A hole in a cavity kept at a constant temperature is supposed to approximate the behaviour of a black body. If an external heat source is keeping the body at a constant temperature wouldn't the heat permeate through the walls of the cavity and be radiated into the centre meaning that there is radiation inside the cavity whether there is radiation coming in through the hole. My understanding was that you put some radiation through the hole and measure the spectrum of radiation that comes out as a result which should approximate a black body spectrum which is modelled by the equation $$U(\nu ,T)=\frac{8\pi h \nu^3}{c^3}\frac{1}{e^{\frac{h\nu}{kT}}-1}$$ My question is does the theory take into account the radiation that is already present inside the cavity, if so where and how?

  • $\begingroup$ Any light entering the hole is reflected indefinitely or absorbed inside and is unlikely to re-emerge, making the hole a nearly perfect absorber from en.wikipedia.org/wiki/Black_body $\endgroup$ – user179430 Dec 29 '17 at 12:01

You have got slightly the wrong impression about a cavity radiator.

Black body radiation is a statistical phenomenon. It relies on the system attaining statistical equilibrium i.e. all the possible modes of the system are all exchanging energy with each other. A cavity is used because in a cavity the radiation from the surface one side is reabsorbed by the surface on the other side so the whole internal surface is continuously exchanging energy with itself and can therefore reach equilibrium. It is the heat being supplied from the outside through the cavity walls that heats the inner surfaces of the cavity.

To produce a radiator we then make a small hole through the walls of the cavity. The intention being that the hole is too small to perturb the equilibrium inside the cavity. The radiation emerging from this hole has a spectrum very close to a perfect black body radiator.

So you don't heat the cavity through the hole. You heat it through the walls and that heat eventually emerges as black body radiation through the hole.

| cite | improve this answer | |
  • $\begingroup$ Thank you for writing such a well explained answer, it makes much more sense now. But when you see a diagram of a cavity you see the light coming in through the hole on the outside, and they say that the hole is the black body because all the light that enters it will enter the cavity and find it hard to get out. If the hole is just there to observe the radiation then why would they give this information. (e.g [link]quantummechanics.ucsd.edu/ph130a/130_notes/node48.html $\endgroup$ – cal Dec 29 '17 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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