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Speed of a mass near black hole

  1. Jun 21, 2014 #1
    By theoretical calculations of super dense black holes show their gravity exceeding to a limit that can pull in light into the heavenly body. This means that the gravitational pull, which are actually acceleration as shown by the Relativity theory, is greater than the speed of light. So won't a part of mass be accelerated greater than the speed of light near the gravity of Black holes.
     
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  3. Jun 21, 2014 #2

    Orodruin

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    First: Acceleration and speed have different units so it makes no sense to compare the two.

    Second: general relativity identifies gravitation as curvature of space-time. One of the consequences is that velocity, acceleration, and many other things must be measured locally. This means that it makes little sense to talk about how fast something moves close to the black hole unless we have something to relate it to. When it comes to objects not subject to external forces (note that gravity in this context is not a force), they will travel along lines in space-time which are straight and are therefore not accelerating. The reason they are attracted to the black hole is that its mass curves space-time (compare with something moving in a straight line on a sphere - after some time it comes back to where it started although it moved straight).
     
  4. Jun 21, 2014 #3

    adjacent

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    You can't have a straight line on a sphere. :confused:
    It can be a tangent but the surface of a sphere cannot be straight.
     
  5. Jun 21, 2014 #4

    Orodruin

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    By straight I mean geodesic, which is as close as you get to straight in a curved space. My assumption was that those who ask this type of question typically are not familiar with the term and thus I decided to "cheat" with nomenclature. On the sphere, geodesics are great circles. If earth was a perfect sphere and you started walking straight forward on the surface, you would be walking in a great circle.
     
  6. Jun 21, 2014 #5

    PeterDonis

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    No, it isn't. I assume you are referring here to the equivalence principle, but that's not quite what the equivalence principle says. It says that *being at rest* in a gravitational field is equivalent to being accelerated (by a rocket, for example) in empty space. It also says that freely falling in a gravitational field is equivalent to freely floating in empty space. But if you're freely falling, you feel no acceleration; whereas if you're at rest in a gravitational field, or being accelerated by a rocket, you do. So these cases are *not* equivalent, and relativity does not say they are.

    No. An object freely falling into a black hole feels no acceleration, and everything looks the same to it, locally, as it would if the object were freely floating in empty space. That includes all the principles of special relativity, such as nothing moving faster than the speed of light.

    Pop physics discussions of black holes are often very sloppy in talking about these things; when they say things like "an object falling inside a black hole moves faster than light", the "speed" they are attributing to the object is not a physical speed; it's a calculated number based on a certain system of coordinates used to describe the black hole. But that calculated number doesn't correspond to any speed that any observer will actually measure.
     
  7. Jul 3, 2014 #6
    But still, can a black hole accelerate a mass faster than light?
     
  8. Jul 3, 2014 #7

    A.T.

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    A great circle is extrinsically straight with respect to the intrinsically curved 2D sphere surface. A great circle is extrinsically curved with respect to the intrinsically flat 3D space in which the 2D sphere surface might be embedded. But such a intrinsically flat embedding space has no physical relevance in GR.
     
  9. Jul 3, 2014 #8

    PeterDonis

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    For the meanings of the terms "accelerate" and "faster than light" as they are used in GR, no.
     
  10. Jul 3, 2014 #9

    Dale

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    I agree with both your use of the word "straight" in this context and also your motivation for using it.
     
  11. Jul 3, 2014 #10

    pervect

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    If I re-interpret this question (which is hard to analyze exactly as written) as "is the escape velocity of a black hole at the event horizion equal to c, and beyond the event horizion greater than c", I would say yes.

    However, if you have a spaceship falling into a black hole carrying with it a device that measures the speed of light (using a rigid bar type meter standard with a mirror, and an onboard atomic clock which measures the round trip time), the device will not measure any change in "c" inside, at, or outside the event horizon.

    If you shine a beam of light into a black hole, and shoot a fast moving massive particle (we'll call it a bullet) into a black hole from the same starting location at the same time, the light beam will always reach the event horizion first, so the bullet will never go "faster than light".

    So the subtle and confusing point here is that GR says that the speed of light is equal to C locally, as measured by local rulers and local clocks colocated with the observer who is measuring the speed of light. It doesn't say anything about the speed of light if it is not measured locally.
     
  12. Jul 3, 2014 #11

    PeterDonis

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    There are a couple of objections to this interpretation. First, the term "escape velocity" is not really appropriate if escape is not possible, as it isn't from any point at or inside the horizon. Yes, it's natural to extend the term "escape velocity" in this way, but that doesn't make it any less prone to misconceptions. IMO it's better to just say straight out that the concept of "escape velocity" does not make sense at or inside the horizon.

    Second, even leaving aside the above, the term "escape velocity" is really "escape velocity relative to a stationary observer", and there are no stationary observers at or inside the horizon. So even if we allow in principle the extension of the concept of "escape velocity" so that it can be equal to or greater than c, we have no way of defining an observer that this "escape velocity" is relative to at or inside the horizon. So again it's better IMO to just say straight out that this is another reason why the concept of "escape velocity" does not make sense at or inside the horizon.

    I would say that similar remarks apply to any way one could come up with to interpret the OP's question; basically the question is based on assumptions about how spacetime works that are only valid outside the horizon.

    I agree with the rest of your post.
     
  13. Jul 5, 2014 #12
    You might find some of the equations in this post of interest.
     
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