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just want to try to clarify a few things about the nature of White Dwarfs and degenerate gases.

So if a white dwarf is help up by an electron degenerate gas, the electrons are unable to lose energy by emitting photons, correct? They basically are trapped at whatever energy level they occupy and that's that. The nuclei that make up the majority of the mass of the star are still constituents of a "normal" gas however, correct? This means that these nuclei and emit photons to lose kinetic energy? How does that work exactly? I understand that electron can emit photos inside an atom to drop to lower energy levels and that they can emit photons of any energy level if they are free (particularly if they're in a magnetic field, synchroton radiation, yeah?). To protons and neutrons do something similar? I ask because I never read anything about protons/nuclei emitting photons. If this basic picture is correct, is this the mechanism by which a white dwarf cools down over time?

Final bit, since most white dwarfs are quite hot, the peak wavelength is going to be in the UV. If I wanted to determine the percentage of such a star's luminosity that is given off in the optical, do I just need to integrate Planck's blackbody equation over the range of wavelengths I'm interested in and then compare that to the total luminosity given by $L = 4\pi R^2 \sigma T^4$?

Thanks and Cheers!

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The main method by which heat is transferred in a white dwarf is thermal conduction. Degenerate electrons have very long mean free paths and can conduct heat excellently.

White dwarfs lose energy by emitting photons from their outer, non-degenerate surface layers. They may also emit neutrinos from their interiors when they are really young and hot.

Gamma rays and X-rays are of course emitted by nuclei. This works just the same as atomic emission - charged particles (protons) change energy levels within the nucleus.

Most white dwarfs are not hot, they are similar to the Sun or cooler. Hot ones are easier to see, but only young white dwarfs are hot. But yes, your scheme is approximately (in that white dwarfs are not blackbodies) correct.

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  • $\begingroup$ Thx m8. Appreciate the info. $\endgroup$ Commented Feb 18, 2019 at 21:38

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