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Event Horizon Question

  1. Aug 9, 2014 #1
    Theoretical question.... If an extremely long object approached the event horizon, let’s say a torpedo shaped craft 1 mile long. The craft would get stretched and the observer’s time would become dilated as the craft/object approached the event horizon correct? Would it ever be possible for an object to cross over the event horizon and be both 'in and out' of the black hole simultaneously? I'm sure the answer is no, but just trying to wrap my brain around why not? Once it hits the event horizon the whole object would just immediately be 'sucked in' and there is no actual cross over time? Once the event is crossed, time basically stops in the singularity is that correct?
  2. jcsd
  3. Aug 9, 2014 #2


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    An object that remains intact as it crosses the horizon will be both in and out of the event horizon at some point. This is possible with supermassive black holes because you can cross the event horizon without being pulled apart. I'd prefer not to talk about what happens at the singularity, as that's simply the point where our math breaks down and stop giving useful predictions.
  4. Aug 9, 2014 #3
    So, if the end of the craft that crosses the event horizon tried to send data back to the other end of the craft that had not 'crossed over' and then lets say the craft was designed to separate so that half enters the black hole and the other half speeds back away from the event horizon to avoid capture, would any of that data from the 'front' of the craft within the black hole actually transmit back to the part of the craft that didn't cross over? I'm guessing no, because nothing that crosses over the event horizon would ever escape?
  5. Aug 9, 2014 #4
    I don't think that's right. I understand that the EH from an observers point of view is always closer to the singularity than the observer. The space ship should have no problems sending messages from one side to the other unless it were very close to the singularity.
  6. Aug 9, 2014 #5


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    You are correct. If you look at the outgoing null cones inside the event horizon of a Schwarzschild black hole
    (c.f. outgoing Eddington coordinates) then you will see that there is no causal curve through space-time you could trace within the horizon that exits the horizon.
  7. Aug 9, 2014 #6
    If the event horizon was somehow detectable, then it would pass through the ship with the speed of light.

    No signal from the front of the ship inside the horizon could reach the rear of the ship before the rear is also submerged under the horizon.

    One basic thing about the event horizon is that it is locally moving. It is not some stationary surface like surface of water. Better visualisation is a front of an explosion. For a far observer it is suspended and looks stationary because of GR magic, but it doesn't change the fact that locally it's moving with the speed of light.
  8. Aug 9, 2014 #7


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    As other posters have suggest, the situation is similar to the following.

    Suppose you are in a spacecraft, very long, split into two parts.

    A light beam (which best represents the event horizon, in the ship frame it's better to think of the event horizon as a trapped beam of light rather than a place) strikes the front of the spaceship.

    If the rear of the spaceship starts accelerating after the light beam has hit the front of the ship, but before the light beam hits the rear (in which case it will be too late!), it can outrun the beam of light. (This may be surprising if you aren't already familiar with it, but it's a consequence of relativity, see the wiki article on Hyperbolic Motion

    Lets suppose the acceleration starts "at the same time" as when the light beam hits the front of the ship, where "at the same time" is measured in the ship frame.

    If your spaceship is a light year long, the required acceleration for the rear of the ship to outrun the light beam would be approximately 1g. (The forumula is acceleration * distance = c^2). If it is only a mile long, the required acceleration is about 6*10^12 g. (Good luck with achieving that!).
    This acceleration won't be enough to pull away from the black hole, BTW - it will be just enough to hold station.

    THis is essentially what happens near a very massive black hole, look up "Rindler Horizon" if you want more background.

    If the rear of the ship does not accelerate at 6*10^12 g immediatley as soon as the light beam reaches the front, the rear of the ship will be sucked in too.

    Let' suppose the front of the ship emits a light beam when it crosses the horizon. (THis is a bit redundant, really, because the emitted light beam travels along with the horizon, which is another light beam).

    If the rear of the ship doesn't accelerate, it will see the light beam representing the horizon, and The light beam emitted from the front of the spaceship at the same time that the front of the ship reached the horizon at the same time. In other terms, this is equivalent to saying the rear of the ship will see the front of the ship crossing the horizon when it itself crosses the horizon.

    If the rear of the ship does accelerate, it will outrun both the beam of light from the event horizon, and the beam of light emitted by the front of the ship when the event horizon reached it, so it will see neither one.
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