11. Comparison with Parke

Parke's result for the bounce action is[17]

$$B = B _0\, r \left [ \left( \frac{\bar{\rho} _0}{2 \Lambda... ...) ^{2} , \frac{\Lambda ^{2}}{\lambda ^{2}} \right ] \quad ,$$ (141)

where

$$r \left [ x , y \right ] = \frac{ 2 \left [ \left ( 1 + x ... ...) \left( 1 + 2 x y + x ^{2} \right) ^{\frac{1}{2}} } \quad .$$ (142)

Moreover

$$\bar{\rho} _0 = \frac{3 S _1}{U _{f} - U _{t}}$$ (143)

is the critical radius in the absence of gravity, and

$$U _{f/t} = U \left( \phi _{f/t} \right)$$ (144)

represent the extremal points of the potential $U \left( \phi \right)$ for the scalar field $\phi$, which is minimally coupled to gravity; they correspond to the false and true vacuum respectively. The quantities $\Lambda$ e $\lambda$ are related to these parameters by the following relations⁷

$$\Lambda ^{2} = \left [ \frac{ 8 \pi \left( U _{f} - U _{t} \right)}{3} \right ] ^{-1}$$ (145)

$$\lambda ^{2} = \left [ \frac{ 8 \pi \left( U _{f} + U _{t} \right)}{3} \right ] ^{-1}$$ (146)

and

$$B _0 = \frac{27 \pi ^{2} S _1 ^{4}} {2 \left( U _{f} - U _{t} \right) ^{3}}$$ (147)

represents the bounce action in the absence of gravity.

On the other hand, our result is

 

$$B = \frac{\pi \xi _{N} ^{3}}{2} \frac{y _1 y _2}{y _0} \left\{ \l... ...c{1}{2} } \left( y _1 - y _0 \right) \left( y _1 - y _2 \right) ^{-1} + \right.$$

$$\qquad \qquad \qquad \qquad + \left. \left( \frac{y _2 }{y _2 - 1... ...\left( y _2 - y _0 \right) \left( y _2 - y _1 \right) ^{-1} - 1 \right\}\quad ,$$ (148)

in terms of the parameters defined in eqs.(74-75)-(83)-(88- 90). The correspondence between our cosmological constants and the false vacuum/true vacuum energies is given by the relations:

$$\epsilon _{in} = U _{f} \quad ,$$ (149)

$$\epsilon _{out} = U _{t} \quad ,$$ (150)

with

$$\Lambda _{in} = 8 \pi \epsilon _{in} \quad ,$$ (151)

$$\Lambda _{out} = 8 \pi \epsilon _{out} \quad .$$ (152)

Furthermore,

$$R _0 = \frac{3 \vert \rho \vert }{\vert \Delta \epsilon \vert... ...ac{2}{\vert 4 \pi \rho \vert} \frac{1}{\vert 1 - \gamma \vert}$$ (153)

corresponds to $\bar{\rho} _0$ which, in turn, enables us to identify:

$$S _1 \to \rho$$ (154)

$$\Delta \epsilon \to U _{f} - U _{t} \quad .$$ (155)

Using this translation code, one can verify that Parke's expression for the nucleation coefficient in the absence of gravity corresponds to

$$B _0 = \frac{27 \pi ^{2} S _1 ^{4}} {2 \left ( U _{f} - U _{... ...c{\pi ^{2}}{2} R _0 ^{3} \left \vert \rho \right \vert \quad ,$$ (156)

which is exactly our result. Passing to the more general case, we note that:

$$\displaystyle \Lambda ^{2} \to \left ( \frac{\Lambda _{in} - \Lambda _{out}}{3} \right ) ^{-1} = \frac{1}{\kappa ^{2} \left ( 1 - \gamma \right )}$$ (157)

$$\displaystyle \lambda ^{2} \to \left ( \frac{\Lambda _{in} + \Lambda... ...3} \right ) ^{-1} = \frac{2}{\kappa ^{2} \left ( 2\gamma - 2 + \alpha \right )}$$ (158)

from which we deduce,

$$\displaystyle \frac{\Lambda ^{2}}{\lambda ^{2}} \to \frac{2 \gamma - 2 + \alpha}{2 \left ( 1 - \gamma \right)}$$ (159)

$$\displaystyle \frac{\bar{\rho} _0 ^{2}}{4 \Lambda ^{2}} \to \frac{R _0 ^{2}}{4} \kappa ^{2} \left ( 1 - \gamma \right )$$ (160)

and, as a consequence,

$$\displaystyle x \to \frac{1}{1 - \gamma}$$ (161)

$$\displaystyle y \to \frac{2 \gamma - 2 + \alpha }{2 \left ( 1 - \gamma \right )}$$ (162)

$$\displaystyle x ^{2} \to \frac{1}{\left ( 1 - \gamma \right ) ^{2}}$$ (163)

$$\displaystyle x y \to \frac{2 \gamma - 2 + \alpha }{2 \left ( 1 - \gamma \right ) ^{2}}$$ (164)

$$\displaystyle 1 + xy \to \frac{2 \gamma ^{2} - 2 \gamma + \alpha} {2... ... ) ^{2}} = \frac{\alpha + \gamma ^{2}}{2 y _0 \left ( 1 - \gamma \right ) ^{2}}$$ (165)

$$\displaystyle \sqrt{1 + 2 x y + x ^{2}} \to \frac{\sqrt{\alpha + \gamma ^{2}}}{\vert 1 - \gamma \vert }$$ (166)

$$\displaystyle y ^{2} - 1 \to \frac{\alpha \left ( 4 \gamma - 4 + \al... ...\alpha + \gamma ^{2} \right )} {4 y _2 \left( 1 - \gamma \right ) ^{2}} \quad .$$ (167)

Substituting all of the above in Eq.(141), we find

 

$$B ({\rm Parke}) \to$$

$$\to B _0 \frac{ 2 \left [ \displaystyle \frac{\alpha + \gamma ^{2}} {... ... \right ) ^{2}} \frac{\sqrt{\alpha + \gamma ^{2}}} {\vert 1 - \gamma \vert} } =$$

$$B _0 \frac{8 \left ( 1 - \gamma \right ) ^{4} y _2} {\alpha \left... ... 1 - \gamma \right \vert \left ( \alpha + \gamma ^{2} \right ) } - 1 \right ] =$$ =

$$B _0 \frac{ 4 \left \vert 1 - \gamma \right \vert ^{3} \left( 2\g... ...\vert \sqrt{\alpha + \gamma ^{2}}} {2\gamma ^{2} - 2\gamma + \alpha} \right ] =$$ =

$$4 B _0 \frac{y _2 y _1}{y _0} \frac{\left \vert 1 - \gamma \right... ...rt{\alpha + \gamma ^{2}}} {2\gamma - 2\gamma ^{2} - \alpha} - 1 \right ]\quad .$$ = (168)

Next, we proceed in the same fashion with our own result. First, we have

$$\displaystyle y _1 - y _0 = \frac{ 2 \gamma \left ( 1 - \gamma \righ... ...a ^{2} \right ) } { \alpha \left ( 2 \gamma - 2 \gamma ^{2} - \alpha \right ) }$$ (169)

$$\displaystyle y _1 - y _2 = \frac{4 \left ( \gamma - 1 \right ) \left ( \alpha + \gamma ^{2} \right )} {\alpha \left ( 4 \gamma - 4 + \alpha \right )}$$ (170)

$$\displaystyle y _2 - y _0 = \frac{ 2 \left ( 2 - \gamma \right ) \le... ...amma - 4 + \alpha \right ) \left ( 2 \gamma - 2 \gamma ^{2} - \alpha \right ) }$$ (171)

$$\displaystyle \frac{y _1}{y _1 - 1} = \frac{\alpha + \gamma ^{2}}{\gamma ^{2}}$$ (172)

$$\displaystyle \frac{y _2}{y _2 - 1} = \frac{\alpha + \gamma ^{2}}{\left ( \gamma - 2 \right ) ^{2}} \quad .$$ (173)

Then, in the case in which $\gamma < 0$, or $\gamma > 2$ our expression for the nucleation coefficient becomes

 

$$\frac{\pi \xi _{N} ^{3}}{2 \left ( 4 \pi \rho \right ) ^{2}} \frac{y _1 y _2}{y _0} \cdot$$ B =

$$\: \: \: \cdot \left\{ \frac{\sqrt{\alpha + \gamma ^{2}}} {\vert ... ...} { 4 \left( \alpha + \gamma ^{2} \right) \left( \gamma - 1 \right) } + \right.$$

$$\quad - \frac{\sqrt{\alpha + \gamma ^{2}}} {\vert 2 - \gamma\vert... ...a - 4 + \alpha \right) \left( 2 \gamma - 2 \gamma ^{2} - \alpha \right) } \cdot$$

$$\qquad \qquad \quad \left. \cdot \frac{\alpha \left( 4 \gamma - 4... ...left( \alpha + \gamma ^{2} \right ) \left ( \gamma - 1 \right) } - 1 \right\} =$$

$$\frac{\pi \xi _{N} ^{3}}{2 \left ( 4 \pi \rho \right ) ^{2}} \fra... ...\sqrt{\alpha + \gamma ^{2}}} {2 \gamma - 2 \gamma ^{2} - \alpha} \cdot \right .$$ =

$$\qquad \qquad \left. \cdot \left( \frac{\gamma}{\vert \gamma \ver... ...ight) + \frac{2 - \gamma}{\vert 2 - \gamma \vert} \alpha \right) - 1 \right ] =$$

$$\frac{\pi \xi _{N} ^{3}}{2 \left ( 4 \pi \rho \right ) ^{2}} \fra... ... \alpha + \gamma ^{2}}} {2\gamma ^{2} - 2\gamma + \alpha } - 1 \right ] \quad .$$ = (174)

Finally, comparing the expressions (168)-(174), one sees that their equivalence is related to the equivalence of the terms

$$\frac{\pi \xi _{N} ^{3}}{2 \left ( 4 \pi \rho \right ) ^{2}} ... ... ( \alpha + \gamma ^{2} \right ) ^{\frac{3}{2}}} B _0 \quad .$$ (175)

However, in view of the equivalence between eqs.(156) and (83), the only equality that needs to be proved is that between

$$\frac{8 \pi } { 2 \left ( 4 \pi \rho \right ) ^{2} \left ... ... \right \vert ^{3} \left \vert 1 - \gamma \right \vert ^{3}}$$ (176)

which is trivially true.

In order to complete our discussion, we note that in the case $0 < \gamma < 2$ our result and that of Parke do not coincide. The reason for this discrepancy may be traced back to the fact that in the given interval of variation of $\gamma$, one finds $\sigma _{in} = - \sigma _{out}$. Therefore, as discussed in the text, the shell configuration $R \equiv 0$ is no longer a classical solution so that no quantum tunneling can occur in the first place.