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B Blackbody definition

  1. Apr 29, 2017 #1
    Hello! I am not sure I understand the concept of blackbody. The definition is that it absorbs all the incoming light and doesn't reflect any. However, it emits thermal radiation. Isn't thermal radiation still EM radiation, so technically still light, just at another frequency? So if it emits some kind of EM radiation, why is it black?
     
  2. jcsd
  3. Apr 29, 2017 #2
    Hi silviu:

    It is only a matter of vocabulary usage. A body's blackness just means it does not reflect photons, although it can absorb them. If a black body has a temperature, it will radiate photons according to Planck's Law.

    Pretty much almost everything we see reflects photons, and the reflected photons is what we see as the thing we see.

    Regards,
    Buzz
     
  4. Apr 29, 2017 #3

    Dale

    Staff: Mentor

    A black body which is sufficiently warm will emit visible light
     
  5. Apr 29, 2017 #4
    So it is not really black, right?
     
  6. Apr 29, 2017 #5

    Drakkith

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    Staff: Mentor

    Yes, it is. The color of an object is not determined by the light it emits when you heat it up, but by the light that reflects off of it. A blackbody will reflect no light and truly be black.
     
  7. Apr 29, 2017 #6
    Wait, I am confused. If you heat it up to a temperature such that the peak frequency corresponds to, let's say, blue, won't it appear as blue and not black? So won't its color be blue?
     
  8. Apr 29, 2017 #7

    Drakkith

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    Staff: Mentor

    Sorry, I think there's been a miscommunication here. If you looked at a blackbody at 5,000 kelvin, it would not look black. It would look white and be as bright as the surface of the Sun. But it would absorb all incoming radiation, and when not heated up it would appear black. Does that clarify things a bit?
     
  9. Apr 29, 2017 #8
    Well I am a bit confused by "when not heated up". Does this mean that a perfect blackbody exists only at 0k (as this means not heated up)? Otherwise it would still emit some radiation (even if not in the visible spectrum) so it would still not be black.
     
  10. Apr 29, 2017 #9

    Drakkith

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    Staff: Mentor

    No, the term "blackbody" doesn't mean that the object always looks black. It means that it always absorbs all EM radiation falling on it and when heated above 0k it is a perfect emitter of thermal radiation. That is, it has an emissivity of exactly 1. Don't get caught up in the "black" part of the name.

    From wikipedia's article on a black body: https://en.wikipedia.org/wiki/Black_body

    So on top of being a perfect emitter of thermal radiation, it transmits no light (or other EM radiation) through it, is a perfect absorber of incident EM radiation, and reflects 0% of any incident EM radiation.
     
  11. Apr 29, 2017 #10

    Dale

    Staff: Mentor

    A black body can emit radiation. That is not part of the definition. The defining characteristic is that it absorbs all radiation.

    Many sources of visible light (sun, candles, etc) are approximately black bodies.
     
  12. Apr 29, 2017 #11

    sophiecentaur

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    Gold Member

    Frequency sensitive reflectivity ('colour') and emissivity are two separate issues.
    If it doesn't reflect any EM then it's defined as a black surface.
    A hot body (black or white or green) will radiate EM. If it radiates EM with a spectrum according to Planck's Law (see above) it is defined as a black body.
    There is no point in worrying about any apparent contradiction here. Context is what counts.
    I think there is a practical aspect to this. The reflectivity of a surface at a given frequency will depend on the energy states of its atoms. Get it hot enough to be glowing, you are disturbing the distribution of states; it is no longer the same 'colour'. Take a visibly coloured object at a temperature below 'red heat' (say less than 1kK) its colour under a white illuminant won't be affected.
     
  13. Apr 29, 2017 #12
    See if answer #3 in this thread helps: https://physics.stackexchange.com/a/148627
     
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