I Transformation for Lemaitre coordinates

[Dale]

TL;DR

I am looking for a transformation for Lemaitre coordinates to standard Schwarzschild coordinates.

Lemaitre coordinates are a nice set of coordinates for describing observers free-falling (aka rain observers) in a Schwarzschild spacetime. Such observers have very easy geodesics that are just constant coordinate $\rho$ and the time coordinate $\tau$ is equal to their proper time $\sqrt{-ds^2}$.

The transformation between Lemaitre coordinates and Schwarzschild coordinates is given as: $$d\tau=dt+\sqrt{\frac{R}{r}}\left( 1-\frac{R}{r} \right)^{-1} dr$$$$d\rho=dt+\sqrt{\frac{r}{R}}\left( 1-\frac{R}{r} \right)^{-1} dr$$ where $r$ and $t$ are the Schwarzschild radial and time coordinates and $\rho$ and $\tau$ are the Lemaitre position and time coordinates and $R$ is the Schwarzschild radius.

I would like to get $r$ and $t$ in terms of $\tau$ and $\rho$, but I don't actually know where to go from here. Can I just integrate like $$\int d\tau=\int dt+\int \sqrt{\frac{R}{r}}\left( 1-\frac{R}{r} \right)^{-1} dr$$to get$$\tau=t+2 \sqrt{r R} - 2 R \tanh^{-1} \left( \sqrt{\frac{r}{R}} \right)$$

 

[Ibix]

I think the general approach is to use the differential transforms to write the Jacobian (or is it Jacobean?) and use that to write the Lemaitre basis vectors (trivial in Lemaitre coordinates) in terms of the Schwarzschild basis vectors. The integral curves of those two fields are your coordinate lines.

 

[Dale]

Ok, so the Jacobian would be $$
\left( \begin{array}{cc}
\frac{\partial \tau}{\partial t} & \frac{\partial \tau}{\partial r} \\
\frac{\partial \rho}{\partial t} & \frac{\partial \rho}{\partial r} \\
\end{array} \right) =
\left( \begin{array}{cc}
1 & \sqrt{\frac{R}{r}}\left( 1-\frac{R}{r} \right)^{-1 } \\
1 & \sqrt{\frac{r}{R}}\left( 1-\frac{R}{r} \right)^{-1} \\
\end{array} \right)
$$

I am not sure where to go from here to get the $t$ and $r$ for a given $\tau$ and $\rho$, either closed form or numerically.

 

[renormalize]

I would like to get $r$ and $t$ in terms of $\tau$ and $\rho$, but I don't actually know where to go from here. Can I just integrate like $$\int d\tau=\int dt+\int \sqrt{\frac{R}{r}}\left( 1-\frac{R}{r} \right)^{-1} dr$$to get$$\tau=t+2 \sqrt{r R} - 2 R \tanh^{-1} \left( \sqrt{\frac{r}{R}} \right)$$

Yes you can! Inspired by https://en.wikipedia.org/wiki/Lemaître_coordinates, note that your two equations:$$d\tau=dt+\sqrt{\frac{R}{r}}\left(1-\frac{R}{r}\right)^{-1}dr\,,\;d\rho=dt+\sqrt{\frac{r}{R}}\left(1-\frac{R}{r}\right)^{-1}dr\tag{1a,b}$$can be subtracted to get:$$d\left(\rho-\tau\right)=\frac{r\left(\sqrt{\frac{r}{R}}-\sqrt{\frac{R}{r}}\right)}{r-R}dr\tag{2}$$which integrates immediately to:$$\rho-\tau=\frac{2r^{2}\left(\sqrt{\frac{r}{R}}-\sqrt{\frac{R}{r}}\right)}{3\left(r-R\right)}\tag{3}$$Here I drop the arbitrary constant of integration since it can be absorbed by a choice of origin for the $\tau$ variable. Solving eq.(3) for $r$ yields multiple solutions, the simplest of which is:$$r=R^{1/3}\left(\frac{3}{2}\left(\rho-\tau\right)\right)^{2/3}\tag{4}$$Now as you point out, eq.(1a) integrates to:$$\tau=t+2\sqrt{R}\left(\sqrt{r}-\sqrt{R}\tanh^{-1}\left(\sqrt{\frac{r}{R}}\right)\right)\tag{5}$$(again dropping the constant of integration). Inserting $r$ from (4) and solving for $t$ gives finally:$$t=\tau+2R\tanh^{-1}\left(\left(\frac{3\left(\rho-\tau\right)}{2R}\right)^{1/3}\right)-2\left(\frac{3R^{2}\left(\rho-\tau\right)}{2}\right)^{1/3}\tag{6}$$Eqs.(4) and (6) express the Schwarzschild coordinates $r,t$ in terms of the Lemaitre coordinates $\rho,\tau\,$.

 

[Dale]

Excellent, thanks @renormalize. It was just not occurring to me to eliminate one of the variables algebraically before integrating.

 

[Dale]

$$\tau=t+2\sqrt{R}\left(\sqrt{r}-\sqrt{R}\tanh^{-1}\left(\sqrt{\frac{r}{R}}\right)\right)\tag{5}$$

Playing around with this a little more, it looks like this only works for $r<R$ inside the horizon. For outside the horizon with $R<r$ we have to use

$$\tau=t+2\sqrt{R}\left(\sqrt{r}-\sqrt{R}\ \tanh^{-1}\left(\sqrt{\frac{R}{r}}\right)\right)\tag{5b}$$

The range of $\tanh^{-1}$ is (-1,1).

 

[PeterDonis]

it looks like this only works for $r<R$ inside the horizon.

This will be true in general for any transformation between the Schwarzschild chart and a chart that doesn't have a coordinate singularity at the horizon: something has to change from one region to the other.

 

[Demystifier]

More generally, for two system of coordinates $x$ and $y$, one has $x^{\mu}=f^{\mu}(y)$, so
$$dx^{\mu}=f^{\mu}_{,\nu}(y)dy^{\nu}$$
If it happens that $f^{\mu}_{,\nu}(y)$ depends only on $y^{\nu}$, then it can be integrated directly as
$$x^{\mu}=\int f^{\mu}_{,\nu}(y^{\nu})dy^{\nu}$$
as Dale did in the first post. However, if the function $f^{\mu}_{,\nu}(y^{\nu})$ is singular for some values of $y^{\nu}$, then one has to be careful. Sometimes the integration over singular values is well defined and sometimes isn't. When it isn't, then it may be necessary to define the integral separately for different regions of spacetime separated by the singular values of $y^{\nu}$.

 

[Dale]

Yes, so I think I need to go back and be more careful with both equations 4 and 6. I will try to do that sometime later this week