The Schwarzschild radius involves an expression in terms of Newton's constant $G$, the mass $M$ inside a radius $r$, and the speed of light squared $c^2$. Current estimates of the universe's matter density are about six protons per cubic meter.  But, the $M$ inside a sphere goes up as $r^3$, while the "time curvature" coefficient is $1 - \frac{2GM}{c^2\, r}$.  So this coefficient is bound to hit zero for $r$ large enough.  The $M$ outruns the denominator as a function of $r$.  I calculated that this coefficient hits zero when $r$ equals 13.54 billion light years.  Question:  Is this any evidence for our universe being one very large black hole?
2687
I was surprised to only recently notice that      An object of any density can be large enough to fall within its own   Schwarzschild radius.   Of course! It turns out that supermassive black holes at galactic centers can have an average density of less than water's.  Somehow I always operated under the assumption that black holes of any size had to be superdense objects by everyday standards.  Compare the Earth to collapsing into a mere 9mm marble retaining the same mass, in order for the escape velocity at the surface to finally reach that of light.  Or Mt. Everest packed into one nanometer.  Reading on about this , it increases proportionally with total mass.  Assuming matter is accumulated at a steady density into a spherical volume, the volume's radius will only "grow" at a cube root of the total volume and be quickly outpaced by its own gravitational radius.  Question: For an object the , what would have to be its diameter for it to qualify as a black hole (from an external point of view)?  Would this not imply by definition that:   The Earth, Solar system and Milky Way are conceivably inside this black hole? Black holes can be nested/be contained within larger ones? Whether something is a black hole or not is actually a matter of perspective/where the observer is, inside or outside?
111772
This question appeared quite a time ago and was inspired, of course, by all the fuss around "LHC will destroy the Earth".   Consider a small black hole, that is somehow got inside the Earth. Under "small" I mean small enough to not to destroy Earth instantaneously, but large enough to not to evaporate due to the . I need this because I want the black hole to "consume" the Earth. I think reasonable values for the mass would be $10^{15} - 10^{20}$ kilograms.   Also let us suppose that the black hole is at rest relative to the Earth.  The question is: How can one estimate the speed at which the matter would be consumed by the black hole in this circumstances?
17632
There is something I don't understand about the  $H$, as it seems to clump two concepts together that I can't quite unify in my head. On the one side, we have  $$V = D H$$  which means that for a given distance $D$, there is a certain amount of new space created over time - and $H$ is simply the factor that makes this relationship work. So, for example, say we have two points 1 Mpc apart, this would mean they recede at about 70 km/s away from each other  (given our current approximation of $H$).  Now the thing I can't wrap my head around is that   $$T =\frac{1}{ H}$$  is also the age of the universe. Contrary to claims made on, say, Wikipedia this means that $H$ cannot possibly have been a constant throughout the past 13 billion years, because mathematically $1/H$ means that $H$ must be continually shrinking as the universe ages.  So if $H$ did start out as some huge value and is now shrinking over time, doesn't this mean the expansion of the universe is slowing down? Because if $H$ is shrinking, I'll get a lower value of $V$ today than I'll get tomorrow. Shouldn't the notation then be more like  $$V = DH(t)$$  So which one is it? If $1/H$ is simply the solution for $D=0$, how can we use it as the expansion-velocity-per-unit-of-distance at the same time? What's worse, how can literature say $H$ has probably been more or less constant forever and simultaneously assert that $1/H$ is the current age of the universe? What am I missing?
66497
According to NASA      a black hole is anything but empty space. Rather, it is a great amount of matter packed into a very small area   According to the documentary  about Black Holes     Gravity here is so strong even light cannot escape   Which leads me to believe that Black Holes basically are masses so compact but great which gives them very strong gravitational influence (Newton's Gravitational Law) that light is not allowed to go out of, hence a "Black" hole.  If my above understanding of Black Holes is correct, then why is those Spiral Galaxies pictured using Hubble Telescope is showing a big shiny ball instead of a "Black" hole ? What is the obvious clue that I am missing here ?  Here is the image of Andromeda Galaxy image taken using Hubble.
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I am reading "The Essential Cosmic Perspective" by Jeffrey O. Bennett, Megan O. Donahue, Nicholas Schneider, Mark Voit. In Chapter 14, it is stated that an evidence of the presence of dark matter in our galaxy is that the rotational curve does not match that obtained from calculation. In the calculation, the following formula is used $$M_r=\frac{rv^2}{G}$$ where $M_r$ is the mass enclosed inside a radius $r$ from the galactic center.  My understanding is that the above equation is a result of the shell theorem. But we know that shell theorem applies to spherically symmetric mass distribution only, and most galaxies are disks. So why can we still do this?
41970
The Bekenstein entropy for a black hole is proportional to the surface area $A$ of the black hole $$ S_{BH} = \frac{k_B}{4 l_P^2} A $$ with the Planck length $l_P = \sqrt{\frac{\hbar G}{c^3}}$.  The area is the surface of a sphere with Schwarzschild radius $r_s = \frac{2 M G}{c^2}$, so  $$ A = 4 \pi r_s^2 = 16\pi \left(\frac{G}{c^2}\right)^2 M^2 $$ and the black hole entropy is therefore proportional to the mass of the black hole $M$ squared: $$ S_{BH} = \frac{4 \pi k_B G}{\hbar c} M^2. $$ But this quite unusual for an entropy. In classical thermodynamics entropy is always supposed to be an extensive quantity, so $S\sim M$. But the black hole entropy $S_{BH} \sim M^2$ is obviously a non-extensive quantity. Isn't a non-extensive entropy inconsistent within the framework of thermodynamics? Why is it, that the entropy of a black hole must be a non-extensive quantity? Shouldn't we better define an entropy for a black hole from e.g. the ratio of Schwarzschild radius to Planck length, which would give us an extensive entropy like $$ S_{BH, ext} \sim k_B \frac{r_s}{l_P} \sim k_B\sqrt{\frac{4G}{\hbar c}} M  $$
36033
I've been looking at the  paper, trying to update my toy cosmology simulator with the most recent data.  Most of the interesting values such as $H_0$, $\Omega_m$, and $\Omega_\Lambda$ can be found in Table 2 on page 12, but the one thing I didn't find was an estimate of the energy density of radiation.  Can this be derived from some other parameters in these data?
2463
I understand that a black hole bends the fabric of space time to a point that no object can escape.   I understand that light travels in a straight line along spacetime unless distorted by gravity. If spacetime is being curved by gravity then light should follow that bend in spacetime.  In Newton's Law of Universal Gravitation, the mass of both objects must be entered, but photon has no mass, why should a massless photon  be affected by gravity in by Newton's equations?  What am I missing?
118540
It is well-known that measurements of the velocity profile in galaxies are not compatible with Newtonian laws. A way to circumvent the problem is to assume that galaxies are surrounded by a spherical halo of Dark Matter. The density of mass of this Dark Matter can then be computed from the velocity profile with Newton's laws.   My question is the following: the formation of a galaxy involves essentially only the gravitational interaction. According to Newton (and General Relativity too), the trajectory of a particle does not depend on its mass. As a consequence, I do not understand how usual matter and Dark Matter could have followed different trajectories leading to such different mass distributions nowadays. Can somebody explain this (apparent) paradox ?
84731
Why does there have to be a singularity in a black hole, and not just a very dense lump of matter of finite size? If there's any such thing as granularity of space, couldn't the "singularity" be just the smallest possible size?
111399
I'm writing this question since I haven't find exhaustive enough answers searching on the Net. I'm writing an application for simulating General Relativity orbits calculating the acceleration from Christoffel's symbols, which are already derived from the Schwarzschild metric in isotropic coordinates:  $$\frac{d^2x^\alpha}{d\tau^2}=-\Gamma{^\alpha}{_\mu}{_\nu} \frac{dx^\nu}{d\tau} \frac{dx^\mu}{d\tau} \hspace{7.1em} (1)$$  My problem is: how should $\frac{dx^0}{d\tau}$ be defined from arbitrary spacial components of the four-velocity?  $$u^i=\frac{dx^i}{d\tau}=\text{arbitrary quantity} \hspace{4.4em} (2)$$  In special relativity, the time component of four-velocity, in terms of the spacial components, is:  $$u^0=\frac{dx^0}{d\tau}=\frac{cdt}{d\tau}=\sqrt{c^2+|u^i|^2} \hspace{4em} (3)$$  where $|u^i|^2 = (u^1)^2+(u^2)^2+(u^3)^2$ is the magnitude squared of the spacial components of the four-velocity. This comes from the Minkowskian metric. Then I derived a similar equation from the Schwarzschild metric in isotropic coordinates:  $$u^0=\frac{cdt}{d\tau}=\sqrt{\frac{B}{A}(c^2+B^2|u^i|^2)} \hspace{4em} (4)$$  where $A=(1-\frac{GM}{2R})^2$ and $B=(1+\frac{GM}{2R})^2$. If this is correct, it's quite straightforward to find that the latter definition gives a greater value than the one from special relativity, since $B &gt; 1$ and $B &gt; A$. Does this mean that the four-velocity magnitude is not constant in General Relativity, as it depends also on radial distance? (I knew that the four-velocity always equals the speed of light)
64333
Let's do this thought experiment. You have an insanely powerful rocket and it can accelerate to 0.999999c. Now you fly to a supermassive black hole and hover just above its event horizon, where the inner gravitational pull towards the singularity matches the outward thrust of your rocket.  Consider: 1. The black hole is non spinning, and doesn't distort space-time into swirling around it.   The black hole is so massive that above the event horizon, you are not sphagettified. The rocket can shield the energy of extreme blueshifted photons that bombard you.    At this point over the event horizon, you would see like 1 million or even 1 billion years of the universe in 1 second of your life, right? Can you witness the end of the universe if you hung around for long enough?  And what if the black hole evaporates by Hawking Radiation before that? Will you immediately return to real time?  PS: This is quite different from falling into a black hole. In this thought experiment, we are not crossing the Event Horizon of the black hole in a split second, but we are physically hanging in there for a long time by Counter Force.  Edit: I'm not sure why this question has been marked as a duplicate, but this differs vastly from an observer falling into a black hole, and nowhere else has it been answered so perfectly and succinctly, as done by Mr. Bob. Many thanks to him.
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When solving the Einstein field equations in Schwarzschild metric for an observer falling into a black hole the radial coordinate r of the black hole and time t switch roles in the equations when r&lt;2M.    If we transform into the resting coordinate system of an observer inside a black hole, the timelike geodesics will be along the radial dimension of the black hole. Would an observer inside the event horizon of a spherically symmetric black hole observer the radial dimension of the black hole as time? If yes, is it safe to assume that the laws of thermodynamics would hold inside the black hole, in which case the singularity of the black hole would as a zero entropy state be in the past along the radial "time" axis and the high entropy event horizon would be in the future along the same?  What would the cosmology of a spherically symmetric black hole look like from the perspective of an observer inside the black hole. It seems to me that from the perspective of an observer within the event horizon of the black hole:   The observable universe originates from a singularity (black hole singularity) The observable universe expands along the radial (time) dimension The exterior of the black hole is not observable from within the black hole There would be future boundary conditions defining the faith of the interior (event horizon) The interior in other than radial dimension would be relatively uniform for a static black hole   How does time behave inside a black hole from the perspective of an observer inside the black hole? Could such an observer see the interior of the black hole as a universe relatively similar to ours (assuming the arrow of time would be along the radial axis of the black hole).
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According to the Wiki on the Rs, the Rs of the observable universe is 13.7BLY.  (The observable universe's mass has a Schwarzschild radius of approximately 13.7 billion light-years.[7][8])  The reference for this statement is:   and the Encyclopedia of Distances  Can someone please explain this to me...  Is this simply because to get into the non-observable portion of the universe, you have to go faster than the speed of light?
94190
Note: Although I have a provided an "answer" to the question, I did not resolve all the questions in this post satisfactorily. I invite anyone willing and able to provide a better answer, which I would be glad to accept instead.     Consider the   $$ds^2=-f(r)dt^2+f^{-1}(r)dr^2-r^2d\Omega^2\hspace{2cm} f(r)=1-\frac{2M}{r}+\frac{M^2q^2}{r^2}$$ where I defined the charge-to-mass ratio $q:=Q/M$, which satisfies $|q|\leq 1$ to avoid naked singularities. As is well known, there are two horizons (or only a single one when $|q|=1$) at $r_\pm=M(1\pm\sqrt{1-q^2})$.  I would like to understand the thermodynamic properties of this black hole solution. The temperature is given by $$ T=\frac{r_+-r_-}{4\pi r_+^2}=\frac{1}{2\pi M}\frac{\sqrt{1-q^2}}{(1+\sqrt{1-q^2})^2}$$ I calculated the heat capacity  $$ C:=T\bigg(\frac{\partial S}{\partial T}\bigg)_Q$$ by using $S=\pi r_+^2$ and a bunch of applications of the chain rule one obtains $$ C=2\pi M^2\frac{\sqrt{1-q^2}(1+\sqrt{1-q^2})^2}{1-2\sqrt{1-q^2}}$$ which has some curious behavior that I would like to understand physically. I've plotted $\frac{C}{2\pi M^2}$ as a function of $q$, and the result is shown below.   $\hspace{4cm}$  For small $q$, the heat capacity is negative (this is the expected behavior, just like in Schwarzschild), and as one adds charge, the heat capacity steadily decreases. However, as one can easily see, the denominator has zeros at $|q|=\frac{\sqrt 3}{2}$, and therefore the heat capacity diverges (!) to $-\infty$ at these values, "emerging from $+\infty$" for larger values of $ | q |$. In, once again, strictly decreasing fashion. What I would like to understand is:   "Adding charge" always seems to decrease $C$. Is there a clear reason for this? Physically, should one expect a diverging heat capacity at a certain charge (for fixed mass) at all? Why is it located at $q=\pm\frac{\sqrt 3}{2}$?
31386
It is said that if the Earth were a black hole, it would be the size of a peanut!?  How is this density possible, are atoms really that sparse that they can be compressed so tightly? Is there some other physical dimension that the matter gets sucked into?
90380
If a large star goes supernova, but not enough mass collapses to form a black hole, it often forms a neutron star. My understanding is that this is the densest object that can exist because of the Pauli exclusion principle: It's made entirely of degenerate matter, each particle of which cannot occupy the same quantum state of any other.  So these objects are so massive that they gravitationally lens light. If you make them more massive, they bend the light more. Keep going and going until they bend the light so much that light passing near the surface can barely escape. It's still a neutron star. Add a bit more mass, just enough that light passing just over the surface cannot escape. Now it's a black hole with an event horizon (I think?). Does this mean the neutron star has become a singularity? Isn't it still just a neutron star just beneath the event horizon?  Why are black holes treated as having a singularity instead of just an incredibly massive neutron star at its center? Does something happen when an event horizon is "created?"