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On the nature of the infinite fall toward the EH 
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#1
Dec412, 05:04 PM

P: 266

Observers Alice and Bob are hovering far above the event horizon of a block hole. Alice stops hovering and enters free fall at time T_0. Bob waits an arbitrary amount of time, T_b, before reversing his hover and chasing (under rocketpropelled acceleration A_b) after Alice who continues to remain in eternal free fall.
Question: For any time T_b does there exist an acceleration A_b (however impractical yet physically possible) such that Bob can reach Alice before she crosses the event horizon, therefore rescuing her from doom? 


#2
Dec412, 05:28 PM

PF Gold
P: 4,087

I'm guessing that if Bob can survive any gforce then he could reach Alice before she crosses the EH. However, knowing how tricky relativity is, there is could be some point above the EH beyond which the rescue is impossible. This point would depend on T_b/R0 where r=R0 is the initial position.



#3
Dec412, 05:56 PM

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I would say that if bob emits light flashes at regular intervals, there will be a "last flash" emitted at time T_b1 that Alice can see before she enters the horizon (and another "last flash" at T_b2 that Alice can see before she reaches the singularity).
As a consequence of this, for T>T_b1, not even light emitted by Bob could reach Alice before she reaches the horizon, and since Bob can't ever overtake a light beam, so the answer must be no. 


#4
Dec412, 06:01 PM

P: 266

On the nature of the infinite fall toward the EH



#5
Dec412, 06:35 PM

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PF Gold
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 That Bob can never get a signal from Alice at or inside the horizon. Thus, horizon crossing events are never part of Bob's past light cone. However, it is equally true that:  Alice receives a specific last signal from Bob on crossing the horizon, and another (in the limit) on approach to the singularity. As a result, horizon crossing events are most defininitely in Bob's future light cone  just never in his past light cone. For alice, her past light cone includes events in Bob's history until she reaches the singularity. However, once she passes the horizon, her future light cone is strictly interior to the horizon, and always includes the singularity. For a nonrotating, uncharged BH, Alice's future light cone necessarily includes less and less of the interior until reaching the singularity. FYI: Bob can definitely make such an announcement if he so desires. For example, here is one way: http://physicsforums.com/showpost.ph...0&postcount=23 


#6
Dec412, 06:55 PM

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Wouldn't it be much simpler to replace Bob by a photon from the very beginning? If the photon can't reach Alice before crossing the EH, Bob can't reach her, either



#7
Dec412, 07:01 PM

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#8
Dec412, 07:06 PM

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Shouldn't be too difficult to calculate that 


#9
Dec412, 08:29 PM

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#10
Dec412, 08:49 PM

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If discussing GR+quantum corrections, the view that evaporation prevents matter from crossing a horizon (or from a horizon ever forming) is just one opinion. I believe it is the minority opinion, due to 2009 paper by Padmanabhan (that is, this paper refuted arguments in a 2007 paper that horizon never forms, and no major paper since has refuted Padmanabhan's arguments, that I know of). 


#11
Dec512, 02:07 AM

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#12
Dec512, 03:19 AM

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http://casa.colorado.edu/~ajsh/collapse.html For some detailed calculations: phase 1. Show that in Schwarzschild coordinates for a black hole of mass m=2, the geodesic is given by [itex]\infty < \tau < 0 [/itex] [tex]r = {3}^{2/3} \left( \tau \right) ^{2/3}[/tex] [tex]t = \tau4\,\sqrt [3]{3}\sqrt [3]{\tau}+4\,\ln \left( \sqrt [3]{3}\sqrt [3]{\tau}+2 \right) 4\,\ln \left( \sqrt [3]{3}\sqrt [3]{\tau}2 \right) [/tex] By showing that it satisfies [tex]\frac{dr}{d\tau} = \sqrt \frac {2m}{r}[/tex] [tex]\frac{dt}{d\tau} =\frac{1}{12m/r}[/tex] See for instance http://www.fourmilab.ch/gravitation/orbits/, or your favorite GR textbook. m=2 was chosen to make the expressions more tractable, you may choose to repeat without this attempt at simplification if you prefer phase 2: convert to ingoing Eddington Finklestein coordinates by the transformation [tex]v = t + r + 4\,\ln \left \frac{r}{2m}  1 \right [/tex] (recall that we set m=2 in phase 1). Phase 2a: recall, or derive, that for infalling light, v=constant. Therefore v(tau) gives you the "flash number" you are viewing at time tau. Get [tex]v = \tau4\,\sqrt [3]{3}\sqrt [3]{\tau}+4\,\ln \left( \sqrt [3]{3}\sqrt [3]{\tau}+2 \right) 4\,\ln \left( \sqrt [3]{3}\sqrt [3]{\tau}2 \right) +{3}^{2/3} \left( \tau \right) ^{2/3}8\,\ln \left( 2 \right) +4\,\ln \left( \left \left( \sqrt [3]{3}\sqrt [3]{\tau}+ 2 \right) \left( \sqrt [3]{3}\sqrt [3]{\tau}2 \right) \right \right) [/tex] Use the fact that ln(a*b) = ln(a)+ln(b) to rewrite this and cancel out the apparent singularity in v [tex] v = \tau4\,\sqrt [3]{3}\sqrt [3]{\tau}+8\,\ln \left( \sqrt [3]{3}\sqrt [3]{\tau}+2 \right) +{3}^{2/3} \left( \tau \right) ^{2/3}8\,\ln \left( 2 \right) [/tex] Observe that v is finite (zero) when tau > 0, so that you do not in fact see the entire history of the universe before you reach the event horizon, furthermore that you don't see the entire history of the universe before you reach the singularity. If you don't like the formal cancelllation of the divergent terms note that you can compute the limit of v as you approach the event horizon, and show that the limit exists to answer the original question. (You won't see any results inside the event horizon that way though). Option: recompute the geodesic equations in EF coordinates and show that the modified solution satisfies them to justify the formal cancelation of the divergent terms. If this seems like waaaaay too much work, just study Hamilton's EF plot, or find the plot of an infalling observer in Eddington Finklestein coordinates in your favorite textbook. 


#13
Dec512, 04:32 AM

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There's a rather nice way of summarizing the above results from the infalling observer's point of view. If an observer at infinity is shining a beam of constant frequency downwards, the infalling observer can measure the doppler shift of the radially infalling light as a function of proper time, or as a function of the Schwarzschild r coordinate.
This encapsulates what one would predict an infalling observer would actually see and measure, without getting overly involved in setting up coordinate systems and such. The doppler shift is just [itex] dv / d\tau [/itex]. The equation in terms of r is particular simple compared to the rather messy equations we've seen to date: doppler shift = sqrt(r) / [ sqrt(r) + 2] One can see that the doppler shift starts out at 1 at infinity, and that at the event horizon at r=4, the doppler shift is 1/2, so the incoming frequency is halved. Furthermore, the doppler shift is always < 1, there's always a redshift (assuming you are looking straight behind you), which increases as you approach the horizon, and the recieved frequency tends towards zero as you approach the central singularity  for a Schwarzschild black hole. This makes sense, as the only "gravity" in the freefall frame is tidal forces, and those would tend to be of the sort to cause redshift, not blueshift. 


#14
Dec512, 08:08 AM

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#15
Dec512, 10:23 AM

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#16
Dec512, 11:09 AM

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PF Gold
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http://arxiv.org/abs/grqc/0609024 (published Phys. Rev. D 2007) and here is a claimed refutation, link to abstract (more info) rather than PDF: http://arxiv.org/abs/0906.1768 (published in Phys. Rev. D 2009) 


#17
Dec512, 12:24 PM

PF Gold
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