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I'm a proponent of the thought experiments that Einstein employed. Look at the experimental evidence and discard everyone's conclusions cause they just might be wrong. A singularity is an infinitely dense point in space. It creates a warping in spacetime that we call gravity. With no proof beyond my own thought experiments, it's my assertion that we are ...


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How can I ever fall into the black hole if by any onlookers perspective I never do? Because Oppenheimer's original frozen-star description is the one that's right. Have a read of The Formation and Growth of Black Holes on mathpages, where author Kevin Brown refers to two interpretations: "Historically the two most common conceptual models for general ...


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Accepted science on black holes say that nothing can escape a black hole, not even light. But there are some theories that hawking radiation on the boarder of the event horizon, is made by virtual antimatter matter particle pairs, where either matter or antimatter is swallowed by the black hole, while matter or antimatter escape the black hole as hawking ...


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This is paradoxical but not contradictory because yours and the onlooker's times flow differently, and once you fall under the event horizon there is no way to reconnect to compare times. The paradox only happens because of the implicit intuition about some "absolute time" that applies to both observers, and that you continue to exist under the horizon ...


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Good question. I think the event horizon has to be absolute, because as you suggested, light either gets out or it doesn't. I venture to suggest that isn't in accord with what most here would say is current teaching, but here's a couple of interesting facts: 1) Light is not redshifted when it ascends, and nor is it blueshifted when it descends. You can work ...


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The definition of the event horizon is `the boundary of the past of future null infinity', so it is the surface beyond which nothing can escape to infinity. It isn't defined with reference to any observer. A consequence of the definition is that an observer can never really determine where the event horizon is, since its location depends on all future ...


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Any metric is locally equivalent to a symmetric matrix, which is always diagonalizable. Just taking a reference system co-rotating with the black-hole and you will get a diagonal form to the Kerr metric.


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How efficient would be these energy extraction processes when applied to micro black holes? I fear the efficiency would be zero percent. See the Blandford–Znajek process on Wikipedia where you can read that the power can be estimated as the energy density at the speed of light cylinder times area: $$P=B^2\left(\frac{r}{r_c}\right)^4 r_c c=\frac{B^2 r^4 ...


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In light of the comments above, regarding the time energy relation, I would need to read more about the time energy relation and later rephrase this question. Normally I would delete the question, but it may be of value to other newbies like myself, making the same wrong assumption that the time energy relation is equivalent to the position momentum ...


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It depends. In an isolated system, they live immensely long. However, if (with their powerful gravity) collect enough mass, they will collapse into a black hole. The rate Hawking radiation (black holes radiating energy away) has to do with the mass of black holes. Small blacks will be gone quickly due to Hawking radiation. They, in a brilliant ...


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It is quite easy to make a solution with unequal masses. First start with a single Schwarzschild solution of mass m and extend it however you like for areal coordinate r<3m. Then take a Schwarzschild solution for a mass M in the range m3m and the solution for mass m in the region r<3m and sew them together at the region r=3m. Since M>m this requires ...


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The way the equations are presented seems unnecessarily obscure, as there are only two equations that matter: $$ \frac{d^2r}{dt^2} = \frac{-4M^2+2Mr+(r-5M)r^3}{r^3}\,\dot{\phi}^2 $$ $$ \frac{d^2\phi}{dt^2} = \frac{2(-3M+r)}{(2M-r)r} \, \dot{r}\dot{\phi} $$ These come from the geodesic equation expressed using coordinate time. So you start at some ...


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Assuming no energy input, the lifetime of a black hole is related to its mass by: $$ T = \frac{5120\pi G^2}{\hbar c^4}M^3 $$ There is a nice summary of the derivation of the lifetime on this web site. I make the condition assuming no energy input because for large black holes the Hawking temperature is less than the temperature of the cosmic microwave ...


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I can give you only a partial account because this is an area I'm only partially familar with. If you're feeling brave there is a rigorous analysis of the physics involved in this paper. The Reissner-Nordström geometry is perfectly stable as long as no extra matter is present, so in the conformal diagram you've included there is no instability. The trouble ...


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Why a black hole? Try earth, this very experiment has been done with precision atomic clocks, on at sea level and one in the mountains! When comparing the times shown on the clocks later, the one that was on the mountain advanced faster (cf. experimental confirmation of gravitational time dilation)! This effect even has to be considered for the precision ...


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First off, traveling at constant velocity in flat spacetime is not the same as traveling g in a uniform circular motion. Quite the contrary, free falling towards the gravitational source is actually equivalent to moving with constant velocity in flat spacetime. This is so because the objects are following a geodesic path defined by the geodesic equation. I ...


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The Schwarzschild metric describes the geometry of the spacetime containing a time independant spherically symmetric mass and nothing else. In other words the spacetime has to have existed unchanged for an infinite time and continue to exist unchanged for an infinite time, and there must be nothing else in the universe. Obviously there is no object in the ...


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By "derived the Schwarzschild metric" I assume you mean calculating the exterior solution of the form $$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 +r^2\Omega^2$$ Applications: Describing deflection of light by the sun Precession of the perihelia of the orbits of the inner planets Schwarzschild singularity and ...


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This question is rather unphysical. Problem one: The speed of sound in the cable cannot exceed $c$, therefore its strength is limited by the very principles of relativistic physics (or it has to get denser, but then it will at some point have sufficient energy density to collapse to a black hole itself). (This argument can be made sharp, I think it can be ...


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Black holes need to conserve the angular momentum of all the matter that collapses into them from a spinning star. As the radius of a black hole is considerably less than the radius of the star, it needs a really fast rate of rotation to conserve the angular momentum of all that matter collapsed into a much smaller radius. Here is an account of the ...


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I think this article by Greg Egan is very helpful. The thing to remember is that you will not be "pushing" the stick towards the event horizon. The black hole will be pulling very hard on it. The force gradient is small, but the force itself (interpreted as the thrust that would be required to hover) is very large. Before the stick reaches the event horizon, ...


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Now if I want to show that some other space time is asymptotically AdS, I should expect the metric to agree in the limit $\rho \to -\infty$ which is the limit $z\to 0$. Is that right? Yes, this is correct. In this limit the leadin order of the metric components should agree with the pure AdS metric. There is however a more mathematically sound way of ...


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The reason why it is often stressed that Hawking radiation is in a pure state, is that this is in apparent contradiction to the fact that Hawking radiation is also said to be thermal. The apparent contradiction is solved when one realizes that in a general curved spacetime there is no unique definition of the vacuum state and therefore the whole Hilbert ...


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If one assumes no other matter is supplied to the black hole (which is difficult to describe in a general manner as it depends on the details of the environment), the question if black holes evaporate depends on the difference between the emitted Hawking radiation and the absorbed cosmic microwave background radiation. According to the Stefan-Boltzmann law ...


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You seem to be asking for a distance above the event horizon at which time dilation due to the gravitational acceleration of the black hole would convert 4 billion years to 1 second. As all time is relative, what you really are asking for is the distance above the event horizon at which it would APPEAR to someone on Earth that an event which takes 4 billion ...


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A few things: 1) time dilation is only a relative effect. A clock on earth would tick differently than one far from the milky way, but people on Earth would notice no abnormal effect. 2) To first order, time dialation effects are governed by the gravitational potential energy at a point. It turns out that the time dilation due to the mass of the ordinary ...


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I think you can't do it with this data. because time dilation of our so-called "UNIVERSAL TIME" is lagging behind w.r.to a body in deep void because of mainly due to earth's gravitation, not sun or the center black hole of milky way proof: $$R=\frac{CT}{2\pi}\sqrt[2]{\frac{\Delta t}{t}}$$ where $T$ : time period of revolution of milky way. ...


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how the hell does relativity justify that the character can be thrown in black hole and survive ? This is a big question and scientists (physicists) are asking the question that "what will be the fate of an astronaut falling into a black hole?" ,since the concept of general relativistic black holes have arisen. The answer is the astronaut may not ...


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The Kruskal-Szekeres coordinates have the reputation of being hard to understand, and to some degree this is justified as they are unintuitive compared to the Schwarzschild coordinates. However their huge advantage is that many phenomena can be explained simply by looking at the illustration of a maximally extended black hole in the KS coordinates: Region ...


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There's no "inside" a black hole. The closest thing a black hole has is a horizon. And you can actually get out a part of what falls below the horizon - just not more than 50% (in theory). I'm not sure about the "transverse" part, that's not a term I've encountered in relation to black holes. Are you sure you got the spelling right? There are multiple ...


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I) For a single$^1$ spherically symmetric thin-shell collapse with (rest) mass $m$, the geometry inside the shell is flat Minkowski space, cf. e.g. this Phys.SE post; and outside the shell the geometry is Schwarzschild geometry with Schwarzschild radius $R_s=2E$ in natural units, where $$E~=~ m\sqrt{1+\left(\frac{dR}{d\tau}\right)^2} - \frac{m^2}{2R} ...


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I forgot I had asked this a while back, but I thought I would provide an answer since I have many references now. I realize that when I first asked this question I may have sounded a bit naive and I hadn't put much time into searching for references. Anyway, here are some references (old) related to no-hair theorems. Older references: 1967, 1968, 1971, ...


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From this paper$^1$ we have the equations for a particle of zero total energy on an infalling trajectory in the equatorial plane: $$\begin{align} \Sigma\frac{d\theta}{d\tau} &= 0 \\ \Sigma\frac{dr}{d\tau} &= -\sqrt{2Mr(r^2 + a^2)} \\ \Sigma\frac{dt}{d\tau} &= -a^2\sin^2\theta + \frac{(r^2 + a^2)^2}{r^2-2Mr+a^2} \\ \Sigma\frac{d\phi}{d\tau} ...


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... is this what is responsible for Radioactive Decay? Sort of ... radioactive decay occurs when the final state (result of the decay) has lower energy than the initial nucleus and the barrier is occasionally surmounted. This barrier penetration and the virtual pair at the black hole horizon are both quantum processes, so they have a little physics in ...


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This question touches some very interesting conceptual questions in AdS/CFT that are often omitted in presentations of the holographic dictionary. So what is the usual story of the holographic dictionary (considering a free scalar first in the Euclidean setting for simplicity first) There is a bulk scalar field with an equation of motion ...


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Does it make sense to talk in General Relativity about only 1 body? Yes. But note that this body is a body in space. The energy tied up in the matter of a star "conditions" the surrounding space, altering its metrical properties. You can see Einstein talking about this sort of thing here: "According to this theory the metrical qualities of the continuum of ...


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It's important to be clear that general relativity is a mathematical model intended to predict the results of experiments. Different physicists will have different views about how closely GR corresponds to whatever reality means, but when asking questions about it you need to remember that it is just a model. Now, you ask: Does it make sense to talk in ...


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Just the principle of a universe consisting of only 1 black hole doesn't make sense. You will quickly end up with more than 1 body via Hawking radiation. The 1 body problem may help one in understanding self energy contributions of general relativity, but not anything that would probably be observable. However, I have not looked deeply into the case of the 1 ...


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The swarzchild radius is the distance a spherical mass has to be shrunk before it becomes a black hole. This is r = (2G/c²) m. 2G/c² is approximately equal to 1.48×10⁻²⁷ m/kg Estimates for the mass of the neutrino vary somewhat. One estimate is from around 0.2 eV to 2 eV. Picking 1eV for simplicity, the swarzchild radius is approximately 1.48×10⁻²⁷ m/kg ...


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Let's say you are orbiting a black hole, and some other black hole comes in 'you - your black hole' system, your will for sure notice it: 1: If the incoming black hole is of mass comparable to your black hole, the system now have a new entity in it which for sure will change your trajectory significantly around original black hole. 2: As they merge, the ...


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Photon "spheres" of Kerr metric are not yet found. That needs a lot of work. Two closed photon circles are known.


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Please read our comprehensive paper on photon spheres published in Journal of Mathematical Physics entitled The Geometry of Photon Surfaces, which is available on the Arxiv. For related areas you might like to look at this publication list.


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I dont' see much discussion on the linked article about the photon sphere being a true sphere -- they talk about the photon orbit at $r=3M$, but this is only valid for an equatorial orbit (and I'd expect that it would depend on the value of $a$ and you'd have a different radius for corotation and anti-rotation), and when you go away from the equatorial ...


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The matter doesn't disappear - it still exists, but we can't access it without entering the black hole ourselves, and then we'd be trapped. Due to Hawking radiation, over time (assuming the black hole is small enough) all of the energy that went into the black hole will be released again as the black hole evaporates. The bigger conservation issue at stake ...


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Yes, the force indeed is proportional to the mass. (Since $F_g=\frac{Gm_1m_2}{r^2}$) ( $m_1, m_2$ are masses of the bodies.) But this doesn't really matter, because say, for $m_1$, $$m_1\vec a=\frac{Gm_1m_2}{r^2}$$ If you notice, the acceleration of the body is independent of its mass, like you mentioned. Why do you think this will be any different for a ...


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So as we from the outside of the black hole observe matter approach the event horizon, we will see time slow for the falling matter. Conversely, an observer falling into a black hole would see external events speed up (possibly, not sure about light travelling from external objects). At the event horizon objects will appear stationary for us, frozen in ...


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The singularities are in a sense mere artefacts of your coordinate system, but do describe important surfaces. The reason your metric was ill defined up on those surfaces is that you picked a metric that was nice when you are far far from the black hole (all those $1/r$ terms get small and it looks like the SR metric for a spherical coordinate system). ...


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What local forces cause the damage? Tidal forces. Tidal forces can only be neglected in extremely small regions, and you have an extended body. More details follow. You suppose an extended body, bound together, interacting gravitationally with a black hole. With the center of mass staying outside the photon sphere, moving on a more than barely unbounded ...


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I think that calculating the Riemann tensor manually is not particularly illuminating, but if you really want to do it, then why ask for help from us and not from a book? It's quite probable that any advice we may give you comes from a book anyway. Having said that, the most powerful tensor manipulation package for mathematica is xAct. It requires some solid ...


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There is a relatively fast approach to computing the Riemann tensor, Ricci tensor and Ricci scalar given a metric tensor known as the Cartan method or method of moving frames. Given a line element, $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu$$ you pick an orthonormal basis $e^a = e^a_\mu dx^\mu$ such that $ds^2 = \eta_{ab}e^a e^b$. The first Cartan structure ...



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