4.2 Nucleation Rate, Symmetry breaking and Mass

With the results of the previous subsection in hands, we can finally relate the nucleation rate of vacuum bubbles with the idea of topological symmetry breaking and mass generation. A common procedure for computing the nucleation rate of vacuum bubbles amounts, in our present formulation, to analytically continue the action (49) to imaginary time ¹⁰

$$S_E( F ; g^2 )= {1\over 2}\int d^4x \left[ \Lambda - g^2 \Theta_B(x) \right] -\mu^3 \int d^p\sigma \sqrt{-\gamma}$$ (51)

A semi-classical estimate for the nucleation rate of a spherical bag of radius $R$ can be obtained through a saddle point estimate of $S_E$:

$$e^{-\Gamma} \simeq e^{-[ S( \Lambda ; g )- S( \Lambda ; 0 ) ]} \equiv e^{- S(R)}$$

$$= \exp\left[- \left({\pi^2\over 4}g^2 R^4 -2\pi^2\mu^3 R^3\right)\Big\vert_{R=R_0} \right]$$ (52)

where the nucleation radius $R_0$ is a stationary point

$$\left({\partial S(R)\over \partial R}\right)_{R=R_0} \quad \longrightarrow \quad R_0={6\mu^3\over g^2} .$$ (53)

Then, one finds

$$e^{-B}=\exp\left(-{\pi^2\over 2}\mu^3 R_0^3\right)$$ (54)

which is the original Coleman De Luccia result for the false vacuum decay rate [22].
Apart from confirming the validity of our approach against a well tested calculation, the Euclidean description of vacuum decay shows how a vacuum bubble may materialize in Minkowski spacetime as a spacelike domain at a finite time. This is precisely the ``extremal'', i.e. initial boundary of the membrane world manifold, and leads us to conclude, by the argument of the previous section, that the current associated with the bubble nucleation process cannot be divergence free. Thus, a Minkowskian description of the bubble nucleation process seems to be in conflict with the requirement of gauge invariance.
Against this background, we wish to show that a consistent description, in real spacetime, of the (quantum) nucleation process, can be achieved by restoring the gauge invariance of the original action. However, restoring gauge invariance is tantamount to ``closing'' the world history of the membrane, so that no boundary exists. There are essentially two ways to achieve this: in the Euclidean formulation one bypasses the problem by ``closing the free boundary in imaginary time'', so that the resulting Euclidean world manifold is again without boundary. Somewhat paradoxically, the alternative procedure in real spacetime is to include in the action an additional source of symmetry violation in the form of a mass term for the cosmological field. We hasten to emphasize, before proceeding further, that the inclusion of a mass term is not a matter of choice. As we have seen in $(1+1)$ dimensions, it is actually dictated by the self-consistency of the field equations. There, we have shown how the explicit symmetry breaking due to the presence of mass and the topological symmetry breaking due to the presence of a boundary actually conspire to produce an action which is gauge invariant, albeit in an extended form. Thus, in the last analysis, it is the self-consistency of the theory that forces upon us the introduction of a massive particle.
Consider the coupling of the cosmological field $A_{\mu\nu\rho}(x)$ to a quantum mechanically nucleated relativistic membrane. According to the discussion in the previous section, the history of such an object is spatially closed, but only semi-infinite along the timelike direction because the membrane comes into existence at a finite instant of time. The nucleation event provides a spacelike boundary that consists of a two-surface where symmetry ``leaks out'' and gauge invariance is broken. Therefore, the apparently gauge invariant action

$$S_0=\int d^4x \left( {1\over 2\cdot 4!} F_{\lambda\mu\n... ...]} -{g\over 3!} J^{\mu\nu\rho} A_{\mu\nu\rho} \right)$$ (55)

leads to field equations

$$\partial_\lambda F^{\lambda\mu\nu\rho}= g J^{\mu\nu\rho}(x)$$ (56)

that are inconsistent. This is because the l.h.s. of Eq. (56)is divergence free everywhere due to the antisymmetry of the Maxwell tensor, whereas the membrane current is divergenceless everywhere except at the nucleation event where

$$\partial_\mu J^{\mu\nu\rho}=j^{\nu\rho}\ne 0 .$$ (57)

Here $j^{\nu\rho}$ represents the boundary current localized on the initial two-surface. For a Minkowskian observer the membrane is created ex nihilo, and its current suddenly jumps from zero to a non-vanishing value. Therefore, it cannot be ``conserved'' and the amount of (topological) symmetry breaking is taken into account by $j^{\nu\rho}$. Thus, what we learn from Eq. (56) is that the massless cosmological field cannot couple to the current of a relativistic membrane which is nucleated from the vacuum.
In order to write down a self-consistent model for interacting semi-infinite world histories, the coupling must involve a massive tensor field:

$$S=\int d^4x \left( {1\over 2\cdot 4!} F_{\lambda\mu\nu\... ... -{g\over 3!} J^{\mu\nu\rho} A_{\mu\nu\rho} \right) .$$ (58)

Inspection of the above action tells us that the physical spectrum consists of massive spin-0 particles, in agreement with property b) listed in the previous section. However, in order to extract the full physical content of the system (58) we can proceed as follows:
from the above action we derive the field equations

$$\partial_\lambda F^{\lambda\mu\nu\rho} +m^2 A^{\mu\nu\rho}= g J^{\mu\nu\rho}$$ (59)

$$\partial_\mu A^{\mu\nu\rho}= {g\over m^2} j^{\nu\rho} .$$ (60)

Next, we use the identity

$$J^{\mu\nu\rho}=\left( J^{\mu\nu\rho}-\partial^{ [ \mu}\... ...} \equiv \widehat J^{\mu\nu\rho}+ \widetilde J^{\mu\nu\rho}\\$$ (61)

in order to split the current into two parts

$$\partial_\mu\widehat J^{\mu\nu\rho}=0 ,\qquad \partial_\mu\widetilde J^{\mu\nu\rho}=j^{\nu\rho} .$$ (62)

Evidently, $\widehat J^{\mu\nu\rho}(x)$ is the divergenceless, boundary free current, while $\widetilde J^{\mu\nu\rho}(x)$ represents the pure boundary current.
In a similar fashion we decompose the tensor gauge field into the sum of a divergence free and a curl free part:

$$A^{\mu\nu\rho}\equiv \widehat A^{\mu\nu\rho} + \widetilde A^{\mu\nu\rho}$$ (63)

$$\partial_\mu \widehat A^{\mu\nu\rho}=0 ,\quad \partial_{[ \lambda} \widetilde A_{\mu\nu\rho ]}=0 .$$ (64)

Consequently, the ``Proca-Maxwell'' equations (59), (60) split into the following set

$$\partial_\lambda F^{\lambda\mu\nu\rho} +m^2 \left( \widehat ... ...ht) = g \left( \widehat J^{\mu\nu\rho} + \widetilde J^{\mu\nu\rho} \right)$$ (65)

$$\partial_\mu \widetilde A^{\mu\nu\rho}= {g\over m^2} j^{\nu\rho} .$$ (66)

From Eq. (67) we obtain

$$\widetilde A^{\mu\nu\rho}= {g\over m^2} \partial^{ [ \mu} {1\over \Box} j^{\nu\rho ]} .$$ (67)

Then, Eq. (66) becomes

$$\partial_\lambda F^{\lambda\mu\nu\rho} +m^2 \widehat A^{\mu\nu\rho} = g \widehat J^{\mu\nu\rho}$$ (68)

which gives

$$F^{\mu\nu\rho\sigma} = \sqrt\Lambda \epsilon^{\mu\nu\rho\sigma} + g^2 \partial^{ ... ...a ]} -m^2 \partial^{ [ \mu} {1\over \Box} \widehat A^{\nu\rho\sigma ]}$$

$$\equiv F^{\mu\nu\rho\sigma}_0 -m^2 \partial^{ [ \mu} {1\over\Box} {\widehat A}^{\nu\rho\sigma ]}$$ (69)

where $F^{\mu\nu\rho\sigma}_0$ represents the solution of Maxwell's equation

$$\partial_\mu F^{\mu\nu\rho\sigma}_0= g \widehat J^{\nu\rho\sigma}$$ (70)

obtained in the previous massless case.

Substituting the solution (70) into the action (58), and following step by step the same procedure outlined in the previous subsection for the massless case, we obtain the corresponding result for the massive cosmological field

$$S = \int d^4 x \left[ {1\over 2\cdot 4!} F_ {0 \mu\nu\rho\sigma} F_0^... ... 3!} \widehat J^{\mu\nu\rho} {1 \over\Box} \widehat A_{\mu\nu\rho} \right. + {g^2\over 4 m^2} \left. j^ {\mu\nu} {1\over \Box} j_ {\mu\nu} \right]$$

$$= \int d^4x \left[ {1\over 2} \left( \Lambda - g^2 \Theta(x) \right... ... 3!} \widehat J^{\mu\nu\rho} {1 \over\Box} \widehat A_{\mu\nu\rho}\right. +{g^2\over 4 m^2} \left. j^ {\mu\nu} {1\over \Box} j_ {\mu\nu} \right] .$$

The first term is of the same form as in (48). It represents the bubble ``volume action'' with respect to the constant energy density background that determines the false vacuum decay rate.
The second and third term govern the dynamics of $\widehat A_{\mu\nu\rho}$ according to the equation

$$\left( \Box + m^2 \right) \widehat A_{\mu\nu\rho} = g {\widehat J^{\mu\nu\rho}} .$$ (72)

We emphasize that this massive mode represents the only propagating degree of freedom, exactly as in the $(1+1)$ dimensional case. As a matter of fact, the last term in the action (72) represents a boundary induced Coulomb interaction [21]. Indeed, a direct calculation taking into account $\partial_\mu j^{\mu\nu}=0$ gives

$${g^2\over 4 m^2}\int d^4x j^ {\mu\nu} {1\over \Box} j_ {\mu\nu} = {g^2\over 2 m^2} \int dx^0 d^3x j^ {0 k}(x^0,\vec x ) {1\over \nabla^2} j^{0k}(x^0,\vec x )$$

$$= {g^2\over 2 m^2} \int dx^0 \int d^3x \int d^3y j^ {0 k}(x^0,\ve... ... \nabla^2} \delta^{3)}\left( \vec x -\vec y \right) j^{0k}(x^0,\vec y )$$

$$= -{g^2\over 4\pi m^2} \int dx^0 \int d^3x \int d^3y j^ {0 k}(x^0,\vec x ) {1\over \vert \vec x -\vec y \vert} j^{0k}(x^0,\vec y )$$ (73)

showing that there is no physical particle mediating such an interaction.