Markus Hanke

What Genady means is that the universe is a curved spacetime, as described by General Relativity. The thing now is that some concepts we are used to from old Newtonian physics do not straightforwardly translate to GR - and “mass” is one of them. The question of “what is the mass associated with a given region of spacetime” has no simple answer; there are in fact several different notions of mass that apply to different sets of circumstances, so it really depends. The underlying reasons for this is that the gravitational field in GR is self-coupling and thus itself a source of gravity (unlike in Newtonian gravity); but this type of energy cannot be localised, and is frame-dependent, so it is difficult to account for in an observer-independent way.

This is just combining two sources of gravity to obtain a new spacetime geometry - gravity nonetheless remains attractive in nature, as it always does for ordinary sources. But for the Alcubierre drive you need actual anti-gravity, which is a completely different thing - it can be shown in a general way that the Alcubierre metric in its original form requires exotic matter; unfortunately you cannot “cheat your way there” just by cleverly arranging ordinary sources. Also, even if you could construct an Alcubierre bubble, I think it would be completely unusable as a propulsion method, since it has some pretty nasty side effects and problems. Just a word of warning - these diagrams depicting spacetime curvature are just visual aids to understanding the basic concept, they are not accurate depictions of actual geometry. Spacetime geometry is intrinsic to the manifold, so there is no actual “direction” to curvature, since spacetime is not embedded into any kind of higher-dimensional space. The crucial concept to understand here is the distinction extrinsic vs intrinsic geometry; this is very important if you want to understand GR.

I disagree. While I know that Icelandic distinguishes these sounds in phonology and orthography, English effectively doesn’t, so there’s no point in this at all. Furthermore the vast majority of people here on this forum presumably will use English keyboards, so these letters are not straightforwardly accessible to them, making this not at all efficient to most of us. There are good reasons why English orthography is largely standardised, so deviating from this is unwise, really comes across as silly, and makes it hard for some to read your text. Trust me, this isn’t making a good impression. My advice to you, if you wish to engage in a proper discussion of your ideas and be taken seriously while doing so, is to stick to standard English orthography. You don’t have to agree with it, you just need to use it.

I disagree, I don’t think you have a bad reputation here. You’re still young, so there are a lot of things about modern physics which you haven’t encountered and learned about yet. We all understand this, so there’s no problem. The most important thing is to keep learning, and keep your mind open - don’t allow yourself to fall into the “I’ve figured this all out” trap. Trust me, the universe is far richer than you can even imagine right now. Keep learning

No, not all quantities in nature are relative in the same way as eg speed is. A good counter-example is proper acceleration - it can be measured locally using an accelerometer, and all observers agree on this measurement. It is not relative to anything. Gravity is similar - if you have an ordinary source of gravity, its effect is always attractive for all observers, never repulsive, irrespective of orientations, states of motion etc. True anti-gravity would require exotic matter to generate, which we have reason to believe does not exist. To put it more technical - “gravity” is a tensorial quantity, and tensors are covariant objects; all observers agree on them.

I suspect what you’re looking for does not really exist, since, as others have pointed out, the Einstein tensor is actually quite a complicated function of the metric and its derivatives, so there’s no simple intuitive way to physically interpret any one of its components in isolation. But taken in its entirety, the tensor measures the extent by which geodesics will on average deviate around a small neighbourhood. It represents average Gaussian curvature in space once a time direction has been chosen. The Einstein equations mean just this - average local curvature (of a very particular type) is precisely equivalent to local energy-momentum, up to a proportionality constant. Note that this tensor only captures some aspects of total curvature, but it doesn’t provide a complete description (which is given by the Riemann tensor) - IOW, when you have G=0 then that does not necessarily mean you’re in a flat spacetime! In practice this tensor is seldom directly used in practical calculations - the Riemann tensor, Ricci tensor and Ricci scalar are more useful objects for practical purposes, and they have easier to understand physical meanings.

I’m not aware of any such concept, at least not under this name. Can you explain what you mean by this? I’m afraid that’s not how spacetime curvature works. So long as you start with positive energy-momentum - irrespective of how this is distributed or oriented in space -, you’ll always end up with ordinary attractive gravity. This is not a question of position or orientation in space.

I think this discussion on Physicsforums, particularly post #12, might shed some light, though I need to go through it more carefully myself.

It’s possible to consider discrete models of fluid dynamics, but generally speaking these don’t take the form of neat, closed formulas that can be easily worked on paper - rather, you’ll be dealing with numerical models on powerful computers. An example of this is CFD-DEM. The standard Navier-Stokes equations are then just the continuum limit of such a model.

Are you familiar with the topological arguments put forward in MTW with regards to why the EFE has the form it does? Not sure if one can call it a “derivation” exactly, but it’s nonetheless very interesting.

Yes, but then you’d end up with a different theory of gravity. In GR, the tensor we’re looking for can only contain first and second derivatives of the metric, must be linear in the latter, symmetric, and divergence-free. Lovelock’s theorem guarantees that the only tensor that fulfills these criteria is the Einstein tensor.

It’s best to look at the Einstein tensor as a kind of machine with two slots - if you feed into both slots a unit vector that points into the future time direction, it will give you as result a real number that represents the average Gaussian curvature in a small spatial region around the event you are working at: \[G_{\mu\nu}t^{\mu}t^{\nu}=k\] So the Einstein tensor is a kind of average curvature around some event. I don’t think it is helpful to try and find specific “meanings” for specific components.

Are you sure this is right? Admittedly I don’t know so much about the history of Christianity, but I’m pretty sure there weren’t any churches at the time these texts were written…

! Moderator Note Moved to Speculations, as this is completely at odds with mainstream astrophysics.

It’s called that because the energy-momentum tensor that corresponds to this metric (FLRW) is actually one describing a perfect fluid, with its constituents being the galaxies, clusters etc that make up the cosmos. So FLRW is basically an interior fluid solution on a very large scale. Indeed. This coordinate system is called Gaussian normal coordinates.

While this is of course true, I think it’s very important to remember that such a transformation changes the physical meaning of the time coordinate - it will no longer correspond to a clock co-moving with the cosmological fluid.

Lol…nice analogy 😉

Would it not correspond to a clock that is in motion relative to the cosmological fluid?

Yes, such a transformation would just be a diffeomorphism, which is of course always allowed. The disadvantage though is that it changes the physical meaning of the time coordinate - it then no longer corresponds to a clock co-moving with the cosmological fluid. I also suspect (not sure though) that this would introduce off-diagonal terms into the metric? I think everything considered, the usual Gaussian normal coordinates probably give the simplest and easiest to understand form of the metric.

Measurements of space and time at different places/times are related via the metric, which is the basic dynamic variable in GR. This is the entire point of the model, so you don’t need to introduce anything new. The difference to SR is that frames are no longer related via simple Lorentz transformations, so time dilation and changes in lengths do not necessarily carry the same factors, nor do they even necessarily occur together. No. Spacetime curvature behaves quite differently from elasticity in a medium; the image is just an analogy and visualisation aid.

I don’t know what ”proportional relativistic variable” is supposed to mean, but we know that measurements of distance are observer-dependent in GR, as several people here have already pointed out. So again - where is the “problem”? No, it’s not a contradiction, because in GR there’s a big difference between proper quantities and coordinate quantities. The two observers use different local notions of time, so you’re not comparing like for like. You compare spatial distances just like you do time separations, and you find that they are observer-dependent, as expected. Where is the “problem”?

In curved spacetimes, measurements of spatial distance, just like measurements of time, generally depend on the observer, because those are not tensorial quantities. Only spacetime intervals are. This is well understood, and the mathematics of differential geometry on Riemann manifolds predate GR. Exactly where is the problem with this?

I don’t actually see what the supposed “problem”, that the title of this threads alludes to, is meant to be. In curved spacetimes, observers will, in general, not agree on quantities that aren’t either tensors, or invariants formed from them. Therefore we do not, in general, expect observers to agree on specific measurements of space, time, energy, momentum etc - but they will always agree on the metric, and thus spacetime intervals and everything that derives from this. That’s just basic differential geometry. So could you just simply state what you think the “problem” is with this, preferably without getting lost in the intricacies of some specific scenario?

It doesn’t assume this, it’s how the maths work out. The FLRW metric is a solution to the Einstein equations - so you start by putting in your initial and boundary conditions, and solve the equations; the result is FLRW. The initial assumptions are homogeneity and isotropy, which taken together already constrain the final metric enough to give it its general form. To obtain the exact form, you begin with the standard energy-momentum tensor for a perfect fluid, feed it into the field equations, and solve. The result is that all coordinate dependencies drop out, except the time dependence of the spatial part. FLRW is a Petrov-type O spacetime; metric expansion (meaning: measurements of distance depend on when they are performed) is a general geometric property of many such spacetimes, and by no means exclusive to FLRW. Yes, Lambda-CDM doesn’t provide a perfect fit to all data, but it provides the best fit amongst all currently known cosmological models. But I agree that it probably won’t be the last word on the subject. There is of course always the possibility that new physics exist which we are not yet aware of. If so, our cosmological models will need to change. FLRW provides the best fit to the data, based on the physics (ie GR and fluid dynamics) which we currently know. Again, you can try and find a solution to the field equations that has a different form, yet still fits the available observational data. I would be interested in seeing it. Again, the FLRW metric is a solution to the field equations of GR, it has not just been invented “by hand” - so the best answer to your question is that the laws of gravity determine it to be so. So far as I can see (having worked through this entire solution process myself), every time you start with homogeneity and isotropy as basic symmetries, you’ll end up with a metric of the general form of FLRW, and the precise spatial coordinate functions will depend on the energy-momentum tensor you use in the field equations. For this to come out different, you’d have to amend the laws of gravity itself.

Local relative motion in a static background would be limited to subluminal speeds in accordance with the usual laws of kinematic, so for redshift we’d find z<1 always, whereas with metric expansion there is no such limit. Furthermore, if there is only local motion in an otherwise static space, then some of these objects will recede from one another, whereas others approach each other, like molecules in a gas. We’d see a mix of both blue- and red-shift, unless you want to postulate that we are the Center of the universe, and everything moves radially away from us for some reason, which is not very plausible. But with metric expansion, it’s the space in between that “expands” (I don’t like this term, but it has become standard), so on the largest scales everything will appear to recede from everything else, and it will do so the same way no matter what direction you look at, and irrespective what’s in between here and there. Also - if the rate of apparent recession isn’t constant (which is what seems to be the case), then, if you were to deal with local motion, you would have to have either some mechanism of acceleration, or some explanation as to why everything falls away from us. Overall you’d end up with a model that’s actually much more complicated and much less plausible than metric expansion.