40

The true answer lies in General Relativity, but we can make a simple Newtonian argument. From the outside, a uniform sphere attracts test masses exactly as if all of its mass was concentrated in the center (part of the famous Shell theorem). Gravitational attraction also increases the closer you are to the source of gravitation, but if you go inside the ...


8

Stars generate a great deal of energy through fusion at the core. Basically the more massive a star is, the more pressure the core is under (due to the star's own gravity) and the more energy it can generate (somewhat simplified). That energy of course radiates outward and heats everything outside the core making it a something like a pressure cooker, with ...


4

Or does gravity depend on the density of the object as well? The problem with this question is that it's rather ambiguous as to what you mean by "gravity". An object doesn't have a single number that is its "gravity". If a ship is near a star, the gravitational force that the ship feels depends on the mass of the star, the mass of the ...


3

Roughly speaking, for a star to become a black hole, its physical radius has to become smaller than its Schwarzschild radius. So even the Earth could be a black hole if it shrinks to below 9 milimiters. It is not precise to say that a black hole depends on the density of the object, since a Schwarzschild metric is a vacuum solution of Einstein's field ...


3

An example that covers the whole spacetime with a finite number of charts is Kruskal-Szekeres coordinates. The only reason you need multiple charts for these coordinates is that the angular coordinates describe a two-sphere, and you can't cover a two-sphere with a single chart. Does the central singularity have to be excluded? In the standard black hole ...


2

If the visible matter became enough dense to be concentrated inside its Schwarzschild radius, it becomes a BH. Until their inner pressure withstand the gravitation they stay being stars.


2

Where is the observer whose metric is the one above? I do not think that question makes much sense in general. In GR, we have coordinate maps and atlases, not observers and frames. Coordinate map is arbitrary homeomorphism from some open set in spacetime to $\mathbb{R}^4.$ Moreover, metric is geometric object. There is no observer associated with it. You ...


2

Two points: There is no clear physical interpretation for a non integer number of microstates. Why not? Because for a quantum system the space of possible states is a vector space. When we talk about the number of microstates, we are referring to the dimensionality of that vector space (or subspace), which is an integer. The entropy is not necessarily ...


1

@Eletie I think Alamar's question, with the expectation that the black hole evolves with time, is based more on concepts, such as: if the singularity does not evolve with time, then how can we say that a massive object has collapsed to form a singularity - a time-independent singularity would either exist in that position (according to some "local ...


1

The Schwarzschild radius is just a distance. What you really are asking is what happens if there are two crossed event horizons. An event horizon is a boundary where nothing on one side can affect the other side; in particular things that cross from the other side to that side will never escape. If you have two closed event horizons that overlap, that ...


1

The angular velocity of a black hole is not a very well defined concept. The best we can do is define an angular velocity to the horizon. This essential the rate at which objects get frame dragged along the horizon, and an outside observer should be able to measure it by observing objects falling into the black hole. However, there is no compelling reason to ...


1

You've gotten the components of your Killing vector field $\xi_\mu$ wrong. Since $\xi^\mu=(1,0,0,0)$, lowering that index with the metric $$g_{\mu\nu}=\begin{bmatrix}\frac{2GM}{r}-1&1&0&0\\1&0&0&0\\0&0&r^2&0\\ 0&0&0&r^2\sin^2\theta\end{bmatrix}$$ actually gives us $\xi_\mu=g_{\mu\nu}\xi^\nu=(\frac{2GM}{r}-1,1,0,...


1

Take a shell with outer radius $R$ and inner radius $r$. If the density of the ring material is $d$, The formula to calculate the mass of the shell is $$\frac{4}{3}d\pi(R^3-r^3)$$ The Schwarzschild radius of this shell is $$\frac{\frac{8}{3}dG\pi(R^3-r^3)}{c^2}$$ It follows that if this equals $R$, the shell should collapse into a black hole. Solving for $d$:...


Only top voted, non community-wiki answers of a minimum length are eligible