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By definition: \begin{align}\tag{1} A_{\mu,\nu} &\equiv \frac{\partial}{\partial x^{\nu}} \, A_{\mu} \equiv \partial_{\nu} \, A_{\mu}, \\[1ex] \tag{2} A_{\mu;\nu} &\equiv \frac{\partial}{\partial x^{\nu}} \, A_{\mu} - \Gamma_{\nu \mu}^{\lambda} \, A_{\lambda} \equiv \nabla_{\nu} \, A_{\mu},\end{align} If the connection is symetric: $\Gamma_{\mu \nu}^{...


5

No, it not true (from a pure theoretical point of view, the last word is of Nature as usual). The point is that Dirac relied his statement upon the principles of QM as known more than 90 years ago. There are other viewpoints nowadays where, for instance, ccrs of position and momentun do not play a crucial role and are obtained as a byproduct in Minkowski ...


5

As mentioned in the comments, two metrics related by a conformal transformation have the same light-cone. What is the scale factor equivalent to? It's time to bring up the topic few people discuss, measurement in GR. There are a variety of structures on a spacetime manifold. A structure helps us define an equivalence relation between two spacetimes. If we ...


3

I know topic this topic is 6 years old, but having just came across it, I feel the need to clarify some confusion around it, starting from the question, passing through the accepted answer and the ensuing comments. Most of this confusion sparks from two different but similar notions, that of a conformal transformatrion and that of a conformal isometry. Let ...


3

Then how can we say that it was parallelly transported since it wasn't so in the conventional sense? This is what curvature is, basically. Recall a simpler example, if $A_\mu$ is a (co-)vector field, you can integrate it against a curve: $$ I[\gamma]=\int_\gamma A_\mu\mathrm dx^\mu=\int_{t_0}^{t_1}A_\mu\frac{\mathrm d\gamma^\mu}{\mathrm dt}\mathrm dt, $$ ...


3

Short version : ${\omega_\kappa}^\mu$ is a matrix of (dual) vectors, and it is basically identical to the Christoffel symbol, via each such vector being \begin{equation} {\omega_\kappa}^\mu = ({\Gamma^\mu}_{\kappa 0}, {\Gamma^\mu}_{\kappa 1}, \ldots) \end{equation} In that paper in particular, they use the dual basis $\{ \omega^\nu \}$, so that \begin{...


2

The equation $$\vec{V} = V^\alpha \hat{e}_\alpha = V_\beta \hat{e}^\beta$$ does not make sense, you are equating two different objects, a vector and a covector. If you take a finite dimensional vector space, $V$, the dual space $V^*$ is the set of all linear functions that map elements of $V$ to the field over which it is defined. The metric tensor is a ...


2

The explanations are not the same. The first says a conical piece of a small geodesic ball, but the second says a small geodesic ball. In other words, Ricci has directional element, whereas scalar curvature does not. Indeed the lack of directional element in scalar curvature is the difference between the two. That said, this is not the usual definition, and ...


2

Yes it is true. The probability interpretation requires solution of the (relativistic) Schrodinger equation, and imposes Minkowski space time. The resolution is that quantum mechanics is formulated in a locally inertial reference frame using a single clock. In this frame Minkowski spacetime is a mathematical abstraction, not a physical something. It is a ...


2

As sets, $M$ is the union of the preimages $S_\tau = t^{-1}(\tau)$ because the absolute time map $t : M \to \mathbb{R}$ is a function, and for any function $f: X \to Y$ we have that $X$ is the union of preimages, i.e. $X = \cup_{y\in Y} f^{-1}_y$, simply because every point of $X$ has to lie in one of the preimages. However, the condition $(\mathrm{d}t)_p \...


2

The semikolon indicates a covariant derivative, in General Relativity only covariant derivatives transform like a tensor, therefore it is so important. The simple comma in difference indicates a simple derivative. the difference between covariant derivative and simple derivative is: $$A_{a;b} = A_{a,b} - \Gamma^c_{ba} A_c$$ where $\Gamma$ stands for ...


2

Keep in mind that the second identity of Bianchi is given by : $$\nabla_tR_{irs}^l+\nabla_rR_{ist}^l+\nabla_sR_{itr}^l=0$$ Let $t=l$ the identity become : $$\nabla_lR_{irs}^l+\nabla_rR_{isl}^l+\nabla_sR_{ilr}^l=0$$ Note that: (We'll use these identities) $R_{rsl}^l=-R_{rls}^l$, $R_{ils}^l=R_{is}$ and $R_{is}^{sl}=R_{si}^{ls}$ So, we obtain the following: $...


2

There is no implication. It doesn't mean that they are describing the same spacetime. The Riemann tensor has 20 independent components, so knowing one quadratic scalar combination of them isn’t much of a constraint on the geometry. In general, even a complete set of invariants is insufficient to determine the metric. See Wikipedia.


2

Here's a proof: Let $\varepsilon_n$ be a pre-euclidean space(Espace ponctuel) and $ds^2$ a linear element : $$ds^2=g_{ij}dx^idx^j$$ $n^3$ Christoffel symbols are $n(n+1)/2$ quantities. The definition of $g_{ij}$ is given by: $$g_{ij}=\mathtt{e_i.e_j}$$ By differentiation : $$dg_{ij}=\mathtt{e_i.}d\mathtt{e_j+e_j.}d\mathtt{e_i}$$ The expression of $d\...


2

Let $\Sigma$ be a spacelike hypersurface in the Lorentzian manifold $M$. Define $h_{ab}=g_{ab}+g_{ac}g_{bd}n^cn^d$. It can be seen case-by-case that if $X$ and $Y$ are any vectors tangent to $\Sigma$, then $h(X,Y)=g(X,Y)$ if $X$ is tangent to $\Sigma$, then $h(X,n)=0$ $h(n,n)=0$ However, given any local coordinate system for $M$, the matrix $[h_{ab}]$ is ...


1

Given a generic curved manifold, at each point you can draw a tangent plane. This is the vector space associated with that point in which the vectors live. Now even two nearby points in this manifold has independent tangent spaces(the tangent space of one is independent of the other). So how can you talk about differentiation of vectors? Well you introduce ...


1

If $\pi_{ijk}$ is the ordinary Levi-Civita symbol, then the Levi-Civita tensor has components $\epsilon_{ijk}=\sqrt{g}\pi_{ijk}$ (I am assuming the metric is positive definite, but if not then replace $g$ with $|g|$), where $g$ is the determinant of the metric tensor. Therefore, we have for $C_{i}=\epsilon_{ijk}A^j B^k$ $$ C_1=\sqrt g(A^2B^3-A^3B^2) \\ C_2=\...


1

The derivation you refer to is done locally in a set of coordinates in which $g_{ik,l} = 0$, which can be also characterized as related to a set of Riemann normal coordinates by a constant linear transform. This means, that you can make operations such as this one $$g^{ij} (g_{kl,mn}) = (g^{ij}g_{kl,m})_{,n}$$ Another useful identity is $g^{ij}g_{jk,l} = -g^{...


1

Nice question you got here! Let's go through this step by step: 1) Embedding of the body To understand the coordinates "rigidly fixed" on the rotating body, think of the object as a manifold $B$ embedded in space $M=\mathbb{R}^3$, i.e. an injective immersion $$\imath:B\rightarrow M.$$ Since we want to discribe a motion of this body, we let this mapping be ...


1

I recommend you to first have a good handle of Spinors in GR. I personally took this route in getting a grip on Spinors: A Child's Guide to Spinors : http://www.weylmann.com/spinor.pdf An Introduction to Spinors : https://arxiv.org/abs/1312.3824 Penrose, Roger (1960). "A spinor approach to general relativity" The third one is going to be a smooth if you ...


1

I'll give you hint. Use $R$ and the coordinate vectors of the coordinate system $$R(\partial_k,\partial_l)\partial_j=\nabla_{\partial_l}(\nabla_{\partial_k}\partial_j)-\nabla_{\partial_k}(\nabla_{\partial_l}\partial_j).$$ Note that $$\nabla_{\partial_k}(\partial_j)=\Gamma^{m}_{kj}\partial_{m}.$$ Also, you only have to calculate the first term on the ...


1

Whether a “reasonable” spacetime is geodesically connected depends on the definition of “reasonable”. If we consider globally hyperbolic spacetime as reasonable, then there is a theorem by Avez (1963) & Seifert (1967), that states that globally hyperbolic spacetimes are causally geodesically connected. Note, that the two points here must be causally ...


1

Using formulae from the note itself and setting $d=1$. $$ \delta z_\mu = 2 c^\alpha z_\alpha z_\mu - z^2 c_\mu, \qquad c_\mu = ( 0 , c ) . $$ where $z^\mu = (z,t)$ and the metric is $ds^2 = \frac{1}{z^2} [ dz^2 + dt^2 ]$. Then, $$ c^{-1} \delta z = 2 t z , \qquad c^{-1} \delta t = 2 t^2 - ( z^2 + t^2 ) = t^2 - z^2 . $$ Thus, the Killing vector is $V^\mu = ...


1

As mentioned in the previous answer it is the causal structure of the spacetime described by metric $g_{\mu\nu}(x)$ that is important. The reason why the formula for causally linked points $x$ and $y$ in Minkowski spacetime take the simple form $|x-y|<0$ is because the light cones given by $ds^2=0$ form $45$ degree lines in a usual spacetime diagram with ...


1

The proper notion in terms of causal structure here is that either $x$ chronologically precedes $y$ or $y$ chronologically precedes $x$, which means that there exists a timelike curve connecting $x$ and $y$.


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In general given a metric $g_{\mu\nu}$, an orthonormal basis $e^a$ is such that, $$\mathrm ds^2 = g_{\mu\nu}\mathrm dx^\mu \mathrm dx^\nu = e^a e^b \eta_{ab}$$ where $\eta = \mathrm{diag}(1,-1,-1,\dots,-1)$. So, if you write out the left-hand side, and the right-hand side, you get an equation you have to solve. More explicitly, $$g_{\mu\nu}\mathrm{dx}^\mu \...


1

Proposition. Given a manifold $M$ with 2 metric structures that yield the same lightcone/causal structure. Then the 2 metric tensors are related by a scale factor. Here is a pedestrian proof. Consider an arbitrary point $p\in M$. Use Riemann normal coordinates (which exist locally on a pseudo-Riemannian manifold) to bring the first metric at $p$ on ...


1

Your first reference defines the symbols and makes it clear that the formula refers to the twisting of a wire. Since no lever arm is given, the F must symbolize a torque.


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Singularities are defined properly by the following condition : For a spacetime $M$, such that every standard measurable quantity on it is defined (ie we won't allow the Riemann tensor to be divergent at a point and so forth), that spacetime is singular if there exists inextendible curves of bounded acceleration for which the curve's half-length (measured by ...


1

There is a lot more than one curvature scalar out there. When considering scalars strictly linear in the curvature tensor, the only independent scalar is the Ricci scalar $R$. When considering scalars quadratic in the curvature tensor, one can choose the set of three independent scalars to be the Kretschmann scalar $K_1 = R^{\mu\nu\kappa\lambda}R_{\mu\nu\...


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