swansont Posted March 6 Share Posted March 6 6 minutes ago, Genady said: But why Earth gets near there? Because it follows a geodesic according to the spacetime geometry. That wasn’t the whole point, though. The geometry tells the mass how to move, but does it cause the mass to exist? Link to comment Share on other sites More sharing options...

Genady Posted March 6 Share Posted March 6 2 minutes ago, swansont said: does it cause the mass to exist? I don't know what you mean. Does mass cause geometry to exist? Link to comment Share on other sites More sharing options...

swansont Posted March 6 Share Posted March 6 21 hours ago, Genady said: "Q: The information that gets lost when we go from the Riemann tensor to the Ricci tensor does not affect the energy-momentum tensor nor Einstein’s equations. What is the meaning of this lost information then? A: It means that for a given source configuration, there can be many solutions to Einstein’s equations. They all have the same right-hand side, namely Tμν . But they simply have different physical properties. For example, the simplest case is to ask: what if this energy-momentum stuff is zero? If it is zero, does it mean that there is no gravitation, no interesting geometry at all? No. It allows gravitational waves." Susskind, Cabannes. General Relativity: The Theoretical Minimum. “Allows” doesn’t mean these exist. How would one get a gravitational wave absent energy-momentum? (Maxwell’s equations allow EM waves, but classically you aren’t going to get one in a situation where you don’t have a charge somewhere) 1 hour ago, Genady said: I don't know what you mean. The scenario has two parts, the earth influencing geometry and the location. You only addressed the latter. 1 hour ago, Genady said: Does mass cause geometry to exist? Does mass cause a particular geometry to exist? It’s my understanding that it does. The geometry you have depends on whatever mass (as a first-order approximation) you have. Link to comment Share on other sites More sharing options...

Genady Posted March 6 Share Posted March 6 (edited) 2 hours ago, swansont said: How would one get a gravitational wave absent energy-momentum? Only perhaps if it's preexisting / primordial. 2 hours ago, swansont said: The scenario has two parts, the earth influencing geometry and the location. You only addressed the latter. Got it. 2 hours ago, swansont said: The geometry you have depends on whatever mass (as a first-order approximation) you have. Sure*. In case the mass we have is zero, the geometry is not necessarily flat. * Depends, not uniquely determined. Edited March 6 by Genady Link to comment Share on other sites More sharing options...

MigL Posted March 6 Share Posted March 6 Maybe if we reduce the situation to its simplest form without trying to find other 'connections' between the left and right parts of the EFEs ... Geodesics are basically a description of space-time curvature or geometry. A test mass falling towards mass M, will follow a geodesic which defines the local space-time geometry. A test mass falling towards mass M^{10} will follow a much different geodesic; one defining a different local geometry. So, I would guess mass ( and its equivalent energy ) has some hand in determining space-time geometry. For a more 'in-depth' explanation see here On 3/4/2024 at 1:03 AM, Markus Hanke said: How does a freely-falling test particle under the influence of gravity move? It follows a geodesic in spacetime, which is a particular solution to the geodesic equation. This equation is itself a particular form of the principle of extremal ageing, ie the tendency will be for the test particle to move such that a comoving clock will record an extremum of proper time between any given pair of events along the trajectory. When you actually perform this variational problem, of course all components of the metric are technically involved. However, when you are dealing with situations that are in some sense close to being Newtonian / not too relativistic, such as Earth for example, the tt-component of the metric will be much larger than the rest of the metric, by a factor of ~c^2. It will dominate the calculation - meaning time dilation plays a much larger role than tidal effects. In that sense, it is indeed almost exclusively time dilation that gives rise to our daily experience of “downward gravity” here on Earth. Of course, this would not be true in other situations, like near the EH of a solar mass BH, where tidal gravity plays a major role. This doesn’t mean that the source of gravity isn’t energy-momentum and curvature, it just means that under certain circumstances the tidal components don’t play a major role, leaving mostly just time dilation as the dominating effect. Link to comment Share on other sites More sharing options...

swansont Posted March 6 Share Posted March 6 1 hour ago, Genady said: Sure*. In case the mass we have is zero, the geometry is not necessarily flat. How? Usually there’s only one solution for a given set of boundary conditions. Link to comment Share on other sites More sharing options...

Genady Posted March 6 Share Posted March 6 1 minute ago, swansont said: How? Usually there’s only one solution for a given set of boundary conditions. It can be a non-flat solution for some sets. Link to comment Share on other sites More sharing options...

swansont Posted March 6 Share Posted March 6 4 minutes ago, Genady said: It can be a non-flat solution for some sets. And what are the boundary conditions of those sets? Link to comment Share on other sites More sharing options...

Genady Posted March 6 Share Posted March 6 (edited) 14 minutes ago, swansont said: And what are the boundary conditions of those sets? Any geometry with zero Ricci tensor and non-zero Riemann tensor. Edited March 6 by Genady Link to comment Share on other sites More sharing options...

swansont Posted March 7 Share Posted March 7 1 hour ago, Genady said: Any geometry with zero Ricci tensor and non-zero Riemann tensor. I give up. GR can apparently only be explained using GR Every time i look into GR it’s like pulling teeth; I come away convinced that nobody does a numerical calculation because none of the math is ever presented in a way where that’s possible. I’ve read that GR reduces to Newtonian gravity, but have never been able to find a worked example of that, because nothing is ever presented but the tensor mathematics. Along the line of my frustration, I recall a seminar (on Lie algebra, IIRC) in grad school where the prof was asked what the difference was between contravariant and covariant, and the response was something like “contravariant means the indices are along the top” which is true but doesn’t do anything to advance anyone’s understanding. Link to comment Share on other sites More sharing options...

Genady Posted March 7 Share Posted March 7 14 minutes ago, swansont said: I recall a seminar (on Lie algebra, IIRC) in grad school where the prof was asked what the difference was between contravariant and covariant, and the response was something like “contravariant means the indices are along the top” which is true but doesn’t do anything to advance anyone’s understanding. The following story from Zee, A. Einstein Gravity in a Nutshell (p. 52) is in the same line: Long ago, an undergrad who later became a distinguished condensed matter physicist came to me after a class on group theory and asked me, “What exactly is a tensor?” I told him that a tensor is something that transforms like a tensor. When I ran into him many years later, he regaled me with the following story. At his graduation, his father, perhaps still smarting from the hefty sum he had paid to the prestigious private university his son attended, asked him what was the most memorable piece of knowledge he acquired during his four years in college. He replied, “A tensor is something that transforms like a tensor.” 😉 17 minutes ago, swansont said: I’ve read that GR reduces to Newtonian gravity, but have never been able to find a worked example of that, because nothing is ever presented but the tensor mathematics. Here is one, from MTW's Gravitation. All variables are real numbers. The most important one, \(T_{00}\), is energy density: Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 7 Share Posted March 7 4 hours ago, swansont said: I’ve read that GR reduces to Newtonian gravity, but have never been able to find a worked example of that This short document might help, particularly chapter 3: https://www.sas.rochester.edu/pas/assets/pdf/undergraduate/first_order_approximations_in_general_relativity.pdf Link to comment Share on other sites More sharing options...

DanMP Posted March 7 Share Posted March 7 21 hours ago, Genady said: But why Earth gets near there? Because it follows a geodesic according to the spacetime geometry. So, I would "blame" the spacetime geometry again. In the case of the Earth approaching my spaceship location, yes, you can do that. But remember that you wrote: 22 hours ago, Genady said: the same changes in geometry can occur in other circumstances, e.g., different body or bodies, with different parameters / locations/ movements so we can replace the Earth in the scenario with a huge spaceship, arriving near my location (much closer than the Earth would do) by using its thrusters (not in "free fall"). Again: 22 hours ago, DanMP said: I would "blame" the approaching mass, not the geometry that is affected by it. You would stil "blame" the affected geometry or some other geometry? Why? And I repeat the question: On 3/6/2024 at 4:35 PM, DanMP said: how exactly the Earth mass does that? Link to comment Share on other sites More sharing options...

Genady Posted March 7 Share Posted March 7 (edited) 2 hours ago, DanMP said: You would stil "blame" the affected geometry or some other geometry? Why? Because Earth or other body you mention are far away and your ship cannot be affected directly by them without "an action at a distance." What it can be affected by directly is the geometry of spacetime at that same point where and when it is "here" and "now". 2 hours ago, DanMP said: And I repeat the question: I still don't know what you mean in that question. Does what? Edited March 7 by Genady Link to comment Share on other sites More sharing options...

KJW Posted March 7 Share Posted March 7 (edited) [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] For a stationary object in the Schwarzschild metric: [math]ds^2 = \left(1 - \dfrac{2GM}{c^2r}\right) c^2 (dt)^2 - \left(1 - \dfrac{2GM}{c^2r}\right)^{-1} (dr)^2 - r^2 \Bigl((d\theta)^2 + sin^2\theta (d\phi)^2\Bigr)[/math] [math]g_{tt} = \left(1 - \dfrac{2GM}{c^2r}\right) c^2[/math] [math]g_{rr} = -\left(1 - \dfrac{2GM}{c^2r}\right)^{-1}[/math] [math]a^r = -c^2 g^{rr} \dfrac{1}{\sqrt{g_{tt}}} \dfrac{d\sqrt{g_{tt}}}{dr}[/math] [math]a^r = -\dfrac{c^2}{2} \dfrac{1}{g_{rr}\ g_{tt}} \dfrac{dg_{tt}}{dr}[/math] [math]a^r = \dfrac{1}{2}\dfrac{dg_{tt}}{dr}[/math] [math]a^r = \dfrac{1}{2}\dfrac{d\left(c^2 - \dfrac{2GM}{r}\right)}{dr}[/math] [math]a^r = \dfrac{GM}{r^2}[/math] One might be surprised to see that the acceleration of a stationary object in the Schwarzschild metric is exactly Newtonian. But this is due to way the [math]r[/math]-coordinate is defined. The [math]r[/math]-coordinate is defined such that the surface area of a sphere of radius [math]r[/math] is [math]4 \pi r^2[/math]. Thus, for the acceleration field to be a conserved field, the surface integral of this field over the sphere needs to be constant with respect to the [math]r[/math]-coordinate, requiring that the acceleration field obey an [math]r^{-2}[/math] law exactly. Edited March 7 by KJW 1 Link to comment Share on other sites More sharing options...

Genady Posted March 7 Share Posted March 7 1 hour ago, KJW said: For a stationary object in the Schwarzschild metric What is \(T\) in \[a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}\] Link to comment Share on other sites More sharing options...

KJW Posted March 7 Share Posted March 7 (edited) 27 minutes ago, Genady said: What is T in [math]a^\mu = -c^2 g^{\mu\nu} \dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] [math]T[/math] is the relative magnitude of a timelike Killing vector at a given location in three-dimensional space. In terms of the metric, it is [math]\sqrt{g_{tt}}[/math], but Killing vectors are covariant, and the magnitude of a Killing vector is invariant once its magnitude at some location is defined (if [math]K^\mu[/math] is a Killing vector field, [math]\lambda K^\mu[/math] is also a Killing vector field for arbitrary constant scalar [math]\lambda[/math]). Edited March 7 by KJW Link to comment Share on other sites More sharing options...

swansont Posted March 7 Share Posted March 7 5 minutes ago, KJW said: T is the relative magnitude of a timelike Killing vector at a given location in three-dimensional space. In terms of the metric, it is gtt−−√ , but Killing vectors are covariant, and the magnitude of a Killing vector is invariant once its magnitude at some location is specified. So it depends on M and r. Which means one could say that mass causes gravity. Link to comment Share on other sites More sharing options...

Genady Posted March 7 Share Posted March 7 2 hours ago, KJW said: ar=−c221grr gttdgttdr Perhaps it's obvious but I don't see where \(g_{rr}\) in the denominator comes from, here: \[a^r = -\dfrac{c^2}{2} \dfrac{1}{g_{rr}\ g_{tt}} \dfrac{dg_{tt}}{dr}\] Link to comment Share on other sites More sharing options...

KJW Posted March 7 Share Posted March 7 (edited) 53 minutes ago, swansont said: So it depends on M and r. Which means one could say that mass causes gravity. Yes, [math]T[/math] and [math]\dfrac{1}{T} \dfrac{\partial T}{\partial x^\nu}[/math] depend on [math]M[/math] and [math]r[/math]. I would prefer to say that the curvature associated with mass causes gravity, although the mass is remote from the gravity, unlike gravitational time dilation which is local to the gravity. Whether or not mass causes the curvature associated with it, or is the curvature associated with it is something I can't answer. @Genady, [math]g^{rr} = \dfrac{1}{g_{rr}}[/math] Edited March 7 by KJW Link to comment Share on other sites More sharing options...

Genady Posted March 7 Share Posted March 7 4 minutes ago, KJW said: grr=1grr Yep, it was obvious indeed. Link to comment Share on other sites More sharing options...

KJW Posted March 7 Share Posted March 7 (edited) One thing I should mention: I distinguish between the curvature directly associated with the energy-momentum and the curvature that is away from its source energy-momentum, and tend not to use the term "gravity" to describe the curvature directly associated with energy-momentum. The curvature associated with gravity is called the Weyl conformal tensor. It has the same algebraic structure as the Riemann tensor, but its contraction is zero (—> zero Einstein tensor). In the [math]{C_{\alpha\beta\gamma}}^\delta[/math] form, it is invariant to conformal transformations: [math]g_{\mu\nu} \longrightarrow \varphi g_{\mu\nu}[/math] for arbitrary scalar function [math]\varphi[/math] Edited March 7 by KJW Link to comment Share on other sites More sharing options...

Genady Posted March 7 Share Posted March 7 30 minutes ago, KJW said: One thing I should mention: I distinguish between the curvature directly associated with the energy-momentum and the curvature that is away from its source energy-momentum, and tend not to use the term "gravity" to describe the curvature directly associated with energy-momentum. The curvature associated with gravity is called the Weyl conformal tensor. It has the same algebraic structure as the Riemann tensor, but its contraction is zero (—> zero Einstein tensor). In the Cαβγδ form, it is invariant to conformal transformations: gμν⟶φgμν for arbitrary scalar function φ What happens if the sources are everywhere? Link to comment Share on other sites More sharing options...

KJW Posted March 7 Share Posted March 7 In another thread, I showed that the flat-space Friedmann–Lemaître–Robertson–Walker (FLRW) metric can be coordinate-transformed to a manifestly conformally flat metric. This means that the Weyl conformal tensor is zero, and therefore according to my way of looking at this, there is no gravity (although there is still the curvature associated the Ricci tensor and scalar). Link to comment Share on other sites More sharing options...

Markus Hanke Posted March 8 Share Posted March 8 6 hours ago, KJW said: This means that the Weyl conformal tensor is zero, and therefore according to my way of looking at this, there is no gravity (although there is still the curvature associated the Ricci tensor and scalar). Another way to say this is that, in this kind of spacetime and under geodesic motion, shapes (ie angles) are preserved, whereas volumes and surfaces are not. Link to comment Share on other sites More sharing options...

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