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## New answers tagged differential-geometry

0

The relevant point is the step where you write $\bar{F}(x,v) = (x,\color{blue}{v};\color{red}{v},F(x,v))$ - you have to think about how the third coordinate (highlighted in red) comes about: While the blue $v$ is $\color{blue}{v}\in T_x Q$, the red is just a component of $(\color{red}{v},a)\in T_{x,v}TQ$ - how do you say what "part" of the vector ...

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$k$ is not a measure of the curvature of spacetime, but rather of the spatial sections. $k = 0$ doesn't mean that spacetime is flat (your example isn't: since $a(t) \neq a_0$, it can't be Minkowski spacetime), but rather that the spatial sections are just the Euclidean $\mathbb{R}^3$ space rather than, for example, a $3$-sphere. It is also possible to get it ...

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This issue is discussed somewhat in Steven Weinstein's essay "Naïve Quantum Gravity", published in Physics Meets Philosophy at the Planck Scale: Contemporary Theories in Quantum Gravity (eds. Callender & Huggett, 2001). He begins by noting that [a]n alternative way to conceive of gravity would of course be to follow the lead of other theories,...

1

I think you're making a mistake in seeing $X$ as arbitrary: you should see it as the "basis vector" of $x$, which isn't arbitrary, because it's the specific vector that tells you, from the origin of your normal coordinates, where to go to reach the point $p$. The proof can then be concluded as follows: $$\text{from}\quad \Gamma^a_{bc}X^b X^c=0\quad ... 0 Once you have the metric, you have already set the stress-energy tensor automatically. You'll have G_{\mu\nu}^{(\text{c})} = 8 \pi T_{\mu\nu}^{(\text{EM})}. What you might then attempt to do is to set T_{\mu\nu}^{(\text{EM})} = T_{\mu\nu}^{(\text{F})} and try to solve for the densities and pressures, with the possibility of failure. Otherwise, you'd need ... 0 The arcsine function terms and the root formula terms are expanded into series to the terms with different coefficients, and then computed.$$\frac{1}{2}\sin^{-1}\left(\frac{r}{a}\right)=\frac{r}{2a}+\frac{1}{12}\frac{r^3}{a^3}+o\left(r^3\right)-\frac{r}{2a}\left[1-\left(\frac{r}{a}\right)^2\right]^{1/2}=-\frac{r}{2a}+\frac{r^3}{4a^3}+o\left(r^3\right)$$... 3 The flow of a classical Hamiltonian observable is the flow of its associated Hamiltonian vector field. The differential equation that an integral curve (q(u),p(u)) of this flow/vector field obeys is precisely your eq. (4). You should think of these integral curves as the motions you would get if the associated observable f was your Hamiltonian - for f=H... 0 Because everything is inside a trace, I will omit it for clarity and specify it when needed. The first term is simple $$d(A\wedge dA) = dA \wedge dA$$ because as you stated d^2=0 Next, d(A \wedge A \wedge A) = [dA^a \wedge (A \wedge A)^b - (A \wedge dA)^a \wedge A^b + (A \wedge A)^a \wedge dA^b]T_a T_b \end{... 7 Is that effect due to being around blackhole? No, what he is showing there is the standard coordinate chart for a uniformly accelerating observer in flat spacetime. There is no black hole or spacetime curvature in that diagram. It is standard flat spacetime with accelerated coordinates instead of inertial coordinates. Recall that the equivalence principle ... 2 Your first reason to disagree with the metric you guess is quite good. In fact, notice that for \alpha = \frac{\pi}{2} the cone reduces to a plane, meaning the answer should be flat in this situation. The second reason is correct, but your metric already satisfies it: notice that for r = 0 the metric is degenerate, and hence that is already a coordinate ... 0 It's more useful to work with an explicitly covariant formulation of electrodynamics. Working with the mostly-minus signature (+---), we define the 4-potential A^\mu = (\Phi/c, \vec A) and its covector partner A_\mu = g_{\mu\nu} A^\nu. The Faraday tensor is given by F_{\mu\nu} = \partial_\mu A_\nu -\partial_\nu A_\mu, and the electric and magnetic ... 0 You look at the tangent space to be able to not have to worry about the effects of curvature. You can see that fairly intuitively on a sphere. If you think about segments of a path along the equator, from one point of view they are parallel. From another point of view they are clearly not. So in the tangent space, which is flat, this consideration is avoided.... 1 I found the solution by starting from the Killing vector x^A \partial_A and manipulating this expression starting from the differentials of the Y and the chain rule. It works out perfectly. 1 You can exploit symmetries. The number 6 you quote comes from taking them into account. The way I'd approach it is to try and start "populating" the indices of the tensor, in order, while keeping the symmetries into account. Say we are calling the coordinates 1, 2 and 3. The first index can be anything, let's set it to 1. So, we are looking at R_{... 0 All the objects \gamma^a  and C and so on are just Clebsh-Gordan coefficients intertwining different spin representations, so they are constants and their covariant derivative is zero. For \gamma^\mu= e_a^\mu\gamma^a the e_a^\mu are the components of the identity operator {\bf E}:TM\to TM where {\bf E}={\bf e}_a \otimes {\bf e}^{*a}= {\bf e}_a ... 2 Generally, the Ricci tensor of a maximally symmetric geometry is proportional to its metric tensor. In this case, the Ricci tensor is given by \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad R_{\mu \nu}= \begin{bmatrix} 1/y^2 & 0 & 0 &0 \\ 0& -1/y^2 & 0 & 0 \\ 0&0&1&0 \\ 0&0&0& (\sin{\... 1 You can find a neat derivation of that equation by Thorne and Macdonald themselves at this link. Just check Sec.2.2 and 2.3. Your equation is labelled (2.4). 0 The problem here is with assuming that because n_{\alpha}\sim \delta_{t\alpha}, then the n_{\alpha;\beta} will also vanish for \alpha \neq t. This is not true. Even though a field might have only some component, it does not mean that its covariant derivative does not have any other components. Let us see, what hides behind the n_{\alpha;\beta}. In ... 4 Is there a way to write this using differential forms (such that it reduces to equations involving covariant derivatives rather than partial derivatives)? It turns out that the differential form equations written in my original post are already what I wanted. This was unclear to me, so I will show that the differential form Maxwell's equations reduce to the ... 0 It seems like it is an error, it is corrected in the second edition: However, for some reason some of the chapters of the second edition on the Springer site (https://link.springer.com/book/10.1007/978-1-4757-2063-1) such as Chapter 9 are those of the first edition. I found the full second edition at https://loshijosdelagrange.files.wordpress.com/2013/04/v-... 0 If you are doing a passive transformation where you are merely expressing the same tensors in a new primed basis, then yes the metric changes too in such a way that any scalars don't transform. No matter what metric you have and no matter what the transformation is, this scalar will not change. So it doesn't tell us anything interesting. If you are instead ... 1 Let T be a (p,q)-tensor, so its components have p upstairs indices and q downstairs indices. Then the covariant derivative \nabla T is a (p,q+1)-tensor whose components are usually written$$\nabla_\lambda T^{\mu_1 \cdots \mu_p}_{\qquad \nu_1 \cdots \nu_q} \equiv T^{\mu_1 \cdots \mu_p}_{\qquad \nu_1 \cdots \nu_q ; \lambda}For example, if T ... 0 He is essentially writing L = KI in index notation. Let us pick some A_{ij} \in W. Then notice how \begin{align} A'_{ij} &\equiv \delta^k{}_{[i}\delta^l{}_{j]}A_{kl}, \\ &= \delta^k{}_{[i}A_{kl}\delta^l{}_{j]}, \\ &= A_{[i|l|}\delta^l{}_{j]}, \\ &= A_{[ij]}, \\ &= A_{ij}, \end{align} where I used indices i,j,k,l to make it clear ... 0 There are some other approaches to computing the curvatures that do not involve dealing in detail with the Christoffel symbols. For example, Wald's General Relativity presents on Sec. 3.4b a method employing an orthonormal basis (tetrads) that sometimes can simplify computations considerably. Later on, he exemplifies this method on Sec. 6.1 by finding the ... 4 There are other approaches to doing Quantum Field Theory in Curved Spacetime and I've never seen anyone trying to define a "Hilbert bundle" (so to speak) in order to achieve it. My guess, but this is just a guess, is that the state of the entire field would need to be on \bigotimes_{x\in\mathcal{M}} \mathcal{H}_x and I don't think this is well-... 0 The constant S comes from the experience of Newtonian physics. In reality it can also generate an integral constant in classical mechanics. We can formulate the gravitational field in the vacuum for a spherical symmetric environment as:\phi = \frac{k}{r}$$where this field fulfills$$\nabla^2 \phi = 0$$what is the classical equivalent to the EFE for the ... 0 By matching the solution to the Newtonian theory. This can be done, for example, by comparing the metric of the Schwarzschild solution with weak-field solutions and the standard results of Newtonian gravity. This is often discussed in introductory books in General Relativity, such as Sean Carroll's Spacetime and Geometry, for example. The constant K is ... 2 To add to the confusion there we have also the invariant notation for Lie-algebra valued forms: \alpha [\wedge] \beta := [\alpha \wedge \beta] This is usually defined on simple forms and then extended by linearity. More precisely, say \alpha = \alpha' \otimes X and \beta = \beta' \otimes Y where \alpha, \beta are ordinary forms and X,Y are ... 1 I also used to also find these sorts of things confusing. It helps me to think of the general case: take a vector of N numbers {a_i} and a vector of M numbers {b_i}. So long as N,M are finite, we have$$ \sum_i \sum_j a_i b_j = \sum_j \sum_i a_i b_j \equiv \sum_{i,j} a_i b_j. $$Where the double sum notation \sum_{i,j} is used because the order ... 0 The trick is do just do the sums one at a time. For simplicity, I'll open the expression in two spacetime dimensions t and x, but it works quite similarly in general. We have$$ \Gamma^{\lambda}{}_{\mu\sigma} \Gamma^{\sigma}{}_{\nu\lambda} = \sum_{\lambda,\sigma} \Gamma^{\lambda}{}_{\mu\sigma} \Gamma^{\sigma}{}_{\nu\lambda}. $$Due to associativity and ... 0 You'll have that$$\Gamma^\lambda_{\lambda \sigma} \Gamma^\sigma_{\mu\nu}= \Gamma^\color{red}{t}_{\color{red}{t}\color{blue}{t}} \Gamma^\color{blue}{t}_{\mu\nu}+\Gamma^\color{red}{t}_{\color{red}{t}\color{blue}{r}} \Gamma^\color{blue}{r}_{\mu\nu}+\ldots + \Gamma^\color{red}{r}_{\color{red}{r}\color{blue}{t}} \Gamma^\color{blue}{t}_{\mu\nu} + \Gamma^\color{...

6

The claims in the question are very strange: Many people do particle physics without ever so much as mentioning the word "bundle". It is perfectly common in physics to circumvent the notion of a bundle for the gauge field by talking about the behaviour of the gauge field "at infinity". See e.g. this question for instances of that language....

0

Assuming stress-energy tensor ${\rm{T}}_{\mu}^{\nu}\equiv {\rm{diag}}~\{\varepsilon,-p,-p,-p\}$ and taking trace from both sides of Einstein field equations (EFE) one obtains the relation $$-R=\kappa~(\varepsilon-3 p). \tag{1}$$ If you know your metric component $g_{rr}$, and $g_{00}$ is constant, you can easily calculate the ...

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As already mentioned in some other answers, it is mainly a matter of pedagogical choices. Deep down, one is indeed using all of the fiber bundle machinery to define the connection, but on the particular context of General Relativity one can take a few shortcuts, even when dealing with texts that are mathematically careful and avoid working in components. For ...

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I'm not sure I understand in what sense the technology of bundles is irrelevant in GR. Vector fields on a smooth manifold $M$ are sections of the tangent bundle $TM$, which is associated to the (generally curved) frame bundle $LM$; the connection $\Gamma$ and curvature $\mathrm{Riem}$ are (local representations of) the connection 1-form $\omega$ and its ...

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This is because its takes a substantial amount of mathematics to get there as you have just outlined. They need to learn about topology, smooth structures, manifolds, connections, bundles, frame bundles, principal bundles and so on. Instead, they take the traditional physics language of components which has its advantages although eschewed by mathematicians. ...

1

It is pretty straightforward to write some simple Mathematica code for this. And it would in any case be useful for you if you have to do this kind of calculation more than once. In any case, this would be the result $$R=-\frac{2 \left(r a'+a^2-a\right)}{r^2 a^2}$$

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Griffiths' calculations are a formal manipulation that can be done more rigorously through other approaches. For the first issue, he seems to be defining the solid in this way. The vectors are orthogonal at every point, but, as you mentioned, they changed from point to point. The surface he has constructed has sides of constant values of $u$, $v$, and $w$, ...

2

I can understand the curvature of a sheet which is 2D but i can't understand the 3D curvature of space described in General Relativity. The usual meaning of the word curvature is what is called extrinsic curvature. Examples are any curved surfaces in the space. But it is possible to know if a surface is curved without looking at it from the outside. The ...

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Space is curved if you don't come back to your starting point when you walk around a square. Or equivalently, you wind up at different points if you walk east-then-north vs north-then-east. The surface of the Earth is curved in this sense. It doesn't show for a small square. But try a really large square. Start on the equator. Walk 1/4 of the way around the ...

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The bowling ball in a trampoline model of curved spacetime has a lot of problems. I have looked at it carefully and come to understand that gravity caused by the curvature of spacetime around a planet (or any other mass) works exactly like a low pressure system in the weather. Why is a leaf attracted to the centre of a low pressure system in the air? The ...

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Curvature (using the Riemann tensor) is just a mathematical means to account for the unintuitive weirdness that happens to an object when you move (parallel transport) the object around a closed path in space-time. Suppose you move an object (ie: a vector $V$) around 2 sides of a square path resulting in a vector $V_1$. You are very careful to only ...

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Before picturing curved spacetime, start with Newton’s laws in the absence of gravity. Newton’s first law says that an isolated object travels in a straight line at constant speed. If you plot the object’s path in spacetime it forms a straight line, called a worldline. An isolated object can be measured as one where an attached accelerometer reads 0. So ...

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The metric being diagonal (in some co-ordinates) doesn't sound like enough to take any sort of shortcut. But maximally symmetric spaces (like dS, AdS or their Riemannian counterparts) satisfy $$R_{abcd} = g_{ad}g_{bc} - g_{ac}g_{bd}.$$ Another useful fact is that Ricci scalars add for direct product manifolds.

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Spacetime curvature is well-described by Einstein's equation for gravity, but the math used in it is generally beyond the capabilities of nonprofessionals like me to explain. I invite the professionals here to take a stab at this for you. The way I visualize this (works for me, may not be technically accurate in all respects) is that large chunks of mass or ...

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This is essentially the Weyl Scalar $\Psi_2$ (the only non vanishing Weyl scalar for Schwarzschild metric) . In NP formalism, if we consider an orthonormal null tetrad ($l, n, m, \bar{m}$) , then the Weyl scalar $\Psi_2$ is defined as$$\Psi_2=-C_{abcd}l^am^bn^c\bar{m}^d$$ For Schwarzschild metric, we can consider the null tetrad: $l=\frac{1}{\Delta}(r^2, \... 2 Yes, you can obtain the stress-energy tensor from the Einstein tensor. If you do not specify the stress tensor beforehand, any Lorentzian manifold will provide a solution to the Einstein equations. However, not necessarily it will be a physically interesting solution. As a historical example, we can consider the Alcubierre Warp Drive, which has a geometry ... 1 There is a lot of nonsense written about "vierbein postulate". This "thing" is not a postulate because a postulate can be accepted or rejected, but this formula is always true. It is simply the statement that the covariant derivative of the identity operator is zero. As such, is it is independent of any assumption of metricity or of ... 0 In STEGR there necessarily must be a metric tensor$g$. The geometry is equip with a connection$\Gamma$that has vanishing total curvature, vanishing torsion, but non-vanishing nonmetricity$Q_{abc} = \nabla_{a}g_{bc}\$. (Sometimes this will be phrased as the connection not being metric, which may be where the confusion comes from). Just like curvature can ...

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Since you asked in the comments, let me provide some references to the geometric quantization of Chern-Simons theory. This does of course not answer your second qustion about how to deal with spinors in geometric quantization and I would be interested in an answer to this myself. However, it goes in the direction of applying geometric quantization in field ...

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