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Do all bodies emmit black body radiation?

  1. Dec 6, 2009 #1
    Is this black body radiation considered light?

    PS: Sorry, I spelt emit wrong:(
     
  2. jcsd
  3. Dec 6, 2009 #2

    Integral

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    Yes every body with a temperature greater then absolute zero emits black body Electromagnetic radiation. The center point of a distribution of frequencies is determined by the kelvin temperature of the body. Light is a form of EM, though what we call visible light consists of a very narrow band of the entire spectrum.
     
  4. Dec 7, 2009 #3

    Andy Resnick

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    No real body emits blackbody radiation; some bodies my approximate it better than others. Blackbody radiation is a very special property of radiation- radiation at thermal equilibrium.
     
  5. Feb 22, 2010 #4
    The same question interests me; for instance integrals answer is the usual response but if you have a gas, say N2 with no apparent means of radiating at 280K or below how does it radiate?
     
  6. Feb 22, 2010 #5
    Post #3 is correct;
    try here to see the difference between thermal and black body radiation:
    http://en.wikipedia.org/wiki/Blackbody_radiation

    Does a black body that is at a lower temperature than its surroundings emit any radiation? I don't think so; I think is absorbs, but am not sure. But I don't know if it would still be called a "black body".

    Leonard Susskind notes in THE BLACK HOLE WAR,pg 164:
    "The sun's surface radiates plenty of light, but it reflects none. To a physicst that makes it a black body."
     
    Last edited: Feb 22, 2010
  7. Feb 22, 2010 #6
    Naty I think you are confusing different phenomena.

    There are two meanings to the term in Physics. 'blackbody radiation' and 'a black body'

    Post#2 is correct, all bodies at the same temperature emit radiation (=light etc) of the same spectrum.

    The term blackbody is used because the amount is zero at absolute zero. So the body viewed solely by its emitted light at absolute zero would appear black.

    The amount of radiation varies from body to body and is the product of the surface area and a coefficient known as the emissivity. The emissivity for real material objects is always < 1 so these are called greybodies. A true blackbody has an emissivity of 1 and radiates the theoretical maximum amount of radiation.

    Thermal equilibrium does not enter into the picture.
    A body, grey or black, emits radiation as effieicently as it absobs it. The absorbtion coefficient is the same as the emission coefficient.
    Any body continually emits radiation solely determined by its absolute temperature.
    The same body continually absorbs radiation, solely determined by its surroundings.
    The balance determines whether the body is undergoing radiative heating, cooling or is in radiative equilibrium. This is not the same as thermal equilibrium as the body may contain an internal heat source (eg a star)

    On the other hand radiation by a black body is by definition zero at any temperature.
    A black body is one which absorbs all incident radiation.

    The coeficient appropriate in this case is called the reflection coefficient or albedo.

    I think you will find Susskind's statement to be of the 'for all practical purposes' kind.
     
  8. Feb 23, 2010 #7
    How does N2 gas radiate at say temperatures below 280K when its spectra shows no available lines-what is the mechanism ?
     
  9. Feb 23, 2010 #8
    Have you met the kinetic theory of gasses? (temp. pressure etc)

    All the molecules are in motion and thus have energy. There is a statistical variation in this energy. Some have more than others. Some have enough to radiate.
     
  10. Feb 23, 2010 #9
    I think that the black body cavity theory developed by classical Physics from considerations of solids and relatively high temperatures does not completely explain gases or even very high temperatures.
    See the "ultraviolet catastrophe".
    Its a perfectly good question.
    How does an isolated molecule of N2 radiate if its temperature is say 280K and its spectrograph shows no evidence of lines at IR or below.
    If we set the emissivity constant at zero then the equation works but then there is NO radiation.
     
  11. Feb 23, 2010 #10
    I think that if you really want to discuss these things in depth you should follow the forum rules and start your own thread, not hijack someone elses. Your comments are not germaine to the original question.

    I also think that if you want a discussion you should respond to comments or queries raised by others rather than just repeating and adding to your own.

    Since you did not answer my simple question I have no idea at what level to pitch an answer. However your response indicates that you do not understand the kinetic theory or you would realize that the statement "an isolated molecule at 280" (or any other value) has no meaning.
     
  12. Feb 23, 2010 #11
    Yes I fully understand the Kinetic Theory of Gases and I have no wish to intrude on your query any further.
     
  13. Feb 23, 2010 #12

    D H

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    A black body is an ideal object. There is no such thing as a true black body (the cosmic microwave background radiation comes very close). Think of the ideal gas law. Every real gas deviates from an ideal gas in some way. Similarly, every real object deviates from an ideal black object. Google the terms emissivity and absorptance.
     
  14. Feb 23, 2010 #13

    D H

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    Blackbody radiation is radiation emitted by a black body. A black body absorbs all incoming electromagnetic radiation.

    Post #2 and the above reiteration are wrong. Blackbody radiation is an idealization. No real body emits thermal radiation that perfectly follows the blackbody spectrum.

    Suppose two objects are placed in an evacuated chamber with perfectly mirrored walls. One of the objects is very close to being a black body (i.e., it has emissivity and absorptance nearly equal to one) but the other object is a highly polished silver sphere. The objects will come into thermal equilibrium with each other via radiative heat transfer. The silver sphere has a very high reflectance and does not absorb much incoming radiation. It therefore doesn't emit much radiation either.

    In a nutshell, A good absorber is a good emitter but a good reflector is a lousy emitter. Kirchoff's law. Also see http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/absrad.html.
     
  15. Feb 23, 2010 #14
    My understanding is the same as #13....Thanks DH
     
  16. Feb 23, 2010 #15
    D H: Thank you for your comments I am new here and got little carried away. It is difficult to know at what level to pitch a reply. Consequently I meant to go no further than Prevost's Law, but overstepped by introducing emissivity and grey bodies.

    However my comment about post#2 came before this part and and is correct (as is post#2) in the context of Prevost's law.

    It just shows how easy it is to make an oversimplification which is then modified when more advanced theory is introduced.

    In fact applying the emissivity modification to Prevost's Law stated in your link to your example would still lead to my simplified conclusion. You need still more advanced theory to handle it where the emissivity for both bodies is a different function of temperature.

    Further complications can be added by including Stewart's Deduction to your model.

    You have only mentioned the surface of your silver body. What about the radiation from the interior?
     
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