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The only situation I know of in which the sign convention has physical implications is when defining the Clifford algebra. The full Clifford algebras of $+{-}{-}-$ and $-{+}{+}+$ signature are not isomorphic. One consequence is that when, like Dirac, you try to write the "square root" of the Klein-Gordon equation, which is $\partial^2\phi = -||\hat ... 5 In standard general relativity there is no physical implication. It is merely a sign convention. You arrive at the same predictions either way. My preference is$(-,+,+,+)$. There may be some differences in other contexts. 4 … [I]f this topology is consistent with Einstein's equations of General Relativity? … [C]an there be such a solution that is Ricci-flat? Short answer: Yes. The resulting Ricci flat solution is known as$\mathbb{RP}^3$geon, and is a$\mathbb{Z}_2$quotient of Kruskal–Szekeres extension of Schwarzschild spacetime. Though the nontrivial spatial topology is ... 4$R$and$r$are just coordinates. They're completely equivalent because there's a bijective mapping between them. The$r>0$and$R>α$regions are the same, and so are the$-α<r<0$and$0<R<α$regions. Being negative doesn't make$r$unphysical. Coordinates labeled by the letter$x$can be negative, and so can coordinates labeled by the ... 3 You probably made some trivial error. For example, did you forget to raise the index on$k$? $$k_\alpha=\left(-1,\frac{1}{1-2m/r},0,0\right)\tag1$$ $$k^\alpha=\left(\frac{1}{1-2m/r},1,0,0\right)\tag2$$ $$\sqrt{-g}=r^2\sin\theta\tag3$$ In the calculation of$\Theta$, only the$\alpha=rterm contributes: $$\Theta=\frac{1}{\sqrt{-g}}\left( \sqrt{-g}\,k^\alpha \... 3 For a general metric, ds^2 = g_{ij}dx^i dx^j. If we specify a path and a parameterization, so x^i=x^i(\lambda), then this becomes ds^2=g_{ij}\frac{dx^i}{d\lambda} \frac{dx^j}{d\lambda} d\lambda^2. Taking the square root and integrating yields the path length:$$s=\int ds = \int \sqrt{g_{ij} \frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}} d\lambda$$... 3 In Schwartz’ unfortunate all-indices-are-lower notation, (\partial_\mu A_\nu - \partial_\nu A_\mu)^2 means (\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial_\mu A_\nu - \partial_\nu A_\mu). You contract both indices, \mu and \nu, to get 16 terms (or 64, depending on what you’re counting as a term). Contractions in one term of \mathcal{L} (or ... 3 We have R(r,t)=a(t)r, \therefore \mathrm{d}s^2 in outer region becomes$$-\bigg(1-\frac{2GM}{ar}\bigg)\mathrm{d}T^2+\bigg(1-\frac{2GM}{ar}\bigg)^{-1}(\dot{a}r\hspace{2pt}\mathrm{d}t+a\mathrm{d}r)^2where dot over a means differentiating with respect to t. Now substituting T as a function of (t,r) substituting it in above equation collecting the ... 3 It depends on what OP means. One on hand, if OP literally means going from a spacetime with 3 spatial and 1 temporal directions to a spacetime with 1 spatial and 3 temporal directions, then it obviously has enormous physical consequences. Closed timelike loops for starters, cf. e.g. this Phys.SE post. On the other hand, if OP just means to change \eta_{\... 2 I think this space is a conical defect of order 2 at the center of \mathbb{R}^3-\{0\}. This means that any path traversing an angle 2\pi at some fixed radius has length 4\pi r. This is akin to 2d polar coordinates ds^2 = 4r^2 d\theta^2 + dr^2. Indeed the analogous construction is \mathbb{R}^2 with the disk B_2 removed and the unit circle ... 2 You could, of course, look for a solution of the form (4). In this case you can know you will find one since we know that Schwarzschild is a solution of that form. When you try to solve a PDE via ansatz (i.e. guessing), you can try whatever you want. Not every possible ansatz will lead to a solution though. There's an art to choosing one that's general ... 2 Let me do it for once and for all. Though the question has been answered by spiridon, I would like to give a formal derivation as spiridon's answer involves guess work. We have a situation where we need to calculate the inverse of a partitioned matrix. So let us first derive a general formula for the inverse of partitioned matrices and then we shall apply it ... 2 For this derivation, we first need to calculate the partial derivative of the covarinat metric tensor (which can be expressed, as the dot product of two covariant basis vectors). \begin{align} \partial_{\omega}{g_{{\mu}{\nu}}} =\partial_{\omega}\langle{\boldsymbol{\varphi}_{\mu},\boldsymbol{\varphi}_{\nu}}\rangle= \langle{\partial_{\omega}\boldsymbol{\varphi}... 2 The first and second order perturbations are defined to be tensors on the background spacetime. Therefore you can raise and lower the indices with the background metric. This is why the coefficients of the inverse metric take a funny form. 2 I was finally able to work it out. There were multiple confusing aspects. First of all, it turns out that if we ignore the matching conditions on the interior metric, Weinberg's T(r,t) is just one choice. The integrand is completely arbitrary. It is only after the matching condition is imposed that the particular function is chosen. Since Weinberg did the ... 2 So this actually happens with any black hole. In fact, it happens with any accelerating observer. In special relativity, before you get to general relativity and the equivalence principle, you have the first-order Lorentz boost formula. This says that when you travel with uniform acceleration \alpha, clocks ahead of you by a coordinate x will tick faster,... 2 The solutions are the same: they describe the same spacetime, but using slightly different coordinates. But the early workers, especially Schwarzschild, were not sure what to make of the region inside the horizon, or whether there was such a region. Droste's work showed more insight on this point, I think, and as I understand it his contribution was somewhat ... 2 Einstein used a linearized version of the Schwarzschild metric to calculate the precession of the perihelion of Mercury. This was a measured quantity that was not adequately explained by the quadrupole moment of the sun, and it was an important early check of his general theory of relativity. (Einstein probably could have found Schwarzschild's exact ... 1 b is their notation for the original metric g_{\mu\nu} and b^{-1} is just their notation for the inverse metric g^{\mu\nu}. So {\rm tr} \{b^{-1}h\}=g^{\mu\alpha} h_{\alpha \mu}={h^\mu}_\mu, and so on. 1 Spacetime doesn't change. The gravitational waves from a black hole merger are located on the future light cone of the merger. There's no earlier "meta-time" at which the spacetime is flat there. 1 The second one. The definition is the double covariant derivative. The first covariant derivative acts on a scalar field, and so it is just a derivative. The second is a covariant derivative on the vector field \partial_i \phi, which can be shown to equal the second expression you gave. 1 It's not quite so easy to prove the general case \phi \in (0, \alpha) (the singularity can't be removed in general), but fortunately, for the case \alpha = 2\pi, it is fairly easy. First let's write our metric in a more appropriate form. This spacetime can be described with two coordinate patches, (U_1, f_1) and (U_2, f_2), with \begin{eqnarray} f(... 1 No, the constants are not frame dependent. The speed of light is the same in every frame of reference. In the same way, the exact energy difference in the hyper fine structure of caesium is fixed and so on. Relativity maybe cause some small error in the measurement of those constants, but this can be neglected, since there are other experimental ... 1 Actually, the Reissner-Nordström metric hash(r)=1-2m/r+q^2/r^2\tag1.$$(The charge is squared.) The Ricci tensor is diagonal with$$R_{tt}=-\frac{q^2}{r^4}g_{tt}\tag{2a},R_{rr}=-\frac{q^2}{r^4}g_{rr}\tag{2b},R_{\theta\theta}=\frac{q^2}{r^4}g_{\theta\theta}\tag{2c},R_{\phi\phi}=\frac{q^2}{r^4}g_{\phi\phi}\tag{2d}.$$This can be understood ... 1 Answering: does the change in signature of the metric have a physical meaning. Not for the example you cited. The example you cited is a choice of convention of which physics is independent. However this metric is very different from (+,+,+,+) or (-,-,+,+), as these do not differ by an overall sign. The former is known as Euclidean signature. Oftentimes ... 1 First of all, note that the covariant derivative of a vector is a tensor. This means that if we have two vectors A_\mu and A^\nu, then their covariant derivatives \nabla_\rho A_\mu and \nabla_\rho A^\nu are tensors. These two tensors satisfy the transformation law:$$(\nabla_\rho A_\mu)=g_{\mu\nu}(\nabla_\rho A^\nu)\$ The same transformation can be ...