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Parent questions: Do photons lose energy after radiation pressure is applied to a perfect reflector?

Since radiation loses energy to radiation pressure in a reflection, can the entire energy of a radiation be consumed through multiple reflections?

Radiation pressure lowers a photon's frequency. I asked separately if this could be used to convert all the photon's energy to kinetic energy.

Now the present question is, by using radiation pressure or not, is it possible to convert a photon's entire energy to kinetic energy in a straightforward way? By straightforward, I mean not using successive conversions like a solar panel and an electric motor, whatever, but doing it the most possible direct way.

Winston
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3 Answers3

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It can't be done through direct kinematic interactions unless you set up multiple reflections.

Your choices of direct kinematic interactions are inelastic or elastic.

In inelastic, all the energy of the photon goes into something else, often heat. You could conceivably instead drive a reaction (like exciting a molecule) and then recover the energy from that reaction. But that sounds like not what you're looking for.

The other choice is an elastic interaction. In that case, the photon must still exist and carry energy away from the collision. The lower the energy of the photon in your frame, the less of the fraction that it leaves with, but also the less energy that you can harvest from it.

BowlOfRed
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  • Is it absolutely inconceivable, even in theory, that a photon through an inelastic collision would accelerate a particle in just one direction instead of inducing heat? – Winston Aug 26 '20 at 22:59
  • It will accelerate, but some energy will be lost because momentum must be conserved. If all the input energy were to go into KE of the combined system, it would have more momentum than it started with. – BowlOfRed Aug 26 '20 at 23:24
  • Yet radiation pressure does not produce heat, does it? (I know it does not convert the entire photon's energy) – Winston Aug 26 '20 at 23:37
  • It doesn't produce heat if the reflection is total. But in that case, the reflected photon leaves with most of the incoming energy. The energy transfer rate is very low. The question asked if all of the energy could be converted. – BowlOfRed Aug 27 '20 at 00:04
  • "In inelastic, all the energy of the photon goes into something else, often heat.", I think you mean in inelastic scattering part of the photon's energy is transferred into vibrational, translational, and rotational energies of the molecules (thus heat), but when all of the photon's energy is transformed, it is absorption. – Árpád Szendrei Aug 27 '20 at 00:15
  • @Arpad: is the last loss of energy from a photon to radiation pressure (after many reflections) an absorption? – Winston Aug 27 '20 at 01:09
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    @Exocytosis correct, in the case of reflection, it is all probabilities, there is a high probability for the photon to get elastically scattered, and a little to get absorbed. If there is many interactions, at the end the photon will be absorbed. – Árpád Szendrei Aug 27 '20 at 03:36
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Yes, most of the energy in EM radiation can be converted to kinetic energy. Look up "photon piston" or photo-Carnot engine. https://en.m.wikipedia.org/wiki/Photo-Carnot_engine

S. McGrew
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  • Interesting. However it would seem this consumes extra energy to rebuild coherency so this is not what I am looking for. – Winston Aug 26 '20 at 22:55
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Is there a way to convert directly most electromagnetic radiation to kinetic energy?

What comes to mind for me is the conversion of microwave frequency electromagnetic energy to kinetic energy when a microwave oven heats food. The frequency of the microwave photon is such that it increases the rotational kinetic energy of water molecules due to their dipolar nature (the alternating microwave electric field causes the water molecules to rotate to align with the field.) That rotational kinetic energy is quickly randomized as translational kinetic energy of the water molecules.

For a more detailed explanation, see this: http://hyperphysics.phy-astr.gsu.edu/hbase/mod3.html

Hope this helps.

Bob D
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  • Good point. As you wrote, motion will be randomized. But in theory at least, is it possible to obtain a translation only (not oscillatory) using EM radiation? – Winston Aug 26 '20 at 23:14
  • I’m not aware of any EM frequency that directly causes translational kinetic energy, but I don’t profess to be an expert.The link in my post only describes rotational and vibrational modes for microwave and infrared, with higher energy levels associated with electron excitation and ionization. – Bob D Aug 27 '20 at 06:58