Markus Hanke

To see whether gravity is a force, simply attach an accelerometer to a freely falling test particle. You will find that the instrument reads exactly zero at all times, even though the trajectory of the test particle makes it obvious that it is affected by gravity. And then of course you have other effects, such as gravitational time dilation, that can’t be explained by forces at all. Thus, gravity isn’t adequately described as a mechanical force. It completes the “zoo” of those particles which the Standard Model predicts within the energy ranges that we can probe with current technology. The hypothetical graviton interacts so weakly that it would be extremely difficult to detect it directly. The entire idea of a “graviton” is based on the notion that gravity can be quantised using the usual framework of quantum field theory. It is in fact easy to write down a QFT for gravity - but the problem is that such a QFT is not renormalisable, and exhibits infinities that cannot be removed via any known method. Essentially, the resulting QFT is useless, in that one cannot extract many meaningful physical predictions from it. So evidently, QFT is not the right method to quantise gravity. Based on current knowledge, it would seem that gravity is conceptually different from the other fundamental interactions, and is hence not amenable to the usual quantisation schemes. This puts a huge question mark behind the notion of a “graviton” - treating gravity as the interchange of vector bosons may not be a meaningful concept. But if it is, then it would not be difficult to incorporate it into the Standard Model (you’d just add an appropriate term to the Lagrangian). This is an area of ongoing research.

I should point out to you that you don’t seem to be following the scientific method, which is a big red flag. You appear to have arrived at a conclusion (electron is composed of photons), and now you are working at making everything fit that conclusion. That is not how science is done. In the scientific method, you start with the data - in this case the known dynamics and properties of electrons -, develop a model to describe that data, and then test that model. If the model does not work, you either amend or abandon it. Crucially, any new model must fit in with all the rest of what we know about physics. If you come up with an idea, and then find yourself unable to abandon that idea even in the face of overwhelming evidence that it doesn’t work and cannot work, then you have a problem. I think you should stop wasting your time with this, and reinvest your resources into learning what we already know about the physics of particles. Only when you are familiar with what we already know, can you make meaningful inroads into what we don’t know yet. Just some friendly advice.

Well pointed out. It bears mentioning though that in standard GR the Levi-Civita connection is used, which is torsion free.

GR does not make reference to any concept of mass within the gravitational field equations, it references only \(T^{00}\), which is energy density, as well as \(T^{\alpha 0}\) and \(T^{0 \beta}\), which is momentum density. Note that these are densities. Gravitational self-influences do not explicitly appear in the field equations (they can’t, because the associated quantities are not covariant), but are encoded in the non-linear structure of the equations themselves. In GR, the source of gravity is neither invariant mass, nor relativistic mass, but the stress-energy-momentum tensor. This is a generally covariant object, so when you go into a different frame of reference, the components of the tensor may change, but they will change in such a way that the relationships between these components - and thus the overall tensor - remain the same. So, having relative motion may change how different observers measure individual quantities such as densities, momenta, stresses etc, but it will not change the source term in the gravitational field equations. This is also relevant in vacuum, because distant sources determine vacuum solutions in the form of boundary conditions. So essentially what I am saying is that relative motion between test particle and source has no bearing on the geometry of spacetime due to that source, it only changes how the observer labels events in that same spacetime. This is of course provided that the test particle’s own gravitational influence is negligible (otherwise we have a GR 2-body problem, which is much more complex). Because in the energy-momentum tensor, a change in one component also implies potential changes in all other components. In this example, if momentum density becomes non-zero, then energy density and all other relevant components will also change in such a way as to “compensate” (so to speak) for that change. You are basically just shifting around things within the tensor, without changing the tensor itself. This is wrong, SR says no such thing. In fact, SR is a model of flat Minkowski spacetime, it has nothing to say at all about the gravitational influence of anything.

The equivalence principle states that uniform acceleration is equivalent to the presence of a uniform gravitational field within the rocket. Any region of spacetime with uniform gravity is in fact flat, it has no curvature. This is why you can derive what happens in this rocket from SR, which is a model of flat Minkowski spacetime. The same is not true for the surface of the Earth - the gravitational field of any central mass is not uniform, but tidal. If you had sensitive enough instruments, you could detect geodesic deviation within the rocket (though the effects would be small, so the field within such a small region is very nearly uniform). So these two scenarios are physically distinguishable, at least in principle, given sensitive enough instruments.

Gravity is not an observer-dependent phenomenon; if you have some \(R^{\mu}{_{\nu \alpha \delta}} \neq 0 \) in one frame, then you will find the same in all other frames as well. When an observer and a gravitational source are in relative motion, what changes is only the form of the metric - however, the metric in the rest frame of the gravitational source and the metric in the rest frame of the observer will be related via a simple coordinate transformation, so we are actually dealing with the exact same spacetime. So very simply put, if a gravitational source is in relative motion, the metric might look “distorted” in some way to an observer, but it will produce the exact same physics. Essentially, relative motion just means we are using different coordinates to describe the same spacetime. The relevant solution to the field equations for this scenario is called the Aichelburg-Sexl ultraboost.

Yes, that is indeed true.

There are no calculations in that document that concern photon orbits in curved spacetimes.

We can measure plenty of things that have nothing to do with mass, length or time. This is the “Theoretical Physics” section of a science forum; the above is thus completely off-topic.

Yes, you could indeed say that. Time (in physics) is what clocks measure; since photons do not have a rest frame associated with them, there is no physical clock that could be “attached” to a photon. In essence, there is no meaningful notion of “time passing” that could be attributed to photons. This does not, however, imply that they do not trace out ordinary world lines in spacetime, like any other particle; it’s just that they are confined to the surface of a light cone centred on any given event.

The identity given is the the norm (“length”, in some sense) of the energy-momentum 4-vector. The norm of a vector is always positive, since you cannot have vectors of negative length.

This is true only for particles at rest in a flat spacetime background. How do you know this? It seems like you posit this as a claim in order to make an idea work - I am pretty sure you have not actually worked through the maths of this. Having done a lot of GR maths over the years myself, I can tell you pretty much for a fact that six photons confined into a small region on the order of the electron radius will not travel on circular paths.

I think you mean (general) covariance, not invariance...? The components of the metric tensor do change under coordinate transformations, but they change in such a way that the relationships between said components are preserved, so the overall tensor remains the same one. 3-vectors aren’t generally covariant, but 4-vectors are, since they are rank-1 tensors. Am I being a pain now by pointing out these things?

Time is a purely local thing. When we talk about frequencies, then these will always be measured by an external observer, using his own clock - so it isn’t a really a problem. It would only be an issue if there was such a thing as a photon’s rest frame, but such a thing does not physically exist.

Ok got it. Thank you!

Yes, that’s right. I would simply say that it follows from Fermat’s principle, in free space. It should also be possible to formally and explicitly derive this, by starting with Maxwell’s equations in curved spacetime, and deriving the path a wave vector would follow. The result is a null geodesic.

Photons - just like all massless particles - always travel at c. They cannot do anything else. You need an extremely strong gravitational field to bend null geodesics into spatially closed curves; what’s more, the smaller the radius of this circle, the more gravity you need, so to speak. The only physical example of a scenario where light deflection of this kind happens would be the photon sphere just outside the event horizon of black holes. On atomic scales and for everyday circumstances, gravity is entirely negligible, even for the most massive of particles.

Indeed, you are correct. I did not consider that. Thanks for pointing it out! But one could still say that they are locally diffeomorphic, right? GR is a model of spacetime. You can’t really have one without the other, in my opinion. Reducing GR down to time dilation alone is an extreme simplification, which works only in some highly specific cases, and even then omits some important features of gravity. I honestly don’t think one does a beginner any favours by casting the model in this highly inaccurate light. Actually, it is the longest path in spacetime. Remember that the geometry we are dealing with here is not Euclidean.

Well, massless particles do not accumulate any proper time, since they trace out null geodesics, so for them ds=0. But that’s not strictly a minimum, because all possible paths they can take are null geodesics, so this (I believe) is called an infinum, not an extremum. To get a true minimum (saddle point) I think we would have to go to a region of spacetime in the interior of a mass-energy distribution, rather than vacuum. One can probably set up a relevant scenario there, but even then it’s not trivial. I’d also like to mention that any confusion about maxima and minima can be avoided entirely if one considers not proper time, but rather the action (in the theoretical physics sense) of the system in question. One then applies the principle of least action (of which the principle of extremal ageing is only a special case) - the path that is taken then is always a minimum of the action, both locally and globally.

You can explain it purely geometrically - the world line of one of the twins is a geodesic of spacetime, the other one is not. That’s all there is to it. Physically speaking though, a world line not being a geodesic implies the presence of acceleration at some point; but you don’t need to explicitly cast it in those terms.

Gravity - and hence also gravitational waves - are a geometric properties of spacetime. It has neither elasticity nor density, nor any other material property. It is not a material of any kind. Nonetheless it does have degrees of freedom, but these are of a geometric nature.

SR is well capable of handling accelerated frames as well; it’s just that the relationships between such frames are more complicated than simple Lorentz transforms. Where SR breaks down is when gravity is involved, i.e. when the region of spacetime in question is not flat. All relativistic effects are relationships between frames; you would never observe any such effects, if you looked only at a single inertial frame.

A variation in c, and a variation in a particle’s world line in spacetime, are not equivalent. There is no force field. An accelerometer attached to a test particle in free fall will read exactly zero at all times, so gravity is not a force.

Why are they not diffeomorphic? I am not a mathematician, so it is very possible that I am misunderstanding something. However, it seems clear that all diffeomorphisms are also isomorphisms (but not vice versa); furthermore, the definition of a diffeomorphism I am familiar with is that of an invertible function that maps one manifold into another (1-to-1) such that the function is smooth and differentiable throughout its domain. Why does this not apply to the cylinder and flat sheet example? That is true for extrinsic curvature, but not for intrinsic curvature. Concluding that the higher dimension does not exist is not trivial. There is nothing that rules out per se that such a dimension exists. However, one can look at the physics in such a universe - it is likely that at least some of the laws of physics in such a world are “sensitive” to the presence of the extra dimensions. For example, there are fundamental reasons why in a universe with 3 spatial dimensions we would expect to find inverse square laws of various kinds; if we were to empirically find (e.g.) inverse cube laws instead, then that could potentially be due to the presence of an extra spatial dimension. So there may be ways. However, it is also possible that all laws of physics are confined to within the embedded manifold; in such a world, detecting the extra dimensions might be very difficult. I don’t know if it would be impossible - that’s a good question. A test particle in free fall (gravity only) experiences no forces - an accelerometer falling in the same frame as the test particle will read exactly zero at all times. So clearly, gravity is not a force.

The metrics are not the same, but they are diffeomorphisms of one another. The crucial difference is that, on the surface of a cylinder, geodesics in one direction are closed curves - you can walk all around the cylinder, and end up again at the same spot where you started. On a flat sheet, there are no closed geodesics, they all extend to infinity, so no matter in what direction you walk, you will never get back to where you started. So based on this observation, they could probably figure out that they are indeed on a surface with the topology of a cylinder. But that alone does not allow them to conclude anything about whether or not that surface is embedded in a higher dimensional space. The crucial point however is that, in both cases these world lines are locally straight everywhere. You can smoothly deform one “world” into the other while exactly preserving all path lengths and angles.