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The rotational nature of the Kerr solution leads to an interesting effect known as frame dragging. We imagine dropping a particle in from in nity with zero angular momentum. This particle will acquire an angular velocity in the direction in which the source is rotating. An easy way to describe this phenomenon is to
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Black Holes
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consider the momentum four vector and the components of the metric tensor. Looking at the inverse components of the metric (11.13) consider the ratio 2mar g t = = tt g (r 2 + a 2 )2 a 2 sin2 (11.17)
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Now imagine a massive particle dropped in with zero angular momentum. The angular velocity is given by d /d p d = = t dt dt/d p With p = 0, we have p t = g tt pt and p = g t pt and so this expression becomes, using (11.17), g t 2mar p = tt = = t 2 + a 2 )2 a 2 sin2 p g (r Note that the angular velocity is proportional to terms that make up the metric, and so think of it as being due to the gravitational eld. Therefore, we see that if we drop a particle in from in nity with zero angular momentum, it will pick up an angular velocity from the gravitational eld. This effect is called frame dragging and it causes a gyroscopic precession effect known as the Lense-Thirring effect. In the equatorial plane we have = 0 and so sin2 = (1/2) (1 cos 2 ) = 0, giving a simpli ed expression for the angular velocity given by = 2mar (r 2 + a 2 )2
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Let s consider what happens to light near a Kerr black hole. More speci cally, we consider light that initially moves on a tangential path (so we set dr = 0). Recall that for a null ray ds 2 = 0, and so con ning ourselves to the equatorial plane using (11.14) we nd the following relation for light: a 2 2ma 2 2 2 4ma r d dtd 1 + 2 + 3 r r r
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(11.18)
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Black Holes
To simplify notation, we follow Taylor and Wheeler (2000) and introduce the reduced circumference R2 = r 2 + m2 + 2m 3 r (11.19)
Then (11.18) can be written in the more compact form 0= 1 2m r dt 2 + 4ma dt d R 2 d 2 r (11.20)
Dividing through (11.20) by dt 2 and then by R 2 gives us the following quadratic equation for d /dt: d dt
1 2m 4ma d 2 1 2 dt rR R r
(11.21)
We wish to consider an important special case, the static limit where r = rs = 2m. In this case notice that the last term in (11.21) vanishes: 1 2m 2m =1 =1 1=0 r 2m
Meanwhile, at the static limit R 2 = 6m 2 and the middle coef cient in (11.21) becomes 4ma 4ma 4ma a = = = 2 2 3 rR (2m)6m 12m 3m 2 Putting these results together, at the static limit (11.21) can be written as d dt
a d =0 3m 2 dt
(11.22)
There are two solutions to this equation, given by a d = dt 3m 2 and d =0 dt (11.23)
a The rst solution, d = 3m 2 , represents light that is emitted in the same direction dt in which the black hole is spinning. This is a very interesting result indeed; note
Black Holes
that the motion of the light is constrained by the angular momentum a of the black hole! The second solution, however, represents an even more astonishing result. If the light is emitted in the direction opposite to that of the black hole s rotation, then d = 0; that is, the light is completely stationary! Since no material particle dt can attain a velocity that is faster than that of light, it is absolutely impossible to move in a direction opposite to that of the black hole s rotation. No rocket ship, probe, or elementary particle can do it.
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