Let the Schwarzschild spacetime be given with coordinates $(t,r,\theta,\phi)$. Change coordinates to Kruskal-Szekeres $(T,X,\theta,\phi)$ so that the line element becomes

$$ds^2 = \dfrac{32 M^3}{r}e^{-r/2M}(-dT^2+dX^2)+r^2d\Omega^2.$$

Now, $T$ is a timelike coordinate, its coordinate lines are hence timelike and qualify as observers.

Each such coordinate line is specified by values $X_0,\theta_0,\phi_0$ of the other coordinates. In particular it is a radial worldline with respect to the observer at infinity.

Now let $\gamma(T)$ be one such worldline, corresponding to $X_0,\theta_0,\phi_0$. I want to understand physically the motion of this observer.

We know that


Hence, the horizon is crossed when $W((X^2-T^2)/e)=0$ which means when $T = \pm X$.

The singularity on the other hand is located where $W((X^2-T^2)/e)=-1$ and this means that $(X^2-T^2)/e = -1/e$. This in turn means that $T^2-X^2 = 1$ characterizes the singularity. For fixed $X$ we would have $T = \pm \sqrt{1 + X^2}$.

So for $\gamma$ suppose first $T$ starts at $-\infty$. This means that $r$ starts at $\infty$ and thus the observer comes from infinity.

Suppose $X_0$ is positive. Then, as $T$ increases, it will cross the horizon when $T = -X_0$ at finite proper time.

Next, $T$ will increase until reaching $-\sqrt{1+X_0^2}$ which is the singularity and this would be the endpoint of the observer's worldline.

So it seems: $\gamma$ is one observer starting at infinity falling radially towards the singularity, crossing the horizon at finite proper time and then reaching the singularity further.

The observer seems accelerated (i.e., not freely falling), since

$$\nabla_{\frac{\partial}{\partial T}}\frac{\partial}{\partial T}=\frac{2GM}{r^2}e^{-r/2GM} (r+2GM)[T\partial_T - X\partial _X].$$

What I want to know is: is my analysis correct? Is this the motion of this observer? If not, how do I correctly understand the motion of said observer?

One odd thing is: when $T\to -\infty$, $r\to \infty$ so the observer comes from infinity, and it is in the exterior region. Now, on the exterior region

$$T = X\tanh(t/4GM)\Longrightarrow t = 4GM \tanh^{-1}(T/X).$$

As $T\to -\infty$ we then have $T/X\to \pm \infty$ depending on the sign of $X$ and thus $t \to 2GM\pi i$ which is not real! So this seems extremely odd and makes me question the whole analsysis.

Anyway, what physically is the motion of this observer in a Schwarzschild spacetime?


2 Answers 2


First of all,

So for $\gamma$ suppose first $T$ starts at $-\infty$. This means that $r$ starts at $\infty$ and thus the observer comes from infinity.

This is wrong. The worldline $\gamma$ exists only for $-\sqrt{1+X_0^2}<T<\sqrt{1+X_0^2}$. At the endpoints of $\gamma$ radial parameter $r$ is zero.

The interpretation of the worldline is straightforward. The observer corresponding to $\gamma$ is emitted by the 'white hole' singularity, crosses the white hole horizon (past horizon) into the exterior region achieves its maximum radial distance $r_\text{max}\ge 2GM$ at $T=0$ and then starts falling back (with decreasing $r$), crosses the black hole event horizon and ends his life in the black hole singularity.

There is one case where $\gamma$ is geodesic: when $X_0=0$. This corresponds to an observer just touching the horizon, without emerging into the exterior region.


This may not be a totally detailed and authoritative answer, but here are a few ideas that may be helpful: --

You can't take a limit as $T\rightarrow-\infty$, for fixed $X$, because the Kruskal-Szekeres coordinate chart is limited to $T^2<1+X^2$. This is presumably the reason why you're getting an imaginary $t$ in this limit.

It's certainly true that a constant $X$ world-line is not geodesic. I don't see any reason to refer to this world-line as an observer. In general, coordinate charts don't have anything to do with observers, although it's true that any timelike world-line can be the world-line of an observer if you want to say it is.

Your constant-$X$ world-line will start in region IV, pass into I or III, and then enter region II. Keep in mind that this is not very realistic. Most observers never fall into a black hole -- they probably end up in the elephants' graveyard at future timelike infinity. And of course astrophysical black holes don't have a region IV.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.