joigus

Then \( E=mc^{2}+\frac{p^{2}}{2m} \) once one corrects for the zero of energies by adding the rest energy, right?

Ok, @martillo, you must agree that the description of a mechanical problem doesn't change if we just re-define the energy by a constant shift E0. Right? E-->E+E0 Equivalent problem. Same solutions. Everything the same. Right? Do we agree on that?

I'm in the middle of answering you, but first please clarify this: vp=mc2/v How is that a velocity?

And it is, if only you just admit that in non-relativistic mechanics you can redefine the spectrum of energy to include this new "energy at zero velocity". It is. It is a particular case. What the non-relativistic approach cannot give you is the mc2 term, for obvious reasons. It cannot give you a formula that's wrong. It only slightly generalises it to be right, ie, to include the rest energy. Haven't you noticed that the second term exactly coincides with Schrödinger's term? Am I not being clear?

I insist: There's a physical reason not to expect the Taylor expansion to recover the Schrödinger dispersion relation. And that's the humongous rest mass factor in the phase of the wave function. That is, \[ e^{⁻imc^{2}t/\hbar} \] which the Schrödinger theory cannot fathom. This "shifts" the Schrödinger dispersion relation. IOW, you should not get the dispersion relation, \[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \] But the one corrected for the rest mass, \[ \hbar\omega=mc^{2}+\frac{1}{2}mv^{2}-\ldots=mc^{2}+\frac{\left(\hbar k\right)^{2}}{2m}-\ldots \] \[ v_{p}=\frac{\omega}{k}=\frac{mc^{2}}{\hbar k}+\frac{\hbar k}{2m}-\ldots \] And indeed, that's what you get, if you Taylor expand what I showed you before, \[ v_{p}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq c\frac{mc}{\hbar k}=\frac{mc^{2}}{\hbar k} \] keeping up to terms linear in \( \hbar k \) That is what I can live with. Namely: an otherwise Schrödinger-compatible dispersion relation that only corrects for the rest-mass term. Can't you?

I can live with that. The phase velocity of a Schrödinger wave is not an observable.

Why? Unless you're doing gravitation, the zero of energy is immaterial, and all that matters is differences of energy.

LOL. And again I made a mistake when LateXing the answer. I guess you caught me. The results are OK I think. Don't worry too much about the phase velocity. It's not very substantial (physical). It's very telling that in the relativistic case you get a phase velocity > c. Think about them as vacuum effects that don't do anything but shifting the ground state. The group velocity is a more serious matter. It's related to the particle's actual velocity.

Ok. First of all, sorry I wrote c-1 on a couple of lines where I should have written c. If you go over the equations I wrote you will see that, for v<<c, \[ \sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq\frac{mc}{\hbar k} \] With this approximation, \[ v_{p}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\simeq c\frac{mc}{\hbar k}=\frac{mc^{2}}{\hbar k} \] \[ v_{g}=\frac{c}{\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}}\simeq\frac{c\hbar k}{mc}=\frac{\hbar k}{m} \] So there seems to be a mismatch with the non-relativistic expressions directly obtained from \( E= \frac{1}{2}mv^{2} \). The latter being, \[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \] \[ v_{g}=\frac{d\omega}{dk}=\frac{\hbar k}{m} \] In particular, it's the phase velocity that is to blame. I understand that's your point. Is it. not? Well there's a physical reason for that. All these expressions come from the assumption that it's a free particle we're talking about. Einstein's celebrated formula for the kinetic energy, \[ mc^{2}\sqrt{1-v^{2}/c^{2}}\simeq mc^{2}\left(1+\frac{v^{2}}{2c^{2}}-\cdots\right)=mc^{2}+\frac{1}{2}mv^{2}-\ldots \] gives a humongous rest-energy term mc2. This enormous energy shift is a constant, so it is of no effect when it comes to energy considerations, but produces a contribution to the frequency that kicks the Schrödinger dispersion relation out of whack. Is something like that what has you worried?

Einstein said: 1) The laws of physics are the same in every inertial reference system 2) The speed of light is the same in all inertial frames of reference This gives a group of transformations that, in turn, result in a transformation of velocities that's called Einstein's transformation of velocities, which is consistent with it, as couldn't be otherwise. For systems moving in the x direction, the Einstein transformation of velocities gives, \[ V_{x}'=\frac{V_{x}-v}{1-vV_{x}/c^{2}} \] Now, if you feed into it a light ray moving parallel to this direction (x, called the 'boost' direction), you get, \[ c'=\frac{c-v}{1-vc/c^{2}}=\frac{c-v}{1-v/c}=c \] So indeed the speed of light doesn't change, consistent with Einstein's principle of relativity. Now, what you are suggesting is a different principle of relativity. Let's call it Killtech's principle of relativity, which says, 1) The laws of physics are the same in every inertial reference system 2) The speed of sound is the same in all inertial frames of reference That is a principle of relativity that's perfectly logically consistent. Same with the speed of 1 mile per hour, same with any other speed that you choose. That's not how Nature behaves though, so the speed of sound must change according to Einstein's transformation of velocities, and consequently it changes from reference frame to reference frame. The observation that one could have one different principle of relativity for every which velocity that one choses, and still be logically consistent is trivial, and obviously doesn't lead anywhere useful. Is that what all this is about? Do you have some reason to suspect that sound plays a pivotal role in it all? The speed of light is fundamental for reasons that other members have told you about. You can do all the maths that you want, and you can rephrase it any way you want. You still got it wrong.

"Classical" is an adjective that people use to mean "as opposed to quantum". I don't think you want to say "classical" there, but "non-relativistic". Namely: referring to a spacetime where time is observer-independent. I don't understand your formulas, so I'l tell you what I know. For a wave that is monochromatic (has just one frequency), the wavelength divided by the period is obviously the speed. What other speed is there? So, \[ v_{p}=\frac{\omega}{k} \] Is both the phase velocity and the group velocity. More general waves have many frequencies. They're not monochromatic. When one such wave is concentrated around a particular region of space, ie, it has certain spatial "lumpiness", it is possible to prove it must be made out of a range of frequencies. It's not obvious, but not too difficult to see either, that the centre of such "lumps" move at a speed that's given by the derivative, \[ v_{g}=\frac{d\omega}{dk} \] In quantum mechanics, you must remember that the \( E \), \( p \) mechanical properties are related to the \( \omega \), \( k \) wave properties (frequency and wave number) by, \[ E=\hbar\omega \] \[ p=\hbar k \] If you use the non-relativistic expression for kinetic energy, \[ E=\frac{p^{2}}{2m} \] you get the dispersion relation, \[ \omega\left(k\right)=\frac{\hbar k^{2}}{2m} \] which produces a phase and group velocities, \[ v_{p}=\frac{\omega}{k}=\frac{\hbar k}{2m} \] \[ v_{g}=\frac{d\omega}{dk}=\frac{\hbar k}{m} \] Now, for light waves (which are always relativistic, there's no non-relativistic approximation for photons), you have, \( c\omega=k \), which gives, \[ v_{p}=\frac{\omega}{k}=c^{-1} \] \[ v_{g}=\frac{d\omega}{dk}=c^{-1} \] For relativistic matter waves, on the other hand, if you repeat these calculations, you get, \[ v_{p}=\frac{\omega}{k}=c\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}\geq c \] \[ v_{g}=\frac{d\omega}{dk}=\frac{1}{2\sqrt{k^{2}+\frac{m^{2}c^{2}}{\hbar^{2}}}}\left(2k\right)=\frac{c}{\sqrt{1+\frac{m^{2}c^{2}}{k^{2}\hbar^{2}}}}\leq c \] So the phase velocity is generally greater than c, while the group velocity is less than c. That's what I meant when I said that the relativistic approach is apparently paradoxical. This is very far from a rigorous analysis in terms of quantum fields, but it gives you an idea that the relativistic formalism implies some superluminal modes, which later we learn are non-observable, and the group velocity is the one that seems to correspond to the measurable degrees of freedom. Quantum field have these virtual or non-observable degrees of freedom. They always do.

Ok. It's not that transversal waves must have a dispersion relation consistent with E=pc. Rather, massless waves, those that have a dispersion relation E=pc, must satisfy a transversality in the context of a gauge theory. You can prove this from Maxwell's equations in terms of the vector potential, if I remember correctly.

Here I explained how that is not the case in terms of frequency (inverse period and proportional to energy) and wave number (inverse wavelength and proportional to momentum): Different waves have different dispersion relations. Relativistic wave equations, eg, have still another different dispersion relation that's apparently paradoxical (if one tries to interpret it in terms of a 1-particle theory).

So is it clear now that the sound equation is not Lorentz invariant, as vs is not a universal constant, nor is it a Lorentz scalar, or are we still discussing that?

Sorry. A c4 should be c2 inside the square root of what I wrote there. No. Each wave has a different dispersion relation. E=pc gives you a photon D.R. Matter waves are a different matter. 😬

No, it's not \( E=\left|\boldsymbol{p}\right|c\) for massive particles. As @MigL said, it's, for massive particles. So \( E=\left|\boldsymbol{p}\right|c\sqrt{1+\frac{m²c^{4}}{\left|\boldsymbol{p}\right|^{2}}} \). With this factorisation maybe it's clearer how and why it's not the same for massive particles?

Sorry I didn't read carefully. You actually mentioned the eikonal equation. What is it exactly that makes it non-wavy? The problem with it is when the wave finds inhomogeneities of size the order of the wavelength, then it no longer is a good approximation. But otherwise it's quite wavy isn't it? I mean, if you solve for the amplitude and the direction of the 'rays' you're home free I suppose.

No. It's similar as when you define mass as m=F/a, while you define F=ma. Things like F=-Gmm'/r² get you out of the tautology. In the case of electromagnetism, you have many other relations that get you out of the tautology, like E=hx(frequency), or E(nu,k) is a solution of Maxwell's equations in free space. Etc. It all works out, and you never look like a dog chasing its own tail. Believe me.

\[ \left|\boldsymbol{p}\right|=E/c \] with, \[ E=h\nu \] with h Planck's constant, and \( |nu \) being the photon's frequency in Hertzs. Direction of momentum given by, \[ \frac{\boldsymbol{p}}{\left|\boldsymbol{p}\right|}=\boldsymbol{k} \]

https://en.wikipedia.org/wiki/Eikonal_equation

No reason to expect it does. In order to function, fiction only needs to create enough sense of plausibility for you to temporarily suspend your critical thinking under the implicit assumption that the narrative will be entertaining: https://en.wikipedia.org/wiki/Suspension_of_disbelief

Or... It may sound weird and be a false assumption as well. Example: Seals are grey because polar bears are white. It sounds weird and it is false. "Photons don't carry mass because they always entail a very small disturbance of the vacuum" is as false a statement as can be. X rays are an example. Extremely-high-energy photons are as massless as extremely-low-energy photons. x-posted with @exchemist, who said more or less what I said.

It's just an attempt to focus the discussion on physical determinism, which is what you meant, I think. Is it not? And we've been talking about free will for quite a while now. The words "elbow room" kind of gave it away. PS: Edited.

One thing is physical determinism, and quite a different thing is behavioural, psicological --or what may have you-- determinism. Ie. Suppose some kind of physical determinism has been established and we all agree on it being the basis for the physical world. There would still be a long way to go in order to prove or convince anybody that this has any bearing on the question of free will, as @Eise has argued somewhere else, if I'm not mistaken. Those are two different things.

Might this not be just a tad glass-is-half-full as to determinism and/or predictability? I think we who are trained in science tend to overestimate the scope of determinism. Determinism suggests itself very strongly when you first fall in love with science. Then it hardly ever occurs in practice.