Markus Hanke

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.

You are right in that it constrains the set of all possible global topologies, but what I attempted to point out is that it doesn’t uniquely determine it, at least not in 4D. More precisely, globally different topologies can be compatible with the same local geometry, so local measurements of curvature alone don’t necessarily give this information. The reverse is also true - manifolds can have the same global topology, yet different geometries. So this issue is subtle. The interesting exception is in 2D - here, the Ricci scalar is also the Euler characteristic, so gravity is entirely topological. A bit more info here: https://en.m.wikipedia.org/wiki/Shape_of_the_universe

This is true of course. It should be noted though that if both GR and QFT are valid theories, at least near the event horizon, then the existence of Hawking radiation is inevitable. If it turns out to not exist, then one or both of these models don’t apply in that region.

They didn’t know whether light has mass or not, it was an unanswered question at the time. But since Newtonian gravity can only handle test particles with mass, they had to assume that it did in order to derive any kind of prediction at all. The deflection angle doesn’t depend on the exact value of the mass - it just can’t be zero in Newton’s theory. You have to assume this to derive a deflection angle from Newton’s theory, which was the only theory available back then - it can’t handle massless test particles. You are absolutely right of course - this is one of the areas where Newtonian theory fails, and GR is needed. One must remember though that this wasn’t well understood prior to Einstein. Actually, it doesn’t - they are separate concepts. GR determines only local geometry, but not global topology. For example, the maximally extended Schwarzschild metric could describe both two separate, singly-connected regions of spacetime, or a single multiply-connected spacetime. Geometry is the same in both cases, but the global topology isn’t.

Newtonian gravity has nothing to say about massless particles, so strictly speaking it makes no prediction here. However, if one assumes that photons have a very small but finite mass, then one can use Newtonian gravity to work out how they are deflected around massive bodies. Turns out that deflection angle doesn’t depend on the exact mass of the photon, so long as it is much smaller than that of the central body. The result you get is off by a factor of 2 compared to actual observations - to get the correct angle, one must use GR.

Yes, this is possible; this is in fact one of the standard ways to (in principle) build GW detectors: https://arxiv.org/abs/1501.00996 You need an extended array of clocks for this, since what you are measuring is the dilation between clocks at different positions within a passing GW wavefront. Let’s just say it’s equally hard You’d need an extended array of very precise and perfectly synchronised atomic clocks. This is doable at least in principle with current technology.