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In this answer it is nicely demonstrated that the numerical values $g_{\mu\nu}$ of the metric tensor depend on the coordinate system chosen or, put another way, ugly mathematical expressions can be "f …

Looking at Wikipedia articles about solutions for the Einstein Field Equation (fluid solution, dust solution) I hoped to find a table relating simple forms of the stress-energy tensor to the metric te …

The interior Schwarzschild metric $$c^2 {d \tau}^{2} = \frac{1}{4} \left( 3 \sqrt{1-\frac {r_s}{r_g}}-\sqrt{1-\frac{r^2 r_s}{r_g^3}} \right)^2 c^2 dt^2 - \left( 1-\frac{r^2 r_s}{r_g^3} \right)^{-1} d …

Suppose I have two massive plates of size $l\times h\times w$ mounted parallel to each other with a distance of $d$ and with a mass density of $\rho$. I send a light beam in the middle between them al …

Lets start naive: empty space, define the origin somewhere, start putting mirrors in a distance of $r$ in many directions so that they roughly sample the surface of a ball of radius $r$. Someone outsi …

Or is the question in the title fundamentally wrong? We label each point in space-time with four coordinate values, one of which typically is suggestively called $t$ for time. This made me think that …

In this answer it is stated that the metric tensor elements have no physical unit, i.e. $[g_{\mu\nu}] = 1$. What is the convention to get the physical unit of the line element $ds = g_{\mu\nu}dx^\mu d …

Inspired by this answer, I start toying with the general equation of the metric tensor $$ ds^2 = g_{\mu\nu}dx^\mu dx^\nu .$$ Let $g$ be diagonal, i.e. $g_{\mu\nu}=0$ for $\mu\neq\nu$ and let $x^0=ct …