pduff%ti-eg.csnet@csnet-relay.arpa (08/30/85)
From: Patrick_Duff <pduff%ti-eg.csnet@csnet-relay.arpa> If you spin a non-rigid body (such as the Earth) it will develop an equatorial bulge. Would the event horizon of a rotating black hole be non-spherical? Given a black hole roughly the size (volume, not mass!) of the Earth, if it were spinning at the same rate as the Earth would it have about the same shape as the Earth does? regards, Patrick Patrick S. Duff, ***CR 5621*** pduff.ti-eg@csnet-relay 5049 Walker Dr. #91103 214/480-1659 (work) The Colony, TX 75056-1120 214/370-5363 (home) (a suburb of Dallas, TX)
mike@bambi.UUCP (Michael Caplinger) (09/02/85)
No, even rotating black holes have spherical event horizons. The radius of the horizon is r = M + sqrt(M^2 - Q^2 - a^2), where M is the mass, Q is the charge, and a is the angular momentum per unit mass (S/M). However, the "static limit" of a rotating black hole is non-spherical. The static limit is the surface at or below which an object cannot stay motionless with respect to distant objects (that is, in an inertial frame.) It's at a radius the same as the event horizon's, but with a cos^2(theta) term multiplying the a^2 term above. I'm not sure if that's an "oblate spheroid" or not. By the way, the shape of the Earth is not the same as that of a perfectly fluid object with the same mass and angular momentum - it's "pear-shaped" by some tens of meters in various directions. - Mike The above equations are from Misner, Thorne, and Wheeler's GRAVITATION, page 878-879.