Why do black bodies in thermal equilibrium with their surroundings only emit in non-visible regions of the electromagnetic spectrum?

Question

This question is a follow on from my previous question; Under what conditions can a body be approximated as a black body?.

This question is also about one specific part of an answer given to this question on Why is a black body visible?

In part of the answer @anna v writes:

A perfect black body in equilibrium with the surrounding temperature will be absorbing incident radiation and emitting black body radiation according to the temperature scales of this radiation. At temperatures where our eyes can exist we perceive it as black, because $\color{red}{\text{these are infrared frequencies}}$. We only see visible light reflections on bodies, not their black body radiation.

Notice that this explanation does not explain why "these are infrared frequencies" and not in the visible region. Which is what I would like to know.

So I read other questions and answers on this site, and, in this question on If a black body is a perfect absorber why does it emit anything?

In an answer @Alfred Centauri writes:

A black body in thermal equilibrium emits more energy than any other object (non-black body) in the same thermal equilibrium since it absorbs more energy.

Imagine several various objects, including one black body, in an oven and in thermal equilibrium. The black body will 'glow' brighter than the other bodies.

So according to this answer, black-bodies in thermal equilibrium are visible.

I wanted to check this so I looked at this other question, Black Bodies and appearing black

In an answer to the question @Yaman Sanghavi writes

For a black-body to look white, it will have to emit wavelengths corresponding to the visible region with nearly equal intensities because white light is composed of visible colors but with EQUAL intensities of all the colors.

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In all of the above 3 quotes (and I believe in the whole of their answers) they were talking about a black body in thermal equilibrium with its surroundings.

Now, we were all taught from an early age that every object with a temperature above absolute zero will emit electromagnetic radiation. What I what like to know is why a black body in thermal equilibrium with its surroundings will emit in the infrared (and ultraviolet, etc.) and not in the visible region?

Or, if you prefer, why will there not be some tiny amount of visible color to the black body since (as mentioned in the third answer) the intensity of the visible region emitted will be very low?

Answer 1

I think your confusion is because the authors of the comments saying that only non-visible wavelengths are emitted are using a shorthand for “detectably emitted”.

As can be seen from many sources such as Wikipedia, a black body spectrum is continuous over all wavelengths, and peaks at a different value depending on temperature. As Anna V says, for temperatures at which human eyes exist the peak is in the infrared region. While there is emission in the visible light region it is too weak for our our eyes to detect.

Answer 2

What I what like to know is why a black body in thermal equilibrium with its surroundings will emit in the infrared (and ultraviolet, etc.) and not in the visible region?

Most of the radiation comes out in frequencies near $\omega$, where $\hbar \omega \sim k_B T$. If you plug in $T = 300 \, \text{K}$, solve for $\omega$, and then convert to a wavelength $\lambda = c / \omega$, you get $$\lambda \sim 8000 \, \text{nm}.$$ This is in the infrared region. The frequency would have to get a factor of $10$ higher to start hitting the visible region, and Planck's law contains an exponential suppression in frequency, so for room temperature objects a negligible amount of visible radiation is emitted -- certainly not enough to see. An even smaller amount of ultraviolet radiation is emitted, which is also negligible.

Answer 3

"why a black body in thermal equilibrium with its surroundings will emit in the infrared (and ultraviolet, etc.)". This is not true.

Blackbody radiators emit radiation with a spectrum that follows the Planck function. If the blackbody is hot enough, then it will emit copious amounts of radiation in the visible part of the spectrum. The Sun, which approximates to a blackbody, is an obvious counterexample to the title of your question, and has an effective temperature of about 5800 K and a spectrum that peaks at about 550 nm.

What is true, is that the peak of the Planck function occurs at longer wavelengths for cooler temperatures (known as Wien's law), so that blackbodies below about 2000 K emit a negligible faction of their radiation in the visible part of the spectrum.