Markus Hanke

No they won’t. Same manifold, same connection, same metric even - just expressed in a different coordinate system. No, I’m not talking about connections. I’m talking about a situation where you express the same physical situation (ie spacetime) in a different coordinate system, so that all laws of physics and all tensorial quantities remain the same - including the metric tensor. This is just the usual diffeomorphism invariance of GR. Yes, it’s enough. You end up with a set of parameters that don’t correspond to what clocks and rulers actually read, so for every quantity you calculate from such a model you need to apply a mapping that takes it back to real-world measurements - and that map is just precisely the inverse of the “correction” you applied in the first place. Like I said, lots of extra work for no discernible benefit. Gravity isn’t a force, and can’t be modelled as one - this is precisely the difference between Newtonian gravity and GR. A rank-1 theory such as a vector field model cannot capture all relevant degrees of freedom of gravity. For example, the polarisation modes of gravitational radiation in any force-based model will be inclined by 90°, whereas in reality these modes are at 45°. You really do not at least a rank-2 tensor model, such as GR. No, it’s a lot more than that. As other posters here have correctly pointed out, for there to be radiation at all, the second derivatives wrt time and space of your “waving quantity” need to be related via a very specific form: \[\frac{\partial^{2}}{\partial t^{2}} =c^{2}\frac{\partial^{2}}{\partial x^{2}}\] In the case of light, the relevant quantity is the electromagnetic 4-potential, and the equation thus becomes (in Lorentz gauge) \[\square A^{\mu}=0\] If c isn’t a constant, this relationship is violated - there is no electromagnetic radiation in such a universe, at least not of the form we see in the real world. This has nothing to do with measurements or conventions.

I use it in the formal sense as defined in differential geometry, ie as a structure that allows you to meaningfully define the inner product of tangent vectors at points on the manifold, which in turn gives a meaningful notion of lengths, angles, areas and volumes. Yes. You need to be careful here - the Christoffel symbols and the connection are not the same thing. A connection allows you to relate tangent spaces at different points on the manifold to one another, ie it provides a notion of parallel transport. This is quite independent of any metric, which is to say you can meaningfully have a manifold that is endowed with a connection, but not a metric. The Christoffel symbols then give you the connection coefficients, ie they tell you what effects your connection has in a particular coordinate basis. They do this by describing what happens to basis vectors as you transport them between neighbouring points, which is something you can calculate from the metric and its derivatives. Without a metric you can still do parallel transport, but you can’t tell what happens to lengths and angles when you do it. Long story short - you can have a connection without a metric. See above. Having a different metric changes the Christoffel symbols (they are not tensors!), but not the connection. Ok, but in the context of physics (SR/GR) the term “metric” is most often used in the differential geometry sense. Physically speaking, equivalence then means a diffeomorphism, so that both metrics describe the same spacetime and thus physical situation. But here’s the thing - as explained above, you’re still on the same manifold endowed with the Levi-Civita connection. By changing the metric like this, you’re doing one of two things: 1. You’re describing a different spacetime, ie a different physical situation, since the two metrics aren’t related by any valid diffeomorphism; or 2. You’re describing the same physical situation, but the coordinates you are using no longer have the same physical meaning. I think what you are trying to do is (2). But the thing is that now measurements on your mathematical manifold (ie in the model) no longer correspond to measurements in the real world, so anything you calculate from this - eg the length of a world line - must first be mapped back into suitable physical coordinates to compare them to real-world measurements. Such a mathematical map may or may not exist, depending on the specifics of the setup. This will also change the form of physical laws, so all the various equations etc will be different for each choice of transformation you make. In either case, this creates a lot of additional work and confusion, for no discernible benefit. It would look for differences in the outcomes of experiments if you vary direction of relative motion, as mentioned previously. For example, if a uranium atom decays if you move it in one direction, but doesn’t decay if you move it at a 90° angle to that direction (everything else remains the same), then you have anisotropic space. This has nothing to do with conventions.

Just a few corrections here. The basic object of this framework is a differential manifold. This can initially be “bare”, ie without additional structure, but, as you say, there’s not a whole lot one can do with that. So we can endow the manifold with additional structures - firstly, we can endow it with a connection, which allows us to relate tangent spaces at different points. This is thus equivalent to having a notion of covariant derivative. Given a connection (but no metric yet), you can define things like curvature and torsion (these can be defined purely in terms of the connection), parallel transport, and tensor fields - IOW, you can do differential topology. But what you don’t have yet is a notion of lengths and angles, and you also don’t have a relation between tangent and dual spaces, so you can’t raise or lower indices on tensors. For these things you need to endow the manifold with a metric, in addition to a connection. Now you can use the full machinery of differential geometry. If your metric is positive-definite, it’s called a Riemann metric; if the metric tensor is everywhere non-degenerate and symmetric, it’s a semi-Riemannian metric, which is what is used to model spacetime. Smooth manifolds, connections and metrics are their own independent concepts, they are not defined in terms of each other. I have difficulty making sense of this - see also what I wrote above. I think what you mean is that you have one differentiable manifold endowed with the Levi-Civita metric, as well as to different metrics on that manifold, each of which uses its own notion of time, but both describe the same physical situation? What do you mean by “equivalent” in this context, exactly? Usually, metrics that are equivalent are those related via a diffeomorphism. My understanding so far is that we have only one manifold, which is endowed with the Levi-Civita connection plus two metrics, so the above makes no sense to me. Whether the metrics are equivalent or not, they are always preserved under the Levi-Civita connection; this is one of the defining characteristics of this connection. But you have so far explicitly stated that what we are using is the Levi-Civita connection…? But then you’re directly contradicting experiment, which clearly shows that space is isotopic, at least within the domain we can experimentally probe. So what is the point in all this?

(Bold/italic are mine) You are really contradicting yourself here - so are we working on one and the same manifold, or not? This makes no sense at all - if the connection is Levi-Civita, it always is torsion-free by definition, and it always preserves the metric; those are not observer-dependent. If it doesn’t do those things, it’s not a Levi-Civita connection…but then you explicitly state that it is, so I don’t know what you’re actually trying to say here. No, because c is the conversion factor between time-like and space-like parts of the line element, it remains locally constant irrespective of connection or metric or observer. I don’t even know what you’d have to do to make it appear non-constant…you’d maybe have to parametrise world lines not by proper time, but by some other non-trivial affine parameter that somehow varies in some sense along the curve. I’ve never seen that done, so not sure if that is even mathematically meaningful ( @studiot?). Across an extended region you can then maybe get a “speed” that varies without acceleration. I’m beginning to suspect that what are you referring to is in fact the scheme by which we parametrise world lines. Ordinarily this is done by using proper time, since that way the geometric length of world lines in the mathematical model directly corresponds to accumulated times on a physical clock. But of course you can use other parametrisations too, such as is done for example with null geodesics (where you otherwise would have ds=0). This in effect introduces a new concept of “time” that is not based on what physical clocks actually read. The trouble is that what you have verbally posted is contradictory and ambiguous. It would be much better if you could present your thoughts in mathematical form, so we all understand what it actually is you are talking about. The reality is also that in all experiments we have ever conducted, the laws of physics have never been seen to vary between inertial frames, which implies that c must be invariant at least within that experimental domain, irrespective of its precise numerical value. I therefore don’t understand why you would try to construct a model where this is not the case - at best it creates additional computational work, at worst it will be just plain wrong.

Just to elaborate a bit more. When we speak of the invariance (not constancy!) of the speed of light, what this physically means is that the outcome of experiments is always the same in all inertial frames, ie uniform relative motion has no bearing on the outcome of experiments. This has nothing much to do with units or numerical values. Yes, it is always possible to describe the same physical situation in terms of different “geometries”, if you so will. You can eg forego any reference to curvature completely by choosing a different connection on your spacetime - the geometry is now curvature-flat, and instead contains all information about gravity in the form of torsion. But all this is saying is that one can draw different types of maps over the same territory, like having a topographical map vs a road map over the same region. That way you emphasise different information, but the actual experience of physically crossing that terrain is always the same, irrespective of what map you use to navigate. This is not revolutionary or mysterious, and reveals nothing new about the world. It’s “kind of trivial” as the poster in your screenshot correctly said. So I think if you put enough thought into it, it may perhaps be possible to come up with a mathematical description of spacetime in which c is explicitly a function of something. The reason why no one uses such a description is that any measurements of space and time obtained from this description won’t directly correspond to what clocks and rulers physically measure in the real world - you’d have to first map them into real-world measurements, which means additional work and complications without any discernible benefit. Irrespective of what description you use, the outcome of experiments will still be the same in all inertial frames, and this is what we actually observe in the real world.

No. In your example you’re just using different coordinates on the same manifold, this isn’t a change in “geometry”. You are free to use any definition of time and space you like, so long as these remain valid solutions to the field equations along with the same boundary conditions. Curvature is not a single scalar quantity, and in general there is no global notion of simultaneity in a curved spacetime, regardless of coordinate choices. Locally this simply means you’re in free fall. Globally this isn’t possible, since such a global transformation would by definition not be a diffeomorphism, and thus not the same physical situation.

No, that’s where your mistake lies. The lengths between the electrons are part of the circuit frame, not the electron frame, and hence not contracted. The total electron number (charge) in the wire is conserved. Exactly. Indeed.

I can see your point, and you are of course right. The kind of density I had in mind though was a different one - I took the balls to be extended objects, and mentally considered the ratio between ball radius and tube length, ie which proportion of the tube volume is occupied by each ball. Since both are length-contracted by the same factor, this ratio does not change. In my defense, I tend to have a tendency to seek out invariant quantities when looking at relativistic scenarios. But I don’t think that’s what the OP has in mind, unless I’m still misunderstanding him. He is comparing the same circuit from the same frame, only with current off and on, and argues that because the electrons are in motion, the distance between them decreases, and thus there’s a larger net negative charge in that section of the circuit because there are more electrons in that same length of wire. He never mentions the rest frame of the electrons, nor the EM fields. What I’m saying here is that the distance between the electrons doesn’t change just because you turn on the current, because the observer is still stationary with respect to the circuit; there’s no length contraction of distances in the wire in the observer’s frame. In other words, the total amount of charge is the same, it’s just that this charge is flowing rather than standing still. I have, but I don’t think that’s what the OP had in mind, see my comments above.

Note first that that response of mine you quoted was based on a misunderstanding on my part of what the OP was actually suggesting, so it missed the point of the scenario. But as to your question, I don’t think anything would happen to the ball density in that particular case, since the radius of each ball (in the direction of motion) is length-contracted by the same factor as the tube itself, so the ball density remains unchanged, unless I’m overlooking something. This is provided that the balls are at rest relative to the tube, or else things become more complicated. In general though, a spatial density is not always automatically an invariant quantity - which is why in relativistic scenarios it is wise to use current density J instead, which is a 4-vector.

That’s technically true. However, I’d like to invite you to yourself calculate the gamma factor for a typical electron drift velocity, which is on the order of 10^(-4)m/s, and draw your own conclusions as to the order of magnitude of any length contraction effects this would induce. Not that it matters, see below. There is no such increase, see below. It is also not denied that the total number of rungs on the ladder, and thus the ratio between rungs and (eg) number of boards used to build the barn walls, does not change. If you are merely comparing “current on” vs “current off” in the same stationary circuit frame, then you have nothing to resolve, because is wrong. What is contracted is each electron’s radius in the direction of motion, but distances in the stationary (from the POV of the observer) wire don’t change - it’s only the electrons that are moving in the frame of the observer, not the wire, so a unit of distance in the wire is the same whether current is on or off. So the total number of electrons seen to be moving through that wire does not change either, just the mix of E and B fields resulting from the presence of these charges changes. Applying an electric field to a distribution of charges in a wire, all else remaining the same, does not change the number of charges, it just sets them in motion, relativistic or not. This is a trivial scenario, there’s nothing to be resolved here. It’s much more interesting to compare circuit frame to electron frame (where distances really do become contracted), and that’s resolved the same way as the ladder paradox. Either way, there’s no difference in outcomes, and thus no paradoxes. Like I showed in my first response, there can never be any physical paradoxes arising from SR; any apparent paradoxes always indicate some error in applying the axioms. You can’t get around this.

To be honest, I don’t think it matters too much - the choice of units is arbitrary, in that the laws of physics don’t depend on that choice, and in the age of digital devices, unit conversion is a trivial task. The only important thing is that one is consistent with whatever choice one makes, so as to minimise potential sources of confusion. So yes, one can of course use that system you describe, but I think ultimately people will stick with whatever units are most convenient for whatever task they have at hand.

You can think of the number of electrons in the wire to correspond to the number of steps on the ladder. This number does not change as you switch between frames (so no excess of either electrons or protons), and despite the circuit being length-contracted, the “electron-ladder” still fits in the wire, due to relativity of simultaneity.

I think you might be right, I misunderstood, see below. Ok, I just saw this while re-reading the thread. Apologies, I initially misunderstood your scenario. So the observer is at rest in the circuit hardware frame. The drift velocity of electrons in a DC circuit is typically nowhere near relativistic, so you wouldn’t see any relativistic effects. If that velocity somehow was made relativistic, the resolution is still the same - in the circuit frame, the classical radius of the electrons is length-contracted in the direction of motion; in the electron rest frame, it is the wire length and the classical radius of the protons that is length-contracted by the same factor. Crucially, in both cases, you furthermore have to consider relativity of simultaneity when figuring out how many particles are actually there in total, since the events “reached beginning” and “reached end” of wire are no longer simultaneous in both frames - the adjustment happens by the same factor, thus the ratio between the number of electrons and the number of protons is the same in both frames. This is really just a variation of the classical ladder paradox, and is resolved similarly. As an aside - an experimental example of where you’d see a type of length-contraction of a roughly spherical object into something resembling more of a flattened disk, is the collision of gold ions at the Relativistic Heavy Iron Collider.

No. Charge as well as particle number are conserved quantities under Lorentz transformations, as I have pointed out above. The ratio of electrons to protons doesn’t change just because something is in relative motion. How could it? Inertial motion doesn’t magically create or annihilate particles in a wire. Consider, as a very primitive and purely classical analogy, a length of transparent plexiglass tube filled with - say - 100 ping pong balls, initially at rest. The balls are all perfectly spherical, and their number is constant and doesn’t change. Let’s say it’s a random mix of red and yellow balls. Now you put the tube into motion at relativistic speeds, relative to an observer. What happens? For that observer, the tube becomes length-contracted in the direction of motion, and so does the radius of each ball. The balls no longer appear spherical to him, but instead look like flattened disks; however, both their total number and the ration between red to yellow balls remains the same. It’s similar with the particles in the wire, with the caveat that those aren’t classical objects, so for a precise description you need to use relativistic quantum mechanics, which means fields of bispinors. But it all works out. As I said above, you cannot construct physically paradoxes with the axioms of SR.

I don’t really understand what you are getting at. First of all, it isn’t just one side that gets length-contracted, but the entire circuit in the direction of motion. Secondly, both the circuit itself and the classical electron radius in the direction of motion are shortened by the same factor, thus the total amount of charge (ie number of electrons in the wire) in any section of wire remains conserved. Particle number is a conserved quantity under Lorentz transformations. I don’t see an issue here? SR is just the geometry of Minkowski spacetime; specifically, in the special case of relationships between inertial frames, it concerns hyperbolic rotations of the coordinate system. Let A be some coordinate-dependent description of an inertial frame, and L be a Lorentz transformation parametrised by rotational angle (=rapidity, a measure of relative velocity). The action of L on A is thus simply \[A’=L(\omega)A\] Furthermore, we know that, from definition, the matrix L is a square matrix and must preserve the metric: \[L^{-1}gL=g\] which implies that \(|det(L)|=1\). From linear algebra we know that all square matrices with determinant other than zero are invertible, and thus: \[L^{-1}(\omega)L(\omega)A=A\] So it is mathematically and logically impossible to construct physical paradoxes using just the axioms of SR, no matter what kind of physical scenario you try to concoct. Any purported SR “paradox” is by the above automatically an erroneous conclusion based on some misapplication of SR.

You haven’t shown us any actual model, which, as both KJW and myself have pointed out, needs to take the form of a metric which is a valid solution to the Einstein equations for a physically reasonable energy-momentum tensor. It is not enough to just present a list of claims; that’s not a model.

According to the rules of this forum, you need to present your idea such that readers do not need to follow any links or open any attachments. This is for safety reasons. You could begin simply by posting here the metric you are suggesting, along with a short explanation of what system of coordinates you’re using. We can then take a look at it.

No it wouldn’t. “Relativistic mass” isn’t a source of gravity; this outdated concept isn’t used any longer precisely because it lends itself to this type of misconception.

@Dejan1981 I’d just like to add that there is in fact a way to “split” curvature, though not along the lines of what you suggested. Imagine you have a very small volume - like a very small ball - that is in free fall towards a larger body. One might ask what happens to this ball, and this can be analysed in terms of volume and shape. As it turns out, if the ball free-falls in a vacuum, its total volume remains conserved, and only its shape distorts over time as it approaches the central body. If however the free fall happens inside an energy-momentum distribution (eg in a very strong electromagnetic field), both its shape and its volume will change. The rate at which a test volume in free fall begins to change is encoded in the Ricci tensor Rμν , which, as you might recall, appears directly in the Einstein equations. This is what the vacuum equations physically mean: Rμν=0 means that a small volume in free fall through vacuum is preserved. The shape of the ball is encoded in another object, the Weyl tensor Wμνϵδ , which does not vanish, and which does not appear in the field equations. If the free fall happens in something other than vacuum, the Ricci tensor never vanishes: Rμν=κTμν−κ12gμνT while the Weyl tensor may or may not vanish. EDIT: The last term in the above should have a factor ½, not 12. For some reason I’m unable to edit the LaTeX, and the rendering looks off too. No idea why. The point is this: the full description of the entire spacetime geometry is given by the Riemann tensor, and it so happens that this tensor can always be covariantly decomposed into a linear combination of the Ricci tensor and its trace, as well as the Weyl tensor. This is called the Ricci decomposition. What I wrote above is the geometric interpretation, and you can also give it a more physical one - the Ricci tensor represents curvature due to local sources, whereas the Weyl tensor represents curvature due to distant sources, for example gravitational radiation. So if you want to decompose curvature, this would be a well-defined and observer-independent way to do it.

No, the metric is the unique solution that arises from the field equations for an electrically charged isolated spherically symmetric non-rotating body in an asymptotically flat spacetime. You can’t uniquely decompose this into two separate metrics that somehow add together, at least not if you want those metrics to themselves be valid solutions to the equations. What you can do though is consider the Reissner-Nordström metric as a special case of the Kerr-Newman metric. In GR, if you take two valid solutions to the field equations and add them together, the result will in general not be a new valid solution; and valid solutions are not generally decomposable into the sum of other valid solutions. That’s because the field equations are non-linear.

No probs, it’s my pleasure 👍 And of course you are right in that, it’s a solution with idealised boundary conditions. But my point was simply to show that the Einstein equations admit vacuum solutions - ie that the local (and even global) absence of sources does not automatically imply flatness of spacetime. GR is a purely classical theory, and the energy-momentum tensor can (to the best of my knowledge) not be promoted to any kind of meaningful quantum operator. IOW, general relativity can only handle classical systems. There are no special mechanisms needed - as previously mentioned, this tensor arises from Noether’s theorem, and thus it describes some of the symmetries of a system, but not directly its internal structure or evolution. So it’s not surprising that you can have different systems with the same EMT. Again, GR is a purely classical theory, it has nothing to say about quantum systems. Also, you have to remember that “mass” and “energy” are Newtonian concepts, which don’t always straightforwardly translate to GR, due its non-linear nature. So this question doesn’t seem very meaningful to me. Consider this: rest mass yields curvature. But so does angular momentum, electric charge, stress, strain, shear, and energy density in general (EM fields etc). But the effects of these don’t linearly superimpose, which is to say you can’t take a spacetime geometry and say “this curvature is because of mass, that curvature because of angular momentum,…”. Also remember that some Newtonian concepts don’t have a gravitational effect at all, for example kinetic energy - you can’t turn something into a black hole simply by making it go very fast with respect to yourself, so one must be very careful when talking about energy as a source of gravity. And then you have non-static and non-stationary gravity. In GR, you can have gravitational radiation that propagates, you can have isolated wave fronts, gravitational solitons, all kinds of things. So you see, there’s a lot of subtleties to consider. Been around this whole time, but as a silent reader - we’ve recently had a lot of threads on politics, which I have very little to say about. I try and stick with GR

The Einstein tensor describes a certain aspect of overall spacetime curvature, in the sense that it determines the values of a certain combination of components of the Riemann tensor (it fixes 10 out of its 20 independent components). But unlike the Riemann tensor, it is not a complete description of the geometry of spacetime. Thus, the Einstein equations provide only a local constraint on the metric, but don’t determine it uniquely. Physically speaking, mass and energy have of course a gravitational effect, but neither appear as quantities in the source term of the field equations - there’s only the energy-momentum tensor field \(T_{\mu \nu}\). Here’s the thing with this - the Einstein equations are a purely local statement. So for example, if you wish to know the geometry of spacetime in vacuum outside some central body, the equation you are in fact solving is the vacuum equation \[G_{\mu \nu}=0\] which implies \[R_{\mu \nu}=0\] There is in the first instance no reference here to any source term, not even the energy-momentum tensor, because you are locally in a vacuum. During the process of solving these equations, you have to impose boundary conditions, one of which will be that sufficiently far from the central body the gravitational field asymptotically becomes Newtonian; it’s only through that boundary condition that mass makes an appearance at all. Only in the interior of your central body do you solve the full equations \[G_{\mu \nu}=\kappa T_{\mu \nu}\] wherein the energy-momentum tensor describes the overall distribution of energy density, momentum density, stresses, strains, and shear (note that “mass” is again not part of this). By solving this equation along with boundary conditions you can find the metric. You can work this backwards - you can start with a metric, and calculate the energy-momentum tensor. But here’s the thing: if the tensor comes out as zero, this does not mean that there’s no mass or energy somewhere around, it means only that the metric you started with describes a vacuum spacetime. If it’s not zero, you’re also out of luck, because the energy-momentum tensor alone is not a unique description of a classical system. Knowing its components tells you nothing about what physical form this system actually takes - two physically different systems in terms of internal structure, time evolution etc can in fact have the same energy-momentum tensor. That’s because this tensor is the conserved Noether current associated with time-translation invariance, so what it reflects are a system’s symmetries, but not necessarily or uniquely its physical structure. There’s no unique 1-to-1 correspondence between this tensor and a particular configuration of matter and energy, since all it contains are density distributions. IOW, a given matter-energy configuration will have a unique energy-momentum tensor associated with it, but the reverse is not true - any given energy-momentum tensor can correspond to more than one possible matter-energy configuration. Thus it is not useful to try and define matter-energy by starting with spacetime geometry. Yes, they are of course closely related, but the relationship is not just a trivial equality; there’s many subtleties to consider. A zero cosmological constant does not imply a static universe, only that expansion happens at a constant rate. Matter and antimatter have the same gravitational affects, they are not opposite in terms of curvature.

The source term in the field equations is neither mass (which doesn’t appear at all) nor energy (which isn’t a covariant quantity), but the energy-momentum tensor. This tensor describes the local densities of energy, momentum, stresses and strains, but does not give a unique breakdown or distribution into matter and energy. Yes, you can run the field equations backwards and start with a given metric - but that does not give you a unique distribution of matter and energy, just as having a particular energy-momentum tensor does not give a unique metric. In both cases you also need initial and boundary conditions that you must impose. Physically that just means that the same spacetime geometry can result from entirely different distributions of matter/energy. Also, as has been pointed out, curvature actually requires no local sources - for example, Schwarzschild spacetime is completely empty (T=0 everywhere), but still not Minkowski.

I understand that, which is why I was talking in terms of likelihoods and tendencies. Human beings are very complex systems, so it is difficult to make absolute statements. Indeed, that can happen. And the reverse too - genuine convictions sometimes go away, given appropriate circumstances. My original comment was simply that externally imposing moral codes by means of fear and threats of punishment is not the same as holding a genuine inner conviction. And this goes for better and worse (think extremism etc).

Yes, good question. I think the difference will show itself in how people act when they think no one is looking - in the case of externally imposed moral codes, there’s a much higher probability that those will be set aside if they interfere with other goals, and if they think no one will know. But if someone acts out of a genuine inner conviction, they aren’t likely to disregard those, irrespective of what circumstances they find themselves in. It reminds me a little of the situation in the former communist Eastern Bloc - the vast majority of ordinary people externally toed the party line, parroted the right slogans etc, which enabled them to live some semblance of a normal life. But behind closed doors it was a different story - very few actually genuinely believed the ideology, which is why the whole thing just crumbled in the end; there weren’t hundreds of thousands coming out to do their utmost to preserve their beloved systems. That’s the difference between imposing a narrative with fear, versus genuine conviction.