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Linearity follows from: Translation invariance (where we mean translation in space and time): the image of a vector $\vec{AB}$ joining events $A$ and $B$ under a Lorentz transformation $\mathscr{L}:\mathbb{R}^{1+3}\to\mathbb{R}^{1+3}$ is unaffected by the addition of any offset added to both ends $A$ and $B$; Continuity: The Lorentz transformation $\... 1 A theorem known as Israel's theorem states that, given an asymptotically Minkowski spacetime, the only valid static spacetime solutions (with a number of regularity assumptions) of the Einstein Field Equations (EFEs) are Schwarzschild ones. The technical definition of "static" is excludes rotation, like the Kerr solution. A similar theorem, for "stationary" ... 2 No. When they merge their horizons will change shape, and eventually become the static or stationary shape of a BH horizon. Nothing inside either horizon while this is happening can escape. At all times the timelike curves stay inside, and the deformed horizons are where the lightlike curves end up. In each, and after they merge. The area of each horizon ... 0 Classically speaking, since total BH evaporation takes finite time, yet being at the event horizon stops all time (in your IRF), the black hole will evaporate before you can get to the event horizon. -2 It is not possible to know. The reason they are called holes is that they are effectively holes in space/time so concepts like length, size, distance and time no longer have meaning in the conventional sense. Equally, while they may be the most compact objects known, there is no evidence to suggest they must be uniform. If you are talking about the event ... 18 Yes, they are perfect spheres. But let's understand what is the sphere and why. The spherical surfaces are the horizons. They are surfaces in space that have perfect spherical geometry. Now, that only holds true for various conditions: 1) they are static black holes, and if they arose from matter/energy collapsing they are in their equilibrium states. ... 24 The shape of a black hole's event horizon depends on who is asking. Observers who are moving quickly towards a hole, for example, will see a different shape compared to those who are not. Per @benrg the event horizon of a static black hole is not observer dependent, similarly to how the shape of an expanding flash of light is not. In the coordinates ... 0 Conformal Transformation are those active coordinate transformations (diffeomorphism)$\sigma^a \longrightarrow \sigma'^a=\sigma'^a(\sigma)$that change the metric in the following form: $$g'_{ab}(\sigma')\equiv\frac{\partial\sigma^c}{\partial\sigma'^a}\frac{\partial\sigma^d}{\partial\sigma'^b}g_{cd}(\sigma)=\Omega(\sigma)g_{ab}(\sigma)$$ So a conformal ... 0 You are over complicating what appears to be your interest. You are asking in essence if there is an energy change in a particle as it moves in a gravitational field, i.e., a given spacetime metric. The answer is simple: except for singularities (i.e., what happens there, which is unknown, that's where any particle path ends in general relativity), a ... 2 I have just now finished an article, "Geometry of the 3-sphere", in which at the end of the paper I give a simple derivation of the Riemann curvature bivector for the unit 3-sphere, using (Clifford) geometric algebra. I also discuss the Lie group$SU(2)$and Lie algebra$SU(2)$on the unit 3-sphere, using the powerful, but still rather unknown geometric ... 3 Distance measurements in$n$dimensional flat space follows the same pattern for$n$equal 1,2,3, or higher values. I'm going to assume a straight line, change in position to simplify the math (that is we're measuring what a introductory book would call the "displacement"$s$rather than distance. But then distance is just an accumulation of many magnitudes ... 2 One meter is a unit defined in the "real world" around us – places we can actually visit. Or it is used for the lengths and dimensions of objects we can touch. It only makes sense to use the same "meter" for other worlds if we can actually get to those worlds. If two worlds are completely separated from each other, it makes no sense to apply the units of ... 2 Whatever unit you're using for distance in 1D is still good in any number of dimensions. Kilometers in manifold of dimension n is fine (assuming non-compactified dimensions). 3 from your profile you seem to be an amateur (gifted?, teenager?, still in school?) self-studying GR. Great! So drop the Old Man's Book (the Meaning of Relativity) and get yourself a modern intro to GR or - my suggestion - the wonderful MTW's Gravitation. I did the same when I was 16. It was published in 1973 and Kip Thorne himself told me two weeks ago that ... 2 The answer is simply that not every space-time has a corresponding effective potential in the sense that we have a coordinate$x$such that$\dot{x}=\sqrt{2(E-V_{eff})}$. But this is true even in Newtonian mechanics, consider a problem with a Lagrangian $$L = \frac{m}{2}(\dot{r}^2 + r^2 \dot{\varphi}^2) - V(\varphi)$$ Obviously,$p_r\equiv m \dot{r}$is ... 0 The problem is that the thickness of the torus is related to its external curvature, whereas a metric only gives information about its intrinsic curvature. Think of a sheet of paper: when it is flat, its metric is clearly zero. Now warp it into a cylinder. This in fact does not change its metric, indeed any transformation that does not crease the paper ... 0 Please asks just one question. For the first one: The thickness of a 2 torus does not enter in. It is simply a 2D manifold, it has length and diameter, and only the outside surface matters. At least classically. And by the way a 2 torus is a flat space. It is one of a number of possible compact flat spaces in 2D. See https://en.m.wikipedia.org/wiki/... 2 That$\Delta g_{ij} = 0$as you define it is equivalent to saying that the gradient of all metric components have vanishing divergence $$g_{ij;k}{}^k \equiv g^{k\ell}g_{ij;k\ell} = 0.$$ Here it is important to remember that the indices$i,j$denote functions. To clarify this we will let$g_k$represent the gradient of an arbitrary component function$g_{ij}...
Actually the result is even stronger: Given a timelike geodesic $\gamma$ and a point $p \in \gamma$, there is a neighborhood $U \ni p$ equipped with coordinates, $x^0,x^1,x^2,x^3$ such that in the portion of $\gamma$ included in $U$, exactly along $\gamma$, the derivatives of the metric vanish in the said coordinates. Equivalently the Christoffel symbols $\... 2 This definition doesn't depend on the metric signature convention. Note that in definition of$\gamma\$ the metric doesn't appear anywhere. It is defined purely in terms of "3-vectors" and "3-scalars" measured by particular observer. So it is impossible for metric to appear here explicitly.