exchemist

However, Japan lost the Battle of Midway and were then progressively crushed by the United States.

No, it was their cucumber sandwiches.

That's far too glib. You do not know the temperatures I said you would need to know. It is the temperature of the two heat exchangers (the nature of which you have not so far specified) at the hot and cold ends that you need to know. You cannot just presume they will have the temperature of the medium conveying heat to and from them. (Strictly speaking it's not even that: it is the hot and cold temperatures reached by the working fluid, during the cycle, that determine the thermodynamics.) Apparently one of these media is boiling water, though you did not say so. (This illustrates why videos are not a substitute for a written description of the setup.) But the hot end is transferring heat into the engine, so the temperature of the heat exchanger will not be quite 100C (212F). If you are trying to insulate the cold end, the presumption must be that its temperature is not the ambient temperature of the room but significantly higher. If the insulation is perfect, then what will happen is that the waste heat rejected by the engine will progressively raise its temperature, eventually reaching that of the input temperature. As this takes place, the engine will run progressively slower and eventually stop. What is for sure is that for the cold end heat exchanger to reject heat into the room, its temperature must be quite a lot higher than the ambient temperature. Recall Newton's law of cooling, according to which the rate of heat transfer is proportional to the temperature difference. So, no temperature difference, no heat transfer. You need to know the real temperatures of hot and cold ends before trying to do a thermodynamic calculation.

Because you evidently still have a temperature difference, in spite of your efforts. What you really need to do is attach thermometer probes to the engine's contact with the hot source and its contact with the cold sink, to see what the input and output temperatures experienced by the engine actually are.. If the temperature of both becomes equal the engine will stop. If it does not become equal then you have a heat leak - or else the engine uses so little heat that the cold side takes ages to heat up and stop. As is usual with videos, the video shows us nothing worthy of comment.

I think you understand perfectly well.

Religion comes into it when it comes to censuring people for their preferences, which is the second of the two issues I was trying to disentangle. The first point is the sex drive point, in which by "we" I mean the great majority that has a heterosexual sex drive.

Because receiving an unwanted advance from a woman feels qualitatively different to me from receiving one from a man. I do not imagine I am unique.

Yes. It is also worth noting that the beam splits into 2J+1 parts. So if the atom has J>1/2, you get more than two parts, something that the classical idea of little compass needles can't account for at all.

No. It is never "nearly perfect". That is bullshit. The maximum degree of alignment is determined by the uncertainty principle limitation I referred to previously and that limitation applies in any situation. The magnitude of the angular momentum vector is √(J(J+1))h/2π, but the maximum value of its projection along the field direction is J.h/2π . This projection can have values of J.h/2π, (J-1).h/2π,.....0,..... -J.h/2π , and only those values. That is what space quantisation of angular momentum is all about. And no, I am not going to get into a side discussion about a piece of classical physics.

This may be my last post to you on this, unless you can avoid dragging other side issues into the discussion. From what you write, I wonder if perhaps you may may suffer from one basic misunderstanding. In space quantisation of angular momentum, the angular momentum vector does not completely align with the applied field. Ever. It continues to precess (in the semi-classical picture). It is only a component of the vector that is parallel or anti-parallel to the z direction of the field. It is impossible for the angular momentum to align totally, as that would violate the uncertainty principle*. There is always a component of the vector in the x,y plane, whose orientation is indeterminate - that is the real QM meaning of the "precession". So it carries on in this state of partial alignment, without any radiation occurring. So your analogy of a magnet oscillating, and then radiating until the oscillation ceases, does not apply. That is not what we observe in the Stern-Gerlach experiment. *The operators for angular momentum along x, y and z axes do not commute. Therefore only angular momentum along one axis can be precisely defined at a time, at the expense of definition of the others.

Exactly. But my problem with the term "homophobia" is that it may be closer to my own personal discomfort at the prospect of a sexual advance from another man than it is to any inability by some to respect the sexual preferences of others. So what we stigmatise in society as homophobia is actually not that, whereas a feeling of personal discomfort or distaste, which could perhaps be described as a mild kind of "phobia" (though overstating it), is a natural thing and not reprehensible at all!

Atoms "radiate abundant energy" is irrelevant, as well as being untrue. Atoms in the ground state do not and cannot radiate energy. This flatly contradicts the predictions of classical EM theory. That is one of the chief reasons why QM was developed. But you seem to adopting a wild, scattergun approach to this topic now. Unless you can stick to the point it will soon become a waste of time discussing this with you.

I suspect distaste for homosexuality is inborn in many of us. Given that we have a drive to be attracted to the opposite sex, we find the idea of sex with someone of the same sex a big turn-off. Consequently we may find the idea of a sexual approach from somebody of our own sex rather disturbing. If that is homophobia, then I am a homophobe. It seems to me that the blanket term "homophobia" is thrown around too easily. One needs to draw a distinction between personal sexual taste and the attempts by some to condemn different (minority) tastes in others. It is the latter that society should refrain from.

No, that is wrong for this situation, because it is a bound system and therefore quantised. I pointed out to you earlier that electrons in atoms do not radiate and fall into the nucleus. That's because it is a bound system, which restricts the states the electron can occupy. Forget brehmsstralung. It's irrelevant. That is for free, i.e. unbound, particles. The Wiki article on brehmsstralung makes the point: https://en.wikipedia.org/wiki/Bremsstrahlung. An unbound particle can occupy a continuum of states: they are not quantised. So the particle can radiate and lose energy in a classical manner. Electrons in a bound system cannot do this.

No need for a video, you just need to be able to read and take in what you have already been told, several times now. You get a double displacement reaction when one of the possible combinations is insoluble, because it is more stable in the solid state than in solution, and therefore precipitates out. I've told you this, so has @chenbeier, and it is repeated in the Wiki link I went to the trouble of providing. In your latest example it is Ca(OH)₂ that is much less soluble than the other possibilities, so it will precipitate from any solution with those ions in it.

Why can't it just divert a proportion of the existing angular momentum - and kinetic energy - to the precession axis? My understanding is that is what happens in a lossless precessing gyroscope. (Though maybe a proper physicist can comment on whether that is the case.)

OK, why do you think precession means additional kinetic energy, rather than just partitioning the existing kinetic energy between motion about two axes? What work is done? Take me through the logic.

That, frankly, is a stupid reply, given everything you have been told.

Ah but what "kinetic energy of precession" is this? On reflection I don't think there is any. I think the phenomenon of precession merely re-allocates the existing angular momentum into motion about 2 axes, doesn't it? The energy change results from the change in potential, due to the alignment/anti-alignment of the magnetic moment to the field.

"Magic happens" is not in any textbook I have ever seen. I stress the idea of precession is a semi-classical analogue for what happens quantum-mechanically. As I say, my understanding is that some of the atoms gain energy from the applied field and some lose it to the field, depending on which M state they are in. This is reflected in the idea of kinetic energy of precession but, as that is a semiclassical analogue, you can't take too far. The analogy works intuitively for those states that are raised to higher energy by the field, but doesn't work so well for those that are lowered in energy. If you take a single atom, the field changes the potential experienced by the electrons in the atom. They no longer just experience a spherically symmetrical electric potential from the nucleus, but also a linearly polarising magnetic potential as well. They either gain potential energy from the field or give up potential energy to the field, depending on orientation. This is presumably what is responsible for the force and consequent deflection they experience as they pass through the field. I don't think there is any need to presume they have to radiate.

You have asked this type of question several times now and the answer has been the same in all cases: consider the solubility of the possible combinations of ions. There is an article in Wiki on this: https://en.wikipedia.org/wiki/Salt_metathesis_reaction Note especially the section on counterion exchange, which is what your questions have been about. There is also a link in the article to a quite useful solubility chart, which is helpful if one does not know whether a given ionic compound is appreciably water-soluble or not: https://en.wikipedia.org/wiki/Solubility_chart

This is yet another commercial plug from SCIRP, a Chinese publisher pretending to be located in California that is suspected of predatory practices: https://en.wikipedia.org/wiki/Scientific_Research_Publishing

The mechanism in the Stern Gerlach experiment is as written up in the textbooks on QM, surely? My understanding of this, admittedly from my 1st year as an undergraduate, 50 years ago, is that the atom precesses at the appropriate Larmor frequency for the M state it occupies, and carries on doing so indefinitely, unless it encounters radiation of the right (i.e. Larmor) frequency to stimulate a transition to a different M state. Regarding energy conservation, what I think happens is that those states with an antiparallel component (of magnetic moment: due to the -ve charge on electrons the magnetic moment points in the opposite direction to the angular momentum itself) - acquire energy from the field, while those with a parallel component give energy to the field, the net effect being zero, in line with conservation of energy. However, once the states are split, statistical thermodynamics enters the picture, causing the ones in the higher states to tend to lose energy and drop down to lower states. As I say above, I think this relaxation process will be primarily non-radiative, because of the improbability of spontaneous emission at such low frequencies. Non-radiative relaxation will manifest itself as heat in the material. So no, I do not think there is significant radiation emitted in the Stern-Gerlach experiment. But I'll be interested to see if anyone else has a different view. P.S. Be aware, too, that the notion of precession is a semi-classical one. In QM, only the compment of the magnetic moment or angular momentum along the field z-direction is defined. The orientation of the component in the x,y plane , i.e. the component that "rotates" in the precession model, is undefined. This is a consequence of the uncertainty principle (non-commuting operators)

When a compass needle aligns with the Earth's magnetic field, are you telling me it emits radiation?

Nope, you emphatically do not get classical EM radiation. This is a quantised system. Please read my 2 previous posts. We can get onto spin-echo EPR later, once we have agreed on the basics.