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black holes vs time dilation |
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| Dec5-06, 11:21 PM | #1 |
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black holes vs time dilation
Ok, I’m a layman with an interest in physics. I don’t know any of the math beyond high school physics and first semester college calculus (with trig) a long time ago. I’ve got pretty good handle on special relativity (for a layman) and an introduction to the concepts of general relativity.
Here’s my premise. From the frame of reference of a distant observer not experiencing any time dilation from a black hole, I have read/seen it explained several times that such an observer will never see a space ship, or whatever, cross the event horizon of a black hole since such a ship will be experiencing ever increasing time dilation from the effects of both special and general relativity as it accelerates toward the event horizon. From the frame of reference of said external observer, the ship will appear to freeze at or near the event horizon. Indeed, I have read/seen it said that we have never observed anything cross such an event horizon. Matter may be swirling around at speeds approaching C, but we have not observed any of it cross the event horizon. 1) How can a black hole be said to consume surrounding matter from the frame of reference of a distant observer? Wouldn’t such an observer see it continually accumulate near the event horizon, yet never go in? I guess what I’m getting at is how much time dilation would such a ship experience, relative to a distant observer, prior to crossing the event horizon? The impression I have received is that it would be an infinite amount, i.e. the ship would never cross the event horizon from the distant observer’s frame of reference. 2) Has any theory been put forth for the huge jets of matter sometimes seen being ejected from the center, perpendicular to the accretion disk at speeds approaching C? 3) Is the formation of an accretion disk related to “frame dragging”? Presumably from a rotating singularity. 4) Is "obital procession" related to or caused by "frame dragging"? thanks |
| Dec6-06, 02:44 AM | #2 |
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For your first point, see any of the zillion threads on the same topic. While the observer 'at infinity' never sees the black hole form, the observer falling through the event horizon reaches the singularity in a finite proper time.
Once the analogy between the "black hole" event horizon and the Rindler horizon of an accelerated observer became known, people for the most part aware of the analogy abanonded the "frozen star" idea. Sea any of the aforementioned threads for more info on the Rindler horizon, associated with an accelerated observer, which provides a useful analogy. For your second question, I'm afraid I don't know very much about the jets. I did find http://imagine.gsfc.nasa.gov/docs/as...s/990923a.html, but I'm not sure how helpful it will be. Maybe someone else will know more. For your last question, orbital precession occurs even around a non-rotating black hole (Schwarzschild black hole). While frame dragging could cause addional precession, frame dragging is not needed to cause precession. |
| Dec6-06, 05:07 AM | #3 |
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I think that the jets form because the material accumulates around the black hole faster than it can fall in, the resulting pressure forces infalling material to flow around the edges existing disk or to displace plasma already in the disk. The magnetic fields formed by the plasma of the accretion disc then confines the displaced plasma to near the rotational axis of the black hole. The energy gained through interaction with the disk and the magnetic fields will allow some of the matter to attain escape velocity, which creates the jets.
EDIT: Just read pervect's link, and it seems that although I am partly correct, astrophysicists are still looking for a good model that shows how the magnetic fields confine the jets and for how the material leaves the accretion disc. Interesting read. |
| Dec6-06, 05:13 PM | #4 |
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black holes vs time dilationThe so-called "gravitational time dilation" is a straightforward curvature effect. In any curved manifold, initially parallel geodesics will in general converge (positive curvature) or diverge (negative curvature) as you run along one of them. Near the exterior of the event horizon of a black hole (in the simplest case, this situation is modeled by the Schwarzschild vacuum solution of the Einstein field equation of gtr, or EFE for short), two radially outgoing null geodesics corresponding to signals sent from an infalling observer will diverge. That means that when the signals are recieved by our distant static observer, the time between the two, as measured by an ideal clock carried by this static observer, will be larger than the time between the emission of the two signals, as measured by an ideal clock carried the infalling observer. These two "ideal clocks" are assumed to be absolutely identical and in particular, by definition they always "run at the same rate" under any circumstances (a real clock, even an atomic clock, will be affected by acceleration and so on); the "relativity" in gtr can be taken to refer to the fact that when we compare identical ideal clocks located at different "places", we must expect discrepancies, depending upon the details of the ambient gravitational field, the relative motion of the observers, and the method by which the comparison is made (typically, lightlike signals, but these can in general take more than one path and there are other complications we probably don't want to get into here). Also, the problem of describing optical effects in gtr is interesting and valid, but not the same as the problem of describing clock effects, so be careful here. If you meant to ask what our distant observer, A, would literally -see- if our infalling observer, B, were say pointing a laser beam steadily in A's direction as he falls toward the hole, then A would see the spot of light redden and then very rapidly wink out as B nears the horizon. For a stellar mass black hole, in fact, this would happen in about 10^-5 seconds! Regarding "speeds approaching C", note that even in flat spacetime, there are in fact multiple distinct but operationally significant notions of distance valid in large regions, all of which agree in very small neighborhoods (in gtr, the latter fact can be understood as a consequence of the "strong equivalence principle"). According to gtr, curvature singularities should exist inside the horizon, but this is irrelvant since signals cannot escape from inside the horizon. Geodetic precession or de Sitter precession (see any gtr textbook for "the precession of the perihelia of Mercury") does not involve frame dragging; the classic formula provided by Einstein works the same for a rotating or nonrotating isolated massive object (not just a black hole). You can think of this effect as saying that a small object in a bound orbit around a massive object will exhibit quasi-Keplerian motion, but the long axis of the "almost elliptical orbit" will very slowly rotate over time at a steady rate. This effect has been confirmed in solar system observations (for Mercury, Venus, the Earth, and various asteriods) and also in binary star systems in which one or both "stars" are neutron stars or black holes, most notably the Hulse-Taylor binary. According to gtr, a gyroscope orbiting a rotating object will experience an additional small precession called "Lense-Thirring precession". This effect does involve "gravitomagnetism" and the Standard "Gravity Probe B" satellite experiment has been testing it. Chris Hillman |
| Dec6-06, 10:11 PM | #5 |
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Wow. Ok, I clearly know even less about gtr than I thought. And apparently I know nothing about black hole geometry except what I've heard in popular science.
Ok, I didn't understand anything between "gravitational time dilation" and "observer will diverge". I think I know what a "Schwarzchild black hole" is though. A stationary, non-rotating black hole that has already collapsed. Basically a black hole with the simplest possible math. So, are you saying that there is only the appearance of time dilation under gtr? Ok, take two ideal clocks "at infinity". One travels to the black hole, gets near the event horizon, then safely returns to it's starting point next to the other ideal clock. Won't the traveling clock have experienced very large time dilation as compared to the stationary clock, even after we calculate out the effect from str? I'm not sure how to rephrase my original question and set aside the observational distortions caused by gtr, but here goes. A space ship travels to a black hole and crosses the event horizon. When it crosses the event horizon won't it be in the far, far, far distant future of the observer "at infinity"? The astronaut aboard the ship will experience time normally from his "point of view" at least until he is killed by the tidal forces, etc. But if he had a "crystal ball" and was able to observe the ideal clock of the observer "at infinity", wouldn't the clock at infinity be millions or billions of years ahead of his own clock? Thanks for taking the time to answer my questions. You obviously took quite a bit of time writing a response. I've been looking for some of the other posts that Pervect referred to and have not yet seen an explanation that I understand. I will have to study the material you both linked to, but the math and geometry involved is way beyond anything I have studied. It is fascinating though. I was once a physics major, so I love this stuff, but I just don't have the foundation necessary to understand a lot of it. |
| Dec6-06, 11:37 PM | #6 |
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I enthusiastically second pervect's recommendation of the popular book by Geroch. In fact, I think it will prove the perfect book for you!
Enjoy! |
| Dec10-06, 06:28 PM | #7 |
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Yes, I hope it will, but in the mean time, could you help to clear up some of my misconceptions?
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| Dec15-06, 06:40 PM | #8 |
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anyone? anyone?
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| Dec15-06, 08:06 PM | #9 |
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Or think about moving walkaways, for a black hole you are standing on them and they go faster and faster the closer you get to the singularity. At one point there is no escape, since you obviously cannot walk faster on them than c. Does that make a bit more sense? Remember that in SR and GR there is no absolute time, no universal clock. Each mass object has its own sense of elapsed time. And this eapsed time depends on the path it traveled in space-time. |
| Dec15-06, 08:07 PM | #10 |
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I thought we addressed at least some of them? What are you still puzzled by?
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| Dec31-06, 09:37 AM | #11 |
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I hope this question belongs in this thread. If someone on the surface of a star which just started collapsing into a black hole is receiving signals sent every second from an external point in "flat space", then the signals the surface receives will be very blue shifted -- As the density of the object increases, won't the time dilation with respect to the external signal source approach infinity before the surface can collapse completely? It seems like a singularity (a point of infinite density) could never form in the sense that the signals from the external observer would continue to strike the non-singular object that is in the process of collapsing throughout all future history.
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| Dec31-06, 01:23 PM | #12 |
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The two effects work in opposite directions. If you imagine someone somehow "hovering" at a constant r coordinate, the blueshift becomes infinite at the event horizon. But this is not an infalling observer, it's a station-holding observer. Calculating the red/blue shift depends on the trajectory. I'd have to look up the thread where this was discussed, but if you assume an observer free-falling from infinity (i.e. he has zero velocity at infinity) into a black hole, the total shift at the event horizon is a redshift which halves the frequency of light falling in "radially". The redshift/blueshift also depends on direction in which he looks, the factor of 2 in the above example is for the observer who looks directly "up" at radially infalling light. Besides the past thread, there's some discussion in http://casa.colorado.edu/~ajsh/singu...l#redshift.map (I think there's even more discussion elsewhere on this webpage about the issue). |
| Dec31-06, 02:24 PM | #13 |
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The radius of the star will continue to shrink and eventually after it reaches a certain density the event horizon will be outside of what is left of the star. Then everything inside is trapped and cannot escape. The star will continue to collapse to a singularity in finite proper time. But for everything inside the black hole that is of no matter, everything inside will collapse to a singularity in finite proper time. Note that there is no such thing as absolute time in GR so even if some event would take say a few billion years for one observer it could be ten minutes for another observer. |
| Dec31-06, 02:35 PM | #14 |
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Even if there is no absolute time in GR, it seems to make sense to say that since any signal sent from flat space time into any black hole will (and always will) strike a collapsing, but not collapsed object, we can say that the universe contains no singularities. By the proper time of the surface of the collapsing objects, they will receive signals (and collisions with other objects such as black holes) from the outside (flat space-time) for all future history before they can complete their collapse, even if it only takes 10**-5 seconds by their own proper time to complete their collapse -- They "experience" those collisions all at once, just as they begin their collapse. Even by their own proper time, I'd say they never get through that 10**-5 seconds.
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| Dec31-06, 11:02 PM | #15 |
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Take a look at the Eddington-finklestein diagram for a pressureless dust collapsing into a black hole, for example. For instance, http://casa.colorado.edu/~ajsh/colla...ml#finkelstein The infalling light rays (yellow lines at 45 degree angle) will eventually strike, not the collapsing sphere of dust (the white curving line), but the singularity, a totally collapsed object (the vertical cyan line). Thus a signal sent from flat space-time can and will strike a collapsed object (i.e. the singularity). |
| Jan1-07, 07:01 AM | #16 |
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Pervect -- Thanks for the response and the link to the diagrams. I am still not seeing how it can be the case that the white line ever turns Cyan if those yellow lines are drawn as shown -- i.e., if you've got density that is approaching infinity, then the gravitational potential will approach infinity and the time dilation with respect to the source of those yellow lines ought to go to infinity too. I would think that either the yellow lines ought to be drawn never hitting the zero radius point or the white line ought to just go up the vertical axis and never hit the zero radius point.
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| Jan1-07, 02:12 PM | #17 |
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GR doesn't really have any concept of "gravitational potential". But it is true that the metric coefficients become singular inside the singularity. g_00, which represents time dilation outside the event horizon, becomes negatively infinite, while g_rr goes to zero. Your notion of time dilation becoming infinite can be made to make some sense by observing that r and t "switch roles" inside the event horion - i.e., if you look at the Schwarascihld metric, inside the event horizon r becomes a time coordinate (because it has a negative metric coefficient) while outisde the event horizon r is a spatial coordinate. Hence we look at g_rr instead of g_tt for time dilation inside the horizon, and we see that g_rr goes to zero - which is "infinite time dilation". However, it simply doesn't follow that g_rr going to zero means that events "never happen" as viewed by the perspective of an infalling observer. To really get into the detials, you'll probably need a textbook. The textbook "Gravitation", authors Misner, Thorne Wheeler (abbr. MTW) for instance, talks about the collapse of a pressureless dust on pg 859 Eddingtion Finklestein coordinates are talked about on pg 828 You'll probably be better off with a book like "Exploring Black Holes" (by some of the same authors as MTW) rather than MTW itself, though I can't guarantee that they'll go into the "dust collapse" model in detail. I would suggest trying to understand the Eddingtion-Finklestein diagram of the Schwarzschild geometry first, then worry about the refinement of the collapse of the dust shell later. Online, you can look at http://casa.colorado.edu/~ajsh/schwp.html for some of this information (but a textbook would still be a better bet). If you look at the Schwarzschild metric: ds^2 = -(1-2M/r) dt^2 + 1/(1-2M/r) dr^2 + r^2 (d theta^2 + sin^2(theta) dphi^2) you can solve for the path of light by setting ds=0. (The lorentz interval of a ligthbeam is always zero) For radially infalling light, dtheta=dphi=0 so you get -(1-2M/r) dt^2 + 1/(1-2M/r) dr^2 = 0 This gives you some of the information you need as to how to assign coordinates by rescaling time (EF coordinates) so that light always appears to travel at 45 degree angles, which is the main point of an EF diagram. |
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