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-1 votes
1 answer
53 views

If an area in 2D cannot be curved and finite is the same regarding the space of our pressumed 3D universe?

Is the sentence in the title right that our universe is infinite? And if so does it mean that stars are not evenly distributed along our universe but they all move from a populated centre to a fairly ...
Krešimir Bradvica's user avatar
1 vote
1 answer
57 views

Does the Weyl tensor amount to tidal effects of gravity?

The Ricci tensor, for the spacetime surrounding the Earth, is zero, so the spacetime around the Earth is Ricci-flat. The Riemann tensor though is not zero since spacetime certainly is curved. This ...
Il Guercio's user avatar
1 vote
2 answers
85 views

What is Dirac's reasoning when saying parallel displacement creates vector field with vanishing covariant derivative?

Section 12 of Dirac's book "General Theory of Relativity" is called "The condition for flat space", and he is proving that a space is flat if and only if the curvature tensor $R_{\...
Lewis Kirby's user avatar
4 votes
4 answers
826 views

Why do we call the Riemann curvature tensor the curvature of spacetime rather than the curvature tensor of its tangent bundle?

I was studying the mathematical description of gauge theories (in terms of bundle, connection, curvature,...) and something bothers me in the terminology when I compare it with general relativity. In ...
eomp's user avatar
  • 123
1 vote
0 answers
84 views

Definition of asymptotically flat spacetime

Following the definition in Wald's book on general relativity, in page 276 asymptotically flat spacetimes are defined using conformal isometry with conformal factor $Ω$. Then one of the requirements ...
ziv's user avatar
  • 1,734
2 votes
2 answers
189 views

Can $\mathbb{R}^4$ be globally equipped with a non-trivial non-singular Ricci-flat metric?

I'm self-studying general relativity. I just learned the Schwarzschild metric, which is defined on $\mathbb{R}\times (E^3-O)$. So I got a natural question: does there exist a nontrivial solution (...
Victor 's user avatar
  • 107
6 votes
3 answers
2k views

Is source of space-time curvature necessary?

Einstein field equations have vacuum solutions that (probably) assumes the source of curvature (either energy-momentum tensor or the cosmological constant term or both) is elsewhere. Like, in ...
Nayeem1's user avatar
  • 1,161
1 vote
1 answer
464 views

Einstein tensor in 2d [duplicate]

Is the Einstein tensor in 2D or 1+1D always zero? If so, why? I recently installed EinsteinPy and started playing wing different metrics - for the 2D cases the result turned out to be always zero.
Nayeem1's user avatar
  • 1,161
-1 votes
1 answer
106 views

When doing general relativity in practice, how do we choose the appropriate manifold describing the scenario?

The theory only deals with the local curvatures, not the global topology. Hence any manifold with an allowed metric is allowed. These can be infinitely many, especially for negative curvature space-...
Cathartic Encephalopathy's user avatar
0 votes
1 answer
71 views

Is curvature localised in General Relativity?

Is the curvature of spacetime in General Relativy localised?
Manuel's user avatar
  • 476
8 votes
2 answers
826 views

How does the covariant derivative satisfy the Leibniz rule?

In Carroll's "Spacetime and Geometry", he states on page 95 (section 3.2) that the covariant derivative, $\nabla$, is a map from $\left(k, l\right)$ tensor fields to $\left(k, l+1\right)$ ...
MBar2269's user avatar
  • 103
4 votes
2 answers
638 views

Characterising Minkowski spacetime as a flat manifold with some other property?

It is known that flat manifolds can be characterized as follows If a pseudo-Riemannian manifold $M$ of signature $(s,t)$ has zero Riemann curvature tensor everywhere on $M$, then the manifold is ...
Ishan Deo's user avatar
  • 1,588
2 votes
3 answers
221 views

What is Dirac's reasoning when showing the curvature vanishing implies we can choose rectilinear coordinates?

In section 12 of Dirac's book "General Theory of Relativity" he is justifying the name of the curvature tensor, which he has just defined as the difference between taking the covariant ...
Lewis Kirby's user avatar
0 votes
0 answers
144 views

Does this theorem holds out for spacetime?

The theorem: Let $F$ and $C$ be two finite geometric figures (those defined by two continuous functions in a given region $D$), where $F$ belongs to an $n$-dimensional Euclidean space and $C$ is the ...
Antoniou's user avatar
  • 495
1 vote
0 answers
39 views

When is the Weyl tensor applied on null vectors a null vector?

Let $C^{\rho}_{~\alpha \beta \gamma}$ be the Weyl tensor of a spacetime $(M,g)$, that is a solution to Einstein's equation. Let $X^\alpha, Y^\alpha, Z^\alpha$ be null vector fields, i.e. $X_\alpha X^\...
warpfel's user avatar
  • 384

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