About the temporal coordinate $t_*$ in Eddinton-Finkelstein coordinate

Question

For the Schwarzschild solution $ds^2=-(1-\frac{2GM}{r})dt^2+(1-\frac{2GM}{r})^{-1}dr^2+r^2(d\theta^2+sin^2\theta d\phi^2)$, if we set $dr^2_*=(1-\frac{2GM}{r})^{-2}dr^2$(in other words, $r_*=r+2GMlog{|\frac{r-2GM}{2GM}|}$), then for light rays travelling in the radial direction (which means the solid angle part is 0), we have $\frac{dr_*}{dt}=\pm1$. We see that null, radial geodesics are given by $t\pm r_*=const.$ We introduce a pair of null coordinates $$ v=t+r_*\qquad and \qquad u=t-r_*\quad. \qquad (1) $$ We can see that $$ v=const.\qquad and \qquad u=const.\qquad (2) $$ represent ingoing and outgoing light rays respectively.

Replace $t$ with $t=v-r_*(r)$ in Schwarzschild metric, we have $$ ds^2=-(1-\frac{2GM}{r})dv^2+2dvdr+r^2d\Omega^2_2.\qquad (3) $$ The condition $u=const.$ now can be written as $$ v=2r_*+const.=2r+4GMlog{|\frac{r-2GM}{2GM}|}+const.\qquad (4) $$ Since $r_*$ sits in the range $r_*\in(-\infty,+\infty)$ outside the horizon, and in the range $r_*\in(-\infty,0)$ inside the horizon, we may not use $r_*$ and $t$ to dipict the graph of $v=const.$ and $u=const.$. Instead, we define $$ v=t+r_*=t_*+r \qquad (5) $$ and dipict the graph of $v$ and $u$ by $t_*$ and $r. It looks like this:

where the ingoing null geodesics $v=const.$ are shown in red, outgoing $u=const.$ in blue. We know that the light cone is define by the area bounded by ingoing and outgoing, future-pointing null geodesics, so we can predict that if some particle enters the horizen, it can never come out again since in the light cone inside the horizon, $r$ does not cannot increase.

Above are from Section 6.1.3 in David Tong's note. My question is, why can we directly replace $t$ with $t_*$? I admit that when defining time with $t_*$, we can get some reasonable results very naturally. However, I think that at least we should show that $t_*$ is an time like coordinate first. I used (5) to calculate the derivative $\frac{dt}{dt_*}$, and outside the horizon, the derivative is positive, but in the horizon, $$ \frac{dt}{dt_*}=\frac{-r}{2GM-r},\qquad (6) $$ which is negative. This means that inside the horizon, when we go along the $t_*$ axis in the positive direction, we are in fact going in the negative direction of the $t$. Then we shouldn't get the above conclusions if I am right. The light cones inside the horizon should also point downward.

Also, if we take (5) into (3), we can find that $t_*$ inside the horizon, just like $t$, is still not a timelike coordinate, due to the coefficient $-(1-\frac{2GM}{r})$. Then why can we still consider it as time even if it is not well defined?

I know these two questions are in contradiction with each other a little bit. I appreciate very much if someone can explain this.

Answer 1

The Schwarzschild metric is important because it has certain physical properties: it solves the vacuum GR field equation, it's spherically symmetric, and it's asymptotically Minkowskian. The Eddington-Finkelstein metric also has those properties, and that's enough to make it a good solution of GR. You shouldn't think of it as replacing $t$ with $t_*$; the latter stands perfectly well on its own, and the problems you're having are actually problems with $t$.

You're right that the $\hat{t_*}$ axis is spacelike inside the horizon. You can see that in the diagram in your question: a vector pointing straight upward lies outside the light cone. But $t_*$ is a good time coordinate because any causal (timelike or lightlike) worldline on the manifold can be parametrized by $t_*$. The Schwarzschild $t$ has that property only in the $r>2GM$ region (a worldline inside the horizon can reverse direction in $t$), so $t_*$ is a better time coordinate than $t$ is.

You're getting $dt/dt_*<0$ inside the horizon because the infalling geodesic reverses direction in $t$:

This is actually just a convention. The Schwarzschild coordinates are singular at $r=2GM$ and they have a $t \leftrightarrow -t$ symmetry, so you can independently choose the direction of $t$ inside and outside the horizon, and may as well choose it so that $dt/dt_*>0$ on both sides. But the geodesic is still double-valued in $t$, and it jumps from $t=+\infty$ to $t=-\infty$ at the horizon. $t$ isn't a good time coordinate.

Answer 2

Lagrange629 asked: "Then why can we still consider it as time even if it is not well defined?"

Just like the Schwarzschild-Droste $t$ is pointed in the future direction of local stationary observers with $v=0$ (relative to the central mass or black hole, which in the latter case can only exist outside the horizon) and the Gullstrand-Painlevé $t$ in the future direction of local observers free falling raindrops with $v=-\sqrt{2/r}$, the Eddington-Finkelstein $t$ points in the future direction of local observers decelerating to $v=-2/r$, for which they need proper acceleration aka. force.

In contrast to the force-free Gullstrand-Painlevé raindrop observers or the stationary Schwarzschild-Droste observers aka. Fidos who also need to experience force in order to stay stationary, the required force for the local Finkelstein observers is finite above and below the horizon and only diverges at the singularity, so it is a physical time coordinate just like the other ones, you just have to keep in mind in what state of motion your local clocks and rulers are in the coordinates you are using.