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?

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.

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.

@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).

@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.