<blockquote>
  <p>So then why would gas giant cores be electron degenerate, but not become hot enough for nuclear fusion?</p>
</blockquote>

<p>Degeneracy isn't an on/off switch. It's a quantum mechanical aspect of pressure that is always present, just as is thermal pressure. A substance is highly degenerate if degeneracy pressure completely dominates over thermal pressure, non-degenerate if degeneracy pressure is negligible compared to thermal pressure.</p>

<blockquote>
  <p>Shouldn't they start fusing before they can become electron degenerate, as stars do?</p>
</blockquote>

<p>While gas giants are too small for gravitational collapse to heat the core to a temperature sufficient to result in fusion, they're not too small for the gravitational collapse to make degeneracy pressure a very significant aspect of pressure. The density at the center of Jupiter is about half that at the center of the Sun, but the temperature is only about 1/600 of that at the center of the Sun. This means that degeneracy pressure dominates at the center of Jupiter but that thermal pressure dominates at the center of the Sun.</p>

CC BY-SA 3.0
2016-05-31T02:30:14.917Z
How is degeneracy pressure always prevalent? Isn't a degenerate gas, by definition, when all the particles are occupying the lowest possible energy states up to the Fermi Energy?
Sir Cumference
CC BY-SA 3.0
2016-05-31T03:18:04.800Z
@SirCumference -- That's the pure fiction of a fully degenerate gas, which only occurs when the gas is at absolute zero. Nothing real ever reaches at absolute zero, and no real gas is a gas at absolute zero.
David Hammen
CC BY-SA 3.0
2016-05-31T03:23:37.000Z
So then what separates a Fermi gas from a normal gas?
Sir Cumference
CC BY-SA 3.0
2016-05-31T09:45:30.377Z
It's a difference of degree rather than of kind, the same as is the case with a plasma. A Fermi gas is one in which the Fermi energy $E_f$ is many, many orders of magnitude larger than the thermal energy $kT$. In such a case, essentially all of the pressure is from degeneracy rather than from temperature. The Fermi energy is ever present, just as is thermal energy (we have yet to take a substance down to absolute zero).
David Hammen
CC BY-SA 3.0
2016-05-31T16:36:12.417Z
I dislike drawing a distinction between "thermal pressure" and degeneracy pressure. There is only one pressure and it is due to the momentum distribution of the particles. Also, $kT$ isn't the thermal energy (or even close) if there is even a hint of degeneracy. A Fermi gas is a gas of ideal fermions.
ProfRob
CC BY-SA 3.0
2016-05-30T22:59:16.513Z
2016-05-30T22:59:16.513Z
David Hammen
<p>I will take the great examples of the well-known objects in the Universe to demonstrate an example to answer your question: The Sun and Jupiter.</p>
<h1> The Sun </h1>
<blockquote>
<p>However, as the core became denser and denser, it reached temperatures
at which fusion could take place, but it did not become dense enough
for the matter to become electron degenerate, right? Only later in its
life, when the star's hydrogen ran out, would it become dense enough
for that.</p>
</blockquote>
<p>That is because, despite the fact that the Sun's core is nearly a tenth of a tonne per m<sup>3</sup> dense (150 g/cm<sup>3</sup>), the Sun's core is still balanced outwards by thermal pressure. The Sun fuses hydrogen to helium, and that liberates a lot of energy equivalent to nearly a trillion H-bombs going off every second. This exerts tremendous amounts of upward pressure, against the enormous gravity of the star.</p>
<p>You do not actually need electron degeneracy pressure to resist the gravity of the star. It's like gilding the lily, or in physics term, mixing uranium with antimatter. It's useless.</p>
<p>Only later in the Sun's life, when the hydrogen runs out and the energetic Triple-Alpha process takes over, then the core of the Red Giant Sun collapses into a electron-degenerate white-dwarf seed with a density of 1000 tons/m<sup>3</sup>. Despite this, the core still has a lot of helium and some carbon in it. The Triple-Alpha process liberates a ridiculously high amount of energy, nearly as much as the CNO cycle that goes on in massive stars. <a href="https://i.sstatic.net/VGKRc.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VGKRc.png" alt="Image credit-Wikipedia" /></a></p>
<p>This means that while the core will be extremely dense (though nowhere near as dense as a neutron star), it would be so hot, that it would cause the surrounding material to swell up and become puffy, a Red-Giant. This <strong>white-dwarf seed</strong> is so hot that the Sun actually manages to fuse some carbon into oxygen, via alpha capture, where a carbon nucleus captures a helium nucleus and turns to oxygen. However this process is extremely slow as the Sun is simply not massive enough to fuse it on a large scale, it merely occurs as a trace reaction.</p>
<p>However let's move onto something that's dull and unimpressive compared to stars- Gas Giants</p>
<h1> Jupiter </h1>
<blockquote>
<p>So then why would gas giant cores be electron degenerate, but not
become hot enough for nuclear fusion? Shouldn't they start fusing
before they can become electron degenerate, as stars do? Am I
misunderstanding this concept entirely, or is there more to it than
what I described?</p>
</blockquote>
<p>Gas giants are called failed stars for a reason. They are simply not massive enough to fuse hydrogen to helium, you would need a core temperature of at least 3 million kelvins to initiate a proper fusion reaction.</p>
<p>However, do not underestimate gas giants. They are also called <strong>giants</strong> for a reason. Jupiter is literally so massive, that its core is crushed to a ball that is 20 times more massive than the Earth, yet has only 2 times the diameter. It is more denser than osmium (25,000 kg per cubic meter). This means that despite the fact that the core does not have the conditions adequate for fusion to occur, it is so dense that there has to be something that is exerting pressure from the outside. And it is electron degeneracy.</p>
<p>Put simply, Jupiter's core is a miniaturized version of a white dwarf, it is incredibly dense <em>(though as dense as cotton candy to a white-dwarf)</em> and incredibly hot <em>(though it is next to freezing to a white dwarf)</em>. The core is crushed to such degrees that there has to be a upward push to prevent Jupiter's core from just turning into a black hole or something of that sort.</p>
<p>Basically, the Pauli Exclusion principle forbids two electron with the same spin to occupy the same &quot;quantum state&quot;, which is fancy word for &quot;place&quot;. So this principle is responsible for providing the degeneracy pressure that exerts a counterpressure against the massive gravity of Jupiter.</p>
<p>TL;DR</p>
<p>Stars are supported by thermal pressure as there is a lot of energy from fusing atoms together.</p>
<p>Gas giants are supported by electron degeneracy as they don't have anything else to hold them up.</p>

CC BY-SA 4.0
2022-10-27T14:36:27.960Z
@ProfRob edited
Arcturus
CC BY-SA 4.0
2022-10-27T15:45:18.450Z
Better, but equating "quantum state" with "place" is incorrect. The PEP does not forbid as many electrons as you like being in the same "place" and having arbitrarily high densities.  As I commented on David Hammen's answer - "thermal pressure" and "degeneracy pressure" aren't two different things. They are the same phenomenon in two different regions of the density/temperature plane.
ProfRob
CC BY-SA 4.0
2022-10-26T16:01:44.477Z
2022-10-27T14:36:08.153Z
2022-10-27T14:36:08.153Z
Arcturus
<p>The test to see whether degeneracy pressure is going to be significant is to compare $kT$ with the Fermi energy $E_F$ </p>

<p>The Fermi energy is the energy level up to which all energy states would be occupied in a completely degenerate fermion gas. It is given by (for non-relativistic conditions)
$$ E_F = \frac{h^2}{2m}\left(\frac{3}{8\pi}\right)^{2/3} n^{2/3},$$
where $m$ and $n$ are the mass and number density of the fermions (in this case, electrons).</p>

<p>If $E_F \gg kT$, then the gas can be considered completely degenerate and the electrons exert a non-relativistic degeneracy pressure. If $E_F \sim 10 kT$ the electrons are partially degenerate.
In <em>both</em> these circumstances, the pressure exerted by the electrons is much higher than they would have in a perfect gas, because a large fraction of electrons occupy high energy (and momentum) states. </p>

<p>Degeneracy can be achieved either by having a low temperature or a high number density of fermions. In a white dwarf, electrons are completely degenerate because the number density of electrons is extremely high. In a gas giant, the electron density is nowhere near as high, but then the temperature is a lot lower and so they reach a state of partial degeneracy.</p>

<p>The pressure of a partially degenerate gas is higher than that of a perfect gas at the same density and temperature. More importantly, the pressure is only very weakly dependent on temperature. As a gas giant radiates away its potential energy, it is able to do so whilst only contracting very slightly and the core does not become hot enough to initiate nuclear fusion.</p>

<p>Note that degeneracy pressure and thermal pressure are not two different things. Both arise merely as a consequence of particles having a distribution of momenta and obeying (in this case) Fermi-Dirac statistics to determine how they occupy the possible energy states. The expression $P = nkT$ is simply a convenient approximation for fermions that holds only when $E_F \ll kT$ and the quantum nature of the particles is not apparent.</p>

CC BY-SA 3.0
2016-05-31T11:17:51.270Z
Thanks. But in a partially degenerate gas, will all the fermions still occupy the lowest energy levels up to the Fermi energy?
Sir Cumference
CC BY-SA 3.0
2016-05-31T11:43:49.380Z
Also, you said degeneracy can be achieved by low temperatures, but it is also weakly dependent on temperature. So if matter becomes degenerate in a low-temperature environment, and we were to raise the temperature to 2000°K, would it still remain degenerate?
Sir Cumference
CC BY-SA 3.0
2016-05-31T16:20:26.647Z
@SirCumference You need to read a textbook on statistical physics. In the meantime play with this applet I wrote. It is for white dwarfs but the slider ranges encompass giant planet conditions. The energies in this app include rest mass energy. https://tube.geogebra.org/m/rWmcMENt?doneurl=%2Fsearch%2Fperform%2Fsearch%2Frobdjeff#material/ACZt5h7a
ProfRob
CC BY-SA 3.0
2016-05-31T08:06:58.980Z
2016-05-31T16:38:13.103Z
2016-05-31T16:38:13.103Z
ProfRob
<p>After a Sun-sized protostar forms, its core will become denser over time due to radiation. The core eventually gets dense and hot enough for hydrogen fusion to take place. In the late phases of the star's life, the core will continue fusing hydrogen until it runs out and outer layers begin fusion hydrogen, while dumping helium into the core. The core will get massive enough that it will begin to contract, and it will eventually become electron degenerate due to the high densities.</p>

<p>However, as the core became denser and denser, it reached temperatures at which fusion could take place, but it did <em>not</em> become dense enough for the matter to become electron degenerate, right? Only later in its life, when the star's hydrogen ran out, would it become dense enough for that.</p>

<p>So then why would gas giant cores be electron degenerate, but not become hot enough for nuclear fusion? Shouldn't they start fusing before they can become electron degenerate, as stars do? Am I misunderstanding this concept entirely, or is there more to it than what I described?</p>

<star><gas-giants><density><protostar><degenerate-matter>
Why is the core of a gas giant supported by electron degeneracy pressure instead of nuclear fusion?
Sir Cumference
# Why is the core of a gas giant supported by electron degeneracy pressure instead of nuclear fusion?

(Asked by: username_0 on 2016-05-30)

After a Sun-sized protostar forms, its core will become denser over time due to radiation. The core eventually gets dense and hot enough for hydrogen fusion to take place. In the late phases of the star's life, the core will continue fusing hydrogen until it runs out and outer layers begin fusion hydrogen, while dumping helium into the core. The core will get massive enough that it will begin to contract, and it will eventually become electron degenerate due to the high densities.

However, as the core became denser and denser, it reached temperatures at which fusion could take place, but it did *not* become dense enough for the matter to become electron degenerate, right? Only later in its life, when the star's hydrogen ran out, would it become dense enough for that.

So then why would gas giant cores be electron degenerate, but not become hot enough for nuclear fusion? Shouldn't they start fusing before they can become electron degenerate, as stars do? Am I misunderstanding this concept entirely, or is there more to it than what I described?

**Tags:** star, gas-giants, density, protostar, degenerate-matter


---

## 3 answers

### Answer by username_2 on 2016-05-31. Score: 10 | Accepted answer

The test to see whether degeneracy pressure is going to be significant is to compare $kT$ with the Fermi energy $E\_F$

The Fermi energy is the energy level up to which all energy states would be occupied in a completely degenerate fermion gas. It is given by (for non-relativistic conditions)
$$ E\_F = \frac{h^2}{2m}\left(\frac{3}{8\pi}\right)^{2/3} n^{2/3},$$
where $m$ and $n$ are the mass and number density of the fermions (in this case, electrons).

If $E\_F \gg kT$, then the gas can be considered completely degenerate and the electrons exert a non-relativistic degeneracy pressure. If $E\_F \sim 10 kT$ the electrons are partially degenerate.
In *both* these circumstances, the pressure exerted by the electrons is much higher than they would have in a perfect gas, because a large fraction of electrons occupy high energy (and momentum) states.

Degeneracy can be achieved either by having a low temperature or a high number density of fermions. In a white dwarf, electrons are completely degenerate because the number density of electrons is extremely high. In a gas giant, the electron density is nowhere near as high, but then the temperature is a lot lower and so they reach a state of partial degeneracy.

The pressure of a partially degenerate gas is higher than that of a perfect gas at the same density and temperature. More importantly, the pressure is only very weakly dependent on temperature. As a gas giant radiates away its potential energy, it is able to do so whilst only contracting very slightly and the core does not become hot enough to initiate nuclear fusion.

Note that degeneracy pressure and thermal pressure are not two different things. Both arise merely as a consequence of particles having a distribution of momenta and obeying (in this case) Fermi-Dirac statistics to determine how they occupy the possible energy states. The expression $P = nkT$ is simply a convenient approximation for fermions that holds only when $E\_F \ll kT$ and the quantum nature of the particles is not apparent.

> **Comments:**
> **username_0**: Thanks. But in a partially degenerate gas, will all the fermions still occupy the lowest energy levels up to the Fermi energy?  
> **username_0**: Also, you said degeneracy can be achieved by low temperatures, but it is also weakly dependent on temperature. So if matter becomes degenerate in a low-temperature environment, and we were to raise the temperature to 2000°K, would it still remain degenerate?  
> **username_2**: @username_0 You need to read a textbook on statistical physics. In the meantime play with this applet I wrote. It is for white dwarfs but the slider ranges encompass giant planet conditions. The energies in this app include rest mass energy. https://tube.geogebra.org/m/rWmcMENt?doneurl=%2Fsearch%2Fperform%2Fsearch%2Frobdjeff#material/ACZt5h7a  

---

### Answer by username_1 on 2016-05-30. Score: 3

> So then why would gas giant cores be electron degenerate, but not become hot enough for nuclear fusion?

Degeneracy isn't an on/off switch. It's a quantum mechanical aspect of pressure that is always present, just as is thermal pressure. A substance is highly degenerate if degeneracy pressure completely dominates over thermal pressure, non-degenerate if degeneracy pressure is negligible compared to thermal pressure.

> Shouldn't they start fusing before they can become electron degenerate, as stars do?

While gas giants are too small for gravitational collapse to heat the core to a temperature sufficient to result in fusion, they're not too small for the gravitational collapse to make degeneracy pressure a very significant aspect of pressure. The density at the center of Jupiter is about half that at the center of the Sun, but the temperature is only about 1/600 of that at the center of the Sun. This means that degeneracy pressure dominates at the center of Jupiter but that thermal pressure dominates at the center of the Sun.

> **Comments:**
> **username_0**: How is degeneracy pressure always prevalent? Isn't a degenerate gas, by definition, when all the particles are occupying the lowest possible energy states up to the Fermi Energy?  
> **username_1**: @username_0 -- That's the pure fiction of a fully degenerate gas, which only occurs when the gas is at absolute zero. Nothing real ever reaches at absolute zero, and no real gas is a gas at absolute zero.  
> **username_0**: So then what separates a Fermi gas from a normal gas?  
> **username_1**: It's a difference of degree rather than of kind, the same as is the case with a plasma. A Fermi gas is one in which the Fermi energy $E_f$ is many, many orders of magnitude larger than the thermal energy $kT$. In such a case, essentially all of the pressure is from degeneracy rather than from temperature. The Fermi energy is ever present, just as is thermal energy (we have yet to take a substance down to absolute zero).  
> **username_2**: I dislike drawing a distinction between "thermal pressure" and degeneracy pressure. There is only one pressure and it is due to the momentum distribution of the particles. Also, $kT$ isn't the thermal energy (or even close) if there is even a hint of degeneracy. A Fermi gas is a gas of ideal fermions.  

---

### Answer by username_3 on 2022-10-26. Score: 0

I will take the great examples of the well-known objects in the Universe to demonstrate an example to answer your question: The Sun and Jupiter.

# The Sun

> However, as the core became denser and denser, it reached temperatures
> at which fusion could take place, but it did not become dense enough
> for the matter to become electron degenerate, right? Only later in its
> life, when the star's hydrogen ran out, would it become dense enough
> for that.

That is because, despite the fact that the Sun's core is nearly a tenth of a tonne per m3 dense (150 g/cm3), the Sun's core is still balanced outwards by thermal pressure. The Sun fuses hydrogen to helium, and that liberates a lot of energy equivalent to nearly a trillion H-bombs going off every second. This exerts tremendous amounts of upward pressure, against the enormous gravity of the star.

You do not actually need electron degeneracy pressure to resist the gravity of the star. It's like gilding the lily, or in physics term, mixing uranium with antimatter. It's useless.

Only later in the Sun's life, when the hydrogen runs out and the energetic Triple-Alpha process takes over, then the core of the Red Giant Sun collapses into a electron-degenerate white-dwarf seed with a density of 1000 tons/m3. Despite this, the core still has a lot of helium and some carbon in it. The Triple-Alpha process liberates a ridiculously high amount of energy, nearly as much as the CNO cycle that goes on in massive stars. [![Image credit-Wikipedia](https://i.sstatic.net/VGKRc.png)](https://i.sstatic.net/VGKRc.png)

This means that while the core will be extremely dense (though nowhere near as dense as a neutron star), it would be so hot, that it would cause the surrounding material to swell up and become puffy, a Red-Giant. This **white-dwarf seed** is so hot that the Sun actually manages to fuse some carbon into oxygen, via alpha capture, where a carbon nucleus captures a helium nucleus and turns to oxygen. However this process is extremely slow as the Sun is simply not massive enough to fuse it on a large scale, it merely occurs as a trace reaction.

However let's move onto something that's dull and unimpressive compared to stars- Gas Giants

# Jupiter

> So then why would gas giant cores be electron degenerate, but not
> become hot enough for nuclear fusion? Shouldn't they start fusing
> before they can become electron degenerate, as stars do? Am I
> misunderstanding this concept entirely, or is there more to it than
> what I described?

Gas giants are called failed stars for a reason. They are simply not massive enough to fuse hydrogen to helium, you would need a core temperature of at least 3 million kelvins to initiate a proper fusion reaction.

However, do not underestimate gas giants. They are also called **giants** for a reason. Jupiter is literally so massive, that its core is crushed to a ball that is 20 times more massive than the Earth, yet has only 2 times the diameter. It is more denser than osmium (25,000 kg per cubic meter). This means that despite the fact that the core does not have the conditions adequate for fusion to occur, it is so dense that there has to be something that is exerting pressure from the outside. And it is electron degeneracy.

Put simply, Jupiter's core is a miniaturized version of a white dwarf, it is incredibly dense *(though as dense as cotton candy to a white-dwarf)* and incredibly hot *(though it is next to freezing to a white dwarf)*. The core is crushed to such degrees that there has to be a upward push to prevent Jupiter's core from just turning into a black hole or something of that sort.

Basically, the Pauli Exclusion principle forbids two electron with the same spin to occupy the same "quantum state", which is fancy word for "place". So this principle is responsible for providing the degeneracy pressure that exerts a counterpressure against the massive gravity of Jupiter.

TL;DR

Stars are supported by thermal pressure as there is a lot of energy from fusing atoms together.

Gas giants are supported by electron degeneracy as they don't have anything else to hold them up.

> **Comments:**
> **username_3**: @username_2 edited  
> **username_2**: Better, but equating "quantum state" with "place" is incorrect. The PEP does not forbid as many electrons as you like being in the same "place" and having arbitrarily high densities.  As I commented on username_1 username_1's answer - "thermal pressure" and "degeneracy pressure" aren't two different things. They are the same phenomenon in two different regions of the density/temperature plane.  

---

