# Line element in a non-static system of coordinates

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 $$$$r=(9 m / 2)^{1/3}\left(r^{\prime}-c t^{\prime}\right)^{2/3}, \quad d t^{\prime}=d t-\frac{(2 m / r)^{1/2}}{1-2 m / r} d r$$$$

and the line element takes the form

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

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

$$$$1-\frac{2 m}{r_{0}}=0$$$$

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.

• 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 ? Commented Jul 31, 2019 at 10:59
• 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. Commented Jul 31, 2019 at 23:48
• This seems to be Lemaitre coordinates. Commented Aug 1, 2019 at 0:24
• 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 Commented Aug 1, 2019 at 0:32
• 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. Commented Aug 1, 2019 at 2:08