leaving black holes
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From: N:dlzc D:aol T:com (dlzc) on 1 Oct 2009 21:22 Dear eric gisse: "eric gisse" <jowr.pi.nospam(a)gmail.com> wrote in message news:ha3eg8$935$3(a)news.eternal-september.org... > dlzc wrote: > [...] > >> You hyperboilically fly around a black hole within (I believe) >> 3M and > > Last stable orbit. Photon sphere is 3M. Last stable orbit for matter is 6M. > Arbitrary acceleration can let you get arbitrarily close > to r = 2M. If you choose a different coordinate system. This leaves you still 1M out from the event horizon. http://casa.colorado.edu/~ajsh/orbit.html .... "3 down to 1.5" Schwarzchild radii (Sr), where I have the Sr at 2M. So 6M down to 3M is unstable, meaning you are going in or leaving with thrust properly applied. David A. Smith
From: Tom Roberts on 2 Oct 2009 00:53 N:dlzc D:aol T:com (dlzc) wrote: > "eric gisse" <jowr.pi.nospam(a)gmail.com> wrote in message > news:ha3eg8$935$3(a)news.eternal-september.org... >> dlzc wrote: >>> You hyperboilically fly around a black hole within (I believe) >>> 3M and >> Last stable orbit. > > Photon sphere is 3M. Last stable orbit for matter is 6M. [This is all for a Schwarzschild black hole. r is the usual Schw. coordinate.] There is an unstable circular null geodesic at r=3M that orbits the black hole, which might be what you are trying to say, but "photon sphere" does not mean that to me. What I would call the "photon sphere" is right at the horizon, r=2M -- that is, outgoing light at the horizon remains at the horizon (an unstable situation). The smallest stable timelike orbit is indeed at r=6M, and I'm pretty sure it is true that if an ORBIT goes inside r=6M it will destabilize and fall in or fly away. But there are hyperbolic-like timelike trajectories that come in from far away, approach between r=3M and r=6M, and then leave; these are not orbits, and none come closer than r=3M. It is also true that any timelike or null geodesic that comes from far away and gets below r=3M will enter the horizon (in some cases it can circle the horizon several times before entering). But light that is headed outward and emitted by a timelike object from 2M<r<3M will get away; that's different from light coming in from far away. >> Arbitrary acceleration can let you get arbitrarily close >> to r = 2M. > > If you choose a different coordinate system. This leaves you > still 1M out from the event horizon. Not really. The horizon is at r=2M. Coordinate choice has nothing whatsoever to do with the locus of the horizon or the location a given spaceship is able to hover outside it with a given thrust [#]. With arbitrarily-high thrust, a spaceship could hover arbitrarily close to the horizon at r=2M. This is an unstable situation if its r is close to 2M. [#] Coordinate choice only affects your DESCRIPTION, not the physical situation. > http://casa.colorado.edu/~ajsh/orbit.html Hmmm. This page says "The orbit at 2 Schwarzschild radii corresponds to zero kinetic energy at infinity, so it is possible to fall freely into this orbit from infinity without rocket power." -- this is wrong, and it is NOT possible to enter an orbit at ANY radius without thrusting, when one starts from infinity and is free-falling [@]. Here by "orbit" I mean a trajectory with a permanent spatial path circling the black hole. This is easy to see, as the spatial path of any timelike geodesic can be traversed in the opposite direction, within the region outside the horizon -- an incoming trajectory that falls into into an orbit, when reversed, would mean it is not an orbit. [@] And gravitational radiation is neglected, as for test particles like rocket ships compared to stars. Tom Roberts
From: "Juan R." González-Álvarez on 2 Oct 2009 08:04 Tom Roberts wrote on Thu, 01 Oct 2009 23:53:35 -0500: (...) > [This is all for a Schwarzschild black hole. r is the > usual Schw. coordinate.] (...) >>> Arbitrary acceleration can let you get arbitrarily close to r = 2M. >> >> If you choose a different coordinate system. This leaves you still 1M >> out from the event horizon. > > Not really. The horizon is at r=2M. Coordinate choice has nothing > whatsoever to do with the locus of the horizon or the location a given > spaceship is able to hover outside it with a given thrust [#]. As you point above r denotes the "Schw. coordinate". It is possible to introduce other coordinates and then the horizon is not at 2M but at a different place. Concretely renormalized coordinates give r=1M and r=2M is just 1M away from the horizon. Notice this renormalized r is not the Schwarzschild r. Don't get confused by symbols Tom! -- http://www.canonicalscience.org/ BLOG: http://www.canonicalscience.org/en/publicationzone/canonicalsciencetoday/canonicalsciencetoday.html
From: dlzc on 2 Oct 2009 10:28 Dear eric gisse On Oct 1, 4:37 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > dlzc wrote: > > [...] > > > You hyperboilically fly around a black hole within > > (I believe) 3M and > > Last stable orbit. No. With the event horizon at 2M (as previously described in this thread), the closest possible stable orbit for matter is 6M. > Arbitrary acceleration can let you get arbitrarily > close to r = 2M. You mangled the original newsgroup listing, so Tom's excellent and thorough response did not make it on the sci.astro branch of this thread. Please be careful how you apply Occam's razor here... David A. Smith
From: Tom Roberts on 2 Oct 2009 10:34
Juan R. González-Álvarez wrote: > Tom Roberts wrote on Thu, 01 Oct 2009 23:53:35 -0500: >> [This is all for a Schwarzschild black hole. r is the >> usual Schw. coordinate.] >> The horizon is at r=2M. Coordinate choice has nothing >> whatsoever to do with the locus of the horizon or the location a given >> spaceship is able to hover outside it with a given thrust [#]. > > As you point above r denotes the "Schw. coordinate". It is possible to > introduce other coordinates and then the horizon is not at 2M but at > a different place. You use words in an unusual way -- this is not a different "place" (which denotes location in the manifold), but rather a different VALUE OF "r" corresponding to THE SAME PLACE IN THE MANIFOLD. Using different coordinates does NOT "move" the horizon, or make it be at a "different place", it makes the (fixed) locus of the horizon be at a DIFFERENT VALUE OF THE COORDINATE YOU HAPPEN TO LABEL WITH THE SAME SYMBOL, "r". > Concretely renormalized coordinates give r=1M and > r=2M is just 1M away from the horizon. Sure -- you can "renormalize" coordinates however you wish. Another well-known set of coordinates puts the horizon at r=0. But that is IRRELEVANT to what I said, as my statements all used the Schw. r, not any of these other ones. > Notice this renormalized r is not the Schwarzschild r. Don't get confused > by symbols Tom! It is not I who is confused here. Tom Roberts |