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It is easiest to directly derive the form of the vector fields from the boundary conditions for asymptotically flat spacetimes. See for example this paper http://arxiv.org/abs/1001.1541 or this one http://arxiv.org/abs/1106.0213 Infinitesimally diffeomorphisms act on the metric via lie derivative $\mathcal{L}_{\xi}g_{\mu \nu}$. In the case of BMS, you ...


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Matt Visser's How to Wick rotate generic curved spacetime is a great reference on this subject, which basically summarizes a lot of folklore on the subject. Addendum (Summary of Paper). This turns out to be an important problem in quantum gravity and QFT in curved spacetime for the obvious reason ("How do we know the usual tricks still work in curved ...


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For $1.$, you have, by applying the Leibnitz rule for covariant derivatives : $\nabla_{d}(\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}) \\= \nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} \delta_{c'}^{c}) \\= \nabla_{d}(\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l} g^{ca}g_{ac'}) \\=(\nabla_{d}\mathcal{A}^{a_1...c'...a_k}_{b_1...c...b_l}) ...


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A manifold should be thought of as a topological space that locally looks like a $\mathbb R^n$. This is made precise by saying that every point has a neighborhood that is homeomorphic to an open subset of $\mathbb R^n$. If you don't require anything else (except maybe Hausdorffness and the existence of a dense countable subset to make everything easier ...


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This is the way the different "pieces" (i.e subsets/neighborhoods/atlases $\lbrace{O}_{\alpha}\rbrace$) can be "stitched" together to provide a smooth/uniform cover of the whole manifold. Also this gives a way to measure/compare a function or vector on one subset (neighborhood/atlas) with a copy of it on a neighboring atlas. A necessary condition to ...


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This axiom is a "smoothness" property of manifolds. You could define a manifold without this property simply as a space which is locally Euclidean. However, then something which looks smooth in one chart may be highly non-smooth in another chart. For example a smooth curve in one chart may be discontinuous in another chart (since the representation in the ...


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Since an $(k,\ell)$ tensor can be viewed as linear combinations of pertinent tensor products of $(1,0)$ tensors (=vectors) and $(0,1)$ tensors (=covectors), and by linearity and Leibniz rule for the covariant derivative, it is enough to consider the latter. The condition (1) in manifestly covariant notation then reduces to $$ ...


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I'm not thinking in dimensions in this answer, so perhaps there will be a flaw: Force as gradient On any Riemnannian manifold $\mathcal{M}$ (think of $\mathbb{R}^n$ with the dot product), the gradient of a scalar field $U : \mathcal{M} \to \mathbb{R}$ is the vector field constructed by the dual of its exterior derivative $\mathrm{d}f$, so it is a vector ...


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First, the definition above is a right definition to include things like (the surface of) a smooth torus – and any similar ... manifold ... with an arbitrary topology, but exclude objects that wouldn't be locally equivalent to smooth space. If you changed something important in the definition, you would get a difference concept that doesn't agree with our ...


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Richens & Berry [Physica 1D, 495-512 (1981)] give beautiful examples of such systems (with phase-space that is a surface of genus > 1), which they call pseudointegrable; their examples are invariant manifolds of billiards shaped as polygons with rational angles. These systems are interesting because there are two constants of motion, so invariant ...


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@benrg I got it. Follow your first half suggestion. Pick $s\in I^+(r)$, so $r\in I^-(s)$. $I^-(s)\cap I^+(p)\ne\emptyset$. Choose any point $q\in I^-(s)\cap I^+(p)$. There is a trip from $q$ to $s$, and at the same time, there is trip from $p$ to $q$, so glue the 2 trips to get a 3rd one from $p$ to $s$. Therefore, $s\in I^+(p)$. $s$ is an arbitrary point in ...


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An equivalent way of defining the 3-covariant derivative $D$ is to say it projects the 4-covariant derivative $\nabla$ onto the hypersurface. This is worked out explicitly in an appendix to Carroll's Spacetime and Geometry, where he writes $$ D_\mu V^\nu = P^\alpha{}_\mu P^\nu{}_\beta \nabla_\alpha V^\beta $$ (D.48), given the projection tensor $$ P_{\mu\nu} ...


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First of all, the first equation for $ds^2$ is only valid if $f$ is nothing else than the azimuthal angle $\phi$. Second, if you are evaluating $X_i X^i$, the squared distance from the origin without any infinitesimals, then it is exactly equal to $-t^2+r^2$ and nothing else. The polar coordinate $r$ is chosen as $\sqrt{x^2+y^2}$ so its square already ...


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Usually people call configuration space, $\mathcal{M}$, to the space of all posible coordinates needed to determine your system (although,I would include the velocities too). In non-relativistic theories The coordinates are three... and we need to provide three coordinates per (point) particle, i.e., for $n$ particles one needs $3n$-coordinates describing ...


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OP is essentially asking about terminology. As usual, be prepared that different authors call different notions differently. Well, here is a suggestion: Call the configuration space before (after) the constraints are implemented for the extended (physical) configuration space, respectively. More generally, if an author is talking about a configuration ...


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I think that your description that the points of the configuration manifold are possible states of the system is as close to a precise definition as one will find. So for $n$ particles in three dimensions, the configuration manifold is just $(\mathbb{R}^3)^n$. As for how this relates to constraints, consider the simplest example: two particles attached with ...


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We interpret OP's question (v3) as essentially asking Can a Lorentzian manifold (with Minkowski signature) by coordinate transformations be redefined as a Riemannian manifold (with Euclidean signature)? The answer is No since the metric signature of a pseudo-Riemannian manifold is invariant under general coordinate transformations. This follows e.g. ...


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Ok, here's my own (pedantic) argument that the answer is "No." Three background steps: Take $(M,g_{ab})$ and $(\tilde{M},\tilde{g}_{ab})$ to be two pseudo-Riemannian manifolds related by an isometry $\varphi:M\rightarrow\tilde{M}$. I will consistently use a tilde ($\tilde{}$) to refer to objects on the second manifold and no tilde to refer to objects on ...


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No, you also need an initial condition for the field, and a boundary condition for the field. A plane wave solution to the Maxwell equations has the same 4-current as vacuum, after all.


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In these particular cases, the authors are interested in the conformal structure, i.e. lightcone structure, of the manifold. A conformal structure can be defined by an equivalence class of metrics, all of which are related to each other by a conformal transformation, $$g_{ab}\sim e^{\omega(x)}\bar{g}_{ab}$$ A nice way to characterize a conformal structure ...


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Consider any given spacetime $M$ and a given timelike geodesic $\gamma$ therein, exiting from an event $p\in M$, and parametrized by means of its proper time $\tau$ with origin fixed at $p$ itself. Next consider the coordinate system constructed as follows (it is possible to prove that it is well defined, see e.g., O'Neill's textbook). Fix a pseudo ...


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I think Jerry's point is that if you have a metric that looks like: $$ ds^2 = a(dx^0)^2 + f(dx^1, dx^2, dx^3) $$ where $a$ is a constant and $f$ is any function, then an observer moving in the $x^0$ direction follows a geodesic of the type you describe. An obvious example is the FLRW metric where a static observer follows the geodesic $t = \tau$, $x = y = ...


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Consider an $n$-manifold $M$. A curve is simply a continuous map $\gamma : \mathbb{R} \to M$. For simplicity, suppose $M$ is covered by a single coordinate chart (diffeomorphism) $\varphi : M \to \mathbb{R}^n$. Putting these together, we have \begin{eqnarray} \mathbb{R} & \stackrel{\gamma}{\longrightarrow} & M & ...


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Let me first refer you to three references pedagogically treating Instantons in quantum mechanics: 1)Riccardo Rattazzi's lecture notes treating instantons in nonsupersymmetric quantum mechanics. In these notes the anharmonic oscillator model is elaborated with great detail 2) Philip Argyres lecture notes treating instantons in supersymmetric quantum ...


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Just like $(r,0)$ are "multilinear functionals" assigning a group of $r$ co-vectors i.e. $r$ $(0,1)$ tensors a scalar, and you seem to accept this visualization, the $(r,s)$ tensors are "multilinear functionals" assigning a group of $r$ co-vectors i.e. $r$ $(0,1)$ tensors and $s$ vectors i.e. $(1,0)$ tensors a scalar. This is just one way among many how to ...


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Check out V.I. Arnold's Mathematical Methods of Classical Mechanics. This book is pretty terse and can have hard to follow notation. However, it is rigorous and contains mathematical explanations and proof of a wide array of topics in mechanics. It is also filled with very interesting examples. He introduces the concepts needed from differential geometry; ...


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$$\nabla_{\mu}\left[U,V\right]_{\nu} = \nabla_{\mu}\left(g_{\nu\lambda}\left[U,V\right]^{\lambda}\right)$$ You have two terms inside the parentheses, and you have to apply the derivative to both of them. Myself, I'd just remember that: $$\left[U,V\right]^{a} = U^{b}\nabla_{b}V^{a} - V^{b}\nabla_{b}U^{a}$$, so, we have: $$\begin{align} ...


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What we want of a law of nature is that is has the same form for every equivalent observer. Therefore, these laws should be construct with geometrical objects which transform into themselves up to multiplicative factors. This is also known as an homogeneous transformation under certain group (typically Lorentz or diffeomorphism). The geometrical object ...


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Let me try to explain a vox populi secret. If you have a two sphere, $S^2$, and define a vector on a point $p$ (you can imagine a vector as an arrow), the vector does not lie on the sphere!. In fact the set of all possible vectors at that point generate (or live) in a flat and tangent space to the sphere at $p$, denoted as $T_p S^2$. First note: On $T_p ...


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Every single comment and answer was very useful. I think i have found an answer in an old paper by Clark Glymour (Minnesota studies in philosophy of science, volume III, pp.50-60): "It has recently been noted (Ellis, 1971; Dautcourt, 1971; Ellis and Sciama, 1972; Glymour, 1972; Trautman, 1965) that in some general relativistic cosmologies various global ...


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As has been discussed in many questions around here (e.g. here), relativity tells us only about local properties and behavior of a space-time. There are some exceptions when we make global assumptions - if we have a space of globally and strictly constant positive curvature, non-trivial topology is imminent because the space has to be the 3-sphere ...


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I'm not going to provide a full answer here, because I don't know the answer, but I want to give some statements that illustrate quite nicely the kind of problems one would face when determining topology of anything: We know spacetime is a manifold. That means, locally, it looks just like $\mathbb{R}^4$. That's already a bummer. We can't do jack at one ...



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