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

Escape velocity outside event horizon

  1. Aug 19, 2004 #1
    Escape velocity at particular radius from the gravitational source means the initial speed that an object needs at that raduis in order to coast without limit ("to infinity") without ever falling back to the gravitational source. If the escape velocity at a radius is so large that nothing inside that radius can ever move outside that radius, then the escape velocity at that radius is infinite.

    Suppose that a light beam is directed radially outward from a source in the vicinity of a black hole event horizon.

    If the light source is exactly on the event horizon, then the beam will become infinitely red-shifted in zero distance. In other words, it won't go anywhere. This is because escape velocity at the event horizon is infinite.

    If the light source is a short distance outside the event horizon, then the beam will travel a short distance before it becomes infinitely red-shifted. This is because the escape velocity at that radius is finite but much larger than the speed of light, so that light cannot escape to infinity.

    At some distance from the event horizon, the escape velocity will be equal to the speed of light. By definition, this is the Schwarzschild radius. By LaPlace's classical physics derivation, this distance would be 2GM/c-squared. When the light source is at that radius, the beam will travel forever without becoming infinitely red-shifted. If the light source was just inside that radius, the beam would travel a very long, but finite, distance before it became infiinitely red-shifted.

    Thus the event horizon is not located at a spatial distance of 2GM/c-squared from the center of mass. It is separated from the center of mass by zero spatial distance and by a time interval of 2GM/c-squared. The center of mass is empty space with null gravity until, after a time interval of 2GM/c-squared (in the free-fall reference frame at the center point), the singularity arrives with all the mass-energy within the spatial Schwarzschild radius falling in on the spatial center point. (That last sentence makes no sense, but it's as close as I can come to visualizing something that happens nowhere after forever.)
  2. jcsd
  3. Aug 19, 2004 #2


    User Avatar

    Staff: Mentor

    Not quite. Since there is an ultimate speed limit in the universe (the speed of light), as long as the escape velocity is above that speed, nothing can ever escape.
    Good until the last sentence: the escape velocity is not infinite at the event horizon, its C.
    No, thats not what an event horizon is. The event horizon is the distance at which the escape velocity is C.
  4. Aug 20, 2004 #3
    "Not quite. Since there is an ultimate speed limit in the universe (the speed of light), as long as the escape velocity is above that speed, nothing can ever escape. "

    No, you're misunderstanding the meaning of "escape velocity."

    When the escape velocity is above the speed of light, nothing can coast to infinity. That's what escape velocity means - the speed needed to coast outward forever without the ever-decreasing pull of gravity ever taking away all of the object's energy (kinetic energy for moving mass, or the energy associated with frequency for a light beam). At a point where the escape velocity equaled the speed of light, the frequency of the outgoing light beam would approach zero as its distance approached infinity.
  5. Aug 20, 2004 #4


    User Avatar

    Staff: Mentor

    Are you proposing a new definition of "event horizon"?

    It sounds like what you are saying is that an object fired outward from the event horizon at any speed would take a parabolic path back to the event horizon and a beam of light fired from the event horizon would be infinitely red-shifted as its distance went to infinity. Since both the object and the light can go out past the event horizon, that isn't the "real" event horizon.

    Trouble is - neither of those to cases are physically possible. Since on the event horizon, time dilation is infinite, you couldn't fire anything - a beam of light or an object - away from it.
  6. Aug 20, 2004 #5


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    Time dilation is only infinite at the horizon for far away observers. For someone actually close to the horizon, and say in orbit, time would not be dilated. This works a little like the speed-related dilations in SR; you may see someone else's coordinates dilated because of their speed relative to you, but they in their rest frame experience no dilation. Here instead of the speed, it's position in the BH gravity field (or spacetime curvature) that makes the difference. GR is even more about changes of coordinates than SR.
  7. Aug 20, 2004 #6
    I see what you're getting at here. Some people believe that an object has to reach the escape velocity of the earth in order to escape from the earth. You correctly point out that this is wrong, an object can ascend as slowly as it likes. In Newton's universe what you say makes sense. However relativity changes things. A light beam can't struggle upwards from the black hole and finally fall back, either it escapes or it goes nowhere.
  8. Aug 20, 2004 #7


    User Avatar

    Staff: Mentor

    Right but what does that mean for the person in orbit of the black hole just outside the event horizon? The entire lifespan of the universe would flash by in an instant, wouldn't it? If he fires a gun directly away from black hole, to an outside observer, the bullet doesn't go anywhere. Heck, to an outside observer, he's sitting motionless at the event horizon, isn't he?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Escape velocity outside event horizon
  1. Event Horizon (Replies: 9)

  2. Event Horizon (Replies: 1)

  3. Event Horizon. (Replies: 8)