leaving black holes
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From: bz on 2 Oct 2009 11:22 carlip-nospam(a)physics.ucdavis.edu wrote in news:ha2lvp$f4d$1 @skeeter.ucdavis.edu: > If she now sticks her hand toward the horizon, she's in trouble. Her > hand needs a huge acceleration to remain at rest, but it doesn't have > a rocket attached; it's accelerating only because it's connected to the > rest of her body, which presumably is in her spacecraft. Bones aren't > strong enough to transmit a near-infinite acceleration; nothing is. What if the black hole were the size and mass of our universe? -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set.
From: dlzc on 2 Oct 2009 12:31 Dear Tom Roberts: original newsgroup list corrected On Oct 1, 9:53 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > N:dlzcD:aol T:com (dlzc) wrote: > > > "eric gisse" <jowr.pi.nos...(a)gmail.com> wrote in message > >news:ha3eg8$935$3(a)news.eternal-september.org... > >>dlzcwrote: > >>> 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. The typical definition is an "Einstein ring" with an infinite number of "laps". http://en.wikipedia.org/wiki/Photon_sphere > 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). Einstein rings and your "salmon swimming upstream" are both unstable, for a structure that will continue to add or expell contents. > 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. Yes, Eric just was thinking "Schwarzchild radii" rather than tying it to mass directly, and I was hoping to unconfuse the issue with the OP. > 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. "A miss is as good as a mile." Eric was selling r_S as 1M. It is off by a factor of 2. And coordinate choice clarity is necessary: http://en.wikipedia.org/wiki/Mars_Climate_Orbiter > >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. An elegant argument. > [@] And gravitational radiation is > neglected, as for test particles like > rocket ships compared to stars. > > Tom Roberts Thank you. David A. Smith
From: "Juan R." González-Álvarez on 2 Oct 2009 15:04 Tom Roberts wrote on Fri, 02 Oct 2009 09:34:18 -0500: > 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". Unfortunately you get totally confused again by your superfitial knowledge of those topics Tom. In my original message I refered *explicitely* to renormalized coordinates, which do NOT describe "the same manifold". There is kind of equivalence between both sets of coordinates. But this is not valid everywhere. In fact for r<<2M there is a large difference between both coordinates (in the renormalized spacetime there is NOT central singularity for instance). >> 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. Above you sniped an relevant part of the message from dlcz to which you said "not really". REINTRODUCING SNIPED PART If you choose a different coordinate system. This leaves you still 1M out from the event horizon. >> Notice this renormalized r is not the Schwarzschild r. Don't get >> confused by symbols Tom! > > It is not I who is confused here. This would be credible if you were not sniping witjout noticing the snip -- http://www.canonicalscience.org/ BLOG: http://www.canonicalscience.org/en/publicationzone/canonicalsciencetoday/canonicalsciencetoday.html
From: eric gisse on 2 Oct 2009 15:19 dlzc wrote: [...] > "A miss is as good as a mile." Eric was selling r_S as 1M. No. [...]
From: Bernhard Kuemel on 2 Oct 2009 15:42
bz wrote: > carlip-nospam(a)physics.ucdavis.edu wrote in news:ha2lvp$f4d$1 > @skeeter.ucdavis.edu: > >> If she now sticks her hand toward the horizon, she's in trouble. Her >> hand needs a huge acceleration to remain at rest, but it doesn't have >> a rocket attached; it's accelerating only because it's connected to the >> rest of her body, which presumably is in her spacecraft. Bones aren't >> strong enough to transmit a near-infinite acceleration; nothing is. > > What if the black hole were the size and mass of our universe? Who do you think might stick her hand into an infinite universe? :) Bernhard |