0
$\begingroup$

Does Planck’s law say -or imply- that (A) there are increasingly more energy levels for fermions to occupy per unit energy interval at higher temperatures so that (B) it takes increasingly more energy to increase the temperature of a star or blackbody with one degree?

$\endgroup$
0
$\begingroup$

Fermi gases behave just like classical ideal gases if the temperature is high enough to make them non-degenerate, i.e., occupation numbers << 1. The same is true of Bose gases if the number of particles is constrained. One cannot say the same for a photon gas, because states with energy < kT will have high occupation numbers.

You can assume that the heat capacity of non-degenerate plasma is ${{C}_{V}}=\tfrac{3}{2}R$ or ${{C}_{P}}=\tfrac{5}{2}R$, per mole of independent particles.

$\endgroup$
  • $\begingroup$ This is to technical for me. If there would be no limit to the mass of a star made out of a single, unspecified particle species (fermions) -no limit to the number of massive particles it contains, then I assume that it's temperature and density increases as the star is more massive. Am I right in assuming that the energy of an increasingly larger part of its particles converges to within a smaller range of (higher) frequencies -so it takes increasingly more energy to raise the temperature of the star by one degree? $\endgroup$ – Anton Mar 30 '18 at 11:43
  • $\begingroup$ What I mean is, do we need increasingly more decimals to express the energy difference between successive (discrete) enegy levels the particles can occupy? $\endgroup$ – Anton Mar 30 '18 at 11:47
  • $\begingroup$ The energy levels are the same regardless of whether it's a Fermi gas or classical, but the exclusion principle makes a difference. If the Fermi gas is degenerate at low T, the only likely excitation involves promotion from a filled state just below the Fermi surface to an empty state just above it. The heat capacity will scale as T^1. In the high-T case, the heat capacity levels off at the values mentioned. $\endgroup$ – Bert Barrois Mar 30 '18 at 14:32
  • $\begingroup$ Looking at the blackbody curve as plotted from Planck's law (Wikipedia) it looks like the curve becomes ever-higher and narrower at higher and higher temperatures, so wouldn't this fit my assumption as described in my question and comments? $\endgroup$ – Anton Apr 1 '18 at 11:34
  • $\begingroup$ No, absolutely not. First of all, Planck's law applies to blackbody radiation, which is a Bose gas. The shape of the distribution does not change with temperature, if you rescale the axes appropriately. $\endgroup$ – Bert Barrois Apr 1 '18 at 13:05

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.