Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Poking someone falling into a black hole with a stick

  1. Jul 15, 2014 #1
    Please excuse me for being a complete and utter pleb, my meager knowledge of Leonard Susskind's holographic principle comes from the second episode of the documentary "Through the Wormhole" (which is available online but I won't link to in case it is copyright infringement?).

    He talks about the difference in perspective from an astronaut falling into a black hole as compared with an astronaut watching her fall.
    I thought I was following it until he got to the difference between a lone astronaut falling into a black hole vs. an astronaut in a plane.

    I don't understand why the moving propeller looks so different from everything else.

    Also, I figured when I was watching it that the woman falling into the black hole really was falling forever, from our perspective, and really was falling for 3 minutes from hers.
    However, talking about this with a friend he thinks that only the photons are left over forever, that that's why we can still see her falling a year later, but she is long gone.

    So imagine she started falling at 1km an hour (I suppose that is impossible.. let's just take these numbers as being relative) and a year later I can still see her, not very far away at all, stuck in space, still falling but not visibly moving at all, and I take out a stick and very quickly poke her with it. Setting aside for one moment whether that is possible, Is there anything there to poke?
    My thinking is that there is something there to poke, and that you are sending the stick back in time. But I don't know, maybe when looking at her we are looking at the past in the same sense as we are when we look at certain distant stars, there is only light reaching us, the source is gone.

    Then there are the logistics of poking someone with a stick like that, would the high gravity mean

    - that no matter how fast I moved the stick it would never catch up with her? Even if it was 500 times faster than her original speed?
    - that if i could poke her I wouldn't feel it, because the gravity would stop the reverberation from traveling up the stick


    Thanks so much in advance for any light you can shed on the subject!
     
    Last edited: Jul 15, 2014
  2. jcsd
  3. Jul 15, 2014 #2

    WannabeNewton

    User Avatar
    Science Advisor

    As an aside, this has nothing to do with BSM; this is a classical GR problem.

    Anyways, you're missing a lot of details in your description but that is only natural given that you are reading about this from a popularization. Let ##O## be an observer radially freely falling into a Schwarzschild black hole. If ##O'## is an observer far away from the black hole who is hovering in place, say using a rocket, and ##O'## attempts to two-way communicate with ##O## through light signals then the closer ##O## gets to the event horizon, the more and more red-shifted the light signals will get that are emitted by ##O## towards ##O'##. They will also take more and more coordinate time to reach ##O'## from ##O## as ##O## falls towards the event horizon.

    The end result is ##O## appears dimmer and dimmer to ##O'## and eventually there comes a point where ##O## is extremely dim and light signals from ##O## take an infinite amount of coordinate time to reach ##O'## so ##O'## sees ##O## "stuck" near the event horizon. But as your friend correctly pointed out, ##O## has long fallen into the event horizon and what ##O'## sees stuck is simply the past image of ##O##.

    Now the key point here is that ##O'## is an observer who is hovering in place at a fixed location using a rocket. That ##O'## is taken to be far away from the event horizon is simply for computational simplicity as the same effect would be observed by any observer who is fixed in place in the spatial part of Schwarzschild coordinates. Now if you got an extremely long pole stretching from your fixed spatial location to the event horizon and pointed it in the direction ##O## fell into the event horizon then the pole won't actually hit anything.

    Finally, you can certainly arrange for the initial conditions of the problem (e.g. initial velocity and position of ##O## relative to a neighboring hovering observer, initial position of ##O'##, and proper length of the stick when it is at rest relative to ##O'##) to be such that ##O'## manages to smack or poke ##O## before ##O## enters the event horizon and there will certainly be sound waves propagating from the end of the pole at ##O##'s instantaneous location during the collision to the fixed location of ##O## but the force felt by ##O## will be blueshifted by the usual gravitational time dilation factor compared to the transmitted force felt by ##O'##. If ##O## is already through the event horizon then obviously ##O'## has no way of telling if ##O## was hit or not because sound waves traveling through the part of the stick within the event horizon will not be able to escape the event horizon and this is assuming the part of the stick inside the event horizon hasn't already broken due to tensile stresses from tidal gravity. And if ##O'## tries to shove the stick just barely into the event horizon so as to pull back on it, the stick will break.
     
    Last edited: Jul 15, 2014
  4. Jul 15, 2014 #3

    ChrisVer

    User Avatar
    Gold Member

    Isn't that supposed to be a problem of the singularities of Schwarzchild's coordinates?
    I mean our coordinates are not the right ones to describe someone who's falling into a black hole, because they have a singularity (coordinate singularity) at the Horizon....
    I am not sure, but if you go to the North Pole of the Earth, and stand there, then you'll complete a day in infinite time....
     
  5. Jul 15, 2014 #4

    WannabeNewton

    User Avatar
    Science Advisor

    No it is not a problem of coordinates because we are not computing observables relative to the observer freely falling into the black hole. For this observer we certainly can't use Schwarzschild coordinates everywhere on the observer's worldline and would have to instead transform to freely falling coordinates centered on the falling observer which are just Fermi-Normal coordinates centered on this observer for zero acceleration and rotation. Rather we are computing observables relative to the observer hovering far away continually receiving light signals from the infalling observer. For this observer the Schwarzschild coordinates are well-defined. The phenomenon of infinite time delay of outgoing light signals from the infalling observer to the hovering observer and that of infinite redshift of these outgoing light signals are very real.
     
  6. Jul 15, 2014 #5

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    I haven't been able to track down anything online, so I don't understand the question about the plane. Maybe someone who has seen the video can answer it.

    That is IMO a good view.
    IMO that's a bad view. From experience on the forums I can say that your friend will be reluctant to abandon it though.

    There are a number of issues with your question, the foremost being the non-existence of sticks that are more rigid than light. But I want to focus on what I think is important, rather than pick nits.

    What I think is at the root of your question is "where is the falling woman now" - the rigid stick is an attempt to define "now" observationally, via expeirment.

    I.e. you are imagining a perfectly rigid stick to represent the idea of "now". Perfectly rigid sticks don't work in special relativity, assuming they do causes mathematical contradictions and immense confusion. Perfectly rigid sticks don't work, and additionally there isn't any other experiment you can do to in the context of special or general relativity that defines "now" in an observer independent manner either.

    It would take another thread to explain the "why" of this. I'm going to continue on to try to explain how this solves your problem, without explaining the "why" of special relativity being like this.

    According to "your now", the natural now of the static observer who is hovering over the black hole, the poor woman falling into the black hole never reaches it.

    According to the woman falling in, who has a different now, she reaches it in a finite time.

    This may be odd, but there's nothing inconsistent about this, as long as one recognizes that "now" is relative. I'll give an example of how this could happen in a more familiar context.

    Consider Zeno's paradox, if you are familiar with that. We have Achillies, chasing a tortise, and an observer named Zeno. Zeno's mapping of time might go something like this (I am being overspecific in an attempt to be clear)

    At time 0, Achilleis is 1km from the tortise.
    At time 1 Achilies is 1/2 km from the tortise
    At time 2, Achilles is 1/4 km from the tortise
    At time n, Achillies is 1/2^n km from the tortise

    Now, according to Zeno, Achilles never reaches the tortise. But this is due to his notion of "now", according to his "now" , the event of Achillies reaching the tortise requires infinite "Zeno Time".

    According to Achilles "now", he does reach the tortise, in a finite time.

    Both statements are correct from the individual viewpoints. We can note that there are some points in Achilles "now" that aren't assigned a time coordinate value in Zeno's "now". The same applies in the black hole case, there are points along the worldline of the falling woman that aren't assigned a time value by the static observer.
     
  7. Jul 16, 2014 #6
    Yes, as far as I understand, she is indeed there to poke. Its not like you are going back in time and poking her, When falling she indeed feels the poke, but may by say for a micro second, but from your perspective you have poked her for very long time say 10 years!!

    I used this analogy to understand this, it might help others too. Imagine if our visual area would be so small that, we could see only as much as we can see through a straw! How difficult it would be to visualize just a table next to us. We had to see each part, keep it in memory and visualize. Since our visual area is wider, we see things in one go. This is exactly how we see through time. Our "now" time is very small. More and more the gravity, wider the "now" gets. Near blackhole it is infinitesimally high, So hundreds of years at earth, pass like minutes(or may be seconds) at blackhole.
     
  8. Jul 16, 2014 #7

    WannabeNewton

    User Avatar
    Science Advisor

    If she fell through the event horizon already then there is nothing to poke outside the event horizon, period. What the distant hovering observer sees outside the event horizon is simply a past image of her. This is very easy to analyze using a Kruskal diagram.
     
  9. Jul 16, 2014 #8
    When you try to poke someone falling into a black hole with a stick, you see the stick approaching the falling person, but never reaching her.

    The stick sees light emitted by the stick approching the falling person but never reaching her. The stick concludes that the falling person does not see the stick approaching.

    The falling person sees the speed of some sticks hovering near the event horizon to approach the speed of light, in such way that the speed of light will be reached in finite time. She concludes that when the sticks reach the speed of light ... if a stick moving at speed of light could emit light, then that light could reach her.
     
  10. Jul 16, 2014 #9

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, to repeat what I said previously with a different slant, the first question is whether the stick is a metaphor for the notion of simultaneity i.e. the stick is being used to answer the question "Where is the infalling woman now". or whether the stick is an actual stick. If the stick is an actual stick, it must be bound by the laws of relativity, and can't be used to send signals faster than light.

    See https://www.physicsforums.com/showthread.php?t=536289 [Broken] "Can I send a signal faster than light by pushing a rigid rod?"

    (To answer the semi-rhetorical quesiton: No, you can't send signals faster than light by pushing a rigid rod.)

    If the "stick" is to be held to the model of something that's compatible with relativity, it would be better replaced with the idea of a radar signal to ensure compliance with the "no FTL" requirement.

    We already know that the outside observer never gets a return radar signal, we can still profitably ask whether the infalling observer sees the entire history of the universe (I'll add briefly this turns out not to be the case and offer to provide some references if there is a quesiton), or whether there is some last instant a signal can be sent if it is to reach the infalling observer before they reach the singularity.

    We should also mention that to answer the later question we need to specify what geometry we are using for the black hole interior. The answers I can provide are for the Schwarzchild interior geometry, and not the true and more complicated geometry of an actual black hole. There are some interesting (and advanced) papers on what the actual interior geometry of a real black hole is, but the topic is under debate and probably always will be due to issues of not being able to probe the interior of a black hole.
     
    Last edited by a moderator: May 6, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Poking someone falling into a black hole with a stick
Loading...