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The action of a massless scalar field is given by: \begin{eqnarray} S(\phi)&=&\int{\cal{L}}dt\\ &=&\int d^{4}x \sqrt{-g}\left(g^{\mu\nu}\phi_{,\mu}\phi_{,\nu}\right) \end{eqnarray} Now choosing a tetrad, i.e., a basis of one form at each spacetime point $\{e^{a}=e^{a}_{\mu}dx^{\mu}\}$ we can rewrite the action as: \begin{eqnarray} ...

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OP's proposal (v2) is a special case of Finsler geometry with $n=3$. The main idea is to replace the quadratic metric tensor $g^{(2)}_{\mu_1\mu_2}$ for pseudo-Riemannian manifolds, which defines (infinitesimal, possibly imaginary) distance on the manifold via $$ds ~=~ \sqrt[2]{g^{(2)}_{\mu_1\mu_2}dx^{\mu_1}dx^{\mu_2}},$$ with (possibly a sequence of) ...

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I think you have all the right pieces to answer the question, here are a few hints that should be of some use. You say that you picked coordinates $\{v^{\mu} \}$. It seems to me that they should instead be called $\{ x^{\mu} \}$, as that is what you're taking partial derivatives with respect to. As you correctly pointed out, you are working with ...

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We know that the Levi-Civita connection satisfies $\nabla_a g_{bc} = 0$ and the product rule. The definition of the inverse metric $g^{ab}$ is $g^{ab}g_{bc} = \delta^a_c$. Therefore, we have: \begin{align} 0 &= \nabla_a \delta^b_c \\ &= \nabla_a (g^{bd}g_{dc}) \\ &= (\nabla_a g^{bd}) g_{dc} + g^{bd} \nabla_a g_{dc} \\ &= (\nabla_a g^{bd}) ... 2 The metric being a rank (0,2) tensor transforms under general coordinate transformations x^\mu \to x'^\mu(x) as g'_{\mu\nu} (x') = \frac{ \partial x^\rho}{ \partial x'^\mu } \frac{ \partial x^\sigma }{ \partial x'^\nu } g_{\rho\sigma} (x) $$Now set x'^\mu (x) = x^\mu + \alpha k^\mu(x) in the above expression and take a limit of small \alpha. ... 2 The (Lie-)group U(1) is the topological space S^1 (what we call a circle together with its standard open subsets) together with a rule how to multiply its points. In its representation as numbers in {\mathbb C} with absolute value 1, we have {\mathrm e}^{{\mathrm i}\alpha}\bullet{\mathrm e}^{{\mathrm i}\beta}:={\mathrm e}^{{\mathrm ... 0 Suppose we start with an orthonormal cartesian coordinate basis with unit vectors \left( \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \right). Suppose we wish to rotate to a new coordinate system (may or may not be a coordinate basis, see the following answer which has a good discussion on the difference) with unit vectors \left( ... 0 This is an example of lie-algebra valued 1-forms. Actually you may write explicitly,  A = A_{\mu} ^a T^a dx^{\mu}. Since the generators also anti-commute so we get the result. And for the same reason sometimes you will find expressions like [A,A] in literature for your term. 6 That's because you are forgetting that A has a Yang-Mills index. You better write this in components, which reads \epsilon^{\mu\nu\rho} g_{IJ} \Big( A^I_\mu \partial_\nu A_\rho^J + \frac{1}{3} f^J{}_{KL} A^I_\mu A^K_\nu A^L_\rho \Big)  This component notation also answers your second question. 0 There are definite advantages of differential forms, geometric calculus, and four vectors over 3d Gibbs vector algebra and vector calculus. Specifically you can solve for the electromagnetic field first ... and then let someone break it into electric and magnetic parts later if they feel like it (if at all). For instance the field due to a non accelerating ... 1 Here's a paper for you to ponder on: Teaching electromagnetic field theory using differential forms Excerpt from the abstract: computational simplifications result from the use of forms: derivatives are easier to employ in curvilinear coordinates, integration becomes more straightforward, and families of vector identities are replaced by ... 1 Killing vector fields correspond to infinitesimal isometry generators of the spacetime manifold and any physical action including the Polyakov action should be preserved under it. In fact, any physical action should be invariant under the (infinitely) larger group of diffeomorphisms of a manifold. Isomotry transformations are just a finite subset of these ... 1 Globally hyperbolic refers to the fact that hyperbolic equations always have locally a well defined Cauchy problem, that is, a unique development given initial conditions. Which means that, given a matter field at a time t_1, there exists a unique solution of that field at a time t_2. It is boosted up to globally hyperbolic if that property holds ... 0 The problem here is that in general case, a metric with some given coordinate dependence, will contain a priori both TT (Traceless-Traverse) fields that correspond to propagating fields, and "non-TT" fields that demand sources to be distributed on the spacetime When one does the process of collapsing a metric to the TT "gauge" (notice the very intentional ... 0 The first expression is a single equation expressing equality between two vectors. The second expression is in Einstein notation or rather index notation. It is a set of 4 equations (for mu=0..3) expressing equality between components of vectors. Both expressions are equivalent ways of working with vectors. Einstein notation can be much faster to manipulate. ... 2 Writing \vec{e}_r = \partial_r, \vec{e}_x = \partial_x transforms the "\vec{e}-notation" into the partial-derivative-notation, so the "relation" is just that \vec{e}_{x^\mu} = \frac{\partial}{\partial x^\mu}. 0 There is a simple derivation: Spacetime curvature is a product of gravitation. Gravitation is a force. The force the gravitational field is declining continually with increasing radius, proportionally to the squared radius. The corresponding function is continuous and differentiable. Once we saw that the function of gravitational force (and thus of ... 2 Horizontal position of indices matters in principle because one might want to raise and lower indices on the Christoffel symbols. If the horizontal position of indices are not observed in a consistent manner, it becomes ambiguous which index was raised or lowered, and so forth, in particular if the connection is not torsionfree. Also note that different ... 0 This is a general foundamental question. Why are manifolds supposed to be differentiable in physics? This is because, with the partial exception of part of quantum physics, we assume that mathematical laws concerning evolution of physical systems are described in terms of differential equations (with either ordinary or partial derivatives). These ... 0 If you had just one coordinate chart, then you wouldn't need to ever transition from one chart to another. So you wouldn't have to worry about whether your transition was continuous, differentiable, C^2 or smooth. But if you have two charts and need two charts, then at some events they will be in the intersection of the two charts. And in the one chart ... 1 1) If the transformation itself is smooth, then the Jacobian will be smooth. This is a desirable property because... 2) The coordinate transformation tells us how the coordinates change, but the Jacobian tells us directly how the coordinate basis vectors change. For example, a transformation from Cartesian to polar coordinates would use the Jacobian to ... 2 The functions are "smooth" because we want to be able to speak about derivatives on our manifold M, and for that, it is convenient to have a smooth structure on M (one could settle for a C^k-structure with k as needed, but physicists rarely care for such details). And of course we want to be able to take derivative because we might be interested in ... 1 Perhaps you could explain what exactly led you to question your understanding... The physical meaning of coordinate invariance is pretty simple. It's just that the laws of physics cannot depend upon your choice of coordinates as long as the reference frame you're working in is inertial. So if you were to perform a coordinate transformation from one inertial ... 1 The metric measures lengths in various directions, and also angles between various directions. For example if \vec{e}_{(1)} is the basis vector in the x^1-direction, it will have length given by$$ \lVert \vec{e}_{(1)} \rVert^2 = g(\vec{e}_{(1)}, \vec{e}_{(1)}) = g_{11}.  If we also have the basis vector $\vec{e}_{(2)}$ in the $x^2$-direction, then the ...

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The metric is an important concept in general relativity. In GR, vectors correspond to weighted directions in spacetime (by "weighted", I mean any scalar multiple of a vector corresponds to the same direction, but weighted differently). The metric tensor can then tell us about the angle between two directions or the magnitude of a given vector, which gives ...

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As has been pointed out in the comments, it's not entirely clear how you intend to specify a metric without the use of some set of coordinates. That said, a couple of common GR texts have non-standard approaches to the Schwarzchild metric that you might find interesting. Misner, Thorne, and Wheeler's Gravitation has a fairly detailed sidebar (Box 23.3, ...

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(8.21) is the inverse of (8.20) up to first order.

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It is simply false, at least written as it stands. The point is that the relation between the topology and the metric is more complicated than in the Riemannian case, where the geodesical balls form a basis of the topology$^1$. As a matter of fact, a (connected) Lorentzian smooth metric $g$ over the time-oriented smooth manifold $M$ does define a ...

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