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Despite what some of the other answers are mentioning, the following equation you have is correct $$\vec{r} \cdot d \vec{r} = r dr$$ You can check this by noting $$\vec{r} = x {\hat i} + y {\hat j} + z {\hat k} \implies d \vec{r} = d x {\hat i} + d y {\hat j} + d z {\hat k}$$ Then $$\vec{r} \cdot d \vec{r} = x dx + y dy + z dz$$ Further note $$r = ... 6 Your observation is correct! The issue is one that is regrettably rather commonly left unexplained in physics texts: A single coordinate system on a manifold does not define the spacetime. Often, the coordinates useful for computations in physics do not even cover the entire space, i.e. they are not defined everywhere on the object of study: The (spacetime) ... 4 One thing you can do with tetrads is express quantities everywhere in terms of what "natural" observers would measure at each point in spacetime. To be more concrete, consider a spacetime foliated by slices of constant timelike coordinate. At each point, one can imagine the "normal observer" whose 4-velocity is the unit timelike normal to the constant-time ... 3 Well, a good example is thinking in term of components. In several areas of physics, the math gets more intuitive when you think in terms of components of the vectors. So, instead of writing the vector \mathbf r for the position of a particle, you write x^i as the i-th component of a vector. The i in the top is to indicate a contravariant vector, ... 3 We consider the metric$$\mathrm{d}s^2=-\mathrm{d}t^2+a^2(t)\mathrm{d}\vec x^2 $$where a(t):= a_0e^{Ht}. To show that these coordinates do not cover the entire spacetime manifold, we consider the trajectory of a freely falling observer, which of course extremizes the proper time$$\tau=\int\mathrm{d}t\sqrt{1-a^2\dot{\vec x^2}} $$Performing the ... 2 The geometry and topology of the relevant phase space is identical for both classical and quantum problems: it is the very same phase space. The scale of the former is the small \hbar limit of the latter. Extended WFs appear like δ-fctn spikes ("points") in the small \hbar limit, once the phase space-variables are suitably rescaled by \sqrt{\hbar}. ... 2 Where does our theoretical prediction of the existence of black holes come from? It comes from the singularity theorems of Hawking and Penrose. Before then, people were aware of solutions to Einstein's Field Equations that had singularities but they required absolutely perfect symmetries, such as perfect radial symmetry for Schwarzschild, or perfect ... 2 I think your last sentence shows you're on the right track. A one form is a linear functional that maps vectors to real numbers. You give it a vector as an input, and, as you say, it returns the rate of change in the implied direction. Let's say we have a path whose tangent at a point is defined by the vector v^j\,\partial_j - the differential operator ... 1 Instead of r\, dr \cos \theta = r\, dr, that line should read r \, ||{d\vec{r}}|| \cos \theta = r\, dr. Since ||d\vec{r}|| \neq dr, the argument does not follow. If you are not sure why ||d\vec{r}|| \neq dr, ask yourself whether ||\frac{d\vec{r}}{dt}|| = \frac{dr}{dt}. 1 By its definition, the Einstein tensor is always defined to be \nabla G = 0, via the Bianchi identity. Here's the proper definition of a spacetime in general relativity : A spacetime is defined by a triplet (\mathcal{M}, g, \nabla) of a manifold \mathcal{M}, a metric upon that manifold g (or alternatively it can be done with a tetrad field) and a ... 1 This isn't too hard. Translations clearly are isometries: if \vec a' = \vec a + \vec c and \vec b' = \vec b + \vec c then |\vec a' - \vec b'| = |\vec a - \vec b|. Consider any isometry f; consider the isometry g(x) = f(x) - f(0) which preserves the origin. It's not too hard to see that this has to be linear and is therefore described by a matrix ... 1 A straightforward advantage is that manifolds are a useful way to talk about things that locally look like Euclidean space. Historically, the notion was developed so that you could do things like talk about surfaces (e.g. spheres) in a way that only made reference to the surface itself; that is without reference to the three-dimensional space they're ... 1 The mistake is in the step where you go from$$\vec{r}.\frac{d\vec{r}}{dt} = r\frac{dr}{dt}$$to$$\vec{r}.d\vec{r} = r.dr$$The integrand is time dependant (and involves a dot product as well) and hence the result is non-trivial. You simply cannot "cancel" off the times as you have done. 1 This may not be exactly what you want, but it does go over some classical field theory and good chuck of differential geometry. http://www.gravity-and-light.org/lectures 1 From a historical perspective black holes weren't predicted. In 1916 Karl Schwarzschild found a solution to Einstein's equations for a spherically symmetric mass. It was only subsequently realised that the Schwarzschild metric is a vacuum solution with an event horizon and a curvature singularity at its centre, and that the metric describes a static ... 1 There is a very geometrical answer. If you assume that the basis 4-vectors (hereafter called vectors) are not just vectors but are multivectors then they are a subspace of dimension four inside an associative, distributive non commutative algebra of dimension 2^4 when vv=\eta_{\mu\nu}v^\mu v^\nu. Such an algebra has scalars, vectors, bivectors ... 1 I) It seems the resolution to OP's question lies in the difference between the Levi-Civita symbol, which is not a tensor and whose values are only 0 and \pm 1; and the Levi-Civita tensor, whose definition differs from the Levi-Civita symbol by a factor of \sqrt{|\det(g_{\mu\nu})|}. II) The 2D Euler-density is$$ E_2~=~ \frac{1}{8\pi} ...