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VCU PHYS 691 - Schwarzschild Metric

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1 Problems with the Schwarzschild MetricThe spacetime outside of a non-rotating star (or planet or whatever) of totalmass m is described by the metric tensords2= 1 2mrdt2+11 2mrdr2+ r2d2This metric solves the vacuum Einstein equations and, according to Birkho¤’sTheorem, is the only spherically symmetric metric that d oe s. It obviously has aproblem at r = 2m. One metric coe¢ cient blows up while another goes to zero.Consider a clock that is holding its position at a constant value of the radialcoordinate. If it does this for an interval t of coordinate time, then the timeelapsed on the clock will be = tr1 2mrFor r near 2m very little proper time will elaps e on the clock even though agreat deal of coordinate time elapses. This result tells us that the t = constanthyp e rsurface s are failing to advance in time near r = 2m.During the early days of general relativity, the problem with the Schwarz-schild solution was regarded mostly as a curiosity of little consequence because,for normal astronomical objects s uch as the sun and the earth, the critical valueof the radius is extremely small. For an object with the mass of the earth, thecritical radius is about a centimeter. For the sun, it is a kilometer. The metricinside an ordinary star is not given by the Schwarzchild vacuum solution andis quite regular everywhere. For the external vacuum spacetime to extend tothe critical radius, the su n would have to be compressed to a radius of a kilo-meter. Until 1939, (Oppenheimer and Volko¤, Oppenheimer and Snyder) noastrophysicist seriously believed that a mass equal to that of the sun could becompressed into an object only a kilometer in radius. Not too many believed itafter that either.From a mathematical point of view, the coordinate-independent nature ofspacetime geometry was not well understood during the early days, so it wassome time before someone considered the possibility that the r = 2m singularitywas simply the result of bad coordinates. One indication that this might be thecase is that none of the scalar curvature invariants such as R; RRandinvariant ratios of lightlike components of curvatures become in…nite at r = 2m.2 The Kruskal Extension2.1 Orthogonal Surface MetricOnce it is realized that the r = 2m singularity is really just a problem with thecoordinates, it is fairly easy to …x. The angle coordinates ; ' are …xed by the1spherical symmetry group, so nothing could be wrong with them. That leavesthe coordinates r; t and the metric2ds2= fdt2+ f1dr2; f = 1 2mron the timelike two-surfaces orthogonal to the group orbits. We need to …ndnew coordinates in which this metric looks more regular.An obvious geometrical feature of this two dimensional metric is the lightcone. On any such surface, it is always possible to …nd coordinates U; V suchthat curves at constant U are lightlike and so are curves at constant V . Forexample, in two dimensional Minkowski spacetime, one can take U = t +r; V =t  r and obtain the metric in the form2ds2= dUdV .In general, so long as the metric takes the f orm2ds2= dUdVthen an interval with either U = 0 or V = 0 will b e lightlike. Thu s, we seekcoordinates U; V such thatfdt2+ f1dr2= dUdVfor some function . Once we …nd such coordinates, we can construct space andtime coordinates u = U  V; v = U + V in which the light cones look exactlylike the ones in Minkowski spacetime. If there is any coordinate system in whichthis metric tensor is regular, th en this has to be the one.2.2 Conditions on Advanced and Retarded Time Coordi-natesRepresent partial derivatives with respect to t; r by subscripts so thatdU = Urdr + Utdt; dV = Vrdr + Vtdtand thereforefdt2+ f1dr2=  (Urdr + Utdt) (Vrdr + Vtdt)2orUtVt= f; UrVr= f1UrVt+ UtVr= 0Divide all of these equations by UrVrand obtain the resultsUtUrVtVr= f2;VtVr+UtUr= 0;which can be solved for the two ratios in the formUtUr= f;VtVr= f:Thus, if the coordinates exist, then they must satisfy these two conditions. Con-versely, the argument can be reversed to show that solving these two conditionsis enough to put the spacetime metric into the desired form where the conformalfactor  can be found from the equation UtVt= f.In more explicit form the conditions to be solved [email protected]@t=1 2mr@[email protected];@[email protected]= 1 2mr@[email protected] Solving the Conditions: Tortoise CoordinateTo see what to do next, recall what these conditions would look like for nullcoordinates in Minkowski spacetime:@[email protected][email protected]@r;@[email protected]= @[email protected];which would yield the result that U depends only on t + r and V depends onlyon t r. To get the conditions into this form, we need a new radial coordinater such that1 2mr@@[email protected]@r From the chain rule for partial derivatives,@@[email protected] @[email protected]@r so we have1 2mr@r @[email protected]@r [email protected]@r [email protected] @r=11 2mr;which can be re-written as [email protected] @r=rr  2m= 1 +2mr  2m3and integrated directly:r = r + 2m ln jr  2mjThis coordinate has been called the “tortoise coordinate”after Zeno’s paradoxand the way that the coordinate creeps up on the value r = 2m; which isachieved on ly as r ! 1. For large values of r, the new radius coordinatebecomes quite similar to r:In terms of the tortoise co ordinate, we have the [email protected]@[email protected]@r ;@[email protected]= @[email protected] ;which simply meanU = U (t + r ) ; V = V (t  r ) :We still have two functions, U (x) ; V (x) of a single variable to choose.One potential di¢ culty needs to be anticipated at this point. The “tortoisecoordinate”r is not c ontinuous across the boundary at r = 2m. When we getthe …nal form of the metric, we need to be sure that it is regular there.2.4 Regularizing the MetricNow all we have to do is …nd th e conformal factor  and se e if there is anyway to choose the functions U; V so that the singularity at r = 2m goes away.Use primes to denote derivatives of functions of a single variable and write thecondition that is to b e solved f or UtVt= [email protected]@tU (t + r ) = U0(t + r )U0V0= 1 2mr=r  2mr:For th is to work, the conformal factor must not vanish at r = 2m or we will stillhave a s ingularity there. Thus, the functions U; V must be chosen so that theyproduce a factor of r  2m. This can be arranged becauseU = U (t + r )= U (t + r + 2m ln jr  2mj)andV = V (t  r )= V (t  r  2m ln jr  2mj)depend explicitly on r  2m. To pull out the over-all factor of r  2m that weneed, choose the

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