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So I am attempting to find the line element in a non-static system of coordinates $r^{\prime}, \theta, \phi, t^{\prime}$ in vacuum, where the transformations are \begin{equation} r=(9 m / 2)^{1/3}\left(r^{\prime}-c t^{\prime}\right)^{2/3}, \quad d t^{\prime}=d t-\frac{(2 m / r)^{1/2}}{1-2 m / r} d r \end{equation}

and the line element takes the form

\begin{equation} d s^{2}=\frac{2 m}{r} d r^{\prime 2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)-c^{2} d t^{\prime 2} \end{equation} which is the line element I am trying to obtain (i.e. the $non$-$static$ system of coordinates line element) by transforming the line element

\begin{equation} d s^{2}=\frac{d r^{2}}{1-2 m / r}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)-\left(1-\frac{2 m}{r}\right) c^{2} d t^{2} \end{equation} in order to remove the singularity at $r=r_{0}$, determined by the equation

\begin{equation} 1-\frac{2 m}{r_{0}}=0 \end{equation}

My attempt:

Consider first the transformation for $dr$, such that \begin{gather*} dr = \frac{\partial r}{\partial r^{\prime}} dr^{\prime} \\ \implies \frac{\partial}{\partial r^{\prime}} \left((9 m / 2)^{1/3}\left(r^{\prime}-c t^{\prime}\right)^{2/3}\right) dr^{\prime} \end{gather*} set $c=1$ for simplicity, then \begin{gather*} \frac{\partial}{\partial r^{\prime}} \left((9 m / 2)^{1/3}\left(r^{\prime}- t^{\prime}\right)^{2/3}\right) \\ \implies (9 m / 2)^{1/3}\frac{\partial}{\partial r^{\prime}} \left(r^{\prime}- t^{\prime}\right)^{2/3} \\ \implies (9 m / 2)^{1/3} \frac{2}{3} \left(r^{\prime}- t^{\prime}\right)^{-1/3} \left(1\right) \\ \implies \frac{2}{3} \left[\frac{(9 m / 2)}{\left(r^{\prime}- t^{\prime}\right)}\right]^{1/3} \implies dr^{2} = \frac{4}{9} \left[\frac{(9 m / 2)}{\left(r^{\prime}- t^{\prime}\right)}\right]^{2/3} dr^{\prime 2} \end{gather*}

Next, consider $\frac{1}{1-2m/r}$ such that \begin{gather*} \frac{1}{1-2m/r} \implies \left(1 -\frac{2m}{r}\right) = \left(1 -\frac{2m}{(9 m / 2)^{1/3}\left(r^{\prime}-c t^{\prime}\right)^{2/3}}\right) \\ \implies \frac{(9 m / 2)^{1/3}\left(r^{\prime}- t^{\prime}\right)^{2/3} - 2m}{(9 m / 2)^{1/3}\left(r^{\prime}- t^{\prime}\right)^{2/3}} \\ \implies \frac{1}{\frac{1}{1-2m/r}} = \frac{(9 m / 2)^{1/3}\left(r^{\prime}- t^{\prime}\right)^{2/3}}{(9 m / 2)^{1/3}\left(r^{\prime}- t^{\prime}\right)^{2/3} - 2m} \end{gather*}

Then substituting back into $d s^{2}=\frac{d r^{2}}{1-2 m / r}$, we have \begin{gather*} \frac{d r^{2}}{1-2 m / r} = \frac{4}{9} \left[\frac{(9 m / 2)}{\left(r^{\prime}- t^{\prime}\right)}\right]^{2/3} \frac{(9 m / 2)^{1/3}\left(r^{\prime}- t^{\prime}\right)^{2/3}}{(9 m / 2)^{1/3}\left(r^{\prime}- t^{\prime}\right)^{2/3} - 2m} dr^{\prime 2} \end{gather*}

No matter what way I look at it I can't obtain the required part of the line element I'm looking for. I've tried cubing across but I just end up getting a mess. My next approach is multiplying by the conjugate of the second term in order to get rid of the subtraction in the denominator. Any help would be greatly appreciated.

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    $\begingroup$ I can't obtain the required part of the line element I'm looking for You do not explain what exactly you're looking for ? What is the problem with e.g. the final expression you give ? $\endgroup$ Commented Jul 31, 2019 at 10:59
  • $\begingroup$ Essentially, what I am attempting to do is to transform the line element which contains $ds^{2} = \frac{dr^{2}}{1-2m/r}$ to $ds^{2} = \frac{2m}{r} dr^{\prime 2}$, therefore, removing the singularity from the first term, yet when I attempt to transform the coordinates I have a difficulty reducing it to the $ds^{2} = \frac{2m}{r} dr^{\prime 2}$ term I am looking for. $\endgroup$ Commented Jul 31, 2019 at 23:48
  • $\begingroup$ This seems to be Lemaitre coordinates. $\endgroup$ Commented Aug 1, 2019 at 0:24
  • $\begingroup$ Yes, you are exactly correct, these are Lemaitre coordinates, and the wikipedia link you provided (thanks by the way) I had actually read already and it isn't too helpful $\endgroup$ Commented Aug 1, 2019 at 0:32
  • $\begingroup$ Should you not be using $dr = \frac{\partial r}{\partial r^{\prime}} dr^{\prime}+\frac{\partial r}{\partial t^{\prime}} dt^{\prime}$ and a similar expression for $dt$. You don't seem to be doing this. $\endgroup$ Commented Aug 1, 2019 at 2:08

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