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1

Ok, so I've finally got to the bottom of this. Basically, the confusion stems from what it means to say the universe is accelerating so let's clear that up. The Hubble parameter is defined to be $H(t)=\dot a(t) /a(t)$ where $a(t)$ is the scale factor. When we say that the universe's expansion is accelerating, we mean that $\ddot a$ is greater than zero. This ...


-1

I would imagine it would be a solid. Gas = atoms that are connected loosely fluid = connected more than gas but still loose solid = compact and held together I would imagine that since the gravity well and escape velocity being so high that not even light can escape, that it has traits of a very dense solid. Hence Singularity would be a very compact solid, ...


4

There are two issues. One issue is that some people try to oversell the "no hair" theorem to areas beyond where it applies, they apply a long term analysis of final states as if it applies in the short term. The second issue is that in the shirt term we have two objects, the black hole and the incoming object. We have to think about both. I'll go into ...


0

The curvature of spacetime is determined by the stress-energy tensor, and in the stress-energy tensor we do not distinguish between matter and energy. The two are treated as equivalent and interconverted using Einstein's famous equation $E = mc^2$. The Schwarzschild radius of a mass is conventionally written as: $$ r_s = \frac{2Gm}{c^2} \tag{1} $$ where ...


-4

Is a black hole's mass uniformly distributed? One hears conflicting answers to this. One article I rather like is the mathspages Formation and Growth of Black Holes. See this bit: "Historically the two most common conceptual models for general relativity have been the 'geometric interpretation' (as originally conceived by Einstein) and the 'field ...


-1

In the mathematical models of black holes that we use, there is a parameter M which is related to the mass-energy of the black hole. There is NO hint of distribution of mass-energy inside the black hole. In particular: a non-rotanting BH is spherically symmetric (these BH probably do not exist in nature, since all astronomical objects rotate and pick up ...


2

A non-rotating black hole can be treated as spherically symmetric using the Schwarzschild metric. A rotating black hole has an axis of symmetry and can be represented with the Kerr metric. Treatments of black holes using either of these would make the assumption that the "test particle" you are considering does not influence the metric (is much less ...


1

So lets start with the first part of your question: Black holes radiate away by the famous Hawking process. Hawking radiation has been interpreted in many ways i.e. as pair creation near the black hole, tunnelling from the black hole and almost every other physicist will have a nice way of explaining this. What is the temperature of a black hole? It ...


3

Actually in most cases you don't see the event horizon, but instead the photon sphere. For example if you are looking from some distance, if light emitted from some star goes inside the photon sphere (where light can travel theoretically in circular orbit, though the orbit is unstable), which is located outside the event horizon, it is more or less doomed to ...


2

Calculating what you would see as you fell into a black hole is straightforward but tedious. Fortunately there are lots of sites that have done this for you. Actually, if you've been to the cinema recently the film Interstellar does a pretty good job of it. Less spectacularly, have a look at this site that has videos of what the journey would look like. ...


0

The event horizon or 'point of no return' is the point where the escape velocity of an object falling into the black hole has to be greater than the speed of light (about 299,792,458 m/s). That concludes that light can't escape it, therefore there is a sharp, clear distinction between the event horizon itself and the background. If you moved towards the ...


0

In general relativity, an event horizon is a boundary in spacetime beyond which events cannot affect an outside observer, i.e. any events separated by an event horizon are space-like. An other way to say this is that any worldline with a start within the event horizon will never cross the boundary of the event horizon. If you are familiar with the light ...


-1

What you're missing is the second paragraph here: That's Einstein saying light curves because the speed of light is not constant$^*$. The force of gravity at some location is related to the gradient in the speed of light at that location. See Baez: "This difference in speeds is precisely that referred to above by ceiling and floor observers". Raise your ...


11

It is a truly unfortunate coincidence that the "escape velocity" at the Schwarzschild radius $R_\mathrm{S}$ is $c$. This leads people to think Newtonian mechanics can predict black holes. It can't. Note that a Newtonian escape velocity is the minimum velocity needed to coast to infinity against gravity. If you have less speed, you still travel away from the ...


3

A particle cannot survive in region II indefinitely. Indeed, there does exist an upper bound on the survival time of the particle. As soon as the particle crosses the horizon into region II, its remaining proper time until it reaches $r=0$ is bounded by $$\tau_\mathrm{max}=\frac{\pi GM}{c^3}$$ We shall now show this. Since Schwarzschild spacetime possesses ...


2

It doesn't just rotate around the accretion disk. As to what actually happens to it isn't clear-cut. Most people will say the infalling matter goes right through the event horizon all the way to the central point-singularity in finite proper time. But check out The Formation and Growth of Black Holes on mathspages: "Historically the two most common ...


3

Matter in the accretion disk is not (yet) inside the black hole. It is orbiting, and it can even escape. In fact, some of it must escape for accretion to happen at all: The disk has too much angular momentum to be accreted, and so some material must be ejected, carrying off excess angular momentum, in order for the rest to fall into the hole. The general ...


3

This problem is dealt with (in the context of classical General Relativity) nicely in Taylor & Wheeler's book "Exploring black holes: An introduction to General Relativity" (2000, Addison, Wesley, Longman). In the section entitled "Project B: Inside the black hole" they perform a calculation for a free-falling observer, based on the Schwarzschild metric ...


0

Answer to question Version 1: No one knows. We can answer this question using general relativity to give a classical description but I think there is now serious doubt that GTR describes the inside of a black hole (i.e. within the event horizon) accurately and that we shall need a full quantum theory of gravity to know what happens there. But the classical ...


2

Taking this as a matter of Fermi estimation, I will take the Newtonian form of gravity. No, this isn't great accuracy, but if anyone has any severe theoretical issues to raise, I will be glad to hear them. I will assume that your body extends 1 m out from its center of mass and that the extremities there will experience 10 g before your fingernails bleed and ...


7

We don't know. Measuring effects of gravity at the microscopic scale is currently out of reach, since gravity is so weak compared to the strengths of all other interactions. This makes getting a clear signal so difficult. We do not have anything near a testable theory of quantum gravity, so all answers must remain speculation. Some well-known physicists do ...



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