Gravitational Tunnelling
of Relativistic Shells¹

Stefano Ansoldi^2 3
Dipartimento di Matematica e Informatica, Universita' degli Studi di Udine,
and I.N.F.N. Sezione di Trieste,
via delle Scienze, 206 - I-33100 Udine (UD), Italy

Lorenzo Sindoni⁴
Dipartimento di Fisica, Universita' degli Studi di Trieste,
via A. Valerio, 2 - I-34127 Trieste (TS), Italy

 

Abstract:

Thin shells in general relativity have been used in the past as keystones to obtain realistic models of cosmological and astrophysical situations. A crucial role for these developments was played by the compact description of their dynamics in terms of Israel's junction conditions. Starting from this geometrical formulation we present a problem related to the WKB regime of shell dynamics and suggest a possible solution.

 

 

General relativistic shells are an interesting system in general relativity and because of the simple geometrical description of their dynamics provided by Israel's junction conditions [1] they became preferred models for many crucial aspects of astrophysical and cosmological situations (see [2] for a more complete bibliography on the subject). Many of these models have been developed under the assumption of spherical symmetry, but (as it happens for instance in the case of gravitational collapse [3]) this does not seem a severe restriction and it is likely that the obtained results can be extended to more general situations. On the other hand, the reduction in the number of degrees of freedom that it is possible to obtain in the spherically symmetric case makes simpler the development of effective models and more transparent the discussion of the interesting subtleties that often appears in the geometrodynamics of shells. Here we are, indeed, going to discuss one of these subtleties that already manifests itself in the spherically symmetric case, where the junction conditions reduce to just one equation⁵

$$[ \epsilon ( \dot{R} ^{2} + f (R) ) ^{1/2} ] \equiv \epsi... ...psilon _{+} ( \dot{R} ^{2} + f _{+} (R) ) ^{1/2} = M(R)/R ,$$ (1)

a first order integral of the second order equation of motion for the shell. In (1) $R$ is the radius of the shell (a function of the proper time $\tau$ of an observer comoving with the shell); $M(R)$ describes the matter content of the shell (i.e. it is related to its stress-energy tensor); $f _{\pm} (R)$ are the metric functions in the two domains of spacetime separated by the shell when the line element is written in the static form adapted to the spherical symmetry; $\epsilon _{\pm }$ are signs (i.e. $0 , \pm 1$). Much of the discussion that follows is centered on these last quantities, $\epsilon _{\pm }$, but, before embarking this program, we also remember that, starting from an effective Lagrangian (the particular form of which is not our concern here), we can compute the second order equation of motion that has (1) as a first integral and also obtain the effective momentum [4] conjugated to the only surviving degree of freedom $R$,

$$P (R , \dot{R}) = R \left[ \tanh ^{-1} \left( \epsilo... ...} + f (R) ) ^{1/2} \right) ^{\mathrm{Sgn}(f(R))} \right ] .$$ (2)

Moreover, equation (1) can be cast in the form of an effective equation [5,4] for the motion of a unitary mass particle with zero energy in a potential $V (R)$, $\dot{R} ^{2} + V (R) = 0$. Then all the solutions of (1) are solutions of this effective equation and viceversa. This solves the problem of obtaining a qualitative description of how the radius $R$ changes as a function of the proper time $\tau$. Of course this is not the full story, since we still have to build up the global structure of the spacetime in which the shell leaves. It is in this process that we need also the information provided by the functions $f _{\pm} (R)$ and by the two signs $\epsilon _{\pm }$. In particular when cutting and pasting the Penrose diagrams to build up the complete spacetime, $\epsilon _{\pm }$ select the sides of the Penrose diagram crossed by the trajectory [5]. Expressions for $\epsilon _{\pm }$ can be obtained with little algebra, $\epsilon _{\pm} = \mathrm{Sign} \left( M(R) \left( f _{-} - f _{+} \mp M ^{2} (R) / R ^{2} \right)\right)$, and the points where $\epsilon _{\pm }$ change from $\pm 1$ to $\mp 1$ are the points in which $f _{\pm} (R)$ are tangent to $V (R)$, if they exist. Since $f _{\pm} (R) \geq V (R)$ always, the signs can change i) when the shell is crossing a region with $f _{\pm} (R) \leq 0$ or ii) along a classically forbidden trajectory⁶, where $V (R) > 0$. It is shown in [4] that integrating the analytic continuation of (2) on the classically forbidden trajectory we can compute WKB transition amplitudes for the tunnelling process through the potential barrier; these amplitudes agree with those already computed by other means⁷ in [7]. In the cases discussed in [4] the signs $\epsilon _{\pm }$ are constant along the forbidden trajectory, but this is not always the case. We are here interested in a more detailed analysis of those cases in which one of the signs, $\epsilon _{\pm }$, indeed changes. Let us then see what happens to the momentum $P (R , \dot{R})$. Since on a forbidden trajectory $V (R) > 0$, then $f (R) > 0$: we can thus forget the weird exponent in (2). Moreover from the effective equation we obtain that $\dot{R} ^{2} < 0$ i.e. $\dot{R}$ is purely imaginary and the momentum $P (R)$ also is purely imaginary, since $\tanh ^{-1} (\imath \dots{}) = \imath \arctan (\dots{})$. Let us now assume there is an $\bar{R}$ along the forbidden trajectory where, say, $\epsilon _{-}$ changes sign. This means that when $R \to \bar{R} ^{\pm}$ we have $(\dot{R} ^{2} + f _{-} (\bar{R})) ^{1/2} \to 0 ^{\pm}$ (or $0 ^{\mp}$) and the argument of the $\arctan(\dots{})$ tends to $- \infty$ on one side and to $+ \infty$ on the other. Correspondingly, choosing the standard branch of the multivalued function $\arctan(\dots{})$, the Euclidean momentum has a discontinuity. We can try to cure this pathology by choosing a different branch of $\arctan(\dots{})$: but then, following the evolution of the now continuous momentum till the second turning point, the offset introduced by the choice of the new branch makes the
momentum non-vanishing there; this seems again a difficult situation to accept. Apparently, we thus face the unpleasant situation of i) having a discontinuous Euclidean momentum that vanishes at both turning points or ii) having a continuous momentum that does not vanish at both turning points (we incidentally point out that if we construct the Penrose diagrams associated to the two spacetimes joined by the shell before and after the transition, some difficulties in their interpretation also occur). This situation is pictured in figure 1 and now, after having stated the problem, we proceed to propose a possible solution, by considering again our Euclidean momentum and following its evolution from the first turning point. It starts from zero and after some path on the $R$ line it reaches $\bar{R}$. At this point we enforce its continuity and keep following it until the second turning point, where we impose that it is zero. We said above that this cannot happen, but we implicitly made an assumption, namely that the Euclidean momentum is a function taking values in the real line. Relaxing this assumption we are going to see that not so much remains of the above problem. Figure 2 shows indeed that if we consider the Euclidean momentum as a function that at each point $R$ along the forbidden trajectory takes values in a circle ( ${\mathbb{S}} ^{1}$) of radius $R$, then we can make the momentum both continuous and vanishing at both extrema! We end this contribution referring the reader to [9] for an extended discussion from the point of view of Euclidean quantum gravity.

$\parbox{4.2cm}{{Graphical representation of the apparent problem wit... ...omentum is vanishing at both extrema (top) \emph{or} it is continuous \dots{}}}$

$\parbox{4.2cm}{\dots{} (bottom). The two... ...eper analysis in the Euclidean sector {\protect\cite{bib:hideki}}.\vspace{1mm}}$

Figure 1

$\parbox{6.2cm}{{Representation of the Eu... ...of the horizontal $R$\ axis in figure {\protect\ref{fig:conzer}}.\vspace{1mm}}}$

Figure 2