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17

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. ...


5

You are presumably thinking of the FLRW metric for a universe with greater than critical density i.e. a closed universe. We normally use comoving coordinates to describe this, in which case the time coordinate is not curved and at every point along this time coordinate the three spatial coordinates have the topology of a 3-sphere. That is, if we draw a ...


3

The Poincaré group is the semi-direct product of the six-dimensional Lorentz group and the four-dimensional translations and hence ten-dimensional (or "has ten parameters" is less precise diction). Since in a global inertial coordinate system you have to have the Minkowski metric by definition, only those transformations (diffeomorphisms) which preserve the ...


2

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)...


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

Since you mention the following in one of your comments I'm less interested in Einsteins historical struggles and would love a more modern perspective on how to get to this insight. I hereby unashamedly ignore history, and offer instead a quick plausibility argument. Let's start with the equivalence principle which, loosely speaking, says that a (...


2

The partial derivatives $\partial_\lambda \delta^\mu_\nu$ are clearly zero because the components of the Kronecker delta are constant functions of spacetime coordinates (one or zero). One may always go to a locally Minkowski frame where the Christoffel symbols vanish and there, the covariant derivative is equal to the partial one and vanishes, too. Because ...


2

I think you are supposed to simply argue that since the metric is "constant" in the sense of Eq. (C.2.3) on $\Sigma_t$, the curvature should be a constant as well. However, here is a more sophisticated point of view. Our definition of homogeneity is that the isometry group $\mathrm{Iso}(\Sigma_t,h)$ is transitive, i.e. given $p,q\in\Sigma_t$ there exists $\...


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}\...


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\...


1

Yes, curved spacetime does change the volume of space. When space is curved by mass it is stretched more in some dimensions than others. Picture a balloon being stretched or squeezed--the volume changes.


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 ...


1

Note that $\delta^i{}_j=\delta^i{}_k\delta^k{}_j$. Then $$\nabla_l\delta^i{}_j=\nabla_l(\delta^i{}_k\delta^k{}_j)=C(k,m)\nabla_l(\delta^i{}_k\delta^m{}_j)=C(k,m)[\delta^i{}_k\nabla_l\delta^m{}_j+\delta^m{}_j\nabla_l\delta^i{}_k]=2\nabla_l\delta^i{}_j$$ where $C(k,m)$ is the contraction of the $k$ and $m$ indices. Thus $\nabla_l\delta^i{}_j=0$.


1

The curvature in general relativity is determined completely by the metric (since the metric determines the Levi-Civita connection). Since the metric is constant on the homogeneous space so too must the curvature be.


1

Consider some path $\gamma^{\mu}(\tau)$, and some vector $x^{\mu}$. Parallel transport is the condition when $\gamma^{a}\nabla_{a}x^{b} = 0$ Torsion is present if, for two paths $\gamma^{a}$ and $\delta^{a}$ that satisfy $\partial_{a}\gamma^{b} = \partial_{a}\delta^{b} = 0$, it is the case that parallel transport of $\gamma$ along $\delta$ produces a ...


1

To some extent Your answered the question already! Look at the basic postulates of cosmology, these are homogeneity and isotropy of space-time. Isotropy implies three Killing vectors (SO(3)) and homogeneity gives another three killing vectors (for translation in three spatial direction). Therefore altogether six Killing vectors. Remember we not considering ...


1

Since no answers have been forthcoming I will summarise what has been discussed in the comments. The usual analysis of geodesic motion around a spherical mass assumes that the spacetime geometry is described by the Schwarzschild metric, and that the test mass is too small to perturb this metric to any significant extent. In that case the motion can be ...


1

You can write the Kronecker delta tensor as a product of the metric tensor $$\nabla_a(\delta^a_b) = \nabla_a (g_{bc} g^{ac}) = \nabla_a (g_{bc} g^{ac}) = g_{bc} \nabla_a g^{ac} + g^{ac}\nabla_a g_{bc} $$ As you may recall, the covariant derivative of the metric tensor is $0$ in general relativity. Version without using the metricity of the connection : ...


1

To add to the many good points already made, I think I agree with #Steve that what you are concerned about is the conundrum that an empty space seems to reflect the presence of matter. The key point is that we are dealing with a field theory, and the solution only describes the region of space outside of the mass, which is modelled as a single point as which ...



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