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This is a curious question. Many black holes are detected and identified due to light emitted from material falling into them. When black holes accrete matter, conservation of angular momentum would usually lead to the formation of an accretion disk. The release of gravitational potential energy as material falls means that this disk can be hot, and it is ...


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Does the gravitational force of a massive, spherical body draw only towards its center of gravity, and to no other area? Yes. This is exactly what Newton's shell theorem says, and also Gauss' law of gravitation. To help illustrate: if you stand beside Mount Everest, there should be a pull towards the mountain even if its too slight to notice. Not ...


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The shell theorem ensures that the gravitational force of any spherically symmetric mass distribution acts towards the centre of mass. However, if spherical symmetry is broken, as in the example you quote, then indeed there can be other components to the gravitational force. For instance you can use Wolfram Alpha (for example) to tell you at what angle to ...


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In order to avoid Christoffel symbols, use the fact that your expression is tensorial to evaluate it in a locally flat inertial frame of reference, where the metric reduces to Minkowski's. Then your equation is simply the time derivative of $u_\mu u^\mu = c^2$, since $u^\mu {u^\nu}_{;\mu} = u^\mu {u^\nu}_{,\mu} = \dot u^\nu$ in this case.


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The reason has to do with time dilation, and specifically, with the resulting red shift. A black hole forms from a collapsing star, which is of course made of brightly glowing matter. The event horizon forms in the centre and moves outwards while the star-matter falls towards it. Because of gravitational time dilation, the infalling matter never crosses the ...


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If the four velocity is normalized $$ U^aU_a=-1 $$ then taking the covariant derivative will give you $$ 2U^bU^a_{;b}U_a=0 $$ Note that the world line need not be a geodesic.


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The answer to your question is that nothing can travel faster than light, and light can't escape through the event horizon. Therefore gravitational waves can't escape either. I give an algebraic proof that light can't escape in my answer to Why is a black hole black?, and a more visual proof in my answer to Would the inside of a black hole be like a giant ...


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First, if you accept that gravitational waves can't travel at fast than the speed of light in regular space, then you can move to the inside of a black hole and then imagine letting the light and the gravitational wave race each other as you fall freely. As you fall freely then over a short time interval and a short distance everything looks normal to ...


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The time dilation is going to be utterly negligable. If I interpret the answer correctly the clock is at the centre of the disk. I would guess you are intended to treat the lowest point of the pendulum swing as exactly over the centre, so the pendulum will swing either side of centre. That means the force on the bob is the restoring force due to gravity ...


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According to classical mechanics the electrons moving outside an infinite solenoid do not feel the magnetic field. This is because the force they experience, according to the Lorentz law, depends only on the fields and not on the potentials. Thus according to classical mechanics the electrons beams passing from the different sides of the solenoid will move ...


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It's the Eddington-Finkelstein metric in the case where $d\phi = d\theta = 0$ i.e. a radial trajectory. The $t$ coordinate is actually the Eddington-Finkelstein $u$ coordinate.


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For example [...] bouncing a photon from [an object] and counting my proper time between firing the photon and receiving it. [...] $L_{[...]} = \frac{c~\tau_{Roundtrip}}{2}$ This would rather be called the "(momentary) chronometric separation" of the "object" under consideration from yourself; or, if you had found equal values of the ping duration ...


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In both SR and GR time is generally treated as a length by multiplying it by $c$, though we often set $c = 1$ so this isn't immediately obvious. So for example the Minkowski metric is: $$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$ and we multiply $dt$ by $c$ to get a quantity $cdt$ that has units of length and can therefore be sensibly added to $dx$ etc. ...


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Let's separate out some definitions: metric(1): Given a set $X$, a function $d : X \times X \to \mathbb{R}$ such that the following axioms hold for all $x,y,z \in X$: $d(x,y) \geq 0$, $d(x,y) = 0 \Leftrightarrow x = y$, $d(x,y) = d(y,x)$, and $d(x,z) \leq d(x,y) + d(y,z)$. pseudo-metric(1): Given a set $X$, a function $d : X \times X \to \mathbb{R}$ ...


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After a bit of research, the key term here is "the secular dynamics of Mercury". With that, you can easily find course notes that cover the whole calculation: https://farside.ph.utexas.edu/teaching/celestial/Celestial/node118.html It's frowned upon to give a link-only answer, but this is a big reference so I think it's appropriate.


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Yeah, you've not yet adapted. That's OK. Let me take you through it. In this conventional world of classical physics we have separate notions of distance and time, with the idea that either two events happen at the same time and therefore have an objective distance between them, or two events happen at different times and therefore have an objective time ...


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I think it might help to think about the spacetime interval $\text{d}s^2$ as a measure of movement in spacetime relative to the speed of light. Let's say that you want to move from a point $p=(0,0,0,0)$ to another point $p'=(t,x,0,0)$. The quantity $\text{d}s^2 = c^2\text{d}t^2-\text{d}x^2$ is then: Positive if $x<ct$, which means that you traversed the ...


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Einstein's equivalence principal states that an accelerated reference point is indecipherable from a reference frame in a gravitational field, so an accelerated reference frame will act in the same way that a gravitational field with the same acceleration would act. As for if all reference points are equally valid, the answer is generally "yes" with some ...


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These equations are often done in units where c=1 to make things easier, in this case: $$ c^2=1 $$


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It's helpful to look at an analogous situation in Classical Electrodynamics where the same issues come up. In General Relativity, the source term is the stress-energy tensor. In Classical Electrodynamics the source terms are the charge and the current. When there are no sources, one possible vacuum solution is a constant uniform nonzero electric field. ...


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If you have a metric, then you also have the manifold itself and all the points in it, and you can use the metric to compute the Einstein tensor at each point, and then multiply by a scalar constant to get the stress-energy tensor (assuming you know the value of your cosmological constant). But the reverse direction is very different. For instance, if you ...


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As you described, if you have a metric, then you also have the manifold itself and all the points in it, and can use the metric to compute the Einstein tensor at each point, and then multiply by a scalar constant to get the stress-energy tensor (assuming no cosmological constant). But the reverse direction is very different. For instance, if you started ...


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Yes, the no-hair theorems are asymptotic, and yes a frozen star or red hole approach has some accuracies associated with it. However, the issue of cosmic censorship is not well understood, and is still relevant to your situation. In particular, if a collapsing system forms a singularity that is not naked, then the event horizon could introduce long range ...


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You can also do the following, which may not be as general as you want it, but the idea might be usefull for other problems. You already know that the given metric is Minkowski metric in different coordinates, so look at the null geodesics. In the usual coordinates $(t',x')$ they are given by $x'\pm t'=const$. Then find the null geodesics in the given ...


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The problem with general relativity is it is too general and allows solutions that it would be very difficult to believe are physical. The first problem is that the metric is defined upon a topology. For example the physically extremely prohibitive condition that a spacetime is empty and with zero cosmological constant is not enough to uniquely define a ...


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In general the answer is no, but if $g_{12}=const, g_{11}=const$ and there is a cosmological constant then yes. To see this recall the Einstein field equations $$ R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}+\Lambda g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu}.$$ Now if the metric is constant then the Ricci tensor and Ricci curvature is trivial and we get $$ \Lambda ...


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In the general case you want the Cartan-Karlhede algorithm. It is an algorithm for producing a complete set of classifying invariants for a metric, expressed as functions of the coordinates. Given the components of the metric $g$ in the coordinates $x_1, x_2, \ldots$, the algorithm produces a list \begin{align} \Lambda & = \Lambda(x_i) \\ \Psi_k & = ...


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If you were to Wick rotate $t \rightarrow i \theta$, the metric would be $ds^2 = dr^2 + r^2 d\theta^2$, which is just flat space in polar coordinates. The standard cartesian coordinates can be obtained by $x=r\cos\theta$, $y=r\sin\theta$. The same procedure works in the original Lorentzian signature metric, but with hyperbolic trig functions instead of sines ...


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"Or is it more correct to say that space is (almost) flat because that is what occurs without any mass at all? In this case, the answer might be: space is almost flat because there is so little mass in the universe." This is correct if you limit yourself to the vicinity of the solar system, spacetime is relatively flat because there is relatively small mass ...


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No. You seem to be implying that the fact that the world-line of the satellite looks curvy, is what is meant by "curvature of space-time". No, that's wrong. The world-line of the satellite looks curvy in the picture, yes. But, it should be clear that you can have a curvy-looking world-line without gravity. Anything that orbits anything for any ...


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Consider a thin spherical dust shell. By the usual arguments, the gravitational field inside the shell is exactly zero -- this is valid both for GR and Newtonian gravity. Over time, the radius of the shell will decrease under the action of gravity until eventually the radius of the shell becomes smaller than its Schwarzschild radius. At that point, the shell ...


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One of the biggest surprises that General Relativity has given us is that under certain circumstances the theory predicts its own limitations. There are two physical situations where we expect for General Relativity to break down. The first is the gravitational collapse of certain massive stars when their nuclear fuel is spent. The second one is the far past ...


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Yes, it is totally possible. You need to use classical GR with continuous matter distributions and it requires energetic continuous matter with zero rest mass, and it is an unstable setup, so you have to do it 100% perfectly with absolutely no margin for any error. Let's see how. You can take a sphere of radius $R$ in Minkowski space, and take a ...


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I am going to answer the question 'When just considering GR without evaporation nor QM, is an empty (containing no matter) black hole possible?' and I will omit the 'anything' part, because is an ambiguous term. And the answer is yes, they are possible. As stated by another person here, Schwarzschild black holes and the rotating and charged versions are ...


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Strictly speaking the two well known uncharged black hole metrics (Schwarzschild and Kerr) are vacuum solutions. This means the stress-energy tensor is zero everywhere except at the singularity where it is undefined. This is what Alfred means when he says "A static (Schwarzschild) black hole 'contains' no matter". However this strikes me as a bit of a cheat ...


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OK,maybe I get where my mistake is. It's very important that $ T^{(1)}_{ab}=G^{(1)}_{ab}=0 $ but $ T_{ab}\not=0 $. From http://www.tapir.caltech.edu/~chirata/ph236/2011-12/lec14.pdf ,we can learn $$ T^{03}=-\frac{1}{16\pi}[(\frac{\partial h_{11}}{\partial t})^2+(\frac{\partial h_{12}}{\partial t})^2] $$ (c=1 and G=1.)Because $$ ...


0

As I understand it your question implies that the "hose" is at some region in $M^4$, this picture is allowed if you consider your $X\times M^4$ embedded in 6-dimensional space, however it doesn't make sense to say the topology of a loop (1 compact dimension) can turn from one loop into two... You can have local changes in the radius of this loop, this is ...


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The formula for gravitational time dilation1 is $$\frac{t_0}{t_f}=\sqrt{1-\frac{2GM}{rc^2}}$$ For a sphere, $$M=V \rho = \frac{4}{3} \pi r^3 \rho$$ So $$\frac{t_0}{t_f}=\sqrt{1-\frac{8G \pi r^2 \rho}{3c^2}}$$ So the greater the density, the greater the time dilation. Has any relation been observed or postulated to exist between the energy-density (or the ...


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We first note that the vanishing of the Ricci tensor does not imply the vanishing of the Riemann tensor. Thus the vacuum equations, $R_{\mu\nu}=0$, do not imply that spacetime is flat. The vacuum equations tell us that certain linear combination of components of the Riemann tensor vanishes. When solving differential equations, one usually has to worry ...


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The fact that the energy-momentum tensor is called the source of curvature doesn't mean that there can't be any curvature where there is no energy-momentum. In fact, even if $T_{\mu\nu}=0$ across all spacetime, there are still nontrivial solutions of Einstein's equations, in the form of gravitational waves. You should remember that $T_{\mu\nu}$ is a ...


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The equations $R_{\mu\nu} = 0$ alone don't define a well-posed problem since you need to add boundary conditions. Moreover the curvature of a manifold is measured by the Riemann tensor and $R_{\mu\nu}=0$ doesn't imply ${R^{\mu}}_{\nu\sigma\rho} = 0$.


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There are different measures of the strength of the gravitational field. One is the curvature, which is a nice local geometric invariant. However, the curvature only becomes "large" in very extreme situations, like near the singularity of a black hole. The curvature is actually not large near the event horizon (ie, you can't do local experiments to tell you ...


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The magnitude of the Ricci scalar places no restrictions on the Weyl tensor. After all, in the Schwarzschild metric the Ricci scalar is zero everywhere (except at the singularity where it's undefined). You may be able to link the Ricci scalar to the curvature in your specific case, but in general your assumption is not a safe one.


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No, you can't send information through an event horizon with gravitational waves. When a compact body contracts to form a black hole it leaves the vacuum spacetime outside the body in a curved manner. That's because certain kinds of curvature can persist on their own, even static ones,they are called vacuum solutions, and they are curved spacetimes that ...


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You can imagine setting up two "Michelson & Morley" experiments, one stationary and one in a moving vehicle. Relativity says you would get the same results, so which experiment is stationary with respect to the ether?


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Who's to say then, that we are not moving with the ether? Could we measure the difference? Yes that was the point of the experiment. They obviously didn't know which direction the ether was coming from so measured the difference through the year when we were at different positions around the sun and moving in a different direction. They also (IIRC) ...


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The story is the following. We start with the simplest Poincare-invariant action that does not depend on the parametrization $$ S=-m\int dl=-m\int\sqrt{-ds^2}$$ here $ds^2$ is the interval. We can rewrite it as $$S=-m\int\sqrt{dX^\mu dX^\nu \eta_{\mu\nu}}$$ here $\eta_{\mu\nu}$ is the Minkowski metric. Now if we suppose that $X^\mu$ depend only on $\tau$ we ...


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I don't remember having seen the specific expression of the proposed "signed arc length" either (anywhere but related to the OP question), nor anything resembling (1) the more abstract expression for determining the sought resemblance. For naming this proposed functional from the set of curves (or rather, arcs) into the set of real numbers (incl. $\mathbb ...


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The accretion of matter onto a compact object cannot take place at an unlimited rate. There is a negative feedback caused by radiation pressure. If a source has a luminosity $L$, then there is a maximum luminosity - the Eddington luminosity - which is where the radiation pressure balances the inward gravitational forces. The size of the Eddington ...


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A 12 billion Solar mass black hole sounds massive, but actually it's not all that big. The radius of the event horizon is given by: $$ r_s = \frac{GM}{c^2} $$ and for a 12 billion Solar mass black hole this works out to be about $1.8 \times 10^{13}$m. This seems big, but it's only about 0.002 light years. For comparison, the radius of the Milky way is ...



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