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You don't need $\pi$ for that formula. Let me provide an alternative definition of $r$ for you. Let $M$ be Schwarzschild spacetime. An orbit of $M$ is a two-dimensional manifold $\Sigma_r$ generated …

You have the order wrong in which you "get" things. You get the energy-momentum tensor from your specific matter theory. You do not know what $g_{ab}$ is. Then, given some general assumptions about yo …

They are exactly the same thing. The only difference is that one usually studies Riemannian metrics in mathematics, whereas the metric of relativity has a signature $(-+++)$, i.e. is not positive def …

We first show that $h=g+V\otimes V$ is a projection operator into the subspace $H_p\mathcal{M}$ orthogonal to $V\in T_p\mathcal{M}$. Idempotence ($h\circ h=h$). A simple calculation gives: $$h^a{}_b …

Apologies for the confusing title, it is late here. I'm wondering exactly what meaning the "time coordinate" has in General Relativity. We always write the line element as $$\tag{1} ds^2=g_{00}(dx^{0} …

A coordinate change just amounts to a diffeomorphism. The geometry is unchanged by diffeomorphisms. Since the geometry is what the physics is, the physics is unchanged by diffeomorphisms as well. Diff …

Let us first argue that ${}_\bot Z^\alpha={}_\bot Z^a$. Now, ${}_\bot\mathbf{Z}$ does not contain the component of $\mathbf{Z}$ along $\mathbf{V}$. Suppose we expand $\mathbf{Z}$ in terms of the the b …

You learned in calculus that for a variable change $x\longrightarrow \bar x$ we have $$d^nx=J\,d^n\bar{x}$$ where $$J=\left|\frac{\partial x}{\partial \bar{x}}\right|$$ Look at the transformation law …

From Wald (1984) The entire content of general relativity may be summarized as follows: Spacetime is a manifold $M$ on which there is defined a Lorentz metric $g_{ab}$. The curvature of $g_{ab}$ i …

When we talk about the geometry of GR, it is understood that the manifold of spacetime is not a Riemannian one, but rather a Lorentzian manifold. This means that the metric is not positive definite. W …

Let $(M^{n+1},g)$ be a Lorentzian manifold. Given $p\in M$, we will show that there is a coordinate system $(x^\mu)$ defined on an open set $p\in U\subset M$ such that $\partial_0$ is a timelike vecto …

If the frame bundle has a global section $\{\theta^j\}_{j=0}^n$, then you obtain a Lorentzian metric by $$g=-\theta^0\otimes\theta^0+\sum_{i=1}^n\theta^i\otimes\theta^i.$$ However, this is overkill. Y …

In general, it should be true if the dimension of spacetime is at least $3$. Suppose we have two points $p,q$ in Minkowski space with $p\ll q$, so there is a smooth timelike curve going from $p$ to …

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\cdo …

Definition 1. A spacetime is said to be spatially homogeneous if there is a one-parameter family of spacelike hypersurfaces $\Sigma_t$ foliating the spacetime such that for each $t$ and for any points …