KJW

The same could be said about any quantum wavefunction because we are in fact talking about the quantum wavefunction of a free non-zero mass particle.

I don't think it would be too difficult to arrange the frames of reference and the rod orientations to give the same length contraction, thereby establishing that the rod is the same length in both frames of reference. The simplest would be for the rod to be oriented parallel to the change in velocity in both frames of reference, and for the third frame of reference to be collinear to the other two frames of reference.

One could measure the clock and the rod in both frames of reference from a third frame of reference relative to which the two frames of reference have equal speeds.

How is the above a direct measurement of electron density? I've already indicated above that x-ray crystallography is not a direct measurement of molecular structure. Producing a detailed image of the electron density of a molecule is not the same as being a direct measurement. That's exactly my point. If a direct measurement is unavailable, then an indirect measurement based on sound principles is acceptable. Thus, although the phase of a De Broglie wave cannot be measured, the phase velocity of a De Broglie wave can be indirectly measured by measuring the wavelength and frequency of the wave.

No, you have misunderstood what I said. If in some inertial frame of reference, I have a clock that ticks away seconds and a rod that is one meter long, and I accelerate to some other inertial frame of reference, then the clock will still tick away seconds and the rod will still be one meter long. This is the principle of relativity in action and neither time dilation nor length contraction would make sense without it.

A lot of science is based on indirect measurements based on some theory. For example, in chemistry, do we really have direct knowledge of chemical structure? Perhaps the most direct measurement of a molecule's structure is through x-ray crystallography, but even that involves the theory of x-ray diffraction along with dealing with Fourier transforms.

I don't know if the frequency is currently measurable, but in principle at least, one could use time crystals.

Or alternatively, by measuring the frequency and wavelength by some form of diffraction experiment.

You need to recheck this.

No, λf does not equal the group velocity. vg = dω/dk whereas λf = ω/k = vp.

It seems to me that the non-relativistic approach is simply incorrect. For De Broglie waves of a non-zero mass particle, one considers wavelength, but I do not recall anyone considering frequency in a non-relativistic context. The frequency of a De Broglie wave is directly proportional the energy of the particle, and this energy quite simply includes the energy-equivalent of the mass. Thus, the relativistic approach is the only correct approach.

But isn't it true that λf = vp? Are you suggesting that neither λ nor f are measurable? I do accept that the phase of a De Broglie wave is unobservable, but it is not clear to me why vp is not at least indirectly measurable.

I agree that the electromagnetic wave equation with the speed of sound instead of the speed of light is invariant to Lorentz transformations with the speed of sound instead of the speed of light. But we know that sound waves do not behave like light waves, and I would like to know precisely where the properties differ. The wave equation you have given agrees with https://en.wikipedia.org/wiki/Acoustic_wave_equation, but I'm not convinced that this wave equation applies to any coordinate system other than the one in which the medium is at rest. This would break the invariance. In the case of light, the Doppler effect depends only on the relative velocity between the source and the observer, whereas in the case of sound, the Doppler effect depends not only on the relative velocity between the source and the observer but also on the relative velocity of the medium. Note that whereas both the speed of sound and the speed of light are independent of the speed of the source, only the speed of light is independent of the speed of the observer. Elsewhere, I have said that c of relativity is not really about the speed of light but rather about the relationship between space and time. The relativistic velocity-addition formula, as measured by the Fizeau experiment, allows one to determine in principle the value of c without measuring the speed of light in a vacuum, thus highlighting the space and time aspect of c. The speed of sound cannot satisfy the space and time aspect of relativity. Only the Lorentz transformations with the speed of light are transformations between different frames of reference in which the space and time coordinates mean the same thing in both frames of reference.

In this case, actual Lorentz transformations (with the speed of light in a vacuum).

I see two problems with this: 1: The wave equation for electromagnetism but with the speed of sound replacing the speed of light may not be the correct wave equation for describing the propagation of sound. In this wave equation there is no dependency on the velocity of the medium, whereas sound waves have a fixed velocity relative to the medium, which may differ from the velocity relative to the observer if the medium is in motion relative to the observer. 2: If the Lorentz transformation but with the speed of sound replacing the speed of light is applied to space and time coordinates, the resulting new coordinates are no longer space and time coordinates. Indeed, it is not obvious what these new coordinates represent. Note that it is Minkowskian spacetime that is invariant to Lorentz transformations, not just the electromagnetic wave equation, so that applying a Lorentz transformation to space and time coordinates results in new coordinates that are also space and time coordinates.

The force between two charges corresponds to the increase or decrease in the total energy of the surrounding electromagnetic field as the distance between the charges is changed. I think what happens to the electromagnetic field of the electron and positron when they annihilate is quite relevant.

It appears to me that you are referring to the energy of the electromagnetic field surrounding the positron-electron pair. The Wikipedia article "Electromagnetic mass" may interest you. It doesn't entirely answer your question because it is about a classical notion. But the notion that the mass of an electron includes the mass equivalent of the energy of the surrounding electromagnetic field is an intriguing one and is connected to the modern notion of renormalization.

Bear in mind that when you go out into the sunshine, you don't feel the force of the light being exerted on you, so it is evidently quite small in magnitude.

As for why a photon is massless, (IIRC) this has something to do with the Higgs field and the symmetry-breaking of the four electroweak bosons.

In Minkowski spacetime, using natural units where c = 1, there is the four-dimensional energy-momentum vector. Energy is the time-component and momentum is the spatial components of this vector. The invariant mass is the magnitude of this vector. That is: m² = E² – p² Just as a tangent vector of a lightlike trajectory in spacetime is a null vector, so is the energy-momentum vector of a photon. That is, E² = p² and thus m = 0. As mass is an invariant, the mass of a photon is zero in every frame of reference.

I'm not aware of any problems with special relativity. If you think that quantum entanglement and experiments such as the delayed-choice quantum eraser experiment are a problem for special relativity, then it would seem that you do not correctly understand the nature of quantum entanglement. Quantum entanglement is often presented as the communication of the result of a measurement of one particle to its distant quantum entangled partner. But this is incorrect. According to the no-communication theorem, it is not possible to use quantum entanglement to communicate information. In other words, there is nothing that a person can do to their quantum state that will have any effect on the result of any measurement performed by another person on the corresponding entangled partner state. This absence of communication means that the correlation between the entangled particles, even if it appears to be faster than light or backwards in time, does not violate special relativity.

It's my experience that a lot of people think Special Relativity is about light, with c being called "the speed of light in a vacuum". I prefer to think of c as a ratio of a distance in space over an equivalent distance in time. What I'm offering is an in-principle method of determining c without measuring the speed of light in a vacuum.

Fizeau performed the measurements of the speed of light in moving water compared to the speed of light in stationary water, but my purpose is to obtain the value of c without measuring the speed of light in a vacuum, thus disconnecting c from the properties of light. If the dispersion of light in water is ignored, the relativistic velocity-addition formula can be rearranged to express c in terms of the two velocities of light in water and the velocity of the water.

Special Relativity is embedded in Maxwell's equations. Maxwell's equations indicate that electromagnetic waves propagate at the speed of light, expressed in terms of the permittivity of free space and the permeability of free space. Maxwell's equations are invariant to Lorentz transformations, indicating that the speed of light is also invariant to Lorentz transformations. That is, the speed of light is the same for all inertial observers regardless of their velocity relative to some specified frame of reference. Thus, the notion of an absolute frame of reference is denied by Maxwell's equations. However, Special Relativity does go beyond Maxwell's equations and electromagnetism in general. Indeed, the speed of light in Special Relativity should not be regarded as specifically about light, but rather about the relationship between space and time. In principle, underwater creatures could determine the speed of light in a vacuum without measuring the speed of light in a vacuum by measuring the speed of light in both stationary and moving water, applying the relativistic velocity-addition formula. In this case, the speed of light in water is an "ordinary" speed, distinct from the c that appears in relativistic formulae.

Good day, Markus. Yes, it is me from The Science Forum. It seems that The Science Forum has finally died. It did seem like an inevitability for quite a long time. But I did stick with it till the very end. Anyway, I joined this site earlier today, posted some stuff. I notice you are still using the generalised Stokes theorem as your signature. 😉