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Gravitoelectromagnetism: How far does the analogy go?

Yes, Gravitoelectromagnetism rank 2 tensor transforms is like the electromagnetic tensor $F_{\mu\nu}$ I'm not aware of any relation. The tensors you listed depend on 2nd derivatives of the ...
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Zero mass Kerr metric

As mentioned in @Umaxo's comment, according to Boyer-Lindquist coordinates - Line element: The coordinate transformation from Boyer–Lindquist coordinates $r,\theta,\phi$ to Cartesian coordinates $x,y,...
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Electromagnetic tensor in a FRW metric

OP misunderstood the equation $(31)$ of the first paper. Component $E_i$ and $B_i$ are meant to be the components of electric and magnetic fields that would be measured by a comoving observer at that ...
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3 votes

Zero mass Kerr metric

The transformation is given in page 15 of this paper: The Kerr spacetime: A brief introduction BTW there is a 2017 paper that claims that the mass zero Kerr metric is not actually equivalent to ...
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Does Mach's principle imply that the gravitational field has a non-zero curl?

Mach's principle and Newton's bucket Mach's principle was a philosophically motivating factor for general relativity, but it is not a necessary postulate of GR. As the first sentence of the Wikipedia ...
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Relativistic Euler-Lagrange equation

Assuming that we are talking about a massive point particle, we know that the arclength $$s~=~c\tau\tag{A}$$ is the speed of light $c$ times the proper time $\tau$ (up to an additive constant), and ...
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Does Mach's principle imply that the gravitational field has a non-zero curl?

You're talking about three things here: (1) Mach's principle, (2) frame dragging, and (3) the gravitational field. 2 is an effect in general relativity. 3 is something that exists in Newtonian gravity ...
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Does Mach's principle imply that the gravitational field has a non-zero curl?

Indeed, a massive rotating mass, will induced a n angular motion to freely falling objects. A ballerina falling freely towards the rotating mass will get a rotational motion when stretching her arm in ...
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Black hole metric of reflected shell of incoming light

The incoming light has a large mass, and therefore exerts a large impulse on the mirror when it's reflected. The impulse integrated over the whole mirror is zero, but force can't be transmitted ...
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1 vote

Geodetic Precession of a Gyroscope in Hartle's GR Book

After some thoughts, I may have come up with an answer, which is now posted here for confirmation and future reference. Given a tetrad $\{\hat{e}_\alpha\}$, the spatial angle between 4-vectors $u,v$ ...
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Why does a non-gaussian distribution imply noise in LIGO?

A statistician to whom I posed a question almost identical to this one gave me a very wise answer, as follows: It is natural to expect that a process subject to small, random perturbations will ...
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Time-dependent Christoffel symbols

Yes there are. It is well known that the Schwarschild metric can be generalized to fulfill the field equations with dark energy, called the Schwarzschild-de Sitter metric: $$\mathrm d s^2 =\left(1-\...
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1 vote
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Cosmic expansion from De-Sitter global coordinates metric

$\mathrm{d}\Omega_3^2$ is a $3$-sphere, which is homogeneous and isotropic. Hence, the metric you're exhibiting is a FLRW solution with $a(\tau) = R \cosh\left(\frac{\tau}{R}\right)$, hence it does ...
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Rotation of astrophysical black holes

The decomposition of angular momentum into orbital and instrinsic depends on the point about which we compute the angular momentum. The Earth has orbital angular momentum about the sun and "...
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Rotation of astrophysical black holes

Albert_RD asked: "Where is the orbital angular momentum in the astrophysical black hole created after the collapse?" If an orbiting body plunges into a black hole its orbital angular ...
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Confusions about Schwarzchild Geodesic Deviation

Before answering I need to point out a mistake in the question: for the Schwarzschild-Droste black hole, the number of one-metre-separated shells between the horizon and any given location is finite ...
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Why would GR's prediction for galaxy rotation curves differ from Newtonian gravity's in spite of prevailing intuition from dimensional analysis?

I read Copperstock & Tieu (2006) and will discuss it a bit. Before getting started, we should remember that galactic rotation curves are not the only evidence for dark matter. Gravitational ...
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What are the properties of metric tensor?

The graviton only has two degrees of freedom, which are essentially its two helicity states (like the photon). Your first step is correct; 6 components are not independent due to the metric being ...
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Throwing and catching a clock in general relativity

I'm trying to create a function that shows the time on a clock B that's been thrown up from and caught in position A, with respect to the mass of the object it's on (the function would be TB(m)). ...
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Is accretion disc caused by frame dragging?

Accretion disk is not caused by frame dragging. Accretion disk is formed due to the gravitational attraction of matter surrounding a black hole. In general, black hole accretion is a transonic process....
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Why would GR's prediction for galaxy rotation curves differ from Newtonian gravity's in spite of prevailing intuition from dimensional analysis?

The Cooperstock-Tieu paper doesn't really justify this in any detail. They show that the metric is proportional to $G^{1/2}$, where $G$ is the gravitational constant, so they say that this makes it a ...
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2 votes

Can we use entangled particles to find out what is going on inside a black hole?

This is a version on steroids of the famous Einstein–Podolsky–Rosen Paradox. In General Relativity, to get out of a black hole requires you to move faster than light, so your question can essentially ...
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-3 votes

Why would GR's prediction for galaxy rotation curves differ from Newtonian gravity's in spite of prevailing intuition from dimensional analysis?

You state: GR itself might reduce the need to invoke the existence of dark matter to the extent that is often done . . . have gone beyond the dimensional analysis argument and actually pursued GR's ...
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3 votes

Vector $V^\mu$ vs the Lorentz contravariant components $V^a(x)$

This is just a choice of basis of vector fields in a smooth manifold really. Let $(M,g)$ be a $D$-dimensional smooth manifold with Lorentzian metric. Let $U\subset M$ be one open set and let $x:U\to \...
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Vector $V^\mu$ vs the Lorentz contravariant components $V^a(x)$

The $e^{a}_{\mu}$ objects are called vierbein field (or tetrads, or frame fields) and connect a general frame of coordinates (in which vectors are expressed as $V^{\mu}$) to a local frame of reference....
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Is it possible to have a spacetime described by a piecewise metric?

You're definitely free to consider whatever you like. Here, you considered a discontinuous $(0,2)$ tensor field on the manifold $\Bbb{R}^2$, so there's probably nothing much you can say about how this ...
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Addition of gravitational fields in general relativity

According to Newtonian gravity, it doesn't account for the strong and weak equivalence principle, where you cannot just aggregate all the sum of their gravitational fields and derive a solution, it ...
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-1 votes

What is the average recessional velocity of an object in the universe?

If I understand your assumption that all of the matter objects are similar, and all on a large scale have similar densities, then what determines one's velocity from us is determined by the equation: $...
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Strong gravitational force induced by static electromagnetic fields?

This is a good illustration of what happens when one forget that Correlation does not imply causation. Yes, the author of the cited paper correctly describes a family of solutions of Einstein field ...
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6 votes

Lie derivative acting on a function

For a smooth function $f:M\to\Bbb{R}$, we have $\phi_t^*f:=f\circ \phi_t$. So, by the definition of Lie derivative, pullback and by the chain rule, \begin{align} (\mathcal{L}_Vf)(p)&:=\frac{d}{dt}\...
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2 votes

Do massless particles have gravity?

It is not strictly true that the curvature of space-time is due to massive objects. The curvature of spacetime depends on the stress-energy tensor, which includes both mass-energy and momentum. The ...
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Do massless particles have gravity?

And the curvature of space-time is due to massive objects. This is not exactly correct. The source of space-time curvature is the stress-energy tensor. The stress-energy tensor includes energy ...
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4 votes

Do massless particles have gravity?

Any energy or momentum-containing particle generates curvature in space-time. This is due to the fact that Einstein's field equation depends on the energy-momentum tensor $T_{\mu \nu}$ as $$ G_{\mu \...
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6 votes
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What is the relative acceleration composition law in General relativity?

Precisely this question has been asked and answered in the following paper: Bini, D., Carini, P., & Jantzen, R. T. (1995). Relative observer kinematics in general relativity. Classical and ...
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6 votes
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Throwing and catching a clock in general relativity

I think it is a bit messier than you would like. You cannot ignore the "Special Relativity part" because the time dilation caused by your launch speed will be of a similar order of magnitude ...
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Why does the graviton polarization satisfy $\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k},\lambda') = 2 \delta_{\lambda\lambda'}$?

The 2 is conventional, but it makes sense when you think of the simplest form these tensors take when $\hat{k} \propto \hat{z}$: $$\epsilon_{ij}^+ =\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 &...
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Are all maximally symmetric spacetimes constant curvature spacetimes?

This is an old question but I'll expand Valter Moretti's answer. Yes, maximally symmetric pseudo-Riemannian spaces (in the sense that they have the maximum number of Killing vector fields) are ...
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What is the relative acceleration composition law in General relativity?

What are you trying to achieve with this? (Four-) Acceleration is essentially the curvature of the world-line of an object. If your question is about four-acceleration, then you are asking how to add ...
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When connections are fundamental objects, is LQG the only possible theory?

What he's probably referring to is called the LOST theorem: There's only one cyclic representation of the holonomy-flux algebra that has a diffeomorphism-invariant cyclic vector (the Ashtekar-...
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3 votes

"Lorentz gauge" or "Lorenz gauge"?

"Lorentz gauge" is a very popular error. arXiv:0803.0047 [physics.hist-ph] : The Lorenz condition/relation and gauge are named in honour of the Danish theoretical physicist Ludwig Valentin ...
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How does General Relativity explain escape velocities?

Escape velocity is the rate at which space-time is falling into the surface of the earth. The fabric of space-time is literally being absorbed by the earths mass. According to General relativity ...
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1 vote

Addition of gravitational fields in general relativity

GR is a non-linear theory, and that is enough to answer the OP's question. Regarding how a test particle possessing mass affects the gravitational field in GR, search the literature on "Back ...
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-1 votes

Time is the only dimension that has an arrow, and the only dimension which contributes an opposite sign to the metric. Is that just a coincidence?

The minkowskian representation of space-time is 'not complete', the euclidean metric also has a role to play: either an observer who approaches an event orthogonally, i.e. $s^{2}=c^{2}t^{2}+x^{2}$ , (...
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Gravitational Redshift around a Schwarzschild Black Hole

Let me take a crack at your subquestion. Earlier in the text, Carroll states that an observer with velocity $U^{\mu}$ measures the energy of a particle along a geodesic to be $E = -p_{\mu}U^{\mu}$. We ...
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Cross product of basis vectors in general relativity

The parallelogram spanned by two vectors - and more generally the generalized volume spanned by $n$ vectors - is expressed by the wedge product of these vectors. The cross product in three dimensions ...
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Time is the only dimension that has an arrow, and the only dimension which contributes an opposite sign to the metric. Is that just a coincidence?

Whenever you are solving a problem in Mechanics which involves finding time of an interval and if you end up getting time in negative, it doesn't mean that that interval never existed. Yes, the answer ...
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2 votes

Action of gradient on tangent vector gives total derivative?

Recall that a parameterized curve through spacetime is specified by the coordinates, $x=(x^0,x^1,x^2,x^3)$, as a function of a parameter, for example, $x^{\mu}(\lambda)$. A tangent vector to this ...
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The Same Origin Requirement of Reference Frames

two inertial reference frames must have the same origin for the rules of special relativity to apply... No, that's not true. Special relativity still applies, but the arithmetic gets a little messier ...
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The Same Origin Requirement of Reference Frames

Do not transform locations. Transform distances between locations. You and Ben will agree about the distance from the oak tree to the Eiffel Tower. Likewise, transform not times, but differences ...
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3 votes

Why does the graviton polarization satisfy $\epsilon_{ij}(\mathbf{k},\lambda)\epsilon^{ij}(\mathbf{k},\lambda') = 2 \delta_{\lambda\lambda'}$?

It is a definition, there is nothing deep going on. We do this very often. If you have some object $a$, you can choose to expand it in some basis $u$ such that $a=\sum_\alpha a_\alpha u_\alpha$. It is ...
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