Comparing Gullstrand-Painleve & Lemaitre Coordinates

[PAllen]

TL;DR

Gullstrand-Painleve coordinates and Lemaitre coordinates are based on the same foliation, but you would never know that looking at their metric expression. This thread explores that difference.

For reference, the wikipedia entries are adequate for this discussion:

https://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates
(henceforth, GP coordinates)

https://en.wikipedia.org/wiki/Lemaître_coordinates
(henceforth, LM coordinates)

Both of these coordinates are based on a foliation by hypersurfaces orthogonal to the congruence of radial infaller's from infinity (with asymptotically zero velocity at infinity). Both coordinate systems have the following properties:

1) The foliation hypersurfaces (surfaces of constant time coordinate in each system) are flat Euclidean 3-spaces.
2) Radial infaller's tangent 4-velocities are orthogonal to the foliation surfaces
3) For a radial infall world line, the proper time between events on it is equal to the coordinate time difference between those events.

So why do they look so different, despite using the same foliation?

GP coordinates
-----------------

The r, θ, and ϕ coordinates are the same as in both the Shwarzschild exterior and interior patches. As result, t coordinate paths (holding these coordinates constant) are the same as the integral curves of the kvf; timelike stationary world lines outside the horizon, and spacelike 'axial' paths inside the horizon. For r=2M, the t coordinate acts as a parameter of a lightlike path of the horizon. Important to note is that the t coordinate values along these kvf paths are completely different from the Schwarzschild t coordinate, in order to achieve property (3) above.

As a result, the metric must have off diagonal elements because the t basis vector is orthogonal to the Schwarzschild foliation (being always in the direction of the kvf), and thus not orthogonal to the GP spatial foliation.

Desirable results of this choice are that the metric components are independent of the t coordinate, and the flatness of spatial slices is manifest.

Undesirable result are that the coordinate expression of a radial infall world line is not trivial (you want r(t) satisfying $dr/dt= -\sqrt {2M/}r$ ). Given this expression, it is easy to then show the tangent to such a world line is orthogonal to the r basis vector. It is a good bit more work to show that property (3) holds for such a world line.

Lemaitre coordinates
-----------------------

The angular coordinates are, of course, the same as Schwarzschild case, but both the radial and timelike coordinate are different. In place of the radial coordinate defined by area of surfaces of spherical symmetry, we introduce ρ as simply an identifier of a particular radial infall trajectory. A world line of constant ρ as well as angular coordinates is a radial infall trajectory, with the LM time coordinate marking proper time along this infall trajectory.

Because this coordinate system has timelike basis orthogonal to the foliation, there are no off diagonal metric components. The metric is orthgonal, and the 3+1 coordinate distinction remains the same outside, on and inside the horizon. The changing nature of the extra kvf has no impact, because no coordinate is based on the kvf. On the other hand, because a given differences in ρ must correspond to smaller proper distances as the foliation surfaces get closer to the horizon (radial infallers get closer), the metric components must depend on the LM time coordinate.

Desirable features of Lemaitre coordinates are that the radial infall trajectories have trivial coordinate representation (all 3 spatial coordinates constant), and that proper time being coordinate time along such a trajectory is trivially obvious by inspection from the metric, and the orthogonality of such world lines to the foliation is also trivial (simply absence of off diagonal metric components). Further, the feature that all 4 coordinates maintain their spacelike/timelike character throughout the coordinate coverage is nice.

Undesirable features are that the flatness of the spatial slices is totally obscured. Further, all metric components depend on the LM time coordinate (except the tt component which is 1).

----

In my next post, I will ask a specific question I would like help with.

 

[PAllen]

I would like to verify that induced 3-metric on a Lemaitre slice is indeed a surface with vanishing curvature tensor, thus Euclidean 3-space in atypical coordinates. However, since I have never set up software for tensor computations, and (while I know all the formulas well) hand computing a curvature tensor is notoriously error prone, I would greatly appreciate if someone with such software would verify vanishing 3 curvature tensor for the following metric (with R and $\tau$ being considered constants):

$$\begin{bmatrix}
\left[ \frac {2R} { 3 (\rho-\tau)} \right]^{(2/3)} & 0 & 0 \\
0 & \left[ (9/4)(\rho-\tau)^2 R \right]^{(2/3)} & 0 \\
0 & 0 & \left[ (9/4)(\rho-\tau)^2 R \right]^{(2/3)} sin^2 \theta
\end{bmatrix}
$$

 

[PeterDonis]

I would like to verify that induced 3-metric on a Lemaitre slice is indeed a surface with vanishing curvature tensor, thus Euclidean 3-space in atypical coordinates.

I think this can be done more easily than cranking through the computation of the curvature tensor. From the Wikipedia page for Lemaitre coordinates we have the following useful relations (using your notation $R$ instead of $r_s$):

$$
d\tau = dt + \sqrt{\frac{R}{r}} \frac{r}{r - R} dr
$$

$$
d\rho = dt + \sqrt{\frac{r}{R}} \frac{r}{r - R} dr
$$

For the metric of the 3-surface we are interested in, we have $d\tau = 0$, which allows us to eliminate $dt$ in the equation for $d\rho$:

$$
d\rho = \left( \sqrt{\frac{r}{R}} - \sqrt{\frac{R}{r}} \right) \frac{r}{r - R} dr
$$

Now it's just algebra:

$$
d\rho = \sqrt{\frac{r}{R}} \frac{r - R}{r} \frac{r}{r - R} dr
$$

$$
\frac{R}{r} d\rho^2 = dr^2
$$

So the line element for $d\tau = 0$ is the Euclidean line element when written in terms of $r$. That is sufficient to show that those slices are flat.

 

[Demystifier]

So why do they look so different, despite using the same foliation?

Did you try to write down the explicit coordinate transformation that relates GP to L?

 

[PeterDonis]

why do they look so different, despite using the same foliation?

I would say it's because the foliation is not all there is to it. One also has to look at the integral curves of the timelike basis vector of the coordinate chart.

In L, those integral curves are everywhere orthogonal to the foliation. That's because in L, those integral curves are just the radial infallers' worldlines. Which in turn means that along those integral curves, the metric is not constant. In the line element, this is shown by the fact that the spatial metric coefficients are functions of $\tau$ as well as $\rho$, and $\tau$ is of course not constant along the integral curves in question, even though $\rho$ is.

In GP, those integral curves are the same as the integral curves of the timelike Killing vector field of the spacetime. Which means that along those integral curves, the metric is constant. In the line element, this is shown by the fact that the metric coefficients are functions of $r$ only, if they are functions of anything, and $r$ is constant along the integral curves in question.

To put it another way, the timelike "grid lines" are drawn very differently in these two charts, even though they both use the same foliation.

 

[PAllen]

I would say it's because the foliation is not all there is to it. One also has to look at the integral curves of the timelike basis vector of the coordinate chart.

In L, those integral curves are everywhere orthogonal to the foliation. That's because in L, those integral curves are just the radial infallers' worldlines. Which in turn means that along those integral curves, the metric is not constant. In the line element, this is shown by the fact that the spatial metric coefficients are functions of $\tau$ as well as $\rho$, and $\tau$ is of course not constant along the integral curves in question, even though $\rho$ is.

In GP, those integral curves are the same as the integral curves of the timelike Killing vector field of the spacetime. Which means that along those integral curves, the metric is constant. In the line element, this is shown by the fact that the metric coefficients are functions of $r$ only, if they are functions of anything, and $r$ is constant along the integral curves in question.

To put it another way, the timelike "grid lines" are drawn very differently in these two charts, even though they both use the same foliation.

That was meant as rhetorical question. The rest of my post basically explained what you have written here.

 

[cianfa72]

Sorry, just to check my understanding: in GP coordinates $(T,r,\theta,\phi)$ the worldlines $r=\text{c}, \theta=\text{c}, \phi=\text{c}$ are not the worldlines of radially infalling observers from infinity whereas in Lemaitre (L) coordinates $(\tau, \rho, \theta, \phi)$ they are.

The timelike KVF of the exterior Shwarzschild spacetime is not orthogonal to the foliation by hypersurfaces orthogonal to the congruence of radial infaller's from infinity. Indeed in GP coordinates the vector field ${\partial} / {\partial_T}$ is not orthogonal to the hypersurfaces $T=\text{c}$ (i.e. the metric in GP coordinates is not diagonal).

So in GP coordinates the path of a radially infalling observer from infinity at fixed $(\theta, \phi)$ has a parametric expression $r(T)$ such that $dr / dT = - \sqrt {\frac {2M} {r}}$ -- i.e. $r(T)$ is the integral of the last expression for the coordinate velocity of a such observer.

 

[PAllen]

Sorry, just to check my understanding: in GP coordinates $(T,r,\theta,\phi)$ the worldlines $r=\text{c}, \theta=\text{c}, \phi=\text{c}$ are not the worldlines of radially infalling observers from infinity whereas in Lemaitre (L) coordinates $(\tau, \rho, \theta, \phi)$ they are.

The timelike KVF of the exterior Shwarzschild spacetime is not orthogonal to the foliation by hypersurfaces orthogonal to the congruence of radial infaller's from infinity. Indeed in GP coordinates the vector field ${\partial} / {\partial_T}$ is not orthogonal to the hypersurfaces $T=\text{c}$ (i.e. the metric in GP coordinates is not diagonal).

So in GP coordinates the path of a radially infalling observer from infinity at fixed $(\theta, \phi)$ has a parametric expression $r(T)$ such that $dr / dT = - \sqrt {\frac {2M} {r}}$ -- i.e. $r(T)$ is the integral of the last expression for the coordinate velocity of a such observer.

That is all correct.

 

[cianfa72]

So the line element for $d\tau = 0$ is the Euclidean line element when written in terms of $r$. That is sufficient to show that those slices are flat.

So, just to write down all the steps, starting from the line element in Lemaitre coordinates $$ds^2=d\tau^2 -\frac{R} {r}d\rho^2 - r^2 (d\theta^2 + sin^2\theta d\phi^2)$$ we get for $d\tau = 0$
$$ds^2=-\frac{R} {r}d\rho^2 - r^2 (d\theta^2 + sin^2\theta d\phi^2)$$
and then using the expression you worked out as $\frac {R} {r} d\rho^2 = dr^2$ we get:
$$ds^2=-dr^2 - r^2 (d\theta^2 + sin^2\theta d\phi^2)$$
That is the Euclidean metric in polar coordinates.