Markus Hanke

Well yes, essentially you are understanding the problem - namely that the outcome of experiments performed in u-time depend on where (and also when) they are performed. So it becomes very difficult to relate predictions from the model to physical outcomes in the real world. In u-time, no two clocks can be dilated wrt one another, so you always have \(g_{\mu 0}=g_{0\nu}=\pm 1\). This creates another issue, consider the following: Suppose you have two u-clocks that start off together at the same place near some very massive, rotating object, like a pulsar. They are initially synchronised and at relative rest. Now these clocks travel along a closed trajectory around the pulsar’s equatorial plane, and come to rest again at the same place afterwards. Both clocks travel along the exact same spatial trajectory with the exact same speed profile, but in opposite directions. Because of the way you defined your u-time, at the end of the experiment both clocks read the exact same amount of total accumulated u-time. Unfortunately, in the real world this isn’t what happens. If you perform this experiment with ordinary t-clocks, their readings will differ when they come together again. So in order to relate predictions calculated from your model to the real world, you need to account not just for location and time, but also for the history of the physical system in question. The spacetime isn’t flat, unless you want to demand that your metrics must be isometric to the Minkowski metric. Is that what your doing? Furthermore I suggest we stick to established classical physics for now, and not introduce unnecessary complications and speculations. Im afraid this makes no sense. Like I have said several times now, it’s not the connection itself that changes, only the connection coefficients.

Yes, this is what I said in my post (I’m quoting my self): But my main point was rather that once a metric is established to be of type Levi-Civita, then all its characteristics are uniquely determined, so you can’t have two “different” connections that are both LC. What changes according to the metric are only the connection coefficients, not the connection itself. That’s an important difference. Yes, I get what you are trying to do. Unfortunately I don’t think you have grasped the concerns I have tried to level at this idea, perhaps because they got buried in technical arguments. So let me try a more practical approach. In the first instance, consider this simple scenario - let’s say you have a box that contains a quantity of muons (a bit contrived, I know, but bear with me). The box is locally in an inertial frame, and otherwise isolated from any external influences. There’s no spatial motion in the frame of the observer, the box just sits there and ages in time. I’d like to use your own earlier example of a metric here, where the 00-component is unity, and the notion of time is your own adapted “new time”, not SI seconds; I will be using the letter u for this, to distinguish it from ordinary time t. In this spacetime, the geometric length of the muons’ world lines between two events A and B then is \[s\prime =\int\limits_{B}^{A} ds=\int_{B}^{A} \sqrt{g_{\mu \nu}dx^{\mu}dx^{\nu}}=\int\limits_{B}^{A} \sqrt{g_{00}} du=\int\limits_{B}^{A} du=\bigtriangleup u+C\] so it is just simply the difference in u’s (we can choose C=0 for simplicity). Let’s say the two events are 1 second u-time (not t-time!) apart, and at u=A the box contains X muons. My question is: how many muons are left at u=B, ie after 1 second u-time? All I’m after is a percentage of the original number of particles X, so nothing to do with any units. I’m interested to see how you go about solving this - which, in ordinary physics, would be an almost trivially simple problem. Like so: \ [ Latex code \ ] just without the space between backslash and angle bracket.

But you were explicitly saying that, I quote, we should “forget the old world entirely”, with old world referring to real-world measurements with clocks and rulers. The choice of units has no impact at all on the physical outcome of experiments, or on the form of physical laws that are written in covariant notation. You can measure lengths and angles in meters and radians, or you can use fingers and degrees, but the apple will always fall when it is ripe, and it will do so radially downwards. No redefinition you do changes this physical process. So you might as well work with the simplest mathematical description of it, and safe yourself unnecessary complications. Of course you can - like a grandfather clock for example, which is influenced by local gravity conditions. Nothing wrong with that, but the question is how useful that is. No it isn’t. It’s a reflection of the fact that relative uniform motion has no influence on how electromagnetism works. Your laptop works in your living room just the same as it would in a rocket at 99% light speed wrt Earth. Of course you can always write down a model where this is not so, I just fail to see the point why you would do that, as you then aren’t describing what actually happens in the real world. Are you talking about free fall here? There is no proper acceleration in the rest frame of a free falling particle. If you change the standard definition of time in a manner that makes it explicitly dependent on location, then \(a\left( t \right) \neq \ddot{s} \left( t \right)\), and thus Newton’s laws are no longer valid in their usual form F=ma. All laws of physics will take on a different form in this case. As stated previously, you can of course do this if you really want, I’m questioning only the point and usefulness in that. For the connection to be of type Levi-Civita, it needs to preserve the metric and be torsion free, otherwise it can’t be said to be Levi-Civita (there are infinitely many possible types). But take careful note that you can have a connection (of any type) on a manifold without there being any metric, it’s just that you then can’t formally say that that connection is of type Levi-Civita. So it doesn’t make much sense to say the connection is explicitly defined via the metric, only that the connection is of that specific type in the presence of a metric. If there’s only one smooth manifold, then there’s only one LC connection, though you can have as many metrics as you want. But you can’t have two “different” connections that are both Levi-Civita, that makes no sense. If your formal manifolds are both Riemann, an LC connection must preserve any metric on either one of these manifolds. If it doesn’t do that, then it’s either not an LC connection, or one of the manifolds is not Riemann, or it’s not a valid metric in the first place (not every notion of inner product is automatically a valid metric, there are conditions here too).

Not it isn’t the same set of points, because in order to flatten it, you had to cut a piece out of the surface (as you correctly stated yourself), so you have lost information in the process. This depends on what exactly you mean by “geometry”. In the context of GR, this means the various tensor fields that describe the distribution of energy-momentum and the associated effects this has on the world lines of test particles. And these very much are invariant under diffeomorphisms, which is to say you can label the same physical events in different ways. If you change neither the physical meaning of coordinates nor the form of equations expressed with them, then you can’t have a different geometry, unless you describe a different physical situation. You can’t have it both ways. If that is what you want to do, you should take a look a teleparallel gravity. This model has no curvature (ie all geodesics remain parallel even globally), and all the information about gravity is contained in the form of torsion along world lines. Einstein himself investigated this in some detail. But then this model is entirely useless, because you cannot compare anything calculated from it against quantities physically measured with real-world instruments. You might as well be talking about invisible pink unicorns, for all the use it has in modelling the real world. Physics makes models that describe aspects of the physical world around us - as such we must be able to extract predictions from those models and compare them to real-world measurements. If you ask us to just forget about the real world, then I’m sorry to say we’re not interested, because it’s of no use to us when solving practical issues like eg calculating the orbit of a satellite around a gravitating body, and knowing how a clock on that satellite relates to a clock on the surface. In standard GR, this comes right out of the model, because what clocks physically read is always identical to the geometric (mathematical) length of the world line it traces out in spacetime, so the problem is rather straightforward, if not always simple. In what you propose that is not so, and you’re asking us to just ignore this…? Im sorry, but I’m still completely failing to see the actual point in all of this.

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.