Markus Hanke

So energy is not locally conserved? How does this fit in with Noether’s theorem? And how does simply transforming the Lagrangian like this yield a spatial shrinkage of the system? Remember also that, in order to get equations of motion for your system, you insert this Lagrangian in the Euler-Lagrange equations (or make it stationary via variational calculus). But this equation depends on the Lagrangian itself, as well as derivatives with respect to its own time and space derivatives. Hence, if you transform the entire Lagrangian as you suggest, the solution of the Euler-Lagrange equation will not be the same - which means different physics. So how does this work - you keep L the same, but then get no shrinkage. You transform the Lagrangian, but then get different physics. So how does this work?

This will eventually be true in the distant future, assuming an accelerated rate. Right now, even for free space the expansion only becomes apparent on scales of ~MPc, so it isn’t detectable within galaxies (I assume you mean empty space between stars). What mechanism keeps metric expansion at exactly zero? This would imply that redshift is the same for all distant objects, since it depends only on our local rate of shrinkage. But this is not what we see at all.

“Rate of time” is a meaningless concept. Locally, time always ticks at 1 second per second, and lengths always measure at 1 meter per meter. You can compress the physical length of a platinum bar by packing its atoms more tightly, which in itself is not the same as a rescaling, because the atoms themselves do not change size. There is, however, a limit to this - once the compression force becomes strong enough to begin affecting atomic structure, then the platinum bar will eventually cease to be platinum. This is what happens at the formation of the neutron star itself - ordinary matter becomes degenerate because the relevant limits get exceeded, leaving only neutrons and a quark-gluon plasma. That’s precisely my point - if you increase energy levels (which is what happens when shrinking atomic structure), you end up with new states of matter that are different from the original state. This is because QFT doesn’t scale - it couples explicitly to a well defined energy scale. You can’t shrink eg an ordinary star and expect it to remain an ordinary star. It simply doesn’t happen. You would obtain the same result - nothing changes for a local observer, so far as lengths measurements are concerned. Except of course that the bar (and yourself) will be flattened into a thin sheet by the gravity of the neutron star, so you’d know you aren’t in an ordinary environment. I commend you for also considering possible problems of your idea - we don’t see this often here. Kudos 👍 By what mechanisms do these change? What determines the rates of change? What mechanism ensures that all changes are fine tuned exactly such that everything remains consistent? Using natural units is standard in all of modern physics, since it simplifies the calculations. This has no physical significance, it just saves you from writing out all the constants all the time. And yes, the Lagrangian has units of energy. Well, you will have to show this mathematically, while taking into account all already known physics. At the moment you are proposing a large number of new physical mechanisms, while assuming that these will produce precisely the results you think they’ll do. It will be up to you to show this mathematically; there’s too much in your post to actually address it all. So can you provide a mathematical formalism that shows the mechanisms for shrinkage in the framework of the Standard Model (in a way that preserves the known laws of QFT), and show that this reproduces all available cosmological observations (not just redshift)? Even for simple redshift I don’t really understand your thoughts here - if redshift was down to your local rate of shrinkage, then wherever we look, all distant objects should exhibit nearly the same redshift, or at the very least there should be no correlation with distance. Clearly this is not so, and we know that there is a direct relationship between redshift and distance of observed object. It’s like a rabbit hole - the more you look at this, the more assumptions you need in order to make it seem even remotely plausible. I really fail to see the point in all this, as it offers no advantages whatsoever compared to standard physics. And it’s not like shrinking matter is a new idea - it’s been around for as long as I can remember, and pops up regularly on forums.

What is the question, exactly?

I agree that the question as to whether they could have been different cannot be scientifically tested, at least not based on current knowledge. However, asking if at least some constants were different in the past is something that can be done - for example using natural fission reactors. Of course there’s some conceptual overlap between the above.

No. As I have already pointed out earlier, while some laws of physics may be scalable in that way, most are not scale invariant, most notably the laws of quantum physics don’t behave well under rescalings. You cannot ‘shrink’ atomic structures and composite particles and expect the physics to remain the same. For one thing, none of the fundamental interactions can be scaled, irrespective of how you fudge the fundamental constants; the whole concept of shrinking matter is pretty much dead right there. Even if that weren’t so, the wave equations that govern atomic structure do not scale as well - and neither do their solutions. And again, even if such rescalings were possible somehow, you’d come up against other issues. For example, if you shrink an atom while keeping its orbitals intact, the position of its electrons becomes more and more localised over time - which of course increases the uncertainty in their momenta. Eventually that uncertainty becomes large enough that electrons can jump orbitals (and fall back), leading to molecules becoming unstable, and ordinary matter emitting a continuous ‘glow’. Still further in the future, all atoms would become ionised; and still further, the hadrons within the nucleus would ‘dissolve’ into a quark-gluon plasma. Needless to say, we observe none of those things. Lastly, we actually have ways to check whether at least some of the fundamental constants might have had different values in the past (~2 billion years) - for example using natural fission reactors, such as at Oklo. The available data indicates that that was not the case. So no matter how you look at this, it simply doesn’t work. Even if it did, the model would generate many more problems and explanatory gaps than it solves.

That is a rescaling. I don’t understand this question...can you explain? Well, that’s exactly what actually happens to light from distant sources...it’s in free fall, after all. Suppose you have a system described by a hypothetical Lagrangian of the form \[L=\frac{a}{r^2} - \frac{b}{r}\] wherein a and b are dimensionless constants. What happens to the Lagrangian when distances shrink by half, ie you perform a rescaling r’=1/2r? This is simply to demonstrate the principle, obviously real-world Lagrangians don’t look like this. The wavelengths aren’t proportional to the size of the atom, they are determined by the structure of the quantum mechanical orbitals - which are, again, not scale invariant, since the potential term in the Schrödinger equation isn’t scale invariant (never even mind the QFTs underlying this). You didn’t address my previous objection - redshift increases as the observed object gets farther away. They depend on distance, not any local quantity. Actually, that’s pretty much what quantum field theory is in fact saying, since none of the beta function of real-world quantum fields vanish. Scale invariance is quite a complicated topic, but a very basic overview can be found here: https://en.m.wikipedia.org/wiki/Scale_invariance https://en.m.wikipedia.org/wiki/Beta_function_(physics)

That makes no sense - if there’s no rescaling of size, there is no shrinking matter. You can’t have it both ways. One is physically possible, the other one isn’t. It’s much more than an assumption - it’s a necessary consequence of the laws of gravity, which are exceedingly well tested. They are not. To give one example - metric expansion is a function of distance, so the further out you look, the higher recession velocities are. This is true for all directions. How do you replicate this with ‘shrinking matter’, which depends only on the local rate of shrinkage? This is precisely the issue I’m pointing out to you - they do not cancel out. If you rescale, you end up with a different Lagrangian; this is why the idea doesn’t work. I’m not just claiming this for no reason - it can be mathematically shown that these interactions are not invariant under rescaling. We know this. The only example of a real-world QFT that is actually invariant under rescaling would be QED without the presence of charged particles (ie sources are far away). Even full QED with coupling sources isn’t invariant under rescaling.

And this is the problem, because, in natural units, the coupling constants in the weak and strong Lagrangians are dimensionless. So if you rescale lengths, the relative strengths of the various terms within the Lagrangian changes, and the whole thing breaks down. The Lagrangian of such a system consists of more terms than just the potential; and the relationship between those terms that is the issue. Yes they do - they would need to be invariant under rescaling, which they are not.

Unfortunately neither the weak nor the strong interaction are invariant under rescaling, so no ‘shrinking matter’ model - irrespective of its details - can ever work, on fundamental grounds.

No, because the geodesic depends chiefly on the orientation of the light ray, not its spatial trajectory. You can see this in one of the other examples I gave - the satellite orbiting on the same trajectory in and against the direction of rotation, with different outcomes. So it doesn’t matter what kind of field you define in space, you can’t capture this behaviour. What does it even mean for the gradient to be shaped like a spiral? What kind of scalar field would give rise to such a gradient? But yes, feel free to investigate further. That’s how science is done, after all, and that’s how one learns 👍

Ok, fair enough. Well, there are very many other solutions to the field equations where that is the case. For example the FLRW metric - the notion of “time dilation field” doesn’t even make sense here, since this spacetime isn’t asymptotically flat, so no Schwarzschild observer exists at infinity to function as reference clock. You might find the book “Exact Solutions to Einstein’s Field Equations” by Stefani helpful, if you have access to it. It’s a nice survey of known analytic solutions - some very remarkable spacetimes here, which aren’t common knowledge. It’s quite mathematical though. They most certainly do in spacetime - ie the geodesics differ (there is a difference in frequency shift at least). Whether their purely spatial trajectory differs I’m not 100% certain, but I suspect it does, as the light ray will get “dragged along” by the spinning mass on one side (just as a massive test particle would), so it should experience more deflection when oriented along the direction of rotation. A quick search yields this : https://arxiv.org/pdf/1910.04372.pdf Underneath equation 16, there are plots for “effective potentials” (a mathematical term within the equation of motion); as you can see, these terms differ in the Kerr case between direct and retrograde geodesics - so there is a difference in deflection angles between these cases. The exact expression is given in equation 60, which unfortunately is very complicated, and can only be treated numerically (it’s an elliptic integral); but you find plots of typical cases a bit further down in figures 8 and 9 (dashed is Kerr-direct, dotted is Kerr-retrograde) - proving that the angle is indeed different depending on whether the deflection is direct or retrograde, as I suspected. The difference is in fact a lot larger than I would have suspected. For comparison it also shows the Schwarzschild, Reissner-Nordström and Kerr-Newman cases (we haven’t spoken about electric charge here, but that adds yet another degree of freedom).

If you really think you have free will, you obviously haven’t been owned by a cat 🐈

I pointed out only that they lie in the same equatorial plane, and pass the body at the same minimum distance. But I also made it clear that they experience different frequency shifts, so the geodesics through spacetime are not identical. In practice, the total deflection angle would likely also be different, despite the same closest approach distance (I’d have to check this first). Which is precisely the first of my conditions - asymptotic flatness. Ok, thank you for clarifying, I was indeed confused on this. It makes more sense now. Now, if you demand geodesics to be approximately determined by time dilation and its gradient alone - which is the g00 component of the metric -, that means the other metric components should be negligible within the geodesic equation (which you retain, as you say). This is precisely the other three conditions in my list, plus an extra assumption of low velocity and weak fields (so that g00 dominates over g11 by a factor of c^2). So we have recovered the necessity of my boundary conditions. Your proposal may work, but only if all these conditions are met. The rotating body eg violates spherical symmetry and is not stationary, so the geodesic cannot depend on g00 alone. Each of the other examples I gave violate one or more of these conditions. GR on the other hand makes no assumptions about the metric components - it treats them all equally, and accounts for them all in the geodesic equation. For interior spacetimes (which we haven’t even spoken about yet) these components are all wired up to the various components of the energy-momentum tensor via the field equation, providing a comprehensive account of gravity and its various sources. In such scenarios, tidal effects in both time and space are important, so it goes far beyond time dilation alone. So if you can just acknowledge that your idea is useful in some circumstances, but limited in its domain of applicability, then we can be all good. After all, you can’t replaced a rank-2 tensor with a single scalar field (its 00-component), and expect to not loose any information in the process - that should make intuitive sense, no?

But this isn’t what GR predicts - the geodesic in fact is determined by all components of the metric, plus boundary conditions. I thought you said your idea is meant to replicate the results of GR? Also, what exactly do you mean by “scalar time dilation field”? Time dilation is a relationship between clocks - so assigning a single value to each point in space isn’t enough, you need to also fix your reference clock somehow. Well then there is a contradiction with your model, because in the real world frame dragging does very much affect the geodesics of light (and massive objects) around rotating objects. Real-world geodesics also depend on more than just the radial coordinate, unless all of the conditions I listed apply. As I mentioned earlier, for rotating objects there will be off-diagonal terms in the metric, making geodesics depend on at least two coordinates (radius and colatitude). I do not see how you propose to have this arise from a single scalar field? The gradient only tells you direction and slope of change at each point, it doesn’t add any extra degrees of freedom.

Do you mean a scalar gradient - which is a scalar quantity -, or the gradient of a scalar field (which is a vector)? These are very different. Yes, frame dragging is the GR effect. However, note that this arises from off-diagonal terms in the metric tensor - I do not see how you can self-consistently model this using a ‘scalar gradient’ (clarification required as per above) alone. This is why this effect does not (and cannot) exist in Newtonian gravity.

Sure, we can’t deduce the exact law, in particular not the constants. However, I think we can deduce the general form it needs to have - that is just a consequence of the generalised Stokes Theorem.

If you want to know the gravitational potential for 2D, just solve Laplace’s equation - you’ll find that the potential is logarithmic, as it needs to be, since the force now follows a 1/r law. However, the story is actually more complicated - the more general law of gravity isn’t Newton, but GR. If you apply GR two a 2D universe, you find that the Weyl tensor identically vanishes; in vacuum, the Ricci tensor vanishes as well, as per the Einstein equation. Since the Riemann tensor decomposes into Weyl and Ricci, the result is that in 2D there is no gravity in vacuum at all, outside a mass distribution. You only have gravity in the interior of masses.

Yes, they are apparent in the LIGO data. Searching for additional polarisation states (vector and scalar states in addition to the tensor states) is also under way, as this provides a way to test for modifications to the laws of gravity. I’m sorry, I don’t quite understand what you mean by this? Gravitational waves are always transverse; their dynamics are also nonlinear, so I doubt it is possible to decompose them in simple ways. This is fine - but again requires that such a unique decomposition is possible, which brings us back to the gravitational potential issue. So yes, your approach works - but only given the boundary conditions I mentioned. Another way to look at this is via the action, which is the difference of kinetic and potential energies (Langrangian), integrated over time. In Newtonian gravity, the Lagrangian is just a sum of three terms, so you can easily see the kinetic and potential parts. In GR however the action is of the form \[S=\frac{1}{2\kappa} \int{R \sqrt{-g}}d^4x \] How do you decompose this into T and V parts? You can’t, except under very special circumstances. And again, this is only sort-of true under special circumstances. In general, in the geodesic equation, the geodesic depends on all components of the metric tensor in a rather non-trivial way, each one of which may depend on several coordinates. If, however, you introduce the symmetry conditions I mentioned, and in addition assume low velocities and weak fields, then the geodesic depends as an approximation only on initial conditions and g00 - which, in effect, plays the role of a potential. Well, you then have to acknowledge that in the general case, the kinematics of test particles depend on more than a single quantity. How does that work, exactly? Remember that it isn’t position that determines the difference in that example, but orientation (ie motion in direction of rotation, or against it).

If the source of the field were a 4-potential (or any combination of vector+scalar potentials), then gravitational radiation would have polarisation states offset by a 90 degree angle - just like electromagnetism. This is not the case though, as in reality the two polarisation states are offset by 45 degrees, meaning such radiation fields can only couple to rank-2 tensors as source. I have mentioned this further down - because the work done to go between points in the field must be independent of the path taken, or else the difference between those points cannot be a single unique quantity (=potential). This is how gravitational potential energy is defined - as a path integral between points within the field. If the value of this integral explicitly depends on the path, the definition becomes meaningless. This is true also in Newtonian theory. The potential must also vanish at infinity, or else you are left with a degree of freedom that cannot be fixed from the theory. Taken together, that gives you the symmetry requirements I mentioned. Why would it do that? The distance to the central body is the same, so your gravitational potential would be the same. Also, relativistic mass is not a source of gravitational time dilation. No, what I am saying is that your refraction model only works in a subset of particular spacetimes with particular symmetries. It cannot be generalised to describe all gravitational degrees of freedom, as GR does. So it’s not wrong as such, just limited. I’m afraid I lost you now - I thought your entire idea is based on the concept of a gravitational potential? If no such thing is required, and you don’t accept GR (ie no curvature), then what is it about spacetime that yields your refractive index? Hm...Im not so sure. How would you show this mathematically? On a very general note, I’m curious - why do you feel the need to replace GR in this way? It works perfectly well as it stands, and we have already known for a long time that relativistic gravity cannot be a scalar or vector theory, or any combination thereof (ref Misner/Thorne/Wheeler). I also don’t see how your idea would locally yield Special Relativity.

The problem isn’t the mathematics, but the basic premise - namely that you assume the existence of the notion of ‘gravitational potential’, as a generally applicable concept. If such a potential exists, then the following boundary conditions must apply: 1. The spacetime is asymptotically flat 2. It is spherically symmetric 3. It is stationary 4. It is static And vice versa - if any of these conditions do not hold, you cannot meaningfully define a gravitationally potential in a self-consistent way. This is basically to say that the work done to get from one point in a gravitational field to another must not depend on which path you take through spacetime, or else the difference between these points can’t be consistently described by a single number. You are right in one respect - if, and only if, you are in a spacetime that admits a consistent definition of gravitational potential, then you can describe gravitational light deflection as a refraction-like process, analogous to some variation of Snell’s Law. An example would be any system that can approximately be described as Schwarzschild. The issue is that it doesn’t generalise; violate any of the above conditions, and it will no longer work. A rotating body is the simplest example. So what I am saying is not that you are categorically wrong; it’s just that your formalism works only for a small subset of gravitational scenarios. It is not a general description of gravitational degrees of freedom. But for some special cases, I grant you that such an approach may come in handy. Two more examples to illustrate the point; each of these violates one or more of the above conditions: 1. In free space far from other sources, emit two parallel rays of light in the same direction - despite the energy-momentum they carry, they remain parallel and don’t gravitationally deviate. Now repeat the experiment, but emit the same parallel rays in opposite directions - they now gravitationally converge! (This is an example of a pp-wave spacetime) 2. Consider two intersecting beams of light at right angles, but in the same plane (a gravitational wave detector). Now expose this setup to a gravitational wave front - as the wave passes, the relative lengths of the beams will contract and expand relative to one another, even though their distance to the source (which is very far away!) is identical. Furthermore, comparing two or more such setups at different orientations in space (ie at different points on Earth) reveals the nature of these waves to be quadrupole, with two polarisation states at 45 degree angles - which necessarily implies that the field must couple to a rank-2 tensor. This is why gravity needs at least a rank-2 tensor description. I could give more examples where the gravitational potential approach doesn’t work, but I think you can see my point - ‘gravitational potential’ can only be meaningfully defined under certain conditions, it is not a general concept.

That’s not true. Instead of introducing a quantised field on smooth spacetime, you can quantise spacetime itself, giving you a set of models that do not require a graviton. One example of this is Loop Quantum Gravity. Higgs bosons do not have the required properties to be carriers of gravity, as Grenady correctly pointed out.

Suppose you have a rotating massive body of some kind. Further suppose you have two identical rays of light that pass this body in the equatorial plane, such that one passes along the central body’s direction of rotation, while the other one passes opposite its direction of rotation. The light rays are identical in all other aspects, ie they pass the rotating body at the same distance. GR predicts that these light rays are frequency-shifted by slightly different amounts as they pass the body, even though they both pass at the same distance, and in the same equatorial plane. Alternatively, you can put two sensitive clocks in the same orbit around the rotating body, but let them move in opposite directions. After one orbital period, even though they will have traversed the exact same region of space on the exact same orbit, their clocks will have recorded slightly different times. How do you explain these using Snell’s Law, and a single scalar field, respectively? I have said it many many times on here before - you can not describe all degrees of freedom of gravity using either scalar or vector fields alone, or any combination of these, for fundamental reasons. You need at least a rank-2 tensor field.

It’s the conserved current associated with spatial translation invariance under Noether’s theorem - as Grenady has pointed out. I say it again - it is not helpful to speak of the ‘mass of an EM field’.

...which of course vanishes for photons. Given all that has been said, in what way is it meaningful or consistent to talk about the ‘mass of an EM field’? Such a concept creates far more problems than solutions. It’s simply not helpful.