_e ,c^2}{k\_b},$$
we find that to avoid electron degeneracy then the temperature inside the white dwarf must exceed $2.75 \times 10^{9}$ K (and be much higher in the centre of the white dwarf where the density and Fermi energy are larger). If temperatures were this high inside a white dwarf and it was made of something like carbon or oxygen then nuclear fusion would take place (rapidly) and the luminosity of the white dwarf would not be as low as it is.

But what if it were made of iron? There would be no nuclear fusion, even at these high temperatures. OK, leaving aside that if the interior really were that hot, it could not be hidden very easily by an atmosphere, then this would not be a stable situation. The star would be losing copious heat via neutrino losses. If the gas were non-degenerate, the star would rapidly contract and become even hotter and even denser, and if you are not going to allow for degeneracy (i.e. you postulate that quantum mechanics does not exist), then rather quickly the white dwarf ends up being a black hole.

A back of the envelope estimate would be to divide the gravitational potential energy by the current luminosity. Taking Sirius B as an example, the timescale is
$$ \tau \sim \frac{GM^2}{RL},$$
where $L$ is the luminosity. Taking the current luminosity from photons at the surface this timescale would be a few $10^{10}$ years (plausibly stable). But if the interior were at $\sim 10^{10}$ K, the neutrino losses would be many orders of magnitude greater and the timescale would be just millions of years.

We know that this isn't happening to white dwarfs because we know they are stable and long-lived. We see them in clusters of stars of known age and associated with stars that are much older than millions of years. For instance we know that stars like Sirius A are hundreds of millions of years old.
We can calculate using the theory of electron degenerate structures that the faintest white dwarfs we see have been cooling for 10 billion years without collapsing - precisely the age expected for the galactic disk. The cooling ages of white dwarfs in clusters match the ages of those clusters very well.

Degenerate matter is very common on earth. The [conduction electrons](http://farside.ph.utexas.edu/teaching/sm1/lectures/node86.html) in a metal form a degenerate gas. If you mean the kind of degenerate matter in white dwarfs, then no, at least not in any stable state. The pressures required to constrain such energy densities are too large to create in the lab.


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