# Tag Info

## Hot answers tagged differential-geometry

23

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. In the coordinates appropriate to very distant "inertial" observers, the event horizon of a nonspinning uncharged black hole in equilibrium is spherical. If the hole is ...

21

In a question like this you need to ask what does the volume change relative to. So it's a little bit ambiguous. However, the answer to your question is "yes" in the following restricted sense. Imagine having a "swarm" of test objects, with mass so small that their effect on the spacetime around them is negligible. Assume that they are in freefall, i.e. ...

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. ...

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For Riemannian manifolds, I believe the best result currently known is that a manifold of dimension $n$ can be isometrically embedded in a euclidean space of dimension $2(2n+1)(3n+7)$. So, for example, a 3-dimensional spacelike slice of spacetime can be embedded in a flat euclidean space of at most 224 dimensions. Maybe in low-dimensional cases like this ...

9

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 $\... 8 Start by considering the ordinary Newtonian gravity. This tells us that the acceleration of a test mass due to our planet of mass$M$is: $$a = \frac{GM}{r^2}$$ The acceleration is the rate of change of velocity with time. A fast moving object spends less time near the planet than a slow moving object so its velocity changes less. That means fast moving ... 7 This is an afternote to WillO's answer which cites: Robert E. Greene, "Isometric Embeddings," Bull. AMS 1969 which addressed known bounds on the dimension required of flat Euclidean / Minkowsian space if it is to be an embedding for a solution of the Einstein field equations, which of course is a four-dimensional signatured manifold. It's worth noting ... 7 It is not true that there is a unique geodesic through every point. To understand it, imagine a point on a sphere (where geodesics are just great circles) or even on a plane (here geodesics are straight lines). Through that point you can draw infinitely many geodesics, e.g. infinitely many straight lines passing through this point. However, if you restrict ... 6 This theorem can be used to prove Archimede's Principle in a region with a non-uniform gravitational field. The weight of the displaced fluid is $$\vec W=\int_\Omega \rho \vec g(\vec r)~\mathrm d\Omega.$$ Let us consider a body fully immersed. Then the buoyancy force is given by $$\vec B=-\oint_\Gamma p(\vec r)~\mathrm d\vec \Gamma =-\int_\Omega\vec\nabla p~... 5 General relativity in four dimensions does not need to be embedded in a larger space of any sort. Curvature in general relativity is completely defined according to curvatures that are intrinsic induced by parallel translation of vectors. One does not need to have the spacetime in four dimensions embedded in some higher dimension spacetime. There is general ... 4 Doesn't Sean Carroll's book give recommendations? It has an extensive bibliography, and recommends Schutz among others. The Preface explains what pre-requisites are useful, and that "building a mathematical framework is the goal" of the early chapters (2 Curvature, 3 Manifolds). It contains 8 mathematical appendices. So you are unlikely to need any ... 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 ... 3 I can answer some of it, and in such a way that it has invariant general relativistic meaning. However, not a general answer. You do have to, and can, treat curvature and some measures of volume invariantly. There are two questions. 1)Does negative/positive curvatures have more volume, that some (in some sense) equivalent spacetime with no curvature? And 2)... 3 The term gauge transformation refers to two related notions in this context. Let P be a principal G-bundle over a manifold M, and let \cup_i U_i be a cover of M. A connection on P is specified by a collection of \mathfrak{g}=\mathrm{Lie}(G) valued 1-forms \{A_i\} defined in each patch \{U_i\}, together with G-valued functions g_{ij} : ... 3 If you simply want to verify that dx_adx_a=\frac{1}{2}\text{Tr}(dg^{\dagger}dg) in this specific case, you can do the following: Think dg as a matrix valued one form, i.e.in your parametrisation, dg=A_adx^a except now the coefficients takes matrix values instead of real values. Then a direct computation can show$$dg=\begin{pmatrix} -\sin(\theta) e^{... 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 ... 2 Often$X$is a coadjoint orbit of a Lie group. These have a natural symplectic structure; see https://en.wikipedia.org/wiki/Symplectic_reduction 2 The two papers talk about very different things. In Kapustin's paper, he considered non-orientable space-time manifold to classify SPT (i.e. the partition function of the phase on these manifolds). To do that, one has to first Wick rotate to Euclidean space-time, where time-reversal becomes a mirror reflection, but with a sign change. In Watanabe et. al. ... 2 Within the Schwarzschild metric, the volume does change. It is the rectangle formed by the radial dimension and time which is invariant: The dilating effect of the Schwarzschild metric $$\mathrm ds^2 = -\left(1 - \frac{2GM}{c^2 r}\right) c^2 ~\mathrm dt^2 + \frac{1}{1 - \frac{2GM}{c^2 r} }~\mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2 \theta~\mathrm d\... 2 That \Delta g_{ij} = 0 as you define it is equivalent to saying that the gradient of all metric components have vanishing divergence$$ g_{ij;k}{}^k \equiv g^{k\ell}g_{ij;k\ell} = 0. $$Here it is important to remember that the indices i,j denote functions. To clarify this we will let g_k represent the gradient of an arbitrary component function g_{ij}... 2 I wish someone had recommended Paul Renteln's Manifolds, Tensors, and Forms. An Introduction for Mathematicians and Physicists. Unlike many mathematically inclined differential geometry textbooks, it works with an indefinite metric the whole way through. Chapters one and two aren't very necessary and primarily form a review of linear algebra. It also ... 2 I learnt from Schutz: Geometrical Methods of Mathematical Physics, combined Choquet-Bruhat, DeWitt-Morette (& I think one other): Analysis, Manifolds and Physics, and I found it a good combination. GMoMP is mildly rigorous, and covers most of the material you need to get a pretty good handle on GR. AMaP is a much more serious approach to a larger ... 2 For D=d+A,with respect to the usual inner product on \mathbb{R}^2 and the ones induced by it on differential forms, one has D^{*}_{A}=-*D_{A} * where * stands for the hodge star operator. For example,$$D^{*}_A (f_1dx_1+f_2dx_2)=-*D_{A} *(f_1dx_1+f_2dx_2)=-*D_{A} (f_1dx_2-f_2dx_1)=-*(\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}+... 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). 2 The answer is obvious: $$x^{i_{1}i_2,\cdots i_n}:= \dot{x}^{i_{1}}(0)...\dot{x}^{i_{n}}(0)$$ is completely symmetric, so that when computing a total contraction, $$x^{i_{1}i_2,\cdots i_n} T_{(i_{1}...i_{n})}= x^{i_{1}i_2,\cdots i_n}T_{i_{1}...i_{n}}\:,$$ for every covariant tensor$T$of order$n$, nomatter the symmetry properties of its indexes. 1 The components of$\text{Ric}$transform during coordinate change$x^\mu\mapsto \tilde{x}^\mu$as$\tilde{R}_{\mu\nu}=\frac{\partial x^\sigma}{\partial \tilde{x}^\mu}\frac{\partial x^\rho}{\partial \tilde{x}^\nu}R_{\sigma\rho}$. This is just the usual transformation rule for coordinate-components of tensors. Contracting over the two indices gives $$\tilde{... 1 Komar mass is only well defined, i.e. Invariant, in a stationary spacetime, i.e., one admitting a timelike Killing vector. Your derivation seems to be good only for a static spacetime, i.e., no rotations, so a Kerr metric for instance would not be admitted. Not only that, you seemed to assume a diagonalized metric also in the space coordinates, not sure if ... 1 To start with, a manifold is not always able to be embed in higher dimension, especially when singularity (black hole) involves. I would more agree if it is described by a 3-d gravity-free field theory. This is similar to the idea named AdS/CFT duality. Of course here is not AdS space, but the spirit is similar, I think. But I'm not an expert in this, so.... 1 As @Erik_Jorgenfelt says, a timelike four velocity will have \vec{u}\cdot\vec{u}=-1. Remember that \vec{u} is proper velocity,$$ \vec{u} = \frac{\mathrm{d}\vec{x}}{\mathrm{d}\tau},$\$ not coordinate velocity. It's perfectly okay to have a component of proper velocity be greater than one in geometrized units as long as the vector remains timelike. To ...

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