I was thinking something Let's consider radiation incident on a gas

If it were a bunch of electrons in place of photons, then incident electrons would increase the kinetic energy of each and every atom individually through elastic scattering. And since temperature is the average kinetic energy, increasing the KE of all the atoms would indeed increase the temperature.

For bunch of incident photons instead, I think the following cases occur when photons falls on individual atoms/molecules of the gas:-

  1. Light of specific frequencies gets absorbed by electrons of atoms changing its electronic state / gets absorbed by rotation and vibration of molecules changing its rotational and vibrational state (PHOTOELECTRIC - LIKE). Now although the change of the state of an atom/molecule does indeed change its energy for few moments of time, but ultimately the entire energy is released either by directly emitting a photon of the absorbed energy or by emitting multiple photons in steps. Conclusion:- No change of energy of the gas hence, no heating of the gas

  2. Light of any arbitrary frequency is not absorbed at all. It just passes by the atom without interacting. In other words there is no elastic scattering of photons with atoms/ molecules. Elastic scattering is the most favorable means of heat transfer. It happens in conduction and convection. Lack of elastic scattering makes it very difficult for any sort of heat transfer. Conclusion:- Yet no energy could be transferred by radiation.

  3. Very energetic photons like X-rays will knock out valence shell electrons (COMPTON). But that I think doesn't cause much change to the system. Because there are lots of free electrons roaming around which will soon occupy such vacant sites. However, one might claim that the free electrons will undergo Bremsstrahlung (emission of radiation by free electrons moving in the presence of an attractive potential) due to the positively charged atoms. But that that would indeed worsen the situation, since emission of radiation would actually cool the gas, rather than heating it up. Conclusion:- Compton effect is no good in heat transfer.

  4. High energetic gamma rays in the vicinity of the nucleus may undergo pair production. But the free electron and positron produced would most of the times, immediately annihilate each other. Even if it does interfere with some atoms, it wouldn't be able to render much energy since only few electrons would be able to interact with only few atoms/molecules. Also, any radiation can heat up a gas. It need not be gamma rays. And photons of energies less than gamma would not be energetic enough to undergo pair production. Conclusion:- Pair Production is for limited cases and even where it occurs, it cannot affect the system much.

Thus I find, literally there is no way the bunch of photons are increasing the KE of evan a single atom, let alone the entire gas.

So the question - How does radiation heat a gas (or any other stuff) than?

  • 2
    $\begingroup$ 'gets absorbed by rotation and vibration of molecules changing its rotational and vibrational state' - It is not mandatory for molecules or crystals to emit a photon once they have absorbed light, you can excite phonons. $\endgroup$ Jun 29, 2020 at 23:46
  • 1
    $\begingroup$ @corcholatacolormarengo Indeed, especially in most solids illuminated at IR wavelengths or longer, phonon scattering is the dominant relaxation mechanism, and directly results in the heating of the lattice. $\endgroup$
    – Puk
    Jun 30, 2020 at 3:42

1 Answer 1


How does radiation heating work at the fundamental level?

In the question for

1.You forget that in matter the relaxation photon can interact further with atoms in the gas, there are $~10^{23}$ atoms per mole.

for 2. this is wrong : "In other words there is no elastic scattering of photons with atoms/ molecules." There is always scattering of photons. See also here.

3 and 4 are consideration for stars and cosmic plasma, and again threre are interactions and they will contribute to the temperature.

In a gas, photons will scatter with individual atoms and molecules and transfer their energy by inelastic scattering increasing the kinetic energy . The photons coming out of the inelastic scattering can further scatter etc, so depending on the frequency and the thickenss of the material traversed, all of the energy of the photon can be absorbed into kinetic energy of the atoms. The mathematics is complicated as you can see in the links, and approximate models are made.

In solids and liquids more states of matter exist complicating the story.

  • $\begingroup$ Yeah I had thought about the answer you replied to point 1, but actually what I cannot digest is that in order to have your temperature risen, every atom/molecule should at least retain their Kinetic energy right, only then the average value would not fluctuate but obtain an equilibrium value. On the other hand, the typical time period for atoms remaining in an excited state is $10^{-8}$ s. So, even if the bunch of photons are exciting multiple atoms, they would soon lose their momentary energy. $\endgroup$ Jun 30, 2020 at 10:40
  • $\begingroup$ So then is the claim you are making goes something like this that photons impart momentum to atoms either by direct absorption or via some other process and these momentum-gained atoms collides with other atoms, thus leading to increase in their Kinetic energy? That is to say, light is not directly responsible in transferring energy, rather collisions between atoms/molecules due to the imparted momentum of light increases the KE? $\endgroup$ Jun 30, 2020 at 10:41
  • $\begingroup$ If you look at the links i gave, inelastic photon atom scattering exists and is calculable. in the end in a gas that is the interaction that the hν energy of the photon ends up into kinetic energy of the atoms/molecules. $\endgroup$
    – anna v
    Jun 30, 2020 at 10:51

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