Two metrics staticity and spherical symmetry are not

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Unformatted text preview: In the massless case, AðrÞ could be removed by a coordinate gauge transformation, and the last equation was redundant; it was a consequence of the first two. With nonzero m, there is no diffeomorphism invariance, so no such coordinate change can be made, and the last equation is independent. We expand these equations around the flat space solution B0 ðrÞ ¼ 1; C 0 ð r Þ ¼ 1; A0 ðrÞ ¼ 1: (5.22) We introduce the expansion (5.23) AðrÞ ¼ A0 ðrÞ þ A1 ðrÞ þ 2 A2 ðrÞ þ Á Á Á : Plugging into the equations of motion and collecting like powers of , the Oð0Þ part gives 0 ¼ 0 because B0 , C0 , and A0 are solutions to the full nonlinear equations. At each higher order in  we obtain a linear equation that lets us solve for the next term. At OðÞ we obtain 2ðm2 r2 À 1ÞA1 þ ðm2 r2 þ 2ÞC1 þ 2rðÀ3A0 þ C0 À rA00 Þ ¼ 0; 1 1 1 (5.24) AðrÞr2 d2 : (5.18) 7 À 2B2 C2 m2 rA4 À 2B2 C2 ðB þ C À 3Þm2 rA3 pffiffiffiffiffiffiffiffiffiffiffiffiffi À A2 BCf2C0 B2 þ ½rB0 C0 À 2CðB0 þ rB00 ފB CðrÞ ¼ C0 ðrÞ þ C1 ðrÞ þ 2 C2 ðrÞ þ Á Á Á ; We now look at static spherical solutions. We specialize to four dimensions, and for definiteness we pick the action (5.1) with the minimal mass term. We attempt to find spherically symmetric solutions to the equation of motion (5.2), in the case where the absolute metric is flat Minkowski in spherical coordinates, ÀBðrÞdt2 4ðB þ rB0 ÞA2 þ ½2r2 A0 B0 À 4BðC À rA0 ފA þ Br2 ðA0 Þ2 A2 BC2 r2 2ð2A þ B À 3Þm2 pffiffiffiffiffiffiffiffiffiffiffiffiffi À ¼ 0; (5.20) A2 BC BðrÞ ¼ B0 ðrÞ þ B1 ðrÞ þ 2 B2 ðrÞ þ Á Á Á ; B. Spherical solutions and the Vainshtein radius dx dx 687   1 1 rðA0 þ B0 Þ À C1 1 1 ¼ 0; À B1 m2 þ 2 À m2 A1 þ 2 r r2 (5.25) rA1 m2 þ rB1 m2 þ rC1 m2 À 2A0 À B0 þ C0 1 1 1 À rA00 À rB00 ¼ 0: 1 1 (5.26) One way to solve these equations is as follows. Algebraically solve them simultaneously for A1 , A0 , and A00 in terms of B1 ’s 1 1 and C1 ’s and their derivatives. Then write the equations ðd=drÞA1 ðB; CÞ ¼ A0 ðB; CÞ and ðd=drÞA0 ðB; CÞ ¼ A00 ðB; CÞ. 1 Solve these two equations for C1 and C0 in terms of B1 ’s 1 derivatives. Then write ðd=drÞC1 ðBÞ ¼ C0 ðBÞ. What is left is 1 À 3rB1 m2 þ 6B0 þ 3rB00 ¼ 0: 1 1 (5.27) Kurt Hinterbichler: Theoretical aspects of massive gravity 688 There are two integration constants in the solution to Eq. (5.27); one is left arbitrary and the other must be sent to zero to prevent the solutions from blowing up at infinity. We then recursively determine C1 and A1 . Thus the whole solution is determined by two pieces of initial data.8 The solution is B1 ðrÞ ¼ À 8GM eÀmr ; 3 r (5.28) C 1 ð rÞ ¼ À 8GM eÀmr 1 þ mr ; 3 r m 2 r2 (5.29) A1 ðrÞ ¼ 4GM eÀmr 1 þ mr þ m2 r2 ; 3 r m 2 r2 (5.30) where we have chosen the integration constant so that we agree with the solution (3.15) obtained from the Green’s function. We can now proceed to...
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This document was uploaded on 09/28/2013.

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