1. Introduction
Recently, based on the theory of the post-Newtonian (PN) approximation [1-3], we calculated the 1PN metric of a self-gravitating and collapsing thin spherical shell in the weak-field and slow-motion limits. We showed that the spherical shell’s metric is time-dependent beyond the Newtonian order, implying that the non-static spherically-symmetric systems may produce time-dependent gravitational fields [4].
In this work, we double check the correctness of the 1PN metric via substituting it into Einstein field equations, and successfully retrieve the leading-order components of the gravitational source’s energy-momentum tensor. This verifies that the time-dependent metric is the solution to Einstein field equations for the self-gravitating and collapsing thin spherical shell under the harmonic-coordinate conditions at the 1PN order.
The rest of this paper is organized as follows. Section 2 gives the leading-order energy-momentum tensor and the 1PN metric of the collapsing thin spherical shell, both of which are time-dependent. Section 3 retrieves the former from the latter. A summary is given in section 4 .
2. The energy-momentum tensor and the metric of the collapsing thin spherical shell
Let $\bar{M}$, $\bar{v}$, and $\bar{r}$ represent the typical values of mass, velocity, and distance in a non-relativistic system. In the theory of the PN approximation, the 1PN metric of any non-extremely relativistic systems can be formulated as [2]
$\begin{eqnarray}\begin{array}{rcl}{{\rm{d}}{s}}^{2} & = & \left(-1+\mathop{{g}_{00}}\limits^{2}+\mathop{{g}_{00}}\limits^{4}\right){{\rm{d}}{t}}^{2}\\ & & +2\mathop{{g}_{0i}}\limits^{3}{{\rm{d}}{x}}^{i}{\rm{d}}{t}+\left({\delta }_{{ij}}+\mathop{{g}_{{ij}}}\limits^{2}\right){{\rm{d}}{x}}^{i}{{\rm{d}}{x}}^{j}\,,\end{array}\end{eqnarray}$
where ${\mathop{g}\limits^{N}}_{\mu \nu }$ denotes the terms in gμν of order ${\bar{v}}^{N}$. δij is the Kronecker sign. Throughout this paper, the natural units in which G = c = 1 are adopted.We assume that the infinitely-thin spherical shell has rest mass M and radius R, and radial velocity ui = uni with ni being the radial unit vector. The energy-momentum tensor of the self-gravitating and collapsing thin spherical shell at the Newtonian order can be formulated as [4]
$\begin{eqnarray}\mathop{{T}^{00}}\limits^{0}=\displaystyle \frac{M}{4\pi {R}^{2}}\delta (r-R)\,,\end{eqnarray}$
$\begin{eqnarray}\mathop{{T}^{0i}}\limits^{1}=\displaystyle \frac{M}{4\pi {R}^{2}}{{un}}_{i}\delta (r-R)\,,\end{eqnarray}$
$\begin{eqnarray}\mathop{{T}^{00}}\limits^{2}=\displaystyle \frac{M}{4\pi {R}^{2}}\left(\displaystyle \frac{1}{2}{u}^{2}+{\rm{\Phi }}\right)\delta (r-R)\,,\end{eqnarray}$
$\begin{eqnarray}\mathop{{T}^{{ij}}}\limits^{2}=\displaystyle \frac{M}{4\pi {R}^{2}}{u}^{2}{n}_{i}{n}_{j}\delta (r-R)\,,\end{eqnarray}$
where $\mathop{{T}^{\mu \nu }}\limits^{N\,\,}$ denotes the terms in Tμν of order $(\bar{M}/{\bar{r}}^{3}){\bar{v}}^{N}$. δ(s) is the Dirac delta function of s. ${n}_{i}=\tfrac{{x}_{i}}{r}$ with $r=| {x}^{i}{x}_{i}{| }^{\tfrac{1}{2}}$. ${\rm{\Phi }}=-\tfrac{M}{R}$ is the Newtonian potential at any point of the shell induced by the other parts. The shell collapsing implies that $u=\tfrac{{\rm{d}}{R}}{{\rm{d}}{t}}\lt 0$. At the Newtonian order, we have $\tfrac{{\rm{d}}{u}}{{\rm{d}}{t}}=\tfrac{{\rm{\Phi }}}{2R}$.We have obtained the 1PN metric of the collapsing thin spherical shell as follows [4]
$\begin{eqnarray}\mathop{{g}_{00}}\limits^{2}=2\displaystyle \frac{M}{r}\theta (r-R)+2\displaystyle \frac{M}{R}\left[1-\theta (r-R)\right]\,,\end{eqnarray}$
$\begin{eqnarray}\mathop{{g}_{{ij}}}\limits^{2}=2\displaystyle \frac{M}{r}{\delta }_{{ij}}\theta (r-R)+2\displaystyle \frac{M}{R}{\delta }_{{ij}}\left[1-\theta (r-R)\right]\,,\end{eqnarray}$
$\begin{eqnarray}\mathop{{g}_{0i}}\limits^{3}=-\displaystyle \frac{4}{3}\displaystyle \frac{{MR}\,u\,{x}_{i}}{{r}^{3}}\theta (r-R)-\displaystyle \frac{4}{3}\displaystyle \frac{M\,u\,{x}_{i}}{{R}^{2}}\left[1-\theta (r-R)\right]\,,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\mathop{{g}_{00}}\limits^{4} & = & 2\displaystyle \frac{M}{r}\left(-\displaystyle \frac{M}{r}+\displaystyle \frac{11}{6}{u}^{2}+\displaystyle \frac{7}{6}{\rm{\Phi }}\right)\theta (r-R)\\ & & +\,2\displaystyle \frac{M}{R}\left[-\displaystyle \frac{M}{R}+\left(\displaystyle \frac{3}{2}+\displaystyle \frac{1}{3}\displaystyle \frac{{r}^{2}}{{R}^{2}}\right){u}^{2}+\left(\displaystyle \frac{5}{4}-\displaystyle \frac{1}{12}\displaystyle \frac{{r}^{2}}{{R}^{2}}\right){\rm{\Phi }}\right]\\ & & \times \,\left[1-\theta (r-R)\right]\,,\end{array}\end{eqnarray}$
where θ(s) is the Heaviside function of s, which equals 1 for s > 0 and 0 otherwise. δij is the Kronecker delta sign which equals 1 for i = j and 0 otherwise. Notice that both R and u are time-dependent.3. Retrieval of matter energy-momentum tensor
In this section, we substitute the time-dependent metric given by equations (6 )-(9 ) into Einstein field equations, to prove it is the correct solution for the spacetime of the self-gravitating and collapsing thin spherical shell. In the meantime, we retrieve all components of the shell’s energy-momentum tensor.
Einstein field equations can be written as
$\begin{eqnarray}{{ \mathcal R }}_{\mu \nu }=-8\pi \left({T}_{\mu \nu }-\displaystyle \frac{1}{2}{g}_{\mu \nu }{T}_{\lambda }^{\lambda }\right)\,,\end{eqnarray}$
where ${{ \mathcal R }}_{\mu \nu }$ is the Ricci tensor. Under the harmonic-coordinate condition, Einstein field equations to the 1PN order reduce to [2]
$\begin{eqnarray}{\mathop{{ \mathcal R }}\limits^{2}}_{00}=\displaystyle \frac{1}{2}{{\rm{\nabla }}}^{2}\mathop{{g}_{00}}\limits^{2}=-4\pi \mathop{{T}^{00}}\limits^{0}\,,\end{eqnarray}$
$\begin{eqnarray}{\mathop{{ \mathcal R }}\limits^{2}}_{{ij}}=\displaystyle \frac{1}{2}{{\rm{\nabla }}}^{2}\mathop{{g}_{{ij}}}\limits^{2}=-4\pi {\delta }_{{ij}}\mathop{{T}^{00}}\limits^{0}\,,\end{eqnarray}$
$\begin{eqnarray}{\mathop{{ \mathcal R }}\limits^{3}}_{0i}=\displaystyle \frac{1}{2}{{\rm{\nabla }}}^{2}\mathop{{g}_{0i}}\limits^{3}=8\pi \mathop{{T}^{0i}}\limits^{1}\,,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\mathop{{ \mathcal R }}\limits^{4}}_{00} & = & \displaystyle \frac{1}{2}\left[{{\rm{\nabla }}}^{2}\mathop{{g}_{00}}\limits^{4}-{\partial }_{t}^{2}\mathop{{g}_{00}}\limits^{2}-\mathop{{g}_{{ij}}}\limits^{2}{\partial }_{i}{\partial }_{j}\mathop{{g}_{00}}\limits^{2}+({\partial }_{i}\mathop{{g}_{00}}\limits^{2})({\partial }_{i}\mathop{{g}_{00}}\limits^{2})\right]\\ & = & -4\pi \left(\mathop{{T}^{00}}\limits^{2}-2\mathop{{g}_{00}}\limits^{2}\mathop{{T}^{00}}\limits^{0}+\mathop{{T}^{{ii}}}\limits^{2}\right)\,,\end{array}\end{eqnarray}$
where ${\mathop{{ \mathcal R }}\limits^{N}}_{\mu \nu }$ denotes the term in ${{ \mathcal R }}_{\mu \nu }$ of order ${\bar{v}}^{N}/{\bar{r}}^{2}$. ${\partial }_{t}\equiv \partial /\partial t$ and ${\partial }_{i}\equiv \partial /\partial {x}^{i}$. ∇2 is the Laplace operator. Firstly, we show the 1PN metric satisfies the harmonic-coordinate conditions at the 1PN order, which can be written as [2]
$\begin{eqnarray}\displaystyle \frac{1}{2}{\partial }_{t}\mathop{{g}_{00}}\limits^{2}-{\partial }_{i}\mathop{{g}_{0i}}\limits^{3}+\displaystyle \frac{1}{2}{\partial }_{t}\mathop{{g}_{{ii}}}\limits^{2}=0\,,\end{eqnarray}$
$\begin{eqnarray}\displaystyle \frac{1}{2}{\partial }_{i}\mathop{{g}_{00}}\limits^{2}+{\partial }_{j}\mathop{{g}_{{ij}}}\limits^{3}-\displaystyle \frac{1}{2}{\partial }_{i}\mathop{{g}_{{jj}}}\limits^{2}=0\,.\end{eqnarray}$
From equations (
$\begin{eqnarray}{\partial }_{t}\mathop{{g}_{00}}\limits^{2}=-2\displaystyle \frac{{Mu}}{{R}^{2}}\left[1-\theta (r-R)\right]\,,\end{eqnarray}$
$\begin{eqnarray}{\partial }_{t}\mathop{{g}_{{ii}}}\limits^{2}=-6\displaystyle \frac{{Mu}}{{R}^{2}}\left[1-\theta (r-R)\right]\,,\end{eqnarray}$
$\begin{eqnarray}{\partial }_{i}\mathop{{g}_{0i}}\limits^{3}=-4\displaystyle \frac{{Mu}}{{R}^{2}}\left[1-\theta (r-R)\right]\,,\end{eqnarray}$
$\begin{eqnarray}{\partial }_{i}\mathop{{g}_{00}}\limits^{2}=-2\displaystyle \frac{{{Mx}}_{i}}{{r}^{3}}\theta (r-R)\,,\end{eqnarray}$
$\begin{eqnarray}{\partial }_{j}\mathop{{g}_{{ij}}}\limits^{2}=-2\displaystyle \frac{{{Mx}}_{i}}{{r}^{3}}\theta (r-R)\,,\end{eqnarray}$
$\begin{eqnarray}{\partial }_{i}\mathop{{g}_{{jj}}}\limits^{2}=-6\displaystyle \frac{{{Mx}}_{i}}{{r}^{3}}\theta (r-R)\,,\end{eqnarray}$
where we have replaced $\tfrac{{dR}}{{dt}}$ with u. With these equations, we can easily verify that equations (Secondly, we verify that the 1PN metric satisfies Einstein field equations about ${\mathop{{ \mathcal R }}\limits^{2}}_{00}$, ${\mathop{{ \mathcal R }}\limits^{2}}_{{ij}}$ and ${\mathop{{ \mathcal R }}\limits^{3}}_{0i}$, which are given by equations (
$\begin{eqnarray}\begin{array}{rcl}{\mathop{{ \mathcal R }}\limits^{2}}_{00} & = & \displaystyle \frac{1}{2}{\partial }_{i}{\partial }_{i}\mathop{{g}_{00}}\limits^{2}=\displaystyle \frac{1}{2}{\partial }_{i}\left[-2\displaystyle \frac{{{Mx}}_{i}}{{r}^{3}}\theta (r-R)\right]\\ & = & -\displaystyle \frac{M}{{r}^{2}}\delta (r-R)=-4\pi \mathop{{T}^{00}}\limits^{0}\,,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\mathop{{ \mathcal R }}\limits^{2}}_{{ij}} & = & \displaystyle \frac{1}{2}{\partial }_{k}{\partial }_{k}\mathop{{g}_{{ij}}}\limits^{2}=\displaystyle \frac{1}{2}{\partial }_{k}\left[-2\displaystyle \frac{{{Mx}}_{k}}{{r}^{3}}\theta (r-R){\delta }_{{ij}}\right]\\ & = & -\displaystyle \frac{M}{{r}^{2}}\delta (r-R){\delta }_{{ij}}=-4\pi {\delta }_{{ij}}\mathop{{T}^{00}}\limits^{0}\,,\end{array}\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}{\mathop{{ \mathcal R }}\limits^{3}}_{0i} & = & \displaystyle \frac{1}{2}{\partial }_{j}{\partial }_{j}\mathop{{g}_{0i}}\limits^{3}=\displaystyle \frac{1}{2}{\partial }_{j}\\ & & \times \,\left[4{MuR}\left(\displaystyle \frac{{\delta }_{{ij}}}{3{R}^{3}}-\displaystyle \frac{{\delta }_{{ij}}}{3{r}^{3}}+\displaystyle \frac{{x}_{i}{x}_{j}}{{r}^{5}}\right)\theta (r-R)-\displaystyle \frac{4{Mu}\,{\delta }_{{ij}}}{3{R}^{2}}\right]\\ & = & 2{MuR}\left[\left(\displaystyle \frac{{n}_{i}}{{r}^{4}}+3\displaystyle \frac{{x}_{i}}{{r}^{5}}+\displaystyle \frac{{\delta }_{{ij}}{x}_{j}}{{r}^{5}}-5\displaystyle \frac{{x}_{i}}{{r}^{5}}\right)\theta (r-R)\right.\\ & & +\left.\left(\displaystyle \frac{{\delta }_{{ij}}}{3{R}^{3}}-\displaystyle \frac{{\delta }_{{ij}}}{3{r}^{3}}+\displaystyle \frac{{x}_{i}{x}_{j}}{{r}^{5}}\right){n}_{j}\delta (r-R)\right]\\ & = & \displaystyle \frac{2{Mu}\,{n}_{i}}{{R}^{2}}\delta (r-R)=8\pi \mathop{{T}^{0i}}\limits^{1}\,,\end{array}\end{eqnarray}$
where we have made use of ${x}_{j}={{rn}}_{j}$, ${\partial }_{j}{x}_{j}=3$, ${\partial }_{j}{x}_{i}={\delta }_{{ij}}$, ${\partial }_{j}r={n}_{j}$ and $\tfrac{d}{{ds}}\theta (s)=\delta (s)$. Notice that R and u are the functions of time t only. Therefore, equations (
Thirdly, we verify the 1PN metric satisfies the Einstein field equations about ${\mathop{{ \mathcal R }}\limits^{4}}_{00}$, which is given by equation (
$\begin{eqnarray}\phi =-\displaystyle \frac{1}{2}\mathop{{g}_{00}}\limits^{2}\,,\end{eqnarray}$
$\begin{eqnarray}\psi =-\displaystyle \frac{1}{2}\mathop{{g}_{00}}\limits^{4}-{\phi }^{2}\,,\end{eqnarray}$
then equation ( $\begin{eqnarray}{{\rm{\nabla }}}^{2}\psi ={\partial }_{t}^{2}\phi +4\pi \left(\mathop{{T}^{00}}\limits^{2}+\mathop{{T}^{{ii}}}\limits^{2}\right)\,.\end{eqnarray}$
Plugging equations (
$\begin{eqnarray}\phi =-\displaystyle \frac{M}{r}\theta (r-R)-\displaystyle \frac{M}{R}\left[1-\theta (r-R)\right]\,,\end{eqnarray}$
$\begin{eqnarray}\begin{array}{rcl}\psi & = & -\displaystyle \frac{M}{r}\left(\displaystyle \frac{11}{6}{u}^{2}+\displaystyle \frac{7}{6}{\rm{\Phi }}\right)\theta (r-R)-\displaystyle \frac{M}{R}\left[\left(\displaystyle \frac{3}{2}+\displaystyle \frac{1}{3}\displaystyle \frac{{r}^{2}}{{R}^{2}}\right){u}^{2}\right.\\ & & +\,\left.\left(\displaystyle \frac{5}{4}-\displaystyle \frac{1}{12}\displaystyle \frac{{r}^{2}}{{R}^{2}}\right){\rm{\Phi }}\right]\left[1-\theta (r-R)\right].\end{array}\end{eqnarray}$
Taking the second time derivative of Φ, from equation (
$\begin{eqnarray}{\partial }_{t}^{2}\phi =-\displaystyle \frac{M}{{R}^{3}}\left(2{u}^{2}-\displaystyle \frac{1}{2}{\rm{\Phi }}\right)\left[1-\theta (r-R)\right]+\displaystyle \frac{M}{{R}^{2}}{u}^{2}\delta (r-R)\,,\end{eqnarray}$
where we have replaced $\tfrac{{\rm{d}}{R}}{{\rm{d}}{t}}$ with u and made use of $\tfrac{{\rm{d}}{u}}{{\rm{d}}{t}}=\tfrac{{\rm{\Phi }}}{2R}$. Applying the Laplace operator on ψ given by equation (
$\begin{eqnarray}\begin{array}{rcl}{{\rm{\nabla }}}^{2}\psi & = & {\partial }_{i}\left({\partial }_{i}\psi \right)\\ & = & {\partial }_{i}\left\{\left(\displaystyle \frac{11}{6}{u}^{2}+\displaystyle \frac{7}{6}{\rm{\Phi }}\right)\displaystyle \frac{{{Mx}}_{i}}{{r}^{3}}\theta (r-R)\right.\\ & & -\,\left.\displaystyle \frac{M}{{R}^{3}}\left(\displaystyle \frac{1}{3}{u}^{2}-\displaystyle \frac{1}{12}{\rm{\Phi }}\right)2{x}_{i}\left[1-\theta (r-R)\right]\right\}\\ & = & \left(\displaystyle \frac{11}{6}{u}^{2}+\displaystyle \frac{7}{6}{\rm{\Phi }}\right)\displaystyle \frac{M}{{R}^{2}}\delta (r-R)-\displaystyle \frac{M}{{R}^{3}}\left(2{u}^{2}-\displaystyle \frac{1}{2}{\rm{\Phi }}\right)\\ & & \times \,\left[1-\theta (r-R)\right]+\displaystyle \frac{M}{{R}^{2}}\left(\displaystyle \frac{2}{3}{u}^{2}-\displaystyle \frac{1}{6}{\rm{\Phi }}\right)\delta (r-R)\\ & = & \mathop{\underbrace{-\displaystyle \frac{M}{{R}^{3}}\left(2{u}^{2}-\displaystyle \frac{1}{2}{\rm{\Phi }}\right)\left[1-\theta (r-R)\right]+\displaystyle \frac{M}{{R}^{2}}{u}^{2}\delta (r-R)}}\limits_{{\partial }_{t}^{2}\phi }\\ & & +\,\mathop{\underbrace{\displaystyle \frac{M}{{R}^{2}}\left[\left(\displaystyle \frac{1}{2}{u}^{2}+{\rm{\Phi }}\right)+{u}^{2}\right]\delta (r-R)}}\limits_{4\pi \left(\mathop{{T}^{00}}\limits^{2}\,+\,\mathop{{T}^{{ii}}}\limits^{2}\right)}.\end{array}\end{eqnarray}$
Notice that ${\partial }_{t}^{2}\phi $ is given by equation (Conversely, it can be retrieved from equation (
$\begin{eqnarray}\mathop{{T}^{00}}\limits^{2}+\mathop{{T}^{{ii}}}\limits^{2}=\displaystyle \frac{M}{4\pi {R}^{2}}\left(\displaystyle \frac{3}{2}{u}^{2}+{\rm{\Phi }}\right)\delta (r-R)\,.\end{eqnarray}$
Considering we have obtained $\mathop{{T}^{00}}\limits^{0}=\tfrac{M}{4\pi {R}^{2}}\delta (r-R)$ and $\mathop{{T}^{0i}}\limits^{1}=\tfrac{M}{4\pi {R}^{2}}{{un}}_{i}\delta (r-R)$ from the 1PN metric in the above, we can easily write down $\mathop{{T}^{{ij}}}\limits^{2}=\tfrac{M}{4\pi {R}^{2}}{u}^{2}{n}_{i}{n}_{j}\delta (r-R)$, which is just equation (
4. Summary
We have retrieved the leading-order energy-momentum tensor from the 1PN time-dependent metric of the self-gravitating and collapsing thin spherical shell. This verifies that the latter is the correct solution to the Einstein field equations at the 1PN order.