New answers tagged

1

The quote you give from Carroll about the covariant derivative is right: it quantifies the rate of change of a tensor field relative to parallel transport. The covariant derivative of a tensor at a point doesn't make sense. However, the commutator of covariant derivatives acting on a point does. The situation is analogous to the vector field commutator. ...


1

Consider a region in the manifold where a tensor field is everywhere well defined. Consider a point $x$ and a neighbouring point $x+dx$. The tensor field, say $V^\mu$ is given at both points as $V^\mu(x)$ and $V^\mu(x+dx)$. We can parallel transport the tensor(vector in this case) from $x$ to $x+dx$. This parallel transported vector $V_P^\mu(x+dx)$ is in ...


0

The attractivity effect depends on the time during the two masses are at proximity. You can experiment it in sending a ball through a water fall or a water flow at different speeds and look the direction the ball takes after crossing the water. The representation of space time curvature by a deformated plane around a planet is a farce. Try to represent a ...


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

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 ...


1

This depends on how you do the analysis -- If you do it in the Kruskal spacetime, which is the more physical way, It's hard to talk about what "$r = 0$" is, and the question becomes somewhat ill-posed. If you naïvely do it in Schwarzschild coordinates, you'll find that $R$ has terms involving constants multiplied $\nabla^{2}\frac{1}{r}$, which an analysis ...


1

If we want to be formal, at $r=0$ there is a manifold singularity. So no tensor/scalar quantities are well defined there (i.e. our mathematical formalism does not work there). On the other hand, as physicists, we can look for different examples that behave in a similar manner. Also the (classical) electromagnetic field generated by a point-like charged ...


-1

When you solve the Einstein equations to find the black hole the energy-momentum tensor is vanishing everywhere, even at $r = 0$. At no moment you are imposing that it is a point source $T_{\mu\nu} \sim \delta(r)$, this would change the solution. The interpretation of the solution as a black hole with a central singularity comes by inspecting the properties ...


1

Suppose there is a flash event that we can represent as a light cone as the flash expands. There are three shutters with detectors around this flash event. The shutters open and close only once and this is almost instantaneous. One shutter opens in spacetime outside the light cone. One Shutter opens in spacetime inside the lightcone and the third opens ...


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 ...


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{...


2

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 ...


0

Let us consider four dimensions with coordinates $(t,x,y,z)$. If the metric tensor is diagonal, then its inverse is also diagonal. Let the inverse be $g^{\mu\nu}=\{A,B,C,D\}$, where I have listed only the diagonal elements. Now, we know that $R_{\mu \nu \theta \phi}$ is antisymmetric in the first two indices. That is $R_{\mu \nu \theta \phi}=-R_{ \nu \mu \...


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

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

In classical mechanics a system is described by a Lagrangian $\mathscr{L}\colon TQ\to \mathbb{R}$, with $Q$ being the configuration space and $TQ$ its tangent bundle, namely the union over $q\in Q$ of all tangent spaces $T_qQ$: $TQ = \cup_q T_qQ$. A local chart on $Q$ looks like $(q_1, \ldots, q_n)$, the $q_k$ being the degrees of freedom of the system. The ...


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 ...


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~...


15

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 ...


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 ...


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....


0

I think the best way to think of it is as follows.(It's not too different from what everyone has said, but may be put into better perspective). Choosing a frame of reference is a completely different job than setting up of coordinates. To observe an event in spacetime you must belong to some frame of reference(or equivalently, you create a frame of ...


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}+...


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 ...


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} : ...


1

The requirement for the velocity to be timelike, i.e lower than the speed of light, is that $u^iu_i < 0$ ($u^iu_i = -1$ is desired in your case since the transformation should retain normalization, which the values given in the comments matches to the 13th decimal), regardless of the numerical value of independent components. To extract the observed 3-...


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 ...


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

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

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}...


1

It's just like a directional derivative in standard Cartesian coordinates. In $\mathbb{R}^3$, if I have a function $f : \mathbb{R}^3 \rightarrow \mathbb{R}$, the gradient $\nabla$ gives me a map $\nabla f : \mathbb{R}^3 \rightarrow \mathbb{R}^3.$ For any vector $\vec{v} \in \mathbb{R}^3$, I can define a map $\vec{v} \cdot \nabla f : \mathbb{R}^3 \rightarrow ...


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 $\...


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. ...


0

OK, I can not show the math, but anything you try to fit inside the box, will itself be curved/stretched/contracted along with the space. Therefore, you should not be able to fit more stuff inside one box as compared to the other of the same size. Even saying same size involves space and its curving. So, even your box will be curved as there is nothing that ...


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\...


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)...


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. ...


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


1

Certain condensed matter systems show emergent behavior that is similar to general relativity: see this for example. Also, in fluid mechanics, sound waves can become trapped behind an "event horizon" called an acoustic black hole. Finally, the Einstein field equations are essentially the only possible classical equations of motion for a massless spin-two ...


1

General relativity is a theory of gravity; as such, it makes predictions about gravity. However, general relativity does make predictions about time and physical entities such as black holes. Some of the predictions general relativity did make: Gravitational waves exist (proved by LIGO last year) Black holes exist Light bends (proved in 1919 by an ...


1

Recall that the (global) conformal group is given by $${\rm Conf}(p,q)~\cong~O(p\!+\!1,q\!+\!1)/\{\pm {\bf 1} \},\tag{1} $$ cf. e.g. this Phys.SE post. Using the embedding $\imath: \mathbb{R}^{p,q}\hookrightarrow \overline{\mathbb{R}^{p,q}}$ into the conformal compactifification $\overline{\mathbb{R}^{p,q}}$, one may show after a short calculation that the ...


2

Comments to the post (v2): Ref. 1 is considering the $d$-dimensional real Euclidean space $(\mathbb{R}^d,|\cdot|^2)$ with the standard norm $$|x|^2~:=~\sum_{\mu=1}^d (x^{\mu})^2~=~\sum_{\mu,\nu=1}^d x^{\mu}\eta_{\mu\nu}x^{\nu}, \qquad \eta_{\mu\nu} ~=~{\rm diag}(1,\ldots, 1),\tag{A}$$ and inner product $$\langle x ,y\rangle~:=~\sum_{\mu,\nu=1}^d x^{\mu}\...



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