Markus Hanke

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).

Of course not. But the phenomena I listed in my post are all gravitational phenomena.

So if I’m at rest with respect to that black hole, I won’t experience time (ie I will stop ageing)? Is that what you are saying?

It doesn’t. What happens is that the relationship in spacetime between clocks nearer to the black hole and reference clocks far away changes. Time dilation is a relationship between frames, not something that happens locally. They are the exact same as everywhere else in the universe, because nothing changes locally. This is why all classical laws of physics can be written in a form that remains the same irrespective of the geometry of the underlying spacetime.

I don’t know what you consider “something special”, but we observe quite a number of phenomena that have nothing to do with speed of gravity - such as gravitational time dilation, gravitational red shift, geodetic precession, Shapiro delay, Thirring-Lense precession, tidal stretching, and gravitational light deflection. That’s just the ones that immediately come to mind. All of these are correctly predicted by standard GR, and they’re either absent or wrong in Newtonian gravity.

Yes, this is what I meant.

Motion with respect to what? Motion is a relationship between frames, and not an inherent property. I’m at rest with respect to the floor I stand on, but I’m moving at nearly the speed of light with respect to the many billions of neutrinos that penetrate this body every second. Both of these are true simultaneously, so how do you define “my” time as motion through space in a consistent manner? What is your reference point? Are you advocating some kind of absolute frame? And if you do, then, if I’m at rest with respect to whatever frame you propose, will I stop experiencing time, ie will I stop ageing?

As I said, there is no proper acceleration for free fall motion, so no forces are acting on the test particle. There is only coordinate acceleration as calculated by any specific outside observer using his own set of coordinates, but this does not correspond to any physical force, since no accelerometer exists that reads this quantity. It’s merely a frame-dependent accounting device. Either way, if you wish to present and discuss your idea, it will be best to open your own thread in “Speculations”. This here is not the right place for it.

A test particle in free fall under the influence of gravity does not experience any forces - which is to say that an accelerometer comoving with such a test particle reads exactly zero at all times. This needs to be true in all potential models of gravity, since this is what we observe in the real world. If that weren’t so, the motion wouldn’t be inertial, and thus the test particle wouldn’t trace out a geodesic.

What do you mean by “essence of gravity”?

No. You’d have to account for quantum effects, since those can’t be ignored on scales of subatomic particles.