Markus Hanke

I don’t really have one, as I have chosen to live in unconventional ways. Currently I am resident in a monastery, and preparing to ordain as a monk in a contemplative tradition, which should happen sometime next year, all going well. I also freelance as an online translator (I speak several languages) on an as-needed basis, to cover the very few expenses I have. In case you meant academic qualifications - I don’t have any, since I never went to university. The things I say here in these discussions reflect my own understanding of the subject matter; it is always up to the reader to verify any information given by consulting established textbooks, before taking them as fact. Online forums in themselves are never valid sources of scientific information.

As has been mentioned earlier on this thread, the concept of ‘gravitational potential’ can only be meaningfully defined in spacetimes that admit a time-like Killing vector field, which is only a small subset of solutions to the field equations - it does not generalise to arbitrary spacetimes.

Yes, and as it happens I am already familiar with some of these sources from my own studies. All of these papers work with highly symmetric, static and stationary spacetimes, mostly Schwarzschild. None of them makes any claim to the effect that the metric can be replaced with a scalar field, in the general case. If you are asking if you can have scenarios where there are gravitational effects without gravitational time dilation being present between reference clocks, then we have already given you several examples. A lot of interior solutions are of this kind, as are some pp-wave vacuum metrics. You can also set up such scenarios in symmetric spacetimes such as Schwarzschild, by looking at geodesics that are not purely radial. Plus many more. The point is simply this - on a 4-dimensional spacetime manifold, you can have ‘curvature in time’ (gravitational time dilation), and ‘curvature in space’ (tidal gravity). Crucially, both of these can (but don’t necessarily have to) be present simultaneously and be mutually dependent in complicated ways - for example, tidal effects don’t need to be static, they can be time-dependent and propagate, and the time dependence can itself by non-trivial. A real-world example would be spacetime in and around a binary star system. It’s due to this inherent complexity and nonlinearity that the 2-body problem does not have a closed analytical “on paper” solution. Thus, in the general case you will need more than a single number to accurately model the situation. That this is so - i.e. that geodesic deviation on this kind of manifold requires a rank-2 tensor - is not specifically linked to GR, it’s just basic differential geometry. As I have pointed out several times already - yes, you can make this work for certain sets of limited and restricted circumstances. The issue is, though, that it doesn’t generalise, so it’s not a “causal mechanism for gravity”, to quote the title of this thread. The only way for you to know for sure is to write down a mathematical model for your idea, and then investigate what kind of predictions it makes in cases other than purely radial in-fall in Schwarzschild spacetime, and comparing those to available data. I can’t stress this enough, and it is the best advice I can give you. I could keep trying to explain things until the cows come home (as they say here where I am), but until you see things with your own eyes in your own mathematical model, you won’t be able to make progress either way. At this point in time, I do not feel I really have anything further of value to add to this discussion.

Apologies, I need to correct myself, I omitted an index. This should have been \[\xi {^{\alpha }}{_{||\tau \tau}} =-R{^{\alpha }}{_{\beta \gamma \delta }} \thinspace x{^{\beta }}{_{|\tau }} \thinspace \xi ^{\gamma } \thinspace x{^{\delta }}{_{|\tau }}\]

This is irrelevant, as it still cannot model tidal effects, for reasons already explained numerous times. The necessary information content just isn’t there in a scalar field, and it won’t magically appear by taking the gradient. I didn’t mention anything about cosmology, the FLRW metric describes the interior of any matter distribution that is homogenous, isotropic, and only gravitationally interacting. Of course it is most often used as a cosmological model, but doesn’t have to be. The frustrating part about this is that you are simply ignoring most of the things we say to you, which makes me feel like I’m wasting my time with this. Also, claiming that you have “explained” something when in fact you haven’t, is also really frustrating. The other thing is that you still haven’t presented an actual model, you just keep verbally describing an idea in your head - there is nothing wrong with that in itself, it is in fact commendable that you spend time thinking about these issues. Nonetheless, until you write down a mathematical model, you can’t be sure just what the implications are - you obviously think you are right, but you won’t know either way until you actually run some numbers. Then I don’t think you really understand what the term “gravity” actually means, because if you did, you would immediately see yourself that this idea of yours cannot work in the general case, and why. Just this one point is already enough; gravity is geodesic deviation. I’ll write it down formally for you: \[\xi {^{\alpha }}{_{||\tau }} =-R{^{\alpha }}{_{\beta \gamma \delta }} \thinspace x{^{\beta }}{_{|\tau }} \thinspace \xi ^{\gamma } \thinspace x{^{\delta }}{_{|\tau }}\] wherein \(\xi^{\alpha}\) is the separation vector between geodesics, and \(x^{\alpha}\) is the unit tangent vector on your fiducial geodesic. Can you find a way to replace the dependence on the metric tensor in these equations with a dependence on just a scalar field and its derivatives, in such a way that the same physical information is captured? If, and only if, you can do so, then you might be onto something with your idea.

Simplify the expression all the way to the end, given the relations you posted earlier: \[g=\frac{c^{2}}{r}\left( 1-\left(\frac{t_{0}}{t_{f}}\right)^{2}\right) =\frac{c^{2}}{r}\left( 1-\left(\frac{t_{f}\sqrt{1-\frac{2GM}{rc^{2}}}}{t_{f}}\right)^{2}\right) =\frac{2GM}{r^{2}}\] As r->0, the gravitational acceleration increases without bound, and diverges at r=0. This is clearly not what we physically observe, since a test particle at r=0 experiences no net acceleration at all; yet it is still time dilated wrt to some external reference clock at infinity. I’ve been thinking about this some more, and I was actually wrong on something, and need to go back on it - even in Schwarzschild spacetime, you cannot specify all aspects of gravity with time dilation alone; you need at least a vector field of some kind. Consider two test particles (with their own gravitational influence being negligible) which fall freely side by side, but separated by some distance, towards a central mass. They fall at the same rate, so at every point their radial distance to the central mass is the same, hence they experience no gravitational time dilation with respect to each other. However, as they fall, their trajectories will start to converge, i.e. they approach each other as they fall towards the central mass, and eventually collide near r=0. There will be relative acceleration between the test particles perpendicular to their radial in-fall, even though they are not time dilated wrt to one another. This is because even in Schwarzschild spacetime there is tidal gravity - all radial free fall geodesics converge at r=0. You can capture purely radial in-fall via time dilation alone, but not these tidal effects. So even in simple Schwarzschild spacetime this idea ultimately fails; if you use a single scalar field to model gravity, you do not obtain the correct free-fall geodesics which we observe in the real world (unless the free fall is purely radial, which is trivial anyway). In fact, if you write the proper equations of motion for light using only a scalar model, you will find that there is no gravitational bending of light around massive objects, which is of course contrary to observational evidence (see Misner/Thorne/Wheeler, Gravitation, §7.1).

As I have attempted to explain at length, this is true only in Schwarzschild spacetime, since that is a 1-parameter family of metrics. It does not generalise to any other case. I don’t think you have understood much of what I spent considerable time trying to explain. Neither time dilation nor gravitational acceleration are variables in the field equation, and for good reason. Gravity, in GR, is geodesic deviation - the failure of initially parallel world lines to remain parallel in the presence of gravitational sources. It’s a geometric property of spacetime. In 4-dimensional spacetime, you cannot describe geodesic deviation by just a scalar quantity, it requires a higher rank object. This is nothing to do with GR specifically, it’s just basic differential geometry. You can write a scalar field model for the case of Schwarzschild spacetime (simply define a gravitational potential as function of r), but that is only because it is a highly symmetric case - this does not generalise to gravity as an overall concept. So if Schwarzschild spacetime is all you are interested in, then there is not actually an issue; you just can’t claim it is a causal mechanism for gravity in the general case, because it evidently isn’t, for all the many reasons already pointed out in previous posts. As for your last question, I already gave an example earlier - in FLRW spacetime, you have relative acceleration between test particles due to expansion or contraction of the spatial part of the metric, but no gravitational time dilation between those same test particles. Any metric where the temporal part is constant, but the spatial part is not, will be of that nature.

Just to be extra clear - the scenario can of course be treated via GR, it’s just that it’s not possible to do so via pen-and-paper methods. You would need to feed this into specialised software, and let a computer run the numbers. I do not have access to such software, so I can’t give you an outright answer to your original question. You are right, it is a pretty fundamental problem - but many fundamental problems in physics can only be solved numerically. Even in simple Newtonian gravity, if there are more than 2 gravitating bodies, the system can only be treated numerically. It is actually not surprising to me at all that this can’t be done on paper, given that the Einstein equations are a system of 16 highly nonlinear, coupled, partial differential equations. It’s more surprising to me that it can be done if one of the two masses is negligible, giving the Aichlburg-Sexl ultraboost solutions. Spacetime curvature overall is a rank-4 tensorial quantity, the Riemann curvature tensor - it describes how geodesics deviate in any arbitrary 4-dimensional spacetime. Time dilation is only a subset of that geometrical information; essentially, you can think of time dilation as ‘curved time’, and tidal gravity as ‘curved space’. Unless you have very special, highly symmetric circumstances (as e.g. in Schwarzschild spacetime), you cannot separate these two aspects - which is why, after taking account of all the various index symmetries, there are a total of 20 functionally independent components in the Riemann tensor, and you need them all to uniquely determine all aspects of a spacetime’s geometry in the general case. Time dilation alone is not enough, i.e. you can’t replace a rank-4 object that has 20 functionally independent components with just one scalar quantity, and expect to be able to capture the same information. So the answer is no, for the general case you cannot separate time dilation from the rest of your spacetime’s geometry in any meaningful way. This being said, as you introduce symmetries into your spacetime, the amount of information required to uniquely determine its geometry decreases. For Schwarzschild spacetime, you are dealing with a highly special case that is spherically symmetric, static, stationary, a vacuum, and asymptotically flat. Because it admits a time-like Killing vector field, you are able to define the notion of ‘gravitational potential’ here - the Schwarzschild geometry then simply is a family of nested surfaces (spheres) of gravitational equipotential. So for this special case, you can in fact write down a scalar field that is simply a gravitational potential with respect to some reference point (usually the center of the gravitating mass). But this is only possible because Schwarzschild spacetime is so highly symmetric - this does not generalise to more general spacetime, and most certainly not to the set of all possible spacetimes. And even then, the simple-looking form of the Schwarzschild metric is somewhat deceptive, because once you do actual calculations with it (e.g. how long it takes for a test particle to fall along a certain trajectory), things can become fairly complicated fairly quickly, since you need to integrate the relevant parts of the line element.

It’s momentum flux, not energy flux. What you are describing here is a relativistic 2-body problem, for which there is no closed analytical solution to the field equations; you can only treat this case via numerical methods. I don’t know what exactly happens here in terms of GR; I have never done this simulation myself. However, if we slightly change the scenario, then I can give you a definitive answer: let’s say there is only one (spherically symmetric) gravitating body plus an observer whose own gravitational influence is negligible. Spacetime around this mass is simply the Schwarzschild metric. If we now introduce relativistic motion (i.e. mass and observer move at nearly the speed of light with respect to one another), how will that change the gravity exerted by the mass? The appropriate solution to the Einstein equations for this case is called the Aichlburg-Sexl Ultraboost - at first glance this metric looks very different from the Schwarzschild metric, however, closer inspection reveals that these two metrics are actually just diffeomorphisms of each other. In other words, we are dealing with the same physical spacetime, it’s just that events in it are labelled differently. All curvature invariants are the same (this can be explicitly shown, though it is tedious) between these two solutions. Thus, relative motion does not increase gravity; you are still in the same spacetime with the same geometry, it is just “seen” differently (roughly analogous to how different inertial frames in SR are related by a simple rotation of the coordinate system about some hyperbolic angle in spacetime). If this weren’t so, you could construct unresolvable paradoxes just by introducing relative motion, and the model would not be internally self-consistent. I should also remind you that, if we are looking at the vacuum outside the mass, the energy-momentum tensor is always zero there. It is only non-vanishing in the interior of the mass distribution. Therefore, whether there is relative motion or not, you are actually solving the same equation: \(R_{\mu \nu}=0\); the only thing that changes are initial and boundary conditions.

Kinetic energy is an observer-dependent quantity, so it is best understood as a relationship between the two reference frames in spacetime. It is in itself not a source of gravity. Neither one of these are in themselves sources of gravity. What enters into the field equations as part of the energy-momentum tensor are momentum density and momentum flux. These are neither linear nor angular (the distinction is just a convention anyway). If there is any kind of momentum present in a gravitational source, then it will contribute to one or both of the aforementioned quantities, but the way it does so is not always trivial; in fact, finding the energy-momentum tensor for a given distribution of matter-energy can be a very difficult task, particularly if the distribution is not static or stationary. If the kinetic energy is evenly (statistically) spread out over the entire distribution, then you can sometimes simplify things by letting it enter as a contribution to another component of the tensor, the energy density. This is just the last case I mentioned above - refer to equation (16) in that paper. The kinetic energy becomes a contribution to the energy density term of the tensor. Physically this means you are describing a different system (one that has a higher temperature as compared to a reference system), not the same system in motion. That’s because you haven’t produced a model yet, you have thus far only described an idea you have had, and how you yourself understand that idea. The next step from here would be for you to actually write down a model - i.e. a field equation for the time dilation field you are proposing -, and then see what kind of predictions that model yields, and how they compare against experiment and observation. Remember, it is always good to have ideas, but it is for yourself to investigate the scientific value of that idea - you can’t just assume your idea is “right”, and then ask for others to show you wrong. Yes, that is the right approach

Gravitation in GR is geodesic deviation, and thus a geometric property of spacetime; all free-falling test particles experience gravity (they must follow geodesics in spacetime), regardless of whether they have mass or not, and regardless of their internal composition or size. Remember also that within the Standard Model, all fundamental particles are point-like, i.e. any mass distribution is simply a collection of point particles. Relative motion is not a source of gravity; the source term in the field equations is the stress-energy-momentum tensor, which, as being a tensor, is covariant under Lorentz transformations. If that were not so, the theory would not be internally self-consistent. It is important to reiterate that there are two physically distinct types of time dilation - there is kinematic time dilation due to relative motion (which also happens in flat Minkowski spacetime), and there is gravitational time dilation due to curvature of spacetime (which only happens when gravitational sources are present). These two effects can be present simultaneously, but they are nonetheless physically distinct effects.

I am honestly not sure if I follow your thought process correctly, since such a notion as “time dilation gradient” does not make much sense to me. But nonetheless, the aforementioned case of an orbit around a rotating mass should be an example. Another scenario that immediately comes to mind would be two parallel beams of light (or any other pp-wave spacetime, for that matter) - if you fire two parallel beams of light in the same direction, there will be no gravitational attraction between them, even though they carry energy. But if you fire the same two beams of light so that they are initially parallel, but travel in opposite directions (i.e. you let emitter and receiver trade places for one of the beams), then they will indeed experience a gravitational attraction. There will of course be time dilation between a clock inside the volume of dust, and some other reference clock outside of the dust cloud; but there is no time dilation between two clocks that are both located inside the dust cloud. I should have been more clear on this, as I was initially thinking of the cosmological case, where there is no “outside”. As I said, I don’t immediately recall where I saw that proof, it was a few years back when I came across it, and it was in a printed textbook. However, I can offer an outline (!) of my own attempt at proving this, for whatever it is worth. For this, allow me to go back to the basics, and consider what it actually means for a manifold (such as spacetime) to have curvature, and how to capture this mathematically. Imagine you choose some arbitrary point P on your manifold, and pick out an arbitrary tangent vector attached to that point. Now you parallel-transport that tangent vector around a small (i.e. infinitesimal) loop that starts and ends at your point P. The question is - will the initial vector before the parallel transport operation coincide with the final vector at the end of the procedure, regardless of the specific curve the loop describes, and what direction I travel on that loop? On a flat manifold, using standard calculus, the answer is obviously yes (I use single bars “|” to denote ordinary derivatives), since ordinary derivatives commute: \[A_{\mu |\nu \gamma } -A_{\mu |\gamma \nu } =0\] However, if we allow the manifold to not be flat, then the situation changes; following the standard prescription for this (refer to any textbook on differential geometry), we must now replace ordinary with covariant derivatives, which do not in general commute. The degree to which they fail to commute is (I use double bars “||” to denote covariant derivatives): \[A_{\mu ||\nu \gamma } -A_{\mu ||\gamma \nu } =R{^{\delta }}{_{\mu \nu \gamma }} A_{\delta }\] The object \(R_{\mu \nu \gamma \delta}\) is called the Riemann curvature tensor, and it uniquely specifies all aspects of the geometry of a given manifold. The question then becomes how you explicitly calculate the components of the Riemann tensor, i.e. what kind of object is it a function of? For this you need to only remember that GR uses the Levi-Civita connection, which is torsion free; this implies symmetry in the lower indices of the Christoffel symbols: \[\Gamma {^{\gamma }}{_{\mu \nu }} -\Gamma {^{\gamma }}{_{\nu \mu }} =0\] This being the case, you can then work out an explicit coordinate expression for the Riemann tensor from the above equations. I won’t typeset it here now since it is tedious to write in LaTeX notation (you can easily Google it, if you are interested) - I will simply point out that the Riemann tensor turns out to be a function of the connection coefficients and their derivatives only, which in turn are functions of the metric tensor and its derivatives only. So in other words, and that is the point of this whole exercise, given the fact that GR uses the Levi-Civita connection to describe parallel transport, a unique description of all relevant aspects of a manifold’s geometry under GR (i.e. the Riemann curvature tensor) arises from a rank-2 tensor, being the metric tensor. A simple accounting of the indices in the above expressions show that there is no mathematical possibility of any lower rank object (such as a scalar or vector) doing the same job. Which is what we wanted to show. The above is obviously only an outline - you could fill in the details and actual calculations yourself, using any standard textbook on differential geometry. I don’t know how rigorous the above really is, but that’s how I would approach such a proof - quantify the failure of derivatives to commute on curved manifolds; then, given a connection, check what kind of object the coordinate expression for the curvature tensor depends on. It seems pretty simple and logical to me. But if someone here who is actually an expert in the area can think of a better, more rigorous way, or can point out an error in the above reasoning, then I would definitely be interested in seeing it!

Well, you can consider a hollow sphere made from a thin shell of matter. The exterior of the shell looks like any other spherically symmetric body, so it is described by the usual Schwarzschild metric. The hollow interior of the shell however is a different story - no tidal gravity is detected therein, meaning a test particle placed anywhere into the interior remains at rest. At the same time though, if you place a clock into the interior, and somehow compare its tick rate against a reference clock far away on the outside, you will find that it is time dilated, even though no forces (which would cause it to move) are detected locally where the clock is. So this would be an example of time dilation, but no tidal forces. Another even simpler example would be a uniformly accelerated frame in an otherwise empty region of spacetime; again, an accelerated clock is dilated, but there is no tidal gravity. A real-world example of a case where you have tidal gravity in the spatial part of the metric, but no time dilation, would be a region of spacetime that is uniformly filled with dust, in a way that ensures homogeneity and isotropy. The FLRW metric - on which our current understanding of cosmology, the Lambda-CDM model, is based - is an example of this. In this metric the temporal part is constant, but the spatial part is not. I should also mention here that within metrics, each coordinate coefficient can depend on all coordinates, including time. So not only can things vary as you move in space, they can also vary with time, and with any possible combination of the two. So you can get quite complicated spacetimes that are neither static nor stationary, with highly non-intuitive geometries. And if that wasn’t enough, then it needs mentioning that the dynamics of GR are highly non-linear, meaning gravity self-interacts; hence (at least in principle) you can have topological constructs that are formed and held together purely by their own gravitational self-energy, in the complete absence of any “traditional” sources. I think you are beginning to see now that the dynamics of spacetime are very rich and varied - they can’t be captured by just assigning some scalar field. I actually seem to remember having once seen a formal proof that a rank-2 tensor is the lowest rank object required to capture all dynamics of GR, I just can’t remember where I have seen it. If I come across it, I will post it here.

Because - as I have attempted to explain - time dilation is a relationship between distant clocks, whereas a field assigns a particular object (a tensor of spinor of any rank) to each local event in spacetime. You cannot point to an event in spacetime and say “I am going to assign time dilation factor X to this event”, without any further qualification - this does not make any physical sense. The most fundamental entity in GR (and the solution to the Einstein field equations) is the metric tensor field - it assigns a metric tensor to each event in spacetime. To put it in the simplest possible terms, the metric tensor field allows you to quantify how each event in spacetime is related to all other events - both in spatial terms, and in terms of time. It does so by defining a mathematically precise relationship between neighbouring events, so that, by integrating along curves, you can calculate relationships between more distant events, e.g. the length of a world line connecting them. Time dilation in GR is a geometric property of world lines, in that it is the ratio between the lengths of world lines between the same events - the total time a clock accumulates between two given events is equivalent to the geometric length of the world line traced out by that clock. And how long that world line will be depends on the geometry of the spacetime it is in, and what kind of world line it is. Take for example a rotating spherical body, such as a planet. If you let a test clock orbit the planet once in its direction of rotation, starting and finishing at some point P, then that orbit will take a total time T1. If you now start at the same spot P, but orbit in the opposite direction (counter the planet’s direction of rotation, but along the same orbit, with all other initial and boundary conditions remaining equal), you will get some orbital time T2, which will be ever so slightly different. That’s because, even though you start at the same point P, and traverse the same spatial distance along the same orbit, the geometry of spacetime is such that the lengths of the two world lines will differ. The ratio between these two geometric lengths is one example of gravitational time dilation - the value of that ratio depends on where the point P is, the initial and boundary conditions of the clock kinematics, and the global geometry of the underlying spacetime. How would you capture all this by assigning a single value to point P, as you seem to want to do with your “time dilation field” idea? Again, on closer consideration, in order to capture all relevant degrees of freedom so that all aspects of gravity can be correctly modelled, independently of the precise circumstances, at least a rank-2 tensor field is necessary. That’s what GR does.

Time dilation is a relationship between distant clocks (not a property of them), wherein ‘distant’ just means that the clocks are separated in space and/or in time. Time dilation arises from the metric, which is a particular solution to the field equations for given initial and boundary conditions. No - again, because it is a relationship between clocks, not a fixed value that can be assigned to a given event.

GR uses the Levi-Civita connection, so there is no torsion. Also, if there were any vector fields involved, then those would be 4-vectors, not 3-vectors; and two separate vector fields still do not capture the necessary degrees of freedom. The field equations - like all physical quantities in GR - need to be covariant, so no, you can’t make any kind of explicit reference to an observer. One could also think of it in terms of gravitational radiation fields. These fields extend to infinity, and wave fronts in free space propagate at the speed of light; furthermore you have two distinct polarisation modes. In terms of field quanta, this automatically implies massless spin-2 bosons - which, mathematically speaking, can only “couple” to rank-2 tensors. So this is the lowest rank object that is needed to fully capture all relevant degrees of freedom of gravity.

Ok, so the gradient would give you a vector field - which is still insufficient to capture all the necessary degrees of freedom. As I said, at the very least you need a rank-2 tensor field to adequately describe gravity. Time dilation is a relationship between clocks, it’s not a covariant quantity, and it isn’t local either. So, such a thing as a “time dilation field” does not make much physical or mathematical sense. A solution to the Einstein equations is given by a metric - this is the primary and most fundamental mathematical object in GR. Once you have the metric, you can then calculate the relationship between given clocks in spacetime from this.

The concept of “gravitational potential” can only be meaningfully defined in some spacetimes with very specific symmetries; it, too, does not generalise.

It is not possible to capture all of gravity’s degrees of freedom with a scalar field theory (or even a vector field); you do require at the very least a rank-2 tensor field to do so. Even if you take just the next “baby step” up from Schwarzschild spacetime to Kerr spacetime, you will find that this concept no longer works. At least in principle you can still derive closed expressions for the relativistic optics in such a spacetime (though the maths are anything but trivial), but they no longer correspond to anything resembling a scalar time dilation field. I invite you to try it out, but be warned - some heavy maths ahead! And the whole thing most certainly does not generalise to arbitrary spacetimes.

This is valid only for spherically symmetric, non-rotating and uncharged gravitational sources in an otherwise empty universe, i.e. in spacetimes that are approximately Schwarzschild. It cannot be generalised to any other case, which is why it is not suitable as a general model of gravity. General Relativity on the other hand represents a general constraint on the metric, i.e. it constrains what form the geometry of spacetime can take, given appropriate initial and boundary conditions. It thus works as a model for gravity regardless of the precise nature of its sources, in any given purely classical scenario.

Don’t forget though that GR as a model does not stand in isolation - the large-scale physics of the universe need to remain compatible with the small-scale physics of the Standard Model. Unfortunately the QFD and QCD parts of the Standard Model Lagrangian are not scale invariant, so you cannot replace a universal expansion with a contracting observer, without breaking some crucial physics in the process.

Surely you meant to write \(g_{\alpha \beta}\), since \(G_{\alpha \beta}\) denotes the Einstein tensor, which is a different quantity.

Recasting the theory of electromagnetism into a geometrical form is straightforward if you use the differential forms formalism. It then becomes simply \[dF=0\] \[d\star F=4\pi \star J\] This has already been known for a long time. For a (very) detailed discussion on the similarities and differences between electromagnetism and gravity when it comes to their respective formalisms, I refer you to Misner/Thorne/Wheeler, Gravitation, chapter 15, most especially box 15.1. All of this has already been recognised and worked out in detail. Essentially, the form of both models shares a common underlying principle, being the topological principle that “the boundary of a boundary is zero”; but because the basic objects involved are different ones, you end up with two models that also have a lot of differences. In spite of any similarities, electromagnetism does not work the same way as gravity does. I would really urge you to consult the above reference, since it seems to me that what you are trying to do is something that has already been done long ago.

Indeed not, but it is an essential requirement in order for said topological space to be considered a spacetime manifold, i.e. a model we can extract quantifiable physical predictions from. Yes, absolutely. In GR, the connectivity (i.e. relations between tangent spaces at different points) is given by the Levi-Civita connection, and the metric provides a way to define measurements. I am unsure whether we are talking about the same thing here now. In order for a given manifold to be a spacetime manifold in the sense of GR, it has to be endowed with both a connection and a metric, or else we are no longer doing GR. Of course, purely mathematically speaking, you can have manifolds without a metric, and these can be studied (ref differential topology), but then you can’t assign a consistent notion of length to curves on this manifold. This makes them rather useless, in terms of extracting physical predictions from them, other than general statements of topology. Well, I guess that depends on what it is you are trying to model with these manifolds. Within GR, we want to be able to study relationships between events, and quantify those in a consistent manner. For that purpose, you do need both a connection and a metric. For other purposes, a connection alone might be sufficient.

You do have to impose a coordinate system to define a foliation (which mathematically is just a set of functions of the metric), but you are free to choose whichever coordinate system works best for the problem at hand. There is no physically preferred one. So different observers are free to choose different foliations for the same scenario, but these will be related via diffeomorphisms, so they describe the same spacetime. This is the exact same situation as standard GR, just written differently. I agree that we need ‘sticks to connect the events’ - that’s really what I was trying to say all along, just in different words. You need to endow your manifold with a connection and a metric, before you can define a (quantifiable) notion of separation between events. Without that extra structure (connection & metric), you have a set of events, but no way to meaningfully define separations in time and space, nor indeed any kind of causal structure. So it wouldn’t be spacetime as we experience it, because it would lack any structure, geometry, or topology. In GR, this is done by endowing the underlying manifold with the Levi-Civita connection, as well as the metric as dynamic variable constrained by the Einstein equations. That is why, when we perform actual calculations in GR to do with separations in time and/or space, these are always based on the metric. All I am really trying to say here is that a collection of events alone does not constitute ‘spacetime’ - you need a connection and a metric structure as well (which would correspond to the ‘sticks’ you mentioned) to define meaningful relationships between these events. You need sticks to connect events, in your words. Without this, I’m pretty sure you wouldn’t even have a manifold in the mathematical sense, because there is no locally defined affine structure to the set (open to correction on this point, though). It seems to me that we are actually in agreement on this point, we are just explaining it in different ways. Only if we have a manifold endowed with a connection and a metric, otherwise not. So we need that extra structure.