Markus Hanke

No it wouldn’t. “Relativistic mass” isn’t a source of gravity; this outdated concept isn’t used any longer precisely because it lends itself to this type of misconception.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

Whatever about the rest, but I completely agree with this particular statement. Genuine morality and ethics cannot be imposed from the outside, they have to come from within, or else the facade simply crumbles once the going gets tough.

That’s both your prerogative and your loss.

Deepseek is a Large Language Model - given some initial query text, it uses an algorithm to compute the probability of each subsequent word as it appears in the “response”, based on a very large body of previously learned text material. And that’s all it is - a mathematical algorithm, albeit a computationally complex one. As such, it has no understanding, sentience, or consciousness; it doesn’t “know” what you are saying, or what it is responding. Thus, I wouldn’t attribute too much significance at all to this.

We don’t know yet what kind of initial conditions were present to give rise to the universe, since our current best models don’t extend as far back as the actual BB event itself. Note though that “before the universe” is pretty much a meaningless concept, much like saying “north of the North Pole”.

That’s of course your prerogative.

My point was about outcomes, not tools. The laws of the universe don’t depend on anyone’s god concept, or lack thereof. Take for example a simple table-top set-up, such as the Cavendish experiment. Anyone can do it, even at home in your own living room; and the outcome of the experiment will always be the exact same, regardless of the experimenter’s religious conviction - you always find the same Newtonian inverse square law, with the same gravitational constant. Concepts of a religious nature are simply irrelevant to the observed laws of nature - which is why you cannot draw conclusions about god (one way or the other) from the natural sciences.

But my point was that whether you are Christian, Muslim, Buddhist, Hindu, atheist, something else entirely, or nothing at all, makes no difference - it’s always the same universe, so experiments performed on it have the same outcome, meaning you always find the same laws of nature, irrespective of any god concepts.

Lol +1 Oh how I love being among nerds…

The universe that’s being studied by science is the same now as it was before the 20th century, so I don’t quite understand this comment.

Yes, you answered someone else’s thread with your own personal theory. This needed to be pointed out.

We don’t know the mass of the universe. And what about other sources of gravitation, like radiation? You haven’t accounted for that at all. We don’t know this either. We know only the size of the observable part of it. Right now the evidence is more consistent with the universe being spatially flat on cosmological scales. No. You can’t. Gravity isn’t a force at all, except as an approximation in the Newtonian limit. It says no such thing. Free fall is always along geodesics, irrespective of spacetime geometry, meaning a=0 at all times. No. You can’t. If the topic was locked, then you shouldn’t bring this up again. We’re done here.

Also, this here: is of course incorrect, irrespective of numbers. If you are surrounded by a spherical shell of matter, assuming an approximately uniform distribution, the net Newtonian gravitational force acting on you is exactly zero, irrespective of your state of motion, and irrespective of where within the shell you are located.

No. The most accurate accelerometers we have are around ~10E-7m/s^2. Please show in detail how you arrived at that number. There is no such thing as “inertial acceleration”. An inertial frame is defined to be one where there is zero proper acceleration; if you measure acceleration, no matter how small, then you’re not in an inertial frame. So even if the universe were somehow more gravitationally attractive in one particular direction (which it isn’t), you’d still be locally in free fall in that direction, and your accelerometer reads zero.

It’s an expansion factor. It tells you how the spatial distance between the same two arbitrarily chosen points change over time. Think of it as a placeholder for “we don’t know yet what happens there”. That’s because if you go far enough in time, both gravity and quantum physics become relevant simultaneously, and we don’t yet have a theory of quantum gravity. Hence we don’t know (yet) just how exactly the universe started off, we only know how it evolved after a certain very early point in time. Without DE the rate of expansion would be a constant (but the equations still work). However, the data we have suggests that this rate isn’t constant, but accelerating - so DE is a mechanism that actively pushes the universe apart. The exact physical nature of DE is as of yet unknown. This is not easy to answer in a non-technical way. Think of a tensor (as it is most commonly found in GR) as a function that takes as input vectors, and produces as output a new vector, a real number, or sometimes another tensor. That depends what kind of background knowledge you have…? Perhaps the Wiki article on General Relativity is a good starting point.

If you put an accelerometer in free fall (inertial motion), it reads exactly zero at all times. This is an experimental fact. Thus, due to F=ma with a=0, no force acts on freely falling test particles.

The one that the Christian far-right in the US advocates. Not in their eyes, unfortunately. They are fully convinced that their own interpretation of the faith is the only correct one. What values are those, in your opinion? I’m not a Christian, but I’ve read the Bible, and there’s a very wide range of differing views and values advocated in there. There’s some good things that I can somewhat stand behind, but there’s also some pretty horrific stuff that I wouldn’t want to go near with a barge pole. Some really disgusting stuff. So who gets to decide what “Christian values” actually are? Modern science doesn’t say anything about God (Christian or otherwise), since the concept is not amenable to the scientific method.

I wonder what the US would actually end up looking like if somehow their brand of Christian Nationalism was in fact fully implemented. Seems to me it wouldn’t be too far off a Republic of Gilead type situation. How could anyone actually want to live in that type of society? This is just beyond me. Maybe they simply don’t understand the ramifications of what it is they are advocating for. Even in Atwood’s novels, many of the main architects of the RoG are portrayed as secretly loathing what they had created. As they say, be careful what you wish for, you might just get it.