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Markus Hanke

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Markus Hanke last won the day on January 27

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  1. ! Moderator Note Moved to Speculations section.
  2. Good question. This depends on the metric - in Schwarzschild spacetime there should be no net drift for this type of motion, but in other spacetimes that don’t admit time-like Killing fields (eg Kerr), there will still be a non-zero net drift, because the two sections are of different arc lengths in spacetime (!). Actually proving this won’t be very easy, since this path isn’t everywhere smooth. No, perihelion precession is different, because only the orientation of the orbit changes, which follows directly from the geodesic equation; deSitter and Lense-Thirring on the other hand affect the orientation of the spin axis of the body (these two precessions are in different directions). There’s also a fourth kind called Pugh-Schiff precession, which is essentially spin-spin coupling. Yes, rotating bodies in orbit within a gravity well experience this. I don’t know if it has been measured for Mercury, but there’s data on this for the moon: https://www.academia.edu/49380218/Measurement_of_the_de_Sitter_precession_of_the_Moon_A_relativistic_three_body_effect But again, this isn’t the same as perihelion precession, they are distinct effects.
  3. Ok, so basically the question is whether or not an existing entanglement relationship is effected by changes in gravity (eg through moving one of the entangled subsystems into a different gravitational environment)? That’s a truly excellent question, and I will admit that I don’t know the answer for sure. Based on basic principles, I would have to say yes - an entangled system such as the one described here is mathematically a sum of tensor products, and in curved spacetime such products would explicitly depend on the metric, which should change the correlations as compared to Minkowski spacetime. A quick online search seems to confirm this, and experiments have been proposed to test this effect: https://iopscience.iop.org/article/10.1088/1367-2630/16/5/053041 Whether or not there is path-dependence will probably depend on the background metric.
  4. Yes, that’s correct. Whether or not there is gyroscopic drift will depend on the background metric, and, depending on absence or presence of relevant symmetries, also on the path taken by the gyroscope. The total precession will be a combination of two separate effects, called deSitter precession and Lense-Thirring precession. At least the deSitter precession part is never zero in a curved spacetime, so there will always be some gyroscopic drift.
  5. I’m afraid I don’t think I understand your question. Spin as a property is relevant only on subatomic or at most atomic scales - on those scales spacetime will, for all practical purposes, be flat, unless you are in an extreme gravitational environment. Were you referring to such exceptional scenarios? If not, then you are only comparing two small patches of spacetime that are locally flat, so there should be no difficulty. It should also be remembered that the components of the spin vector do not commute, so you can only ever know for definite one of the components at a time, plus the squared magnitude of the vector (these do commute). So when we speak of the ‘orientation’ here, we really are talking about eigenvalues of the respective spin operator, rather than a classical, well defined vector with known components. So the only difference can be a sign, since the squared magnitude is immutable. The spectrum of these operators shouldn’t depend on the metric of the background spacetime.
  6. What is correlated in this entanglement relationship is the relative orientation of the spin vectors, so it doesn’t matter how one locally defines the direction of spin measurement. The crucial point is that whatever direction you measure it in, the correlated particle will come out as opposite in terms of relative orientation. In some sense it’s only the “sign” that’s important.
  7. When the new EHT image appears on screen for the first time, astrophysicists be like...
  8. Yes - this is in fact a fundamental symmetry of nature, called unitarity. Colloquially speaking, information should be conserved when a system evolves, at least in principle. There is of course a precise mathematical definition for this, but for now you get the idea. For example, if you burn a book, the information contained therein becomes inaccessible for all practical purposes; however, if you somehow knew everything there is to know about the final state of the burning, ie all details of every single ash particle left behind etc, then in principle it would be possible to reconstruct the original book, so the information has been preserved, albeit in different form. Unitarity is very important particularly in quantum mechanics. Crucially, the black hole information paradox would be an example where unitarity is violated - this is why it is so problematic, and requires resolution. No, ordinary physical processes should be unitary, ie information only changes its “form” and “location”, so to speak. In the BHIP, information enters the event horizon. Quantum field theory combined with GR tells us that the event horizon carries entropy and radiates; this Hawking radiation is perfect black body radiation and thus carries no usable information. At the same time, the BH shrinks and eventually evaporates completely, leaving no remnant other than its Hawking radiation. The final state of BH evolution is thus simply a black body radiation field that contains no relevant physical information - meaning it is impossible to reconstruct whatever information entered the event horizon previously, based on what’s left of the original BH. The information is lost, not just for practical purposes, but also in principle - a violation of unitarity. As a side note - in practice (as opposed to in principle) determinism does not imply predictability. For example, a GR 3-body problem is fully deterministic, but in general only predictable for limited amounts of time (Lyapunov time), due to chaotic dynamics.
  9. In a way, yes. To be more precise - if you know all relevant details of the radiation field, you can reconstruct the masses of the two black holes prior to their merger. This is how gravitational wave observatories know the approximate masses of objects undergoing a merger. So the information is preserved in this sense.
  10. To understand this, you have to remember how the “M” parameter enters into the black hole metric in the first place. What happens is that, because the Einstein equations are differential equations, you need to provide boundary conditions in order to obtain any particular solution; for the case of Schwarzschild spacetime (I presume this is what your question is referring to), one of these boundary conditions is asymptotic flatness - meaning that, sufficiently far away from the black hole, gravity behaves like a Newtonian inverse square law. Matching up the proportionality constants thus introduces the parameter “M”, which, accordingly, is interpreted as total mass. It’s important to realise though that this is a property of the entire spacetime, not just of any particular subregion. Now, if you reduce the event horizon area of the black hole and in its stead add a gravitational radiation field in just the right way, this will still be true - the “overall curvature” of the entire spacetime is in some sense a conserved quantity. It will only be distributed differently, and the new metric will be something more complicated than Schwarzschild. But nothing will be lost as such. The information loss paradox arises only once we consider quantum effects at the event horizon, but not in the purely classical realm of GR.
  11. The in-fall time is a function of the black hole’s mass, and if that is very large, then it takes a longer time to reach the singularity. For example, for a 15 billion solar mass supermassive Schwarzschild black hole, the proper in-fall time from horizon to singularity would be 71.2 hours, so nearly three days.
  12. It seems you are just ignoring all the points I have raised earlier. ‘Gravitational potential’ is not a generally applicable concept in GR, and it is not the same as gravitational self-energy. It came out that way because that is the only possible form the equations can take, given the necessary mathematical consistency conditions. He initially tried to equate the Ricci tensor with the source term, but that didn’t work because the Ricci tensor is not, in general, divergence-free, so there was an inconsistency. There is also a deeper reason why they are of this form, which has to do with topology, but I won’t get into this here, and Einstein himself wouldn’t have known about it.
  13. It’s not that simple, I’m afraid. The source of gravity isn’t just mass, but energy-momentum; things like massless particles, electromagnetic fields, stress, pressure etc etc all have a gravitational effect too. What’s more, gravity also couples to itself, making it non-linear. Well, in principle there’s no guarantee that there is such a thing as quantum gravity - at present there is no experimental evidence for its existence. The issue however is that if gravity is always classical, then we run into all manner of inconsistencies that are difficult to resolve - for example the formation of different kinds of singularities, or violations of unitarity at event horizons. It’s not dissimilar to the kind of problems that historically lead to the development of quantum mechanics. Thus, most physicists assume that gravity can be quantised somehow. One important point though - quantisation of gravity does not necessarily imply the existence of gravitons. Gravitons arise if we attempt to quantise GR in the same way as the other fundamental interactions, namely by using the framework of quantum field theory. We pretty much know by now that this approach doesn’t really work out very well, so current research is looking at other options. Depending on how things turn out in the end, the graviton may or may not be a part of the solution.
  14. I never mentioned anything about ‘accuracy’ in my posts...? I don’t known what you mean by ‘perfect’, by I very much doubt that any knowledgeable physicist, Ohanian included, would make such a claim. For one thing, GR only works in the classical realm, so its domain of applicability is naturally limited. We’ve known this for nearly a century. But quantum gravity isn’t the topic of this thread. In the quote you then give, Ohanian is referring to linearised gravity, which is a simplified approximation to full GR. It is sometimes useful because its mathematics are much simpler than full GR. Singularities arise if one applies the model to cases that are beyond its domain of applicability - in this case, there are quantum effects that classical GR cannot account for. That’s not quite true. GR handles the strong field regime very well (unlike Newtonian gravity), up to a certain point. Where it breaks down is when quantum effects become non-negligible. Standard GR does already account for gravitational self-interaction. That’s kind of what I was trying to explain.
  15. Not relativity in general - but it is definitely an integral part of relativistic gravity, ie GR. Yes, absolutely. Self-coupling (meaning that the gravitational field is its own source, in some sense) is reflected in the Einstein equations being non-linear - which clearly distinguishes Einstein gravity from Newtonian gravity.
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