Is the radiation from closed boxes blackbody radiation irrespective of the material of the box?

Question

Suppose I have a box made of silver at temperature T, in thermal equilibrium with the surroundings. Silver has very low absorptivity and hence it will have the same low emissivity to maintain thermal equilibrium. The box is closed so it will have no radiation from outside coming in. So the radiation inside the box would be purely from the radiation emitted by silver. But this radiation is not blackbody radiation as silver is not a perfect absorber and wont be a function of its temperature alone. So if we poke a hole in this silver box small enough to not disturb the equilibrium we should not be getting black body radiation.

But many textbooks say the radiation is black body irrespective of the nature of the box .example-

This is from Astrophysics for Physicists by ARNAB RAI CHOUDHURI.

What am I missing here?

Answer 1

As naturallyInconsistent says the answer is yes, but it's interesting to go into why this is.

The reason is simply that for any material the emissivity and the absorptance are identical. This is Kirchhoff's law of thermal radiation. So if you make your box out of silver the walls will be bad at emitting radiation but they are equally bad are absorbing it again.

With a perfect black body any photon emitted is reabsorbed at the first collision with the wall of the cavity. With silver that photon may reflect off the walls many times before it is reabsorbed, and during that time the silver is emitting more photons. That means at equilibrium the intensity of radiation inside the box is the same with silver as it would be for a perfect black body. This remains true even for materials having an emissivity that varies with wavelength.

This is strictly only true if the light cannot escape the box, as then every photon emitted by the silver is (eventually) reabsorbed by the silver. That's why when we make a reference black body we use a cavity with a hole that is small compared to the size of the cavity. This means the amount of radiation escaping the cavity is negligibly small.

Answer 2

The answer to your title question is yes. Even if you used silver, i.e. low emissivity materials, to make the inside of the box, as long as you wait long enough for thermal equilibrium to be established, the box will essentially act as if the emissivity is exactly equal to one. All the light that fell inside the blackbody cavity will be absorbed, and all the light that comes out of the cavity will be blackbody radiation.

Answer 3

The box is closed so it will have no radiation from outside coming in.

Even if the silver box is closed, it is not perfectly reflecting. Radiation of high-enough frequency goes through, and thus there can be no thermodynamic equilibrium at those frequencies. Metals are transparent to X-rays and gamma radiation, for example.

So the radiation inside the box would be purely from the radiation emitted by silver.

Silver has thermal emission at almost all frequencies. It is true that at some intervals of them, the emission is very weak. But for achieving thermodynamic equilibrium with eventually, this does not matter. Due to reflections, intensity of radiation at these weak frequencies will increase until it reaches equilibrium state at the maintained temperature $T$, where the spectrum is the unique spectrum of equilibrium radiation at given temperature (like in blackbody thermal emission).

What am I missing here?

Your text says "kept at thermodynamic equilibrium". Then that implies the radiation inside is equilibrium radiation. It does not matter what the walls are made of. If they are perfect reflecting walls, they do not emit, but they still reflect the equilibrium radiation present.

In real world, there are no perfectly reflecting walls. So the walls emit and allow some external radiation at high frequencies to come in, and there cannot be a perfect thermodynamic equilibrium at arbitrarily high frequencies.

Answer 4

A "black body " is an idealised concept, like a geometric, infinitely thin line. As stated in Wikipedia, "a black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence."

No actual body can absorb all radiation, any real body will always reflect or transmit some of the radiation. A black box is not made out of any material. Some materials or constructs, like a box with a hole, may approximate the concept, but nobody can make an actual black body.

Answer 5

The system being referenced is a two-state system, described by the Einstein Coefficients as quoted and follows:

Einstein A and B Coefficients In 1917, about 9 years before the development of the relevant quantum theory, Einstein postulated on thermodynamic grounds that the probability for spontaneous emission, A, was related to the probability of stimulated emission, B, by the relationship

$A/B = 8πhν^3/c^3$ From the development of the theory behind blackbody radiation, it was known that the equilibrium radiation energy density per unit volume per unit frequency was equal to $ρ(ν) = 8πhν^3/c^3$ Einstein argued that equilibrium would be possible, and the laws of thermodynamics obeyed, only if the ratio of the A and B coefficients had the value shown above. This ratio was calculated from quantum mechanics in the mid 1920's. In recognition of Einstein's insight, the coefficients have continued to be called the Einstein A and B coefficients.

Note that not all systems are two-level, for instance a laser as quoted further:

The implication of the Einstein A and B coefficients is that these two processes will occur at equal rates, so that no population inversion can be attained in a two-level system like that depicted here. In the helium-neon laser, for example, this limitation is overcome by collisional transfer from the helium gas to the neon gas, achieving the necessary population inversion for laser action.

However, degeneracy still plays a factor. Also, although it seems like Silver does not absorb well, a cubic cavity made with Silver walls is able to absorb very well photons, considering especially that the hole is so small that it does not disturb the thermal equilibrium that is not just stored within the states of the Silver atoms, but also the Electric and Magnetic Fields inside the cavity.

The article "If the Rayleigh-Jeans Law is derived considering waves contained in a cavity, how does it give us the emitted wavelengths of a blackbody" has more information on this point.