# Tag Info

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You fall right across the event horizon without even knowing it is there unless you are paying attention. In classical General Relativity, spacetime at the event horizon is locally Minskowskian, just like it is everywhere else in the universe where there is no curvature singularity. Thinking that something locally-bizarre happens at the event horizon — like ...

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A basic way to put this is that the points of a fall can define the distance. If you fall through a cloud then you might fall for 3 minutes along a distance of 60,000 feet with the underside of the cloud being point A and the ground being point B. Point A remains where it is. Therefore there is a spatial distance. Once inside the Event Horizon everything is ...

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This is my self-answer based on looking around at papers and on the web. I've noted some issues that I'm uncertain about, and would be grateful if others could help with these. Thanks, safesphere and Javier, for helpful comments and for the clarification provided by this question on math.SE. The topology of the maximal extension of the Schwarzschild ...

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As can be seen from the Schwarzschild metric for an r-coordinate r < 2M the term (1-2M/r) becomes negativ. This means that r- and t-coordinate interchange which does not mean that space and time interchange. This has two consequences: The spacetime within the horizon is not stationary because r has the character of time, r decreases inevitably. The r-...

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It is in general impossible to find coordinates where multiple Killing fields are simple partial derivatives. I mean, if you have Killing fields $K_1,...,K_r$ with some relation $$[K_i,K_j]=C^k_{ij}K_k,$$ then unless $C^k_{ij}=0$, you cannot take all the $K_i$ to be partial derivatives, since those always commute. But. For the result OP is asking about, ...

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Is it possible to calculate if at the event horizon a hypotethical diffraction bending of a laser beam tangential to the black hole can counterpose a gravitational bending of the same beam?..... No, you can't evade the event horizon this way. The solution of an initial-value problem for a wave equation in GR depends only on the initial conditions that lie ...

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No. The right side of your equation is meaningless. You have to contract indices two at a time, not four at a time. Otherwise you don’t get a Lorentz-invariant result.

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The gravitational redshift near a black hole is $\sqrt{1-R_s/R}$ (that is, frequencies get downshifted by this factor). The innermost stable circular orbit is $3R_s$, so the redshift from there is $\sqrt{2/3}=0.81$. Not enormous. Would turn a 1 GHz signal into a 810 MHz signal, blue light into blue-green light. The redshift from expansion on the other hand ...

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This isn't terribly easy to check, but here are some various thoughts : First, let's assume an arbitrary metric (ie, we're not worrying about the matter terms). From our various theorems concerning topology change, assuming everything well-behaved (causally closed, time-orientable and causal), our spacetime doesn't change topology. Therefore, if we ...

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As a third, but equivalent, perspective to the existing answers: When in doubt, you can always put the summation back explicitly. (Just take it back out once you understand the answer and before getting evil looks from your peers for not having used Einstein summation!) In your case, you would have $$\sum_a \sum_b g^{ab} g_{ab} = ? = \sum_a \sum_a g^{aa} ... 2 As @G.Smith pointed out in his answer, g^{aa}g_{aa} doesn't make sense. You can calculate g^{ab}g_{ab} as a special case of raising and lowering indices. Contracting the b indices first you get$$g^{ab}g_{ab}=\delta^a_a=...$$I leave out the final step as an exercise for you. 2 This question is quite old, but let me give it a shot. In general, as you pointed out, all coordinate charts are "equally good", as long as the physical processes that you're interested in take place in the domain of your chart. If you want to calculate the orbit of body around a black hole, the precession or mercury, the gravitational redshift and so on, ... 2 Indeed, there is such a manifold conformally equivalent to the AdS. As explained for example here in chapter 2, we find \text{AdS}_2 to be equivalent to a strip, \text{AdS}_3 to a cylinder and in general \text{AdS}_{n+1} equivalent to \mathbb{R}\times D_n. Here, D_n denotes the n-dimensional open ball. However, the metric of this space is not the ... 2 Other answers already stated that in standard covariant formulation of general relativity there is only one metric. Nothing in MTW quotes suggests otherwise, only different notation for spacelike / timelike separation. At the same time I would like to mention that in the nonrelativistic, Newtonian limit, spacelike and timelike displacements become ... 2 If you do not assign a 3D spacelike submanifold where to compute the variations you say (and there are infinite inequivalent choices), the infimum is always 0. In fact, you can always continuously deform every given spacelike curve joining two spacelike related points to a light-like curve. This fact implies that the spatial distance between two spacelike ... 2 There are very different descriptions, some say time and space swap, some say they do not Space and time do not swap, what swaps is the sign of the g_{rr} and g^{tt} components of the metric tensor, but that is only the case in Droste coordinates, not in Raindrop or Finkelstein coordinates. some say spacetime bends extremely For the curvature you ... 1 Coordinate charts are always one-to-one from the manifold to some open subset to \mathbb{R}^n. This is because we use charts as local homeomorphisms to define what a manifold is (it can be defined in other, more abstract ways, e.g. locally ringed spaces, but let's not get into that). Let's remark another important point: in general relativity we only know ... 1 Yes, you can talk about there being two metrics, which are coincident for a proper relativistic spacetime - but whose distinction becomes important when considering the classical limit. This is similar to the case for quantum mechanics, where the classical limit requires us to bifurcate the wave function to separate positional and momental wave functions, so ... 1 g_{\mu\nu} is a matrix used to calculate "intervals". The sign of a given interval depends on what it is about: either it is a distance in space or it is a duration of particle motion. 1 This interpretation is wrong. There is only one metric. It is not positive-definite like a Euclidean metric is, so the squared spacetime interval can be positive, zero, or negative. 1 Yes. However, the last term is a bit 'icky' because it looks like$$ \frac{dx_\nu}{d\tau} = \frac{d}{d\tau} x_\nu $$while what is actually meant is$$ \frac{dx_\nu}{d\tau} = \left(\frac{dx}{d\tau}\right)_\nu $$It doesn't really make sense to 'lower' the index of a coordinate of the base manifold. However, while I wouldn't recommend it, I think I've seen ... 1 Can Bob be in the same reference frame as Alice while freefalling in gravitational fields of different strengths No. General relativity doesn't have global frames of reference. See How do frames of reference work in general relativity, and are they described by coordinate systems? 1 Sanity check: Am i correct that at this point Bob should observe Alice's watch ticking slower than his own as a result of the difference in their accelerating reference frames? Yes. What relative rate will Bob observe Alice's watch ticking at now that they are both in free-fall?$$ \Delta\tau_B = \frac{\sqrt{\frac1{r_s} - \frac1{r_{B0}}} - \sqrt{\frac1{...

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Well, the metric you started with is assuming that $k$, the spatial curvature is zero, $k=0$. So, even when you switch to global coordinates, the spatial curvature is bound to be zero. One way to see it is, the factor (scale factor) sitting in front of the spatial terms has changed, this scaling has affected all three spatial terms equally. Meaning $d\chi$ ...

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What it is increasing is the distances between the galaxies measured in the comoving rest space (the 3-space orthogonal to the world lines of the galaxies). This phenomenon at large scale is homogeneous and isotropic in this sense it is an universal expansion at large scale. It does not make sense instead to think of an universal expansion of everything at ...

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There is a difference though. With a bomb, there would be a shock wave. In the universe, we don't observe such a shock wave. With a bomb, the velocities of objects would diminish as one goes further away from the point where the bomb exploded due to the decrease in energy further away from the explosion. In the universe by contrast, we see that the ...

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Locally, there is no problem. The neighborhood $\Omega\subseteq\mathbb{R}^d$ of the chart is only isomorphic to (a subset of) the manifold. The neighborhood $\Omega$ imports the metric (and curvature) from the manifold via the isomorphism. The standard flat metric on $\mathbb{R}^d$ plays no role in the construction. Globally, there is a problem. The action ...

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According to the No-hair theorem (or, more generally, conjecture), linear momentum is conserved in a black hole along with angular momentum, mass-energy, and electric charge. Also, nothing in the theory of dark matter contradicts the conservation of momentum as a Noether current due to the translation symmetry of space (see the Noether theorem) plus the ...

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