If we are able to see an object in a certain color,that means it reflects the color.A blackbody is one which has zero reflectivity.Then how come it is not black always?
2
-
1$\begingroup$ This isn't an engineering question. I think it would get a better reception on Physics SE. I think there is also a Science History SE in the mill that would could help. This was a really big deal in physics a while back (they though they had everything figured, just a couple little niggling issues, one of which was black body radiation). Explaining it motivated the theories of general relativity and quantum mechanics. That's the language you'll need to understand it quantitatively. $\endgroup$– Phil SweetApr 14, 2018 at 3:12
-
$\begingroup$ here's an intro - chem.libretexts.org/Textbook_Maps/… $\endgroup$– Phil SweetApr 14, 2018 at 3:13
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Your assumption is incorrect. The color we see is the combination of reflected light, if any, and radiated light. All objects radiate photons of one wavelength or another. All objects absorb photons of one wavelength or another. These two spectral curves (absorption and radiation) are the same, but the spectral power curve of the irradiance obviously depends on external sources, not on the material or temperature of the black body.
answered Apr 13, 2018 at 17:21
$\begingroup$
$\endgroup$
1
There is no Perfect black body (including black hole see hawkings radiation for details) that is why you can see a small light coming up from every black body and hence no body have emissivity of 1. This means true black is not yet discovered.
-
1$\begingroup$ Your answer implies that the perfect black body would not emit light, or, black bodies only emit light because they are not perfect black bodies. That is not correct, every object in thermal equilibrium needs to emit as much as it absorbs $\endgroup$ Jun 15, 2018 at 15:17