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On the consistency of the Oppenheimer-Snyder solution for a dust star. Reply to Marshall’s criticism

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Abstract

The recent alternative to the Oppenheimer-Snyder (OS) solution for a dust star proposed by Marshall in the paper “Gravitational collapse without black holes” (Astrophys. Space Sci. 342:329, 2012) is analyzed. It is shown that this proposal leads to a non-diagonal metric, with which the Einstein equations become practically unsolvable. Any ansatz proposed as their exact solution turns out to be arbitrary and may be unlimited number of the such solutions. This is due to the fact that an auxiliary function \(y(R,r)\), introduced by OS as \(t=M(y)\), is unambiguously fixed by the diagonality condition and the matching on the surface, and thus in the non-diagonal case it remains arbitrary. It is also shown that the OS solution, as a description in terms of the Schwarzschild coordinates, leads to a frozen star (or frozar) picture not only for the surface, asymptotically freezing outside the gravitational radius, but for interior layers too which also freeze near their own asymptotes. At most of the inner region these asymptotes are located almost equidistantly and only for layers initially close to the surface they become denser. The reason for the such densifying is not “a gravitational repulsion”, but their later freezing and higher spatial contractions, while they remain be uniform and free falling in the comoving frames.

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Acknowledgements

I would like to thank to anonymous referee for useful remarks and suggestions which help me to improve and to make more clear for understanding the revised version of the article.

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Correspondence to Zahid Zakir.

Appendix: The derivation of the OS expression for \(y\)

Appendix: The derivation of the OS expression for \(y\)

Here the derivation of the OS form of \(y\) is presented according Zakir (2017b). The solution of Eq. (32):

$$ \frac{y'}{\dot{y}} = - \frac{(r_{g,R}r)^{1/2}}{R} $$
(41)

one can search among the functions of the form

$$ y(R,r) = A(R) + B(R,r), $$
(42)

where \(A\) does not depend on \(r\), and the partial derivative of \(B\) with respect to \(R\) vanishes:

$$ y' = A' + B' + \frac{\partial B}{\partial r}r' = A'. $$
(43)

This gives

$$ B' = - \frac{\partial B}{\partial r}r', \qquad \dot{y} = \frac{\partial B}{ \partial r}\dot{r}. $$
(44)

The diagonality condition (32) then takes the form:

$$ \frac{y'}{\dot{y}} = A' \biggl( \frac{\partial B}{\partial r}\dot{r} \biggr) ^{ - 1} = \dot{r}r' $$
(45)

which gives:

$$ A' = \dot{r}^{2}r'\frac{\partial B}{\partial r} = \frac{r_{g,R}}{R}\frac{ \partial B}{\partial r}. $$
(46)

Since the left-hand side of (46) does not depend on \(r\), the right-hand side should not contain \(r\), which implies that \(B\) has the form \(B = q(R)r\) and from (44) we obtain for \(q\):

$$ B' + \frac{\partial B}{\partial r}r' = q'r + qr' = 0. $$
(47)

Taking into account (30), this equation takes the form

$$ \frac{q'}{q} = - \frac{r'}{r} = - \frac{1}{r} \frac{r}{R} = - \frac{1}{R}. $$
(48)

In its solution

$$ \ln q = - \ln R + \ln w $$
(49)

the constant \(w\) we find from the matching condition on the surface:

$$ q(R_{b}) = \frac{w}{R_{b}} = \frac{1}{r_{g}}, \qquad w = \frac{R_{b}}{r_{g}}, $$
(50)

which gives

$$ q = \frac{w}{R} = \frac{R_{b}}{Rr_{g}}, \qquad B = qr = \frac{R_{b}}{Rr_{g}}r. $$
(51)

Substituting this into (46) we obtain

$$ A' = \frac{r_{g,R}}{R}q = \frac{R}{R_{b}^{2}},\qquad A = \frac{1}{2}\frac{R ^{2}}{R_{b}^{2}} + p, $$
(52)

where the constant \(p\) is also found from the matching condition:

$$ A(R_{b}) = \frac{1}{2} + p = 0,\quad p = - \frac{1}{2}. $$
(53)

Finally, this gives the OS solution:

$$\begin{aligned} &A(R) = \frac{1}{2} \biggl( \frac{R^{2}}{R_{b}^{2}} - 1 \biggr) , \end{aligned}$$
(54)
$$\begin{aligned} &y(R,r) = \frac{1}{2} \biggl( \frac{R^{2}}{R_{b}^{2}} - 1 \biggr) + \frac{R _{b}}{Rr_{g}}r. \end{aligned}$$
(55)

The derivatives of \(y\) are equal to:

$$ y' = \frac{R}{R_{b}^{2}}, \qquad \dot{y} = \frac{R_{b}}{Rr_{g}}\dot{r} = - \frac{R ^{2}}{R_{b}^{2}}\frac{1}{(r_{g,R}r)^{1/2}} $$
(56)

and the diagonality condition (32) is satisfied.

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Zakir, Z. On the consistency of the Oppenheimer-Snyder solution for a dust star. Reply to Marshall’s criticism. Astrophys Space Sci 363, 30 (2018). https://doi.org/10.1007/s10509-018-3246-9

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  • DOI: https://doi.org/10.1007/s10509-018-3246-9

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