New answers tagged

0

It is only infinite for an infinitesimaly short coordinate length, so the integral is of course finite: $\int_{r_{\rm s}}^{r_{0}} \sqrt{|g_{rr}|} \, dr =\int_{r_{\rm s}}^{r_{0}} \frac{1}{\sqrt{1-\frac{r_{\rm s}}{r}}} \, dr = \sqrt{(r_{0}-r_{\rm s}) r_{0}}+\ln \left(r_{0}+\sqrt{(r_{0}-r_{\rm s}) r_{0}}-\frac{r_{\rm s}}{2}\right) = {\rm finite}$ For the ...


2

The "special" surfaces of spacetime defined by Killing horizons are null hypersurfaces. A null hypersurface which is a hypersurface whose normal vector at every point is a null vector (with respect to the local metric tensor). The "boring" and trivial example is a light cone, as already mentioned. EDIT: from the comments, it is true that this statement is ...


-1

I’m no physicist, but I’ve always liked to think that the square of Planck’s length is a real-life pixel in a simulated universe. The entire universe as we know it is a huge, pixelated depository of information. Where that information came from, I haven’t the slightest clue. I am not a big fan of the big bang theory as it simply does not make sense to me ...


3

My best guess is that, as I cross the surface of the sun, I start accumulating solar mass on the opposite side of me from the center of gravity that’s attracting me, and that prevents the formation of an event horizon. Is this reasoning correct? Yes, that reasoning is correct. The simplest solution to the Einstein field equations in General Relativity is ...


0

From Earth perspective and gravitationally speaking, the main difference between the Sun and a black hole of the same mass would be its size. A black hole is basically an enormous amount of mass concentrated in a small volume. The horizon you speak of is located at what is called the Schwarzschild radius $R_s$ and any mass condensed in a sphere of radius ...


5

You’re essentially correct in the sense that when you fall into the Sun its gravitational pull on you decreases. The parts of the Sun that are at a greater radius from the center than you are no longer exert any net force on you. This is called the Shell Theorem. A black hole, on the other hand, doesn’t have its mass distributed in a ball like the Sun does, ...


Top 50 recent answers are included