We begin with the most general form [Kramer] of a spherically symmetric, time-dependent spacetime, in our usual coordinate chart {r, c, φ, t}:

dsThe velocity vector of an observer^{2}= B(r, t)^{2}dr+ R(r, t)^{2}^{2}dc/ (1 - c^{2}^{2}) + (1 - c^{2}) R(r, t)^{2}dφ- A(r, t)^{2}^{2}dt^{2}

u = {0, 0, 0, 1 / A(r, t)}and setting the pressure equal to zero, we find that the only nonzero component of the stress-energy tensor is

Twhere ρ(r, t) is the as-yet unknown fluid density.^{t}_{t}= - ρ(r, t)

We examine the conservation equation first, because it is first order in derivatives of the stress tensor. It is obviously going to be much simpler than Einstein's Equations, which of course are second order in derivatives of the metric. The nonzero components of the conservation equation are

Tand^{a}_{r; a}= ρ(r, t) A'(r, t) / A(r, t)

Twhere here and below we denote derivatives with respect to r with primes and derivatives with respect to t using dots. We see from the first equation that A(r, t) does not depend on r, and so with a simple rescaling of the time coordinate we can set g^{a}_{t; a}= - (R(r, t) ρ(r, t) B^{. }(r, t) + 2 B(r, t) ρ(r, t) R^{. }(r, t) +B(r, t) R(r, t) ρ^{. }(r, t)) / (B(r, t) R(r, t))

tp -> Integrate[ A[t], t]followed by a redefinition of the remaining metric functions in terms of the new time coordinate. We can use Mathematica to find a solution to the other conservation equation:

where cons[[4]] is the time component of the conservation equation and "C[1][r]" denotes an unknown function of r. We will call this function B[r] and work now from the metricDSolve[ cons[[4]] == 0, B[r, t], {r, t}]{{B[r, t] -> C[1][r] / (R[r, t]^{2}rho[r, t])

dsWe now compute the Einstein Tensor and look at its simplest components first. By setting the "r r" component equal to zero, we find an equation for B(r) as a function of ρ(r, t) and R(r, t) and its derivatives, and by setting the "r t" component equal to zero we have a differential equation involving ρ(r, t) and R(r, t). Solving these equations we obtain^{2}= B(r)^{2}dr/ (R(r, t)^{2}^{4}ρ(r, t)^{2}) + R(r, t)^{2}dc/ (1 - c^{2}^{2}) + (1 - c^{2}) R(r, t)^{2}dφ-^{2}dt^{2}

B(r)and^{2}= R(r, t)^{4}ρ(r, t)^{2}R'(r, t)^{2}/ (1 + R^{. }(r, t)^{2}+ 2 R(r, t) R^{.. }(r, t))

ρ(r, t) = ρ(r) / (R(r, t)The former equation is accompanied by an implicit constraint: since B(r) is not a function of t, the time derivative of the right hand side must vanish. This gives us a constraint on R(r, t):^{2}R'(r, t))

2 R(the justification for this nomenclature will appear below) and so^{. }(r, t) R^{.. }(r, t) + R(r, t) R^{... }(r, t) = 0= (R^{. }(r, t)^{2}+ 2 R(r, t) R^{.. }(r, t))^{. }/ 2== E(r)

^{. }

B(r) = R(r, t)We now have the metric solely as a function of R:^{2}ρ(r, t) R'(r, t) / (1 + 2 E(r))^{1/2}

dssubject, of course to the constraint on R(r, t). The only remaining equation to solve is the "t t" component of Einstein's Equations, which has the solution^{2}= R'(r, t)^{2}dr/ (1 + R^{2}^{. }(r, t)^{2}+ 2 R(r, t) R^{.. }(r, t)) +R(r, t)^{2}dc/ (1 - c^{2}^{2}) + (1 - c^{2}) R(r, t)^{2}dφ-^{2}dt^{2}

ρ(r) = - R(r, t) (2 RThis solution too comes with an implicit constraint, which is satisfied so long as the previous one is. This means that^{.. }(r, t) R'(r, t) + R(r, t) R'^{.. }(r, t)) / (4 π)

ρ(r, t) = - (2 Rso that^{.. }(r, t) R'(r, t) + R(r, t) R'^{.. }(r, t)) / (4 π R(r, t) R'(r, t))= - (R(r, t)^{2}R^{.. }(r, t))' / (4 π R(r, t)^{2}R'(r, t))

4 π R(r, t)Therefore, if we treat R(r, t) as the physical radial coordinate,^{2}ρ(r, t) dR(r, t) = - d(R(r, t)^{2}R^{.. }(r, t))

M(r, t) = - R(r, t)is the mass contained in a ball of radius r at time t. r then becomes a label coordinate, labeling an infinitesimally thick spherical shell of dust. If we write the sum of the kinetic and potential energies of one such shell, we have^{2}R^{.. }(r, t)

E = ε(r,t) Rwhere ε(r,t) is the mass of the shell. Hence E(r) is the energy per unit mass of the shell at coordinate label r.^{. }(r, t)^{2}/ 2 - ε(r,t) M(r, t) / R(r, t)= ε(r,t) (R^{. }(r, t)^{2}/ 2 + R(r, t) R^{.. }(r, t))= ε(r,t) E(r)

We now have only to solve the constraint on R(r, t) in order to have a solution. We observe that both terms of the constraint have time derivatives of order 2 or larger. Hence one solution is any function linear in t. We therefore try a slightly more general solution of the form

R(r, t) = (fand substitute into the constraint to obtain_{1}(r) + f_{2}(r) (a t + b))^{d}

a (d - 1) d (3 d - 2) fIf you use the Mathematica ReplaceAll and Rule functions to naively substitute this ansatz for R(r, t) into the constraint equation, you will be disappointed: ReplaceAll is absolutely literal, and as a result, none of the derivative factors will be changed. The following code can help here:_{2}(r) (f_{1}(r) + (a t + b) f_{2}(r))^{d - 1}= 0

The argument "rlis" torules[rlis_List] := Block[ {args}, (args = Table[ rlis[[1, 1, i]], {i, Length[ rlis[[1, 1]]]}];FullSimplify[ conseq /.Return[ Flatten[ {rlis, Table[ Thread[ D[ rlis, args[[i]]]], {i, Length[args]}],

Table[ Table[ Thread[ D[ D[ rlis, args[[i]]], args[[j]]]],{j, Length[args]}], {i, Length[args]}]}]])];rules[rules[ {R[r,t] -> (f[r] + f2[r] (a t + b)) ^ d}]]]

Returning to the constraint, we see that it is solved for d = 1 or d = 2/3. For d = 1 (the solution linear in t), we find that ρ is zero. But for d = 2/3, we have a nontrivial solution. The mass is then

M(r, t) = 2 aintroducing the^{2}f_{2}(r)^{2}/ 9= M(r) == f(r) / 2,

R(r, 0) = rwhich gives us

fPutting this all together we have the_{1}(r) = r^{3/2}- b f_{2}(r)

dswith density function^{2}= (2 (r f(r))^{1/2}- t f '(r))^{2}dr/ (4 f(r) (r^{2}^{3/2}- 3 t f(r)^{1/2}/ 2)^{2/3}) +(r^{3/2}- 3 t f(r)^{1/2}/ 2)^{4/3}dc/ (1 - c^{2}^{2}) +(1 - c

^{2}) (r^{3/2}- 3 t f(r)^{1/2}/ 2)^{4/3}dφ-^{2}dt^{2}

ρ(r, t) = f(r)(We are ignoring the solution t -> -t, which represents expanding dust.) E(r) is zero for this solution, which indicates that its shells are marginally bound. If we note that for this solution,^{1/2}f '(r) / (r^{3/2}- 3 t f(r)^{1/2}/ 2) (8 π ((r f(r))^{1/2}- t f '(r) / 2))

R(r, t) = (rwe can put the metric in the form^{3/2}- 3 t f(r)^{1/2}/ 2)^{2/3}

dswith density function^{2}= R'(r, t)^{2}dr+ R(r, t)^{2}^{2}dc/ (1 - c^{2}^{2}) + (1 - c^{2}) R(r, t)^{2}dφ-^{2}dt^{2}

ρ(r, t) = f '(r) / (8 π R(r, t)^{2}R'(r, t))

It may be useful to note that

Γ ^{ r}_{r r}R''(r, t) / R'(r,t) Γ ^{ r}_{r t}R' ^{. }(r, t) / R'(r,t)Γ ^{ r}_{c c}- R(r, t) / ((1 - c ^{2}) R'(r,t))Γ ^{ r}_{φ φ}- (1 - c ^{2}) R(r, t) / R'(r, t)Γ ^{ c}_{r c}R'(r, t) / R(r, t) Γ ^{ c}_{c c}c / (1 - c ^{2})Γ ^{ c}_{c t}R ^{. }(r, t) / R(r, t)Γ ^{ c}_{φ φ}c (1 - c ^{2})Γ ^{ φ}_{r φ}R'(r, t) / R(r, t) Γ ^{ φ}_{c φ}-c / (1 - c ^{2})Γ ^{ φ}_{ φ t}R ^{. }(r, t) / R(r, t)Γ ^{ t}_{r r}R'(r, t) R' ^{. }(r, t)Γ ^{ t}_{c c}R(r, t) R ^{. }(r, t) / (1 - c^{2})Γ ^{ t}_{φ φ}(1 - c ^{2}) R(r, t) R^{. }(r, t)

R'(r, t) = ((r f(r))As expected from the spherical symmetry of the solution, the r-t hypersurface is geodesic, as is the r-c-t hypersurface. The structure of the Christoffel Symbols leads us to expect that the coordinate chart breaks down at the zeroes of both R(r, t) and R'(r, t). Zeroes of R'(r, t) are unphysical: if the physical radius does not depend on the shell label, then multiple shells are at the same physical radius.^{1/2}- t f '(r) / 2) / (f(r) R(r, t))^{1/2}R

^{. }(r, t) = - (f(r) / R(r, t))^{1/2}R'

^{. }(r, t) = (r^{1/2}f(r) - (r^{3/2}- t f(r)^{1/2}) f '(r)) / (2 R(r, t)^{2}f(r)^{1/2}), andR''(r, t) = (-((r f(r))

^{1/2}- t f '(r) / 2)^{2}+R(r, t)^{3/2}(2 (f(r) / r^{1/2}) + t (f '(r)^{2}- 2 f(r) f ''(r)) / f(r)^{1/2}) / 4) /(2 f(r) R(r, t)

^{2})

To investigate the zeroes of R(r, t), we will examine the equation for **null geodesics**

dsWhen applied to radial geodesics (dc = dφ = 0) it implies^{2}= 0

dr / dt = 1 / R'(r, t)For such geodesics, the change in physical radius with time is

dR(r, t) / dt = R'(r, t) dr / dt + RHence if f(r) approaches zero at the same rate as R(r, t) approaches zero, an asymptotic observer will not see the singularity ar R(r, t) = 0. Otherwise it will be visible as a^{. }(r, t)= 1 + R^{. }(r, t)= 1 - (f(r) / R(r, t))

^{1/2}

The nonzero components of the Riemann Tensor are

and the nonzero components of the Ricci Tensor are

R _{ r c r c}R(r, t) R'(r, t) R ^{. }(r, t) R'^{. }(r, t) / (1 - c^{2})R _{ r φ r φ}(1 - c ^{2}) R(r, t) R'(r, t) R^{. }(r, t) R'^{. }(r, t)R _{ r t r t}- R'(r, t) R' ^{.. }(r, t)R _{ c φ c φ}R(r, t) ^{2}R^{. }(r, t)^{2}R _{ c t c t}- R(r, t) R ^{.. }(r, t) / (1 - c^{2})R _{ φ t φ t}- (1 - c ^{2}) R(r, t) R^{.. }(r, t)

Many of the invariants are less than meaningful, but those which are, are given below, along with the denominators of the others:

R _{ r r}R'(r, t) (2 R ^{. }(r, t) R'^{. }(r, t) + R(r, t) R'^{.. }(r, t)) / R(r, t)R _{ c c}(R'(r, t) R ^{. }(r, t)^{2}+ R(r, t) R'(r, t) R^{.. }(r, t) + R(r, t) R^{. }(r, t) R'^{. }(r, t)) / ((1 - c^{2}) R'(r, t))R _{ φ φ}(1 - c ^{2}) (R'(r, t) R^{. }(r, t)^{2}+ R(r, t) R'(r, t) R^{.. }(r, t) + R(r, t) R^{. }(r, t) R'^{. }(r, t)) / R'(r, t)R _{ t t}- (2 R'(r, t) R ^{.. }(r, t) + R(r, t) R'^{.. }(r, t)) / (R(r, t) R'(r, t))

It is obvious from the above that the singularity at R(r, t) = 0 is essential.

R 8 π ρ(r, t) R ^{a b}R_{a b}64 π ^{2}ρ(r, t)^{2}R ^{a b c d}R_{a b c d}~ 1 / (R(r, t) ^{4}R'(r, t)^{2})R ^{a b}R^{c}_{a}R_{b c}128 π ^{3}ρ(r, t)^{3}R ^{a b c d}R_{a c}R_{b d}384 π ^{3}ρ(r, t)^{3}R ^{a b c d}R^{e}_{a}R_{b c d e}~ 1 / (R(r, t) ^{6}R'(r, t)^{3})R ^{a b c d}R^{e f}_{a b}R_{c d e f}~ 1 / (R(r, t) ^{6}R'(r, t)^{3})R ^{a b c d}R^{e}_{a}^{f}_{c}R_{b e d f}~ 1 / (R(r, t) ^{4}R'(r, t)^{2})R ^{a b c d}R^{e}_{a}^{f}_{c}R_{b f d e}~ 1 / (R(r, t) ^{6}R'(r, t)^{3})R ^{a b; c}R_{a b; c}~ 1 / (R(r, t) ^{6}R'(r, t)^{6})R ^{a b; c}R_{a c; b}~ 1 / (R(r, t) ^{6}R'(r, t)^{6})R ^{a b}_{; a}R^{c}_{b; c}~ 1 / (R(r, t) ^{6}R'(r, t)^{6})R ^{a b c d; e}R_{a b c d; e}~ 1 / (R(r, t) ^{6}R'(r, t)^{6})R ^{a b c d}_{; a}R^{e}_{b c d; e}~ 1 / (R(r, t) ^{6}R'(r, t)^{6})Euler class 32 R ^{. }(r, t) (2 R^{.. }(r, t) R'^{. }(r, t) + R^{. }(r, t) R'^{.. }(r, t)) / (R(r, t)^{2}R'(r, t))ε ^{a b c i ...}ε^{e f g}_{i ...}R_{b c e}^{h}R_{f g a h}~ 1 / (R(r, t) ^{3}R'(r, t)^{2})

dsThis is not obviously related to the Schwarzschild metric, and so we need to find a coordinate transformation which makes explicit the relationship between the two. To do this, we examine the Kretschmann Invariant for both metrics. For the^{2}= 8 M rdr/ (8 M (r^{2}^{3/2}- 3 t (M / 2)^{1/2})^{2/3}) +(r^{3/2}- 3 t (M / 2)^{1/2})^{4/3}dc/ (1 - c^{2}^{2}) +(1 - c

^{2}) (r^{3/2}- 3 t (M / 2)^{1/2})^{4/3}dφ-^{2}dt^{2}

Rwhile for the Schwarzschild metric it is^{a b c d}R_{a b c d}= 48 M^{2}/ (r^{3/2}- 3 t (M / 2)^{4}

REquating these and solving for r^{a b c d}R_{a b c d}= 48 M^{2}/ r_{S}^{6}

coordxform[ gs, {rs, c, ph, ts}, {(rfrom the Schwarzschild metric to the^{3/2}- 3 t (M / 2)^{1/2})^{2/3}, c, ph, h(r, t)}, {r, c, ph, t}]

(2 rthe t t component becomes^{3/2}- 3 2^{1/2}M^{1/2}t) -> u(r, t),

(4 M + (8 M - 8 MSetting this equal to -1 and using^{2}/ (u(r, t) / 2)^{2/3}- 2^{1/3}u(r, t)^{2/3}) h^{. }(r, t)^{2}/ (2^{1/3}u(r, t)^{2/3}- 4 M)

h(r, t) = Integrate[ u(r, t)Once again, Mathematica is unable to perform this integration after the substitution for u(r, t). But we can cheat a little: we do not need h(r, t), but only^{2/3}/ (2^{5/3}M - u(r, t)^{2/3}), t] + h(r)

f ( u(r, t) ) dtand

( f ( u(r, t) ) uare identical if u(r, t) is linear in t. And of course it is. The latter integral, after substitution of our u(r, t), is^{. }(r, t) dt ) / u^{. }(r, t)

rRetaining the terms in t, we have the desired coordinate transformation relating the^{3/2}/ (3 (M / 2)^{1/2}) - t + 2^{7/6}M^{1/2}(2 r^{3/2}- 3 (2 M)^{1/2}t)^{1/3}-4 M Tanh^{-1}((2 r^{3/2}- 3 (2 M)^{1/2}t)^{1/3}/ (2^{5/6}M^{1/2}))

The next section discusses several computations involving BPS membranes in D=10 and D=11 supergravity.

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©2005, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

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