Bob_for_short

If we calculate only vacuum, zero point energy of the quantized electromagnetic field, then the plates, disconnected from this fild have nothing to do with it, why should they attract or interact? Therefore it is charge interaction effect that includes the quanitzed electromagnetic filed, attached to each charge. Such an interaction exists not only for conductors but for any material so it is not due to "boundary conditions". As I said, there is an atom-atomic interaction at long distances in which the QEMF contributes. Even in an isolated atom there is the Lamb shift which is due to taking into account the coupling to QEMF (an additional QM charge smearing).

An electromagnetic field, as a plane wave, has a known space-time dependence: E = E_0⋅cos(ωt-kx). In a transparent medium it is the same except for involving the refraction indices n. Now, let us look at the field in a moving reference frame - that with v = c/n. What solution is for the wave in such a frame? Isn't it a E' = E'_0⋅cos(ω't') ? Merged post follows: Consecutive posts mergedI meant E' = E'_0⋅cos(k'x').

How did you manage to write such a long sentence without a stop mark or any other delimiter?

If it is a position measurement, then one point suffices. Why then to talk about the space wave function? In my opinion, the system state information is obtained in a series of measurements, so each measurement is not a collapse but retrieval of a piece of the wave function. The state wave function is thus built from pieces since one measurement contains too little information about it.

Do you measure the electron or photon momentum (or coordinate) in a double or one-slit experiment?

According to Severian, "Quantum mechanical wavefunctions are only quantized if you set boundary conditions. And this is not really a 'quantum' phenomena, as ordinary waves are similarly quantized (pluck a guitar string and you get one note). The unique thing about QM is that observations change the state of the system, so it would have been better to incorporate that into a name (Observer modified mechanics or something)." I would like to clarify this question: a single measurement do not modify the system state ψ but supplies a bit of information about the state. Only an infinite number of measurements cover every possibility contained in the QM state, for example, the ψ(x) profile in a double slit experiment. The same is valid in Classical Mechanics: we exchange with many-many photons with a body to finally obtain an average - the "deterministic" center of inertia coordinates (think of one-slit QM experiment). Without sufficient exchange the state infirmation is not complete in both CM and QM. Let us note that in both cases the exchange means elementary interaction with a system so observing includes and only possible due to interaction (or "change of the state" in a narrow sense).

Yes, the formula for s.

Forget about time. There is a law of velocity adding, a relativistic law. It is more complicated than just v3 = v1 + v2. According to this law (discovered by H. Poincaré) whatever v1 and v2 are, the resulting v3 is smaller than or equal to c.

If a device uses a laser or some other essentially quantum effect in its functioning, then it is OK. I admit though there may be some exaggerations and abuse mentioned in previous posts.

Me neither. Well, I am disappointed and tired. I quit this conversation. Cheers.

If the light propagation depended on number of photons, it would be a non-linear effect. It exists indeed but at very high intensities. So the linear, usual theory of light propagation is valid for any number of photons, even for one.

Look in opto-electronics books on propagation of EMF in optical wave guides. You may learn interesting things like acceleration and deceleration of light in non homogeneous media, with variable n.

Again, speaking of absorption and re-emission as a reason for delay is wrong and it is you burden to learn when you were told so.

No, I am not going to. It is your interest, not mine, sorry.

Were not it you who spoke about uncertainties? If the coherent states are "enriched" (pumped), non coherent ones are suppressed, the total number of photons being conserved (equal in stimulated and spontaneous emission scenarios).

First, it is a common knowledge available in textbooks. Next, why should I make a search for you?

You references demonstrate a low educational level of those posters. I remind you that in a transparent medium there are bound charges that participate in creation the resulting field. It is impossible to separate (to distinguish) the incident and the induced fields - they both appear in sum in the charge motion equations as a unique force. The resulting field is different from EMF in vacuum, it should be clear. In transparent media there is no dissipation (absorption), do not fool yourself.

Read post #4 at http://www.physicsforums.com/showthread.php?t=104715, for beginning.

Yes, it is my opinion and it is the experimental fact. Your "explanation" is completely wrong. A transparent medium polarizes and takes part in EMF propagation. There is no absorption and re-emission, like in a fog. On the contrary, there is a collective effect that determines the actual EMF in a medium. The photon velocity in a medium is smaller: v =c/n, where n > 1. Shame on you.

You forgot to teach me the phase velocity too.

I do not need to prove that. Rather, it is you who should learn, pass exams, solve practical problems in order to judge professionally. So far you demonstrated your lack of knowledge and misled the OP author. Merged post follows: Consecutive posts mergedHave you ever heard of Cherenkov's radiation? Do you understand its physics?

You do not know the basics of physics here. Learn before expressing your guess.

The frame of a photon makes sense in a transparent medium where v < c.

The frame of a photon makes sense in a transparent medium where v < c.

No, two such photons are like one with a doubled amplitude.