dimensional surfaces the induced metric is found by formally setting d r and r

# Dimensional surfaces the induced metric is found by

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dimensional surfaces the induced metric is found by formally setting dr0andrr*so thatds23-surface=dt2+ (r2*+a2cosθ2) dθ2+ (r2*+a2) sin2θdφ2+2mr*bracketleftbigdtasin2θdφbracketrightbig2(1 +a2cos2θ/r2*).(105)Now we could calculate the determinant of this 3-metric directly. Alterna-tively we could use the the block-diagonal properties of the metric in Boyer–Lindquist coordinates plus the fact that for the full (3+1) dimensional metricwe have already observeddet(gab) =sin2θ(r2+a2cos2θ)2,(106)to allow us to deduce that on the 3-surfacer=r*:det(gij)3-surface=sin2θ(r2*+a2cos2θ) (r2*2mr*+a2).(107)Now forr*> r+orr*< r-the determinant det(gij)3-surfaceis negative,which is a necessary condition for these 3-surfaces to be (2+1) dimensional[two space plus one time dimension].Forr*(r-, r+) the determinantdet(gij)3-surfaceis positive, indicating the lack of any time dimension — thelabel “t” is now misleading and “t” actually denotes a spacelike direction.Specifically at what we shall soon see are the inner and outer horizons,r*=r±, the determinant is zero — indicating that the induced 3-metric[gij]3-surfaceis singular [in the sense of being represented by a singular ma-trix,notin the sense that there is a curvature singularity] everywhere onboth of these 3-surfaces.In particular, since [gij] is a singular matrix, then at each point in eitherof the 3-surface atr=r±there will be some 3-vectorLithat lies in the3-surfacer=r±such that[gij]Li= 0,(108)which implies in particular[gij]LiLj= 0.(109) The Kerr spacetime: A brief introductionMatt Visser26Now promote the 3-vectorLi[which lives in 3-dimensional (t, θ, φ) space] toa 4-vector by tacking on an extra coefficient that has value zero:LiLa= (Lt,0, Lθ, Lφ).(110)Then in the (3+1) dimensional sense we havegabLaLb= 0(Laonly defined atr=r±)(111)That is: There is a set of curves, described by the vectorLa, that lie preciselyon the 3-surfacesr=r±and which do not leave those 3-surfaces. Further-more on the surfacesr=r±these curves are null curves andLais a nullvector. Note that these null vector fieldsLaare defined only on the inner anouter horizonsr=r±and that they are quite distinct from the null vectorfieldaoccurring in the Kerr–Schild decomposition of the metric, that vectorfield being defined throughout the entire spacetime.Physically, the vector fieldsLacorrespond to photon “orbits” that skimalong the surface if the inner and outer horizons without either falling in orescaping to infinity. You should then be able to easily convince yourself thatthe outer horizon is an “event horizon” [“absolute horizon”] in the sense ofbeing the boundary of the region from which null curves do not escape toinfinity, and we shall often concentrate discussion on the outer horizonr+.Indeed if we define quantities Ω±byΩ±=a2mr±=ar2±+a2,(112)and now defineLa±= (Lt±,0,0, Lφ±) = (1,0,0,Ω±),(113)then it is easy to check thatgab(r±)La±Lb±= 0(114)at the 3 surfacesr=r±respectively. That is, the spacetime curves  #### You've reached the end of your free preview.

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