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2

There's an error in the notes you posted. The tortoise coordinate is usually defined via $$ \frac{dr}{dr_*} = 1 - \frac{2m}{r} = f \neq \frac{1}{f}. $$ Note that the correct definition is given in eq. (42) of your link. I suspect this will fix your problem.


1

What would happen to light passing through a narrow space between the event horizons of two equal-mass black holes? Would it deviate or follow a straight path? Like iharob and JohnnyMo1 said, the light goes straight. But something else happens to it. See this screenshot of Irwin Shapiro's seminal paper: See where he said the speed of light depends on ...


1

I'm going to assume you mean that the light travels on the precise center line between the holes, as iharob did. This sort of symmetry question is very common in physics. Here's a similar question in classical electrodynamics. "If I place a positive charge at the center of a perfect equilateral triangle of equal negative charges, will it move?" Let's say it ...


2

I think the answer is very simple if you ask another question, you are implying that both black holes generate the same gravity and that the light passes exactly through the middle between them, so the question is If the light deviates, where will it deviate to? Since given the conditions it's not possible to give an answer to this question, it means ...


0

@RandyWelt Note that, first, theta cannot be pi/2 for a this specific problem. There exist no $theta, J, M, r$ that solve the time dilation equation for $theta = \pi/2$. Even in my calculations, I was careful to take for example $\pi/2 < \theta \leq \pi$ or $0 \leq \theta < \pi/2$.Also, as an aside, for a Kerr metric, it is singular at points ...


0

I think there is some confusion on your part here. a = 0.998 represents the "speed" at which the black hole is rotating, which is taken to be by Kip to be 99.8% the speed of light, i.e., it is a = 0.998*c. The "a" in the calculations for the Kerr metric, is very different. It is a length scale. This can be seen for example by considering a spherical shell ...


0

$M=10^8M_\odot,$ so $r_s=$ $2GM/c^2=$ $2.95\times 10^{11}m.$ And $\alpha = J/Mc,$ has units of distance, and we can set it to $\alpha =0.998 r_s/2,$ which is very close to the critical value of $r_s/2.$ Now you can put $d\tau=$ 1 hour, $dt=$ 7 years, and $ \theta=90° $ into $$\left(\frac{d\tau}{d\ t}\right)^2=1-{\frac{2GMr ...


2

Really, to answer this carefully, we have to really think through what a horizon is. And for a general spacetime, there are several different notions of horizon, and "event horizon" is probably the most difficult of them to work with. The formal definition of "event horizon" says "Let's go to the distant future, take every freely-falling path that ...


1

There is already a good answer from John Rennie. Here we will just mention that if the spherically symmetric metric (2) is supposed to be a vacuum solution to Einstein field equations with $\Lambda=0$, then Birkhoff's theorem shows that the metric (2) [after a possible reparametrization of the time coordinate $t$] is exactly the Schwarzschild metric. See ...


11

Let's start with what we mean by a horizon: The event horizon of an asymptotically-flat spacetime is the boundary between those events from which a future-pointing null geodesic can reach future null infinity and those events from which no such geodesic exists. A null geodesic is the path followed by a light ray, so the horizon marks the surface at ...


-2

You are standing on surface of a planet of mass $M$ and radius $R$. With which velocity $v$ you need to throw away from the planet an object of mass $m$ that it will not return? The gravitational force is $F=G\frac{mM}{r^2}$. The work to be done to move the object in the gravitational field of the planet from distance R to infinity is $A = ...


-4

There is an amusing "trick" (so is not meant as an answer to your question) in Newtonian gravity (plus a bit of special relativity) to derive that the Schwarzschild radius should be "special" Since self-gravitational energy of massive bodies is always negative (at least in Newtonian gravity) $$ - \frac{G M^2}{R} $$ And massive bodies have positive rest ...


5

The geometry of spacetime is described by an equation called the metric. This is analogous to Pythagoras' theorem but with some key differences. Start with a 2d plane, where we identify positions of points by their $(x, y)$ coordinates. Suppose you move a distance $dx$ then a distance $dy$, then the distance from your starting point, $ds$, is given by ...


1

The best evidence for the black hole at the center of the Milky Way comes from the simple Keplerian motion of nearby stars. Using orbital data deduced by scientists, I made a simulation of their motion. Please note that this simulation needs a browser that supports WebGL.


0

Your questions are indeed related. First let's describe spacetime as like a stack of photos, the photos being different times. Real things are in some region of one of the photos and in a possibly different region of a different photo but don't jump discontinuously from one region to another in between two individual photos. Note this is an analogy and lots ...


1

What the author means is as follows. Consider the (un-normalized) vector field $\partial_r$ where $r$ is the radial coordinate; $\partial_r$ is thus just the vector field orthogonal to the level sets $r = \text{const.}$ or, equivalently, it is the vector field foliating said level sets. As an aside, note that Schwarzschild coordinates are perfectly valid ...


-1

He says that there is nothing funny going around for r < rs because both dt² and dr² term changes sign. I'm perfectly fine with this You shouldn't be. When you aren't happy with some conclusion you should go back over everything with a fine tooth comb. Examine your assumptions, check your postulates, look closely at the things you've taken for granted. ...


2

The Schwarzschild metric as you've written it is only one particular coordinate system and the fact that $r$ and $t$ switch roles at the evnt horizon is an artifact of that coordinate system. There are other coordinate systems which make certain properties of the metric more intuitive. The ones that might be most useful for you are ones which can be drawn as ...


1

If you plot a continuous curve in spacetime, it could be a path of a body if the tangent to the curve exists and has positive squared interval. General relativity is a geometrical theory, so everything is written in a geometrical way and the geometrical generalization is the predictions the theory makes. So outside the event horizon your curve has to have ...


2

You are right that the $(\Omega_H-\omega)^2$ term doesn't contribute. This is because it is the square of something that vanishes at the horizon: when you take the derivative, there remains a vanishing factor. As for the other term, since $\Delta$ vanishes at the horizon, this term vanishes except when the derivative hits $\Delta$. This yields the last ...


0

For trying to give a conclusive answer it seems necessary to rigorously characterize the geometry (the "conicidence structure", incl. the "light-cone structure") of the region under consideration (arguably with exception of "the singularity itself"). Unfortunately, this seems complicated (as may be gathered from efforts to tackle related problems at least ...


1

What you learned is correct. More simply, it's a consequence of the "time reversal symmetry" of most of fundamental physics. This symmetry is still present in general relativity. But, it's obscured by the standard system of coordinates. When you transform these coordinates into the Kruskal coordinate system, you not only have a black hole, you also have ...


-1

I learned in school (very long ago) that the path followed by light is independent of its orientation on this path. There's 2 things going on there. One is that our knowledge has changed since you went to school (or more correctly when the textbook author went to school), and the other is that these statements tend to hold true only at human timescales ...


8

I think the problem in understanding this is the idea of "space being sucked into a black hole." The reality is matter is "sucked" into a black hole. Space is warped around the black whole, but space is not "sucked" into anything. Here's the issue. What is space? You can't touch space (or better, the space-time continuum). So, one view is that space is ...


5

This is really Kip Thorne's interpretation of the black-hole collapse. Many other scientists and theorists would disagree and in many quantum gravities the interior of the black hole is not empty and the central singularity is somehow regularized. In any case, what you really feel pulling on you is not the mass very far away, what you really feel is the ...


1

I think you may be referring to this interview: http://www.space.com/17086-bizarre-black-holes-kip-thorne-interview.html. If not, possibly something of the same idea. He didn't say that the inside of a black hole is empty. He said that matter which falls into a black hole is destroyed. When it's destroyed, its mass attribute becomes an energy attribute, ...


3

Yes, a sufficiently massive neutron star can prevent collapse to a black hole, up to a point. The most massive neutron stars are thought to be spinning very rapidly, and they exceed the mass of the theoretical limit of a non-rotating neutron star. However, most models predict that a neutron star couldn't spin much faster than 1 kHz because they would break ...


0

No, you can't form a black hole by moving faster. The source term in general relativity includes mass-energy (rest energy) kinetic energy other energies and also momentum and stress. So when you just go faster your momentum increases too which affects the curvature too. What does the curvature look like? If you moved fast near a massive body the massive ...


0

Imagine putting some soap on the ring singularity then that whole disk is what is enclosed inside the event horizon and if you passed through that ring you'd end up in a strange unstable place where you can travel in time just by moving around. So its kind of nice that the whole thing is protected by an event horizon, two actually just to be on the safe ...


0

If we calculate the schwarzschild radius (and below is a really convenient link if you don't want to calculate yourself. http://physics.unl.edu/~klee/flash_astro/bhole_sim010.swf And if we look at the size of an atomic nucleus: 1.6 fm (1fm= $10^{−15}$ m=0.000000000000001 m) for a proton in light hydrogen to about 15 fm for the heaviest atoms, such as ...


3

It is worth stressing that black holes are predictions of classical General Relativity models. Our experimental data have established that the underlying level of nature is quantum mechanical. There is a large body of research on quantizing gravity and unifying the three forces studied with particle physics experiments with the gravitational force. String ...


1

The smallest possible black hole that could be observed would be one with a mass on the scale of the Planck mass (~ 22 µg) and a radius on the scale of the Planck length (really small). Thermodynamics makes it impractical to pack multiple particles into such a small space, so the best bet would be to accelerate elementary particles to have relativistic mass ...


0

In 2008 Ulf Leonhardt of University of St. Andrews UK led the first research group to make an artificial black hole. Its purpose was to look for experimental evidence of "Hawking Radition," photons and neutrinos emitted by black holes. "Hawking Radiation" couldn't be detected from astrophysical black holes because the signal would be overwhelmed by the CMB ...


3

The answer you your question is no. There is a radius, larger than the Schwarzschild radius at which a neutron star, quark matter, whatever its equation of state, cannot be supported against collapse. There are limits imposed by causality and General Relativity on the structure of compact stars. In "Black Holes, White Dwarfs and Neutron Stars" by Shapiro ...


1

I agree that you would not "notice" or "feel" anything as you "fall" into a black hole. As far as matter is concerned, the event horizon is not a line of demarcation were "strange" things happen on one side but not on the other. As far as you are concerned, all your molecules would be accelerated the same before and after crossing the event horizon, and ...


0

Classically a black hole made from collapsed matter would look (to people on the outside) exactly the same as a black hole made from collapsed anti-matter. They would absorb things the same and the curvature outside would be the same so everything would orbit it the same way and get pulled in the same way. What's interesting is the word black hole refers ...


1

I think I found a coordinate transformation that shines more light on this. Instead of looking for a conical singularity in $g_{tt}$ alone I must look at conical singularities in any part of the metric (apart from $g_{rr}$) First of all the metric can be expressed in terms of $r_\pm$ and Wick rotated $t\to it_E$ such that $$ ds_{E}^2 = \frac{ ...


-2

According to the relativistic doppler effect light shifts into blue or red depending on the relative velocity of object and observer. Thus, looking at a redshifted light ray which source is close to the event horizon of a SMBH it must be possible to shift its light back into blue just by accelerating the speed of the observer by the relative amount. ...


2

For a non-rotating uncharged black hole (the Schwarzschild metric) once you have crossed the event horizon there is no timelike trajectory that will increase the distance between you and the singularity. By this I mean that to increase your distance from the singularity would require you to be moving faster than light, which is of course impossible. So all ...


3

If we take the Milky Way as an example, the black hole at the centre, Sagittarius A$^*$, has a mass of about 4 million times the Sun. However the mass of the Milky Way is somewhere around a trillion Suns. So the central black hole makes up 0.0004% of the total mass. So even if our central black hole was pure anti-matter it wouldn't come close to accounting ...


4

This is largely the same answer as Rob's, though rather than use Eddington-Finkelstein coordinates I'm going to use Kruskal-Szekeres coordinates because I think this makes the argument easier to understand. This is what the situation looks like in Kruskal-Szekeres coordinates: For the non-nerd the Kruskal-Szekeres coordinates seem formidably complicated, ...


0

Therefore I'd like to ask a related question in which pings are plainly the main point... OK. I would refer you to Einstein talking about the speed of light varying with gravitational potential. And to Irwin Shapiro, who was involved in pinging radar signals to Venus and back, saying "the speed of a light wave depends on the strength of the gravitational ...


4

The answer must be closely related to my answer in Thought experiment - would you notice if you fell into a black hole? You can certainly use a similar Eddington-Finkelstein coordinate diagram to consider it (the E-F coordinates transform away the coordinate singularity at the event horizon). NB: This considers only GR and a non-rotating black hole (and ...


0

If you or or indeed any observer wants to make measurements of how some object moves we need to choose a coordinate system to do it. This isn't as complicated as it sounds. If I have a clock and a ruler then I can use this these to mark out a grid in space and time, then I can plot the trajectory of a falling object as a sequence of points, $(t, x, y, z)$, ...


2

The maximal analytic continuation of the Kerr metric extends the $r$ coordinate from the $r=0$ "ellipsoid" to negative values of $r$ which involves literally making another copy of your coordinates and switching from one to another when you pass through the disk $r=0.$ But there is another sense where there is another universe for the Kerr solution. As long ...


0

The ring singularity would be in the plane of rotation, therefore it would not form a toroid. As Horus mentions, the shape would be an oblate spheroid. How "flat" it is, depends on the angular speed of rotation. To find its volume, find its elliptical area and "rotate" it (integrate) around an axis.


1

There are a lot of review articles and books on these topics. But most of them require some knowledge of basic General Relativity and QFT. For Black hole thermodynamics I found the reviews by Jacobson and Ross very useful. On EE from vondensed matter (QFT) point of view there are papers by Cardy and others. But for those you need CFT background. Here is a ...


0

I may be missing your point here, and apologies if I am, but how can we contemplate the universe from outside it? Even if that ever moves from the, to me anyway, currently purely philosophical/opinion type question to a definite physics based, engineering type solution, (if ever) how do we then communicate between universes?     Once matter crosses the ...


1

I think your considerations are led by the daily experience that what you see actually happens as you see it. However, this is an illusion. All you perceive are photons, but they can (1) take their time and (2) can be redshifted or absorbed. Of course, the object quickly passes the event horizon (in its own frame it doesn't even notice the event horizon), ...



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