Markus Hanke

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.

In some sense, yes. But it isn’t really a physical change that one would notice - spacetime remains smooth, regular, and locally Minkowskian everywhere (outside the singularity). What changes is mostly the physical meaning of the coordinates we use, relative to the exterior region. They now become spatial in nature, along the radial direction. Future means going “down” radially, past means going “up”. So the physical meaning of the r and t coordinates trade places. But again, it’s not something you would notice; spacetime looks just the same below as above the horizon. The only difference is its causal structure - below the horizon, all physically possible world lines (whether geodesics or not) terminate at the singularity. Locally, everything looks perfectly normal there, at least up to the point where tidal forces become noticeable. No, nothing special happens at the horizon at all - spacetime is perfectly regular there, and if you were to fall through it, you wouldn’t locally notice anything out of the ordinary. It’s really just a mathematical concept, not a physical entity. He can, because he’s in free fall. Visualise it like this (though it’s not really correct) - a photon emitted radially outwards very close to the horizon has a very slow radial (!) velocity wrt to the event horizon. On the other hand though, Pinocchio falls through the horizon at nearly the speed of light (wrt some outside reference), and thus meets the photon on the way. So it’s not like the photon necessarily propagates to his eyes, but rather that his eyes fall right to where the photon is. It’s kind of like jumping upwards in an elevator - you can put relative motion between yourself and the elevator floor, but both you and the elevator continue to move down regardless (maybe a stupid example, but you get my drift hopefully). Note that the situation is different if you’re not in free-fall. If Pinocchio, after the tip of his nose crosses the horizon, somehow fires magical thrusters that arrest his fall before he reaches the EH, then his nose will visually disappear for him (and get ripped away).

The diagram is necessarily correct (it shows a valid solution to the EFE), and I think so is Genady’s interpretation of it. The thing is that, at the event horizon, something very unintuitive happens - the physical meaning of the coordinates we use is no longer the way we are accustomed to. Imagine an astronaut in free fall, just as he crosses the horizon - let’s for simplicity’s sake say his feet emit light. Once his feet have crossed the horizon, and always assuming free fall, this light signal is now no longer “below” the eyes, but in their future. Light below the horizon is perfectly free to propagate in all spatial directions, yet it can still never leave the BH, because the singularity is in the future, and the horizon is in the past. It is no longer a question of up, down, above or below, once you’re past the horizon. Thus, for the astronaut, the light leaves his feet, and his eyes will necessarily “meet” it, because he’s in free fall. Both age towards the singularity, their relative velocity remains c (so everything is locally Minkowskian), yet their geodesics must intersect, just as the diagram shows. Thus he sees his feet like normal, perhaps slightly redshifted and dimmed. He will otherwise never notice anything special at the horizon. And he can’t, because locally everything must look Minkowskian. This is probably the biggest mistake people make when trying to visualise black holes - they think that, past the horizon, the singularity is “down”; but it’s not, it’s in the future. Likewise, the horizon isn’t “up”, but in the past. This is extremely important to understand, or else there’ll be all sorts of misunderstandings. It’s the other way around, see also above - light past the horizon remains past the horizon, but the eyes which see that light are falling inwards, and can intersect that light in the future.

It seems to be you who’s disagreeing with standard textbook stuff, such as spin, not me.

I’m a bit confused here - over on the other thread on cosmology you seemed to be implying that the theory of relativity is not a good model; yet here you talk about spin, which is a relativistic phenomenon? In the case of spin, this principle says that you cannot measure more than one component of the spin vector simultaneously with arbitrary precision. You can, however, measure one component plus the overall magnitude of the spin vector simultaneously without problems.

The question is one of scale, not balance. If you use GR to model any gravitational mechanics on a scale of the solar system, or some few multiples of it, you get the correct results to very high levels of accuracy - so the equation isn’t “flawed” in any meaningful sense. Remember that we have tested it very extensively locally here in the solar system. Rather, what happens is that on large scales, systems behave as if they contain much more matter than is visible in the electromagnetic spectrum. Fundamentally, this can mean one of three things: 1. There’s extra stuff there which we can’t see (dark matter) 2. There’s nothing extra there, but the laws of gravity have to be modified on larger scales; GR remains perfectly valid on solar scales 3. There may be some kind of other interaction happening, over and above gravity, which we don’t know about. So whatever happens, GR will remain a valid and good model; at most, its domain of applicability might become more limited.

Oh yes, there is a fundamental connection between these, given by Noether’s theorem - translation invariance in time corresponds to a conserved quantity, which is precisely the energy-momentum tensor. Without time, there would be no meaningful notion of energy-momentum. This is wrong - it’s called mass-energy equivalence, because there’s no distinction between them; mass is just a specific form of energy, they are equivalent to one another. It’s the other way around - energy-momentum arises (as a meaningful concept) from the continuous symmetries of this spacetime, in this case time-translation invariance and rotational invariance, via Noether’s theorem.

I call them by their usual names, Dark Matter and Dark Energy. Had you clearly stated that this is what you were referring to, your posts would have been easier to decipher. Nonetheless, the answer is the same as with the quantum gravity issue - right now we’re not sure about the precise nature of these entities, but it’s being worked on. Such things take time and effort to understand. Perhaps also the answer might be a modification of the laws of gravity (also being worked on); though, considering latest results, the air seems to be getting a bit thin for that option. Like I said, science is an ongoing process.

But we already have this? It’s called the Navier-Stokes equations, and they work pretty well.

What do you mean by this, exactly? The universe is just there - all we do in physics is to find models that provide the best possible descriptions of aspects of it. Most of it “unites” just fine, it’s only gravity that is a problem right now. But we’re working on this - physics, like any other science, is a process.

This is absolutely untrue, on all levels.