# Painleve chart for FRW spacetime

1. Dec 2, 2011

### Staff: Mentor

In a thread a while back, Mentz114 posted a Painleve chart for FRW spacetime; here's the link to the post:

https://www.physicsforums.com/showpost.php?p=2985307&postcount=60

He posted the metric in Cartesian coordinates, and I've derived a corresponding metric for polar coordinates. (I'm doing this so I can then see what the worldlines of "comoving" observers look like in this chart.) [Edit: originally I thought my answer looked different than what I expected based on Mentz114's post, but I made an error in deriving it; the error is now corrected below.]

Here's what I'm getting: I start with the FRW metric for k = 0 (i.e., flat spatial slices) and with a matter-dominated equation of state, so the scale factor is proportional to $t^{\frac{2}{3}}$. (Mentz114 didn't say so, but it looks to me like that's the equation of state for the metric he wrote down.) I pick units so that the constant of proportionality for the scale factor is 1 (i.e., a(t) = 1 at t = 1), so

$$ds^{2} = - dt^{2} + t^{\frac{4}{3}} \left( dr'^{2} + r'^{2} d\Omega^{2} \right)$$

We want a coordinate transformation that will make the purely spatial part of the metric static (i.e., independent of t). I try this:

$$r' = t^{- \frac{2}{3}} r$$

(leaving all other coordinates the same), which gives

$$dr' = t^{- \frac{2}{3}} dr - \frac{2 r}{3} t^{- \frac{5}{3}} dt$$

Substituting into the metric gives, after some algebra,

$$ds^{2} = - dt^{2} \left( 1 - \frac{4 r^{2}}{9 t^{2}} \right) - \frac{4 r}{3 t} dt dr + dr^{2} + r^{2} d\Omega^{2}$$

If the above is correct, then the worldlines of comoving observers are easy. We haven't changed the t coordinate so for comoving observers we want $ds^{2} = - dt^{2}$. That gives:

$$\frac{dr}{dt} = \frac{2 r}{3t}$$

Does all this look correct?

Last edited: Dec 3, 2011
2. Dec 3, 2011

### Mentz114

It looks OK to me.

From the line element I get the metric as (just checking)

$$\pmatrix{-1+\frac{4\,{r}^{2}}{9\,{t}^{2}} & -\frac{2\,r}{3\,t} & 0 & 0\cr -\frac{2\,r}{3\,t} & 1 & 0 & 0\cr 0 & 0 & {r}^{2} & 0\cr 0 & 0 & 0 & {r}^{2}\,{sin\left( \theta\right) }^{2}}$$

and this gives the right Einstein tensor.

Last edited: Dec 3, 2011
3. Dec 3, 2011

### Staff: Mentor

Thanks! So it seems like the general principle of a Painleve-type chart is that the "Painleve observers" move at velocity $\beta = \frac{dr}{dt}$, and the line element looks like:

$$ds^{2} = - dt^{2} \left( 1 - \beta^{2} \right) - 2 \beta dt dr + dr^{2} + r^{2} d\Omega^{2}$$

Also that the coordinate time in this chart *is* the proper time for Painleve observers, which in the FRW case are the "comoving" observers (I had speculated in the other thread that this was *not* the case, but it appears I was wrong).

4. Dec 3, 2011

### Mentz114

It's ingenious to derive it by requiring the S3 hyperslices. I tried doing it by calculating the acceleration vector for a general 4-velocity and setting it to zero and failed. This was some time ago and I don't remember why.

You're right about the general principle, which is to replace the spatial coord by one corrected with $\beta \tau$ so it becomes comoving and $t=\tau$.

[I corrected the typo in my post , g00 had the wrong sign.]

It is possible to derive the metric by the geodesic route. Requiring that $dt/d\tau=1$ means that our observer has 4-velocity $u^\mu=(1,\beta,0,0)$ (mixing tensor and vector notation). Calculating the acceleration $u_{\mu;\nu} u^\nu$ gives a first-order simultaneous differential equation.
$$\frac{d}{d\,t}\,\beta =-\beta\,\left( \frac{d}{d\,r}\,\beta\right) -\frac{2\,r}{9\,{t}^{2}}$$
The solution is your result for $\beta$. But this much longer than your derivation.

Last edited: Dec 3, 2011
5. Dec 3, 2011

### Staff: Mentor

Do you mean the FRW spacetime with k = 1 (i.e., closed universe, assuming zero cosmological constant)? For that case the "comoving" spatial hyperslices are not flat. Does that spacetime admit any slicing with flat spatial slices?

6. Dec 4, 2011

### Mentz114

I'm sorry my remark "It's ingenious to derive it by requiring the S3 hyperslices" is not relevant and has sowed confusion.

It's interesting that in the comoving coords you derived the 'gravitational' redshift is the familiar $1/\sqrt{1-\beta^2}$.

7. Dec 4, 2011

### Staff: Mentor

Yes, it looks like that is a general feature of Painleve charts since g_tt always assumes the same form in terms of beta. Another way of expressing this would be that "static" observers in this chart (i.e, observers who hold station at a constant r, theta, phi) will have to accelerate to hold station, with an acceleration that increases with r, and the "gravitational redshift" they experience can be thought of as due to their acceleration relative to the observer at r = 0.

Also, the "static" observers will see the "comoving" observers falling past them with a speed beta. Since beta goes to 1 when r = 3/2 t, there is a "horizon" there, and beyond that horizon there are no static observers, i.e., no observers holding station at constant r, theta, phi (because they would have to move faster than light), and no observers beyond the horizon can send signals back to the region "above" it. All very interesting parallels to a black hole spacetime.

8. Dec 4, 2011

### Staff: Mentor

On thinking this over, I realized it's not correct. I should have realized that it can't be, because there are "comoving" observers who emerge from "behind" the horizon! This follows easily from the formulas I derived: the horizon is the line r = 3/2 t, but "comoving" worldlines have the equations

$$r = r_{0} t^{\frac{2}{3}}$$

where r_0 is the radius of that particular worldline at t = 1. (You can see from my derivation that r_0 is also the "r" coordinate of that particular "comoving" observer in the standard FRW chart.) For any value of r_0, there is some value of t at which the corresponding comoving worldline intersects the horizon; before that time t, that comoving observer is behind the horizon, but after it, he has emerged from it.

So the horizon is not a complete causal boundary in the spacetime the way a black hole horizon is.