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Space drag-drag question.

  1. Jun 13, 2012 #1
    Hi, Sorry if this has been asked before.

    Gravity probe B showed that a rotating mass actually drags space along with it. I was wondering, does dragging space actually cause drag on the rotation velocity of an Object?

    If it does, I would assume this dragging of space effect also effects any object in motion, but instead of rotating space, would act like resistance, eventually slowing any object in motion.

    If it does not, then warping space does not require any energy, thus making any ideas of warp drives for propulsion futile :(

    I'm hoping my logic is wrong on that last one.
    Last edited: Jun 13, 2012
  2. jcsd
  3. Jun 13, 2012 #2


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    The framedragging is not the same or similar to "drag" in the sense of aerodynamics. The frame-dragging literally "drags along" the "inertial frames" (as defined by a free gyroscope). The effect is that a free gyroscope which measures what "inertial" motion is (i.e. no torques) will tend to precess with respect to the fixed background stars. This does not lead to some analogue to aerodynamic drag in space time as far as I know.

    However, orbiting objects CAN emit gravitation waves (if it has a quadrupole or higher gravitational moment) and this CAN lead to decays in the orbit and loss of energy somewhat analogously to aerodynamic drag.
  4. Jun 13, 2012 #3


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    There's frame dragging, but there's also dragging of the orbit. That is, in the Kerr solution, orbits at a given value of r have different periods depending on whether they are clockwise or counterclockwise.
  5. Jun 13, 2012 #4
    The Wiki article on the Kerr metric states:

    "A moving particle experiences a positive proper time along its worldline, its path through spacetime. However, this is impossible within the ergosphere, where gtt is negative, unless the particle is co-rotating with the interior mass M with an angular speed at least of Ω. Thus, no particle can rotate opposite to the central mass within the ergosphere."

    I find this interesting, because it seems reasonable that almost all black holes will have some none zero angular momentum and so there will be a direction close to the event horizon that particles can not circulate. Lets say that a given black hole is rotating almost clockwise so slowly that the rotation is almost imperceptible, but particles will not be able to circulate anti-clockwise.

    Wiki continues:

    "A rotating black hole has the same static limit at its event horizon but there is an additional surface outside the event horizon named the "ergosurface" given by in Boyer-Lindquist coordinates, which can be intuitively characterized as the sphere where "the rotational velocity of the surrounding space" is dragged along with the velocity of light. Within this sphere the dragging is greater than the speed of light, and any observer/particle is forced to co-rotate.

    The region outside the event horizon but inside the surface where the rotational velocity is the speed of light, is called the ergosphere (from Greek ergon meaning work). Particles falling within the ergosphere are forced to rotate faster and thereby gain energy. Because they are still outside the event horizon, they may escape the black hole. The net process is that the rotating black hole emits energetic particles at the cost of its own total energy. The possibility of extracting spin energy from a rotating black hole was first proposed by the mathematician Roger Penrose in 1969 and is thus called the Penrose process. Rotating black holes in astrophysics are a potential source of large amounts of energy and are used to explain energetic phenomena, such as gamma ray bursts."

    This last paragraph definitely implies positive "dragging" and acceleration of particles over and above centrifugal precession.
  6. Jun 14, 2012 #5


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    One way to look at it is that the dragging effect near a rotating mass tips the light cones. To see this, look at the equatorial plane for the Kerr solution in Boyer-Lindquist coordinates:

    ds2 = (1-2m/r) dt2 + 4ma/r dt dφ - (r2 + a2 + 2ma2/r) dφ2

    Consider a curve in the (t, φ) plane having tangent vector v = (1, ω) where ω = const. Its norm is

    v·v = (1 - 2m/r) + 4ma/r ω - (r2 + a2 + 2ma2/r) ω2

    Setting this to zero is the condition for the curve to be null.

    First for a = 0 (Schwarzschild), the condition reduces to

    ω2 = (r - 2m)/r3

    In this case the two roots are equal and opposite, showing that the null curves are inclined symmetrically, clockwise and counterclockwise. In the limit r → ∞ we have ω = ± 1/r and the tangential velocity rω is 1 (i.e. c), while at the horizon r = 2m the two null curves coincide, showing that the horizon is a null surface.

    For a ≠ 0, the two roots for ω will be unequal, showing the tipping effect. ω = 0 will be a root (φ = const is a null curve) at r = 2m. This is the outside of the ergosphere. The two roots coincide when r2 -2mr + a2 = 0. This is the horizon, again a null surface.
  7. Jun 14, 2012 #6
    Ok, I am far away from a rotating black hole and I drop a test object directly downwards towards the black hole. Is it possible the test object could complete several orbits before it disappears?
  8. Jun 17, 2012 #7
    Hi yuiop

    This would depend on the amount of spin. Even a black hole with a spin parameter of just a/M=0.05 would have an ergosphere (no matter how slight) where spacetime is being dragged around at c, the question is, would the free falling object do a few revolutions of the BH before it passed the EH (keeping in mind that this close to the BH, the infall velocity is approaching c). With a black hole where the spin parameter is a/M=0.95 then there's no doubt that the infalling object would manage a few revolutions before entering the black hole, probably before reaching the ergosphere.

    For the observer from infinity, regardless of the spin parameter, the object will appear to orbit the black hole close to the horizon for infinity.

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