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Schwarzchild radius and escape velocity?

  1. Feb 4, 2013 #1
    We know that the escape velocity at the schwarzchild radius is c.

    Since the escape velocity is defined as the velocity needed to escape from the gravitational field, to reach a total energy of 0 at infinity:

    Doesn't this mean that an object falling from infinity starting at rest, to the schwarzchild radius of a black hole will then have a velocity of c? This is obviously wrong, but I can't see where my mistake was.
  2. jcsd
  3. Feb 4, 2013 #2


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    I think you're right. The infaller from infinity reaches a velocity [itex]dr/d\tau= c[/itex] on the horizon.

    ( Welcome to the forum - you're the first Klingon I can remember here).
    Last edited: Feb 4, 2013
  4. Feb 4, 2013 #3
    I thought an object with mass could not reach the speed of light in any frame of reference though? Or is that invalid in a curved spacetime?
  5. Feb 4, 2013 #4


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    In any local inertial frame, the relative speed of two massive objects must be less than the speed of light. In the flat spacetime of special relativity, a local inertial frame can cover the entire universe.... But not so in curved spacetime, as you suspected.
  6. Feb 4, 2013 #5


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    In GR, there are possibly global coordinate systems, but only local frames of reference. For the infaller, they will be move at less than c for compared to any nearby material body. Compared to a distant observer, the correct statement is that relative velocity is undefined because you can't compare vectors at different points in curved spacetime in any unique way.

    As for the meaning of dr/dτ, you the relevant issue is that the r coordinate is changes from spacelike to timelike across the horizon. So dr/dτ does not represent a spatial velocity at the horizon. It does represent the r component of a 4 velocity in a particular coordinate basis; however, to discuss a spatial velocity, you must define some local spatial basis vector to take the dot product with, of the 4-velocity of the infaller. This will always be less than c, all the way to singularity.
  7. Feb 4, 2013 #6


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    A simplified version of what Mentz said:

    First, let's describe a static observer. A static observer is one who maintains a constant Schwarzschild "R" coordinate. Or if you don't mind a bit of distortion, a static observer "hovers" at a constant distance above the black hole. (If you want to be precise, the static observer hover's at a constant R coordinate value, which is not really a distance). Static observers exist at all values of R ABOVE the event horizon. Static observers do not exist at the event horizon.

    If you look at the infalling velocity (measured with local clocks and rulers of a static observer), it will indeed approach c as a limit.

    However, static observers can't exist at the event horizon, so while the velocity will exist in the limit, and approach 'c', no physical observer will ever see such a speed.

    From the point of the view of the infalling person, the event horizon is light-like. It's a surface traced out by light. For someone familiar with SR, who realizes that light can't have a point of view, this is another demonstration of why you can't have an observer at the event horizion. The event horizon is like a light beam, it's what we call a "null surface". As such, while light can "hover" at the event horizon, no physical observer can.

    If you're not familiar with SR, there are some FAQ's on the topic of why light doesn't have a "point of view". Hopefully they will help, I get the feeling some people still come away confused by this. But the FAQ is the simplest resoucre I know.

    It's perhaps not critical to understand how the two things (light not having a POV and the lack of static observers at the evet horizon) are related - it's sufficient to know that the later doesn't exist.

    If you want a detailed mathematical analysis, https://www.physicsforums.com/showpost.php?p=602558&postcount=29 has one, though it does use geometric units where c=1.

    Another poster has a similiar analysis in the next post in the thread (#30) which comes to the same result.
  8. Feb 5, 2013 #7


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    The Schwarzschild horizon is Inseln a light-like surfte, i.e. each point on the horizon defines s light-like trajectory. The horizon is the surface from which photons (emitted radially outwards) neither escape not fall into the singularity; so they stay at the horizon. So it's trivial that every (massive) object crosses the horizon whith v = c w.r.t. the horizon.

    But the Lorentz transformation becomes singular for v = c which means that it doesn't allow us to introduce a reference frame moving with v = c w.r.t. a physical reference frame defined by a massive object. So the horizon is not a valid, physical reference frame.

    (Think about a photon as a reference frame; a massive particle moves with v = c w.r.t. this frame; this s problematic, but not due to the particle but due to the frame)
  9. Feb 7, 2013 #8
    this radius is computed with the classical formula mv*v/2 equals Gm/r when v equals c.
    Hower if one uses the relativistic expression of the cinetic energy this radius turns out to be zero.
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