Markus Hanke

You need only look at the very first equation: a=0. When you are in free fall, you experience no acceleration and thus no forces act on you - and yet you are under the influence of gravity. That’s why gravity isn’t fundamentally a force in the Newtonian sense. That’s all there is to it. The other equations are simply different ways to write that same statement; so there’s more than one way to look at it, but ultimately it always comes back to the geometry of spacetime, which is the fundamental “object” underlying all of this. No, because both matter and anti-matter have positive energy density (roughly speaking, technicalities aside for now) - so gravitationally they behave the exact same. To get repulsive gravity, you would need to create a region of negative energy density - this is called exotic matter (which isn’t the same as anti-matter), and there is no evidence that such a thing exists. No, the photon is its own anti-particle. It’s kind of difficult to explain why this is so (and must be so) in a non-mathematical way; just suffice to say here that the fact can be shown mathematically. They are not. Neutrinos are much more closely related to electrons than to photons, and they form their own class of particles. There are three types of neutrinos and three types of anti-neutrinos, and (as being fermions) they have no direct relationship to photons. The Standard Model is in excellent agreement with experiment and observation, so unfortunately its level of complexity is necessary.

Yes, but that is only because you are in a non-inertial frame. The important point here isn't the table itself, but the fact that its presence prevents the test particle from remaining in its natural state of motion, being free fall. To put it differently, gravity "acts" on freely falling test particles (by determining their trajectories) even though a co-moving accelerometer reads exactly zero everywhere and at all times - meaning gravity isn't a force in the Newtonian sense. There are four main ways to look at this: 1. Kinematics. An accelerometer in free fall reads zero everywhere, so the equation of free fall motion is simply: \[a^{\mu}=0\] If we set up a local coordinate system, we can denote the position vector of our test particle as \(x^{\mu}(\tau)\), and the above then becomes \[x{^{\mu }}{_{||\tau \tau }} =0\] wherein the || denotes covariant differentiation, since we are in a curved spacetime. The solutions of this equation of motion are geodesics of spacetime, being free fall world lines. No need to reference the concept of "force" anywhere here - which would be difficult, since a=0 implies F=0. 2. Geometry. Rewrite the above equation as \[\nabla _{u}\vec{u} =0\] This is the same equation as above, just written in terms of 4-velocity instead of acceleration. The geometric meaning of this is that free fall world lines are curves in spacetime that parallel-transport their own tangent vectors. Again, the concept of "force" does not come into this at all, it is purely geometric. 3. Pseudo-forces. Start with the equation in (1), and write out the covariant derivative fully in component form: \[\frac{d^{2} x^{\mu }}{d\tau ^{2}} =-\Gamma ^{\mu }_{\alpha \beta }\frac{dx^{\alpha }}{d\tau }\frac{dx^{\beta }}{d\tau }\] Again, this is the same equation, only written out fully. The left hand side is a Newtonian acceleration, the right hand side can be interpreted as pseudo-forces, which originate from the fact that the background spacetime is not flat. This is a valid way to look at this for massive particles; however, the right hand side of the equation taken on its own is not covariant, so it depends on the observer, and one can always locally eliminate the Christoffel symbol by going into the rest frame of the falling particle. Of course, this interpretation actually fails in the case of photons, since the very concept of Newtonian "pseudo-forces" is meaningless for massless particles. 4. Least action. Free fall world lines are those for which total proper time is an extremum, i.e. they are the longest (or shortest, depending on sign convention) possible world lines between two given events: \[\tau =\int _{C} ds=\int ^{B}_{A}\sqrt{-g_{\mu \nu } dx^{\mu } dx^{\nu }} =\text{extremum}\] This again does not require any notion of force. If you look at all of the above ways to consider the geodesic equation (which is what this is), then you will find that the common factor in all of them is not force, but geometry. I have parametrised the test particle's world line by proper time here, but everything remains valid if you replace this by a more general affine parameter, in order to include massless test particles as well. Do note though that the "pseudo-forces" way to look at it is highly problematic in the case of massless particles. The point I am trying to make is that it is best to look at gravity as a geometric property of spacetime, since that is the most general description that works for any test particle, and any observer, and does not require any extra concepts.

The natural state of motion for all bodies is free fall - this is the state where no forces act on the body, i.e. it is a state in which a co-moving accelerometer reads exactly zero. In this scenario, the weight is prevented from freely falling by the surface of the table, so it is the table which exerts a force on it, not the other way around.

The source of gravity in GR isn't mass, it's the energy-momentum tensor. All forms of energy (including, but not limited to, mass) contribute to the local geometry of spacetime, and all test particles are affected by that, not just massive ones. The equivalence principle applies only locally in small enough reference frames; on larger scales, gravity is tidal in nature. You can do this locally in a small enough region for massive test particles, but it is really much better to take gravity for what it actually is - geodesic deviation. This concept does not rely on any restrictions of scale, and applies to all cases, not just massive test particles. In my opinion, resorting to local definitions of pseudo-forces unnecessarily confuses things, it is much better and simpler to just stick to a purely geometric picture.

This makes no sense, because gravity in General Relativity is not a force, and does not obey any kind of simple inverse square law. General Relativity is a model of gravity, it has nothing to say about the dynamics of electrons within atoms. Again, gravity is not a force - even though its description can be approximated by Newtonian mechanics in the weak-field, low velocity regime.

Can you provide a reference to look at?

! Moderator Note I’m moving this to Speculations for now, as that is the correct forum section for personal theories. Why would accelerated expansion be an issue? It’s a natural geometric property of this type of spacetime, and thus fully consistent with the gravitational field equations. Gravity is a geometric property of spacetime; to be more exact, it is geodesic deviation, i.e. the failure of initially parallel geodesics to remain parallel. Using the mathematical tools of cosmology, it is possible to construct a universe that - starting from a Big Bang - first expands, then slows, stops, and re-contracts to end up in a Big Crunch again, only for the cycle to repeat over and over again. The problem is that this is not consistent with what we actually observe in the real world. We already have a very detailed model of (classical) gravity, being General Relativity, which works extremely well - what you seem to propose is not very consistent with what we already know about gravity.

This, and almost everything else in that post, is most certainly not mainstream, whatever you might think.

For the more technically minded people with an interest in (classical) gravity - Misner/Thorne/Wheeler “Gravitation”. Hands down one of the best texts I have ever read on the subject. There are some real gems in there that simply aren’t found in any other text - like their excellent presentation of exterior calculus and Ricci calculus, with a visualisation of the electromagnetic field as a honeycomb-like structure in spacetime; and their derivation of the form of both Maxwell and Einstein equations from the same topological “boundary of a boundary is zero” principle, to pick just two examples. I have learned more about geometrodynamics from this 1500+ pages book than from all other books I’ve read on the subject combined. Caveat: it dates back to the 1970s (I think), so some information given about cosmology and gravitational waves is outdated. The other thing is that it has been long out of print, so it can be hard to find copies at an affordable price; I picked mine up off eBay, and I’ve also seen it on Amazon. Anyway, this text is considered the gold standard in General Relativity for a reason. Very highly recommended!

As an external observer (I live in Ireland) who isn’t usually interested in politics, I have to say even I am absolutely gobsmacked at what has happened here. A direct attack on the democratic foundations of a once great country, and in my mind it very nearly qualifies as a coup attempt. It’s just scary, and I do not understand how he isn’t being removed immediately. I feel very sorry for the vast majority of decent American people - no single person has ever done as much damage to both the country’s democratic institutions, as well as its international reputation, as Trump has. It will take a generation to recover from this. At least I am relieved to see that the democratic process still seem to function, as the elections results have just been ratified as I am typing this.

Because there is a general principle in nature that says that all systems will always tend towards that state which represents the lowest energy level, and/or the most stable configuration (in technical terms, the “least action”). This is called the principle of least action, and it is a formal mathematical statement that underlies both the macroscopic world (GR) and the microscopic realm (quantum field theory). The principle of extremal ageing is just a special case of this. In the case of test particles under the influence of gravity, the resulting world lines are the simplest possible ones (so it isn’t “elaborate”) - they are geodesics of spacetime, i.e. world lines where proper acceleration vanishes at every point (hence “free fall”), or equivalently the “straightest possible” world lines. Because it is mathematically inconsistent. The object that describes the curvature of spacetime, and hence gravity, the Riemann curvature tensor, can be thought of as “made up of” two parts - its trace-free part, the Weyl curvature tensor (which roughly speaking encodes tidal effects, i.e. distortions in shape of a test volume in free fall); and its trace, the Ricci tensor (which encodes the volume itself). If you have only 3 spatial dimensions, the Weyl tensor identically vanishes (a basic result in differential geometry, which can be straightforwardly proven), so there’s no tidal gravity. In fact, since in vacuum the Ricci tensor vanishes as well due to the Einstein field equations, you would have no gravity at all in the exterior of massive bodies - which is clearly not consistent with what we observe. So the universe cannot have only three spatial dimensions, since this is logically at odds with what we see in the world around us. I can easily formalise this argument mathematically, but I think you understand what I am trying to point out, so there shouldn’t be any need. Nice one 😄

Since creation and annihilation are not simultaneous in the rest frame of the particle, what you have observed is an interval of time. You don’t need another external clock for this. I didn’t say anything about biological ageing, which is a different matter. The principle of extremal ageing means that a test particle under the influence of gravity will tend to trace out that world line in spacetime which represents an extremum of proper time, i.e. that world line which represents an extremum of geometric length. So gravity is intrinsically linked to time being a dimension within spacetime. If this wasn’t so there would be no gravity, at least not in the way we observe it in the real world. No, they are issues of physical reality. Again, you cannot have Einsteinian gravity (which is what we observe in the real world) without time being a dimension. If there was only 3-space but no time dimension, there could not be any tidal gravity; clearly, the presence of tidal gravity is what we observe. Feel free to open a new thread and ask your questions there - we, as a community (there are others here who know more than I do), will be happy to attempt to answer them.

No. What actually happens is that the relationship between these frames in spacetime changes - clocks don’t “slow down”, and rulers don’t “shorten”, only the way they are related changes (in the case of SR, that’s mostly just a hyperbolic rotation in spacetime about some angle). All clocks always tick at “1 second per second”, and all rulers always measure “1 meter per meter”; its only when you compare two of these in relative motion, that you find that they no longer share the same concept of simultaneity, and thus that their frames are rotated with respect to one another. But because the laws of physics precisely are about relationships between events in spacetime, this has measurable physical consequences, which are quite real. Also no. Experiment and observation tell us precisely the opposite, namely that Lorentz invariance is a fundamental local symmetry of the world; this has been so thoroughly tested (see above) that it is beyond any reasonable doubt. Given this, spacetime has to be locally Minkowskian, which implies that the metric is invariant. Therefore the “number of events” - which is the geometric length of the clock’s world line in spacetime - is also invariant. It does not change and cannot change, and all observers agree on it - this is mathematical fact, and not subject to any “interpretations”. The only thing that changes with relative motion is the way two frames are related to one another.

I don’t get your point, because that is precisely what we directly observe. When you observe an elementary particle come into existence and then decay, you have observed an interval in time (and in most cases also in space). Whether you can freely move in that direction or not is immaterial, as that ability does not form part of what makes a dimension. When you feel ordinary gravity holding you down, then this happens because of the principle of extremal ageing - objects tend to follow world lines that extremise proper time, so again this wouldn’t happen if time wasn’t part of spacetime. When a photon gets frequency-shifted along a radial trajectory towards Earth, then this happens only because the spacetime manifold has a time dimension - this is as real as space, because without the time dimension, there would be no tidal gravity (the Weyl tensor vanishes identically in 1,2,3 dimensions). And so on. On a more direct level, we know from experiment and observation beyond any reasonable doubt that the world has local Lorentz invariance as a fundamental symmetry - which would of course not be possible if time wasn’t part of spacetime. So saying time is real, but denying that it is a geometric dimension within spacetime, is both physically and mathematically meaningless.

That’s a bit like saying “there is no evidence for the existence of light” while staring at your computer screen and seeing an image displayed there. Without time there would be no relative velocity between frames, no Lorentz invariance, no tidal gravity, and no notion of energy or momentum. Since we know that all these things exist, your statement is evidently meaningless. “Events” don’t exist without time. ”Frequency” does not exist without time. None of this makes any sense without a notion of time. Now, it is legitimately debatable just how fundamental the notion of time is in the overall framework of nature’s laws, but on the macroscopic level of relativistic physics it is both real and required.

Like so, to quote just one experiment amongst many, many that have been done to date: https://www.nature.com/articles/s41567-019-0663-9 And they didn’t even do it with just one atom (which is nearly considered ‘easy’ these days), but with a few thousand of them simultaneously. Needless to say the superposition didn’t last for all too long (for reasons I pointed out in my post), but it does demonstrate the principle nicely.

This is what I wrote, using a square root. The photon is its own antiparticle, since it is a massless spin-1 boson without electric charge. The neutrino on the other hand has non-vanishing rest mass, and it also differs in handedness - all neutrinos are left-handed, and all antineutrinos a right-handed, so they are distinct particles. Neutrinos are elementary particles, they are not composed of quarks, and do not carry colour charge, so they are not subject to the strong interaction. Of course not. Only the energy-momentum vector is a 4-vector. I don’t know what you mean, since obviously \[\lim _{r\rightarrow 0}\frac{GM}{rc^{2}}\rightarrow \infty\] To be honest, the rest of your post was so garbled that I can make little to no sense of it at all, hence I can’t comment.

The full energy-momentum relation (which is simply the relationship between the temporal and spatial parts of the 4-momentum vector) is \[E=\sqrt{m^2c^4+c^2p^2}\] For massive particles at rest you have p=0 and thus \[E=mc^2\] For photons you have m=0, and thus \[E=pc\] Particles do not need to be stable in order to be elementary. For example, the muon is elementary, but has only a short lifetime. Protons don't decay, so it is "more stable" than the neutron - even though both of them are quark triplets. Neutrinos naturally arise from the way the weak interaction works, since energy and momentum need to be conserved. Protons are not fundamental, they are composed of quark triplets, same as neutrons. They do interact with matter, it's called the Mikheyev-Smirnov-Wolfenstein effect. They also interact gravitationally, if you have enough of them. Gravity is not a force (though it can be approximated as such in the Newtonian limit) - as is easily seen by going into free fall while carrying an accelerometer.

Sure, there are many ways to do this. Momentum is defined to be the derivative of the Lagrangian with respect to velocity, i.e. it describes how kinetic energy relates to relative velocity, as a function of rest mass - this is simply a generalisation of the good old p=mv, as we all know it from our school days: \[p^{\mu } =\frac{\partial L}{\partial v^{\mu }} =mu^{\mu }\] with the Lagrangian for relativistic motion being \[L=-mc\int ds=-mc\int \sqrt{\eta _{\mu \nu } dx^{\mu } dx^{\nu }}\] Energy, on the other hand, is defined as \[E=p^{\mu } v_{\mu } -L\] which is in essence a restatement of the fact that energy is the conserved quantity that arises from time-translation invariance in spacetime (see Noether‘s theorem for formal proof). Put these together and rearrange to get \[E=\sqrt{m^{2} c^{4} +p^{2} c^{2}}\] as stated above. You could also simply look at the general form of a 4-momentum vector, and see immediately that its temporal and spatial parts are related as above. This would be a standard way to derive this, but there are many, many other ways to do this, both informally and in very formal ways. For massive particles at rest you have m<>0 and p=0 - insert into the above to get \[E=mc^2\] as requested. You can also derive all of this from first principles - the symmetries of Minkowski spacetime - in a very formal way by using Noether‘s theorem; this gives you the energy-momentum tensor as a conserved quantity, and from its vanishing divergence you can derive the above expression as well. The problem here isn‘t any of the above, because you can find all of this in pretty much any standard undergrad text on Special Relativity. This is basic stuff, and it is not in contention. The problem though is that your response to this will be along the lines of: “You don’t understand anything, you are just parroting what you have read!”. Am I right?

First of all, this particular relation is valid only for massive particles at rest (p=0); it’s the limit case of the full energy-momentum relation \[E^2=m^2c^4+c^2p^2\] This relation follows from the fact that the inner product of the 4-momentum with itself (i.e. its Minkowski norm) is an invariant: \[\eta _{\alpha \beta } p^{\alpha } p^{\beta } =m^{2} c^{2}\] So the validity of the energy-momentum relation is a direct result of the transformation properties of the 4-momentum under Lorentz transformations. Since you have already stated that Lorentz invariance is indeed a given, you can’t then go on and dispute the energy-momentum relation, especially not if you try to replace it with an expression that is dimensionally inconsistent and not itself Lorentz-invariant.

Just to add to what has already been said: there is, in principle, no law of nature that stops macroscopic systems from behaving quantum-mechanically. The problem is only that you need to prevent decoherence from occurring - meaning you need to prevent the system in question from interacting with its environment. This is relatively easy to do for very small systems, but becomes exponentially harder the larger the system in question becomes. Putting a single atom into a superposition of states isn't too difficult, given a suitable setup; doing the same with (e.g.) an elephant is - for all intents and purposes - a virtual impossibility.

I think his is a philosophy of acknowledging the overwhelming experimental and observational evidence for the validity of SR within its domain of applicability. Anyone who comes along and claims that SR isn't a valid model will thus have some serious explaining to do. Because of this sheer amount of observational evidence, claims to the contrary will generally be dismissed pretty much by default, on the simple principle that extraordinary claims require extraordinary evidence - it's thus your responsibility to show yourself right, not ours to show you wrong. And if you don't mind me saying so, for the case of SR your evidence will need to be exceptionally extraordinary indeed, given how well-tested and verified a model this is. This means you need to show that your own model can replicate all the results of SR, can function as a local approximation to GR, and can make verifiable predictions that standard theory can't. I dare say you will find this hard to do.

Against what, exactly?

Yes, this is a better way to look at it than thinking of space as a thing that expands. Yet another way to look at it is to say that distance measurements in this type of spacetime are time-dependent - so the outcome of a distance measurement between two otherwise fixed points explicitly depends on when that measurement is taken. There needn't be any reference to motion of any kind.

Fixed that for you.