The energy of the photon is related to the linear momentum, which does not change in this experiment. Total angular momentum is also conserved, this leads to a torque on  and transfer of angular momentum to the crystal, which was measured by Beth.  You may also want to look at the paper by Moothoo et al, Beth’s experiment using optical tweezers, Am. J. Phys., 69, (2001), 271.

Thank you very much Gert Van der Zwan,

But:

Observation of Angular momentum of the plate means that is has obtained rotational energy connected to the angular momentum -2hbar. Where does it come from if the energy of the photon doesnt change?

Interesting question, to which there probably is a simple answer but I don't see it immediately. A similar situation occurs when you reflect a photon on a mirror. The momentum is reversed (p -> -p), and the change in momentum is taken up by the mirror ("radiation pressure"). Since the mirror acquired  momentum  it acquired kinetic energy, where does it come from? Not from the photon (the magnitude of the momentum is unchanged). My first guess is that it is a relativistic effect: from the moving frame of the mirror the photon appears red shifted, but how to do this correctly for a rotating system?

Remi, I don't think the change in angular momentum tends to zero. If you look at the AJP paper I referred to, the authors state that:

"We use circularly polarized light to set birefringent calcite particles into rotation. Rotation speeds of about 10 Hz were readily achieved."

You're right, I should have said: I don't think the rotational energy tends to zero.

Actually I now think the mirror (and disc) gets colder. If you do the momentum and energy balance for a photon absorbed by a body, you need to take into account the internal energy.* The body both gets a certain amount of momentum (and kinetic energy as a whole) but the absorption also adds internal energy. If I view the reflection process as absorption and emission (of a photon with the same energy) the energy balance shows that the final internal energy of the system is lower, so it has a lower temperature. I will try to work this out in more detail, but don't have the time to do it today.

* In a paper by Bartell (J. Chem. Educ. 78, (2001), 1068) the author states in this context  that: "It is only a small skip from the concept of pressure exerted by radiation to illustrate Einstein’s favorite way of demonstrating that E = mc2 . Unfortunately Bartell does not give a reference for Einstein's derivation.

I attached a derivation for the energy and momentum balance for the absorption of a photon, related to the question Bartell asks (and my disagreement with what he wrote).

I wonder, would want to imply, that the chaotic movement of the molecules of the birefringent body  responsible for its heat content, is reorganized into highly organized circular movement, when a photon is absorbed? This seems highly unbelievable.

Using a classical physics analogy, when a small ball hits a greater object in free space non-centrally, the object will acquire translational and rotational motions. The energies are not where to be taken from, but from the translational energy of the small ball.

So I think the situation is the same, photon looses energy.

All these questions look different, and are perhaps easier, when the large particle is moving. Then definitely energy gets transferred: if you throw a ball toward a moving train, it will get back to you with finite larger energy. This come from the fact that, in that case, the relative change in kinetic energy caused by a small change delta V in the large  object's velocity is linear in delta V, and the same holds for momentum.

@Ilian Peruhov  If you phrase it that way the same is true for the reflection  of a photon:  disordered heat motion is converted to momentum, and kinetic energy, of the body. Do you agree that for absorption, in order to satisfy both momentum conservation and energy conservation, it is necessary to take the internal energy of the object into account?  Reflection of a photon can be viewed as absorption followed by emission in the opposite direction from a body that is now moving faster than before absorption. The two processes will cool the body a bit, and give it momentum, and kinetic energy. In the same way you can view the Beth experiment as the absorption of a chiral photon, followed by the emission of a photon of opposite chirality from a body that is now rotating at a different rate than before absorption.

But I agree, it almost seems to violate the second law of thermodynamics: heat from a single reservoir is converted to work. However,  the universe has also changed, a photon is now moving in the opposite direction, or in Beth's case has now opposite chirality.

There is no need for the beam to loose a tiny tiny bit of energy. Also the mass of the body does not need to be large, I can imagine a molecule with two optically active centers of opposite chirality,  between which energy can be transferred. One of the centers absorbs, transfers the energy to the second, which emits. This should set the molecule in rotational motion. Actually the disks in the AJP paper I referred to are already very small, and there is no particular reason they could not go to the nanoscale.

In any case, I attached a simple calculation which indeed does show that no conservation laws need to be violated. I don't see any problems with mechanisms either. Unless you want to discuss what exactly happens in absorption and emission, in a reflection, or in conversion of a right handed photon into a left handed.

Well mathematically it may seem OK, but I don't think it is justified at all to connect a single photon with thermal energy. The absorption of the photon can not happen at zillion of molecules, but at one single molecule. Then all what changes would be the KE of that particular molecule. QM may let the photon take all possible paths , but never let the interaction to take all possible places at once. Of course this is contrary to experiment.

Another thing I forgot to mention is that your suggestion also brakes the Third law of TD. You can easily reach 0 K by reflecting a number of photons from a body just by covering it with a good reflecting surface.

I think if you did a full relativistic treatment (after all the body goes faster and faster, so in the end you need to take that into account) no violation of the third, or any other law would take place. I don't have the time to work this out this weekend, but will come back to it later.

Of course an absorbed photon can be connected to thermal energy, this is what happens to most absorbed photons: they heat up the system. It may initially absorbed by a single atom or molecule, but the energy is rapidly dissipated as heat. And, by the way, in a reflection the interaction is with the collective electron modes of the system, which are not localized. These in turn interact with the phonons.

Let me ask you one question: do you agree that absorption of a single photon gives momentum to the system (which  can be any size, for heavy particles you will just not see it move) and that the absorbed energy heats it up? And that the energy and momentum balance are as I stated in the attached notes?

I have not yet worked out the relativistic case but it occurred to me that you probably don't even need that in order to save the third law: the body moves away faster and faster after every photon, so the next photon will be more redshifted than the previous one. You can see that from the expressions in the derivation. Therefore the temperature steps become smaller and smaller, exactly as argued by Nernst. It is a little unfortunate since I hoped for a short while that relativity is necessary to save the third law, but it probably isn't.

@Gert Van der Zwan,

I thought that you are trying to persuade me that the energy of the photon is dissipated between many molecules and than recollected and emitted by another molecule somewhere in the bulk.

But I see now you want to say that every molecule has thermal energy Ui to be regarded separately of the kinetic energy of the whole body mv2/2. So there is no need the molecule to transfer its energy to another molecule (although it can happen before emission of the photon). This would only makes thing worse but I return to it later (better in another post) because to clear the simplest case: a molecule absorbs and than emits a photon.

Now I believe you agree that there is a chance (say 10%) that the molecule that absorbs also emits. In fact I think it must be 60 % or one would hardly see refection.

To make things simpler let the body is at rest V=0. So the absorbing molecule has only thermal energy Ui. This means in has some velocity of unknown value and direction. As we know that the photon is absorbed and emitted again by some electron and this electron then has this same 'thermal' velocity.

What you say is: the new photon has the old energy (why exactly the old?) and the electrons velocity in the necessary direction  is diminished appropriately. Nevertheless the electron gained a velocity against the direction of the photon of necessary value too.

I think this should be a case of Compton scattering. Then of course the wavelength of the emitted photon is changed. And the reduction of the energy of the photon is compensated by an energy increase of this electron. In fact for metals one exactly has free electron cloud and hence Compton scattering.

I can't  see what an elementary process can do so as you say?? Of course there is the Moessbauer effect but there one has a grid and very low temperatures and there is no energy which rests in the grid (as far as I remember).

A relativistic treatment does not give much additional insight, except that it shows that once the body reaches a certain velocity, reflection is no longer possible. The reason is that for high enough velocity the energy loss by the system to produce a photon of the same frequency but  moving in the opposite direction as the incoming photon becomes larger than mc2 where m is the rest mass. I don't see a need to write down a detailed derivation,  it is rather simple, unless someone asks for it.

I think my previous answer to the alleged violation of the third law is still valid.

If you don't like the idea of single photons, it is also possible to do the whole derivation with plane waves with a certain energy density, as Einstein does in his: "Does the inertia of a body depend upon its energy content" .  Just use plane waves with circular polarization for the Beth problem.

So I stick to my answer to the original question: it is internal energy loss of the body that is set in motion which supplies the kinetic energy of the body as a whole. In a non-relativistic treatment this is loss of internal energy (or heat), in a relativistic treatment it is loss of mass, but that is obviously the same thing. To quote Einstein: "The mass of a body is a measure of its energy content" (The Principle of Relativity, Dover, p71). This includes thermal energy.

Just absorption of a chiral photon will also give the body angular momentum. I don't think anyone would have a problem seeing or believing that part of the energy of the photon goes to work (as kinetic energy) and part into heat. If there is a mystery here, it is how nature manages to flip the spin direction of the photon during its passage through  the system. Or, in the reflection problem, how the electrons in the system conspire to reverse the direction of the photon. But that was (fortunately) not the question.

@Gert Van der Zwan

Because of the quick response I didn't understand if you read my last post.

I would like to understand why you don't accept that it is Compton scattering that happens? Maybe with a (much) higher effective mass of the scattering electron which makes energy loss of the energy of the photon experimentally difficult to observe.

Apart from this the initial goal of my question is to understand by Beth while there is no energy ascribed to a photon due to rotation of the EB vector (energy of a photon is connected only with its impulse but not its spin) nevertheless when its changes from left to right (or v.v) it makes some work (rotation of the plate).

About third law: I think that the mechanism you propose can put the body into zero temperature simply because when when Ui is as low as possible and the body is at rest you can chose a photon or a flux of photons (plane wave possibly) with that E=hv  to reflect and rearrange  the whole UI into KE of the body.== >just  Nhe=Ui - this is of course difficult but not impossible. I can not see here necessity to consider relativistic effects at all.

I see that reflection can not be accounted for as a one electron - photon interaction. But to convert heat to kinetic energy by reflection diminishes entropy in a closed system.