Active questions tagged kerr-metric - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-21T18:16:10Z https://physics.stackexchange.com/feeds/tag/kerr-metric http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/444650 4 Kerr Black hole EH and Ergosphere embedding Alessandro Rovetta https://physics.stackexchange.com/users/193994 2018-12-02T09:29:27Z 2019-08-07T22:08:49Z <p>Goodmorning everyone. I would like to share with you a question that has been gripping me for some time, but which I have never been able to give a convincing answer. When representing the ergosphere or the external event horizon of a black hole, it is often not taken into account that the coordinates used (if space-time is Kerr, the most usual are those of Boyer-Lindquist) have no physical meaning, in the sense that they do not allow us to "see" what the real form of such spatial hypersurfaces would be if they could be "spied" from the earth.</p> <p>Now, I tried to formulate the embedding, so that the line element of the metric was the Euclidean one IE <span class="math-container">$$ds ^ 2 = dx ^ 2 + dy ^ 2 + dz ^ 2;$$</span> the problem (which I found also in the literature) is that this process is not always possible (for example if the spin of the black hole exceeds a certain critical value).</p> <p>My question is: imagining a rotating black hole with very high angular velocity (t.c. angular momentum at = 0.99 in natural units), what should I see? And how do I understand analytically what geometric shape would have external horizon and ergosphere (spied from the earth) if I cannot embed them in a 3D space? </p> https://physics.stackexchange.com/q/490819 2 Zero mass Kerr metric Professor Kirby https://physics.stackexchange.com/users/232485 2019-07-10T11:12:17Z 2019-07-10T11:12:17Z <p>When mass in Kerr metric is put to zero we have <span class="math-container">$$ds^{2}=-dt^{2}+\frac{r^{2}+a^{2}\cos^{2}\theta}{r^{2}+a^{2}}dr^{2}+\left(r^{2}+a^{2}\cos^{2}\theta\right)d\theta^{2}+\left(r^{2}+a^{2}\right)\sin^{2}\theta d\phi^{2},$$</span> where <span class="math-container">$a$</span> is a constant. This is a flat metric. What exactly is the coordinate transformation that changes this into the usual Minkowski spacetime metric form <span class="math-container">$$ds^{2}=-dt^{2}+dx^{2}+dy^{2}+dz^{2}?$$</span></p> https://physics.stackexchange.com/q/397067 0 Simulating a Test Particle in a Kerr Spacetime $(M,\mathcal{O}, \mathcal{A},\nabla^{L.C.})$ Sergio Charles https://physics.stackexchange.com/users/131313 2018-04-01T18:50:31Z 2019-07-08T04:56:13Z <p>The equations of motion for a test particle in a Kerr spacetime $(M,\mathcal{O}, \mathcal{A},\nabla^{L.C.})$ are dictated by four degrees of freedom (i.e. invariant mass $m$ in $p^\mu g_{\mu\nu}p^{\nu}=-m^2,$ the energy $E$, the Carter constant $Q,$ and the orbital angular momentum $L_z=-p_{\phi}$ in the spin direction). On the <a href="https://en.wikipedia.org/wiki/Kerr_metric" rel="nofollow noreferrer">wikipedia page</a> for the Kerr metric, there is a simulation on the right-hand side of the trajectory equations section (see simulation <a href="https://en.wikipedia.org/wiki/File:Orbit_around_a_rotating_Kerr_black_hole.gif" rel="nofollow noreferrer">here</a>). By any chance, does anyone know of such a program (for Python) that I can use to simulate this, as shown in the aforementioned program implementation?</p> https://physics.stackexchange.com/q/487729 4 Metric for a rotating star AoZora https://physics.stackexchange.com/users/226311 2019-06-23T20:38:07Z 2019-06-25T15:18:43Z <p>If we want to describe a static spherically symmetric star we can use a metric which matches the Schwarzschild solution with correct mass on the outside of the star but differs from Schwartzschild in the inside of the matter distribution.</p> <p>Basically we solve the Einstein equations with a source <span class="math-container">$T_{\mu\nu}$</span>, for instance <span class="math-container">$$T_{\mu\nu}=(\rho+p)u_{\mu}u_{\nu}+p\,g_{\mu\nu}$$</span> where <span class="math-container">$u_{\mu}$</span> has zero spatial components, meaning it is the velocity in a static fluid (this can also be seen as a consequence of Einstein equations).</p> <p>Can we do something similar for a rotating star using the metric for a Kerr black hole?</p> <p>I heard that it is a much more difficult problem and I would like to understand how difficult it is (Is it possible?) and what makes it so difficult.</p> https://physics.stackexchange.com/q/483965 0 Numerical Solutions for Equatorial Orbits in the Kerr Black Hole Ícaro Lorran https://physics.stackexchange.com/users/98310 2019-06-03T00:13:43Z 2019-06-09T15:47:34Z <p>Currently, I am trying to find timelike orbits in the Kerr metric around the equator. The problem is that no matter which parameters I choose or the method I use I can't seem to get to physically sound orbits. The solutions I got so far diverge to infinity, make some crazy nonsensical spirals or oscillate in a circular path.</p> <p>Here's what I'm using currently for the differential equations:</p> <p><span class="math-container">$$r''=\frac{-1}{2r}\left(r'^2+\frac{1}{r}\left(2r(1-T^2)-r_0+\frac{r_0}{r^2}\left(aT+R\right)^2\right)\right)$$</span></p> <p>and</p> <p><span class="math-container">$$r'^2 = -\frac{\Delta(r)}{r^2}(1+T\dot{t}+R\dot{\phi}).$$</span></p> <p>The symbols <span class="math-container">$T$</span> and <span class="math-container">$R$</span> are constants, whereas for <span class="math-container">$\Delta$</span>,<span class="math-container">$\dot{t}$</span> and <span class="math-container">$\dot{\phi}$</span>:</p> <p><span class="math-container">$\Delta(r)=a^2+r^2-rr_0$</span></p> <p><span class="math-container">$\dot{\phi} = \frac{1}{\Delta}\left[\left(1-\frac{r_0}{r}\right)R-\frac{ar_0}{r}T\right]$</span></p> <p><span class="math-container">$\dot{t} = \frac{1}{\Delta}\left[-\left(r^2+a^2+\frac{a^2r_0}{r}\right)T-\frac{ar_0}{r}R\right]$</span></p> <p>I chose <span class="math-container">$\phi$</span> as my azimuthal angle.</p> <p>I tried to use these on a code I made in Python and another one in Java. Both of them seem to give the same crazy solutions.</p> <p>If you need to take a look at the code: <a href="https://github.com/icarosadero/black_holes/blob/master/geodesic.java" rel="nofollow noreferrer">https://github.com/icarosadero/black_holes/blob/master/geodesic.java</a> <a href="https://github.com/icarosadero/black_holes/blob/master/script.py" rel="nofollow noreferrer">https://github.com/icarosadero/black_holes/blob/master/script.py</a></p> <p>With all of that said I would like to ask whether or not those equations are right. I don't have many people near me available to help at the moment.</p> https://physics.stackexchange.com/q/484270 0 Light-like normal vectors Souradeep https://physics.stackexchange.com/users/203830 2019-06-04T15:05:22Z 2019-06-04T16:15:30Z <p>Can someone please show me how to mathematically establish that the normal vector to the event horizon of a Kerr Black Hole is light-like?</p> https://physics.stackexchange.com/q/483806 1 How to compute Kerr geodesics? almost https://physics.stackexchange.com/users/214908 2019-06-02T07:02:32Z 2019-06-02T09:31:04Z <p>How would I start to numerically compute trajectories of Kerr geodesics with constants of motion like in this <a href="https://en.wikipedia.org/wiki/Kerr_metric#Trajectory_equations" rel="nofollow noreferrer">wikipedia page</a>. I want to recreate trajectories like in this picture in Matlab. <a href="https://i.stack.imgur.com/2Uvgw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Uvgw.png" alt="enter image description here"></a></p> https://physics.stackexchange.com/q/220638 4 Time independent Kerr metric imranal https://physics.stackexchange.com/users/94395 2015-11-26T09:31:59Z 2019-05-10T10:01:35Z <p>The Kerr metric expressed in terms of polar coordinates $r,\theta,\phi$, such that $x = r\sin(\theta)\cos(\phi)$, $y = r\sin(\theta)\sin(\phi)$, $z = r\cos(\theta)$. Then the <a href="http://en.wikipedia.org/wiki/Kerr_metric" rel="nofollow noreferrer">Kerr metric</a> is given as \begin{align*} ds^2 = &amp;-\left(1 - \frac{2GMr}{r^2+a^2\cos^2(\theta)}\right) dt^2 + \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2\\ &amp;+ \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 - \left(\frac{4GMra\sin^2(\theta)}{r^2+a^2\cos^2(\theta)}\right) d\phi\, dt \end{align*} where $a \equiv S/M$ is the object's angular momentum per unit mass, and $G$ is the gravitational constant. This is an exact solution for the empty-space Einstein equation.</p> <p>Say, If we are to consider the metric for a constant time, $t_0$. Is it then possible to define the Kerr metric on a submanifold of spacetime, say only in space? If so how can I accomlish this? Is it as simple as dropping the time dependent terms, i.e \begin{align*} ds^2 = &amp; \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2\\ &amp;+ \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 \end{align*} or do I need to use the <a href="https://en.wikipedia.org/wiki/Induced_metric" rel="nofollow noreferrer">induced metric</a> to describe the metric on the submanifold?</p> <p><strong>Edit :</strong> I solved the geodesic differential equations using a "time independent" Kerr metric, with <code>a = 0</code> (i.e this reduces Kerr metric to the Schwarzschild metric), and the Schwarzschild radius to define the other parameters :</p> <p><a href="https://i.stack.imgur.com/0I7yw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0I7yw.png" alt="Geodesics for a time independent Kerr metric"></a></p> <p>Most plots I got spiraled around a singularity at the origo.</p> <p>Here is a plot where I set $\phi$ to a constant, the z-axis becomes the "time" :</p> <p><a href="https://i.stack.imgur.com/CpcZc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CpcZc.png" alt="Geodesics for $phi$ independent Kerr metric"></a></p> <p><strong>Update :</strong> I have found the following figure which seem to verify my first figure. <a href="https://i.stack.imgur.com/DKaYt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DKaYt.png" alt="enter image description here"></a></p> <blockquote> <p>Strategies for Direct Visualization of Second-Rank Tensor Fields by Werner Benger and Hans-Christian Hege</p> </blockquote> https://physics.stackexchange.com/q/190795 8 Physical motivation for mathematically extending solutions to Einstein's equations Javier https://physics.stackexchange.com/users/5788 2015-06-23T00:16:17Z 2019-05-06T21:03:25Z <p>Sorry if this question gets a little long; I want to explain why I'm asking it.</p> <p>The usual Schwarzchild metric</p> <p>$$ds^2 = -\left(1-\frac{2M}{r}\right) dt^2 + \left(1-\frac{2M}{r}\right)^{-1} dr^2 + r^2(d\theta^2+\sin^2\theta\ d\phi^2)$$</p> <p>makes sense when $t,r$ lie in the ranges $-\infty&lt;t&lt;\infty$, $0&lt;r&lt;\infty$. We can change to <a href="https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates" rel="noreferrer">Kruskal–Szekeres coordinates</a> $u,v$ (which Wikipedia calls $X$ and $T$), which are nonsingular at the horizon. The corresponding range for them is $u+v&gt;0$. But if we draw spacetime on the $u,v$ plane, we can see that there is no problem extending it to all possible values of the coordinates. This adds a symmetric half to spacetime, with a past singularity that behaves like a white hole: any object or light ray will eventually exit the horizon.</p> <p>When I learned this in General Relativity class, the professor said that the usual physical reasoning for doing this is that the good old Schwarzschild metric is geodesically incomplete: if we imagine a particle falling into the black hole and try to trace back its path into the past, it looks like (as long as it doesn't have too much energy) it should have come up from the black hole, stopped, and then proceeded to fall in. I wasn't too convinced of this for two reasons: One, the black hole hasn't existed for all eternity so whatever is falling into it can have its origin somewhere else. Two, we haven't actually observed any white holes.</p> <p>This remained a mathematical curiosity until we analyzed a Penrose diagram of the Reissner-Nordstrom metric for a charged black hole. This spacetime has two horizons and a singularity inside. But now there are timelike curves that end at some point in the future without hitting a singularity. To me this seems like a much bigger deal, since I can perfectly well imagine something falling into a charged black hole as a physically realistic situation.</p> <p><img src="https://i.stack.imgur.com/LU95L.png" alt="Kerr metric Penrose diagram taken from Wikimedia"></p> <p>The extension in this case is something much weirder: an infinte chain of universes. You can enter a black hole here and come out at some other universe, and proceed to do the same until you get tired and settle down on some planet on whatever universe you happen to be on. This is as far from physically realistic as it gets, and yet it seems unavoidable if we want to have a charged black hole (I think the same thing happens for a rotating black hole too).</p> <p>Let me state my question, then: is the incompleteness of timelike curves in a charged black hole a real thing? Is it a problem? Is the infinite tower of universes the only way to make the problem go away, and if so, wouldn't that imply that it exists in the universe, since charged and rotating black holes do exist?</p> https://physics.stackexchange.com/q/473450 11 How does the Penrose diagram for a spinning black hole differ in realistic scenarios (formed by stellar collapse)? user1379857 https://physics.stackexchange.com/users/157704 2019-04-18T01:57:44Z 2019-04-24T00:07:44Z <p>The Penrose diagram for a non-spinning Schwarzschild black hole is <a href="https://i.stack.imgur.com/tFZaU.png" rel="noreferrer"><img src="https://i.stack.imgur.com/tFZaU.png" alt="enter image description here"></a></p> <p>Notably, there is a second universe "on the other side" of the black hole. However, actual black holes form by stellar collapse, and the collapse process leads to a different Penrose diagram:</p> <p><a href="https://i.stack.imgur.com/jfrNS.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jfrNS.png" alt="enter image description here"></a></p> <p>A spinning Kerr black hole famously has the following crazy Penrose diagram:</p> <p><a href="https://i.stack.imgur.com/oqi2d.png" rel="noreferrer"><img src="https://i.stack.imgur.com/oqi2d.png" alt="enter image description here"></a></p> <p>The Kerr spacetime has two horizons, in natural units <span class="math-container">$r_\pm= M \pm \sqrt{M^2 - a^2}$</span> where the angular momentum is <span class="math-container">$J = aM$</span>. <span class="math-container">$r_+$</span> is the regular event horizon. Inside of that is <span class="math-container">$r_-$</span>, where you can begin navigating to leave the black hole and enter an entirely new universe. (You can also pass through the ring singularity, or the "ringularity", and enter an "antiverse.")</p> <p>My question is, what is the Penrose diagram for a realistic black hole that forms via stellar collapse, and not an idealized "eternal" Kerr black hole? What happens to the parallel universes, <span class="math-container">$r_-$</span>, and the antiverses? Do any of these features remain?</p> https://physics.stackexchange.com/q/137610 7 What happens to a particle in the exact center of a Kerr black hole? HDE 226868 https://physics.stackexchange.com/users/56299 2014-09-28T00:48:59Z 2019-04-14T08:32:41Z <p>Kerr black holes (and Kerr-Newman black holes), instead of the "point" singularity theorized in spherically symmetric black holes, instead have a "ring" singularity, spread along the equatorial plane of rotation. any particle inside a spherically symmetric black hole will tend to go towards the center. But what if a particle is in the exact center of a Kerr black hole? Will it go towards any point in the ring singularity, or will it stay where it is, unless it is perturbed in one direction?</p> https://physics.stackexchange.com/q/471419 2 Metric diameter of a ring singularity Yukterez https://physics.stackexchange.com/users/24093 2019-04-09T00:40:54Z 2019-04-12T17:35:57Z <p>In the Kerr metric the ring singularity is located at the coordinate radius <span class="math-container">$r=0$</span>, which corresponds to a ring with the cartesian radius <span class="math-container">$R=a$</span>.</p> <p>So the center of the ring singularity in cartesian coordinates is at <span class="math-container">$r=-a, \ \theta=\pi/2$</span>. </p> <p>But the center in cartesian coordinates is also at <span class="math-container">$r=0, \ \theta=0$</span> (at <span class="math-container">$r=0$</span> all <span class="math-container">$\theta$</span> are in the equatorial plane, at least in Boyer Lindquist and also in Kerr Schild coordinates).</p> <p>To calculate the physical diameter to see how much fits through the ring (in one example it is a <a href="https://www.youtube.com/watch?v=oESYLX6IHe4&amp;t=12m32s" rel="nofollow noreferrer">tiger</a>, in another one <a href="http://yukterez.net/org/alice-vs-sponge-bob-kerr.ringsingularitaet.gif" rel="nofollow noreferrer">Alice &amp; Bob</a>), would I integrate</p> <p><span class="math-container">$$(1) \ \ \ \ \theta=\pi/2 , \ \ d =2 \int_{-a}^0 \sqrt{|g_{rr}|} \ \ {\rm d}r = 2 \sqrt{(2-a) a}+4 \arcsin \left(\sqrt{\frac{a}{2}}\right)$$</span></p> <p>in the equatorial plane, or is it rather</p> <p><span class="math-container">$$(2) \ \ \ \ r=0 , \ \ d =\int_{-\pi/2}^{\pi/2} \sqrt{|g_{\theta \theta}|} \ \ {\rm d}\theta = 2a$$</span></p> <p>since that should also cover the distance from one side of the ring to the opposite. </p> <p>Approach <span class="math-container">$(2)$</span> gives exactly the diameter in cartesian coordinates, but I don't know if that's supposed to be so, or only a coincidence, since otherwise the metric distance is not nescessarily the same as the coordinate or cartesian distance.</p> <p>So which one is it, <span class="math-container">$(1)$</span> or <span class="math-container">$(2)$</span>? Or is it done in a completely different way?</p> <p>The coordinates I used are <a href="http://yukterez.net/org/ringsingularity.proper.radius.png" rel="nofollow noreferrer">Kerr Schild coordinates</a>, which should cover the inside with the relevant components</p> <p><span class="math-container">$g_{r r}=-\frac{2 r}{a^2 \cos ^2 \theta +r^2}-1 \ , \ \ g_{\theta \theta }= -r^2 - a^2 \cos^2 \theta$</span></p> <p>I guess it is approach <span class="math-container">$(2)$</span> since no one can enter the ring singularity from the equatorial plane, but I'd like to hear a 2nd opinion on that</p> https://physics.stackexchange.com/q/164636 0 Why can't a particle rotate opposite to the central mass within the ergosphere? draks ... https://physics.stackexchange.com/users/6292 2015-02-12T01:21:49Z 2019-04-06T12:19:46Z <p>Wiki says about the <a href="http://en.wikipedia.org/wiki/Kerr_metric" rel="nofollow">Kerr metric</a>:</p> <blockquote> <p>A moving particle experiences a positive proper time along its worldline, its path through spacetime. However, this is impossible within the ergosphere, where $g_{tt}$ is negative, unless the particle is co-rotating with the interior mass $M$ with an angular speed at least of Ω. Thus, no particle can rotate opposite to the central mass within the ergosphere.</p> </blockquote> <p>I don't get it. Can it be somehow seen that $g_{tt}$ is negative from</p> <p><a href="http://en.wikipedia.org/wiki/Kerr_metric#Wave_operator" rel="nofollow">\begin{align} g^{\mu\nu}\frac{\partial}{\partial{x^{\mu}}}\frac{\partial}{\partial{x^{\nu}}} = &amp; \frac{1}{c^{2}\Delta}\left(r^{2} + \alpha^{2} + \frac{r_{s}r\alpha^{2}}{\rho^{2}}\sin^{2}\theta\right)\left(\frac{\partial}{\partial{t}}\right)^{2} + \frac{2r_{s}r\alpha}{c\rho^{2}\Delta}\frac{\partial}{\partial{\phi}}\frac{\partial}{\partial{t}} \\ &amp; - \frac{1}{\Delta\sin^{2}\theta}\left(1 - \frac{r_{s}r}{\rho^{2}}\right)\left(\frac{\partial}{\partial{\phi}}\right)^{2} - \frac{\Delta}{\rho^{2}}\left(\frac{\partial}{\partial{r}}\right)^{2} - \frac{1}{\rho^{2}}\left(\frac{\partial}{\partial{\theta}}\right)^{2} \color{red}{?} \end{align}</a></p> <p>And why can't <em>no particle <strike>can</strike> rotate opposite to the central mass within the ergosphere</em>?</p> https://physics.stackexchange.com/q/470501 1 Quadrupole moment of Kerr spacetime supercoolphysicist https://physics.stackexchange.com/users/124838 2019-04-04T11:42:48Z 2019-04-04T13:35:01Z <p>In this paper <a href="https://journals.aps.org/prd/pdf/10.1103/PhysRevD.99.044005" rel="nofollow noreferrer">this paper</a>, the Kerr black hole is described as having quadrupole moment of <span class="math-container">$q=J^2/M$</span> (which means <span class="math-container">$q=a^2M$</span> using <span class="math-container">$J=aM$</span>) whereas in <a href="https://arxiv.org/abs/0909.4150." rel="nofollow noreferrer">this paper</a> it says in the abstract that the limiting case of Kerr is <span class="math-container">$q=0$</span>. and finally <a href="https://arxiv.org/abs/1012.2007" rel="nofollow noreferrer">this paper</a> says <span class="math-container">$q=-a^2M^3$</span> (I think this is due to a different definition of <span class="math-container">$a$</span> though, as they say <span class="math-container">$a=J/M^2$</span>). Which one is correct? Perhaps in the second paper <span class="math-container">$q=0$</span> in the approximation they take?</p> https://physics.stackexchange.com/q/468964 1 For a given mass, how big can a Kerr black hole get? Rick https://physics.stackexchange.com/users/189760 2019-03-27T13:01:38Z 2019-03-29T15:09:53Z <p>We know that in a Kerr Black Hole the singularity is in the form of a 1 dimensional ring. If we have a 25 solar mass black hole, how big would the Kerr Ring be, width wise? </p> <p>Also, I read the Wiki on Kerr Black Holes, is the ring shape due to centrifugal force?</p> https://physics.stackexchange.com/q/455247 4 Is a stable orbit possible inside the ergosphere of a Kerr (spinning) black hole? user841495 https://physics.stackexchange.com/users/220303 2019-01-19T08:27:55Z 2019-03-16T14:18:59Z <p>I have heard that it's "impossible to hover" inside of an ergosphere, but everywhere I read this seemed to be speaking in the context of "relative to a stationary observer outside of the ergosphere". The explanation for this is that objects inside the ergosphere are forced to "co-rotate" with the black hole, although I'm unsure if this means</p> <ul> <li>"objects begin to orbit (revolve around) the black hole in the same direction the black hole spins around its axis" </li> </ul> <p>or </p> <ul> <li>"objects begin to spin around their own axes (rotate), parallel with the black hole's axis, in the same direction the black hole spins around its axis" </li> </ul> <p>or both.</p> <p>Wouldn't, however, an object with this forced motion around the black hole be able to orbit stably inside the ergosphere as it might around a non-rotating black hole?</p> <p><a href="https://link.springer.com/article/10.1007/BF00764017" rel="nofollow noreferrer">One paper on the topic</a>'s abstract includes phrases like</p> <p><code>For certain parameter values there are also orbits inside the inner horizon not reaching the center.</code></p> <p>and</p> <p><code>All negative energy orbits enter the horizon of the black hole.</code> (As opposed to orbits that do not?)</p> <p>which seems to indicate stable orbits are possible.</p> https://physics.stackexchange.com/q/181288 7 Why is the photon-sphere around a Kerr Black Hole spherical and not ellipsoid? Randy Welt https://physics.stackexchange.com/users/60901 2015-05-05T19:37:33Z 2019-03-11T19:22:16Z <p>Kerr Black Holes have usually (excluding extrema <span class="math-container">$a=0$</span>, <span class="math-container">$a=1$</span>) due to their spinning activity an ellipsoidal ergosphere. </p> <p>So why does the photon-sphere does not have an ellipsoidal form?</p> <blockquote> <p>On the possibility of observation of the future for movement in the field of black holes of different types. Yu.V. Pavlov. <a href="https://doi.org/10.1007/s10714-012-1453-1" rel="nofollow noreferrer"><em>Gen. Relativ. Gravit.</em> <strong>45</strong>, 17 (2013)</a>, <a href="http://arxiv.org/abs/1203.4000" rel="nofollow noreferrer">arXiv:1203.4000</a>.</p> </blockquote> https://physics.stackexchange.com/q/338441 2 Area of the event horizon of a rotating black hole nightmarish https://physics.stackexchange.com/users/59234 2017-06-09T21:37:50Z 2019-03-10T10:25:19Z <p>The Kerr metric for a black hole of mass $M$ and angular momentum $J = aM$ is</p> <p>$$ds^{2} = - \frac{\Delta(r)}{\rho^{2}}(dt-a\sin^{2}\theta d\phi)^{2} + \frac{\rho^{2}}{\Delta(r)}dr^{2} + \rho^{2} d\theta^{2} + \frac{1}{\rho^{2}}\sin^{2}\theta (adt - (r^{2}+a^{2}) d\phi)^{2},$$</p> <p>where $\Delta(r) = r^{2} + a^{2} - 2Mr$, $\rho^{2} = r^{2} + a^{2} \cos^{2}\theta$ and $- M &lt; a &lt; M$.</p> <hr> <p>The event horizon is at $r_{+} = M + \sqrt{M^{2} - a^{2}}$. This is obtained by solving the equation $\Delta(r) = 0$.</p> <hr> <p>How do you use this to compute the area of the horizon?</p> <p>My idea is to simplify the metric to obtain</p> <p>$$ds^{2} = - \left( \frac{r^{2} + a^{2} - 2Mr - a^{2} \sin^{2}\theta}{r^{2} + a^{2} \cos^{2}\theta} \right) dt^{2} + \left( \frac{r^{2} + a^{2} \cos^{2}\theta}{r^{2} + a^{2} - 2Mr} \right) dr^{2} - \left( \frac{4aMr \sin^{2}\theta}{r^{2} + a^{2} \cos^{2}\theta} \right) dtd\phi + \left( r^{2} + a^{2} \cos^{2}\theta \right) d\theta^{2} + \sin^{2}\theta \left( \frac{(a^{2} + r^{2})^{2} - a^{2} \sin^{2}\theta (a^{2}-2Mr+r^{2}) }{r^{2} + a^{2} \cos^{2}\theta} \right) d\phi^{2}.$$</p> <p>Then, I think that the area of the horizon is given by</p> <p>$$A = \int d\theta\ d\phi\ g_{\phi\phi}g_{\theta\theta}|_{r=r_{+}}.$$</p> <p>Am I wrong?</p> https://physics.stackexchange.com/q/464635 0 Can anyone tell me how can draw shadow of black hole like in presented in Intersteller movie? Is there any code for it in Mathematica or in Python? [closed] Bturimov https://physics.stackexchange.com/users/185595 2019-03-05T17:24:10Z 2019-03-05T20:08:36Z <p>Equation of motion for photon <span class="math-container">$$\Sigma \frac{dt}{d\lambda} = aL\left(1-\frac{r^2+a^2}{\Delta}\right) + \omega\left(\frac{\left(r^2+a^2\right)^2}{\Delta}-a^2 \sin ^2\theta\right)\ ,$$</span> <span class="math-container">$$\Sigma\frac{dr}{d\lambda} = \sqrt{R(r)}=\sqrt{\left( \omega\left(r^2+a^2\right)-aL\right)^2-K\Delta}\ ,$$</span> <span class="math-container">$$\Sigma\frac{d\theta}{d\lambda} = \sqrt{\Theta(\theta)} = \sqrt{K-\left(\frac{L}{\sin\theta}-a\omega\sin\theta\right)^2}\ ,$$</span> <span class="math-container">$$\Sigma\frac{d\phi}{d\lambda} = L\left(\frac{1}{\sin^2\theta}- \frac{a^2}{\Delta}\right)+a\omega \left(\frac{r^2+a^2}{\Delta}-1\right)\ ,$$</span> where <span class="math-container">$K$</span> is the Carter constant of the motion.</p> https://physics.stackexchange.com/q/382588 0 The Killing vector $\chi=\partial_t+\Omega_H\partial_\phi$ doesn't look normal to the Killing horizon for a Kerr BH StudyHard https://physics.stackexchange.com/users/182909 2018-01-27T14:46:10Z 2019-02-07T03:03:05Z <p>As mentioned in Carroll's Spacetime and Geometry p. 244, a Killing vector is normal to its Killing horizon. With some help from the other forum, I could check this is true. (FYI, here the Killing horizon $\Sigma$ of a Killing vector $\chi$ is defined by a null hypersurface on which $\chi$ is null.)</p> <p>But when I try to apply this general statement to a Kerr BH, something weird thing happens: in a Kerr BH, we consider a Killing vector $$\chi=\partial_t+\Omega_H\partial_\phi,$$ where $\Omega_H$ is designed to make $\chi$ to be null on the event horizon, $$\Sigma:r=r_H=M+\sqrt{M^2-a^2}.$$ So by definition $\Sigma$ is the Killing horizon of a Killing vector $\chi$. Then according to the general statement, this $\chi$ must be normal to $\Sigma$ but it doesn't look like satisfying this condition. </p> <p>To be clear, note that we can write the normal vector of $\Sigma$ as $$n_\mu=\nabla_\mu(r-r_H)=(0,1,0,0).$$ But this $n$ is not parallel to $\chi$ at all. Equivalently, tangent vectors on $\Sigma$ which is orthogonal to $n$ is not orthogonal to $\chi$. This means $\chi$ is not normal to $\Sigma$...?!?!</p> <p>I have no idea at this point... If you see what is going wrong here, please help me out with this nonsense! </p> https://physics.stackexchange.com/q/457995 5 Gravitational lensing redshift around a Kerr black hole safesphere https://physics.stackexchange.com/users/164879 2019-01-31T09:11:31Z 2019-01-31T16:07:50Z <p>Light from a source passes by a Kerr black hole on two sides at the equator and converges at the observer. The axis of rotation of the black hole is perpendicular to the direction of light. Two rays of light pass through the spacetime regions of a significant frame dragging, on one side along and on the other side against the direction of light.</p> <p>Would frame dragging cause a red shift of one ray and a blue shift of the other? Or would both rays come to the observer with the same frequency?</p> https://physics.stackexchange.com/q/456941 0 Beyond Kerr Carter constant? riemannium https://physics.stackexchange.com/users/22916 2019-01-26T21:22:53Z 2019-01-27T01:08:50Z <p>What are the most symmetrical black hole spacetimes whose motion is completely integrable with a <a href="https://en.wikipedia.org/wiki/Carter_constant" rel="nofollow noreferrer">Carter constant-like</a> and hidden symmetry superintegrability condition? Do type D-spacetimes have a Carter like constant? Is there any non-Kerr-like black hole having Carter constant/s?</p> https://physics.stackexchange.com/q/245285 5 Closed timelike curves in the Kerr metric Johnny https://physics.stackexchange.com/users/111781 2016-03-24T11:25:49Z 2019-01-22T17:02:01Z <p>I just read in Landau-Lifshitz that the Kerr metric admits closed timelike curves in the region $r \in (0, r_{hor})$ where $r_{hor}$ is the event-horizon ( I am talking about the case $|M|&gt;|a|$ (subextremal case) here ). Now, unfortunately they don't give an example of such a curve. Could anybody of you write down explicitly such a CTC so that I could go through the computation once by myself. I would really like to see this once. </p> <p>If anything is unclear, please let me know.</p> https://physics.stackexchange.com/q/452815 2 Visualization of $dtdx$ and $dxdy$ term in metric tensor Angela https://physics.stackexchange.com/users/181828 2019-01-08T02:10:18Z 2019-01-09T14:57:35Z <p>For the sake of simplicity, lets take a 2+1 dimensional spacetime. Lets take the metric </p> <p><span class="math-container">$$ds^2 = g_{tt}dt^2 + g_{xx}dx^2 + g_{yy}dy^2 + g_{tx}dtdx + g_{xy}dxdy$$</span> </p> <p>What is the visualization or physical interpretation of the <span class="math-container">$g_{tx}$</span> and <span class="math-container">$g_{xy}$</span> terms of the metric? Does <span class="math-container">$g_{tx}$</span> mean motion of space i.e. on object in this spacetime point will be moving w.r.t an observer at infinity? What would <span class="math-container">$g_{xy}$</span> mean? </p> https://physics.stackexchange.com/q/450787 -2 Centrifugal force on spinning black hole? user6760 https://physics.stackexchange.com/users/75502 2018-12-28T06:08:05Z 2018-12-28T07:57:09Z <p>I saw the term spinning black hole popping up everywhere so my question do spinning black hole behave similarly to say a planet where it bulge in the equatorial and compress at the poles? what fundamental force is causing the bulging because in the case of planet it is the electrostatic force?</p> https://physics.stackexchange.com/q/434269 3 Periodicity trick for Kerr Black Holes blackhole1511 https://physics.stackexchange.com/users/75467 2018-10-13T12:34:18Z 2018-10-19T08:06:15Z <p>I am slightly confused concerning the euclidean section of a Kerr black hole. In page 5 of the following paper <a href="https://arxiv.org/abs/hep-th/9908022" rel="nofollow noreferrer">https://arxiv.org/abs/hep-th/9908022</a> it is said that in order to get the euclidean section, we need to set <span class="math-container">$t \to i \tau$</span> and <span class="math-container">$a \to i a$</span>. (They consider general Kerr-Newman-AdS black holes but I am simply interested in Kerr asymptotically flat.) This makes sense because we want to keep the <span class="math-container">$dt \otimes d\phi$</span> components of the euclidean metric real. What confuses me is that if we do the analysis of the conical singularities as they mention, we will get the following periodicity for <span class="math-container">$\tau$</span> and <span class="math-container">$\phi$</span></p> <p><span class="math-container">\begin{equation} \tau \sim \tau +\beta \end{equation}</span> <span class="math-container">\begin{equation} \phi\ \sim \phi+i\beta\Omega_H \end{equation}</span> with <span class="math-container">$\beta$</span> the inverse temperature and <span class="math-container">$\Omega_H$</span> the angular velocity of the event horizon, namely <span class="math-container">\begin{equation} \Omega_H=\frac{a}{r_{+}^2+a^2} \end{equation}</span> where <span class="math-container">$r_{+}$</span> is the event horizon and <span class="math-container">$a$</span> is the rotation parameter of the black hole. What is strange to me is that if we take <span class="math-container">$a \to 0$</span> in Boyer-Lindquist coordinates, we get that <span class="math-container">\begin{equation} \phi \sim \phi \end{equation}</span> because <span class="math-container">$\Omega_H$</span> vanishes. This becomes a trivial identification and it does not tell us anything about the periodicity of the <span class="math-container">$\phi$</span> coordinate. However, we also know that if we take the <span class="math-container">$a \to 0$</span> limit, we get the Schwarzschild black hole in Schwarzschild coordinates. In Schwarzschild Euclidean, we should take the <span class="math-container">$\phi$</span> coordinate to have period <span class="math-container">\begin{equation} \phi \sim \phi+2\pi \end{equation}</span> and even though the Boyer-Lindquist <span class="math-container">$\phi$</span> is different than the <span class="math-container">$\phi$</span> in Schwarzschild, they match in the limit I am considering <span class="math-container">$a \to 0$</span>. What does this imply? Does this mean that even though Kerr goes to Schwarzschild in the limit <span class="math-container">$a \to 0$</span> as a lorentzian geometry, their euclidean sections are not connected continuously somehow?</p> <p>Edit1: I also have the notion that in lorentzian Kerr, the <span class="math-container">$\phi$</span> coordinate has periodicity <span class="math-container">$2\pi$</span>. When we go to Euclidean, we seem to get this other periodicity: but shouldn't the periodicity of <span class="math-container">$2\pi$</span> be preserved as well? At least that is what happens in Schwarzschild. So we would have both <span class="math-container">\begin{equation} \phi\ \sim \phi+i\beta\Omega_H \end{equation}</span> <span class="math-container">\begin{equation} \phi\ \sim \phi + 2\pi \end{equation}</span> It also confuses me that this manipulations are usually done based on the coordinate systems and therefore it is harder to get a notion of what it means to 'euclideanize' in a coordinate invariant way. If someone has a coordinate invariant way to talk about this analytic continuation, I would like to hear it.</p> <p>Edit2: If we see what really is the expression in the identification of <span class="math-container">$\phi$</span>, we get <span class="math-container">\begin{equation} i\beta \Omega_H=i4\pi \frac{r_{+}a}{r_{+}^2\left(1-\frac{a^2}{r_{+}^2}\right)} \end{equation}</span> By doing the analytic continuation <span class="math-container">$a \to ia$</span>, we have <span class="math-container">\begin{equation} i\beta \Omega_H=-4\pi \frac{r_{+}a}{r_{+}^2\left(1+\frac{a^2}{r_{+}^2}\right)} \end{equation}</span> we see that it is alway less then <span class="math-container">$2\pi$</span> because <span class="math-container">\begin{equation} r_{+}=a+\sqrt{2}a \end{equation}</span> defines extremality assuming the fact that we set <span class="math-container">$a \to ia$</span>. So it seems to make the <span class="math-container">$\phi$</span> direction smaller in general. But if I try to compute the action on-shell <span class="math-container">\begin{equation} I=\int_{\partial \mathcal{M}}K-K_0 \end{equation}</span> I have to integrate from <span class="math-container">$0$</span> to <span class="math-container">$2\pi$</span> along <span class="math-container">$\phi$</span> to get the right result mentioned in <a href="https://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.2752" rel="nofollow noreferrer">https://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.2752</a> because since we are sending the boundary to infinity only the leading order of <span class="math-container">$1/r$</span> matters which is the same as in Schwarzschild. So I am confused what kind of geometry we have along <span class="math-container">$\phi$</span>.</p> https://physics.stackexchange.com/q/434321 -1 What would happen to the Earth, if the moon was a black hole? [closed] Árpád Szendrei https://physics.stackexchange.com/users/132371 2018-10-13T18:12:48Z 2018-10-13T19:38:08Z <p>Would it be a feasible scenario?</p> <p>I have read this question:</p> <p><a href="https://physics.stackexchange.com/questions/61422/what-would-happen-to-the-moon-if-earth-is-turned-into-a-black-hole">What would happen to the Moon if Earth is turned into a black hole?</a></p> <p>Where Lubos Motl says:</p> <blockquote> <p>The extremal Kerr J=GM2/c∼RbhMc. Now, the Earth-mass black hole has radius 9 mm or so so we get about 1030 Js. The Earth's spin, actual angular momentum now, is indeed over 1040, ten orders of magnitude too much. This discrepancy is of course linked to Earth's too low density that makes the collapse de facto impossible. </p> </blockquote> <p>So based on this, the moon needs a radius of 2 mm. So let's disregard that the moon's low density makes it practically impossible to collapse, but let's instead say that it already is a black hole, size 2 mm, and nothing else, changes, it is revolving around the Earth, same speed, trajectory. </p> <ol> <li><p>Would this be possible? Can a planet have a black hole moon at all?</p></li> <li><p>Would anything on Earth change?</p></li> </ol> https://physics.stackexchange.com/q/430384 0 Proof that the Kerr metric may be written in orthogonal form N. Steinle https://physics.stackexchange.com/users/201856 2018-09-23T15:07:20Z 2018-09-26T02:54:08Z <blockquote> <ol> <li>Prove or disprove that the Kerr metric can be expressed in a set of orthogonal coordinates over some coordinate chart.</li> </ol> </blockquote> <p>Motivation for this question stems from my understanding that a metric can always be written in orthogonal coordinates if it exists on a flat spacetime. A metric <a href="https://en.wikipedia.org/wiki/Orthogonal_coordinates" rel="nofollow noreferrer">written</a> in an orthogonal set of coordinates has no off-diagonal terms. </p> <p>As an example, the Kerr metric always seems to have at least one off-diagonal term. I understand physically why the off-diagonal terms are <a href="https://arxiv.org/pdf/0706.0622.pdf" rel="nofollow noreferrer">present</a> in specific coordinates, but people seem to take it as a fact that it can never be transformed away or they say it is <a href="https://physics.stackexchange.com/questions/315661/the-physical-meaning-of-the-cross-term-of-kerr-metric">obvious</a> that it can be. </p> <p>So, since the Kerr spacetime has curvature, it makes sense that the Kerr metric cannot be written in a global set of coordinates. However, why can the metric not be written in a set of orthogonal coordinates?</p> <p>Is it as simple as using a co-rotating timelike observer near infinity to be able to transform away the off-diagonal term(s) of the Kerr metric?</p> <p>This question was similarly asked <a href="https://physics.stackexchange.com/questions/73514/kerr-metric-in-orthogonal-form">here</a>, but the answers were unsatisfying because they are all stated as fact without citation or proof. </p> <p>Lastly, would this be better for the <a href="https://math.stackexchange.com/">Mathematics</a> Stack Exchange? Is it purely a question of differential geometry?</p> https://physics.stackexchange.com/q/150446 7 Derivation of Kerr metric, is there any reference? phy_math https://physics.stackexchange.com/users/57252 2014-12-04T15:28:52Z 2018-09-25T01:51:19Z <p>In studying general relativity, many text deals with the derivation of Schwarzschild metric starting from generic metric form. After that impose static, spherical symmetry and obtain the desired Schwarzschild metric. </p> <p>But I haven't find any reference for above process in <a href="http://en.wikipedia.org/wiki/Kerr_metric" rel="noreferrer">Kerr metric</a>. (Add a condition of rotation.) In many textbooks on General relativity and black hole textbook they just state the form of Kerr metric, and do some calculation. </p> <p>Is there any reference (textbook or paper) contain explicit derivation of Kerr metric?</p> https://physics.stackexchange.com/q/420773 2 Kerr metric in BMS (Bondi-Metzner-Sachs) coordinates Nomenomen https://physics.stackexchange.com/users/203021 2018-08-02T19:20:36Z 2018-08-02T19:20:36Z <p>I am trying to put the Kerr metric into the famous Bondi gauge, which is given for instance by the formula (6.2.10) at page 154 of the following paper: <a href="https://arxiv.org/abs/1801.01714" rel="nofollow noreferrer">https://arxiv.org/abs/1801.01714</a>. Now, Barnich and Troessaert in their paper BMS charge algebra (<a href="https://arxiv.org/abs/1106.0213" rel="nofollow noreferrer">https://arxiv.org/abs/1106.0213</a>) did the calculations, which can be found in Appendix D and F of the paper. My questions are: 1) In the original metric the determinant of the angular part is different from the unit sphere metric determinant, hence they redefine the radial coordinate. I managed to work out all the new metric coefficients except for $g_{u\theta}$; in particular, I don't understand from where the new leading order coefficient $\frac{a\cos{\theta}}{2\sin^2{\theta}}$ comes from. Any help to understand this transformation would be very appreciated. 2) As one can see in Appendix F (it is pretty straightforward to calculate thiz), $C^{AB}C_{AB}=\frac{2a^2}{\sin^2{\theta}}$. However, in the Bondi gauge one has that the coefficient of the power $r^{-2}$ in $g_{ur}$ should be $\frac{C^{AB}C_{AB}}{16}=\frac{a^2}{8\sin^2{\theta}}$, and in the metric found by Barnich the same coefficient is actually $a^2(\frac{1}{2}-\cos^2{\theta})$. These are very different, unless I am missing some property of sinusoidal functions. Can anyone help me to solve this incongruence? </p> <p>Thanks for any help or hints!</p>