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Thermal equilibrium in stars

  1. Aug 8, 2009 #1
    Hello physicists,

    I'm trying to understand thermal equilibrium, I used to know that a body is in thermal equilibrium if it has a uniform spatial temperature distribution.

    This definition couldn't hold up after I knew that stars emit black body radiation, i.e. stars are in thermal equilibrium, while we know that stars don't have a uniform distribution of temperature, like our sun, surface's temperature 6000, core more that 20x10^6.

    And so the question is, is there any more complicated thermodynamic or statistical definition for thermal equilibrium?

    Thanks you
     
  2. jcsd
  3. Aug 8, 2009 #2

    Ygggdrasil

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    I don't think the sun is in thermal equilibrium, but rather it operates in a relatively steady state. For a system at steady state, energy flows through the system, but the amount of energy that enters the system is equal to the amount of energy that leaves the system. This allows the properties of the system to remain constant in time, without having to be in equilibrium.
     
  4. Aug 8, 2009 #3
    All stars are in thermal equilibrium, this is a fact that astronomers never deny, and this is why we get black body radiation from the sun with highest intensity in the visible range.

    The definition of the steady state you're talking about is the concept I'm looking for, but I need some more systematic details, this steady state is called thermal equilibrium, but I still don't understand that mathematically, let's say in terms of Entropy or Gibbs free energy for example.

    Thank you for your try :)

    Any suggestions?
     
    Last edited: Aug 8, 2009
  5. Aug 8, 2009 #4

    Mapes

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    What these astronomers call "thermal equilibrium," most scientists would call "steady state." A star is not in thermal equilibrium in the sense that its temperature is uniform, its entropy is maximized, or its energy or Gibbs free energy is minimized. It's a terminology difference only.

    In steady state (called "thermal equilibrium" in this context), the energy rate entering a region equals the energy rate leaving a region. The energy storage term is zero by definition (for stars, we must specify the thermal energy storage, since the nuclear energy storage is changing). The thermal energy is constant, but is not necessarily minimized or maximized.
     
  6. Aug 8, 2009 #5

    Ygggdrasil

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    Mathematically, most systems in steady state are described by a set of rate equations (essentially a system of differential equations). These rate equations describe the change of numbers of particles in each state with respect to time. At steady state, these rates of change are set to zero (i.e. they do not change with time).

    A good example of a mathematical treatment of a system in steady state is a laser. Unfortunately, I cannot find a good reference online, but you can look up the laser rate equations in an advanced optics text.

    Note that thermodynamics really only applies to systems at equilibrium. Very often, applying thermodynamic concepts (i.e. entropy, free energy) to systems out of equilibrium can be problematic. For example, the basis of a laser is population inversion: a case where a higher energy level has a larger population than a lower energy level. If one were to apply the thermodynamic definition of temperature to this system, one would find that a system in population inversion has a negative temperature!
     
  7. Aug 9, 2009 #6
    Well, then let me rephrase the question. Why do stars show black bodies' radiation while they have non-uniform temperature, while the black body's radiation is for bodies in thermal equilibrium? This definitely means that there is a different definition for thermal equilibrium.

    Do we say that if the entropy is maximised along the whole body (spatially) then it's in thermal equilibrium?
    Please don't give me the answer unless you're sure, because this answer may be involved in an oral exam.

    Thank you
     
  8. Aug 10, 2009 #7
    Oh my god! in the whole forum no one can say what thermal equilibrium is!?!?? come on guys!
     
  9. Aug 10, 2009 #8

    Vanadium 50

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    But their surfaces do have (almost - ignoring things like sunspots) uniform temperature. It's the surface that's radiating.

    Then shouldn't this be in homework/coursework?
     
  10. Aug 10, 2009 #9

    vanesch

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    Things like stars are in a good approximation in *local* thermal equilibrium. That means that, if you cut out sufficiently small pieces of star, they can approximatively be considered to be in thermal equilibrium within that box.

    What counts for black body radiation, is that the "emitting part" is in thermal equilibrium. What is emitting a star's radiation (mostly) is the outermost layers of gas, and in as much as these are in thermal equilibrium, you can assume that they emit black body radiation. That is, the depth from which radiation can escape directly into space from the star, is small enough, the shell of gas is thin enough, that we don't have to consider temperature variations over this thickness.

    Radiation from deeper, and hotter, layers, cannot escape directly into outer space, because the layers on top of them are opaque. So this radiation doesn't contribute to the seen radiation at the outside, but will rather heat the upper layers.
     
  11. Aug 10, 2009 #10
    http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/thereq.html#c2

    In short if there is no net heat transfer between two systems they are thermal equilibrium.
    The Zeroth Law takes this a bit further to say if two objects A and C are in thermal equilibrium with a third object C, then they would be in thermal equilibrium with each other i.e thermal equilibrium exists between A+C and B+C as well as A+B

    As for a star my understanding when you start talking about Gibbs free energy and entropy you are starting to discuss thermodymanic equilibrium. Thermodymanic equilibrium exists when the a thermal equilibrium exists (no heat transfer), when a mechanical equilibrium exists (pressure in the two systems are the same) and when the chemical potentials of the two systems are the same. Obviously more than two systems can be involved.

    So for a sun earth system, thermal equilibrium does not exist as their is heat transfer from the sun to the earth after all we can see the sun, that would not be possible if the Earth was in thermal equilibrium with the Sun. The Sun is not in thermal equilibrium with itself either as there is a temperature gradient between the core and surface which means convection occurs which is useful work.

    http://spiff.rit.edu/classes/phys440/lectures/lte/lte.html

    Take the photosphere, density is 2.5 g of hydrogen per/cc this give a mean free path of 1.5X10-4 m. With a box of 1.4x10-4 m there are 2.2x1019 hydrogen atoms. This volume should be in thermal equilibrium i.e no heat transfer with that volume. And will be probably in equilibrium with a neighbouring volume.
     
  12. Aug 10, 2009 #11
    Thank you guys, I got the answer. What I understood from all of you is that it's because only the surface contributes in black body radiations, while inner shells' radiations damp easily while trying to escape, this makes sense for me because when we use black body's radiation law we only consider the surface's temperature. If there is anything to add, or any further discussion, would please me so much :)

    Thanks to all of you :)

    and about not putting this in homework/coursework, is that because I'm a master student not a bachelor or high-school student, and I need a scientific discussion, not purely an answer, and as I could see my question could not have been answered that fast ;).

    Thanks again :)
     
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