freecharly

@ArthurOfNavases Our first natural instinct would be, of course, to look for a causal explanation. This guides us successfully in our usual environment. But it seems that there are objects with non-local connections in the microscopic world that are part of nature we have to accept, We have the theories to describe them. Thats usually the goal of physics.

@ArthurOfNavases Re "...there is nothing in physics that says no information is exchanged between particles." Information is exchanged between observers, not particles. Bohms theory is interesting, has the same results as QM, introduces this nonlocal pilot field, but has problems with relativistic extension.

@ArthurOfNavases "There is no experiment that tells you which measurement outcome will occur". Thats just typical QM! By whose definition is this "not complete"? QM has shown very successfully that this is how nature works. The observed measured non-local correlation effects, can be "superluminal", but the sequence of events can be reversed depending on the reference frame. This is contrary to the commonly accepted concept of causality in physics. Bohms theory seems to be causal, but introduces exactly the same random effects by "unknown" initial random variables.

@KDP Very vivid thought experiment! It corresponds to the Bell experiment measuring the two spins (or photon polarizations) in parallel directions. This parallel detectors case, as opposed to certain inclined ones, could still be explained by local hidden variables or angular momentum conservation and thus be causal. The problem with the causality of non-local correlations measured faster than light is that the SR time order of events can be reversed from a suitable reference frame. The usual definition of causality requires that the time order of events is the same for all reference frames.

@ArthurOfNavases The no-communication theorem of quantum information theory states that measurements on one quantum mechanical (qm) subsystem cannot be used to transmit information to another subsystem, in particular, by the non-local correlations in entangled subsystems. It follows that no superluminal communication is possible by these correlations. Indeed, no superluminal information transmission has ever been found experimentally. All experiments are explained by standard QM. Thus there is no need for a "more complete“ theory. Research on theories (e.g, unification of GE & QM) continues.

@KDP This is a very interesting discussion! I have to leave now and will continue tomorrow. Thank you!

@KDP Thank you for your comment. I wonder whether the observed nonlocal correlation falls under the usual definition of causality. Even in classical science correlation is usually no proof of causality. A would appreciate, if you coiuld give an expert reference for that.

@ArthurOfNavases The experimental results on one part of the entangled system are random, you cannot chose them. Thus the later observed correlation cannot be used to send a signal. Bell has shown that there is no local hidden variable theory (which could give a causal explanation) that explains the results. As far as I know, the Bohmian theory is an explicitly nonlocal (hidden variable) theory which yields the same results as quantum mechanics and doesn't allow transfer of information either. Experiments and QM theory show that entanglement doesn't allow superluminal signal transfer.

I will do it. Thank you...

TEM means that there are only electric and magnetic field orthogonal to the propagation direction

The question is simply whether there are spherically propagating TEM waves or not.

That's also OK with me...

Jackson excludes possible TEM waves by considering only solutions where not both E and H are orthogonal to r.

Without additional boundary conditions in space.

I was thinking of free-space spherical wave solutions, i.e., solution of the homogeneous spherical EM wave equations, or Helmholtz equation if you want.

@EmilioPisanty - I thought that you might also discuss the general topic as you are pretty expert in the field. If you don't want to discuss it, it's also OK with me. Maybe its a good suggestion to eventually pose a question on the main.

@EmilioPisanty - My questions are driven by pure interest in physics not by any any supposed or perceived personal intransigence. I don't want to continue on this, but as you bring it up, my answer really assumed that the question was related to a plane wave only. I didn't think at all that the question could refer to general EM waves. I don't take your criticism of my answer at all personal, I only disagreed with the assessment that it claimed a general orthogonality assertion.

I have the following thought experiment. Take two parallel metal plates with a standing transverse EM wave in between. You bend the plates to two concentric spherical shells with the standing transverse wave mode inside. Is this possible and is it still a transverse wave? Then you suddenly eliminate the metal shells and let the EM wave propagate outwards. Is it still a transverse EM wave?

@EmilioPisanty - Emilio, I really don't want to bother you, but I have the impression that you are a theoretically rather knowledgeable contributor on this site. Therefore the question whether there are some solution of the EM wave equations that are the spherical analogue to transverse plane waves. In Jackson, the considered Helmholtz equations are for the variables $\vec r·\vec E$ and $\vec r·\vec H$ which makes possible transverse EM wave solutions to trivial zero solutions.

@EmilioPisanty - Emilio, am familiar with the mentioned Jackson chapter. Thank you for pointing it out and for the interesting link to the "final question".

@EmilioPisanty _ You are missing my point. I am not looking for more counter-examples to a general orthogonality misconception, which I never adhered to, I am looking for a possible counter-example to the idea that said orthogonality only occurs in plane waves.

@EmilioPisanty - It seems that, in contrast to a purely scalar wave equation, there seem to exist no EM wave solutions that are completely spherically symmetric. There has always to be at least a $\theta$ dependence, which in itself is already rather intriguing. Whether the $\theta$ dependent solutions can still have the property of $\vec E$, $\vec B$, $\vec S$ (or $\vec k$) to be mutually perpendicular is not clear to me. In this context, I also wonder if an atom can emit a photon as a spherically symmetric EM wave with identical probabilities to detects anywhere on a sphere.

@EmilioPisanty - Thank you for the link to the articles. I have to read them in detail but they seem to refer to rather complicated situations. I am thinking of the much simpler question whether there is a free space spherical analogue to a plane wave where the EM fields and local propagation are still mutually perpendicular. Simply put, are there spherically propagating solutions of the homogeneous EM wave equations for $\vec E$ and $\vec B$ in polar coordinates where this still holds.

@ZeroTheHero -Thanks for the interesting link.

@ZeroTheHero - But I would like to know whether only plane free space waves have mutually perpendicular $\vec E$, $\vec B$ and propagation direction. Especially also, whether it doesn't hold, as stated, for the free spherical solutions of the homogeneous equations.

@ZeroTheHero - This seems to describe a TE guided wave propagating between two parallel metal plates in z-direction which has a B component in z-direction. It is well known that such waves can be described as the superposition of two plane TEM waves, that their $E$ and $B$ field are not perpendicular, and that they have $B$ components in propagation direction. This has already been stated by Emilio. But these waves are not free space waves, but waves determined by the metal boundary conditions. I don't doubt at all that such superposition solutions of Maxwell's equations exist.

Dear Emilio - You have not given a proof of your claim that "the wave number $\mathbf k$ and the electric and magnetic fields $\mathbf E$ and $\mathbf B$ are not perpendicular to each other", with the exception of plane waves. The (linked) counter-examples are mostly waves determined by metal boundary conditions, superpositions of plane wave solutions or solutions of the inhomogeneous wave equation. I also wonder where it is shown that $\vec E$ and $\vec B$ and $\vec k$ are not mutually perpendicular in spherical wave solutions of the homogeneous electromagnetic wave equation.

You provided the usual derivation for plane waves. But you didn't give a proof that it doesn't hold in general. And it is pretty trivial that it doesn't hold for an arbitrary superposition of plane waves.

@EmilioPisanty - Take it easy! You should probably learn to accept that somebody is not of your opinion. I am not aware of any "off-topic" posts or "rants" of mine, much less "rude or abusive" ones. The single comment you are complaining about was actually on topic, admittedly with a humorous twist, but definitely not rude and abusive. When you are frequently very rough with others, you should yourself not be a mimosa.

@EmilioPisanty - I was only responding to your comment above where you insist that "this answer is incorrect" which I obviously don't agree with. In the end I suspect that you are not so fond of this classification either. As suggested, I will put my comment into my answer.

@EmilioPisanty - In my opinion, and I speak as a researcher and academic teacher, such an reductionist classification of photon-matter interactions is at best useful for a few theorists studying QED and a mere curiosity. For the general physicist, especially students, such a classification is mostly harmful. It doesn't help at all to understand, much less quantitatively describe, most of the important physical effects squeezed into it.

@knzhou _ I think that you are right that here the definitions of the three interactions are essential. Therefore, it would be interesting to know the actual definitions of the three interactions that supposedly encompass all interactions. The question is then whether these still correspond to the original meaning of Compton scattering, photoelectric effect and pair production, or whether they are just over-extensions of these concepts.

@Trevor_G - In the CB junction, for $V_{CB} \lt 0$, the current coming through the base by diffusion flows against the applied potential difference. But, due to the built-in voltage being larger than the opposing applied $V_{CB}$ , the electrical field in the junction is still directed towards the base so that electrons are transported to the collector producing a positive electrical current flowing into the collector terminal.

@Trevor_G - The base is not creating so much current. It is a current that flows through the base mainly by diffusion and in the CB junction for $V_{CB} \gt 0$ against the applied voltage. Therefore, in a circuit diagram the output circuit usually includes a controlled current source.

@Trevor_G - I am really sorry, it was definitely not my intention to insult you! I sincerely apologize if you got this impression. I am not thinking that you are an idiot! Communication of technical issues is sometimes not easy. The whole problem obviously stems from the unclear formulation of the question. You were assuming a load circuit which was not at all obvious to the reader before you added the circuit diagram. My honest intention is to understand your problem and to give you a satisfactory answer.

@ThePhoton - I am not sure whether the OP really wanted to hear this answer considering the inner working of the bipolar transistor. Actually his question was based on a certain circuit which he entered only in a later editing. Thus the original question was completely unclear and one had to make intelligent guesses, as yours, which was very founded, what the question was actually about. Now it turns out that the voltage drop in the collector circuit plays a major role in his thinking.

@Trevor_G You have also keep in mind that the CB junction is not a simple lumped resistance in this case.

@Trevor_G - I see your circuit diagram now. You should have included this from the beginning to avoid guessing what your actual question is. With large enough positive collector current (flowing into the collector terminal), the voltage drop on the resistor can become easily so high that $$V_{CE}=V_2 - I_{C} R_1 \lt V_{BE}$$ This is simple circuit theory.

@ThePhoton - Thus you can have a positive electric current entering the collector terminal in spite of the negative $V_{CB}$, or equivalently, for $V_{CE} \lt V_{BE}$

@ThePhoton - Let's talk about the electron current (which is opposite the usual current arrow). When $V_{CB}$ becomes negative, the CB junction becomes forward biased so that an electron injection current flows from the collector into the base (and emitter) which is superimposed opposite to the electron injection current from the emitter into the base and collector. For not too large negative $V_{CB}$, this electron current injected from the collector into the base is usually smaller than the usual electron current injected from the emitter reaching the collector ($n_E >> n_C$).

@ThePhoton - So he doesn't formulate the question in an understandable way. I have been teaching this subject for decades in EE courses as a professor at a university.

@Trevor_G - So you assume that there is a load resistance connected to the collector which is not shown in the diagram? Then the question is circuit related and you should draw the whole circuit to make clear what your question actually is.

@Trevor_G - The transistor doesn't "generate" any voltage. Usually, the collector-emitter voltage is defined by the applied voltage or equivalently by the external circuit. If your problem is the direction of the collector current, of course it can change direction when the collector-base voltage becomes negative and high enough, when the electron injection current from the collector into the base becomes larger than the electron injection current from the emitter into the base.

@ThePhoton - The question is not so sophisticated. The OP simply cannot understand why the collector-emitter voltage can be smaller than the base-emitter voltage because the collector is drawn above the basis in this picture. He has to realize that the collector-emitter voltage can take any voltage value and sign you apply between the collector and emitter terminal.

@Trevor_G The voltage you are applying defines the collector emitter voltage! There is no reason why you should not be able to apply any arbitrary voltage to the collector with reference to the emitter. That the collector is drawn above the base doesn't mean that the voltage cannot be lower than the base.

@kludg - There might be (or exist already) more results in this field that will be experimentally testable. Just because ideas are not mainstream doesn't mean that they should not be pursues scientifically. I don't know which high priests of physics have decreed that. Einstein's meddling with the meaning of space and time or gravitation was definitely not mainstream at his time. Also think of the work of Aharonov and Bohm, who showed that the vector potential $\vec A$ of the EM field is not just a useful mathematical trick, but is a physical quantity itself that can be tested experimentally.

@kludg - I cannot agree with your assertion that the interpretational question "is simply not physics"! A good example is the work of John Bell who had doubts in the Copenhagen interpretation of QM, influenced also by Bohms causal interpretation. He developed his now famous Bell's inequalities for local hidden parameter interpretations. This work triggered a still ongoing wave of experimental and theoretical investigations. Although presently unimportant for practical calculations, the question of interpretation of QM could lead to further advances, it definitely belongs to physics.

@Steve -You describe it accurately. I can only agree with you! Such a presumptuous show-off is unworthy of serious scientists.

@EmilioPisanty - I really don't want or need to change your subjective assessment. In my opinion, it is reasonably correct and meaningful for a short ad hoc answer. Maybe it could have been worded better, but it is definitely not "wrong". I think that many physicists would probably agree on that.

@EmilioPisanty - A little moderation would probably be better for a civil discussion. If you see something "wrong" and "misleading" you should clearly point it out and give an improved explanation, what you have not done. You seem not to have really read my answer. In the last sentence, even though I avoided the high level jargon you are using, I pointed out the deterministic "unitary" evolution of the wave function according to the Schrödinger equation which determines the "projective" measurement probabilities.