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Frame dragging help:

  1. Apr 30, 2012 #1
    Can someone please explain to me (it can only be brief, I'll try to do the dirty work myself ^.^):

    Frame dragging in terms of:

    What is a frame, What are it's dimensions (2d/3d(x,y,z))

    What does this have in relation to the Schwarzschild radius in terms of astrophysics;

    With that answer, is it possible to use the metrics to work out the dragging of matter and space(/time) per second according to the mass and gravitational force of the blackhole?

    E.g. using the Kerr-Newman metric:

    Can I use the metric to work out how much the mass would stretch from a distance of 5km, and how much more it would be stretched per 1km closer the object is to the center?

    In terms of the metric, what does the angles mean? In the metric it shows E.g: dθ^2, what does this mean in terms of frame dragging?

    Thank you for reading, I really need this to be answered by anyone who can! Please! :)

    Thank you so much! :)
  2. jcsd
  3. Apr 30, 2012 #2


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    Staff: Mentor

    Unfortunately I don't think it is easy enough to do for someone who hasn't been formally educated in GR and the prerequisite courses.
  4. Apr 30, 2012 #3
    No I mean't, I just wanted a basic explanation, nothing hard, I just wanted to know short answers to my questions, no hardcore explanation =)
  5. May 6, 2012 #4
    In the equatorial plane, the Schwarzschild boundary (2M) is in the same place as the outer boundary of the ergosphere for a spinning black hole. This remains unchanged regardless of how much spin the black hole has, this doesn't apply at the poles. the coordinate radius for the event horizon within the ergosphere reduces the greater the spin.

    ergosphere (re)-

    [tex]r_{e} = M + \sqrt{M^2 - a^2 \cos^2 \theta}[/tex]

    where [itex]M=Gm/c^2,\ a=J/mc[/itex] and [itex]\theta[/itex] is the plane angle (i.e. equatorial plane = 90 degrees)

    the event horizon (r+) is-

    [tex]r_{+} = M + \sqrt{M^2 - a^2}[/tex]

    Regarding the second question, the frame dragging rate as observed from infinity is-

    where [itex]\omega[/itex] is the frame dragging rate in rads/s, [itex]\Sigma^2=(r^2+a^2)^2-a^2\Delta \sin^2\theta[/itex] and [itex]\Delta= r^{2}+a^{2}-2Mr[/itex]

    For the local frame dragging rate, you multiply [itex]\omega[/itex] by the redshift [itex](\alpha)[/itex] where-


    where [itex]\rho=\sqrt{r^2+a^2 \cos^2\theta}[/itex]

    http://www.lsw.uni-heidelberg.de/users/mcamenzi/CObjects_06.pdf [Broken]
    Last edited by a moderator: May 6, 2017
  6. May 10, 2012 #5


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    Gold Member

    In addition to the environment near black holes frame-dragging is also evident in rotating spiral galaxies and in dynamic galaxy clusters. Of course, the effects are less because the masses are less, but the effect is cumulative over the eons of galaxy/cluster formation. Here are some simplified statements about their evolution:

    • The geometry of a rotating rigid disc, or that of a virialized rotating spiral galaxy, is not Euclidean because of the Lorentzian contraction.

    • Identical clocks in a spiral galaxy run at different rates, depending on their location along the radius. A clock at the circumference runs more slowly than one at the center.

    • Emitters that are accelerated or in a gravitational field have their wavelengths shifted according to the strength of that field. Emitters in a rotating system are subject to two accelerating fields: Centripetal and Coriolis.

    • The inertial frame of the rotating galaxy is twisted by Lense-Thirring frame-dragging, an effect predicted by relativity, has been measured by the Gravity Probe B satellite.

    • The total gravitating action created by the galaxy or cluster depends on its total energy, that is, the total ponderable energy plus the gravitational energy.

    • The energy of the gravitational field itself contributes to the space-time curvature.
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