KJW

This is an appeal to your own authority. I think that means you lost the argument.

Why would I open a new thread? You wrote something I disagreed with in this thread. It was not my intention to discuss topological aspects of GR. It was my intention to argue that the metric tensor and the Kronecker delta are not equivalent. In spite being numerically identical, not even the Euclidean metric tensor is equivalent to the Kronecker delta because the Euclidean metric tensor contains information that the space is flat, whereas the Kronecker delta does not. This is a false analogy. In the case of the Riemann tensor, it is not what you do with the metric tensor that matters because what you do with the metric tensor to form the Riemann tensor doesn't change. But if I change the metric tensor, I may change the Riemann tensor. A more correct analogy is DNA sequencing where the process of converting the DNA to a sequence of letters "A", "G", "C", and "T" is a procedure that doesn't change with the sample. The letters "A", "G", "C", and "T" are not the same as the DNA molecule, but we assume an informational connection between the two. There are many possible values of the metric tensor, leading to many possible values of the Riemann tensor. But there is only one Kronecker delta (for a particular number of dimensions). I don't think it is a misconception. I think you are operating under the common misconception that all the various subscripted/superscripted versions of a given tensor are necessarily equivalent. But for the same reason that the Kronecker delta is not equivalent to the metric tensor, raising or lowering indices may alter the information content of the tensor. However, the reversibility of raising or lowering indices is due to the metric tensor or the inverse of the metric tensor retaining any information that might have been lost as a result of raising or lowering indices. This does imply that the given tensor is dependent on the metric tensor or its inverse. Otherwise, raising or lowering indices can't alter the information content, and all the various subscripted/superscripted versions are equivalent.

Of course. The metric tensor does encode curvature in the sense that one can mathematically obtain the Riemann tensor from it. One can't mathematically obtain the Riemann tensor from the Kronecker delta. Therefore, the Kronecker delta is not mathematically equivalent to the metric tensor. By definition... it need not be covariantly constant. That's because it has been replaced with a different quantity to represent the anholonomy. But it's not a different story. It's saying that the metric tensor contains information to mathematically obtain the Riemann curvature tensor.

BTW, I run Microsoft Office 2010 on a Windows 10 system. But Microsoft 365 has piqued my interest because Excel has dynamic arrays, which seems to be a fundamental improvement over CSE arrays.

But if the decomposition depends on the particular frame, that is giving up covariance. Is this not a form of non-locality? Basically you’re saying that whether or not a charge radiates in some local region depends on the existence of potentially distant sources (=external field). The response of the charge is to the electromagnetic field at the location of the charge, and is therefore local to the charge. We are not interested in the source of the external electromagnetic field. The problem then becomes decomposing the electromagnetic field at the charge into the field from the charge and the field not from the charge. But, the response of the charge is to the total electromagnetic field, so the hypothesis is that the field from a point charge is zero at the charge regardless of its motion. However, if a charge can be accelerated non-electromagnetically, then one can test if it radiates in the absence of an external electromagnetic field.

But they don't encode the same information. The metric tensor encodes the spacetime curvature whereas the Kronecker delta does not. But the inverse of the metric tensor does encode the same information as the metric tensor.

From https://en.wikipedia.org/wiki/Microsoft_Office_XP I'm guessing that "not officially supported" means that it might install and run anyway. Or perhaps you could install a Windows XP emulator to install and run Microsoft Office XP. This seems like a lot of effort to me.

I guess what you mean is that the radiation field (and the EM field in general) will always be much larger than the local free-fall frame. There’s also the issue of the field “back-reacting” with the charge, which would make true free-fall impossible in the first place. These are good points, and I’m not sure how they influence the analysis of this situation. I should point out that I don't have a complete answer to the question. I'm just suggesting plausible possibilities as well as pointing out particular problems as I see them. That is, I'm not claiming that the radiation from a charge is a non-local phenomenon, but merely offer it as something that may nullify the equivalence principle. But it is also worth noting that equations in integral form may also have a differential form, a local form of otherwise non-local equations. For a charge in an external electromagnetic field, the acceleration of the charge must be accompanied by a back-reaction on the field. And this is both covariant and local. I’m struggling to understand this - why would the absence of a magnetic field contradict the charge not radiating? It would seem that I misspoke. What I meant was that an argument supporting the notion that a charged object sitting on a table doesn't radiate is the absence of a magnetic field. But this argument applies regardless of whether the observer is in the same accelerated frame of reference as the charge or is in an inertial frame of reference that is at instantaneous rest relative to the charge. I completely agree, and this insight should be all that’s needed to understand why some observers see radiation and others don’t. That's true. Give up covariance and the whole problem goes away. Seems like a Faustian deal to me. That’s fair enough - how would you yourself evaluate and understand this situation? As I said above, I don't have a complete answer. But I do see two possibilities that are not necessarily consistent: (1): Radiation only occurs if an accelerated charge is in an external electromagnetic field. That is, there is no radiation purely from an accelerated charge's own electromagnetic field. Such a hypothesis is not invalidated by cyclotron radiation, bremsstrahlung, nor Cherenkov radiation, but would explain in a covariant manner why some accelerated charges do not radiate. (2): The divergence of the quadratic expression in terms of the electromagnetic field tensor of the energy-momentum tensor gives not only the Lorentz force, but also some expression of the source in terms of an accelerated charge. This may actually be inconsistent with (1), but I think it is fair to assume that the Lorentz force law (along with its equivalent alternative expressions) is consistent with Maxwell's equations.

Yes. As a light-emitter goes from the event horizon to further away, the outwardly-directed solid angle of the beam of light that escapes the black hole increases from zero at the event horizon to a hemisphere at the photon sphere and continues to increase further out.

One thing that is not generally recognised is when you see an image representing a black hole, the outer edge of the black disc is not the event horizon. In fact, it is the photon sphere at [math]r = \dfrac{3r_s}{2}[/math], where [math]r_s[/math] is the Schwarzschild radius (the radius of the event horizon). Note that the photon sphere is the minimum radius for light to escape transversely away from a black hole.

Because this would provide a way to locally test whether you’re in free fall in a gravitational field, or just in an “ordinary” inertial frame - which is a violation of the equivalence principle. Either way, I think the answer to this has been worked out mathematically by different authors, for example here. Actually, my question was semi-rhetorical because I knew you had to answer the way you did. But there is the possibility that radiation from a charge is a non-local phenomenon, in which case the equivalence principle doesn't apply. Not if the detector is comoving wrt to it. However, if the detector is in a locally inertial frame (ie freely falling past the charge), then radiation is detected. One argument against the notion that a charged object sitting on a table doesn't radiate is the absence of a magnetic field. A locally inertial detector can be at instantaneous rest relative to the accelerated charge, so that even though they have different accelerations, there is no magnetic field detected from the charge. That can’t be the case, since the electromagnetic field is a tensorial quantity. The electromagnetic field is a tensor, but this is about the splitting of the electromagnetic field into the radiation field and the Coulomb field, and it is this splitting that might not be covariant. So far in this discussion, I've not seen any explanations that are covariant. This is the first time I've seen the Rindler horizon invoked to resolve this issue. At present, I don't accept the Rindler horizon explanation for two reasons: 1: What is happening in the spacetime region between the accelerating charge and its Rindler horizon? Note that the proper distance between a constantly accelerating charge and its Rindler horizon is [math]\dfrac{c^2}{\alpha}[/math]. 2: Rindler horizons don't exist in reality. A Rindler horizon in Minkowskian spacetime occurs as a result of an object outrunning a pulse of light after being given a head start. This requires that the object asymptotically approach the speed of light, the head start given corresponding to the proper distance to the Rindler horizon. But in reality, the object will eventually stop accelerating, the pulse of light will inevitably catch up, and there was no Rindler horizon after all.

In the topic I'm remembering, the person was claiming to be the victim of a targeted attack with electromagnetic radiation.

The OP evokes in me a sense of déjà vu. But this was from the other forum, so maybe @exchemist or @geordief could weigh in on this topic.

The way I see it, if one expresses the electromagnetic energy-momentum tensor in terms of the electromagnetic field tensor, then the divergence of this quadratic expression, which gives the Lorentz force, should also be the source of this quadratic expression similar to the way the charge current is the source of the electromagnetic field. That is, instead of considering the divergence of the energy-momentum tensor itself, which is a force, one considers the divergence of a particular quadratic expression of the electromagnetic field, which should be part of Maxwell's equations rather than requiring the Lorentz force explicitly.

But when a charge accelerates, there has to be a back-reaction on the electromagnetic field causing it to accelerate. So, I don't think it is as straightforward as invoking derivatives of the acceleration. On the other hand, radiation from a charge can be obtained from the solution to Maxwell's equations.

I fully agree with this. One possibility that I'm currently unwilling to accept is that radiation from a charge is not covariant, that it does depend on the frame of reference from which it is observed.

What are you talking about? Changing your IP address that external world sees, or changing your IP address within your local network? What you've written above seems to be about your local network. Not long ago, I had a problem in this area so I do have some experience, although it did take a change of modem to completely fix the problem. Your IP address that the external world sees is allocated to you by your ISP, and if you want to present a different IP address to the websites you visit, you could use a VPN.

Why do you say that freely falling charges do not radiate? A charged object sitting on a table doesn't radiate even though it is accelerating. I think the circumstances under which a charge radiates is not straightforward. This is a topic that interests me a lot and I've even been considering this issue quite recently. Part of the difficulty as I see it is that the mathematics that says an accelerated charge radiates is limited to the Minkowskian metric and needs to be generalised to arbitrary metrics. Another difficulty as I see it is how the Lorentz force relates to Maxwell's equations. I'll discuss this in more detail in the hopefully-not-too-distant future.

I think many websites are adopting this as their business model for ads:

What is a water wave without the water? In the case of an electromagnetic wave, without the medium, one still has the electromagnetic wave.

I knew this (that the distinction between x-rays and gamma radiation is how they are produced and not their wavelength, frequency, or energy), but had forgotten this (that the kinetic energy alone of the electron striking the nucleus could provide gamma radiation).

Hmmm. I'm actually somewhat surprised that in the fusion from hydrogen to iron, the first step to helium provides about 80% of the energy (or am I interpreting the diagram incorrectly?).

Do beta processes (any type) emit gamma radiation?

Thank you for your effort, but I was hoping for something I didn't already know (although a clarification of the distinction between global and local symmetries pertaining to Noether's theorem would be helpful). My particular difficulty is about the connection between a wavefunction, which describes probability of general objects, and electromagnetism, which is a more specific notion than the objects to which wavefunctions apply. And of course, there are the weak and strong forces, with their own symmetries. The symmetries of the electromagnetic, weak, and strong forces are often described as "internal symmetries", but I find this term unsatisfying. Another thing: How does [math]A'_{\mu} = A_{\mu} + \partial_{\mu} \phi[/math] from classical electrodynamics relate to [math]\psi\, ' = e^{i \phi} \psi[/math] from quantum mechanics?

What I meant was: to what symmetry does charge conservation correspond?