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5

I believe it can be useful to define the following concepts (I won't be very formal here for pedagogical reasons): Any event can be described through four real numbers, which we take to be: the moment in time it happens, and the position in space where it takes place. We call this four numbers the coordinates of the event. We collect these numbers in a ...


4

You're right. $\eta_{\mu\nu}\rightarrow \eta_{\mu^{'}\nu^{'}}=\Lambda^{\alpha}_{\,\,\mu^{'}}\eta_{\alpha\beta}\Lambda^{\beta}_{\,\,\nu^{'}}$ just says that the metric transforms as a tensor as you would expect from its indices; there's nothing special about that. Being invariant means that when you make the transformation you get back the same matrix: ...


3

The convention we pick here will interact with the convention we have for matrix multiplication in the following way: If we have matrices $A$ and $B$ and we use the usual convention that the matrix multiplication $AB$ multiplies the rows of $A$ with the columns of $B$ then we have either $$(AB)_{ij}=\sum_kA_{ik}B_{kj}\tag{#}$$ or ...


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

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

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, ...


2

The lapse function is not defined by the metric alone, but instead depends on both the metric $g_{ab}$ and its slicing into timelike hypersurfaces. One way to "slice" a spacetime $\mathcal{M}$ into timelike hypersurfaces is to define a timelike coordinate $f$, which is just a function $f: \mathcal{M} \to \mathbb{R}$ such that $\nabla_a f$ is a timelike ...


2

Let's go step by step as it seems you're missing some fundamentals. We know from (linear) algebra, that a symmetric bilinear form can be transformed to a diagonal matrix with elements $e$ on the main diagonal $e\in \{0,1,-1\}$. The tripel counting the amount of times each number appears is called signature. If you didn't know that, check this. Now, a ...


1

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 ...


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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