KJW

The familiar gravity that we all experience on earth is caused by gravitational time dilation. I have devised "the slinky analogy" to provide some insight: Note that as the slinky is bent, the side away from the direction of bending is longer than the side towards the direction of bending. Furthermore, the sharper the bending, the greater the relative difference in the length of the two sides. There is a geometric relationship between the relative change in the length of an arc with respect to the radial distance and the curvature of the arc. Gravitational time dilation is the change in the length in spacetime of a timelike worldline with respect to height and is geometrically related to the curvature of the worldline, which corresponds to the acceleration we experience standing on the ground on earth. [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] where: [math]a^\mu[/math] is the upwardly directed acceleration of an observer at rest on the ground. [math]T[/math] is the relative spacetime length of a timelike Killing vector at a given location in three-dimensional space. [math]\dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] is the gravitational time dilation. [math]g^{\mu\nu}[/math] has signature [math](1, -1, -1, -1, -1)[/math] [math]c[/math] is the speed of light in a vacuum (used to relate the curvature of a worldline to its acceleration) The rubber sheet analogy that is frequently used to explain gravity is incorrect even as an analogy. Of the various problems it has, the one that is relevant to this post is that the familiar gravity that we all experience is not caused by the curvature of three-dimensional space, as suggested by the rubber sheet analogy. It is specifically time dilation that gives rise to such gravity.

You could say that, but in fact I was making a point. The subscripted form of the partial differential operator is the natural form of this operator, whereas the superscripted form is obtained from the subscripted form by raising the index with the inverse of the metric tensor. The superscripted form of the partial differential operator therefore implicitly contains the inverse of the metric tensor which has to be explicitly taken into account in [math]\partial_\mu \partial^\mu \phi[/math]. Thus: [math]\partial_\mu \partial^\mu \phi = \partial_\mu (g^{\mu\nu} \partial_\nu \phi) = \partial_\nu (g^{\nu\mu} \partial_\mu \phi) = \partial_\nu g^{\nu\mu}\, \partial_\mu \phi + g^{\nu\mu}\partial_\nu \partial_\mu \phi = \partial_\nu g^{\nu\mu}\, \partial_\mu \phi + \partial^\mu \partial_\mu \phi[/math] A similar problem occurs with [math]dx_\mu[/math]. Is [math]dx_\mu = g_{\mu\nu} dx^\nu[/math]? Or is [math]dx_\mu = d(g_{\mu\nu} x^\nu)[/math]? I generally avoid this problem by not using [math]dx_\mu[/math]. However, in the case of: [math]u^\nu = \dfrac{dx^\nu}{ds}[/math] it seems unambiguous to me that: [math]u_\mu = g_{\mu\nu} u^\nu = g_{\mu\nu} \dfrac{dx^\nu}{ds}[/math] and not: [math]u_\mu = \dfrac{dx_\mu}{ds}[/math] The point is that the various subscripted/superscripted forms of a given quantity are not created equally and that there is often one form that is the natural form from which all the other forms are derived, and sometimes it does require one to know which is the natural form.

Hmmm, I nearly misread this. But it seems you are not giving me the credit to know that the identity matrix can only be a tensor if it has one superscript and one subscript. But the metric tensor and its inverse are not only frame-dependent quantities, but they also depend on the intrinsic manifold. By contrast, the Kronecker delta is not only frame-independent, but also independent of the intrinsic manifold. That is, the metric tensor or its inverse tells you both what intrinsic manifold you have, and also the frame within that intrinsic manifold. The Kronecker delta tells you neither. Quantities that are frame-independent but tell you what intrinsic manifold you have would be very useful, but the Kronecker delta is not that quantity. Neither is the metric tensor, but the Riemann tensor is frame-independent on a flat manifold. Only the metric tensor and its inverse can tell you about orthogonality. The ϵ tensor can only tell you about the weaker condition of linear-independence. I suspect you know this, but are underestimating me. I am fully aware that the concept of a tensor is dependent on the group of transformations. In my own research, I explored tensors within the context of the general linear group. This requires that the definition of the partial differential operator be extended beyond the chain rule. It introduces a new quantity to deal with the anholonomy. At the time, it was unclear to me about how to deal with the oxymoronic "anholonomic coordinate systems". Only later did I realise that they were a generalisation of the vierbein. What is special about the differentials of the coordinates is that they are differentials of the coordinates. What more needs to be said? I'm perfectly aware of the modern trend of getting rid of the entire notion of coordinate systems. But I reject this trend, regarding it as throwing the baby out with the bathwater. The notion of covariance is about putting all coordinate systems on equal footing by ensuring that equalities are the same in all coordinate systems. But somehow, even the notion of a coordinate system is an anathema. Even though tensors can be reformulated without coordinate systems such as I mentioned above, there is still an underlying aspect of the notion of coordinate systems within this formulation. It underlies the concept of curvature, regardless of how curvature is defined in the modern formalism.

[Expression 2] is incorrect. [Expression 1], although not incorrect (but see below), is not how I would choose to write it. I generally write partial differential operators in covariant (subscripted) form. Thus, I would use [Expression 2] in the corrected form: [math]\partial_\mu \left( \displaystyle \sum_{n} \dfrac{\partial\mathcal{L}}{\partial(\partial_\mu \phi_n)} \partial_\nu \phi_n - \delta^\mu_\nu \mathcal{L} \right) = 0[/math] At this point, I must admit that I have made an error. I took something you said out of context and placed it into another context in which I said you made an error. I apologise to you and Genady for this. However, I will ask you this: In general, does [math]\partial_\mu \partial^\mu \phi = \partial^\mu \partial_\mu \phi[/math] ?

I'm not aware of any experiment where charge is seen to radiate in some frames but not others. I don't think it requires the absence of electromagnetic acceleration, only the presence of an acceleration due to a non-electromagnetic force. Then one can examine if the radiation is accounted for by the total force or just the electromagnetic force. A proton-proton interaction experiment might be a suitable if the energy is large enough to overcome the electromagnetic repulsion so that the strong force is felt, but not too large to cause a transmutation. 👍 The problem I see is not a disagreement but an inability to see the principle behind the electromagnetic energy-momentum. It seems to come out of thin air... AFAICT, it does not come from Maxwell's equations. Actually, it has occurred to me that it cannot come from Maxwell's equations for dimensional reasons (no mass), but if it can be shown that the quadratic expression in terms of the electromagnetic tensor has energy-momentum-like properties, then this may satisfy me.

Actually, your example of [math]ds^2 = dx^i g_{ij} dx^j[/math], [math]ds^2 = dx_i g^{ij} dx_j[/math], and [math]ds^2 = dx_i \delta^i_j dx^j[/math] only serves to strengthen my point. Of the three metrics, only in the case of [math]ds^2 = dx^i g_{ij} dx^j[/math] are both vectors differentials of the coordinates. The differentials of the coordinates vector is naturally contravariant as a result of the chain rule. The [math]dx_i = (dr \ \ \ \ r^2 d\theta)[/math] vector is not a differentials of the coordinates vector as it is clearly seen to contain components of the metric tensor. Indeed, this vector required lowering the index of [math]dx^i = (dr \ \ \ \ d\theta)[/math] with the metric tensor. One can't lower indices with the Kronecker delta, another difference between the Kronecker delta and the metric tensor. In all three cases, the metrics [math]ds^2 = dr^2 + r^2 d\theta^2[/math] are the same. Note that the coefficients of the differentials of the coordinates quadratics are components of the metric tensor [math]g_{ij}[/math]. There is not a Kronecker delta in sight. Although the rules of index manipulation do allow the three expressions [math]ds^2 = dx^i g_{ij} dx^j[/math], [math]ds^2 = dx_i g^{ij} dx_j[/math], and [math]ds^2 = dx_i \delta^i_j dx^j[/math], [math]ds^2 = dx^i g_{ij} dx^j[/math] is the natural expression. At [math]r = 0[/math], there is a (removable) singularity that is a natural part of the polar coordinate system. If you want to remove this singularity, you change the coordinate system to something like the Euclidean coordinate system. The Kronecker delta tells you nothing. Its invariance guarantees that. You chose a flat space for your example. How about choosing the spacetime of a Schwarzschild black hole, where the central singularity is not removable. I don't think that what I'm talking about is my interpretation of tensors. I see it as more about general mathematical logic.

I am not going to download your file, and I doubt anyone else will. Please provide something that at least initially does not require us to download your file.

I need to think about this one first Have you ever derived the relativistic Lorentz force law from the divergence of the electromagnetic energy-momentum tensor? If you haven't, it may be difficult to appreciate where I'm coming from. I won't do it here, but basically one starts with the energy-momentum tensor on the left side of the equality, and a particular quadratic expression in terms of the electromagnetic field tensor on the right side of the equality. Perform the divergence of both sides. On the left, one has a force, and on the right after a number of steps, the product of the electromagnetic field tensor with the charge current vector (or charge times velocity). If one is willing to accept without question the electromagnetic energy-momentum tensor, then the force on the left becomes mass times acceleration, and we have an expression that says that the source of the particular quadratic expression in terms of the electromagnetic field tensor, divided by the mass of the charge, is the acceleration of the charge. This is analogous to the expression that says that the source of the electromagnetic field tensor, divided by the charge, is the velocity of the charge. Let's consider something more basic: The Coulomb field around a charge drops with [math]\dfrac{1}{r^2}[/math]. Enclose the charge in a sphere and perform a surface integral of the field to obtain the charge as the source of the field, noting that the surface area increases with [math]r^2[/math], cancelling the [math]\dfrac{1}{r^2}[/math] drop of the field. But the radiation field drops with [math]\dfrac{1}{r}[/math], so that one has to square the radiation field to obtain a field that drops with [math]\dfrac{1}{r^2}[/math] and whose surface integral obtains the source of this radiation field by cancelling with the [math]r^2[/math] increase of the surface area. For me personally, I don't accept without question the electromagnetic energy-momentum tensor. Thus, instead of obtaining acceleration of the charge from the left side of the equality as above, I would like to obtain it from the right side of the equality, thereby disconnecting the acceleration of the charge from the energy-momentum tensor. It is worth noting that due to the antisymmetry of the electromagnetic field tensor, the product of the electromagnetic field tensor with the charge current vector is orthogonal to the charge current vector, just as the acceleration vector is orthogonal to the charge current vector. But that doesn't imply that these two vectors that are orthogonal to the charge current vector are parallel to each other. Nevertheless, the orthogonality of the product of the electromagnetic field tensor with the charge current vector to the charge current vector is a right side of the equality result.

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.