1
$\begingroup$

I have came across the following paragraph in Wikipedia

A perfectly insulated enclosure which is in thermal equilibrium internally contains blackbody radiation, and will emit it through a hole made in its wall, provided the hole is small enough to have a negligible effect upon the equilibrium.

So my question is, if I have a blackbody in my room but this black body isn't in equilibrium with with my room. However, it's in equilibrium with its internal structure, let's just imagine I have a perfect glowing blackbody in my room, Does the emitted radiation resembles that of a blackbody?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
0
$\begingroup$

Yes, radiation of a blackbody that is not in equilibrium with its environment is still blackbody radiation, with properties of blackbody radiation.

Blackbody need not be in equilibrium with its environment, and its emission need not be equilibrium radiation! In fact usually radiation of a compact finite-sized blackbody can't be, as equilibrium radiation has no macroscopic energy flux in any direction, but blackbody emission produces energy flux from the body away.

The concept of blackbody was not invented as something that exists only when in thermodynamic equilibrium. The premise is that it is the body that absorbs every incoming radiation, and reflects none of its; but at the same time, it also produces its own radiation (emission), and the emitted radiation is characterized by the blackbody temperature and geometry.

The cavity with perfectly reflecting walls with equilibrium radiation inside is not a blackbody, instead the hole that is made in that wall through which radiation escapes, is a realization of blackbody surface (absorbs all incoming radiation, and emits its own blackbody radiation).

Improve this answer
$\endgroup$
3
  • $\begingroup$ if i am talking here about producing the bell shaped Plank curve, the blackbody is to be brought into equilibrium with its internal structure but then i must make sure i am insulating it from the ambient, or either, i should bring it into an equilibrium with its ambient, and in that latter case it will be in an equilibrium internally as well, is that right? $\endgroup$
    – Jack
    Commented Feb 17, 2023 at 8:25
  • 1
    $\begingroup$ Blackbody can't lack equilibrium with its internal structure, it is a theoretical concept, it always produces blackbody radiation. You do not need to bring it to equilibrium, with anything. $\endgroup$
    – Ján Lalinský
    Commented Feb 17, 2023 at 9:28
  • 1
    $\begingroup$ Thinking about it more, the blackbody can have internal part that is at different temperature than the outer shell, and then it is not in internal equilibrium. But this changes nothing in the emitted radiation - that has spectrum of blackbody at temperature of the outer shell. The central part at different temperature does not change the radiation emitted to the outside. $\endgroup$
    – Ján Lalinský
    Commented Feb 17, 2023 at 10:21
0
$\begingroup$

The internal structure will not be in equilibrium either. It will change in time and so will the radiation output. However emmitted radiation would represent that of BB.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What you're saying is if the BB isn't in equilibrium with the ambient it can't be with equilibrium with its internal structure hadn't we isolated it ? $\endgroup$
    – Jack
    Commented Feb 16, 2023 at 23:57
  • $\begingroup$ You have isolated it true but total isolation may not be possible in practical terms. $\endgroup$
    – SAKhan
    Commented Feb 18, 2023 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.