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8

Warning: this answer takes the perspective of the second table, rather than the first, in a question @knzhou linked. We need to first explain what's analogous to $A_\mu$. The Christoffel symbols are. We can then ask what's analogous to $F_{\mu\nu}$; the Riemann tensor is. I suspect someone else's answer will provide a rationale for why that differs from what ...

5

No, this is not what it means. There are usually two (equivalent) ways of describing tensor fields. Let $M$ be a smooth manifold, and $T^r_s(TM):= \bigcup\limits_{p\in M}T^r_s(T_pM)$ be the $(r,s)$ tensor bundle (for any vector space $V$ over $\Bbb{R}$, $T^r_s(V):= V^{\otimes r}\otimes (V^*)^{\otimes s}$ is the space of $(r,s)$ tensors over $V$, which is ...

5

First let build up some intuition by looking at the divergence in 1D. Imagine you have a river which has a very uniform flow so you can approximate the flow as 1 dimensionsal. The flow is also static in time. This particular river flows in the positive x direction so $v(x)$ is positive and assume at a particular location we have that $\frac{\partial v}{\... 5 (continued from ANSWER - Part I) ANSWER - Part II$\boldsymbol\S$D. Parallel Transport on the plane - The Christoffel Coefficients To determine the Christoffel coefficients in equations \eqref{C-02a}-\eqref{C-02d} it's necessary to introduce the first fundamental form of the theory of surfaces. The infinitesimal displacement$\mathrm d\mathbf x$(see ... 5 I am not sure if this is what you are looking for, but nevertheless, let me say the following: In full generality, a "vielbein" on a d-dimensional spacetime is defined to be a vector bundle isomorphism $$e:T\mathcal{M}\to\mathcal{T},$$ where$T\mathcal{M}=\coprod_{p\in\mathcal{M}}T_{p}\mathcal{M}$denotes the tangent bundle of the spacetime ... 4 ANSWER - Parts I & II & III$\texttt{PARALLEL TRANSPORT ON THE PLANE}\texttt{C O N T E N T S}\boldsymbol\S\texttt{ A. Preliminary}\boldsymbol\S\texttt{ B. Curvilinear Coordinates}\boldsymbol\S\texttt{ C. Using Curvilinear Coordinates}\boldsymbol\S\texttt{ D. The Christoffel Coefficients}\boldsymbol\S\texttt{ E. Polar Coordinates}...

4

Since $f = f(\theta,\phi) = Y^l_m(\theta,\phi)$ is a scalar function, $\nabla_a f$ is a covariant vector, and so it's covariant derivative is that of the covariant derivative of a covariant vector, thus we have \begin{align} \nabla^3 \nabla_3 f &= g^{33} \nabla_3 \nabla_3 f \\ &= g^{33} (\partial_3 \nabla_3 f - \Gamma_{33}^a \nabla_a f) = g^{33} (\...

4

The problem with the divergence of the fields you wrote is that it is ill-defined in the origin. So whatever you find is valid only if $r\neq0$ and we need to manually add the value of the divergence in the origin. How do we do it? We use the fact that the integral of the divergence in the volume is equal to the flux of the vector field on a surface which ...

4

This answer says roughly the same things as @JG's answer, but phrased slightly differently. The analogy between the Riemann tensor $R$ and the electromagentic field strength tensor $F$ is not so apparent when we dress them up with indices $F_{\mu\nu}$ and $R^\rho_{\,\,\mu\nu\sigma}$. However, things become more apparent if we look slightly more abstractly. $... 4$A^\mu(\lambda) \equiv \frac{dx^\mu}{d\lambda}$are the components of a$(1,0)$-tensor$\mathbf A = A^\mu \left.\frac{\partial}{\partial x^\mu}\right|_{x(\lambda)}$attached to the point$x(\lambda)$. The derivative$\frac{d}{d\lambda} \mathbf A$involves computing the difference between two tensors -$\mathbf A(\lambda + d\lambda)$and$\mathbf A(\lambda)$... 3 I think it's worth saying that there is, in a way, not a great 1-1 analogue here: the Maxwell action is proportional to$F^{ab}F_{ab}$, while, while JG is right that in a lot of ways, the "analogue" to choose for GR is$R_{abcd}$, the Hilbert action is proportional to$g^{ab}R_{ab}$not$R^{ab}R_{ab}$. The coupling to matter happens via a$g^{ab}$... 3 From the perspective of Principal Lie-group (G) bundle, the curvature 2-form $$\textbf{F}=\textbf{dA}+\frac{1}{2}[\textbf{A,A}]$$corresponding to Lie group$G=SU(N)$is essentially the Yang-Mills field$F_{\mu\nu}$. For$G=GL(n;C)$or$GL(n;R)$, the components of curvature 2-form are the Riemann tensor components$R_{\alpha\beta\gamma\delta}$. The ... 3 You can prove this by moving to Riemann normal coordinates at a point$p$. In these coordinates $$g_{ij}(p) = \eta_{ij} , \qquad \partial_k g_{ij}(p) = 0 , \qquad \partial_k \partial_l g_{ij}(p) = \frac{1}{3} [ R_{ikjl}(p) + R_{jkil}(p) ] .$$ It is then clear that anything that contains 2 metric derivatives can be re-written in terms of the Riemann tensor.... 3 As you may know, the metric tensor is a bilinear 2-form. It accepts two vectors from vector space$V$and gives back a real number in$\mathbb R$. It is linear in both arguments, hence 'bilinear'. The metric tensor is interpreted as a linear operator in the sense that it maps one of its arguments (either one; doesn't matter because it's symmetric) to a dual ... 3 The metric tensor is bilinear by definition, since tensors are multilinear maps from vector spaces (and their dual spaces) to real numbers, which means that they must be linear in all slots. In general, on a$n$-dimensional pseudo-Riemannian manifold, the metric tensor cannot be put into diagonal form everywhere by a coordinate basis. It can only be done if ... 3 If the$\nabla$were$\partial$, the equation would simplify by the chain rule to$m\frac{d}{dt}p^\mu=0$, so the momentum is "conserved" in the sense of being unchanged by a change in$\tau. This notion of free motion is just the intuition of Newton's first law, in relativistic language. But since a geodesic can be described with any choice of ... 3 I'll take a stab at answering this. Short answer: it does not play a major role in neither the canonical or covariant LQG. A bit about boundary terms Terms like that aren't at all mysterious, in fact, they are ubiquitous in all first-order systems. Take for example the action from Hamiltonian mechanics $$S[x, p] = \int dt \left( p \dot{x} - H(x, p) \right). ... 3 Yes, they are independent of each other. For an example that is time-orientable but not space-orientable, start with \mathbb{R}^4 with the metric dt^2-(dx^2+dy^2+dz^2) and then twist it by identifying (t,-x,y,z) with (t+1,x,y,z). This is still time-orientable (the vector field \partial/\partial t is nonzero and timelike everywhere), but it's not ... 2 This is not really an answer, but aims to clear up some of the confusions that can happen with this problem. There are three levels of structure that are at play here : an underlying topological space (usually \mathbb R^4, but in the case of a Schwarzschild BH, we need to remove the worldline of the singularity for everything to work out properly later) ... 2 A projection operator satisfies two properties$$ 1)~P^2 = P, \qquad 2) ~ P \cdot n = 0 $$where n is normal to the subspace that P projects to. Alternatively the second condition can also be written as$$ 2') ~ P \cdot t = t where t is any vector that lives entirely in the subspace that P projects to. In other words, t \cdot n = 0. You can ... 2 Say (M,g) is a 2-dimensional oriented Riemannian manifold, and let \star be the Hodge star operator. Given a vector field F on M, we define \begin{align} \text{curl}(F)&:=\star d(g^{\flat}F) \end{align} i.e we take our vector field F, convert it to a covector field g^{\flat}(F), take its exterior derivative (so we now have a 2-form) and ... 2 Birkhoff's theorem guarantees that any spherically symmetric interior solution you write down will have the Schwarzschild geometry as its exterior field, so all you have to do is ensure that the interior stress-energy looks reasonable for matter. In the rotating case, a no-hair theorem guarantees that uncharged black holes have the Kerr form, but there's no ... 2 I will add a different point of view by using tetrad formalism. As @J.G. said, the Christoffel symbols are roughly the analogs of the gauge connection. In fact, the tetrad formalism has a more accurate analog: the spin connection \omega. This spin connection is \mathfrak{spin}(1,3) valued, just like a gauge connection is valued in some gauge group (for ... 2 (continued from ANSWER - Part II) ANSWER - Part III \boldsymbol\S F. Parallel Transport on the plane - Geodesics in Polar coordinates The geodesic curves on smooth two-dimensional surfaces have the following characteristic properties each one of which could be used as equivalent alternate definition Osculating plane : Geodesic on a surface is any curve ... 2 Your equation will only be a tensor if you are in flat Minkowski spacetime with an affine coordinate system. The reason is that the partial derivative is simply not a tensor. The velocity of a point moving along a curve x^\mu (\lambda) is given by\mathbf{v}=v^\mu\mathbf{e}_\mu = \frac{\text{d} x^\mu}{\text{d}\lambda}\mathbf{e}_\mu$$and the acceleration ... 2 Unfortunately there are too many very small errors (signs, factors of 2, 4! etc...) in the question statement and other answers to just leave comments. I will follow the conventions of ([1] Sections 3.1, 7.5, 7.6). Given$$F = \frac{1}{2} F_{\mu_1 \mu_2} dx^{\mu_1} \wedge dx^{\mu_2}$$and assuming \varepsilon_{0123} = + 1 so that ([1], 7.6)$$*(dx^{\... 1 As commenters already suggested, a Killing horizon is necessary to associate a temperature to a spacetime. This also means that spacetime needs timelike Killing vector field (KVF), then we can define a Hartle–Hawking state on this spacetime background regular across this horizon so that observers static with respect to that KVF would register a thermal bath. ... 1 The "temperature" of the Schwarzschild metric is not exactly a temperature. If Hawking radiation is exactly thermal then the unitarity of quantum mechanics will be violated. There is growing consensus that quantum gravity is unitary and information is conserved. To preserve unitarity the radiation emitted cannot be uncorrelated as you would expect ... 1 A geodesic, when parametrized by the arc lenght, can be expressed as the parallel transport of the tangent vector being zero.U^\mu=\frac{dx^\mu}{d\tau}$is exactly the tangent vector with such parametrization. $$U^\lambda \nabla_\lambda U^\mu=0$$ Multiplying by$m$the$U$'s we get the expression for the momentum. For a visualisation, we could change from ... 1 Along time-like or null geodesic we can associate "rate of change" wrt some affine parameter$\tau$as $$\frac{d}{d\tau}=P^a\nabla_a$$ where$P^a$is tangent vector along geodesic. Then the geodesic equation can be stated as$\frac{dP^a}{d\tau}=0$. The intuition is that, if we work in a flat space-time and in non-relativistic limit (assume$P^a=mU^...

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