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1 vote

Reducing Tensor-rank by fixing an argument

Yes you can but only once. In one chosen coordinate system and at on point in time. If after that you want to transform things to a new coordinate system, or if there is a transformation because a ...
Jos Bergervoet's user avatar
0 votes
Accepted

Conformal Transformation of Torsion

This perhaps isn't a very useful answer, but it simply depends on whether you define it to or not. Usually conformal transformations act on the metric alone, and so torsion is invariant. However, if ...
Eletie's user avatar
  • 3,213
0 votes

When we apply these concepts to physics, where do we put the UNITS in vector spaces and manifolds? Do units have a clear mathematical meaning?

It is good that you are already familiar with the concepts introduced by GR. Then I can just directly tell you that we are supposed to simply lay any coördinates (that work), and forget about trying ...
naturallyInconsistent's user avatar
5 votes
Accepted

Why does $\oint_C \vec{E}\cdot d\vec{\ell}=0$ imply $\nabla\times \vec{E}=\vec{0}$?

The first equation is only for all closed loops, not for all contours. That’s why you can’t conclude $\vec{E}=0$. The only thing you can conclude is that $\vec{E}=\text{grad}(V)$ for some function $V$....
peek-a-boo's user avatar
  • 6,100
2 votes

Why does $\oint_C \vec{E}\cdot d\vec{\ell}=0$ imply $\nabla\times \vec{E}=\vec{0}$?

For the very reason you say at the end of your question: while the second relation holds for all surfaces, the first one only holds for all closed loops, and not any arbitrary line. Let's do some ...
basics's user avatar
  • 8,231
0 votes

Comparing vectors on a sphere

Maybe you can try something like this: Reinterpret the two points $p_1$ and $p_2$ on the surface of the sphere as 3D points, which means they correspond to two unit vectors $ \vec{p}_1$ and $ \vec{p}...
Futurologist's user avatar
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0 votes
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Independent Components of the Riemann Curvature Tensor

Your symmetry properties do not make sense. The correct ones are $$ R_{\rho\sigma\mu\nu} = - R_{\sigma\rho\mu\nu} $$ $$ R^\rho{}_{\sigma\mu\nu} = - R^\rho{}_{\sigma\nu\mu} $$ $$ R^\rho{}_{\sigma\mu\...
Prahar's user avatar
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1 vote

How can a triangle have a sum exceeding 180 degrees in a curved space?

You can imagine putting mirrored square tiles on a disco ball. No matter how hard you try, you end up leaving awkward gaps between the tiles or having to cut parts off the tiles to completely cover ...
KDP's user avatar
  • 2,165
3 votes
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How can a triangle have a sum exceeding 180 degrees in a curved space?

Here are three diagrams to illustrate that the angle sum of a triangle can differ from $180^\circ$. The Wikipedia article Spherical Geometry might be of interest? It has a nice illustration relating ...
Farcher's user avatar
  • 95k
2 votes

How can a triangle have a sum exceeding 180 degrees in a curved space?

The angles are the same in 2d and 3d! Your example of the sphere is perfect. If you take the North pole and two points on the equator then a triangle with 3 x 90 degrees is formed. If, for this ...
Jos Bergervoet's user avatar
1 vote

How can a triangle have a sum exceeding 180 degrees in a curved space?

So in essence a straight line for us (in respect to a 3D space) is in fact a curved line on a 2D space, this essentially means that the lines are "stretched" over the curve of our spherical ...
user394147's user avatar
2 votes

Comparing vectors on a sphere

The two-sphere, i.e. $S^2$ is an example of a smooth manifold and the thing about smooth manifolds is that the usual notions of vectors are naturally lacking because of the decidedly non-linear ...
Albertus Magnus's user avatar
0 votes

No torsion with calculating the commutator of the covariant derivatives

I prefer the notation $\nabla_X$ to the apparently more explicit $X^\mu\nabla_\mu$ because in the presence of torsion there is a dangerous ambiguity lurking in the notation $\nabla_\mu$. To see ...
mike stone's user avatar
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0 votes

No torsion with calculating the commutator of the covariant derivatives

For arbitrary vectors $V,W$ and $X$, we have \begin{equation} \begin{split} \nabla_V \nabla_W X &= \nabla_{V^\lambda \partial_\lambda}(\nabla_{W^\tau \partial_\tau}X^k e_k) \\ &=\nabla_{V^\...
Sean's user avatar
  • 848
7 votes
Accepted

Clarification about Wald's notation on his General Relativity Book

Greek indices are used to label elements of an ordered set. For example an arbitrary element from a set of basis vectors $\{e_0, e_1, e_2, e_3\}$ would be labeled $e_\mu$ so that $\mu \in \{0,1,2,3\}$ ...
Er Jio's user avatar
  • 1,181
2 votes

Clarification about Wald's notation on his General Relativity Book

$(\mathrm{d}x^1)_a = (\mathrm{d}x)_a$, $(\mathrm{d}x^2)_a = (\mathrm{d}y)_a$, $(\mathrm{d}x^3)_a = (\mathrm{d}z)_a$. In summary, $\mu$ labels which coordinate he is talking about while $a$ is the ...
Níckolas Alves's user avatar
0 votes

On general covariance

Newton's law can be formulated as a covariant law, i.e., valid for all Cartesian coordinate transformations inertial and non-inertial, if the ordinary time derivative is replaced by the so-called ...
Peter Zipfel's user avatar
1 vote

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

The answer depends on how Dirac chooses to define curvature tensor. If he takes it by definition as that tensor which gives $$ R^a_{\lambda c d} v^\lambda = \nabla_c \nabla_d v^a - \nabla_d \nabla_c v^...
Andrew Steane's user avatar
0 votes

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

I will try an intuitive approach valid for 2D. If we have a perfect flat floor, we can fill it with square tiles. We start from a point in the bottom left and go to the bottom right, each tile ...
Claudio Saspinski's user avatar
1 vote

Isomorphism of the tangent space and the space of directional derivatives

In the following answer I will assume you are familiar with germs of functions and derivations at a point. I will provide an argument starting with the definition of tangent space of a manifolds in ...
Sir Crackpot's user avatar
  • 1,763
0 votes

How to find the double covariant derivative of a general vector?

I find it very hard to disentangle such expressions without writing out the tensors completely. See this lecture by Frederic Schuller to see how to take (and define) the covariant derivative. Using ...
Ben H's user avatar
  • 920
0 votes

When one discusses the "boundary" of Anti-de Sitter space, what do they mean precisely?

TL;DR: The conformal boundary of (Euclidean) $AdS_{d+1}$ is easiest to analyse in stereographic coordinates (3), cf. Ref. 1. The bulk of $AdS_{d+1}$ is isomorphic to an open ball $B_{d+1}=\{y\in\...
Qmechanic's user avatar
  • 200k
2 votes

Application of Stokes theorem in closed thermodynamic cycles

Although Stokes' theorem, neither in its original formulation for three-dimensional vector fields nor in its generalized form, applies directly to the cited statement, you are correct in seeing some ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
0 votes

Is this case a failure of Stokes' theorem?

You don't really have to be worried about the non-analyticity of the gauge potential on the Dirac string, when deducing Dirac's quantization: With your choice of potential ${\bf{A}} = -g\frac{(1+\cos\...
Mudit Jain's user avatar
4 votes

Are there reasonable models of Earth's surface as the space $\mathbb{R}P^2$?

Often, we treat the Earth's surface as being embedded in 3-dimensional space. Since the real projective plane cannot be embedded in 3-space, the Earth's surface is clearly not equivalent to it in any ...
Sandejo's user avatar
  • 5,468
1 vote

Finding killing vector

Quoting Sean Carroll's conclusion ("An Introduction to General Relativity - Spacetime and Geometry", Chapter 3, "Curvature", Section 3.8, "Symmetries and Killing vectors")...
Martin Nava-Callejas's user avatar
2 votes
Accepted

Is the Godel universe Wick rotatable?

Yes, it can be Wick-rotated to a negative-definite Riemannian metric on the (Berger sphere)xR. See the my more detailed answer at https://mathoverflow.net/questions/462704/is-the-g%c3%b6del-universe-...
Sigbjørn Hervik's user avatar
0 votes

When we apply these concepts to physics, where do we put the UNITS in vector spaces and manifolds? Do units have a clear mathematical meaning?

A similar question was asked on Maths Stack Exchange, and the top answer was that dimensional object live in a tensor product of 1-dimensional graded vector spaces. This builds on an essay by Terence ...
Anders Sandberg's user avatar
6 votes
Accepted

Metric tensor calculation tool?

Solving the Einstein field equation is notoriously hard. When Einstein originally calculated the GR contribution to Mercury's perihelion precession (the famous $\sim 43$ extra arc seconds per century)...
0 votes

When we apply these concepts to physics, where do we put the UNITS in vector spaces and manifolds? Do units have a clear mathematical meaning?

I think taking the formal tensor product of (say) $\mathbb{R}\otimes m$ for a length (meter) is good enough, in analogy how one can complexify vector spaces. Now use the natural convention of $(\...
Confuse-ray30's user avatar

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