Do atoms recoil when emitting a photon?

[sol47739]

Go read the second paragraph of my post #22 again.

You don't seem to be reading responses very carefully. I strongly suggest that you stop and take a step back and think for a while about what you have been told.Yes. Once again, the reference you yourself gave in post #5 clearly shows that. Did you actually read what it said?This is just another way of saying that temperature is the average kinetic energy of the atoms.

It does not, however, say that photons are "bound to the atoms", which is what I told you was nonsense.No, convection and conduction are always present whenever systems are in contact. Whether any net heat transfer occurs by these modes will depend on the temperatures of the two systems. But the processes themselves are taking place (the atoms at the contact surface are still interacting and exchanging energy) even when the two systems are at the same temperature and aren't exchanging any net heat.

I do read your responses. The problem I think is that I am too critical and I do not just accept something. And you are so to speak textbook correct. But since I doubt textbooks before accepting something, my questions might sound stupid.

Thanks for the exchange. I will think about it.

 

[PeterDonis]

you are so to speak textbook correct

My answers are not just repeating what textbooks say. I am using my understanding of the subject matter and ordinary common sense reasoning. You should be able to do the same to judge what I am saying. But you need to have a valid understanding of the subject matter to base your reasoning on.

 

[Vanadium 50]

I do read your responses.

About half of your messages are posted less than three minutes after the one you are responding to, and half of those are posted in a minute.

Is that enough time to read, think about and pose a thoughtful response?

 

[Dale]

those degrees are still “converted” in to photons or were originally exicited by photons.

This is not generally true. The degrees of freedom include atomic electron orbitals, molecular electron orbitals, linear kinetic energy, rotational kinetic energy, molecular vibrations, longer range vibrational modes, and probably others I missed. These modes can be excited by photons of the appropriate energy, but to assume that they always are, originally are, or even usually are is incorrect. They can also be excited through collisions, phonons, relaxation, and other similar processes. Matter simply doesn’t work as simplistically as that.

So the amount of photons bound to a system is directly proportional to the temperature.

I suspect that your concept of bound photons is probably wrong. When a photon excites some degree of freedom (regardless of what kind of degree of freedom it is) that photon ceases to exist. It is not bound, it is destroyed. If it were not, then its energy would be unavailable to excite the degree of freedom.

 

[hutchphd]

Also it is even incorrect to think that the "photon" interacts only with the electron. It interacts with the atom because the system has separate charges and therefore dipole moment that allows conservation of angular momentum and energy. The spectrum of transitions is complicated (as @Dale mentions) because the atom has other degrees of freedom. But the system atom+photon conserves energy,momentum,angular momentum, charge,..........etc

 

[vanhees71]

How is that consistent with the Blackbody spectrum if an atom in its ground state can move around?

Let's argue within non-relativistic QT for simplicity (and it's entirely justified for not too high $Z$). Due to translation invariance the total momentum of an atom is conserved, i.e., the center-of-mass motion separates from the relative motion of the electrons and the atomic nucleus. The bound energy eigenstates of the latter define the intrinsic states of the atom. Indpendently from this the atom can move with any momentum relative to your (arbitrarily chosen) reference frame.

If you know have the atom in some excited state you can always Galilei-boost to its rest frame, i.e., to the frame, where the total momentum of this atom is 0. Now it will at some random time relax to a lower state by spontaneously emitting a photon (due to the vacuum fluctuations of the em. field). This photon has a momentum $\vec{p}_{\gamma}=\hbar \vec{k}$, and since momentum is conserved (due to spatial translation invariance of the closed system consisting of the atom + (quantized) radiation field) the total momentum of the atom after emission of the photon must be $\vec{P}_{\text{atom}}=-\vec{p}_{\gamma}$.

This should be in any textbook on atomic physics. I've no time to look for a specific reference though ;-)).

 

[Frabjous]

I do read your responses. The problem I think is that I am too critical and I do not just accept something. And you are so to speak textbook correct. But since I doubt textbooks before accepting something, my questions might sound stupid.

Thanks for the exchange. I will think about it.

You need to apply some of that criticality to yourself. You are taking a hodge podge of facts and are trying to paste them together. If you glue an apple to an orange, it does not mean that an “orple“ is a useful concept.

 

[berkeman]

I doubt textbooks before accepting something

How is that learning strategy working out for you so far? Maybe more importantly, how is it working out for your teachers and tutors (like us)?

 

[Chas Tennis]

The text Optics, 3rd Ed by E. Hecht (1998) has the answer to the OP question. Page 51, Figure 3.16. Yes, atoms recoil. I don't see a reference.

The experiment shown is of
1) Source of atoms.
2) Through 1st aperture to form directed atoms.
3) Excitation is added.
4) Through 2nd aperture forming beam of excited atoms.
5) Later atoms radiate and the beam spreads.

Search for that unidentified experiment and similar experiments.