9. Hamiltonian for $A_{in} {\buildrel > \over <} 0$, $A_{out} {\buildrel > \over <}0$

For the sake of notational simplicity, let us set G_N=1. We start with the expression of the momentum in the case⁶ A_in > 0 and A_out > 0:

$$\frac{P_R}{R} \tanh ^{-1} \left( \frac{ \dot{R} \left( \sigma _{in} \beta _{in}... ... { \sigma _{in} \sigma _{out} \beta _{in} \beta _{out} - \dot{R} ^{2} } \right)$$ =

$$\frac{1}{2} \ln \left [ \frac{ \left( \sigma _{in} \beta _{in}-\d... ... \dot{R} \right) \left( \sigma _{out} \beta _{out} - \dot{R} \right) } \right ]$$ =

$$\ln \left [ \frac{ \left( \sigma _{in} \beta _{in} - \dot{R} \rig... ...ft( \sigma _{out} \beta _{out} + \dot{R} \right) ^{2} } \right ] ^{\frac{1}{2}}$$ =

$$\ln \left [ \sqrt{\frac{A_{out}}{A_{in}}} \frac{\vert \sigma _{in... ...n} - \dot{R} \vert} {\vert \sigma _{out} \beta _{out} - \dot{R} \vert} \right ]$$ =

$$\ln \left [ \sqrt{\frac{A_{out}}{A_{in}}} \frac{ \left( \sigma _{... ... { \left( \sigma _{out} \beta _{out} - \dot{R} \right) \sigma _{out} } \right ]$$ =

$$\ln \left [ \sigma _{in} \sigma _{out} \sqrt{\frac{A_{out}}{A_{in... ...} \beta _{in} - \dot{R}} {\sigma _{out} \beta _{out} - \dot{R}}\right ] \quad .$$ = (111)

Then, enlisting the equalities

$$e ^{\frac{P_R}{R}} - \sigma _{in} \sigma _{out} \sqrt{\frac... ...\beta _{out}} {\sigma _{out} \beta _{out} - \dot{R}} \right)$$ (112)

$$e ^{- \frac{P_R}{R}} - \sigma _{in} \sigma _{out} \sqrt{\fr... ...out}} {\sigma _{out} \beta _{out} + \dot{R}} \right) \quad ,$$ (113)

so that

$$1 + \frac{A_{out}}{A_{in}} - 2 \sigma _{in} \sigma _{out} \sqrt{\frac{A_{out}}{A_{in}}} \cosh \left( \frac{P_R}{R} \right) =$$

$$\ = \left ( e^{\frac{P_R}{R}} - \sigma _{in} \sigma _{out} \sqrt{... ...{P_R}{R}} - \sigma _{in} \sigma _{out} \sqrt{\frac{A_{out}}{A_{in}}} \right ) =$$ (114)

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \frac{1}{A_{in}} \left( \frac{H - \kappa R^{2}}{R} \right) ^{2} \quad ,$$ (115)

we find the explicit form of the Hamiltonian quoted in the text (Eq.55):

$$H = \kappa R ^{2} - R \left [ A_{in} + A_{out} - 2 \sigm... ... \left( \frac{P_R}{R} \right) \right ] ^{\frac{1}{2}} \quad .$$ (116)

We now repeat the same steps for the case A_in > 0 and A_out < 0. Letting $\bar{A} _{out} = - A_{out}$, we find

$$\frac{P_R}{R} \tanh ^{-1} \left( \frac{ \sigma _{in} \sigma _{out} \beta _{in} ... ... \left( \sigma _{in} \beta _{in} - \sigma _{out} \beta _{out} \right) } \right)$$ =

$$\frac{1}{2} \ln \left [ \frac{ \left( \sigma _{in} \beta _{in} - ... ...ta _{in} \right) \left( \dot{R} - \sigma _{out} \beta _{out} \right) } \right ]$$ =

$$\ln \left [ \frac{ \left( \sigma _{in} \beta _{in} - \dot{R} \rig... ...\sigma _{out} \beta _{out} - \dot{R} \right) ^{2} } (-) \right ] ^{\frac{1}{2}}$$ =

$$\ln \left [ \sqrt{\frac{\bar{A} _{out}}{A_{in}}} \frac{ \left\ver... ...vert } { \left\vert \sigma _{out} \beta _{out} - \dot{R} \right\vert } \right ]$$ =

$$\ln \left [ \sqrt{\frac{\bar{A} _{out}}{A_{in}}} \frac{ \left( \s... ... _{in} } { \left( \sigma _{out} \beta _{out} - \dot{R} \right) ( - ) } \right ]$$ =

$$\ln \left [ - \sigma _{in} \sqrt{\frac{\bar{A} _{out}}{A_{in}}} \... ... \beta _{in} - \dot{R}} {\sigma _{out} \beta _{out} - \dot{R}} \right ] \quad .$$ = (117)

Enlisting now the equalities

$$e ^{\frac{P_R}{R}} + \sigma _{in} \sqrt{\frac{\bar{A} _{out}... ..._{out}} {\sigma _{out} \beta _{out} - \dot{R}} \right) ( - )$$ (118)

$$e ^{- \frac{P_R}{R}} - \sigma _{in} \sqrt{\frac{\bar{A} _{ou... ...out}} {\sigma _{out} \beta _{out} + \dot{R}} \right) \quad ,$$ (119)

so that

$$1 - \frac{\bar{A} _{out}}{A_{in}} - 2 \sigma _{in} \sqrt{\frac{\bar{A} _{out}}{A_{in}}} \sinh \left( \frac{P_R}{R} \right) =$$

$$\qquad \qquad = \left ( e ^{\frac{P_R}{R}} + \sigma _{in} \sqrt{\... ...- \frac{P_R}{R}} - \sigma _{in} \sqrt{\frac{\bar{A} _{out}}{A_{in}}} \right ) =$$ (120)

$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad = \frac{1}{A_{in}} \left( \frac{H - \left( 4 \pi \rho \right) R^{2}}{R} \right) ^{2} \quad ,$$ (121)

we find

$$H = \kappa R ^{2} - R \left [ A_{in} + A_{out} - 2 \sigma... ... \left( \frac{P_R}{R} \right) \right ] ^{\frac{1}{2}} \quad .$$ (122)

Evidently, we can follow the same procedure in the cases A_in < 0, A_out < 0; A_in < 0, A_out > 0. The corresponding expressions are

$$H = \kappa R ^{2} - R \left [ A_{in} + A_{out} - 2 \sqrt{A_{in} A_{out}} \cosh \left( \frac{P_R}{R} \right) \right ] ^{\frac{1}{2}}$$ (123)

$${\rm if} \quad A_{in} < 0 \, , \ A_{out} < 0$$

$$H = \kappa R ^{2} - R \left [ A_{in} + A_{out} - 2 \sigma _{out} \sqrt{- A_{in} A_{out}} \sinh \left( \frac{P_R}{R} \right) \right ] ^{\frac{1}{2}}$$ (124)

$${\rm if} \quad A_{in} <0 \, , \ A_{out} > 0 \quad ,$$

or, in a more compact notation

$$\kappa R^{2} - {\rm Sgn} ( \rho ) R \cdot$$ H =

$$\cdot \left[\, A_{in} + A_{out} - 2 \sigma _{in} \sigma _{out} \l... ...\vert A_{in} A_{out} \vert} } } \right) ^{\frac{1}{2}} \right] ^{\frac{1}{2}} .$$ (125)

The Hamiltonian for the shell of dust quoted in Section IV can be derived along the same steps with little change, and one obtains

$$m - {\rm Sgn} ( m ) R \cdot$$ H =

$$\cdot \left[\, A_{in} + A_{out} - 2 \sigma _{in} \sigma _{out} \l... ...\vert A_{in} A_{out} \vert} } } \right) ^{\frac{1}{2}} \right] ^{\frac{1}{2}} .$$ (126)