[math-fun] dumb question about general relativity
This question is about "classical" "infinitesimal" GR -- i.e., no quantum stuff like Hawking radiation & no Planck length. The laws of GR are symmetric w.r.t. time, so we have the following conundrum: If an object falls towards a black hole, from the perspective of an outside observer, time on the object gets slower & slower & in the limit it stops. I've been told that as a result of this time dilation, everything that "falls into" the black hole doesn't really fall in, but ends up getting painted onto its surface (from the perspective of the outside observer). But if the laws are truly symmetric w.r.t. time, then it should be possible for the black hole to "burp" out objects, as well. I've read expositions of what happens to an observer _on_ the object falling into the black hole, and time supposedly doesn't stop. Also, if the black hole is large enough, then the tidal forces near the surface aren't large enough to destroy even a human observer, so one can ask the question about what such an observer would see. All of this presupposes some sort of invertible transform in going from the perspective of the outside observer to the observer on the moving object. But clearly, such an invertible transform has a singularity at the "time" when the object gets painted onto the black hole (from the outside observer's perspective). Is there an analogy here with Taylor & Laurent series in the complex plane, where you can extend a function beyond its usual convergence by utilizing a different type of series?
On Fri, Jul 24, 2009 at 3:15 PM, Henry Baker <hbaker1@pipeline.com> wrote:
This question is about "classical" "infinitesimal" GR -- i.e., no quantum stuff like Hawking radiation & no Planck length.
The laws of GR are symmetric w.r.t. time, so we have the following conundrum:
If an object falls towards a black hole, from the perspective of an outside observer, time on the object gets slower & slower & in the limit it stops. I've been told that as a result of this time dilation, everything that "falls into" the black hole doesn't really fall in, but ends up getting painted onto its surface (from the perspective of the outside observer).
But if the laws are truly symmetric w.r.t. time, then it should be possible for the black hole to "burp" out objects, as well.
I don't think this is right. I think the right conclusion is that it should be possible for a different sort of object, a time-reversed black hole, that burps out objects, to exist. But it's wildly unlikely that any such object will come to exist, for entropic reasons. An analogy: if you break an egg, the fact that the laws of physics have time-reversal symmetry (or close enough for this example) doesn't mean that the pieces of eggshell will reform an egg. They mean that if the situation with all the egg fragments having exactly the negative of their current velocities happened, then they would form an egg. But in practice, we see eggs breaking, and not fragments reuniting, just as we see black holes but not white holes.
All of this presupposes some sort of invertible transform in going from the perspective of the outside observer to the observer on the moving object. But clearly, such an invertible transform has a singularity at the "time" when the object gets painted onto the black hole (from the outside observer's perspective).
I don't think there is such a time. I think that from the outside observer's perspective, the object gets exponentially closer to the event horizon with time, but never actually reaches it (Of course, when it gets within the Planck length, quantum-mechanical considerations will become relevant; but we're excluding those). So there is no singularity in finite time from the perspective of the outside observer. -- Andy.Latto@pobox.com
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Andy Latto Sent: Friday, July 24, 2009 3:18 PM To: math-fun Subject: Re: [math-fun] dumb question about general relativity On Fri, Jul 24, 2009 at 3:15 PM, Henry Baker <hbaker1@pipeline.com> wrote:
This question is about "classical" "infinitesimal" GR -- i.e., no quantum stuff like Hawking radiation & no Planck length.
The laws of GR are symmetric w.r.t. time, so we have the following conundrum:
If an object falls towards a black hole, from the perspective of an outside observer, time on the object gets slower & slower & in the limit it stops. I've been told that as a result of this time dilation, everything that "falls into" the black hole doesn't really fall in, but ends up getting painted onto its surface (from the perspective of the outside observer).
But if the laws are truly symmetric w.r.t. time, then it should be possible for the black hole to "burp" out objects, as well.
I don't think this is right. I think the right conclusion is that it should be possible for a different sort of object, a time-reversed black hole, that burps out objects, to exist. But it's wildly unlikely that any such object will come to exist, for entropic reasons.
Kruskal spacetime is the extension of a Schwarzschild Black Hole, I think. Note that, for a wormhole, you need spin or (possibly) charge in your black hole.
An analogy: if you break an egg, the fact that the laws of physics have time-reversal symmetry (or close enough for this example) doesn't mean that the pieces of eggshell will reform an egg. They mean that if the situation with all the egg fragments having exactly the negative of their current velocities happened, then they would form an egg. But in practice, we see eggs breaking, and not fragments reuniting, just as we see black holes but not white holes.
All of this presupposes some sort of invertible transform in going from the perspective of the outside observer to the observer on the moving object. But clearly, such an invertible transform has a singularity at the "time" when the object gets painted onto the black hole (from the outside observer's perspective).
I don't think there is such a time. I think that from the outside observer's perspective, the object gets exponentially closer to the event horizon with time, but never actually reaches it (Of course, when it gets within the Planck length, quantum-mechanical considerations will become relevant; but we're excluding those). So there is no singularity in finite time from the perspective of the outside observer.
The person falling in falls in (in his reference frame) quickly. An outside observer would see him asymptotically approach the event horizon, red-shifting more and more, but would not see him cross it. --
Andy.Latto@pobox.com
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Every (circular) cone including an ellipse fixed in 3-space (say, z=0, x^2/a^2 + y^2/b^2 = 1) has its apex on a hyperbola (in this case, y=0, x^2/(a^2-b^2) - z^2/b^2 = 1) thru the foci of the ellipse, and every cone including that hyperbola has its apex on the original ellipse, which, of course, pierces the foci of the the hyperbola. http://gosper.org/conethm.png . Coxeter promptly stomped off to the library and found it in a 1912 book by Salmon. It seems to me that a time-variant formation of low-flying jets could superpose their shockwaves in a moving, destructive hyperbola. --rwg
It seems that when x = pi sqrt(rational), (sum(n/(%e^(n*x)-1),n,1,inf)+1/(4*x)-1/24)/eta(%e^-x)^4 inf ==== \ n 1 1 > -------- + --- - -- / n x 4 x 24 ==== e - 1 n = 1 -------------------------- 4 -x eta (e ) is algebraic, but disappointingly messy (sqrt cubic surd) for rational = Beegner > 7, as you can judge by the rational = 44 case in http://gosper.org/nuetas.html . Rational = 11 isn't much better. --rwg
It seems that when x = pi sqrt(rational), (sum(n/(%e^(n*x)-1),n,1,inf)+1/(4*x)-1/24)/eta(%e^-x)^4
inf ==== \ n 1 1 > -------- + --- - -- / n x 4 x 24 ==== e - 1 n = 1 -------------------------- 4 -x eta (e )
is algebraic, but disappointingly messy (sqrt cubic surd) for rational = Beegner > 7, as you can judge by the rational = 44 case in http://gosper.org/nuetas.html . Rational = 11 isn't much better.
Understatement! Best I could do with it is Sum[(k/(E^(Sqrt[11]*k*Pi) - 1)), {k, 1, Infinity}]== ((29442373*Gamma[1/22]*Gamma[3/22]*Gamma[5/22]* Gamma[9/22]*Gamma[15/22])/(4224*(-11*2^(1/ 3)*(-88327119*Sqrt[3]*(999716798*Sqrt[11] - 2713716189) + 51490609396526601*Sqrt[11] - 58077417204452290)^(1/ 3) - ((88*2^(2/3)*(16292869731*Sqrt[11] - 38005314908))/(-88327119* Sqrt[3]*(999716798*Sqrt[11] - 2713716189) + 51490609396526601*Sqrt[11] - 58077417204452290)^(1/3)) + 8*(8747*Sqrt[11] - 63877))*(Sqrt[Pi])^7)) - (1/(4*Sqrt[11]* Pi)) + 1/24 {N[%], N[%, 33]} {False, True} FullSimplify has been over a day on it... --rwg
From: Henry Baker <hbaker1@pipeline.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, July 24, 2009 12:15:14 PM Subject: [math-fun] dumb question about general relativity This question is about "classical" "infinitesimal" GR -- i.e., no quantum stuff like Hawking radiation & no Planck length. The laws of GR are symmetric w.r.t. time, so we have the following conundrum: If an object falls towards a black hole, from the perspective of an outside observer, time on the object gets slower & slower & in the limit it stops. I've been told that as a result of this time dilation, everything that "falls into" the black hole doesn't really fall in, but ends up getting painted onto its surface (from the perspective of the outside observer). Yes. There is a last time at which a signal emitted by the infalling body can propagate to infinity. The final cycle of that signal is red shifted infinitely, and exponentially fades out. But if the laws are truly symmetric w.r.t. time, then it should be possible for the black hole to "burp" out objects, as well. White holes do not contradict classical general relativity, but perhaps they cannot exist in quantum gravity due to the vacuum becoming unstable. The big bang is something like a white hole, but quantum gravity will be needed to explain it. Then also, black holes do exhibit a whiteness, in the form of Hawking radiation, but again, it's a quantum effect. I've read expositions of what happens to an observer _on_ the object falling into the black hole, and time supposedly doesn't stop. Also, if the black hole is large enough, then the tidal forces near the surface aren't large enough to destroy even a human observer, so one can ask the question about what such an observer would see. An observer on the infalling body observes nothing in particular upon passing the event horizon, but reaches the central singularity in a finite proper time, which is about 0.1 ms for a solar mass black hole. Tidal forces become infinite as the singularity is approached. While falling in, the observer continues to receive signals from the external universe. The backward light cone from the point at which the singularity is reached delineates the final time for which external signals are received. I am not sure about whether such signals are red or blue shifted. Only a finite history of the universe can be viewed before the observer reaches his end-of-time. All of this presupposes some sort of invertible transform in going from the perspective of the outside observer to the observer on the moving object. But clearly, such an invertible transform has a singularity at the "time" when the object gets painted onto the black hole (from the outside observer's perspective). Is there an analogy here with Taylor & Laurent series in the complex plane, where you can extend a function beyond its usual convergence by utilizing a different type of series? Many of these questions can be answered by staring at Penrose diagrams. In particular, it sure does look like the central singularity is some kind of natural boundary to spacetime. The event horizon, however, is no more than a coordinate singularity, and spacetime is smooth here. -- Gene
participants (5)
-
Andy Latto -
Cordwell, William R -
Eugene Salamin -
Henry Baker -
rwg＠sdf.lonestar.org