joigus

The ruler's length? Invariant under what? Rotations? It is. Lorentz transformations? It is not. General coordinate transformations? (Thereby including gravitational fields, either static or dynamic) It is not. You see. It's not as simple as something is or is not invariant. It takes a little preliminary work to even say what you mean. At least to me. Other people have the priceless ability to grasp the inner logic of fuzzy statements. Not me.

No, my reason to be confused is what I said: Two observers seeing two different things see... well two different things! What's strange with that? As @Markus Hanke has made perfectly clear. GR tell you how different the different things appear to be provided you carefully specify the conditions. If they see the same thing, how come, if they are apart? @swansont said that. There are other inconsistencies or unspecified conditions: How do they synch their clocks 14 billion years ago? You didn't say or wrote a diagram. @studiot pointed it out. You're spreading all your argument with "they see", "they measure" and the like. That's not good enough, not in relativity, never mind special or general, as a good read of Einstein's original papers or a good modern relativity book --Ray D'Inverno's book is a nice example-- makes clear from the beginning. Something like this: The diagonal lines are photons going from one observer to the other, telling each other exactly where and when they "saw" something or actually "seeing" something. That's what "see" means in relativity. The non-diagonal lines are inertial observers or objects moving at v<c. By the way, how do two distant observers "see" the same photon? Because then it's just one photon we're talking about, and two distant observers somehow catch the same photon. Now it's Saturn's orbit. You seem to keep moving the goalposts. You have to state your problem clearly, and try not to change the conditions, unles it is to refine the statement of the same problem. Stating the problem clearly is part of the art of doing physics. So I'm still confused. It was many years ago that I understood there is no blame for being confused during a physics class. The real problem is being confused and not being able to tell why.

Well they would agree on the speed of light, wouldn't they? They would disagree on the frequencies and wave numbers, or IOW, the number of cycles a particular physical interval contains. I'm a little confused with the initial setting of the problem, as I see no reason why Alice and Bob should agree on anything, as they see different things. But I take it that the OP is trying to reformulate the problem for just one photon. But maybe I misunderstood the whole thing, so I'll take some more time tomorrow.

Here's a wild idea: https://www.google.com/search?q=applications+of+zener+diode+in+smps And take it from there.

Absolutely. I reacted to your pointing this out from the very beginning. I elaborated (or tried to) a little more on that. But to no avail. The funny thing is, when I first saw this thread, I interpreted the title literally, "analogies for relativistic physics". I said to myself. "oh, that's interesting". Nothing further from the truth. Not interesting, not enlightening, not even funny at all or in any sense. --funny as in, "That's funny, this member doesn't seem to understand the principle of special relativity: It has two parts; the second one destroys his reasoning".

That surely is the Laplacian operator, which is nabla dot nabla. "dot" meaning the 3D "dot". 4D nabla dot 4D nabla is called the d'Alembertian, and it's a square (at least in physics). https://en.wikipedia.org/wiki/Laplace_operator https://en.wikipedia.org/wiki/D'Alembert_operator

Ok. Yes, thank you for pointing this out. The word "wave" is sometimes used very loosely, for these and other reasons. People also use the word "wave" for solitons, which are objects somewhere in between. I did go too far when I said "totally wrong". My apologies to @KJW.

You were waxing philosophical, "everything is an abstraction", "it's only in our minds". So Ok, I guess in a way you're right. Ultimately, the only thing we see, the only thing we measure, is positions of measuring scales, numbers flashing in LED lights, and the like. So everything is position, colour, time, what may have you. But there's always a theoretical layer beneath that. My point was that there are concepts that seem to demand you to consider them from the theory, although they cannot be measured, directly or indirectly. Example: phase velocity of a quantum wave.

I think it's probably always fair to say that theories are models in our minds. But not just any models. They are under the obligation to explain and predict the world. And that's a very stringent constriction. As to the phase velocity, the wave function, and its local phase, all indications are that there are part of a "factor" of the world we can move through using mathematics, but quite invisible for us. You seem to feel some discomfort because of this fact, that there are quantities we seem to be forced to talk about but cannot see. But that's part of physics since the inception of QM. But maybe it was always this way. We cannot directly see energy, or angular momentum. Think about it.

Yes, thank you for your appreciation of this point. That's what I meant when I said, The very moment you trim down a quantum state from the idealisation of an infinite monochromatic wave to a short pulse, you are introducing infinitely many frequencies, and you need the group velocity to describe its motion. No other velocity makes any sense. In the relativistic case, it's actually superluminal, as I proved from simple relativistic-dynamical and quantum-dynamical constraints.

I'm afraid we have different understandings of what "explicit" means, at least for this case. "Explicit" as in "He gave me very explicit directions on how to get there." (Taken from Oxford's dictionary). Yes, I'm sure about that. https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/QED/GaugeInvariance_2.html Read under the heading "Local Gauge Invariance". IOW: You can re-define the local phase at will as long as you accompany such change by the corresponding gradient in the gauge field. What is "the" phase now? No wonder phase velocity is not an observable. You can concoct situations in which phase and group velocity are essentially the same. I'd venture to say that for those cases you can "measure" the phase velocity. What you're doing (secretly) is, of course, measuring the group velocity, and using the theory to deduce a value of the phase velocity consistent with it and with the choice of quantum state (or "ray") that you have obtained in that particular gauge. No. You got this totally wrong. Quantum waves are not sinusoidal. For starters, they are not real functions. They typically go like complex exponentials. No. I can write A=XY/Z, the relation be totally right, and neither A, nor X, Y or Z be measurable. In fact, what you both are saying contradicts the principles of mainstream quantum mechanics --except for free particles. If the particle is interacting, the so-called Hamiltonian has a potential energy term, and the momentum (the inverse wavelength) does not commute with it, so the principles say no, you can't measure both at the same time. They're called "incompatible." You both are confusing a component (among infinitely many) of the quantum state with the whole quantum state. More on this: https://www.mathpages.com/home/kmath210/kmath210.htm And so on, and so on, and so on...

Again. How? Be explicit, please. That would contradict gauge invariance, which we know to be exact. In a nutshell, gauge invariance tells us that quantum states don't have "a" phase, meaning an unambiguous local phase. Let alone a phase velocity. The phase velocity of ripples on a pond can be measured. The phase of the wave function cannot. You can measure interference patterns. But those have nothing to do with phase velocity. You talk about frequency and wavelength as if there was just one frequency and one wavelength for matter waves. It is precisely because all "realistic" matter waves package many frequencies and wavelenths that the phase velocity is rendered all but meaningless in QM. Only the group velocity makes physical sense. I've just had a déja vu. Didn't I say something like this before? It's either right or wrong. What's sure not to be is irrelevant to the question at hand.

I don't think that's what the text means. I think it's a reference to the group velocity. Whenever group velocity is different from phase velocity, we call the medium "disperssive", and that's because every monochromatic component travels at a slightly different speed. v=v(f)

This sounds all very reasonable, and you do have a point. Nevertheless... We don't have a handle on the wave function itself. X-crystallography and the like is based on phase differences. Interference positions --and thereby wavelengths--... pretty much the same. No way to see the phase itself though. In fact, I'm working with some advantage here. I happen to know there is a very robust principle of physics --the gauge invariance principle-- that tells us it is impossible to know what that phase might be, as I can always gauge away any phase prescription that you take by locally re-defining the phase. So I have a pretty solid understanding of why what you claim cannot be true. Gimme any phase you like and I will "gauge it away" without breaking any known rules of quantum field theory, or classical electromagnetism, etc.

The particle's velocity is the group velocity. What laboratory measurement gives you the phase velocity? Didn't I say this before? And didn't you?

So if I have a "theory" that relates the size of the universe with the size of my nose --never mind how crazy that theory is--, when I'm measuring the size of my nose, can I claim to be measuring the size of the universe? Electron diffraction also measures electron wavelengths. How do you measure both wavelenght and frequency? --Thereby phase velocity. Huh? I'll be patiently waiting for any of your answers. This should be fun.

How?

Really? Please, tell me of an experiment to measure the phase velocity of the wave function of an electron.

Monochromatic-wave solutions of the non-relativistic Schrödinger equation.

They are incompatible, only if you interpret they are both exact. They are compatible, if you interpret one is approximate and the other is exact. If you just say, Then you're misrepresenting what I said. If you juxtapose an approximate statement and an exact one and you take both at face value, of course you would be expected to find contradictions! x-posted with @swansont

I think you are misrepresenting what I said. One is fully relativistic (vp=c²/v); the other (vp=v/2) is a non-relativistic approximation. I see now I must have done a very poor job of explaining that. I thought it was clear for you when you said, Is ex=1+x for small x incompatible with exe-x=1? No, if instead of ex=1+x you write \( e^{x} \simeq 1+x\) instead of just saying they are equal.

You can do as you please, of course. I will just finish by telling you that a theory that fixes a problem that doesn't exist is not a better theory. The phase velocity is not an observable quantity, so any theory that "fixes" the phase velocity to conform to a particular theoretical prejudice of what it should look like is deeply misguided. And please understand that I mean well. I don't want you to waste your time in something that I think is hopeless, and that's why I'm telling you.

Ok, let's take it from there then. This \( \frac{1}{2}mv^{2}=p^{2}/2m \) is what should be identified with the Einstein quantum relation ℏω (Planck's constant times the angular frequency), as the time-independent Schrödinger equation says, \( \left(P^{2}/2m\right)\psi=E\psi \). Then, it's a matter of doing some algebra after plugin in De Broglie's \( p=\hbar k \), \[ \hbar\omega\simeq\sqrt{\left(\hbar k\right)^{2}c^{2}+m^{2}c^{4}}-mc^{2} \] \[ \hbar\omega\simeq mc^{2}\sqrt{\frac{\hbar^{2}k^{2}}{m^{2}c^{2}}+1}-mc^{2} \] Taylor expanding the square root in powers of the small quantity \( \frac{\hbar^{2}k^{2}}{m^{2}c^{2}} \) (as \( mc\gg\hbar k \)), \[ \hbar\omega\simeq mc^{2}\left(1+\frac{1}{2}\frac{\hbar^{2}k^{2}}{m^{2}c^{2}}\right)-mc^{2} \] So that, \[ \hbar\omega\simeq mc^{2}\frac{1}{2}\frac{\hbar^{2}k^{2}}{m^{2}c^{2}}\Rightarrow v_{p}=\frac{\omega}{k}\simeq\frac{1}{2}\frac{\hbar k}{m} \] while, \[ v_{g}=\frac{d\omega}{dk}\simeq\frac{d}{dk}\left(\frac{\hbar k^{2}}{2m}\right)=\frac{\hbar k}{m} \] So indeed, \[ v_{p}=\frac{v_{g}}{2} \] Please, mind the \( \simeq \) in the next-to-the-last line. What about \( v_{g}=c²/v_{p} \) Mind you, this is an exact identity that comes from, \[ v_{p}v_{f}=\frac{\omega}{k}\frac{d\omega}{dk}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\frac{c}{\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}}=c^{2} \] Of course you cannot get that from the Schrödinger dispersion relation, as is clear from the fact that it has no \( c \) in it. So you're mixing an exact identity there with an approximation, which could be part of the reason of your confusion. Does any of that address any of your concerns? Does it make sense?

Ok. So the quantity that's "under the obligation" of giving us back the non-relativistic kinetic energy \( \frac{1}{2}mv^{2} \) is not \( E \), but \( E - mc^{2} \). That is, when \( v \) is very small as compared to \( c \), \[ \frac{1}{2}mv^{2}\simeq\sqrt{p^{2}c^{2}+m^{2}c^{4}}-mc^{2} \] Do you agree so far? I just want to make it very clear that I'm not trying to pull wool over your eyes in any way. This technique of "incremental agreement" I learned from @studiot, by the way.

It's a Taylor-series expansion. Remember?: \[ \frac{mc^{2}}{\sqrt{1-v^{2}/c^{2}}}\simeq mc^{2}\left(1+\frac{v^{2}}{2c^{2}}-\cdots\right)=mc^{2}+\frac{1}{2}mv^{2}-\ldots \] Didn't you understand when I said it, or don't you understand now? It is neither strange nor non-strange. It's a Taylor-series expansion. A Taylor-series expansion is... well. A Taylor-series expansion. What's strange about that?