RevModPhys.84.671

n r where kr and kl are the extrinsic curvatures

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Unformatted text preview: Án À @ n ¼ 0; 2 @  n  þ 1 Áh À Án ¼ 0: 2 (9.36) These should be thought of as constraints determining some of the boundary variables in terms of the others.17 We at this point imagine that we have solved these constraints, and that the action is really a function of the independent variables. The de Donder gauge is preserved by any 5d gauge transformation ÄA satisfying hð5Þ ÄA ¼ 0: (9.38) The component Ä5 must vanish at y ¼ 0 because the position of the brane is fixed. Equation (9.38) then implies that Ä5 vanishes everywhere. The other components can have arbitrary values Ä ðx; 0Þ ¼  ðxÞ on the brane, which are then extended into bulk in order to satisfy Eq. (9.38), Ä ðx; yÞ ¼ eÀyÁ ðxÞ: (9.39) The residual gauge transformations acting on the boundary fields are then h ¼ @  þ @  ; n ¼ ÀÁ ; (9.40) n ¼ 0: The constraints (9.36) are invariant under these gauge transformations. The 4d effective action must and will be invariant under Eq. (9.40). The 5d part of the action reads M3 Z pffiffiffiffiffiffiffiffi S5 ¼ 5 d4 xdyN Àg½RðgÞ þ K2 À K K Š: 2 (9.41) 17 Note that we cannot think of them as determining n , n in terms of h . Acting with @ on the first equation, Á on the second, and then adding, we find the equation @ @ h À hh ¼ 0; (9.37) which is precisely the statement that the 4d linearized curvature vanishes (which is, in turn, the linearized Hamiltonian constraint in general relativity). Thus, we must think of these constraints as determining some of the components of the metric. Rev. Mod. Phys., Vol. 84, No. 2, April–June 2012 We want to expand this to quadratic order in h , n , and n and then plug in our solution. We need the expansion of K to first order, K ¼ 1ð@y h À @ n À @ n Þ: 2 (9.42) Expanding, we have (after much integration by parts in 4d) Z 2 1 S ¼ d4 xdyn@ @ h À nhh þ @ h @ h 35 2 M5 1 1 À @ h@ h À @y h@ n þ ð@ n Þ2 2 2  n þ 1 n hn 1 h hh þ @y h @ 2 4  1 1 1 À @y h @y h À hhh þ ð@y hÞ2 : 4 4 4 Now, in the last line, integrate by parts in y, picking up a boundary term at y ¼ 0, and use Eq. (9.26) to kill the bulk part, Z 2 1 S5 ¼ d4 xdyn@ @ h À nhh þ @ h @ h 3 2 M5 1 1 À @ h@ h À @y h@ n þ ð@ n Þ2 2 2 Z 1 1 þ @y h @ n þ n hn þ d4 x À h@h 2 4 1 þ h @y h : 4 We now insert the following term into the action:  2 M3 Z 1 SGF ¼ À 5 d5 X @B HAB À @A H : (9.43) 2 4 The 5d equations of motion solve the de Donder gauge condition, so this term contributes 0 to the action (thought of as a function of the unconstrained variables) and we are free to add it. However, we write it in terms of the uncontrained 4d variables for now, Z 2 SGF ¼ d4 xdy 3 M5  2 1 1 À @ h À @ h þ @y n À @ n (9.44) 2 2 À  2 1 1 @ n À @y h þ @y n : 2 2 (9.45) Adding this to the previous 5d term, we find that after using the 5d Laplace equations, the entire action...
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