Understanding the uniform thin shell collapse

Question

In section 3.9 of A Relativist's Toolkit by E. Poisson, the author discusses the gravitational collapse of a thin uniform spherical shell.

The spacetime inside is assumed¹ to be Minkowski and the spacetime outside is necessarily Schwartzschild by spherical symmetry. The shell is made of pressureless matter and its surface stress-energy tensor is constrained to have the following form.

$$S^{ab}=\sigma u^a u^b.\tag{3.68}$$

The coordinates on the hypersurface are $\{\tau,\theta,\phi\}$. He continues to write the following components of the extrinsic curvature.

$$K^{\tau}_{\pm \tau}=\frac{\dot{\beta}_{\pm}}{\dot{R}} \text{ & }K^{\theta}_{\pm \theta}=K^{\phi}_{\pm \phi}=\frac{\beta_{\pm}}{R} \tag{1}$$ where $\beta_+=\sqrt{\dot{R}^2+1-2M/R}$ and $\beta_-=\sqrt{\dot{R}^2+1}$.

1. What exactly are the equations of the hypersurface/shell as seen from both the spacetimes?
   Is it $r=R(\tau)$ and $t=T(\tau)$ as seen from both Minkowski and Schwartzschild? If this were true, then the first junction condition would be eq. $(2)$, which is not mentioned anywhere in the section. $$h_{\tau \tau}=F\dot{T}^2 - F^{-1}\dot{R}^2=\dot{T}^2-\dot{R}^2 \tag{2}$$ where, $F=1-\frac{2M}{R}$.
2. The author mentions that he has borrowed the extrinsic curvature results from section 3.8, in which he discusses the Oppenheimer-Snyder collapse, to write eqs. $(1)$. But the first junction condition of FRW-Schwartzschild junction (eq. $(3)$) was assumed to derive the results given in section 3.8. $$F\dot{T}^2 - F^{-1}\dot{R}^2=1 .\tag{3}$$ Eq. $(3)$ doesn't follow from eq. $(2)$. So, how is the author allowed to borrow the results?

I'd greatly appreciate any assistance that helps me better understand this section of the book.

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^₁ Although, according to this answer, it must necessarily be Minkowski inside as well.

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Edit 1

$$n_{+\mu}=\frac{1}{\sqrt{F\dot{R}^2-F^{-1}\dot{T_+}^2}}(-\dot{R},\dot{T_+},0,0)$$ $$K_{+\theta \theta}=\underbrace{-{\Gamma^t}_{\theta \theta}n_{+t}}_{=0}-{\Gamma^{r}}_{\theta \theta}n_{+r}=\frac{1}{2}g_{rr}\partial_rg_{\theta \theta} \frac{\dot{T_+}}{\sqrt{F\dot{R}^2-F^{-1}\dot{T_+}^2}}$$ $$\Rightarrow K_{+\theta \theta}=FR\frac{\dot{T_+}}{\sqrt{F\dot{R}^2-F^{-1}\dot{T_+}^2}}$$ $$K^{\theta}_{+\theta}=K_{+\theta \theta}h^{+\theta \theta}=\frac{F}{R}\frac{\dot{T_+}}{\sqrt{F\dot{R}^2-F^{-1}\dot{T_+}^2}}\neq \frac{\beta_+}{R}$$

Answer 1

The first Israel junction condition is essentially mentioned in the equation

$$ \mathrm{d}s_{\Sigma}^2~=~\underbrace{-(F_{\pm}\dot{T}^2_{\pm}-F^{-1}_{\pm}\dot{R}^2)}_{=h_{\tau\tau}}\mathrm{d}\tau^2+R^2(\tau)\mathrm{d}\Omega^2 $$

before eq. (3.62) in Ref. 1.

Here $F_-=1$ and $F_+=1-\frac{2M}{R}$ are Minkowski/Schwarzschild structure functions from the inside/outside bulk metrics, respectively. Note that the inside/outside time coordinates $T_{\pm}(\tau)$ are different, cf. above comment by Aleksandr Artemev. Similarly let us define $$\beta_{\pm}~:=~\sqrt{F_{\pm}+\dot{R}^2}.\tag{3.63'}$$ OP's formulas (1) for extrinsic curvature [which are listed below eq. (3.68) in Ref. 1] follow from their definitions after relatively long calculations that we will not try to reproduce here.

References:

1. Eric Poisson, A Relativist's Toolkit, 2004; Sections 3.8 + 3.9.

Answer 2

This is a complementary answer to Qmechanic's, which provides somewhat more detailed calculations.

1. The physical quantity identical from the viewpoint of both spacetimes $\mathscr{V}^+$ and $\mathscr{V}^-$ is the respective induced metrics. Intuitively, it is the tangential projections of the spacetime metrics $g_{\mu\nu}^\pm$ onto the surface. This is the start point, otherwise, as we are dealing with two different surfaces due to their distinct (intrinsic) curvatures, how can we even recognize that there is a unique surface $\Sigma$ serving as the boundary for the two manifolds?

   In practice, the easiest way to see this mathematically is to adopt the same coordinate system on the surface, in the present case, $y^a=(\tau, \theta, \phi)$. One then requests that the induced metrics are the same by matching up the corresponding tensor components (such as $h_{\tau\tau}$). This is precisely the first Israel junction condition. However, in general, the coordinates for $\mathscr{V}^+$ and $\mathscr{V}^-$ are not the same. (also see W. Israel's original paper for more discussions) In your case, the functions $R(\tau)$ and $T(\tau)$ are not the same ones for $\mathscr{V}^+$ and $\mathscr{V}^-$.

2. Regarding the calculations for the extrinsic curvature, in the following part, we first derive the normal vector of the surface as well as the four-velocity of a comoving observer on the surface edge and then use these quantities to calculate the extrinsic curvature.

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The equations referred to below are from section 3.8 of the textbook

Eric Poisson, An Advanced Course in General Relativist (Draft January 2002)

We will only consider the spacetime $\mathscr{V}^+$ outside of the shell $\Sigma$, as the generalization to $\mathscr{V}^-$ is straightforward.

For the normal vector, if one already knew the function which defines the surface, namely, $f^+(t,r,\theta,\phi)=0$, the normal vector can be obtained by taking the partial derivatives $$n_\mu = \frac{\epsilon \partial_\mu f^+}{\left|g^{\mu\nu}\partial_\mu f^+\partial_\nu f^+\right|^\frac12} .$$ The present scenario, however, is a bit tricky. So we consider the world-line of an observer who is comoving with the collapsing shell but does not experience any angular displacement on the surface. We further assume that, on the world-line, the first two components of the coordinate $x^\mu=(t, r, \theta, \phi)$ are governed by the functions $r=R(s)$ and $t=T(s)$, while $\theta=\phi=\mathrm{const}$. Since the surface is spherically symmetric, so for the normal vector $$\partial_\theta f^+=\partial_\phi f^+=0 .$$ Now, one can convenient follow the trajectory of the comoving observer, as it is no the surface, its motion satisfies $f^+(t,r,\theta,\varphi)=0$, we find $$\dot{T}\partial_tf^+ + \dot{R}\partial_rf^+ = 0 .$$ One is readily to solve the above equation for the first two components of the normal vector, together with the requirement that $n^\mu n_\mu =1 (=\epsilon)$. We obtain the following simple forms for the four-velocity of the observer and the normal vector $$\begin{align} &u_+^\mu=(\dot{T},\dot{R},0,0) ,\\ &n^+_\mu=(-\dot{R},\dot{T},0,0) . \end{align}$$

Now we proceed to evaluate the extrinsic curvature $K_{\mu\nu}$. By definition, some specific components formally possess the forms $K_{\tau\tau}=-n_\alpha {u^\alpha}_{;\beta}u^\beta$, $K_{\theta\theta}=n_{\theta;\theta}$, and $K_{\phi\phi}=n_{\phi;\phi}$, which can be evaluated by taking into consideration the Schwarzschild metric in question. One notes that some cause should be taken while utilizing these expressions. For instance, the subscript $\theta$ of $K_{\theta\theta}$ indicates the $\theta$ coordinate of $x^\mu=(t, r, \theta, \phi)$ in the manifold $\mathscr{V}^-$, while that of $n_{\theta;\theta}$ represent the coordinate $\theta$ of $y^a=(\tau, \theta, \phi)$ on the surface $\Sigma$. We have been using the same symbol, but the corresponding coordinates and (induced) metrics are distinct.

The relevant non-vanishing Christoffel symbols read $$\begin{align} &{\Gamma^0}_{10}={\Gamma^0}_{01}=g^{00}\Gamma_{010}=g^{00}\frac12(g_{01,0}+g_{00,1}-g_{10,0})=g^{00}\frac12 g_{00,1}=(-\frac1f)\frac12(-f)_{,r}=\frac{1}{2f}(f)_{,r}\\ &{\Gamma^1}_{00}=g^{11}\Gamma_{100}=g^{11}\frac12(-g_{01,1})=f\frac12(-1)(-f)_{,r}=\frac{f}{2}(f)_{,r}\\ &{\Gamma^1}_{11}=g^{11}\Gamma_{111}=g^{11}\frac12 g_{11,1}=f\frac12\left(\frac1f\right)_{,r}=-\frac{1}{2f}(f)_{,r}\\ &{\Gamma^1}_{22}=g^{11}\Gamma_{122}=g^{11}\frac12(-g_{22,1})=-f\frac12\left(r^2\right)_{,r}=-fr\\ &{\Gamma^1}_{33}=g^{11}\Gamma_{133}=g^{11}\frac12(-g_{33,1})=-f\frac12\left(r^2\sin^2\theta\right)_{,r}=-fr\sin^2\theta \end{align}$$

It is noted that for a quantity $X=X(r)$, we have $(X)_{,r}u^0=\dot{X}$. Also, on the surface $f(R)=F$. The extrinsic curvature is calculated as follows $$\begin{align} {u^0}_{;\beta}u^\beta&=\ddot{T}+{\Gamma^0}_{00}u^0u^0+{\Gamma^0}_{10}u^1u^0+{\Gamma^0}_{01}u^0u^1+{\Gamma^0}_{11}u^1u^1\\ &=\ddot{T}+2{\Gamma^0}_{10}u^1u^0 =\ddot{T}+2\frac{\dot{F}}{2F}\dot{T}=\ddot{T}+\frac{\dot{F}\dot{T}}{F}\\ {u^1}_{;\beta}u^\beta&=\ddot{R}+{\Gamma^1}_{00}u^0u^0+{\Gamma^1}_{10}u^1u^0+{\Gamma^1}_{01}u^0u^1+{\Gamma^1}_{11}u^1u^1\\ &=\ddot{R}+{\Gamma^1}_{00}u^0u^0+{\Gamma^1}_{11}u^1u^1=\ddot{R}+\frac12\frac{F\dot{F}\dot{T}^2}{\dot{R}}-\frac12\frac{\dot{F}\dot{R}}{F}\\ n_\mu {u^\mu}_{;\beta}u^\beta&=-\dot{R}{u^0}_{;\beta}u^\beta+\dot{T}{u^1}_{;\beta}u^\beta =-\dot{R}\left(\ddot{T}+\frac{\dot{F}\dot{T}}{F}\right)+\dot{T}\left(\ddot{R}+\frac12\frac{F\dot{F}\dot{T}^2}{\dot{R}}-\frac12\frac{\dot{F}\dot{R}}{F}\right) \end{align}$$

To simplify the last expression, we make use of Eq. (3.8.6) to eliminate $\ddot{T}$ by taking derivative on both sides of $F\dot{T}=\sqrt{\dot{R}^2+F}$, which in turn gives $$F\ddot{T}+\dot{F}\dot{T}=\frac{1}{2}\frac{2\dot{R}\ddot{R}+\dot{F}}{\sqrt{\dot{R}^2+F}}$$ By substituting the above expression, one finds $$-\dot{R}\frac{1}{2}\frac{2\dot{R}\ddot{R}+\dot{F}}{F\sqrt{\dot{R}^2+F}}$$ Further, we use Eq. (3.8.6) again to eliminate $\dot{T}$ by $\dot{T}=\frac{\sqrt{\dot{R}^2+F}}{F}$. Moreover, since $$\dot{F}=\left.(f)_{,r}\right|_{R}\dot{R}=\frac{2M}{R^2}\dot{R}=\frac{1-F}{R}\dot{R} ,$$ it can be used to replace $\dot{F}$ in favor of $F,R,\dot{R}$. We find $$\begin{align} n_\mu u^\mu_{;\beta}u^\beta &=-\dot{R}\frac{1}{2}\frac{2\dot{R}\ddot{R}+\dot{F}}{F\sqrt{\dot{R}^2+F}}+\dot{T}\left(\ddot{R}+\frac12\frac{F\dot{F}\dot{T}^2}{\dot{R}}-\frac12\frac{\dot{F}\dot{R}}{F}\right)\\ &=-\dot{R}\frac{1}{2}\frac{2\dot{R}\ddot{R}+\dot{F}}{F\sqrt{\dot{R}^2+F}}+\frac{\sqrt{\dot{R}^2+F}}{F}\left(\ddot{R}+\frac12\frac{\dot{F}(\dot{R}^2+F)}{F\dot{R}}-\frac12\frac{\dot{F}\dot{R}}{F}\right)\\ &=-\dot{R}\frac{1}{2}\frac{2\dot{R}\ddot{R}+\frac{1-F}{R}\dot{R}}{F\sqrt{\dot{R}^2+F}}+\frac{\sqrt{\dot{R}^2+F}}{F}\left(\ddot{R}+\frac12\frac{(1-F)(\dot{R}^2+F)}{RF}-\frac12\frac{(1-F)\dot{R}^2}{RF}\right)\\ &=-\dot{R}\frac{1}{2}\frac{2\dot{R}\ddot{R}+\frac{1-F}{R}\dot{R}}{F\sqrt{\dot{R}^2+F}}+\frac{\sqrt{\dot{R}^2+F}}{F}\left(\ddot{R}+\frac12\frac{(1-F)}{R}\right)\\ &=\frac{-\dot{R}\frac{1}{2}\left(2\dot{R}\ddot{R}+\frac{1-F}{R}\dot{R}\right)+(\dot{R}^2+F)\left(\ddot{R}+\frac12\frac{(1-F)}{R}\right)}{F\sqrt{\dot{R}^2+F}}\\ &=\frac{F\left(\ddot{R}+\frac12\frac{(1-F)}{R}\right)}{F\sqrt{\dot{R}^2+F}}=\frac{\ddot{R}+\frac12\frac{(1-F)}{R}}{\sqrt{\dot{R}^2+F}} \end{align}$$

From the other side, one calculate the derivative of Eq. (3.8.6) to match the r.h.s. of the desired Eq. (3.8.8), we have $$\frac{\dot{\beta}}{\dot{R}}=\frac12\frac{2\ddot{R}+\frac{\dot{F}}{\dot{R}}}{\sqrt{\dot{R}^2+F}} =\frac12\frac{2\ddot{R}+\frac{1-F}{R}}{\sqrt{\dot{R}^2+F}}$$ Finally, by considering $h^{\tau\tau}=-1$, one finds $${K^\tau}_\tau=h^{\tau\tau}K_{\tau\tau}=(-1)(-n_\mu u^\mu_{;\beta}u^\beta)=\frac{\ddot{R}+\frac12\frac{(1-F)}{R}}{\sqrt{\dot{R}^2+F}} =\frac{\dot{\beta}}{\dot{R}} .$$ This is the first expression of Eq. (3.8.8).

Similarlly, for the angular part we have $$\begin{align} n_{\theta;\theta}&=0-{\Gamma^\mu}_{\theta\theta}n_\mu=0-{\Gamma^0}_{\theta\theta}n_0-{\Gamma^1}_{\theta\theta}n_1 =-{\Gamma^0}_{22}(-\dot{R})-{\Gamma^1}_{22}\dot{T}=-{\Gamma^1}_{22}\dot{T}=FR\dot{T}\\ {K^\theta}_{\theta}&=h^{\theta\theta}n_{\theta;\theta}=\frac{1}{R^2}FR\dot{T}=\frac{F\dot{T}}{R}=\frac{\beta}{R} . \end{align}$$ The derivation for $\phi$ component is mostly identical to that of $\theta$ above except a factor of $\sin^2\theta$ is canceled out in both the numerator and denominator, which also furnishes the second expression of Eq. (3.8.8).

By the end of the day it is mostly patience and some straightforward algebra. :)