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I don't remember having seen the specific expression of the proposed "signed arc length" either (anywhere but related to the OP question), nor anything resembling (1) the more abstract expression for determining the sought resemblance. For naming this proposed functional from the set of curves (or rather, arcs) into the set of real numbers (incl. $\mathbb ... 1 No, a magnetic monopole a la the Dirac string does not "violate" gauge symmetry. Rather, the statement "we have a magnetic monopole" means only that we are forced to consider the gauge theory not on the whole spacetime, but on the spacetime with the location of the magnetic monopole removed. Why? Because, at the location of the magnetic monopole, the curl of ... 1 We have the frame$\{e_\mu\}_{\mu=0,\dotsc,3}$in terms of which the velocity vector is$v=v^\mu e_\mu$. There are a few properties of the affine connection which I would like to summarize: $$\nabla_{fX}Y=f\nabla_XY$$ $$\nabla_X(fY)=f\nabla_XY+X(f)Y$$ $$\nabla_{e_\mu}e_\nu=\Gamma^\lambda{}_{\mu\nu}e_\lambda$$ Using this, let's get to work. We have ... 4 The "signed arc-length" is not used in relativity and I give reasons why. You are free to call and denote it in any way you like.$s$and$\ell$are interchangeably used to denote arc-length of space-like paths$g(\gamma',\gamma')>0$in relativity and$\tau[\gamma]$is used for "proper time" of$g(\gamma',\gamma')<0$time-like paths but the notation ... 2 I have never seen that second definition before. The first definition is standard. For a Riemannian manifold the metric tensor is positive definite, that is $$g(u,u)>0\quad\forall u\ne0$$ We have the standard relation (tensor product omitted) $$\mathrm{d}s^2=g_{ij}\mathrm{d}x^i \mathrm{d}x^j$$ Let$t\in\mathbb{R}$be a curve parameter and ... 0 In a sense, parallel transport, covariant derivative and connection are all synonym for you can recover one from the other. So given a manifold one usually starts by giving one notion (e.g. how a vector field is transported parallel to itself along a family of curves) and then, if needed, the other objects are derived. In physics, when dealing with a ... 0 The linear transformation is the following composition of linear maps: Go from$R^n$to$T_m M$using the natural identification Go from$T_m M$to$T^*_m M$with the symplectic form Go from$T^*_m M$to$T_m M$using the inverse isomorphism given by the metric Go again back to$R^n$(here$M = R^n$and$m = (q,p)$By the way, this transformation of ... 2 Same as with the symplectic form:$\omega(v) = (u_\omega,v)$defines the isomorphism between 1-forms and vector fields. When the metric is Euclidean the dual basis to an orthonormal basis corresponds to the basis itself. 3 Recall that one-forms are defined as linear maps on vector fields to real numbers, so that for every one-form$\alpha$and every vector field$X$,$\alpha(X)$is a scalar function. Hence on a simple tensor$\alpha \otimes X$we can define the contraction by$C : \alpha\otimes X \mapsto \alpha(X)$and extend by linearity. On a tensor with more factors, for ... 2 I would recommend using Mathematica to calculate curvatures, unless there's a good reason to do it by hand (for example, perhaps you want to calculate the curvatures for a metric while keeping the dimension general). It's not hard to write your own code to do this, and I think it's a nice idea actually. I also have found this code to be very useful: ... 1 If by "straight line" you mean "linearly varying the coordinates from the beginning point$(x_0,y_0)$to the endpoint$(x_1,y_1)$," then the trajectory is just that. You can parametrize it as$(x,y) = (1-\lambda) (x_0,y_0) + \lambda (x_1,y_1)$for$\lambda$running from$0$to$1$. Call this path$C$. The distance is $$\int\limits_C \sqrt{ds^2} = \int_0^1 ... 3 The separation vector is a Jacobi field because it obeys the Jacobi equation. Here I will derive geodesic deviation from scratch because I find MTW's derivation hard to follow. (Much like everything else in that book.) Definition 1. Consider a family of timelike geodesics, having the property that in a sufficiently small open region of the Lorentz ... 0 I think one can enter a dispute regarding the notion of "accepted" but the idea is that General Relativity is successfully described by a Pseudo-Riemannian Manifold, subject to Einstein Equations, with free-falling objects following geodesics. Now you look for a set of axioms that give you this structure. One such set, although not entirely rigorous, is ... 2 General relativity can be constructed from the following principles: The Principle of Equivalence Vanishing torsion assumption (\nabla_XY-\nabla_YX=[X,Y]) The Poisson equation (or any other equivalent Newtonian mechanics equation) Explanations: The Equivalence Principle can be used to show that spacetime is locally Minkowskian, i.e. the laws of ... 1 I think the concept of charts might be leading you astray here. A simple, practical and intuitive way to define the dimension of a manifold is the number of numbers you would need to locate a point on that manifold. For the case of a sphere, you need two, commonly written as \theta, \phi. For \mathbb{R}^3, you need 3, x,y,z. Sure you can embed S^2 ... 0 The best reference is Polchinski's textbook (vol.'s 1 and 2). This low-energy action is indeed invariant under diffeomorphisms--all objects appearing in the integrand are geometric invariants. This means that there are no un-contracted indices and also that they transform as scalars under diffeomorphisms. And the last ingredient is present too--the ... 0 Specifically, the word composite refers to a composed form of the {\rm USp}(8) connection Q_{\mu a}{}^{b}. It depends on a vielbein V_{ABab} and a connection A_{\mu IJ}, see eq. (4.3) in Ref. 1 for details. Here the index \mu is a 5-dimensional spacetime index; the indices AB refer to the \bar{\bf 27} of E_{6(6)}; the indices ab refer to ... 0 The important point is that both t^a and v^b are vectoror fields defined (at least) on C. The idea is that if t^a\nabla_av^b=0 and you take v^b at a particular point p on C then v^b at all other points on C is the parallel transported vector from p. Another way of thinking about it is that the covariant derivative in a given direction is ... 2 We are not entirely sure what OP's question (v4) is asking, but here are some comments: I) The Dirac belt trick demonstrates that the Lie group SO(3) of 3D rotations is doubly connected,$$ \pi_1(SO(3))~=~\mathbb{Z}_2. $$II) As for the title question Are spinors somehow connected to spacetime? one answer could be: Yes, in the sense that the mere ... 0 You might equally well ask, "How does the physical belt in the Dirac trick sense the topology?" This question is, when you think about it, no less mysterious than yours. The answer, by experiment, is that it simply does. And ultimately, if something transforms "compatibly" with the Lorentz group, or with SO(3), then there is really only a one-bit question ... 1 If you are working on a complex manifold with a Hermitian metric, then the Hodge star operator should be taken antilinear: \star(\alpha + i\beta) = \star\alpha - i\star\beta. If you work on a real manifold without a complex structure by itself, and you extend the scalars to the complex numbers, it may be that there is no harm in taking it to be linear. 2 This notation arises often in supergravity. Suppose one has a d-dimensional theory. The Hodge \star operator has the usual definition, and \star_p is the Hodge star operation defined on a p-dimensional sub-manifold. The question of which sub-manifold is often either explicitly stated or obvious from the context. For example, in the Klebanov-Strassler ... 1 In general, having an asymmetric matrix for a metric won't really help, because only its symmetric part will contribute to the norm of any given vector. Take some finite-dimensional real vector space V with an inner product \langle\cdot, \cdot\rangle represented by some matrix g_{ij} in a given basis \beta. Then for any vector v with components ... 1 The Hodge star operation acts on differential forms. Numbers, real or complex, transform as 0-forms. The Hodge dual of a 0-form will result in something proportional to the volume form of the manifold. In detail, for a d-dimensional manifold, \star 1 = \text{vol}_d = \sqrt{|g|}dx^1 \wedge ... \wedge dx^d, and the Hodge operation commutes with ... 3 Technically, yes (for loose enough definition of "metric"), but there's very little point to it. Some attempts at unification of gravity and electromagnetism, including several attempts by Einstein and various co-authors, basically amount to some variation of trying to interpret the antisymmetric part g_{[ab]} as the electromagnetic Faraday tensor ... -3 Here is a simple thought experiment to help visualize the shape of a Big Bang universe. All directions point back in time. Theoretically if one could see far enough back in time, one is looking towards the Origin, a single point, the only point we all have in common. Therefore all directions ultimately point toward the Origin, which in some sense can be ... 2 Asymmetric tensors have been considered in the quest for a unified classical field theory. Einstein in particular went through a whole series of candidate theories. His last paper on the subject - co-authored by Bruria Kaufman, submitted 3 months before and published 3 months after his death - is about a field of this type; the theory actually was referred ... 8 A metric on a manifold M is, by definition, a symmetric 2-tensor field g with the property that g_x is positive-definite for every x\in M (plus some smoothness/continuity requirements if M is smooth/topological). This ensures that the norm of a vector in a fibre of the tangent bundle to M is a non-negative number, and that the angle between ... 3 The dual is defined by the map$$\frac{\partial}{\partial x^\mu} \mapsto g_{\mu\nu}\mathrm{d}x^\nu$$and hence the dual of a \partial_t + b \partial_1 is only a \mathrm{d}x^t + b \mathrm{d}x^1 iff the metric is Euclidean flat in the t,1-direction. Note: Do not write "X=" if you mean the dual of X is equal to something. Duality/equivalence is ... 1 Comments to the question (v3): Given a manifold M, if a smooth vector field X\in \Gamma(TM) does not vanish in a point p\in M, then one may choose a local coordinate neighborhood U\subseteq M of p, with local coordinates (x^1, \ldots, x^n), so that X=\frac{\partial}{\partial x^1}. This procedure is sometimes called stratification or ... 4 @Phoenix87 is spot on, but I'll elaborate a bit. Definition 1 A spacetime (M,g) is stationary if there exists a timelike Killing field K, i.e. a vector field K such that \langle K,K\rangle<0 and \mathcal{L}_Kg=0. We shall show that Definition 1 implies the existence of local coordinates for which g_{\mu\nu} is independent of time. ... 3 The rough idea: take the local flow of the vector field and use it to get a new "time" coordinate. In general this will work locally, so you have to patch your manifold with small enough open subsets where you can then define the new set of coordinates where now the Killing vector field corresponds to \partial_t. 0 If you use the ({-}{+}{+}{+}) sign convention then ds is the proper distance between two infinitesimally separated points; if you use the ({+}{-}{-}{-}) convention then it's the proper time. In each case, if you choose points whose separation is such that the proper distance (respectively, proper time) is not meaningful then ds will become imaginary. ... 2 The Einstein equations are equivalent to an relation on the Riemann tensor. Given a, b that are linearly independent vectors,$$R(a \wedge b) = C(a \wedge b) + 4\pi [a \wedge T(b) - b \wedge T(a) - \frac{2}{3} T a \wedge b]$$where T(a) is the stress energy tensor acting on a and T is its trace, and C is the Weyl ("conformal") tensor. This ... 0 The line element for Schwarzschild in the usual static coordinates is ds^2 = -f dt^2 + f^{-1} dr^2 + r^2 d\Omega_2^2, \qquad f= 1-\frac{r_+}{r}. If you were to consider a constant t slice of this, the metric would be simply ds^2_{t=0} = f^{-1} dr^2 + r^2 d\Omega_2^2. Note that this is not the horizon, nor does it contain it. The horizon is a ... 2 Regarding why the Ricci tensor, not the Riemann, appears in the EFEs the answer is in the Newtonian theory.Very heuristically consider Newton's law \ddot{r}=-\frac{GM}{r²} . Now let's try to make this a more local statement. In order to do so I'll divide both sides by r and some numerical factors in order for the volume V=\frac{4\pi r³}{3} of some ... 2 As already mentioned by others, \mathrm ds^2 is used as suggestive notation for the metric tensor$$ g = \sum_{\mu,\nu}\mathrm g_{\mu\nu} \, dx^\mu\otimes\mathrm dx^\nu $$In case of a positive definite metric and given a curve \gamma:[0,T]\to M, it has a precise meaning in terms of either the length function$$ s_\gamma(t) = \int_0^t ... 6 It is a mnemonic notation that indicates that$\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$is the object whose square root is to be used as the infinitesimal line element, traditonally denoted$\mathrm{d}s$, when determining the lengths of worldlines$x : [a,b] \to \mathcal{M}by integrating the line element along them as \begin{align*} ... 2 It is a square of a proper time interval or a square of proper distance (modulo an inessential sign). 1 It is a notational device. Note that in (-+++\cdots) the proper lengthds^2=g_{\mu \nu}dx^\mu dx^\nu$$is negative for timelike dx. Thus ds\equiv \sqrt{ds^2}\in\mathbb{C}. It (the square root) thus has no physical meaning. 1 The general convention is that all repeated ("dummy") indices (and they had better only be repeated in pairs, one up, one down) are summed over. All remaining indices -- the "free" indices -- vary over all possible values. With no free indices, you have a rank-0 tensor (a scalar). With one free index, you have a rank-1 tensor (a vector), which can be ... 10 I think this question is more trivial than you think. You should ask yourself why should the full Riemann tensor appear. I'll sketch a heuristic derivation of the field equations. We know that with small velocities and a static field, the Poisson equation$$\Delta\phi=4\pi G\rho\$ is approximately satisfied. From special relativity we know that the ...