Markus Hanke

That’s because mass never appears in the GR field equations - what is generally called the source term here is the energy-momentum tensor. One must also remember what these equations actually say - they state a local equivalence between a certain combination of components of the Riemann tensor (the Einstein tensor) and the energy-momentum tensor. Nowhere does it claim a ‘causative relationship’, but instead it says that these two are the same thing (up to a constant of course); neither one comes first and ‘causes’ the other.

Indeed. Just wanted to mention this again quickly in case other readers find it helpful to have some geometric intuition what the various aspects of curvature - Riemann, Weyl, Ricci - actually mean.

Another way to say this is that, in this kind of spacetime and under geodesic motion, shapes (ie angles) are preserved, whereas volumes and surfaces are not.

This short document might help, particularly chapter 3: https://www.sas.rochester.edu/pas/assets/pdf/undergraduate/first_order_approximations_in_general_relativity.pdf

Sorry, I have not been able to keep up with these discussions over the last few days, as I’m busy with a large RL project. What was the question here? deSitter spacetime has non-zero Riemann tensor, so it’s not flat.

In fairness, I think this misrepresents what KJW is trying to say. How does a freely-falling test particle under the influence of gravity move? It follows a geodesic in spacetime, which is a particular solution to the geodesic equation. This equation is itself a particular form of the principle of extremal ageing, ie the tendency will be for the test particle to move such that a comoving clock will record an extremum of proper time between any given pair of events along the trajectory. When you actually perform this variational problem, of course all components of the metric are technically involved. However, when you are dealing with situations that are in some sense close to being Newtonian / not too relativistic, such as Earth for example, the tt-component of the metric will be much larger than the rest of the metric, by a factor of ~c^2. It will dominate the calculation - meaning time dilation plays a much larger role than tidal effects. In that sense, it is indeed almost exclusively time dilation that gives rise to our daily experience of “downward gravity” here on Earth. Of course, this would not be true in other situations, like near the EH of a solar mass BH, where tidal gravity plays a major role. This doesn’t mean that the source of gravity isn’t energy-momentum and curvature, it just means that under certain circumstances the tidal components don’t play a major role, leaving mostly just time dilation as the dominating effect.

The kind of wavelengths you get would depend on the specifics of the setup - it’s conceivably possible to get visible light too. For a stationary charge supported in a gravitational field, the result I am familiar with from the literature (see link further up in the thread) would indicate that a comoving detector would not detect any radiation, but another detector freely falling past the charge, would. Good point. But I think given enough charge and enough acceleration, it should be detectable. I must admit I’m not sure what the actual numbers are like, I never looked at this in that much detail.

AFAIK (and can remember) it comes basically from the general definition of the Hamiltonian, with the potential field \(A_{\mu}\) plugged in. I’m a bit pressed for time these days (involved in a big project here at the monastery), so I wasn’t able to immediately find a proper textbook reference; but the second-to-last answer on this PSE thread outlines what I mean: https://physics.stackexchange.com/questions/283519/derivation-of-electromagnetic-stress-energy-tensor-in-curved-spacetime I’ll have to dig through my notebooks when I get a chance, I know I’ve got a proper reference on this somewhere.

Sure - isn’t that already a suitable model for the situation at hand? The charge is seen to radiate in some frames but not in others. Ok - this doesn’t sound like too hard of a test to perform, I wonder if this has been done? I remember having seen this done, so I’m aware of the concept. What problem do you see with this?

But we’re not giving up covariance, are we? We’re simply considering how the EM field - a covariant object - decomposes in a particular frame. 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). I need to think about this one first

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’m struggling to understand this - why would the absence of a magnetic field contradict the charge not radiating? 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 fair enough - how would you yourself evaluate and understand this situation?

You would see radiation regardless of how you move, since this arises overwhelmingly from the interactions between particles within the plasma. This is kind of a different scenario. But if it was only a single isolated charge falling into the BH, then you would not see any radiation from it due to its free fall motion. Again assuming an isolated charge in free fall. You would not see radiation from the particle, but you would see radiation all around you, since the vacuum has now become a “thermal bath” for you (Unruh effect). The detector needs to undergo accelerated motion for there to be a Rindler horizon, so it needs to be supported against gravity. I don’t think detector and emitter need to be perfectly comoving (ie at relative rest), but to be honest I’m not 100% sure on this. The Rindler horizon is a result of accelerated motion, and thus observer-dependent. But yes, essentially.

The fully collapsed version of this is called a kugelblitz, whereas the non-collapsed version would be some form of gravitational geon. The equivalent of Schwarzschild spacetime in the presence of a positive cosmological constant is called deSitter-Schwarzschild spacetime. There’s an upper limit here to how large such BH can be, which is called a Nariai black hole.

What is present in every frame and for all observers is the electromagnetic field due to the presence of the charge (which is what you refer to as “event” above). This is a tensorial quantity, so all observers agree on it being non-zero. However, observers do not necessarily agree on the value of the individual components of the tensor, since these will be functions of space and time, which are observer-dependent concepts. Physically speaking this means that everyone agrees there’s an electromagnetic field, but not everyone agrees what this field “looks like” in terms of its decomposition into E and B components in a given frame, since this decomposition is again observer-dependent. For the field to look like radiation, E and B must be periodic functions of space and time of a specific form, and they must be related in specific ways; this may not be the case in all reference frames. When you do the maths, what you find is that the radiation emitted by a charge supported in a gravitational field is in fact present even for a comoving (=accelerated) observer, but it is located in a region of spacetime that is inaccessible to him (it is beyond the Rindler horizon). On the other hand, the freely falling observer is locally inertial, so there’s no Rindler horizon, and he can detect the radiation. There’s no contradiction, it’s just that one must be careful about frames and their particular conceptions of space and time.

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. I agree, we need to be very careful with reference frames and the form of the laws we 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. That can’t be the case, since the electromagnetic field is a tensorial quantity. However, we must remember that accelerated reference frames have Rindler horizons - so for a comoving detector the radiation is essentially in a region of spacetime that’s inaccessible to it. Everyone agrees (tensor!) that there’s a radiation field, but not everyone has access to it.

Technically quite correct. But I think we’re considering an idealised situation here, or else the Schwarzschild metric can’t be used, and everything gets more complicated.

Yes this sounds correct, for Schwarzschild BH. In that case the answer is yes, such orbits exist, at least in principle.

So long as you are far enough from the BH pair, you can consider this situation as being two charged point particles in free fall. We already know that freely falling charges do not radiate (irrespective of metric), so my educated guess would be that there is no light detected. Given that, I don’t see how the situation could be different in the near field, so my guess is that there’s no radiation anywhere. However, this is an unusual and mathematically pretty involved scenario, so it is possible that I might be wrong. I don’t see how they could radiate though without violating the equivalence principle.

No. The potential energy function of a photon in Schwarzschild spacetime has only a local maximum (photon sphere), but no minimum.

What you used there is the exterior Schwarzschild metric, meaning this is only valid in vacuum outside the central mass. If you want to find the time dilation between a clock on the surface and another clock at the Center of the Earth, you need to use an interior metric along with appropriate boundary conditions. This can of course be done, but is algebraically a bit more involved. As others have said, in practice this dilation factor wouldn’t be large.

What do you mean by “radically different”? In the presence of ordinary matter (ie positive energy), originally parallel geodesics will always converge, never diverge; meaning that gravity is always attractive, and clocks closer to the mass are thus always dilated wrt some external reference clock. There is no “trick” by which this can be circumvented.

What you did there unfortunately does not make much sense. The r in the Schwarzschild metric is not the size of a mass, but a radial coordinate. The parameter M is a global property of the entire spacetime, not just some isolated region; the physical size of the mass has no bearing on the geometry of the vacuum region around it, which is why it is most often assumed to be point-like, and which is why its size does not appear as a parameter in the metric. If you want to consider time dilation, you have to first of all decide on two clocks, one of which will be dilated with respect to the other. Oftentimes this will be a clock in free fall towards the mass, and another one stationary somewhere far away (so gravity is negligible there). You can then use the metric to calculate how these clocks relate to one another - it involves setting up two integrals, which you can then evaluate. In the case of Schwarzschild, this can be done explicitly and in closed form - but it’s more complicated than just inserting numbers.

Time dilation is a relationship between frames, it’s not a property of a single isolated clock. You wrote that you are considering an observer near the mass - is this a free-fall observer, a stationary observer (meaning he applies radial acceleration), or some more complicated motion? And which other clock exactly are you comparing his proper time against?

It goes in all directions, including towards his eyes (roughly the radial direction). So it’s both that the light radiates “upwards” to the degree that it can, given the presence of the horizon, and that he himself falls “down” towards it. The local laws of physics guarantee that the relative velocity remains always exactly c. The crucial bit is that these geodesics (light & eyes) intersect someplace, meaning Pinocchio can see the tip of his nose. If the light was emitted below the horizon, then that’s also where the intersection takes place - unless he somehow manages to arrest his fall before reaching the horizon, then his nose tip of course disappears from sight. Yes, pretty much.

Yes, true - I’ve been neglecting these, to avoid that extra level of complexity. This really is kind of tricky to get one’s head around. That’s one of the reasons why spacetime diagrams, like the one Genady has posted, are so useful - you can see immediately whether geodesics intersect or not.