# 2m r in .NET framework Maker QR Code in .NET framework 2m r

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which means that this term will blow up when 1 2m a 2 + 2 =0 r r
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(11.15)
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Multiplying through by r 2 we obtain the following quadratic equation: r 2 2mr + a 2 = 0 with solutions given by (using the quadratic formula) r = b b2 4ac 2m 4m 2 4a 2 = 2a 2
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(11.16)
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Looking at this result, we see that in the case of a rotating black hole, there are two horizons! First, it is reassuring to note that if we consider a nonrotating black hole by setting a = 0, we get back the Schwarzschild result; that is, 4m 2 0 r = =m m 2 r = 2m or r = 0 2m Now let s consider the case of interest here, where a = 0. If we take the positive sign we obtain an outer horizon while the minus sign gives us an inner horizon. The inner horizon goes by the name the Cauchy horizon. Let s consider the maximum angular momentum that can be found. We can nd this rather easily. Notice the term under the square root sign in (11.16). We have 4m 2 4a 2 = 2 m 2 a 2 . This term is real only when m 2 a 2 0. The inner radius is going to attain a maximum when a = m. In fact, looking at (11.16), we see that in this case we obtain r = m. The inner and outer horizons coincide. Returning to the two horizons associated with the Kerr metric, let s label the inner and outer horizons by r . These are genuine horizons; that is, they are one-way membranes where you can cross going in, but can t come out. As we mentioned above, the two one-way membranes or horizons are given by r = m m2 a2
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We take the outer horizon to be r+ = m + m2 a2
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This horizon represents a boundary between the black hole and the outside world. It is analogous to the Schwarzschild horizon that we found in the case of a nonrotating black hole, and as we noted above, if you set a = 0 then you will get the familiar result r = 2m. Now, turning to the inner horizon represented by r = m m 2 a 2 , note that since it resides inside the outer horizon, it is inaccessible to an outside observer. Earlier we noted that in the Kerr geometry at rs = 2m, the time component of the metric vanishes, i.e., gtt = 0. The solution rs = 2m can be described as an outer in nite redshift surface that lies outside of the outer horizon r+ = m + m 2 a 2 . Particles and light can cross the in nite redshift surface in either direction. But think of the surface represented by the horizon r+ = m + m 2 a 2 as the actual black hole. It is the one-way membrane that represents the point of no return. If a particle or light beam passes it, escape is not possible. Interestingly, however, at = 0, the horizon and the surface of in nite redshift coincide, and so at these points if light or massive bodies cross, they cannot escape. The volume between these surfaces de ned by the static limit and the horizon, that is, the region where r+ < r < rs , is called the ergosphere. Inside the ergosphere one nds the frame-dragging effect: an object inside this region is dragged along regardless of its energy or state of motion. More formally we can say that inside the ergosphere, all timelike geodesics rotate with the mass that is the source of the gravitational eld. In between the two horizons where r < r < r + , r becomes a timelike coordinate. This is just like the Schwarzschild case. This means that if we were to nd ourselves in this region, no matter what we do, we would be pulled with inevitability to the Cauchy horizon r = r in the same way that we all march through life to the future. It is believed that the Kerr solution describes the geometry accurately up to the Cauchy horizon.