Markus Hanke

You didn’t respond to my request for clarification as to what the scenario you are talking about actually is, so no, I didn’t know. But it doesn’t matter, because if we are not in a flat spacetime then this isn’t a Special Relativistic scenario, and you need to use the usual General Relativistic relations between frames. Either way, it is no problem to do this. But then, why do you keep talking about Lorentz transformations? As I have pointed out, we already know the source of the Pioneer anomaly, and it doesn’t have anything to do with gravity or new physics.

To be honest, I did not consider any specific scenario (but the author of the paper I linked earlier did), I was thinking only about general principles with this. So I don’t have any specifics to offer. What I will say though is that, in order to bring one of the particles to rest at a different gravitational potential wrt to the other one, some form of acceleration needs to be applied, which is (assuming constant a) already locally equivalent to a uniform gravitational field. So even before the final state is achieved, the question of what effect gravity has here already arises. So do you mean to say that subjecting an (already) entangled system to the influence of gravity will break the entanglement? Of course entanglement means non-separability of the wave function, so perhaps my earlier comment was misleading - I did not mean that the two parts of the system evolve separately (in that they have separate propagators), only that the 2-particle system as a whole must evolve in a different way than the one that isn’t subject to gravity. Simply on account of them not sharing the same notion of time. I think I didn’t express this very well. I am not clear though what this would really mean mathematically, since the spatiotemporal embedding of such a system would span a region of spacetime that is now no longer necessarily Minkowskian. This should have an impact on the wave function itself (does the tensor product reference the metric?), as well as on its propagator (how to formulate this, if time is a local notion?).

The problem is that both the stone and the mountain are classical objects and as such share the same fundamental properties. The same is not true, however, for the chair I am sitting on and the elementary particles of which it is ultimately composed - you can’t describe the properties and interactions of these particles with Newtonian mechanics, and conversely the chair as a whole will not exhibit any quantum effects. So these are distinct categories of objects, even though there is a definite relationship between them.

I’m struggling to follow you on this one - if you do this, then the system is no longer entangled. It is precisely the non-separability of the wave function that is the essence of what ‘entanglement’ means. I think this is a matter of degrees, i.e. it depends on what ‘significant’ means for a specific scenario. In principle, I would argue the following: let’s say you prepare two identical entanglement pairs, both of which consisting of two entangled particles each. Keep one of these entangled pairs in a locally inertial frame, simply for reference purposes. For the other pair, place the system such that there is a gravitational gradient present between the two particles that make up the entanglement pair, i.e. there is relative acceleration between their geodesic world lines as they age into the future. Both of these pairs will now have wave functions that are of the same form and are both non-separable; however, the evolution of these wave functions must differ, because in the presence of gravity the propagator is a purely local operator, so the two parts of the non-separable wave function subject to gravity will evolve differently, as compared to the reference pair that is not subject to gravity. So clearly, gravity must have some effect on the entanglement relationship. But of course I agree with you in that for most real-world scenarios such effects should be entirely negligible - unless you are in a spacetime with extreme tidal gravity, such as near the event horizon of a microscopic black hole.

Ok, but the problem then is that such a universe would not permit any (tidal) gravity in the vacuum outside of massive bodies. This is contrary to both observational evidence in the real world, as well as Maxwell’s equations. All of these, and many many more. But you are getting this backwards, because, since you are the one proposing a new idea, it is up to you to show experimental evidence that there exists electromagnetic radiation that propagates at v > c. This is simply not true. The EM wave equation follows directly from Maxwell’s equations, and its solutions are precisely the kind of wave forms we find in the real world. The entire field of electrical engineering relies on this, and it evidently works very well - in everything from aircraft avionics to microwave ovens. There is really only one field, the electromagnetic field \(F_{\mu \nu}\); the E and B fields are merely observer-dependent aspects of this, and thus make up the various components of the field tensor. When you look at how these fields transform, you will see that they already contain the gamma factor, so this is nothing new. What is this? You are essentially giving us the finger here, by saying that you are not prepared to look at any evidence that might contradict what you believe. That’s not how science is done.

I have no idea what you mean by this, you need to explain some more. A Lorentz transformation is a relationship between inertial frames; if one of the frames is not inertial, or if spacetime in between the frames isn’t flat, then the relationship will be more complicated. Note also that Special Relativity encompasses not just inertial frames, but any situation so long as the respective region of spacetime is approximately flat. The Pioneer “anomaly” has nothing to do with relativity, it’s simply due to uneven heat loss from the probe. There is no mystery here. What curve?

I think this is an interesting question, and the answer is certainly not obvious. I would expect that, if you were to place one part of an entangled system into a different gravitational potential, then this should have a measurable effect on the entanglement relationship, simple because the two parts of the system no longer evolve in time in the same way, meaning something would need to change in the overall wave function describing that pair. At the same time though I don’t see how this could possibly affect the fundamental non-separability of that wave function, so some notion of entanglement should persist. I have no idea what this would really mean in physical terms, though. I did a Google search, but this was the only thing I could find on the subject. The experiment hasn’t been performed yet, but clearly the author also expects there to be an observable effect of some kind (he talks about “entanglement degradation”). No, because a) entanglement is usually discussed on the premise of the entire system being in the same inertial frame, and b) even if gravity does have an effect, there would still be entanglement, though aspects of it might be subtle different.

Time dilation - both the kinematic and gravitational kinds - is arguably the single most extensively tested phenomenon in the history of physics, and is being directly utilised/accounted for in a large number of engineering applications, some of which are common household items which we all use. Also, some features of our everyday world are direct results of special relativity, such as the colour of some metals for example. Given this, why do you think the idea is “indefensible”? To me, that’s kind of like saying that the idea that the best shape for car tyres is “round”, is indefensible. It doesn’t make any sense to me to claim such a thing.

If you have two inertial frames in spacetime with non-zero relative velocity between them, then these frames will be related via a hyperbolic rotation in spacetime. That’s the meaning of Lorentz transformations - they are rotations (and boosts) in spacetime. The hyperbolic angle of that rotation is \[\varphi =arctanh\left(\frac{v}{c}\right)\] which means that the gamma factor is \[\gamma=cosh \varphi \] So the actual meaning of the gamma factor is that it is an expression of the hyperbolic rotation angle by which the Lorentz frames are related. It is thus fundamentally a geometric entity.

Thanks! I do have a copy of Griffiths “Introduction to Quantum Mechanics” here, he goes through the maths in quite some detail. I’d look at the SE as an eigenvalue equation for the system’s Hamiltonian though, not so much as an equation of motion.

Indeed, and that’s the crux - there simply is no meaningful notion of the “shape” of the atom until such time when it is interacted with in some way. Yes, this makes sense now, and it is essentially what I was thinking about in my last post. Also, it’s important to remember that the wave function is a probability density distribution, so it needs to be volume-integrated first in order to become a probability distribution; and for an isolated atom in free space, one is free to choose the orientation of the volume form in whatever way one wants. Indeed.

This is as interesting as it is confusing to me - maybe I should just stick to my good old simple GR, atomic physics is too complicated Let’s take the normalised wave function of the H atom for example: \[\Psi _{nlm}( r,\theta ,\phi ) =\sqrt{\left(\frac{2}{na}\right)^{3}\frac{( n-l-1) !}{2n[ n+l) !]^{3}}} e^{-\frac{r}{na}}\left(\frac{2r}{na}\right)^{l}\left[ L^{2l+1}_{n-l-1}\frac{2r}{na}\right] Y^{m}_{l}( \theta ,\phi )\] wherein L are the associated Laguerre polynomials, and Y are the spherical harmonics, as usual. When you plot this function for some possible choices of n,l,m (see e.g. Griffiths) then it is pretty obvious that only \(\Psi_{100}\) and \(\Psi_{200}\) are actually spherically symmetric. So when you say that the “shape of the atom” is spherically symmetric, then you can’t mean this analytic wave function. But as you quite rightly say, obtaining (and plotting) this wave function implicitly involves a specific choice of coordinate system; since there are no preferred coordinate choices in the real world, and since the components of the angular momentum and spin vectors don’t commute, the overall atom cannot have any specific shape until we effectively impose a coordinate system by measuring any which one of the angular momentum components as well as the total angular momentum (since each of the vector components commutes with the magnitude of the vector). So to make a long story short, the atom exists in a linear superposition of all possible “shapes” (which would add up to something that is approximately spherical) until we perform a suitable measurement on it that establishes a definite orientation in space - at which point the wave function resolves into a definite shape as in the plots above, which won’t in general be spherical. Is this the right way to look at it? I can’t make the conceptual connection at the moment, so help needed here please.

Fair enough. The point though was to contrast it against “subatomic particles” as mentioned in the OP, which clearly this isn’t. This really isn’t my area of expertise (I’m much more of a relativity guy), but I question if this is actually true. Assuming for a minute that this is a non-relativistic situation, the solutions to Schroedinger’s equation for a 3D potential well with electrons that themselves interact electromagnetically would need to involve products of associated Laguerre polynomials and spherical harmonics, which in the general case don’t yield anything like a spherical distribution. The issue I have is that the analytic expression for this can be derived only for hydrogen, and even then only \(\Psi_{100}\) appears to be spherically symmetric - so how do we know that the distribution is spherically symmetric for something as complicated as strontium? I’m not saying you’re wrong, I’m just trying to understand how you know this.

That’s because once you fix r to any exact value, the associated momentum of the particle in question becomes infinite, because this isn’t a classical system. So attempting to define the field energy in this way is meaningless, which is why it’s not done that way in QFT. You can’t calculate the vacuum energy of a quantum field simply by integrating over a volume, as you would in classical field theory. It’s very much more complicated than that, I’m afraid.

I haven’t read the book myself, but it seems obvious to me that this is a figure of speech; it would never occur to me to grant this title the status of a scientific claim, most especially not since this isn’t a technical text but a pop-sci presentation. I would assume that the actual content of the book makes this abundantly clear. It’s kind of like seeing an ad for the movie “The China Syndrome”, and then complaining that the storyline has nothing to do with either China nor any syndroms. Pretty silly, if you ask me. Pop-sci is full of such figures of speech - they talk about “black holes” (though they are neither black nor are they holes), “wormholes” (no worms involved), “vacuum” (though it’s not empty), “Big Bang” (though it was neither big nor noisy), and any number of other such terms. We could replace all these with more accurate technical terms, but then the general public wouldn’t know what it is we are on about any longer. The other thing of course is that this is a commercial publication, so it needs to sell and make money, otherwise you have a bunch of really unhappy people (not just the author!). As such, marketing is an important consideration, and “A Universe from Nothing” piques people’s interests a lot more than “A Universe From The Hartle-Hawking State, Being A Solution To The Wheeler-deWitt Equation” (the technically correct version, because “something” is just as wrong!) would do. It simply sells better, and that matters if you are in a market economy and need to at the very least recoup the costs of printing and distribution, and hopefully have some left over afterwards. I don’t consider this a malicious intent, or attempt at intentional deception in any way. It’s simply an attempt to capture the target audience’s attention. My question to you would be why this bothers you so much? This seems perfectly harmless to me, especially once you actually read the contents of the book, which, I assume, make it clear what it is the author intents to present.

Yes, that’s true, I don’t deny that at all. But mathematics as a language is fairly objective, in the sense that - even if you don’t understand any of the underlying physics - you can adopt a “shut up and calculate” approach, and still eventually obtain the correct results. With enough effort and time and repetition this will allow you to eventually figure out the underlying physics. This is of course after you learn the mathematical techniques required - so I agree with you on that point. I don’t think that will happen, based on the fact that it hasn’t happened with any other historical scientist either. For example, Isaac Newton is nearly 400 years in the past now, and in his own time his paradigm was as revolutionary as Einstein’s paradigm in the 20th century, and just as difficult for people of his time to understand. And still, Newton hasn’t been made a deity - on the contrary, his results have slowly been assimilated into people’s basic worldview, and nowadays they are essentially taken for granted, and taught in secondary school; they are now very “mundane”. I see no tendency for him (or anyone else in physics) to be deified. Could you give a concrete example of what you are suggesting actually having happened?

This is neat, but it isn’t really what the OP was referring to. For one thing, the object in question here is an entire strontium atom - not an elementary particle. The other thing of course is that this isn’t a visual image of the actual atom, but merely diffuse re-emitted light, after exciting the two valence electrons of the outer shell with a laser. That’s not the same thing at all. The atom itself has 38 electrons in five shells, none of which is spherical - I wasn’t able to find a good 3D diagram of the orbital configuration, but suffice to say it is pretty non-trivial. So the picture in the link is quite an astonishing feat (kudos to the guy who took it) - but it’s not a “photo of the atom” in the sense I understand the OP to mean.

The problem here is that our sense of ‘seeing’ is a purely classical process - it’s light of certain wavelengths being reflected off macroscopic objects that simultaneously have well defined positions and momenta. But subatomic particles are not classical objects in that same way - so your question is, in some sense, a category mistake; quantum objects don’t ‘look like’ anything, because they don’t obey the classical principles which underlie our visual sense. If anything, you’d have to turn the question around and ask: what would the rest of the universe look like if you were somehow able to piggy-back along on an elementary particle? And I’m afraid I don’t have a good answer for that one. Don’t think of it visually at all - think of it as an abstraction, similar to how an emoji can be an abstraction of someone’s mental state. The essence of an elementary particle is that it is a representation of a set of fundamental symmetries, nothing more. In tech speak: it is an irreducible representation of a symmetry group. So the best and most accurate way to think of elementary particles isn’t as ‘things’ at all, no matter how tempting that may be, but as abstract expressions of symmetry.

I understood what you were trying to say, as this is an area I have been researching extensively myself. Yes, it is possible to generalise MOND into the relativistic domain by introducing additional fields into the GR Lagrangian. Explicit examples are TeVeS (Tensor-Vector-Scalar gravity), GVT (gauge-vector-tensor gravity), STVG (scalar-tensor-vector gravity), various bigravity models, and quite a few others. So as you can see, this has indeed been considered, and a number of models have been developed. But as I pointed out before, all these models have problems of one kind or another - some make very good predictions in some areas, but fail in others; and some can be ruled out on observational grounds. None of these models has been successful enough to really replace the Dark Matter paradigm for now.

Mathematics. But once again, if every generation of physicists was to reinvent the wheel, because they didn’t believe what the generation before them has already discovered and ascertained, then science will never get anywhere. It is of course good to be sceptical and subject ideas to continued testing, but at some point one also has to put some trust into the consensus about what is already been well ascertained through the scientific method.

Yes, it is possible to do this - both in the purely Newtonian domain, and as a relativistic theory. This is essentially what’s known as MOND (and relativistic MOND). The trouble is that the modification yields residual effects even on smaller scales, which can be experimentally tested for; the most well known of these effects would be that in most MOND models gravitational waves would propagate slower than the speed of light; but we know from observation that such waves do indeed seem to propagate at c, which eliminates a large number of MOND theories. The remaining MOND models then cannot fully explain the observed motions of galaxies and galaxy clusters, so they offer no real advantage over traditional Dark Matter.

If the acceleration was large enough, and the rocket able to withstand the forces involved, then you could make this happen in principle, since the air in the cabin is not rigidly connected to the rocket. In practice though it is unlikely that any kind of real-world rocket would survive this kind of acceleration. But regardless, your understanding of the basic principle is correct. No, because acceleration isn’t the same as velocity. A plane may go reasonably fast at cruising altitude, but it takes time to reach that maximum velocity, so the rate of acceleration involved is comparatively small - which is fortunate for the passengers, since otherwise they’d get crushed into bloody puddles during takeoff

Having a teacher makes grasping relativity easier, but it is certainly not a required necessity. Given some familiarity with basic calculus and linear algebra, anyone could read the original paper on SR and eventually figure out the basic principles involved by themselves, though it might take some time and effort. The same is true of GR, though it would be more difficult. The advantage of having a teacher is that we don’t have to do this - others have figured it out before us, so it is easier and much faster to tap into the existing consensus on these matters. Why reinvent the wheel over and over again? But if the case arises that there are doubts about what a teacher says, we can always go back to the original source and check for ourselves. That’s the beauty of math.

I’m afraid I don’t understand what you mean here. GR evidently works very well, in that it makes testable predictions. Belief thus doesn’t come into it.

Not necessarily. Identifying DE with the cosmological constant is only one possible option among several. It is also conceivable that DE is the effective result of the interplay between more than one factor, such as the presence of a cosmological constant in conjunction with some background scalar field. There is no consensus about this as of now. The cosmological constant has orders of magnitude of ~10^-52 per meter squared; for localised solutions on small local patches it is thus irrelevant.