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17

There seem to be several confusions here. Massive and massless particles behave qualitatively differently, even if the massive particle is traveling very fast. The minimum radius for a stable orbit for a massive particle is $3 r_s$. Circular orbits above this radius are all stable. Massless particles only have circular orbits at the photon sphere, $(3/2) ... 7 Thanks to the hint given by knzhou I figured out that if one wants to give the particle a proper initial velocity of$v_0$, the initial velocity in terms of Schwarzschild coordinates$v_i$would then be $$v_i = \frac{v_0}{ \color{green}{\sqrt{ 1-v_0^2/c^2}}\cdot \color{blue}{\sqrt{1-r_s/r_0}}}$$ for the transversal component, and the same without the blue ... 4 The Einstein field equations $$R_{\mu\nu}~-~\frac{1}{2}Rg_{\mu\nu}~=~8\pi GT_{\mu\nu}$$ for zero stress energy means that the Ricci Curvature$R_{\mu\nu}$is proportional to the metric with$R_{\mu\nu}~=~\frac{1}{2}Rg_{\mu\nu}$. This is called an Einstein spacetime, and for a constant Ricci scalar$R~=~R_{\mu\nu}g^{\mu\nu}$this is a spacetime of constant ... 4 You can easily see this isn't the case by considering the special case of the stress-energy tensor equal to zero i.e. the vacuum solutions. These include the Minkowski metric, which is flat, but also the Schwarzschild and Kerr metrics and of course gravitational waves. 4 As stated in the comment by Peter Diehr, the question is in principle no different whether you ask it for electromagnetic, gravitational or any other kind of wave. The wave's entropy is simply the conditional Shannon entropy of the specification needed to define the wave's full state given knowledge of its macroscopically measured variables. A theoretical ... 3 The radiation emitted by an accelerated charge depends on the boundary conditions on the fields at infinity. When one takes this into account properly, then accelerated observers will agree with inertial observers about the emitted radiation (after trivial transforms are applied). Any treatment which purports to show that in the accelerated observer's frame ... 3 Indeed,$f$is a symmetric form, since$\omega$and$\omega '$are Grassmann-even: $$(\text dx \wedge \text d y)\wedge (\text d z \wedge \text d t)=(\text d z \wedge \text d t)\wedge(\text dx \wedge \text d y)$$etc.. Now, to calculate the signature, you should find a basis which diagonalizes$\omega$, the dimension of the space is$6$. A basis is given ... 2 I will answer this with a simple example. Let us consider the metric for weak gravity, $$ds^2 = \left(1 - \frac{2GM}{rc^2}\right)c^2dt^2 - dr^2 -r^2d\Omega^2.$$ The$g_{tt}$metric element is largest by a factor of$c^2$and we have $$\Gamma^r_{tt} = \frac{1}{2}g^{rr}\partial_r g_{tt} = \frac{GM}{r^2}.$$ Now let us work with the geodesic equation that is ... 2 The answer is a definite yes and no. Gravitational waves have entropy in that we can think of them travelling from their source to our detector as a channel that sends units of information in the sense of the Shannon formula. The ringing of our detector is then the reception of that information. The Shannon formula$S=-k\sum_n p_n log(p_n)\$ would give a ...

2

Torsion is not frame dragging. Torsion is having an anti-symmetric spacetime connection. As you do parallel transport in general relativity (GR) you drag frames the frames roll as they move. With torsion they would twist. The connection is GR is the Christopher symbols, symmetric in the two bottom indices. The torsion is an anti-symmetric tensor. It will ...

2

First of all we need the equation: $$\require{Amsmath} 2\nabla_{[a} \nabla_{b]} K_{cd} = R_{abc}{}^e K_{ed} + R_{abd}{}^e K_{ce} = 2 R_{ab(c} K_{d)e}\tag{1}.\label{eq:KT}$$ We have: $$R_{d(ba}{}^e K_{c)e} = \frac{1}{3}(R_{db(a}{}^e K_{c)e} + R_{da(b}{}^e K_{c)e} + R_{dc(b}{}^e K_{a)e}) = \frac{1}{3} (\nabla_{[d} \nabla_{b]} K_{ac} ... 2 Yes the Schwarzschild metric describes the spacetime geometry around the Earth, and I describe how to use the geodesic equation to describe objects falling in Earth's gravity in How does "curved space" explain gravitational attraction?. An example of how the Schwarzschild metric describes the Earth's gravitational field is the time dilation of GPS ... 2 Assuming that Classical Electrodynamics (Maxwell's Equations) holds, the answer is that the inertial observer would see the radiation while the non-inertial observer would NOT. The question you are asking is basically the following paradox: https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field This paradox has been analyzed and ... 2 The true meaning of all the trouble is that coordinates are just tools for calculations and don't have any intrinsic meaning in GR. You should never interpret the coordinates alone. These can be very misleading and in particular become singular in some regions of spacetime (because coordinates are local). And never trust the labels of coordinates: Fact that ... 1 Let me answer this question systematically. 1)The first is how to think of causality and locality. Locality (sometimes people use locality to talk about microcausality but that's not very important) is the statement that two events cannot communicate with each other if they are separated by spacelike distances. So what does this mean? Suppose you have a ... 1 There is some misunderstanding here. whether a freely falling charge radiate photons, how strongly and relative to which frame of reference it does or does not radiate if you mean a charge in free fall. In this calculation:, from the conclusion It is found that the "naive" conclusion from the principle of equivalence - that a freely falling ... 1 The conformal transformation g'_{\mu\nu} = e^{-2\sigma}g_{\mu\nu}, sigma = sigma(x) leads to the transformation of the Ricci scalar$$ R' = e^{2\sigma}R - 12e^{2\sigma}(2\sigma_{,\mu}^{,\nu} - 2\sigma_{,\mu}\sigma^{,\mu} $$Since \phi' = e^{\sigma} then$$ \frac{1}{12}\phi'^2R' = \frac{1}{12}\phi^2 R - \phi^2(2\square\sigma - ...

1

Traversable - Overlapping (actually intersecting) region would not be Traversable even if the gravity at some parts of the region may be zero. For exampple, between earth and moon, gravity will be zero at some point. That does not mean something in that region can go out of earth/moon system. As soon as an observer leaves that region, it either falls towards ...

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I do not buy into the view that spacelike geodesics are not physical entities or, at best, can only be understood in tachyon terms. Mathematically spacelike geodesic paths are well understood in 3+1 spacetime. They can be computed in principle given a metric. Given 2 events in a local (but not infinitesimal) part of manifold they can be joined by a unique ...

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