Markus Hanke

That’s because elementary particles - as opposed to composite particles - do not have any internal structure; that’s why they’re termed ‘elementary’.

I’m afraid I’m getting that same impression, so I don’t feel it is worth my while to volunteer any more of my time in this thread. Good luck.

I’m beginning to see where it is that you’re stuck, I think. You see, the form of the geodesics themselves is already completely determined by the geometry of the underlying spacetime, which has nothing to do with velocities. There are really infinitely many geodesics in any spacetime - this is called its ‘geodesic structure’. So, in order to find the correct geodesic for a given particular situation, you need to specify initial and boundary conditions for the problem. And that’s where velocity comes in - it serves as an initial condition to select the correct geodesic out of all possible ones. It doesn’t actually determine what that geodesic looks like - only the geometry of spacetime does that, and that follows from the presence of sources of energy-momentum. In practice, you start by solving the Einstein equations - you input what and where sources of gravity are, and out comes a description of the geometry of spacetime. For our purposes here you can think of that description as a big bundle of free-fall geodesics - all possible ones for all possible cases, and what each of them looks like is already determined in that description. So, as a second step you need to find and select that one geodesic out of that big bundle that applies to your problem at hand; so you need selection criteria. These are your boundary conditions - initial velocity being one of them. But it isn’t a case of velocity determining the geodesic structure of spacetime - it simply helps you find “your” geodesic in an infinite pile of possible ones. The pile itself depends only on the distribution of energy-momentum. Note that velocity alone isn’t enough though, you need at least one more boundary condition. It doesn’t. It simply tells you which geodesic is followed, the ‘shape’ of which is already determined by the geometry of spacetime. See above. Gravity is defined as being geodesic deviation, in GR. It doesn’t really, it presents only one specific aspect of gravity. And it isn’t a model either, it’s just an analogy. The rubber sheet visualisation is what is called an ‘embedding diagram’ - the form that’s usually depicted uses Schwarzschild coordinates, and plots changes in the radial coordinate against changes in proper distance. That’s all - it shows just this one relationship. It doesn’t depict the time coordinate, nor the angular coordinates - so you can’t see the tidal components of gravity (or any other gravitational phenomena) in that plot. Generally it also only shows the region outside the central body, and ignores the interior part. You can deduce some of these things from what you see - but that’s only because you are dealing with the simplest of all geometries, Schwarzschild spacetime, which is highly symmetric. For something even slightly more complicated, such as Kerr spacetime, this kind of visualisation fails badly, since you can’t easily deduce any of the other aspects, such as frame dragging. Are you actually aware that in everything you’ve said so far you are tacitly assuming a very specific spacetime geometry, being Schwarzschild? It’s the simplest of all solutions to the Einstein equations - it’s spherically symmetric, static, stationary, asymptotically flat, and depends only on the mass of the central body. This solution is great for academic purposes, since it’s pretty simple and works well as an approximation. But actually, really world gravity is vastly more complicated - it may involve angular momentum, gravitational radiation, sometimes electric charges, non-linear self-interactions, and a whole host of other things. If you account for these, the geometry of spacetime very quickly becomes vastly more complex. Please don’t think that what you find in Schwarzschild is all there is to gravity. That’s not the case at all.

You are right, an orbit is a geodesic in curved spacetime, where it is the “straightest” possible connection between events, in a certain precise mathematical sense. But the geometry of spacetime (more precisely: geodesic deviation) is gravity. That’s the definition, and the current consensus. I don’t quite understand why you feel there is any kind of contradiction to “convention”? I think you may be referring to Newtonian gravity, which uses forces. This is a much older model than GR, but it still works really well in situations that deal with slow motion and weak fields. In such cases it is often unnecessary to employ the full machinery of GR, which is mathematically much more complicated than Newton. So Newton is a good approximation to Einstein in the right circumstances - which is why it’s still taught in schools.

You had to get that in, didn’t you Either way, I’m familiar with it under its original name Swordy-Abbasi spectrum, from when this was first published - but you seem to be right that it appears to be called “the knee” and “the ankle” nowadays. See, I learned something new today! And no, my area of expertise is General Relativity (mostly its theoretical foundations), not HEPP, which is why I haven’t followed latest developments on this particular issue. Exactly why do you think this is even relevant to theories of gravity? Yes. The fundamental grounds are rock solid.

The speed of light is invariant, not constant. That’s an important difference. Furthermore then it is directly related to the electromagnetic properties of the underlying medium: \[c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}\] Since these are fundamental constants, irrespective of anyone’s state of motion, c has to have the same numerical value in all reference frames.

Sure. But that’s not due to forces, because these stars in free fall. Of course. Coincidentally though this is not a gravitational phenomenon. I don’t know what you mean by “knee shift” - can you provide a source? GR is a purely classical theory, so it does not predict any gravitons. I have explained this in my previous post. Also, GR only indicates that there is an additional source of gravity that is distributed in a certain way - it does not say anything about the nature of this source. In particular, GR does not say that DM is particulate matter. The search for DM is currently underway, but not finding DM particles does not mean that DM isn’t there; there are options other than it being made of new particles. Lastly, if it turns out that DM isn’t there, then that still doesn’t mean that GR is wrong - it means only that GR’s domain of applicability is limited to shorter scales, and needs to be modified for longer scales. Either way, your idea is not a contender, since it is ruled out on fundamental grounds, as I’ve explained. This is what MOND tries to do - unfortunately the resulting model is inconsistent with observational data, so this approach does not work. It also requires extra vector and scalar fields, for which there is no evidence. No, see above. DM is a prediction about there being additional sources of gravity, but GR says nothing about their nature. So it doesn’t predict new particles. How could it? It’s a purely classical model that has no concept of quantum fields.

Well, that’s the problem with your idea, because objects going at different velocities with respect to the galactic center still experience gravity in the same way. The laws of gravity do not depend on any reference frame whatever, and thus they don’t depend on any velocity with respect to anything. This fact is already inherent in the formalism of GR, and borne out by all observational data. Gravity depends solely on sources of energy-momentum, and the non-linear self-interaction of gravity itself. Consider the SMBH at the center of our galaxy - its velocity with respect to itself is zero. So does that mean, according to your ideas, that no gravity exists there? And what about objects outside our galaxy, which don’t move around any central point? Like other galaxies? Or entire galaxy clusters? What does their gravity depend on, according to your ideas? Or consider the Cavendish experiment (which you can do yourself at home) - the gravitational interaction between the balls demonstrably depends only on their masses, but not on the Earth or its relative velocity to anything else.

You posted an idea to a scientific discussion forum, so it is not unreasonable for me to expect you to work within the scientific method. That entails putting down a formalism for your idea, so that one can extract predictions from it and compare these to experiment and observation. All scientific models in physics - without exception - work that way. If there’s no mathematical formalism, then your idea is useless, because it can’t be used to model or predict anything; it’s just a personal opinion. Having a formalism, besides allowing for numerical predictions, also removes ambiguity. That’s really important. Writing a mathematical statement means everybody knows precisely what it is you mean to say, in a way that verbal description can never really do. This is in fact one of the chief problems on this thread, because evidently I’m not the only poster here being confused on what you are actually trying to say. The other problem of course is that if your idea isn’t amenable to the scientific method, then you yourself have no way of knowing whether there’s any value to it or not. You need to be honest with yourself on this. Don’t just assume you are right and everyone else here is wrong - some of us here have spent years studying gravitational physics, so we know precisely what GR says and how it works, and based on that we have the tools to give honest feedback on people’s alternative ideas. Don’t just dismiss us - instead, use us as a valuable resource to further your own understanding. I guarantee that you’ll get much more out of this thread that way. I can see you are frustrated. That’s understandable, but you must realise that I merely gave you the perspective of the current scientific consensus on the subject of gravity (being GR). As far as I can see your own ideas do not conform to that, and, this being a science forum, you will thus naturally be challenged on them. This isn’t personal, that’s just the nature of discussion forums. You are correct, gravity isn’t a force - which is why we have GR to correct the shortcomings of Newton. However, gravity in GR isn’t merely due to time-related effects (which is what you seem to be saying) - you have to consider spacetime in its entirety, and it’s dynamics are a pretty subtle thing. Crucially, you can’t really separate time from space, except perhaps for illustrative purposes under very special circumstances. Even the simplest spacetime geometries (ie Schwarzschild spacetime) have tidal components in the spatial parts, it is not just about time alone.

The phenomena I listed are demonstrably due to the presence of gravitational sources, so of course they are gravitational in nature. You don’t get (eg) a Shapiro delay, or Thirring-Lense precession of gyroscopes, or any of the other examples, without the presence of energy-momentum distributions such as a planet. You claiming that this isn’t so is...’bizarre’ is the only word I can think of. It is easy to show that no forced-based model can ever accurately describe the correct degrees of freedom for real-world gravity, irrespective of its details. This is why Newtonian gravity only works as an approximation in the weak field regime. For one thing, the source term for gravity is a rank-2 tensor (this follows from Noether’s theorem), so whatever dynamical quantity you couple to it has to be rank-2 also. This rules out vector field models such as the one you propose, but is fully consistent with GR. Furthermore, a model for gravity based on forces alone predicts incorrect polarisation states for gravitational radiation. A rank-1 theory means that these polarisation states are inclined at 90 degree angles - which is not what we observe. You need a rank-2 theory to obtain the observed 45 degree angle between quadrupole polarisation states - which is, again, consistent with GR. This is a basic result from general field theory, and not exclusive to just gravity. And then of course there’s the trivial fact that accelerometers in free fall always read exactly zero - so no forces act on freefalling bodies, yet they are obviously still affected by gravity. That massless particles are indeed influenced by gravitational sources is an observational fact. Unfortunately your idea is falsified by the above points - gravity cannot be a force in the Newtonian sense, because that’s simply not consistent with what we observe. I think you got this all backwards - of course GR predicts DM. That’s the whole point of postulating DM in the first place. We can see based on observational data that there are various gravitational dynamics happening that are not due solely to baryonic matter that we can see; hence we deduce, based on GR, that there are additional sources of gravity that don’t appear in the visual or radio spectra. That’s a solid prediction. Now we are in the process of checking this prediction. If, in due time, no such thing as DM is found, then we know that GR isn’t the correct model on larger scales, and we can look at appropriate modifications to the theory. That’s how the scientific method works. However, what you propose as an alternative is trivially wrong on fundamental grounds, so it’s a non-starter. It doesn’t even work locally, never even mind on larger scales.

Honestly, I have no idea what your contention actually is. You seem to be contradicting yourself. Are you rejecting GR and Newtonian gravity? Are you rejecting time as being physically relevant? What, according to you, is gravity then? Clearly your comments on velocity don’t make sense, since it is trivially easy to demonstrate that massive bodies at rest with respect to one another (ie in the same state of motion with respect to some external reference point) also gravitate. You need to succinctly summarise your position here, because I think everyone is confused now.

I’m sorry but I can’t make sense of this. You really need to come up with a mathematical formulation, so that one can extract actual predictions from this, and compare them to experiment and observation. I don’t think you’ve been saying anything even remotely like that.

Good observation! +1 The other issue is that there are no stationary frames or stable orbits inside an event horizon - so we’d be spiralling into the center of the galaxy, which is not the case.

A planet is spherical (ignore angular momentum for now) because the geometry of the underlying spacetime has spherical symmetry. That means that the solutions of the geodesic equation depend only on the r-coordinate, but not on the angular coordinates. Thus, a free fall from rest will tend to be radially inwards towards the center of gravity. I can’t be sure, because you’re using non-standard terminology. Geodesics are those curves in spacetime for which proper acceleration vanishes (a=0 at all times and at all points). These are the curves that are traced out by particles in free fall. You cannot separate time from space in any meaningful sense - but spacetime as a whole can’t be visualised because it is 4D. So I’m really not sure what it is you’re doing, or what it has to do with velocity.

There are different numerical algorithms depending on what you are trying to achieve, and the required level of accuracy - you pick the one that’s appropriate for the task at hand. Linearised GR is not the same as Newtonian gravity - it works with spacetime geometries instead of forces, but treats these as small deviations from flat Minkowski spacetime, so it works only for weak fields. Its advantage is that the dynamics are linear, so the maths are easier. Well, if you have only a single source, or some special case of two or three sources, then often you can solve the GR equations directly. So no need for numerics in these cases. Almost certainly not. They probably did a full numerical solution of the full Einstein equations for the BH merger, using whatever algorithm works for this - hence the need for powerful computers. My guess though is that they probably used “lattice GR”, ie they treated spacetime not as continuous, but as a finite lattice made up of small volumes. That kind of approximation reduces the computational load considerably. But that’s just a guess on my part, I don’t know for sure - I’ve never really studied numerical GR.

! Moderator Note Moved to Speculations for now - conspiracy theories don’t belong into the main forum sections.

I define them as gravitational because they are direct consequences of the presence of gravitational sources. For example, a gyroscope not subject to any other interaction does not precess if there are no gravitational sources - planets etc - nearby. Some of these phenomena will happen regardless of which model for gravity you use, but only GR predicts them all with the correct magnitudes. For example, Newtonian gravity predicts neither frame dragging nor time dilation, and gets both light deflection and perihelion precession pretty badly wrong. Neither GR nor QM have anything to say about this proposition, because it makes no sense. I notice you didn’t answer my question, which I find to be important - what about massless test particles, ie photons? Is radiation subject to your proposed effect? If you base this on F=ma then the answer should be no.

I do not actually know precisely how one would go about doing this in a numerical algorithm. If I was to be tasked with figuring this out, my approach would be to use a linear approximation. I would linearise the field equations, and solve for each source in isolation initially taking into account only lower-order correction terms to keep things simple. Since this is now a linear model, you can simply add up the solutions. I would then redo this in iterations, taking into account more and more high-order correction terms with each iteration. With each step this will become increasingly more complicated - so I’d terminate once the calculation takes too long, or I reach the required accuracy. Another idea would be to “pixelate” my spacetime, ie do a lower-resolution approximation rather than use continuous functions. This is just brainstorming.

Velocity does not factor into the gravitational field equations, because it is irrelevant to the geometry of spacetime - or, to put it differently, relative motion is not a source of gravity. Where velocity does play a role is in determining which of the possible geodesics a test particle in free fall will follow. The geodesic equation is a system of partial differential equations - so, in order to find a particular solution, you need to supply boundary conditions. Initial velocity - as a vector - is usually one of these. It’s like selecting the correct geodesic out of all the possible ones. However, which ones are possible, and how exactly these look, is determined by the metric and the connection - ie the geometry of spacetime. And this has nothing to do with any velocities. Lack of education isn’t an obstacle, as it can be remedied easily - these days, you can learn any topic you like using freely available resources online. This is especially true for maths and physics. What is an obstacle is thinking you can simply dismiss a well-established model that you know little to nothing about, and replace it with an idea of your own based solely on it making sense to you. Surely you can see the problem yourself. You cannot visualise gravity in all its degrees of freedom - even I can’t do that, after spending many years on this. To this day I sometimes get surprised by totally unexpected and counterintuitive results, which one can only find using the maths. That’s how it is.

Do you mean I’m wrong about what GR says? Certainly not - what I told you is a basic fact about the model. I can show you the maths, if you like, or you can just take my word on it that I spent years studying it in detail, and kind of know what I’m talking about. It’s my area of expertise. Or do you mean GR is wrong about gravity being geodesic deviation? Well, you must realise that it is an exceptionally successful model, which has been extensively tested over the past century. It works far to well for its basics to be “wrong” in any meaningful sense. So it’s probably best if you don’t stick out your neck all too far...

You cannot add them. GR is a nonlinear model, which means that, in general, the sum of two valid solutions to the field equations isn’t itself a valid solution. What you’d have to do is solve the equations using a distribution of multiple sources as boundary condition. This is quite difficult, and can, in general, only be done numerically.

Locally, GR reduces to SR, so you are right. Nonetheless, all the specific phenomena I listed are gravitational ones. So that means only massive objects are affected by this, but not electromagnetic radiation?

Gravity in GR is geodesic deviation, meaning the failure of initially parallel geodesics to remain parallel due to the geometry of spacetime. This has nothing to do with velocities, and involves both time and space.

But you are the one who made that claim in the first place? If you can’t answer this, what do you base your claims on? No, the opposite is true - once you cross the event horizon, there are no longer any stationary frames. You can be at rest with respect to the BH by locating yourself along its axis of rotation, and firing your thrusters until you hover above the horizon. You’d be stationary there (no orbiting, no in-fall).

The amount of sunlight the Earth receives will vary if you vary the orbit, which would of course have an impact. Exactly what those impacts are in detail is a question better asked to someone dealing with the Earth sciences (not my area of expertise).