Markus Hanke

1. Write down the transformation functions x=..., y=..., z=... for spherical coordinates 2. Calculate the Jacobian matrix from these 3. Calculate the determinant of the Jacobian matrix, in order to get dV=... 4. Determine the integration limits in your new coordinates - radius and equatorial angle are trivial, but for the polar angle you need to find the intersection of the ball and the cone (hint: eliminate z from the equations in the OP) 5. Write down the volume integral using your limits and volume form, and evaluate

Janus beat me to it! Maybe just as well, since I’m actually somewhat confused now - my understanding of the original scenario was that the distance A-B was to be 1 light-minute as seen from frame C, and that clocks were synchronised (taking into account light travel times) to all be seen as showing zero at the instant when C passes A, again as seen from C. My previous answers were based on that. But reading the last few posts, I may not have understood correctly what the OP meant, as the scenario seems to be different now...?

It’s not that simple. In frame C, the instantaneous tick rate of B is dilated, whereas the distance A-B appears longer for frame B than frame C. We have thus far only concerned ourselves with the final result of the experiment (C accumulating less time than B in total) - if you want to analyse what C visually sees on clock B at every moment of the journey, then things become complicated, because C and B do not share a common notion of simultaneity while there is a spatial separation between them. So you would have to account for relativity of simultaneity as well, and the analysis will lead you to the relativistic Doppler effect. In actual fact, what you visually see is that the hands on the distant clock advance faster as compared to your own clock - that’s because these intervals “tick out” a total spatial distance that’s different from yours. The key issue here is that there’s a difference between instantaneous tick rate, and total accumulated time - the instantaneous tick rate of B is dilated with respect to C, but you’re not sharing the same notion of simultaneity, so if you integrate those small infinitesimals, you end up with longer intervals, and thus more overall time passed as seen by you. It’s like the distant clock projects pictures of itself at you at a steady rate of 1 frame per time unit - but because you are moving towards the clock at high speed, you are encountering each picture at less than one time unit in your own frame, and there will be more of those pictures in total. This is why it visually looks like it’s running fast. But if you were to compare each picture individually to your own clock, using some appropriate concept of relativity of simultaneity, you’d find the instantaneous readings to be dilated with respect to you. The overall times are different, because the overall distance A-B is also different between these frames. I’m sorry I don’t know how to explain it better - this is neither intuitive nor particularly simple, despite the quite basic scenario. The devil is in the details. Nonetheless, once the maths are done correctly (also not as simple as it might at first seem!), they are found to correspond to what we actually find in the real world.

No, C will show less elapsed time at the moment of reading his clock as it passes B, since it has been in constant relative motion with respect to B throughout (we assume there was no starting or stopping). Yes, correct. That’s because observers C and A/B don’t agree on the distance between A and B - a phenomenon called length contraction, which goes hand in hand with time dilation. Also, there’s a difference between total elapsed time, and instantaneous “tick rate”. The train C reckons that the distant clocks A/B run slower, but he also reckons that the distance A-B is shorter as measured in his frame, so it takes less time on his own train clock (C) to reach B, than it does on B’s clock. So what this means is that yes, B is dilated as reckoned in frame C (just as C is dilated in B) - but the distance A-B is also longer in frame B than it is as seen from C (because A is stationary with respect to B, but neither A nor B are stationary with respect to C). So, when the train arrives, then, as seen from the train C, clock B had to have been measuring out a longer distance at a slower tick rate (remember that speed of light is the same in both frames!), meaning it has to have accumulated more total time than C - even though it was seen as dilated at each instant, just as C was dilated as seen from B. A proper analysis of this will need to reference relativity of simultaneity at each point of the journey, but rest assured that everything works out as self-consistent. In the end, they both agree that total time in C is less than B. In other words, both observers C and B agree that at the moment of C’s passing by B, the clock C shows less total elapsed time - but they disagree on the reason why that is so. B will say it’s because clock C ran slower, whereas C says it’s because the distance A-B was shorter and thus it took less time to cross it. But both agree on the physical outcome - which is the crucial bit. Take careful note of the highlighted bit - comparing the clocks needs to happen at a single instant, when the clocks pass by one another at negligible spatial distance, or else you’d need to account for light travel time as well. A real-world example of just such a scenario is atmospheric muons. These are short-lived, very fast moving particles that are produced just outside the earth’s atmosphere. Without relativity they don’t live long enough to have sufficient time to reach the Earth’s surface. But they do, we detect them. Why? Because of relativity. In the frame of the earth, the muon moves very fast, so it’s associated clock ticks slower (=it lives for longer), and thus reaches the surface in time before it decays. In the frame of the muon, earth’s atmosphere is length-contracted, so it has to cross a shorter distance, and thus lives long enough to reach the surface. Both observers agree on the outcome (=no paradoxes), but they disagree on the reason. This is exactly analogous to your train experiment. Ok, agreed 👍

There are a whole lot of technicalities involved, but fundamentally it boils down to two basic issues - background independence, and time. GR is background-independent, in the sense that it does not make reference to any underlying frame or system of coordinates; what’s more, it is the metric (system to measure space and time) itself that is a dynamic variable here. Gravity is geometry of spacetime. QM on the other hands is formulated as wave functions that explicitly depend on space and time, so it is background-dependent. Reconciling this is proving to be very difficult. In GR, time is treated on equal footing as space, and thus forms spacetime. It’s a geometrical dimension, and a fundamental part of the world. Time in QM, however, is just another parameter appearing in the theory, and is not the same as space. Whether we’ll ever bring these together into a fully self-consistent model of quantum gravity that has the proper limits, is anyone’s guess. There are various attempts and candidate models at present, but none of them is universally accepted as being the ‘right’ one. It’s an area of active research, so we’ll see what happens. On a side note, you can formulate quantum field theories in curved spacetime, but that’s not the same as quantum gravity.

I’m having difficulty gauging your understanding, based on what is written here. Are you thinking that (kinematic) time dilation is merely an optical effect, and produces no measurable physical consequences other than what an observer can visually see? He perceives it as dilated (‘slowed’), because they are in relative motion with respect to one another. He also perceives it as dilated, because they are likewise in relative motion. In the formula for kinematic time dilation, the relative speed appears squared, so its relative sign (moving towards or away) is irrelevant. He perceives both A and B to be dilated (slower), because he finds himself in relative motion with respect to both those frames. He perceives his own watch at C to be ticking normally (no dilation), because there is no relative motion between himself and his watch. They are in the same frame. No, because he’s in the same frame as that clock, so there is no relative motion. Kinematic time dilation arises due to relative motion between frames. Observer C himself notices nothing special - his own clock ticks at 1 second per second from his own point of view (no relative motion). However, he sees both A and B going slower - and conversely both A and B see C ticking slower from their own vantage points. That’s because in the frame of the train, both A and B are in motion whereas the train appears stationary; whereas in frames A and B, the train in frame C is in motion, whereas A/B are stationary. In both cases, the respective observer sees the other clock to be in motion, and thus dilated. The observers just trade places. Because kinematic time dilation isn’t something absolute that ‘happens’ locally to a clock - it is a relationship between frames/clocks. Think about it - from the vantage point A, the train is in relative motion with some constant speed v, whereas A itself appears stationary. From vantage point of the train on the other hand, frame A is in relative motion with that same speed, whereas the train appears stationary. In both cases the relationship between the frames is the same one - relative motion at speed v - so they both see the same thing, namely the other frame’s clock being dilated. This is also exactly what the mathematics tell you. The relationship between frames is the same one irrespective of which frame you find yourself in - there’s the same relative motion (v is always the same), thus in each case the clock that’s seen to be moving is dilated with respect to the observer, and never appears to be speeding up; you’re plugging the same v into the same formula to obtain time dilation, no matter which frame you are in. All observers are of course right, even if they don’t agree - but only in their own local frames. This is why measurements of time are not absolute, but depend on which frame they are performed in. This is quite a paradigm shift as compared to our own non-relativistic experience of the world, so it is quite understandable that it seems confusing or even paradoxical at first. You might wonder whether there are quantities that are not frame-dependent, meaning all observers agree on them, irrespective of relative motion; the answer is yes, but to find them you need to account for both time and space simultaneously. Time dilation always goes hand-in-hand with length contraction, and vice versa. Note that what we are discussing here are kinematic effects - if you add gravity, things become more complicated still. So the main points are: 1. Kinematic time dilation is a relationship between clocks (frames), and not something that ‘happens’ locally to a clock. It’s meaningless to say that a single clock is dilated. Nonetheless, this relationship is real (it’s a geometric rotation in spacetime, as it turns out), and thus produces real physical consequences; it’s not just on optical ‘illusion’ based on what you might visually see (though of course optics are affected by this too, so there are corresponding visual effects). 2. Motion is also a relationship between frames, and not an absolute property of an object. 3. Measurements of time or space on their own are observer-dependent. Hopefully this helps.

Clocks always tick at the same rate in their own local frames - what changes is their relationship in spacetime. Speaking of clocks ticking “slower” or “faster” is thus meaningless, we can speak only of how two or more clocks are related. Nonetheless this change in relationship between frames is a real phenomenon that has real, measurable physical consequences; it is not just an ‘illusion’, whatever this even means.

! Moderator Note Moved to Speculations section.

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.

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.

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.

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.

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.

When the new EHT image appears on screen for the first time, astrophysicists be like...

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.

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.

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.

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.

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.

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.

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.

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.

The self-coupling of the field is encoded in the non-linear structure of the field equations themselves - thus, since the Friedmann equations are obtained from the field equations, the self-coupling of gravity is already accounted for in their structure. You cannot easily separate this out in a simple way. As mentioned above, the correct Friedmann equation follows from the Einstein equation, given a suitable energy-momentum tensor; hence the correct one is the standard one which you find (with derivation) in any GR textbook. My advice would be to consider carefully why it is that the self-interaction of the field does not explicitly appear as a source-term in the field equations; you’ll find only ordinary energy-momentum there. That’s because the self-interaction energy cannot be localised, or written down in a way that all observers agree upon. Attempting to explicitly include them in a particular solution (as an analytic term) thus cannot work, because this is inconsistent with the tensorial character of the metric. It is instead encoded in the non-linear structure of the equations themselves, so it influences the relationships between the different components of the metric, rather than directly appear within those components. Conversely, because the cosmological constant appears as a local term in the Einstein equation, it cannot be interpreted as gravitational self-energy (but it certainly contributes to that energy). It is best to look at it as a kind of background curvature that is there independently of ordinary energy-momentum. For certain spacetimes such as FLRW, it thus ends up having an effect on how the metric components depend on the time coordinate (‘rate of expansion’). This is all rather subtle, and easily misinterpreted. Perhaps do some research on the Landau-Lifschitz pseudotensor - though unfortunately this involves some not so basic maths.

I haven’t read Ohanian yet, but of course he’s right - the self-coupling has a gravitational effect, and that effect is qualitatively the same like that of ordinary energy-momentum. This is what I said. It can’t be any different, as otherwise you couldn’t have any stable vacuum solutions. I still don’t get you I’m afraid. First of all, gravitational potential energy can only be meaningfully defined for specific spacetimes with very specific symmetries - it is not a generally applicable concept in GR. For example, it can be defined for Schwarzschild, but not for the FLRW metric in cosmology. Secondly, even in cases when grav potential energy can be defined, it is not the same as the non-linear self-coupling of the gravitational field in GR. These are very different concepts. Because you are mixing it all together in your posts, it is really difficult to give a meaningful reply over and above what I have said already. You are right of course in that accelerated expansion requires a repulsive effect of some sort - but that can’t come from self-coupling. The cosmological constant acts as a sort of ‘background curvature’ that modifies the vacuum equation...we don’t yet know what it is, physically, but we do have pretty good idea about what it can’t be.

I’m having trouble figuring out what it actually is you are trying to say, since parts of your posts contradict each other (one example above). I’ll just offer a few words from a GR perspective, since the issue of gravitational self-interaction is subtle and often misunderstood. The starting point is the law of energy-momentum conservation. This is well understood in Newtonian physics, and easily translatable to SR, so long as spacetime is flat. However, if we try to find such a law for curved spacetime, we run into trouble - replacing ordinary with covariant derivatives in the conservation law leaves us with extra curvature-related terms that do not, in general, vanish. Worse still, these terms aren’t themselves covariant, so they depend on the observer. Not good. One way to try and recover a meaningful conservation law is by taking into account not just the energy-momentum of matter and radiation, but also the energy inherent in their gravitational interactions, as well as that of gravity’s own self-coupling. However, there’s a problem - gravitational self-energy cannot be localised. The mathematical consequence of this is that there is no covariant (observer-independent) object that captures this quantity. The best we can do is use what’s called a pseudotensor, which isn’t quite the same as a full tensor, and thus not usually a ‘permissible’ object in GR. Even then, there is no unique choice of object - the one most commonly used is called the Landau-Lifschitz pseudotensor. So what we do is form a certain combination of the pseudotensor (representing the energy in the gravitational field) with the normal energy-momentum tensor (representing energy and momentum in matter and radiation) - and notice to our pleasant surprise that the divergence of the resulting object is covariant, and vanishes. So we aren’t looking at the field itself, but rather at the density of sources in a combination of gravity and matter/radiation. What does this mean? Suppose you have a small 4-volume that contains matter/radiation, as well as its gravitational field (in the form of spacetime curvature); the above means that the overall source density (divergence) of the combination of energy-momentum in matter/radiation and in the gravitational field within that volume comes out as zero, so that there is no net flow of energy-momentum through the boundary of the volume. Note the highlighted terms - we are talking about overall density of sources of a combination of the two contributions, resulting in no net flow through the boundary (Stokes theorem). The combination itself is a sum of tensors, one of which is a complicated function of the metric; nowhere are we implying that there are any exotic sources involved - we are just saying that one must account for both matter and gravity in order to write down a conservation law. We also aren’t saying at this point that ‘total energy is zero’ - only that the combination of the two is conserved in a certain precise sense. This is a subtle and somewhat counterintuitive matter, and easily misunderstood. That’s what is meant when we say that ‘the energy of the gravitational field is negative’. It’s essentially an accounting device that leads to a covariant conservation law in curved spacetime, and not an ontological statement about its nature. It’s a bit like debits and credits in accounting, which make the balance sheet balance; but the nature of money for each individual entry is of course not affected. The same with gravity - we are balancing the books, but the gravitational effect of the field’s non-linear self-coupling remains gravitationally attractive, as of course it must. You can see this in the fact that we get stable vacuum solutions to the field equations - scenarios where there is only gravitational self-energy, and no ordinary sources (T=0 everywhere). Geodesics still converge in these spacetimes, and there’s no inflation or expansion, unless you permit a non-zero gravitational constant (the above reasoning about conservation still holds even then). A trivial example is the ordinary Schwarzschild metric; a less trivial but far more striking example is the gravitational geon. Plus all other vacuum solutions without ordinary sources. These solutions wouldn’t exist if the self-coupling had repulsive gravitational effects.