What is the fate of a photonic quantum that hits a black wall? How does this work with the conservation of energy and matter?

  • 1
    $\begingroup$ Is "photonic quantum" a weird name for what we usually call a "photon"? What about the obvious answer "It gets absorbed and heats the wall a little" just as in classical electromagnetism doesn't satisfy you? $\endgroup$
    – ACuriousMind
    Dec 20, 2016 at 16:13
  • $\begingroup$ I know the photon gets absorbed, but I'm wondering about the mechanics of the photon while it is absorbed, especially since the photon can eventually leave the black molecule. $\endgroup$ Dec 20, 2016 at 19:26

3 Answers 3


There is no conservation of matter , as photons belong to the special relativity and quantum mechanic frame. Only energy and momentum and angular momentum conservation laws are applicable.

A black wall is black because it absorbs the photons impinging on it. Depending on the frequency of the photon its energy is given up to displacing an electron in an atomic or molecular or lattice level to a different energy level and the relaxation to lower levels releases photons that are further absorbed. Momentum is conserved by the wall mass.

Note that black is a term related to the frequency of light that the photon belongs to. No matter is black for all frequencies, that is why we can have infrared photography.

  • $\begingroup$ Isn't vantablack pretty black in really wide spectrum? I mean, in most if not all frequencies that even make sense to talk about? It is black from UV to deep IR. (I consider if it's worth asking as a separate question, I comment here because I'm not sure if "no matter is black for all" is really valid). $\endgroup$
    – Mołot
    Dec 20, 2016 at 12:33
  • $\begingroup$ @Mołot all ready yourself have set limits. The electromagnetic spectrum starts from the very long wave and ends at gamma rays. general physics statements are about all frequencies en.wikipedia.org/wiki/Electromagnetic_spectrum $\endgroup$
    – anna v
    Dec 20, 2016 at 13:01
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    $\begingroup$ @annav Electron interactions are, just the same, only probable in a very narrow range of EM energy. A microwave, for example, is not going to excite an electron when it is absorbed - it will rather induce some rovibrational mode in the molecule, etc. Radio waves, likewise, will have resonant dielectric interactions and so on. For completeness, there are many ways in which photons will interact during absorption. $\endgroup$
    – J...
    Dec 20, 2016 at 15:20

The black wall would be composed of molecule that absorbs light at the wavelength of the photon. So the photon (quantum state and all) gets absorbed by one such molecule and the energy and momentum gets transferred to the molecule. Now there are a plethora of ways in which the energy and momentum can be dissipated. In general the energy is converted into heat that is dissipated into the wall or perhaps re-radiate back into the atmosphere. The momentum would be adding to the momentum of the wall. However, the amount of energy and momentum that the wall receives from one photon is so tiny that it makes an insignificant difference.


The energy will be conserved as the the wall becomes hotter after having absorbed the photon. Strictly is no such thing as conservation of matter, but there is conservation of mass which will of course be respected as hotter objects weight more than identical cold ones.

Note that the wall will also acquire the photon's momentum, which will push it in the direction in which the photon was travelling.


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