3 ms·

Something I've been wondering. Does the earth's core have negative heat capacity?[1] I've read that it is usually modeled as a self-gravitating system[2], and

• pdonis 10y ago
  If you look at the Wikipedia article in more detail, you will see that the negative heat capacity of a self-gravitating, isolated system arises from the virial theorem. Heuristically, as a self-gravitating system radiates energy away into space, it becomes more tightly bound; that means it contracts. But as it contracts, the "orbital speeds" of its constituents increase (because decreasing orbital radius means increasing orbital speed), meaning its average kinetic energy increases, meaning it heats up. Technically, the virial theorem argument only applies to systems whose constituents are in free-fall orbits. But I believe you can do a similar analysis for an isolated self-gravitating system that is in hydrostatic equilibrium--basically, as it radiates energy away into space, the hydrostatic equilibrium changes to one in which the radius is smaller and pressure and temperature are higher. Unfortunately I don't have a handy link to such an analysis right now, though. As far as whether any of the above would apply to the Earth's core, I don't think it would since the Earth's core is not isolated; but it might apply to the Earth as a whole, to the extent that the Earth can be viewed as being in hydrostatic equilibrium (i.e., pressure balancing gravity everywhere). AFAIK it is common in astronomy to treat stars this way.
  • nonbel 10y ago
    Interesting, thanks. I still am not sure what the technical meaning of "isolated" is here. But wouldn't such an object continue heating up "infinitely" (maybe stopping at some phase-transition temperature) without constant input of energy?
    • deleted 10y ago
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    • pdonis 10y ago
      > I still am not sure what the technical meaning of "isolated" is here. Technically, it means an object of finite spatial extent surrounded by vacuum, and nothing else in the entire universe. Obviously that's an idealization. :-) But it makes the mathematical model tractable. > wouldn't such an object continue heating up "infinitely" Not necessarily; it might reach a stable state where it can't radiate any more energy. Cold white dwarfs, cold neutron stars, and cold planets are possible examples of such states. ("Cold" here basically means "at absolute zero", i.e., in the ground state for its configuration of particles.)
      • evanb 10y ago
        "Cold" in astrophysical settings tends to mean T << µ (the chemical potential) which usually justifies a low temperature expansion in T/µ, although thermal effects can be qualitatively important.
        • pdonis 10y ago
          > "Cold" in astrophysical settings tends to mean T << µ (the chemical potential) Yes, I know, but this in itself does not preclude energy being emitted by radiation. To preclude that, the object needs to be in a genuine ground state--no internal transitions possible that can reduce its total energy. That's a considerably stronger condition than T << µ.
      • hermitdev 10y ago
        > Technically, it means an object of finite spatial extent surrounded by vacuum, and nothing else in the entire universe. Obviously that's an idealization. Reminds me of Fritz Zwicky's "spherical bastards". Although unrelated, my physics prof that taught thermo introduced spherical bastards as an aside after stating we'd basically assume all bodies are spherical when calculating thermo transfers. This prof was where I learned that Zwicky often referred to some colleagues as "spherical bastards", and when when questioned why, he responded: "because no which way you look at them, they're a bastard." I'm sad I can't remember the prof's name; he was quite a character, a short, squat old man. Just add a beard and appropriate clothing, he'd have played a good Santa Clause. He had a minor role in the Manhattan Project, and would often share stories. Lectures were always fun. One of the few classes I never skipped.