# Painleve-Gullstrand coordinates

1. Jul 16, 2006

### pervect

Staff Emeritus
This came up in another thread. I thought I'd make a few notes about them.

It should be spelled Painleve, I can't edit the typo in the title :-(

Google finds http://www.physics.umd.edu/grt/taj/776b/hw1soln.pdf, which, being a homework solution, will probably disappear soon.

The line element is:

$$-{d{{T}}}^{2}+ \left( d{{r}}+\sqrt {{\frac {2M}{r}}}d{{T}} \right) ^{2}+{r}^{2} \left( {d{{\theta}}}^{2}+ \sin ^2 \theta d {{\phi}}}^{2} \right)$$

The following coordinate transformation will map PG coordinates into Schwarzschild coordinates:

$$t = T -2\,\sqrt {2\,M\,r}+4\,M\,\mathrm{arctanh} \left( \sqrt {{\frac {r}{2 \, M}}} \right)$$

arctanh(x) is defined only for x<1, while arctanh(x) = 1/2 (ln(1+x)-ln(1-x)), more work needs to be done to deal with the sign issues that arise when making x > 1.

The above
The metric is not a function of T, therfore $u_0 = g_{0i} u^i =$ = (-1+2M/r) $dT/d\tau$ + sqrt(2M/r) $dr/d\tau$ = constant, a conserved energy-like quantity of the orbit.

Here $u^i$ is the 4-velocity $(dT/d\tau, dr/d\tau, d\theta/d\tau, d\phi / d\tau)$

Similarly, since the metric is not a function of $\phi$

$u_3 = g_{3i} u^i$ = r^2 sin^2 $\theta$ $d\phi / d\tau$= constant

representing a conserved angular momentum-like quantity of the orbit.

Generally, the orbit will be taken to be in the equatorial plane, $\theta = \pi/2$ and the above two conserved quantites plus the metric equation will be sufficient to calculate orbital motion.

Last edited: Jul 17, 2006
2. Jul 17, 2006

### pervect

Staff Emeritus
OK, it looks like

$$t = T -2\,M\sqrt {2}\sqrt {{\frac {r}{M}}}-2\,M\ln \left( \left| \sqrt {{\frac {r}{2\, M}}}-1 \right| \right) +2\,M\ln \left( \sqrt {{\frac {r}{2\,M}}}+1 \right)$$

works for the transformation between Schwarzschild and Painleve-Gullstrand inside & outside the horizon, t being Schwarzschld and T being Painleve.

Another source for the above info, including the conversion equation between Schw and PG:

http://arxiv.org/abs/gr-qc/0509030

less likely to go away and more complete than the homework solution (though it's only periphrial to the paper).

Last edited: Aug 15, 2007
3. Jul 17, 2006

### pervect

Staff Emeritus
Another worthwhile note: the solution for r(tau), T(tau) from a free-fall at infinty:

let Tdot = $dt/d\tau$, rdot = $dr/d\tau$

The energy equation says

(-1 + 2M/r) Tdot + sqrt(2M/r) rdot = constant

If Tdot = 1 (zero velocity) at r=infinty, this constant must be -1

The metric equation says

$$\left( \sqrt{\frac{2M}{r}} Tdot + rdot \right)^2 - Tdot^2=-1$$

(just divide both sides of the Lorentz interval by ds^2, and note that $d\tau^2 = -ds^2$ to get the above eq).

Solving, the energy equation and the metric equation gives us the formula for rdot and Tdot. Taking the minus sign for rdot (for radially infalling particles), we get:

rdot = -sqrt(2M/r)
Tdot = 1

This can be integrated to find
$$r = \left( \frac{9M}{2} \tau^2 \right) ^ \frac{1}{3}$$

and T = $\tau$

We can further note that $d^2 r / d\tau^2 = -M/r^2$

Because T=$\tau$, one can re-write r(tau) as R(T) in the above.

4. Jul 24, 2006

### pervect

Staff Emeritus
Solving the previous two equations, we have a quadratic in tdot = dt/dtau
$$\frac{r-2 m}{2 m} tdot^2 - \frac{E r}{m} tdot + \frac{E^2 r}{2 m} + 1 = 0$$

This has two solutions. Any horizon crossing solution must satisfy

$$\frac{dt}{d\tau} = {\frac {2\,m+{E}^{2}r}{Er+\sqrt {2}\sqrt {m}\sqrt {-r+2\,m+{E}^{2}r}}}$$

The form of the required solution is dicated by the fact that the coefficient of r^2 vanishes at r=2m, therefore we must use the 2c/..... form of the solution of the quadratic, the ...... / 2a solution will have a zero denominator at r=2m

For non-horizon crossing solutions

$$\frac{dt}{dtau} = {\frac {Er+\sqrt {-2\,rm+4\,{m}^{2}+2\,m{E}^{2}r}}{r-2\,m}}$$

is a possible solution as well.

Sticking with horizon crossing solutions for the moment, we have

$$\frac{dt}{dtau} = {\frac {2\,m+{E}^{2}r}{Er+\sqrt {2}\sqrt {m}\sqrt {-r+2\,m+{E}^{2}r}}}$$

and

$$\frac{dr}{d tau} = -{\frac { \left( -2\,rm+4\,{m}^{2}+2\,m{E}^{2}r+Er\sqrt {2}\sqrt {m}\sqrt {-r+2\,m+{E}^{2}r} \right) }{\sqrt {2 r m} \left( Er+\sqrt {2}\sqrt {m}\sqrt {-r+2\,m+{E}^{2}r} \right) }}$$

When dr/dtau = 0, one might need to know d^2 r / dtau^2. Solving the geodesic eq directly when dr/dtau = 0 gives

$$\frac{dr^2}{d tau^2} = - \left( 1-\frac {2m}{r} \right) \left( \frac{m}{r^2} \right) \left( \frac{dt}{d tau} \right)^2+{ \left( r-2\,m \right) \left( \frac{d phi}{d tau} \right)^2$$

note that the above DOES assume that dr/dtau = 0.

Last edited: Jul 24, 2006