7

In coordinates that have the dimension of length, the dimensions of the Riemann curvature tensor are inverse length squared. Therefore at each point the components of this tensor establish length scales which you can loosely think of as radii of curvature. “Local” refers to a region whose length scales are small compared with any of these curvature-based ...


7

Even if you only consider the gauge potentials up to gauge equivalence, you still can't recover a potential from a field in the non-abelian case. Not even locally! This is called the "field copy problem" in the literature, and goes back to Wu and Yang. There are some special cases where the potential can be recovered from the field, e.g. if the ...


5

$$\det\left[\frac{df_h}{dq_k}\right] \neq 0 \tag{1}$$ because we are dealing with a bijective and bi-differentiable transformation of coordinates (see below). Hence, if $$ \left[\frac{\partial L'}{\partial \mathbb q'} (\circ) - \frac{\mathrm d}{\mathrm dt} \frac{\partial L'}{\partial\mathbb{\dot q'}} (\circ)\right]\cdot\frac{\partial\mathbb f}{\partial\...


5

Instead of using those coordinates, you could identify the space around the string with a wedge of Minkowski space, with the usual Minkowski coordinates and metric, and the edges of the wedge identified. Geodesics that don't cross the edges are straight lines. If $α\ge π$ then you can always arrange for any particular geodesic to lie entirely in the wedge, ...


4

Too long for a comment: Note that the geodesic equations are exactly soluble in this case. Specifically, we can derive an analog of the Binet equation. Defining $u = 1/r$, we have $$ \frac{dr}{d\sigma} = \frac{d\phi}{d\sigma}\frac{dr}{d\phi} = \frac{\mathcal{J}}{r^2} \frac{d(1/u)}{d\phi} = - \frac{\mathcal{J}}{r^2u^2 } \frac{du}{d\phi} = - \mathcal{J} \...


4

You can't recover the gauge potential uniquely without specifying more information, because of gauge invariance. $A$ is not uniquely defined by $F$. For any gauge transformation $g: \Sigma \to G$, the transformed connection $A^g = g^{-1}A g + g^{-1}dg$ has curvature $F_{A^g} = g^{-1}F_A g$. It's physically a bit weird to specify $F_A$ in non-Abelian gauge ...


4

The point is that "locally flat", "locally Riemannian", "locally Lorentzian", do not mean that all geometric properties tend to become the corresponding ones in flat space when restricting to a sufficiently small neighborhood of a point. Roughly speaking, this is only valid for those properties that depend on the metric ...


3

I've fixed all of my issues (I think). Here's the $x y$ plane with a typical curve (the starting point is on the right): And here's a view of the cone on which the geometry is defined, with conical deficit angle $\alpha$. There are three typical curves drawn on it, with different impact parameter $b$. I'm a bit surprised that the initial velocity $v$ ...


3

A geodesic is a curve $\gamma$ which extremizes the path length functional $$S[\gamma]= \int d\lambda \ L(\gamma,\dot \gamma) =\int d\lambda \sqrt{g_{\mu\nu}(\gamma)\dot\gamma^\mu\dot\gamma^\nu}$$ The Euler-Lagrange equations are $$\frac{\partial L}{\partial \gamma^\alpha} = \frac{d}{d\lambda}\frac{\partial L}{\partial \dot\gamma^\alpha}$$ $$\implies (\...


3

To supplement the answers that were posted earlier, here's an explicit example from ref 1. This example proves that $A$ is not always uniquely determined by $F$, not even up to gauge transformations, not even locally. Work in four-dimensional spacetime with coordinates $(w,x,y,z)$, and take the structure group to be $SU(2)$. Let $T_1$, $T_2$, $T_3$ be the ...


2

The conifold singularity is the quadric hypersurface singularity given in complex coordinates by $x_1^2+x_2^2+x_3^2+x_4^2=0$ It is also known as the 3-fold ordinary double point. You can smooth this singularity by perturbing the equation: $x_1^2+x_2^2+x_3^2+x_4^2=\epsilon$ (for some $\epsilon\neq 0$). This variety is smooth ("looks the same everywhere&...


2

One way to see that the curvature tensor is zero is to start with the fact that there is no dependence of metric components on null coordinate $v$. So performing Kaluza–Klein reduction along the Killing vector $\partial_v$ we would obtain a Newton–Cartan spacetime with two spatial coordinates ($x$ and $y$), while $u$ now becomes Galilean time coordinate. ...


2

You're getting tripped up in two different ways here. (1) When you reason about the sizes of the terms in the K-S metric, you're ignoring the fact that the expression you're using is written using a mixture of variables from the two coordinate systems. This point is easier to understand with a simpler example. Suppose that $ds^2=e^{-r}dX^2$, where $r=\ln X$. ...


2

Let $M$ be a smooth manifold and $(P,\pi,M,G)$ a principal $G$-bundle over $M$. Let $\omega$ be a connection one-form and let $\sigma : U\subset M\to \pi^{-1}(U)$ be a local section which locally trivializes the bundle by means of $h:U\times G\to \pi^{-1}(U)$: $$h(x,g)=\sigma(x)\cdot g.\tag{1}$$ With $h$ we can pullback $\omega$ to understand how it is ...


2

The surface of a sphere is locally Euclidean. Consider two points on the equator of the Earth maybe a meter apart. Start them on northward trajectories. For a very short time, they will appear to be at constant distance. But by the time they reach 5°N latitude, you will notice that the distance between them is smaller by about half a millimeter. But, ...


1

What G. Smith said is correct, but I want to also point out that there is curvature of spacetime from the metric expansion that contributes to the dynamics of a galaxy is incorrect. There's no such effect. There are various ways to see this. Here's one. FLRW spacetimes have no Weyl curvature, only Ricci curvature. Ricci curvature is nonpropagating: the ...


1

I am not seeing how since the only part of ${\Gamma_j^i}_\mu$ which depends on the coordinates is the index $\mu$ which transform as one form. That's not accurate. You're right that the $\mu$ index refers to the manifold coordinates and the $i$ and $j$ indices refer to the tangent space. However, the tangent space basis vectors also depend on the manifold ...


1

The Christoffel symbols are not the components of a tensor. Their transformation laws involve the second derivatives of the parameter transformation : \begin{align} \bar{\Gamma}^k_{ij}&\boldsymbol{=} \left[\Gamma^\gamma_{\alpha\beta}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\boldsymbol{+}\dfrac{\partial^2 u^\...


1

Suppose we have a geodesic congruence. To describe the relative behavior of nearby geodesics, we "pick out" a one-parameter family of geodesics $\gamma_s(t)$ from it such that for every fixed $s$, $\gamma_s$ is a geodesic with affine parameter $t$, and such that $s$ and $t$ are "good coordinates" on the 2-D submanifold spanned by this ...


1

I don't think there are any $r\to\infty$ asymptotic timelike observers. There seems to be a limit of that sort in some coordinates, like Eddington-Finkelstein $(u,r)$, but it's an illusion. Conformal spatial infinity is a single point with no time dimension, and its causal past only contains past infinity. The $r\to\infty$ limit of stationary worldlines on ...


1

No, time-orientability is not enough. An oriented pseudo-Riemannian manifold needs to be time-oriented (or space-oriented) and have vanishing second Stiefel-Whitney class to admit a spin structure. It's a bit hard to find good references for this in the pseudo-Riemannian and not just the Riemannian setting, but most citations of this fact (e.g. in "Spin ...


1

In this instance the answer is easily obtained from the fact that the spacetime is a direct product of two spaces, thus the number of independent Killing vector fields (KVFs) is $6=3+3$ where each of the $3$s is the dimension of isometry group (and thus also the number of KVFs) of corresponding factor. Without knowing the structure of spacetime beforehand, ...


1

I found a reliable source that addresses my question (Appendix 4.1 - M. Gasperini, Theory of Gravitational Interactions; DOI 10.1007/978-88-470-2691-9), so I will present the resolution to my confusion here for posterity's sake. The issue is not with using the structure equations on-shell (indeed, this is purely a classical theory) and there is no error in ...


1

The article Axially symmetric spacetimes: numerical and analytical perspectives is a very nice reference and treats axially symmetric spacetimes in generality, showing how to construct the "cylindrical" slices (see also this). If your spacetime is not only "axially symmetric" but also "circular" (i.e. there are no meridional ...


1

I'm not a physicist and not great with tensor technicalities but I did have a desire to understand this issue, having encountered the Fisher Information metric in the field of information geometry. So here's a very informal perspective that connects this to a super basic linear algebra procedure. I think it's roughly correct in spirit and definitely helped ...


1

I know this is an old thread. But I think there still is a point missing, so if people still come to this post for reference, let me have a swing at it. I want to argue from a mathematical standpoint that is somewhat geometric. We want to take vectors from some vector space $V$. We will especially not demand that $V$ be an $\mathbb{R}^n$ for some dimension $...


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