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Temperature measured from Wien's Displacement law

  1. Jul 8, 2008 #1
    Can anyone please tell me whether the temperature measured from Wien's Displacement law/Planck's Radiation law/Stefan-Boltzmann law is the same(in every context) as is defined by the zeroth law of thermodynamics?? If not,then how and where do they differ???
    Moreover,how's brightness temperature related to that temperature as is defined by Zeroth law??
    Any relevant books/papers dealing extensively with these topics??
    Any help would be greatly appreciated!!!
     
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  3. Jul 8, 2008 #2

    Andy Resnick

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    Re: Temperature

    We are having a similar discussion on a different thread; my answer is that temeprature is defined only for a particular kind of radiation. That is, the radiation has a very specific spectral property.

    The zeroth law defines temperature in the same way length is defined as something that's measured by a ruler. That is, the zeroth law posits the existence of thermometers which can compare how hot two objects are.

    If someone can provide a decent text, I'd be interested as well.
     
  4. Jul 13, 2008 #3
    Re: Temperature

    hmm i thought of temperature of a body as "something" that exclusively depends on the avg kinetic energy of all the particles constitutiinn the body as seen from the centre of mass frame of refrence
     
  5. Jul 13, 2008 #4

    Andy Resnick

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    Re: Temperature

    That's the interpretation of temperature as per statistical mechanics. That definition apparently leaves open the question on defining temperature for continua, although it is possible to define a equilibrium temperature for a very specific spectrum of electromagnetic radiation. I do not know if it is possible to assign a temperature to an arbitrary spectrum of light.
     
  6. Jul 17, 2008 #5
    Re: Temperature

    But is the Brightness Temperature of a Black Body equal to it's surface temperature??
    If not,how are they related??
    Plz explain..
     
  7. Jul 17, 2008 #6
    Re: Temperature

    Let us assume that we have two ideal blackbodies. Let us also assume that, by the zeroth law, they are in thermal equilibrium. Then by the (Planck/Wien/Stefan-Boltzmann) law, these two bodies must exhibit the same (emission spectra/peak spectral radiance/total irradiance).

    Does this sufficiently answer the question?

    My (limited) understanding of brightness temperature is that it is invoked primarily in observational astronomy and that it is more of an "effective" temperature. The concept is then useful for comparison/categorization of celestial objects. So if this is the case, it would seem to me that this "effective" temperature would not give straight forward connections to thermodynamics if at all.

    Again, this is based on my limited understanding of astronomical methods; please do not hold it as fact unless confirmed.
     
  8. Jul 17, 2008 #7
    Re: Temperature

    What we always have to remember is that there are essentially two kinds of EM radiation: incandescent (i.e. thermal radiation which is usually approximated as blackbody) and luminescent (i.e. non-thermal radiation, e.g. lasers and LEDs).

    I'm not sure if it is still the case, but the literature did refer to "effective" temperatures of luminescent radiation treated as blackbody radiation. One peculiarity that falls out of this treatment is that it is possible to have an effective absolute temperature less than 0.

    I am actually currently reading a very interesting paper on just this subject from the 80's. The author treats luminescent radiation thermodynamically by scrutinizing the chemical potential of that radiation:

    P. Wurfel, J. Phys. C, 15 (1982) 3967.
     
  9. Jul 17, 2008 #8

    Andy Resnick

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    Re: Temperature

    There are lots of ways to define a radiometric temperature- radiation temperature, brightness/radiance temperature, ratio temp, distribution temp, color temp, and effective temperature. Here's some of the defintions:

    Radiation temperature: The temperature of a blackbody that gives the same *total radiance* as the body in question.
    Radiance temperature: The temperature of a blackbody that gives the same radiance as the body in question *at a defined wavelength*.
    Ratio temperature: The temperature of a blackbody that has the same *ratio of radiance* as the body in question at two specific wavelengths
    Color temperature: The temperature of a blackbody that has the same *color coordinates* as the body in question.
    Distribution temperature: The temperature of a blackbody that has the same *spectral distribution* as the body in question
    Effective temeprature: The temperature of a blackbody that has the same *irradiance at a plane* as the body in question.

    Does that help?
     
  10. Jul 17, 2008 #9

    Andy Resnick

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    Re: Temperature

    I disagree- there are not "two kinds" of EM radiation. The field is continuous with an infinite number of degrees of freedom- thus the properties are continuous as well. Apparently, blackbody radiation is the only EM field which possesses thermodynamic equilibrium. Since that radiation field does not exist, it means all the EM fields we experience are out of equilibrium. Clearly, assigning negative temperatures (or infinite temperatures) contradicts the thermodynamic meaning of "temperature", so some other approach is needed.

    I'm downloading the article as I type this, but I am skeptical how a chemical potential can be defined since the number of photons is a non-conserved property.
     
  11. Jul 17, 2008 #10
    Re: Temperature

    Sorry, I was talking more in regards to the body in question along the lines of the OP. I see your post was questioning the attribution of temperature to the field itself.

    I suppose I should have said there are two source of EM radiation, as opposed to "two kinds of". Anyway, I would be curious to hear your comments on the article I suggested.
     
  12. Jul 18, 2008 #11

    Andy Resnick

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    Re: Temperature

    It's an interesting paper. An assumption is that electromagnetic radiation only has a temperature due to interactions with matter- a cavity, for example. And the author claims that radiation may not have thermodynamic properties associated with it.

    I need to read the paper a few times; some of the sections are difficult for me to follow. However, apparently according to the author, if I start with radiation emitted by a blackbody at temperature T and pass it through a dichroic mirror (to separate spectral components), even if the dichroic mirror is also at temperature T, the light downstream from the mirror no longer has thermodynamic properties. This is odd, to say the least.
     
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