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New answers tagged differential-geometry

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If the metric is Riemannian (positive) your conjecture (a maximally symmetric spacetime is a constant curvature spacetimes) is a known theorem: Theorem 3.1 in Transformation Groups in Differential Geometry by S. Kobayashi. From the proof, it seems to me that the result should hold in the Lorenzian case too, but without a closer scrutiny I am not completely ...

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Suppose you're in a coordinate system where the Christoffels don't vanish at some point. To choose a coordinate system where the Christoffel symbols vanish at a given point $p$, you must apply a Christoffel symbol change of variables: $$0={\bar\Gamma}^k{}_{ij} = \frac{\partial x^p}{\partial y^i}\, \frac{\partial x^q}{\partial y^j}\, \Gamma^r{}_{pq}\, ... 0 The metric is telling you how to calculate the proper time along a path of your choosing. If you select a path where the time is everywhere constant then as you integrate along that path dt = 0 and any terms involving dt disappear. It is as simple as that. 0 The principles of space time homogeneity and isotropy are dependent of one another. The reason of dependency of one another refers to the region of space time of being homogenous. Homogeneity in space time results from being symmetric, and what causes space time to be symmetrical is simply the Laws of Nature. Hence our space time, or the shell we are living ... 3 Calculating the sum of the interior angles precisely woud be a big task as we'd need to compute the trajectory of the light ray and there isn't a convenient analytic expression for this. However we can easily calculate an upper limit for the interior angles. The key fact we need to know is that the deflection angle \theta of a light ray in the ... 4 Well, actually you are looking for a one-parameter group of diffeomorphisms (or isometries if referring to the boost vector field). This group is obtained by solving the differential equation$$\frac{dx}{ds}= X(x(s))\tag{1}$$with a generic initial condition z at s=0 in the manifold M (Minkowski spacetime in your example). X is your vector field on ... 0 Manifolds are defined such that locally they look like Euclidean space; this is why we call them smooth manifolds. A riemannian manifold is a manifold that locally has some inner product structure, ie a way of measuring length and angles. Lengths and angles are invariants, hence will have an invariant expression in terms of a local coordinate basis; and ... 0 In Riemannian geometry there is a beautiful theorem which states that a manifold with a symmetric connection is locally flat everywhere if and only if the curvature tensor vanishes. Therefore, in a locally flat coordinates such that \Gamma_{jk}^i=0, g_{ij} is constant throughout the chart and a linear transformation can be used to diagonalize the metric ... 0 If ds^2=\eta_{\alpha \beta}d \xi^{\alpha} d \xi^{\beta} were true for all points of space, we would have no curvature, hence no gravity! Take for example a sphere (the Earth), locally we can measure distances by ds^2=dx^2+dy^2, but this can't hold for two arbitrary points on the sphere. In fact, this coordinate system changes from point to point ... 0 Let \mathcal{M} be the space time manifold, whose local charts (open sets) are described by U_i. A local coordinate frame S_i is a map \xi\colon U_i\mapsto \mathbb{R}^N such that \xi(m) = (x_1,\ldots,x_N) \in \mathbb{R}^N, m\,\in U_i. Let, moreover, g be a (0,2) rank tensor (the metric). A change of coordinates is any smooth invertible map ... 2 What you are confusing here is speed and velocity. Light speed is constant, but the velocity, which takes into account the direction as well as the speed is not. As an example of how something can accelerate without changing speed, consider the case of circular motion, where the acceleration of an object moving at a speed v in a circle of radius r is ... 0 The disconnect is between the first and second clauses of your first sentence: Light speed is constant, therefore experiences no acceleration Yes, the speed of light is a constant, but it experiences no acceleration in its direction of travel. Light definitely accelerates laterally when gravity pulls on it, which is why it curves when passing near ... 0 g_{\mu \nu}(x) means that g is a function of location (x) --- so it varies across the manifold, which is the problem. I think that if g \ne g(x), then necessarily g = \eta ... Hopefully someone else can chime in on that. -1 Actually, the assumption of a psuedo riemannian manifold doesn't require many tacit assumptions. Can you measure time and distances? Can you define a right angle? Ok, you now have a manifold equipped with a metric. Want to include time as a dimension? Now you have four dimensions. You can't turn around in time like you can in space, so you need the time ... 1 Here is a purely geometrical way to think about this Edward says it is possible to cut a wedge out of a flat spacetime and glue the edges together. So in my mind this looks like a paper cone. A cone is flat precisely because it can be created by rolling up a flat sheet. Rolling up preserves the metric on the interior of the sheet (not on the boundaries ... 0 The most intuitive way to express this, as far as I know, is to start by taking the limit of the ratio of the circumference of a circle about the singularity to its radius, with the radius tending to zero. This ratio must be 2\pi for a manifold, and the conical deficit indicates that the space-time is singular at the centre of the circle. You may find more ... 0 but keep getting confused if I should sum over each term separately. It's not particularly clear what you mean here. Your expression uses the Einstein summation convention, which means that every repeated index is summed over. In principle this should be done for each term separately, so your expression reads$$ {R_{00}}= ...

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I posted the question a few hours ago, and realized the answer lies in the fact that ${(\gamma^m)^{\alpha}}_{\beta}$ has two kinds of indices. It is indeed true that so does ${(\gamma^\mu)^\alpha}_{\beta}$. But the fact is that the flat space gamma matrices are invariant tensors of the Lorentz group $SO(d-1,1)$. The fact that ...

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Comments to the question (v1): There are three types of indices at play: (i) spinor indices, (ii) flat (vector) indices, and (iii) curved (vector) indices. The gamma matrices with flat indices are constants. They don't transform under local Lorentz transformations (LLTs). They can be viewed as intertwiners between spinor indices and flat indices. (LLTs ...

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One less well-known but great reference are the classical field theory notes by Deligne and Freed in the '99 IAS lectures. Some good things about them Very elegant treatment written for mathematicians Begins with a nice discussion of ordinary classical mechanics using principal bundles and connections Useful comments on supersymmetric gauge theories ...

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The answer to my question is simpler than I suspected. It is fairly easy to describe the movements of both soldiers mathematically. The first soldier's spear is being transported along $X_1 = [y^2+z^2, -x^2, 0]$. The second soldier's spear along $X_2 = [y^2+z^2, 0, -x^2]$. Both are valid parallel transports relative to the tangent vector field $V=[0, -z, ... 0 What motivates the assumption that a closed timelike curve must cross a spacelike slice an odd number of times? Its not an assumption. And it isn't true of all manifolds. Consider$\mathbb S^2\times\mathbb R^2$as a subset of$\mathbb R^6,$or just $$\{(a,b,A,B,y,z)\in\mathbb R^6:a^2+b^2=A^2+B^2=1\}$$ with the metric ... 0 The solution can be found on this wiki page in section "Example: Null tetrad for Schwarzschild metric in Eddington-Finkestein coordinates". Your metric is exactly in the same form (to see that just calculate the inverse of the given$g^{\mu\nu}$). The value of$F$is slightly different and there is$c$in several places, but because the metric is exactly in ... 1 There's something wrong with your sign permutations in the Hodge star operator calculation. If$F = B + E \wedge dt$, then, in 2D,$F = B dx \wedge dy + E_x dx \wedge dt + E_y dy \wedge dt$, as you wrote yourself. Now, let us take our initial Hodge star as$\star dx \wedge dy = dt$. This means that$\star dt \wedge dx = dy$and$\star dy \wedge dt = dx\$, ...

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Your last expression is not valid, for two reasons: first, any given index can only occur twice per term in the Einstein convention, once as an upper index and once as a lower index. Remember that when an index is repeated, it means you sum over it with the metric: $$T^a T_a = \sum_{a,b} g_{ab} T^a T^b$$ You have terms like ...

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