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One way to establish that $r$ in Schwarzschild coordinates is equivalent to the spherical radial coordinate is its asymptotic behaviour, namely for $r\to\infty$ the metric tends to Minkowski, and the weak field approximation yields Newton's gravitational potential. One way to establish that the parameter $M$ is indeed the mass of the spacetime is to ...

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In special relativity there are two major assumptions: -the laws of physics are the same in all inertial frames -the speed of light that you observe is always the same, (thus independent of the relative motion between the light source and the observer). From this two assumptions follows the famous Lorentz transformations. In these Lorentz transformations ...

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The set of transformations that leaves the speed of light unchanged is the Lorentz group. Representation theory enables us to investigate the irreducible representations of the Lorentz group. The lowest-dimensional representations act on scalars four-vectors However, take note that usually we consider representations of the corresponding Lie algebra ...

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In this nice reference the autor assumes the relativity principle + homogeneity + isotropy and deduce the general coordinate transformations which contains both Lorentz and Galileo transformations. Further he imposes the postulate of the constancy of the speed of light, restricting the transformations to be the Lorentz type.

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In a sense, the two definitions you mention are the same. One of the postulates of special relativity is that the speed of light is the same in all reference frames, so the definition of "the coordinate transformation according to the postulates of special relativity" is the same definition as "the coordinate transformation under which the speed of light is ...

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There are lots of ways of approaching special relativity. My own preferred approach is the invariance of the line element. Suppose you move a small distance in spacetime $(dt, dx, dy, dz)$ then the length of the line element $ds$ is defined by: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 \tag{1}$$ This equation is known as the metric equation and is derived ...

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A general diffeomorphism does not map geodesics to geodesics. Some simple counter examples You can a build diffeomorphism on the Euclidean plane by imagining putting one finger on a tablecloth at point $x$ and dragging it. This map is clearly smooth, a smooth inverse is constructed by dragging your finger back. Any geodesic on the plane (a line) passing ...

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The general story Here's an attempt to formalize how physicists build spacetime manifolds. Let $N$ be one of $\mathbb R^n$ Some dimension $n$ product manifold of $S^1$ and $\mathbb R$ (corresponding to periodic solutions). Pick one such $N$. Now Take stress tensor $T$ defined on some open subset $U \subset N$ and some boundary conditions (e.g. ...

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The concept of 'straight' is a bit ill defined in GR and has no real definition. In fact in a sense the geodesics themselves be seen as 'straight' lines; they are the shortest paths connecting 2 points (this is what in normal Euclidean space would be a 'straight line') In the LC connection they are the integral curves of some vector field $V$ with $... 4 Let us for simplicity work in units where the speed of light$c=1$is equal to one, and assume that there is no cosmological constant$\Lambda=0$. A spherically symmetric vacuum solution to the EFE of the form $$\tag{1} ds^2~=~g_{tt}(r)dt^2 + g_{rr}(r)dr^2 +r^2 d\Omega^2,$$ and such that it asymtotically becomes Minkowski space $$\tag{2} ... 6 Conserved quantities in GR In GR, energy (or mass) is typically an ill-defined concept. In flat spacetime, we define energy as the conserved quantity corresponding to time translational symmetry. Extending this to GR is quite tricky mainly because, what one is calling time is already observer dependent (this is of course also true in flat spacetime, but at ... 5 Is there another way to conclude the Schwarzschild solution has a mass M It's not so much a conclusion as a definition. From Schutz in "A first course in general relativity", section 8.4 "Newtonian gravitational fields", pages 207 - 208: Any small body, for example a planet, that falls freely in the relativistic source's gravitational field ... 6 Your intuition that the Einstein equations are equations for the metric tensor, not for the manifold is mostly on the right track, but the details are wrong. That core bit of intuition is best phrased, I think, as saying that the Einstein equations are local equations for the geometry of the manifold. That is, they tell you that, whatever manifold your ... 1 The square root of the determinant of the metric can be understood as a particular function of the components of the metric g_{ab}$$\sqrt{-g} =f(g_{ab})$$By the chain rule we of course have$$\nabla_a \sqrt{-g} = \nabla_a f(g_{bc}) = \frac{df}{d g_{bc}} \nabla_a g_{bc}$$But we know that \nabla_a g_{bc}=0 so that of course \nabla_a \sqrt{-g} =0. This ... 0 I'd say none. If you are talking about a Riemannian cylinder, its metric (say, induced from its embedding in \Bbb{R}^{3}) is$$ g=dz^{2}+d\phi^{2} $$If the two were conformally equivalent, a conformal transformation (z,\phi)=(z(t,x),\phi(t,x)) would pull back the metric as$$ g'=\bigg(\frac{\partial z}{\partial t}dt+\frac{\partial z}{\partial ... 0 Once you have the metric in u,v coordinates, you notice that$g_{uv}=0$and this shows that the basis vectors$e_u$and$e_v$, respectively tangent to the v=const and u=const lines (no typo, you see why?) are orthogonal. 2 Here's a heuristic calculation: Let$\{E_i\}$be an orthonormal frame ($g(E_i,E_j)=\epsilon_i\delta_{ij}, \epsilon_i=\pm 1$). Then$\mu$is the canonical volume form$\sqrt{g}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$iff$\mu(E_1,\dotsc, E_n)=1\$. Then $$(\nabla_X\mu)(E_1,\dotsc,E_n)=\nabla_X(\mu(E_1,\dotsc,E_n))-\sum \mu(E_1,\dotsc,\nabla_X ... 6 Comments to the post (v2): Note that \sqrt{|g|} transforms as a density rather than a scalar under general coordinate transformations. In particular, the covariant derivative of \sqrt{|g|} does not necessarily coincide with the partial derivative of \sqrt{|g|}. Here is a heuristic explanation using local coordinates. The Levi-Civita connection is ... 0 The straight answer to your question is no, the value of the components of the stress energy tensor do not change according to the metric signature. Qmechanich showed you how the formulas change, depending on the metric signature, just in order to keep the component values the same. 0 First of all, don't think about embeddings: there is no notion (in GR) that spacetime is embedded in some higher-dimensional flat space (let alone a five-dimensional space): it's just what it is. This is not to say embeddings are not interesting, but it is a red herring to worry about them too much. So, straight lines. As you say, in general there is no ... 3 I read a few lines about general relativity and [... an equation for] the eigentime of a time-like curve. I suppose that this is referring to an equation similar to$$\tau A_J^Q := \int_0^1~dt~\sqrt{g[~\dot\gamma, \dot\gamma~]},$$where A denotes a particular participant ("material point", "principal identifiable individual"), the quantity being ... 1 Exactly as an ideal clock at rest with the observer (here pictured as a timelike curve) measures the proper time of the observer, ideal rulers at rest with the observer measure the distances in the rest space of the observer. Mathematically these rulers are pictured as an orthonormal basis made of 3 vectors normal to the unit tangent vector to the ... 6 Parametrize the particle's worldline w.r.t. t:$$~x^\mu (t) = (t,1)$$Its four-velocity is:$$u^\mu =\frac{d x^\mu}{d \tau}$$To evaluate this we use the fact that:$$d \tau^2 = coshx~dt^2-dx^2$$Also use x=1 and dx=0:$$d \tau^2 = cosh(1) dt^2$$Therefore:$$u^\mu = \frac{d x^\mu}{d \tau} = \frac{d t}{d \tau} \frac{d x^\mu}{d t} = ...

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