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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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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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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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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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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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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} ...


2

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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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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The short answer is that calculating the Riemann Tensor is a grind. It will take a while, no matter what way you do it. Presumably you're doing the Schwarzschild metric in the standard (Schwarzschild) coordinates, so you're aided by the fact that the metric tensor is diagonal. This means that $R^\alpha_{\beta \gamma \delta} = g^{\alpha \alpha}R_{\alpha ...


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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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