Markus Hanke

Interesting article, thanks for the reference!

Very true, thanks for the reminder!

I spoke about rotations in the post you quoted me on, as did you yourself in the post I quoted you on. You then said that this has never been studied, which is evidently wrong. As for electrodynamics, of course this can be formulated using quaternions - even Maxwell himself did this. You need to realise that there is nothing special at all about the quaternion formulation - using quaternions just means you are using different language to describe the same physics. Here’s a sample text of how electrodynamics are done using quaternions: https://www.scribd.com/document/91253848/quaternionic-electrodynamics As a matter of fact, quaternions are a very common tool in mathematical physics, and are used extensively in all different areas. To prove that point, here’s a further 1382 (!) references to texts and papers employing the quaternion formalism in one form or another: https://arxiv.org/pdf/math-ph/0510059.pdf Are you still so sure that they have never been studied (and not just in electrodynamics)?

Using quaternions for rotations - as well as comparing that method to other methods - is pretty basic material, and has hence been well studied and understood. You’d be discussing this stuff on an undergrad level of a math degree. There’s even a Wiki page on it: https://en.m.wikipedia.org/wiki/Quaternions_and_spatial_rotation

You are right, I misread this. My bad. In this case, what I had written in the third paragraph of my earlier post applies - as you correctly say, the earth-bound detector will detect radiation.

A couple of points: Vector algebra stipulates no such thing The curl and divergence operators are differential operators; they are not part of vector algebra, but rather vector calculus You are trying to contrast vector algebra with quaternions, but quaternions and their inner product are an early type of vector algebra To avoid any of these types of issues, I have presented the differential forms formalism, which supersedes both quaternions and vector calculus, since it is both covariant and more general than either of these. Differential forms are in fact antisymmetric tensors, so there is a strong link between exterior calculus, and tensor calculus The curl (being a differential operator) does not actually represent any rotating bodies, but infinitesimal rotations of a vector field about some point. You can indeed use quaternions to represent rotations (such objects are called versors), but this method is not any more or less advantageous than more common methods such as rotations matrices. In fact, I’d say an argument can be made that quaternions are more cumbersome, and less transparent, in terms of computational effort.

I would strongly advice any readers of this to not open attachments from unknown and potentially hazardous sources. Doing so presents a substantial security risk. Post reported.

What exactly do you mean by “topological difference”? If you smoothly deform the region in question (no breaks, holes etc), then all topological invariants remain the same, and you end up with something that is topologically equivalent to what you started with. This is just a diffeomorphism, and GR is diffeomorphism invariant, so you will get the same physics in that region. As a practical example - you start off with a region of spacetime that has the topology of a torus. You then transform the “shape” of that region into that of a teacup - they look different at first glance, but all topological invariants remain the same, so this transformation is just a diffeomorphism, which leaves the physics untouched. You do the same physics on a teacup as you do on a torus.

In a general scenario where you combine two spin-½ particles, the overall system would have either total spin 0, or total spin 1, with (I believe) equal probability.

This is not correct. There is no relative motion between detector and electric charge, so, from the perspective of the detector, there is no magnetic field. Without a magnetic field, there can of course be no radiation (which is an oscillating electromagnetic field) - despite the fact that this is a “supported” (i.e. accelerated) charge. Hence, in condition A), the radiation detector detects nothing, because there is no magnetism there to form any radiation. More generally, a “supported” observer in the same frame as a “supported” charge will never detect the charge emitting any radiation. The same is also true when you put the entire assembly into free fall - again there is no relative motion between detector and charge, and hence there is also no radiation detected in condition B). This preserves the equivalence principle - it’s in fact just a simple charge without any forces acting on it locally. It is very important to understand though that, in a curved spacetime background, the answer to the question of whether or not there is radiation detected is a purely local one; if the detector were placed (e.g.) on the surface of the Earth, as opposed to in a co-moving frame with the charge, the outcome would be different. For condition A) you still would not have radiation, but for condition B) you would now detect a radiation field, because there is accelerated motion between detector and charge. Freely falling charges will always appear to radiate from the point of view of a “supported” observer. The paradox is only an apparent one though, because observers in curved spacetimes are not related by Lorentz transformations, so they do not necessarily agree on which aspects of the electromagnetic field they detect, and which ones not. It’s not a contradiction, just a question of perspective. The other thing of course is that an observer who experiences proper (!) acceleration (a supported observer) has a horizon associated with his reference frame, so there are regions of spacetime that are inaccessible to him - simply put, the entirety of the electromagnetic field is always there, but not all observers are able to detect all aspects of it. Hence under the right conditions, some observers see radiation, whereas others do not. Since energy is also observer-dependent, there is also no contradiction in that regard.

This force is the result of dipole-dipole interaction, so the total force between two macroscopic magnets will depend on how these dipoles are spatially distributed. In other words - it depends on the shape, location, and orientation of the magnets. For two single, point-like dipoles, the force indeed falls off with the 4th power of distance; but for other types of dipole distributions the force law will be more complex: https://en.m.wikipedia.org/wiki/Force_between_magnets This is not a new insight, since it directly derives from Maxwell’s laws.

The OP didn’t, but member quiet brought up the question of whether quantum effects can be deduced from Maxwell’s classical equations. The comment was aimed at that particular sub-discussion.

That’s a good question. I would tend to answer yes to this.

What I wrote there isn’t primitive - it’s the full covariant Maxwell equations in 4-dimensional spacetime. It encompasses all of classical electrodynamics. Also, the curl and divergence operators are defined in 3-dimensional (Euclidean) space, they are not 2-dimensional.

No, because the general public does not usually read arXiv, since few people have the technical knowledge needed to understand the documents published there. Also, arXiv is a pre-print server, so these are publications that have not yet been subjected to peer review, they have only undergone a basic screening. Their scientific validity is hence not yet established.

The modern formalism of the Maxwell equations uses differential forms. This formalism is fully covariant, and all of classical electrodynamics reduces to two very simple statements: [math]\displaystyle{dF=0}[/math] [math]\displaystyle{d\star F=4 \pi \star J}[/math]

Just to clarify my earlier comment a bit more - they are scale related in the sense that you won’t see quantum effects on large (i.e. macroscopic) scales. If you have a single quantum object, then quantum effects are always apparent, but if you have a very large ensemble of quantum objects (like a mascroscopic body), then the statistical average of that ensemble’s dynamics becomes classical. In other words - it’s actually classicality that is an emergent, scale-dependent phenomenon, whereas ‘quantumness’ is a fundamental property of the universe.

No, it refers to the fact that quantum objects sometimes behave like waves, and sometimes like particles, depending on what kind of setup you use to observe them. They have both wave and particle characteristics (which are complimentary descriptions of the same entity), hence the name “wave-particle duality”.

Yes. We fully understand the classical behaviour of the electromagnetic field - we can say this with some confidence, because we know not only the Lagrangian from which it arises, but also the topological principles that underlie the structure of the electromagnetic field. And of course we know that it is the classical limit of QED. So we can check its completeness from several different angles. It’s the other way round - you deduce classical electrodynamics from quantum electrodynamics. The quantum description is more fundamental, and exhibit phenomena that do not exist in the classical world. It is a much broader framework, so no, it cannot be deduced purely from Maxwell. What you can do though is work backwards, and apply the tools of quantum field theory to perform a full quantisation of the classical field theory, to arrive at QED. But this quantisation procedure does not follow from Maxwell’s equations, it’s an external tool based on the principles of quantum physics.

The energy-momentum tensor remains the same (more precisely: covariant) regardless of the state of relative motion of your test particle. Also, remember that GR is a purely classical theory, so strictly speaking it has no concept of photons - it would be more correct to speak either of an electromagnetic field, or of null dust. Either way, the energy-momentum tensor is always covariant.

Not really, except of course that it would make conditions on Earth largely unobservable. We would have to rely on indirect evidence (such as spectral analysis of light etc) - which is precisely what we are now doing in the case of all those many exoplanets we have found around other stars. No. What would change is only our observation methods (see example of exoplanets), but there would be no doubt that what we observe is a purely classical system. Quantum effects are intrinsic to the nature of quantum objects, not just an observational artefact.

It would not be the same - what one would see is a very tiny (relative to the observer), but purely classical system. In other words, quantum effects are not just an artefact of observation scale. Also, there are no “problems” in quantum physics - it’s just different from the classical world at larger scales.

Maxwell‘s electrodynamics is the purely classical limit of quantum electrodynamics - hence it does not and cannot account for any quantum effects.

Why would that be a problem?

Gravity is geodesic deviation - and in the geodesic deviation equation, it is not straightforwardly possible to separate the “time” from the “space” contributions. I don’t think such an approach is very meaningful. What we can do, however, is look at the metric itself, and sometimes (like e.g. in the case of Schwarzschild) we see that the time part plays a bigger role than the space part, due to the presence of a factor of c squared. It should be noted though that this is not always true - for example, in the immediate vicinity of a black hole with a very small mass, the spatial contributions would be at least on the same order of magnitude than the time contributions. And then of course you have cases where the metric has non-vanishing off-diagonal terms...