IV. THE BUBBLE SOLUTION We begin by considering the action with metric and scalar field The O (3 , 1) × U (1) symmetry of the bubble of nothing is easily seen within this ansatz. We arrive at the equations and having denoted The Einstein tensor then becomes and the energy momentum tensor is given by S = ∫ d $^{5}$x √ − g$^{(}$ 1 2 κ $_{2}$R − 1 2 ∂$_{M}$ Φ ∂ M ¯ Φ − λ 4 (Φ ¯ Φ − η $^{2}$)2 − Λ$^{)}$ , (23) ds 2 = dr 2 + B $^{2}$( r )( − dt 2 + cosh $^{2}$( t ) d Ω 2 $_{2}$) + r $^{2}$C ( r ) $^{2}$dy2 , (24) Φ( x $^{M}$) = f ( r )e iθ ( y ) = f ( r )e iny . (25) R$_{MN}$ − 1 2 g$_{MN}$R = κ $^{2}$T$_{MN}$ , (26) ∂$_{M}$ ( √ − g∂ $^{M}$Φ) − √ − gλ Φ( | Φ | 2 − η $^{2}$) = 0 , (27) T$_{MN}$ = ∂$_{M}$ Φ ∂$_{N}$ ¯ Φ + g$_{MN}$ ( − 1 2 ∂$_{P}$ Φ ∂ P ¯ Φ − λ 4 (Φ ¯ Φ − η $^{2}$)2 − Λ$^{)}$ . (28) G t t = − 1 B 2 + 2 B ′ rB + B ′ 2 B 2 + 2 C ′ rC + 2 B $^{′}$C ′ BC + 2 B ′′ B + C ′′ C G r r = − 3 B 2 + 3 B ′ rB + 3 B ′ 2 B 2 + 3 B $^{′}$C ′ BC G y y = − 3 B 2 + 3 B ′ 2 B 2 + 3 B ′′ B (29) T t t = − 1 2 f ′ 2 − n $^{2}$f 2 2 r $^{2}$C2 − λ 4 ( f 2 − η $^{2}$)2 − Λ T r r = 1 2 f ′ 2 − n $^{2}$f 2 2 r $^{2}$C2 − λ 4 ( f 2 − η $^{2}$)2 − Λ T y y = − 1 2 f ′ 2 + n $^{2}$f 2 2 r $^{2}$C2 − λ 4 ( f 2 − η $^{2}$)2 − Λ . 8