# Why is the core of a gas giant supported by electron degeneracy pressure instead of nuclear fusion?

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?

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.

• Thanks. But in a partially degenerate gas, will all the fermions still occupy the lowest energy levels up to the Fermi energy? – Sir Cumference May 31 '16 at 11:17
• 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 May 31 '16 at 11:43
• @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. tube.geogebra.org/m/… – ProfRob May 31 '16 at 16:20

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.

• 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 May 31 '16 at 2:30
• @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 May 31 '16 at 3:18
• So then what separates a Fermi gas from a normal gas? – Sir Cumference May 31 '16 at 3:23
• 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 May 31 '16 at 9:45
• 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 May 31 '16 at 16:36