To my understanding, the reason objects appear of a certain color is because it is the color (in visible light) that is absorbed least by the electrons in the material's atoms, and the rest is reflected.

It is also my understanding that those wavelengths are absorbed because of the possible energy levels of the electrons.

Also, the reason why that light is not re-emitted is because while the electrons have a tendency to drop in energy level (to a more stable state) the more immediate way to do so is through heat transfer, and that energy is then lost through heat energy in the bonds between atoms giving off energy as black-body radiation.

Is there anything wrong with this understanding? Are there any details missing that would be good to know?

Why does infrared contribute so much more to heat than other wavelengths such as ultraviolet if it is all transferred to heat anyway?

  • $\begingroup$ For usual temperatures (~300 K), blackbody radiation spectrum peaks around infrared band. And "heat emitted by a body" is just blackbody radiation. Energy follow the following path: photons->heat->emission. Since heat->energy part is only about temperature, most of heat is in infrared, independent of the initial energy source. $\endgroup$
    – acarturk
    Aug 21, 2019 at 16:32
  • $\begingroup$ Re, "Why does infrared contribute so much more to heat than other wavelengths?" Who told you that was true? Planck's Law will tell you how much power is emitted at what wavelengths for a black body radiator. And for most of the "black" bodies you'll encounter in day-to-day life (glowing coals, heated stove), you'd be right: Most of the power is in IR. But that's just coincidence--because of their temperature. The surface of the Sun is much hotter--around 5000K--and it emits about half its power in visible wavelengths. $\endgroup$ Aug 21, 2019 at 17:38
  • $\begingroup$ Then there's lasers: A laser emits 100% of its power in one or a few wavelengths. If they happen to be visible wavelengths, then that's where all the power is. $\endgroup$ Aug 21, 2019 at 17:39
  • $\begingroup$ @SolomonSlow so it's not that infrared is the only wavelength that gives any significant contribution to heat, it's that it is a more common occurrence? That was one of my speculations. If around half of the sun's emissions are visible, does that mean that a significant proportion of the heat you feel from direct sunlight is from visible light rather than almost entirely infrared? What about fire? $\endgroup$
    – User_DJ
    Aug 21, 2019 at 18:22
  • $\begingroup$ @User_DJ, "...significant proportion of the heat you feel from direct sunlight is from visible light...?" Short answer is "yes." A longer answer might account for how well your skin absorbs various wavelengths. Wear a black tee shirt, and the long answer becomes less significant. $\endgroup$ Aug 21, 2019 at 20:37

1 Answer 1


Is there anything wrong with this understanding?

A few minor points:

  • Black-body radiation does not refer to a specific microscopic process, and in particular is not necessarily "the bonds between atoms giving off energy". Black-body radiation is an explicitly macroscopic phenomenon, whose usefulness comes from the fact that it doesn't particularly care about the microscopic structure of the material. For more details, see What are the various physical mechanisms for energy transfer to the photon during blackbody emission?.

  • Absorption and reflection are not the only things that can happen when light hits a medium. There is also transmission, for one. And in higher-energy cases, you can also have decomposition/photodegradation, ionization, and pair production, to name a few.

  • In addition to being dissipated as heat, electromagnetic radiation can also cause an object to accelerate or rotate, if the properties of the object and of the incident radiation are correct.

Why does infrared contribute so much more to heat than other wavelengths such as ultraviolet if it is all transferred to heat anyway?

Well, firstly because not all absorbed radiation is dissipated as heat. If the radiation is high enough in energy, it will break chemical bonds (like UV radiation) and/or ionize the atoms of the material (like X-rays) rather than simply increase the material's temperature. If the radiation is low enough in energy, it will simply be transmitted without interacting significantly with the material. So, in order for radiation to be primarily dissipated as heat, it needs to be around the same energy as a transition that won't actually break the material apart. Many (but not all*) materials have just such a transition in the infrared range, due to interatomic bonds having a typical binding energy of a few eV.

It also tends to help that, for materials around room temperature, the peak of the blackbody radiation spectrum is in the infrared range. This means that the amount of infrared radiation a body emits is correlated with its temperature, and in general macroscopic materials around room temperature will emit and absorb infrared radiation readily.

Of course, if you happen to know a lot about your material, you can use transitions outside the infrared to transfer heat. For example, water molecules have rotational transitions that are very closely spaced in energy; the energy difference between them corresponds to the energy of a microwave photon. So when water is blasted with microwaves, it heats up, which makes microwave ovens possible.

*For example, zinc selenide is effectively transparent to infrared radiation, so you'd have a hard time heating it up that way if you tried.

  • $\begingroup$ What causes the material to absorb certain wavelengths as heat? Is there a difference in how materials absorb infrared versus some visible wavelength in their emission spectrum? You said that the absorption of infrared light can depend on the inter-atomic bonds' binding energies, does this mean that rather than the electrons absorbing the energy, the bonds do? Are absorption as heat and emission of infrared radiation different processes? For example, would infrared-transparent objects such as zing selenide emit black-body radiation mostly in infrared although they don't absorb it? $\endgroup$
    – User_DJ
    Aug 21, 2019 at 18:32

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