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Temperature and spectral lines

  1. Jul 11, 2004 #1

    turin

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    I'm trying to understand this application of thermodynamics to astronomical inferrence. As I understand it, the temperature of a diffuse gas cloud in distant space can be inferred from the relative intensities of the spectral lines (correct?) which indicates the distribution of the population of excited states. However, until I uncovered this suggestion, I had accepted temperature to be proportional to the average kinetic energy of the particles (including rotational and vibrational) by definition (through Boltzmann's constant). So, which is it? Which defines the temperature: kinetic energy or internal energy (or a combination of both)?

    And, if it is kinetic energy, then does a cloud of gas moving through space have a higher temperature than an identical cloud of gas that is stationary; is temperature frame dependent?
     
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  3. Jul 11, 2004 #2

    Integral

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    Because the temperature (average kinetic energy) is higher more atoms will be in excited states due to more energetic collisions. Thus the spectral lines present will be an indication of the temperature.
     
  4. Jul 12, 2004 #3
    I think the definition is made up with the internal energy of the system which is rotational, vibrational energy of the particles and translational energy wrt the center of mass. To get the total energy one has to add [itex]P^2 / 2 M[/itex](total mass/momentum). Like this the definition is frame invariant.
     
  5. Jul 12, 2004 #4
    In what way ? Exactly what is the relationship between temperature and line intensities ?

    Higher temperature -> more high energy photons ?

    I guess it's more complicated than that ?
     
  6. Jul 13, 2004 #5
    when it is hotter the electrons are in higher energy levels. at these different (higher) energy levels different atomic transitions can occur. this gives the different spectral lines.
     
  7. Jul 13, 2004 #6
    Probably the best way to define temperature is the mean free energy per unit quantum, regardless of the nature of the quantum, which could be a free photon, electron, lattice exitation or internal vibrational mode.
     
  8. Jul 13, 2004 #7

    turin

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    What's "mean free energy?"
     
  9. Jul 13, 2004 #8

    Njorl

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    "mean" is essentially average here.

    Free energy is that part of its energy not bound up in being what it is. For solitary massive particles, don't include rest energy. For lattices, don't include binding energy of the lattice.

    Njorl
     
  10. Jul 13, 2004 #9

    Njorl

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    Oh, you can also see temperature distributions in the broadening of transition spectra. A hot gas will broaden a line by red-shifting and blue-shifting around a mean wavelength more than a cold gas will.

    You can also guage temperatures from absorption broadening - broadening of dark lines in a spectrum.

    Njorl
     
  11. Jul 13, 2004 #10
    Wouldn't the intensities of the spectral lines also be indicative of the amount of substance in the gas cloud?
     
  12. Jul 13, 2004 #11

    turin

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    How does this exclude the overall kinetic energy of the bulk (which is free energy by this definition as I understand it)? Doesn't this disallow the distribution of oribital energies as an indication of temperature?
     
  13. Jul 14, 2004 #12

    Njorl

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    When you consider the kinetic energy of the bulk, you have a choice of reference frames. You can zero the bulk kinetic energy by the judicious choice of frame, then attribute gross red or blue shifting to the moving observer.

    Njorl
     
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