Blackbody radiation

[gentsagree]

Hi,

Very basic question.

Blackbodies are ideal emitters: at every frequency, they emit an amount of energy equal to or greater than any other object at the same temperature. Furthermore, they were named blackbodies since, AT ROOM TEMPERATURE, i.e. 300K, whatever radiation they re-emit (after absorbing it) is NOT in the visible (wavelength, hence frequency, change between absorbed and emitted radiation).

Now, the fact that best emitters are black is only true around room temperature, or below, correct? I think I have read somewhere that above 2000K best emitters are no longer black, as they emit in the visible.

Does this mean there are no truly black objects hotter than 2000K ?

Thanks!

 

[jfizzix]

Blackbodies are ideal absorbers of light, though at thermal equilibrium, they emit as much light as they absorb.

Any blackbody with a temperature above absolute zero will emit some form of electromagnetic radiation. The spectrum of different frequencies (colors) depends on the temperature and is called the blackbody or Planck spectrum.

The peak frequency of the blackbody spectrum increases in proportion to temperature.

At room temperature (say.. 300K), most of the radiation emitted is in the infrared or below..
..but above the Draper temperature (about 800 K or 1000 F), enough of the emitted light is in the visible spectrum that we can see it glow.
As the temperature gets higher, the object gets red-hot, then yellow, then white-hot and even blue-white at over 10,000K.

Long story short, there are no truly black objects at any temperature above absolute zero, though especially not above 2000K.

 

[jerromyjon]

Long story short, there are no truly black objects at any temperature above absolute zero, though especially not above 2000K.

Except black holes and maybe some exotic forms of matter, black holes are the only "true" black objects because nothing escapes them.

 

[gentsagree]

Thanks, very clear. Two things:

- They definitely are perfect absorbers, by definition. Wiki, though, states that at any temperature and frequency, they also are the most efficient emitters possible emissivity=1). Is this inaccurate?

Because of this, and because I really meant "black to our eye", so not in the visible spectrum, I gather that the answer to my question would be that these "best emitters", as defined above, are indeed black below 800K and also above roughly 8000-10000K.

I think one could just use Wien's formula to work out the temperature range, given the wavelength range of visible light. I think this gives roughly the temperatures that you mention.

 

[jfizzix]

Except black holes and maybe some exotic forms of matter, black holes are the only "true" black objects because nothing escapes them.

The jury's still out on whether black holes are truly black.
There's good theoretical arguments to be made that they emit some kind of blackbody radiation according to their temperature (this being called Hawking radiation for the guy who figured this out), but I don't think that that will be proven any time soon. The radiation would be incredibly dim because the temperature for even a small black hole would be incredibly small.

 

[jfizzix]

... Because of this, and because I really meant "black to our eye", so not in the visible spectrum, I gather that the answer to my question would be that these "best emitters", as defined above, are indeed black below 800K and also above roughly 8000-10000K.

Although the percentage of the total blackbody radiation in the visible part spectrum goes down as the temperature gets a lot higher (say, past 10000 K), the total intensity of the light in the visible part is the spectrum keeps going up.

As the object gets hotter, it still gets brighter to out eyes, though the total tadiation emitted grows much faster than the part we can see.

 

[gentsagree]

Ah, that makes sense, of course. Thank you.

 

[jerromyjon]

this being called Hawking radiation for the guy who figured this out

I thought Hawking radiation was supposed to be from virtual pairs produced near the event horizon and since they are naturally opposed one of the pair would be absorbed and the other would escape as radiation... leading to the possible mechanism for evaporation. A paper I saw this morning was regarding black holes modeled with LQG and seemed to indicate they could lead somewhere else or to another universe altogether! The jury has been deliberating over twice my life...

 

[jfizzix]

I think what you describe would be the mechanism for producing the radiation, though its spectrum would still be a blackbody spectrum.

 

[jerromyjon]

though its spectrum would still be a blackbody spectrum.

Too shay!

 

[Khashishi]

Even if the object is glowing brightly because it has a high temperature, we still call it a black body if it is close to an ideal absorber/emitter over a wide range of frequencies. Hence, "black" has a technical meaning which has little to do with what looks black to your eye.

Does this mean there are no truly black objects hotter than 2000K ?

If, by truly black, you mean no visible light coming off, and not totally transparent, and no reflected light either, then the answer is yes, you are correct.
But, in principle, you could have something totally transparent to visible light or something totally reflective to visible light at 2000K, and it wouldn't give off visible light. These would not be black bodies.