# On the nature of the "infinite" fall toward the EH

by rjbeery
Tags: fall, infinite, nature
 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 rocket-propelled 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?
 PF Gold P: 4,081 I'm guessing that if Bob can survive any g-force 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.
 Emeritus Sci Advisor P: 7,443 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.
P: 266

## On the nature of the "infinite" fall toward the EH

 Quote by pervect 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.
Are you sure there would be a "last flash"? I'd be curious to see this analyzed mathematically. Reason being, if there were such a flash calculable by Bob then he could announce definitively "when" Alice has crossed the EH, which contradicts my understanding.
PF Gold
P: 4,863
 Quote by rjbeery Are you sure there would be a "last flash"? I'd be curious to see this analyzed mathematically. Reason being, if there were such a flash calculable by Bob then he could announce definitively "when" Alice has crossed the EH, which contradicts my understanding.
There is definitely such a flash. All too often, popular presentation present only half the causal structure of a BH:

- 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 non-rotating, 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
 Sci Advisor P: 5,307 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
PF Gold
P: 4,863
 Quote by tom.stoer 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
But the OP was all about Bob hovering after Alice started falling, and deciding at some point to try to catch Alice. Do you mean replace 'try to catch' with 'try to send a light signal'? If so, that is the essential issue; and I thought that's what Pervect was pointing out. Once a light signal would only catch Alice at or inside the horizon, it is too late for Bob to rescue Alice. Any time before this, it is possible, in principle, for Bob to rescue Alice.
P: 5,307
 Quote by PAllen Do you mean replace 'try to catch' with 'try to send a light signal'?
Yes

 Quote by PAllen and I thought that's what Pervect was pointing out
I overlooked that ...

 Quote by PAllen Once a light signal would only catch Alice at or inside the horizon, it is too late for Bob to rescue Alice. Any time before this, it is possible, in principle, for Bob to rescue Alice.
Exactly

Shouldn't be too difficult to calculate that
P: 266
 Quote by PAllen There is definitely such a flash. All too often, popular presentation present only half the causal structure of a BH: - 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 non-rotating, 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
When you say "definitely", is that taking dissipative effects such as Hawking Radiation into account? If Hawking Radiation exists, in my understanding, Alice would appear to "almost" reach the EH and appear to continue to do so as the BH dissipates and the Schwarzschild radius is reduced.
PF Gold
P: 4,863
 Quote by rjbeery When you say "definitely", is that taking dissipative effects such as Hawking Radiation into account? If Hawking Radiation exists, in my understanding, Alice would appear to "almost" reach the EH and appear to continue to do so as the BH dissipates and the Schwarzschild radius is reduced.
Hawking radiation does not actually apply to the SC geometry. Firstly, the SC geometry never quite forms; secondly, Hawking radiation precludes an exact spherical symmetry. Further, you must distinguish classical GR (which does not include Hawking radiation), from GR + quantum corrections as an approximation to some unknown successor theory. I thought we were discussing classical GR.

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).
Emeritus
P: 7,443
 Quote by PAllen Hawking radiation does not actually apply to the SC geometry. Firstly, the SC geometry never quite forms; secondly, Hawking radiation precludes an exact spherical symmetry. Further, you must distinguish classical GR (which does not include Hawking radiation), from GR + quantum corrections as an approximation to some unknown successor theory. I thought we were discussing classical GR. 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).
Would that be http://arxiv.org/pdf/0906.1768.pdf ?
Emeritus
P: 7,443
 Quote by rjbeery Are you sure there would be a "last flash"? I'd be curious to see this analyzed mathematically. Reason being, if there were such a flash calculable by Bob then he could announce definitively "when" Alice has crossed the EH, which contradicts my understanding.
One can find a web reference by Hamilton (and some diagrams in Eddington-Finklestein coordinates) that show the existence of a last flash.

 Answer to the quiz question 5: False. You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time, and you witness only a finite (in practice rather short) time pass in the outside Universe.
Eddingtion Finklstein coordinates. Yellow lines are light, white line is infalling observer. You can see there is a last flash.

For some detailed calculations:

phase 1. Show that in Schwarzschild coordinates for a black hole of mass m=2, the geodesic is given by

$-\infty < \tau < 0$

$$r = {3}^{2/3} \left( -\tau \right) ^{2/3}$$
$$t = \tau-4\,\sqrt [3]{3}\sqrt [3]{-\tau}+4\,\ln \left( \sqrt [3]{3}\sqrt [3]{-\tau}+2 \right) -4\,\ln \left( \sqrt [3]{3}\sqrt [3]{-\tau}-2 \right)$$

By showing that it satisfies
$$\frac{dr}{d\tau} = \sqrt \frac {2m}{r}$$
$$\frac{dt}{d\tau} =\frac{1}{1-2m/r}$$

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

$$v = t + r + 4\,\ln \left| \frac{r}{2m} - 1 \right|$$

(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
$$v = \tau-4\,\sqrt [3]{3}\sqrt [3]{-\tau}+4\,\ln \left( \sqrt [3]{3}\sqrt [3]{-\tau}+2 \right) -4\,\ln \left( \sqrt [3]{3}\sqrt [3]{-\tau}-2 \right) +{3}^{2/3} \left( -\tau \right) ^{2/3}-8\,\ln \left( 2 \right) +4\,\ln \left( \left| \left( \sqrt [3]{3}\sqrt [3]{-\tau}+ 2 \right) \left( \sqrt [3]{3}\sqrt [3]{-\tau}-2 \right) \right| \right)$$

Use the fact that ln(a*b) = ln(a)+ln(b) to rewrite this and cancel out the apparent singularity in v

$$v = \tau-4\,\sqrt [3]{3}\sqrt [3]{-\tau}+8\,\ln \left( \sqrt [3]{3}\sqrt [3]{-\tau}+2 \right) +{3}^{2/3} \left( -\tau \right) ^{2/3}-8\,\ln \left( 2 \right)$$

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.
 Emeritus Sci Advisor P: 7,443 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 $dv / d\tau$. 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 free-fall frame is tidal forces, and those would tend to be of the sort to cause redshift, not blueshift.
PF Gold
P: 4,863
 Quote by pervect Would that be http://arxiv.org/pdf/0906.1768.pdf ?
Yes. I referenced it in another thread. Published in same journal as Krauss et. al.; refers to that paper explicitly as representative of the position it is refuting. And I didn't find any peer reviewed response to this Padmanabhan paper arguing the position of the 2007 paper. I also see most of the QG field just ignoring the Krauss et. al. argument and continuing discussion of how a horzion and interior behave with quantum, string, or loop corrections, and how the information paradox gets resolved; rather than accepting the view that there is no problem because a BH never forms. The whole recent debate on horizon firewalls between Polchinski and Susskind would be moot if either accepted the Krauss et. al. position.
P: 266
 Quote by PAllen Yes. I referenced it in another thread. Published in same journal as Krauss et. al.; refers to that paper explicitly as representative of the position it is refuting. And I didn't find any peer reviewed response to this Padmanabhan paper arguing the position of the 2007 paper. I also see most of the QG field just ignoring the Krauss et. al. argument and continuing discussion of how a horzion and interior behave with quantum, string, or loop corrections, and how the information paradox gets resolved; rather than accepting the view that there is no problem because a BH never forms. The whole recent debate on horizon firewalls between Polchinski and Susskind would be moot if either accepted the Krauss et. al. position.
Thanks to both of you, would you happen to have a reference to the 2007 paper as well? My only response to the text above is that BH's are interesting mathematical studies; I wouldn't personally take work in this area as evidence of their existence any more than Klein bottles.
PF Gold
P: 4,863
 Quote by rjbeery Thanks to both of you, would you happen to have a reference to the 2007 paper as well? My only response to the text above is that BH's are interesting mathematical studies; I wouldn't personally take work in this area as evidence of their existence any more than Klein bottles.
Here is the 2007 paper:

http://arxiv.org/abs/gr-qc/0609024 (published Phys. Rev. D 2007)