When a Fermi gas of electrons (e.g. in a star) is squeezed by the gravity into a smaller volume, the Pauli exclusion principle requires the electrons to take higher energy levels. This raises the temperature and increases the speed of the electrons to relativistic limits.

Consider the star is under the Chandrasekhar limit and becomes stable in the long term with the fermionic pressure exactly balancing out the gravity. Assume the star has enough time to cool down (whatever billions or trillions of years necessary).

Could someone please clarify the following? On one hand, the Pauli exclusion principle requires the electrons to have high energies and therefore the star to have a high temperature. On the other hand, the star cannot possibly have a high temperature, because it has completely cooled down.

A hot temperature would imply a black body radiation, but radiation cannot last forever without a source of energy. What is the way to resolve the high temperature required by the Pauli exclusion principle against the fact that there is no source of energy to maintain the radiation in the long run?


The temperature does not depend on fermions occupying higher energy levels, it depends on the distribution of fermions across the possible temperature levels. A piece of metal cooled to absolute zeros still has all its electrons in states with energies up to the "Fermi level" which can be pretty high, and no electrons in states higher than the "Fermi level." Because this is the MINIMUM energy configuration possible, it is associated with zero temperature, with 0 K.

So the same would apply to a star.

https://en.wikipedia.org/wiki/Fermi_energy will give you a lot more on this, and states explicitly that a "Fermi gas" of electrons can be at zero temperature if all the electrons are occupying their lowest available energy states, and none are in any energy states higher when there is still an empty state at lower energy left empty.

  • $\begingroup$ Makes sense, thanks! Then why, for example, it is often stated that the universe was very hot at the moment of the Big Bang? Since the entropy increases in time, the entropy at the Big Bang must be minimal and likely zero thus implying the absolute zero temperature. Only the Fermi temperature would be very high (if fermions are created from the start), but very cold to the "touch". Does this make sense? $\endgroup$ – safesphere Sep 6 '17 at 20:50
  • $\begingroup$ Another related question for you :) What happens to the Fermi energy when fermions pair up to form a Bose condensate? Does the Fermi energy become the heat energy in this case? $\endgroup$ – safesphere Sep 6 '17 at 21:03
  • $\begingroup$ When fermions pair, their heat capacity is lowered even further. @safesphere $\endgroup$ – Rob Jeffries Sep 6 '17 at 21:31
  • $\begingroup$ @Rob Jeffries: Well, let me clarify my concern. Consider a fermionic gas is under a critical pressure where fermions are deeply relativistic and the Fermi energy is very high. Now assume these fermions pair up to form boson pairs. Bosons don't obey the Pauli exclusion pronciple and thus don't have the Fermi energy. So my question is, what happens to the kinetic energy of the fermions in the fermionic gas once they all have paired up as bosons? Would this energy (no matter how large or small) become the heat energy and radiate out? $\endgroup$ – safesphere Sep 6 '17 at 21:42
  • $\begingroup$ @safesphere The "pairing" energy may be radiated away, but this is very small compared with the Fermi energy, which in turn is very much larger than $kT$. The paired fermions are still close to the Fermi energy. $\endgroup$ – Rob Jeffries Sep 6 '17 at 22:10

Ideal electron degeneracy pressure does not depend on temperature at all. In the example you use, the thermal energy of the white dwarf is held by the non-degenerate carbon and oxygen ions; almost no thermal energy is in the electrons, which have a heat capacity close to zero.

Your statement that "the Pauli exclusion principle requires the electrons to have high energies and therefore the star to have a high temperature " is incorrect in the sense that in a highly degenerate gas there is no relationship between the maximum electron energy and the gas temperature; it only depends on the electron number density.

As the white dwarf cools, the ions lose thermal energy, but the internal energy and pressure exerted by the electrons hardly changes. The white dwarf can become very cold and still maintain the same radius.

  • $\begingroup$ Thanks Rob, I understand now that my logic was correct that the Fermi energy could not make a start hot "to the touch". Can you possibly clarify my Big Bang concern in my comment above? I'd appreciate it. $\endgroup$ – safesphere Sep 6 '17 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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