# Does someone falling into a black hole see the end of the universe?

This question was prompted by Can matter really fall through an event horizon?. Notoriously, if you calculate the Schwarzschild coordinate time for anything, matter or light, to reach the event horizon the result is infinite. This implies that the universe ages by an infinite time before someone falling into the black hole reaches the event horizon, so could that person see the universe age by an infinite time?

To be more precise, suppose the observer starts falling from rest at time $t = 0$ and some initial distance $r > r_s$. If we wait for some time $T$ then shine a light ray at the falling observer. Will the light ray always reach the falling observer before they cross the event horizon? If not, what is the formula for the longest time $T$ that we can wait and still be sure the ray will catch the observer? If $T$ is not bounded it implies that observer could indeed see the end of the universe.

I can think of a qualitative argument for an upper limit on $T$, but I'm not sure how sound my argument is. The proper time for the observer to fall to the event horizon is finite - call this $\tau$. The proper time for the light ray to release the horizon is zero, therefore the light ray will reach the observer before they cross the event horizon only if $T < \tau$. Hence $T$ is bounded and the observer won't see the end of the universe.

I think a more rigorous approach would be to determine the equations of motion (in the Schwarzschild coordinates) for the falling observer and the light ray, and then find the condition for the light to reach the falling observer at some distance $\epsilon$ from the event horizon. Then take the limit as $\epsilon \rightarrow 0$. In principle this seems straightforward, but in practice the algebra rapidly defeated me. Even for a light ray the radial distance:time equation isn't closed form (Wolfram claims it needs the $W$ function) and for the falling observer the calculation is even harder.

I would recommend steering clear of Schwarzschild coordinates for these kind of questions. All the classical (i.e. firewall paradox aside) infinities having to do with the event horizon are due to poor coordinate choices. You want to use a coordinate system that is regular at the horizon, like Kruskal-Szekeres. Indeed, have a look at the Kruskal-Szekeres diagram: (source: Wikipedia)

This is the maximally extended Schwarschild geometry, not a physical black hole forming from stellar collapse, but the differences shouldn't bother us for this question. Region I and III are asymptotically flat regions, II is the interior of the black hole and IV is a white hole. The bold hyperbolae in regions II and IV are the singularities. The diagonals through the origin are the event horizons. The origin (really a 2-sphere with angular coordinates suppressed) is the throat of a non-traversable wormhole joining the separate "universes" I and III. Radial light rays remain 45 degree diagonal lines on the Kruskal-Szekeres diagram. The dashed hyperbolae are lines of constant Schwarzschild $r$ coordinate, and the dashed radial rays are lines of constant $t$. You can see how the event horizon becomes a coordinate singularity where $r$ and $t$ switch roles.

Now if you draw a worldline from region I going into region II it becomes obvious that it crosses the horizon in finite proper time and, more importantly, the past light-cone of the event where it hits the singularity cannot possibly contain the whole spacetime. So the short answer to your question is no, someone falling into a black hole does not see the end of the universe. I don't know the formula you ask for for $T$, but in principle you can read it off from light rays on the diagram and just convert to whatever coordinate/proper time you want to use.

• @JohnRennie I see now that on the previous question you brought up pretty much this argument yourself (and of course I suspected you already heard of it), but seem to find it unsatisfactory somehow. But I must confess that I don't see what is unconvincing about it, any more so than saying that geometry works just fine at the north pole despite some common coordinate systems giving a different impression... – Michael Brown Oct 29 '13 at 14:40
• Ah, yes, I see it now and it's such a clear argument :-) All radial null geodesics run bottom right to top left, and wherever your infalling observer hits the singularity, any light ray hitting the singularity to the upper right of the observer cannot be seen by them. On thing that's missing is how to construct the trajectory of the infalling observer. Are there any simple rules to help me draw the observer's (time like) path on your diagram? – John Rennie Oct 29 '13 at 16:52
• Right, I've rewritten your answer to help me understand it. Could you have a look at my rewrite and criticise as appropriate (obviously your answer is the accepted one! :-). – John Rennie Oct 29 '13 at 18:05
• I have a small but important objection to this argument (cc @John Rennie): If our coordinates inadvertently expanded a point (in 1+1D) to a line, then hitting different parts of the line doesn't mean the trajectories don't converge. Consider maps of Earth that expand the North Pole to a horizontal line - you wouldn't say someone at (90 N, 20 E) is disconnected from someone at (90 N, 130 W), even though the diagram implies it. Similarly, the point where my worldline intersects $u=v$ looks disjoint from where some null ray intersects this line, but you need to prove that they are distinct. – user10851 Oct 29 '13 at 22:50
• @JohnRennie Just happy to help. :) – Michael Brown Oct 30 '13 at 9:13

This is a rewrite of Michael Brown's answer to help me get my thoughts clear, and possibly to help everyone else who's interested to get their thoughts clear too :-) Michael presents a very simple answer to my question based on the geometry of spacetime around the black hole.

The key point is that the usual radius/time Schwarzschild coordinates are unhelpful because they obscure what's going on. To get round this we use a coordinate transformation to draw the spacetime around the black hole using the Kruskal-Szekeres coordinates $u$ and $v$. This what the result looks like: The $u$ coordinate is horizontal and the $v$ coordinate is vertical.

The problem with these coordinates is that they are highly unintuitive. A displacement in $u$ or $v$ doesn't correspond to any simple physical quantity, unlike a displacement in the usual radial coordinate $r$ or time coordinate $t$. Nevertheless the KS coordinates simplify things drastically as follows:

In these coordinates constant $r$ is a hyperbola as shown by the dashed line. The event horizon is the solid 45° line. You can sort of think as $t$ increasing as you move up - it does, though not in a linear way. The singularity is the red hyperbola (this is a spacetime diagram remember, so the singularity is a curve not a point). The region I've labelled $I$ is the exterior of the black hole and the region I've labelled $II$ is the region inside the event horizon. Ignore the region of the diagram to the bottom left as it isn't relevant to my question.

Finally, the key feature that makes it possible to answer my question is that all radial ingoing light rays are straight 45° lines running from bottom right to top left. I've drawn several such light rays as magenta lines.

Now we can answer my question. We start with a rocket hovering at a constant distance away from the black hole, which is represented by the black dashed hyperbola of constant $r$ (as I mentioned above you can sort of think about time increasing as you move up). At time $t_0$ our observer leaves the rocket and starts falling towards the black hole. The blue line shows the trajectory followed by the observer. The observer hits the singularity at the point where the blue and red lines meet.

At time $t_1$ the rocket shines a light ray at the infalling observer. The light ray, travelling at 45°, reaches the observer before they cross the event horizon - so far so good. At time $t_2$ the rocket shines a second light ray at the observer, and this light ray reaches the observer just as they hit the singularity. At time $t_3$ the rocket shines a third light ray into the black hole, but this doesn't reach the observer because the observer has already hit the singularity and no longer exists. That means the observer never sees the light ray released at time $t_3$. The observer sees any light ray released between $t_0$ and $t_2$, but doesn't see any light ray released after $t_2$. So the dashed magenta line marks the boundary between light rays the observer can see and ones they can't.

And there is the answer to my question. The observer does not see the end of the universe because the last light ray they see is the one released at time $t_2$.

This doesn't give me an easy way to calculate the value of $t_2$, because I'd have to derive an expression for the trajectory of the infalling observer (blue line) and that's hard. Nevertheless is shows that $t_2$ is finite so, using the notation in my question, $T$ is bounded.

• As an aside and related to the interpretation of u (spacelike) and v (timelike) coordinates, note that in the exterior region, diagonal lines through the origin are lines of constant Schwarzschild time coordinate t. And, as you point out, the locus of constant Schwarzschild space coordinate r is a hyperbola. Now, with this picture in mind, take a look at Rindler coordinates. – Alfred Centauri Oct 29 '13 at 20:16
• Looks good!.... – Michael Brown Oct 30 '13 at 9:14
• John Rennie, Michael Brown, I salute you both and bow in deep respect! This is delightful and just the kind of diagrammatic clarity of what happens for which I was most hoping, but afraid to ask (the exact numbers are secondary; it's the conceptual framework that's critical). I may have more comments after I fine-tooth this one (it merits close examination!), but mostly I just want to say thanks. I doubt I'm the only one who has been bothered by this one over the years (and there's statistical evidence to support that claim!) Also, John Rennie, thanks for that nice final figure. Very clear! – Terry Bollinger Oct 30 '13 at 16:58
• @TerryBollinger It's a general fact of differential geometry that any two dimensional space is (locally) conformally flat, so if you suppress the angles and just look at the $r-t$ space you can always find a coordinate transformation that makes (radial) null rays 45 degree diagonals, and bring the metric to the form $\omega^2 (u,v)(-dv^2 + du^2) + r^2 (u,v) d\Omega_2^2$ for some suitable functions $\omega(u,v)$ and $r(u,v)$. Radial light rays becomes $v=\pm u+\text{const}$. You can't always do the same trick in $>2$ dimensions - too many degrees of freedom in the metric. – Michael Brown Oct 31 '13 at 11:05
• It's not clear to me that that's the type of path the infalling observer would take. Can you give any intuition from this coordinate system of the scale of time that passes? I think that was the crux of the question to begin with. For example, is it clear that the path doesn't asymptote to 45deg before touching the singularity? – Inverse Sep 7 '14 at 0:32

(Michael Brown's answer is the correct answer and this is merely to amplify via an added diagram.)

Below is figure 31.4 from page 835 of Gravitation (MTW). Both diagrams are of the Schwarzschild geometry. Note that in the Kruskal-Szekeres coordinates, light cones appear as they do in Minkowski spacetime.

As Michael points out, lightlike radial geodesics are 45 degree lines as can be seen by looking at geodesic B.

Clearly, there are lightlike worldlines that cross the horizon after some timelike worldlines so the worldline of an astronaut falling radially towards the hole does not intersect all the lightlike radial worldlines before crossing the horizon.

Also, it is clear that there are lightlike worldlines that end on the singularity after some timelike worldlines.

Thus, the astronaut does not see the infinite future before crossing the horizon or encountering the singularity.

Also, and this is just an interesting side note to consider, the Schwarzschild solution is the spherically symmetric static (well, outside the horizon at least) solution to Einstein's equations. In other words, there is no "end of the universe" in this solution.

• Ah, we wrote elaborations of Michael's answer at the same time :-) Could you have a look at my version and criticise as appropriate. – John Rennie Oct 29 '13 at 18:06
• A couple of us think MTW's time-like geodesics may be sketched incorrectly in K-S coords in this diagram (AA'' and particularly FF''). Would you have an opinion on this? – Rob Jeffries Apr 4 '18 at 18:07

I agree that for a spacetime that is exactly Schwarzschild, the infalling observer does not see the entire history of the universe. However, this turns out not to be the generic case you would expect for an astrophysical black hole, which formed from collapse of some approximately spherical distribution of matter. This topic is actually being actively researched, and there are some very interesting results about what the inside of a black hole actually looks like. See, for example, this recent paper.

The reason that in Schwarzschild the infalling observer does not see the entire history of the universe is that the singularity is spacelike. This means that there is a range of points where the infalling observer can hit the singularity, and each point can only see part of the universe in its causal past.

But people have known of other kinds of black holes for a long time that do not share this behavior. The best know examples are the Reissner-Nordstrom solution for a charged, spherically symmetric black hole, and the Kerr solution for a spinning black hole. These both have timelike singularities, and hence the situation is quite different. Here is a causal diagram of a Reissner-Norstrom black hole: The vertical jagged lines represent the timelike singularities for this black hole. In this case it is possible to avoid the singularity and emerge into a new universe that you could attach to the top of this picture. In this case, when you cross the inner horizon, you should be able to look back and see the entire history or the universe.

This brings up a problematic point however. The observer passes the inner horizon in finite proper time, yet he is able to see all the light that enters the black hole from the entire infinite history of the universe. Since light has energy, you might think that this pile up of radiation from the outside universe should lead to a great deal of curvature, and indeed it does. This is known as a mass inflation instability of the black hole. Kerr black holes share this feature, although the structure of the singularity in that case is more complicated.

So for generic black holes that are not exactly Schwarzschild, a different behavior is expected. The perturbations tend to change the singularity from being spacelike to behaving like a null surface, i.e. following the trajectories of light. A picture from the above paper shows this situation: The outside universe lives in the bottom right triangle of this picture. The lines labeled $\mathcal{CH}^+$ are the null singularities. The paper found that this situation resulted from perturbing the Schwarzschild solution with scalar field matter. In this case if you fell into the black hole from the outside universe, you would run into the null singularities, and assuming you hit the one on the right, you will see everything the entire history of the universe, in the sense that all that you will have access to light that enters the black hole from arbitrarily late times of the universe's history.

• Do you mean that if you're a massless point and you free fall into a charged or rotating black hole, you actually take an infinite amount of time to reach the singularity but since you continue accelerating though space time ever faster, your time slows down enough to only pass a finite amount? – Timothy Jul 22 '16 at 16:20

The currently accepted answer sidesteps the question about calculating what events can actually be seen using Schwarzschild coordinates. It is possible to find an answer to this question using Schwarzschild coordinates, both numerically and analytically. The answer of course is that the past light cone for the limiting case does not encompass the entire universe outside the black hole and that there is a finite time available to signal to a falling object (even in Schwarzschild coordinates), that depends on where the falling observer was released from.

There are two separate problems, each with two separate cases. The first is to work out whether light intercepts a falling observer before they reach the event horizon. However there is then a small additional correction to be made to work out whether a light signal can still intercept a falling observer after they cross the event horizon but before they reach the singularity.

1. Whether light can intercept an object before it reaches the event horizon

(a) Object falling from infinity

I start off with an observer at a radius $r_0$ (all radii are expressed as multiples of the Schwarzschild radius $r_s$). The observer is passed at time $t_0$ (in Schwarzschild coordinates, which is equal to $\tau =0$ according to the observer's own clock), by an object falling radially inward towards the black hole from infinity (where it started at rest). At some time $\Delta t$ later, the observer fires a laser beam radially inwards. The problem is to work out the maximum $\Delta t$ that will intercept the falling object and then convert this to a $\Delta \tau$ in terms of proper time according to the observer. That there must be a maximum $\Delta t$ and $\Delta \tau$ is conceptually easily established by considering (e.g.) Kruskal-Szekeres coordinates.

The null geodesic (in Schwarzschild coordinates) that the inwardly travelling light follows (in $c=1$ units) is: $$t = -r - r_s \ln \left| \frac{r -r_s}{r_0-r_s}\right| + a + \Delta t\, ,\tag{1}$$ where the constant $a = r_0 + t_0$.

The geodesic followed by a body released at rest from infinity is (e.g. see eq. 25.38 in "Gravitation" by Misner, Thorner & Wheeler, 2017, Princeton University press) $$t = r_s \left( -\frac{2}{3}\left(\frac{r}{r_s}\right)^{3/2} - 2\left(\frac{r}{r_s}\right)^{1/2} + \ln \left| \frac{\sqrt{r/r_s} + 1}{\sqrt{r/r_s} -1}\right|\right) + b \tag{2}$$ The constant $b$ can be chosen to ensure that the object passes through the point $(t_0, r_0)$ - thus: $$b = t_0 - r_s\left( -\frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} - 2\left(\frac{r_0}{r_s}\right)^{1/2} + \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right|\right) \tag{3}$$

By plotting these geodesics and using a bisection method to determine when and if they intersect, I was able to determine the maximum $\Delta t$ ($T$ in the OP, although I started my object in freefall from infinity) that still permits the light to intercept the falling object as a function of where that light is emitted from. The result appears stable to reducing the tolerance (I used $10^{-14}r_s$). An example of the limiting case is shown below. The red curve is the light geodesic whilst the blue curve shows the geodesic of an object falling from infinity and passing through (in this case) $5.8r_s$ at $t=0$. Only events below the red curve could be seen by a falling observer. I then "derived" this curve analytically. Rearranging equation (1) we can write $$r - r_s = (r_0-r_s) \exp((a + \Delta t -r)/r_s) \exp(-t/r_s)$$ and if (close to the limit where it is possible for light to intercept the falling object) we let $t$ become large, then $r \rightarrow r_s$ and we can write $$r - r_s \simeq (r_0 - r_s) \exp((a + \Delta t -r_s)/r_s) \exp(-t/r_s) \, , \tag{4}$$ where we exploit that fact that the limit of $r \exp(-r/r_s)$ as $r\rightarrow r_s$ is just $r/e$.

Rearranging equation (2) in a similar way, we get $$\frac{\sqrt{r/r_s} - 1}{\sqrt{r/r_s} +1} = \exp(-t/r_s)\exp\left(-\frac{2}{3}\left(\frac{r}{r_s}\right)^{3/2} -2\left(\frac{r}{r_s}\right)^{1/2} + \frac{b}{r_s} \right)\, .$$ Again, we argue that around the limiting case $r \rightarrow r_s$ and so we can write $$\sqrt{r/r_s} = 1 + 2\exp(b/r_s - 8/3)\exp(-t/r_s)$$ Squaring this and neglecting the $\exp(-2t/r_s)$ term: $$r - r_s \simeq 4r_s \exp(b/r_s - 8/3)\exp(-t/r_s)) \tag{5}$$

Whether there is an interception point or not is determined by whether the ratio of equations (4) and (5) is less than 1 as $t \rightarrow \infty$. $$\lim_{t\rightarrow \infty} \frac{(r_0 - r_s) \exp((a + \Delta t -r_s)/r_s) \exp(-t/r_s)}{r_s( 1 + 4\exp(b/r_s - 8/3)\exp(-t/r_s))} < 1\,$$ which leads to $$\frac{(r_0 - r_s) \exp((a + \Delta t -r_s)/r_s)}{4r_s \exp(b/r_s - 8/3)} < 1$$ $$\exp(\Delta t/r_s) < \frac{4r_s}{r_0 - r_s} \exp(\frac{b - a}{r_s} - \frac{5}{3})$$ $$\Delta t < \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left(\frac{b - a}{r_s} - \frac{5}{3}\right)r_s$$ Reinserting the expressions for $a$ and $b$ $$\Delta t < \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0$$ This matches what is plotted above.

To turn this into a maximum proper time interval $\Delta \tau$ from the point of view of the observer, the result would be multiplied by $(1 - r_s/r_0)^{1/2}$.

(b) Object falling from rest at $t_0, r_0$

Now the setup is that the observer releases the object from $t_0, r_0$, then waits a (coordinate) time interval $\Delta t$ before signalling.

Equation (1) is still valid in this scenario, however equation (2) needs to be replaced by the following geodesic for an object freely falling from rest at $t_0, r_0$. $$\frac{t-t_0}{r_s} = \ln \left| \frac{ (r_0/r_s -1)^{1/2} + \tan (\eta/2)}{(r_0/r_s -1)^{1/2} -\tan(\eta/2)}\right| + \left(\frac{r_0}{r_s}-1\right)^{1/2} \left( \eta + \frac{r_0}{2r_s}(\eta + \sin \eta)\right). \tag{6}$$ Here the "cycloid parameter" $\eta(r)$ is defined by $$r = \frac{r_0}{2}(1 + \cos \eta)$$

As $r \rightarrow r_s$, the first term in equation (6) grows exponentially whilst the second term, which I will define as $b(r)/r_s$, tends to a constant: $$\lim_{r \rightarrow r_s} b(r) = b_{\rm rs} = r_s\left(\frac{r_0}{r_s}-1\right)^{1/2} \left( \eta_{\rm rs} + \frac{r_0}{2r_s}(\eta_{\rm rs} + \sin \eta_{\rm rs})\right),$$ where $$\cos \eta_{\rm rs} = \left(\frac{2r_s}{r_0} -1 \right).$$

Using the identity that $\tan \eta/2 = \sin \eta/(1 + \cos \eta)$, then $$\tan (\eta/2) = \left( \frac{r_0}{r} - 1 \right)^{1/2}.$$ Substituting this into equation (6) we can set $t_0=0$, exponentiate and find $$\left(\frac{r_0}{r_s}-1\right)^{1/2}\left(1 - \exp\left[\frac{b-t}{r_s}\right]\right) = \left(\frac{r_0}{r}-1\right)^{1/2} \left( 1 + \exp\left[\frac{b-t}{r_s}\right]\right)$$ Squaring this and neglecting terms containing $\exp(-2t/r_s)$ as $t$ becomes large, this can be rearranged to give $$r = r_s\frac{\left(1 + 2\exp[(b-t)/r_s]\right)}{1 - 2\exp[(b-t)/r_s] + (4r_s/r_0)\exp[(b-t)/r_s]}.$$ Again, as we are looking for a limiting behaviour at large $t$, then the denominator can be expanded as a binomial, retaining only the first two terms. Multiplication with the numerator then yields: $$r -r_s \simeq 4r_s \left(1 - \frac{r_s}{r_0}\right) \exp\left[\frac{b-t}{r_s}\right]. \tag{7}$$

To find the limiting $\Delta t$ for which a light beam from the observer will "catch" the falling object, we take the ratio of equations 4 and 7, set $b=b_{\rm rs}$ and demand that this is less than 1. This yields $$\exp\left[\frac{\Delta t}{r_s}\right] < 4 \left(\frac{r_s}{r_0}\right) \exp\left[\frac{b_{\rm rs}}{r_s}\right] \exp\left[\frac{r_s-r_0}{r_s}\right]$$ and hence $$\Delta t < r_s \ln \left(\frac{4r_s}{r_0}\right) + b_{\rm rs} + r_s - r_0$$

The result is plotted below as the red curve (and I have confirmed that it is correct using a numerical bisection method) and compared to case 1 with the free falling object from infinity (blue curve, as in the first picture). As expected the allowed $\Delta t$ is larger when the object is released from rest.

As before, this result is the maximum Schwarzschild coordinate time interval. It must be reduced by the appropriate time-dilation factor $(1-r_s/r_0)^{1/2}$ to yield the maximum proper time interval. An example of the limiting case is shown below. The red curve is the geodesic of light, the blue curve is the geodesic of the falling object. Only events below the red curve (which asymptotes to a gradient of -1) can be "seen" by an object falling into a black hole from rest, from (in this case) about $5.8r_s$. 2. Whether light can intercept an object before it reaches the singularity

The answer above give the maximum (coordinate) time-delay for a signal from a stationary observer to reach a falling object before it reaches the event horizon, $(\Delta t)_{\rm EH}$. But that does not completely answer the (headline) question, because the object can still receive light during the time it takes to reach the singularity after crossing the event horizon. This is most clearly seen in Kruskal-Szekeres coordinates, but again it is possible to solve this (rather easily) in Schwarzschild coordinates.

The condition here is that the coordinate time of the delayed light geodesic must be less or equal to the coordinate time of the falling object geodesic at $r=0$.

This condition is actually rather easy to find. For the case of the object free-falling from infinity, equations (1-3) show that the original $\Delta t$ that I derived should be increased as $$(\Delta t)_{\rm singularity} = r_s \ln \left( \frac{r_s}{r_0-r_s}\right) - r_s\left( -\frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} - 2\left(\frac{r_0}{r_s}\right)^{1/2} + \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right|\right) - r_0$$ Or in terms of the previous result. $$(\Delta t)_{\rm singularity} = (\Delta t)_{\rm EH} +r_s\left(\frac{5}{3} - 2\ln 2\right) =(\Delta t)_{\rm EH} + 0.280r_s$$

For the case of an object falling from rest, we see that $\eta = \pi$ at $r=0$, so that if the coordinate time to be less or equal to the coordinate time of the object at $r=0$ is obtained from equations (1) and (6) as $$(\Delta t)_{\rm singularity} = r_s \ln \left(\frac{r_s}{r_0-r_s}\right) + \pi r_s\left(\frac{r_0}{r_s} -1\right)^{1/2}\left(1 + \frac{r_0}{2r_s}\right) -r_0,$$ which is larger than $(\Delta t)_{\rm EH}$ by an amount that depends on $r_0$, but is asymptotic to the falling from infinity results as $r_0$ becomes large. This new relationship is plotted below - the higher red curve is the maximum (coordinate time) delay that can be tolerated and still send a signal that reaches the falling object before the singularity. The lower plot shows the difference between this and the previous result for the delay to still reach the object before the event horizon. The plot below should make things clearer. It shows the geodesics either side or $r_s$ in the case of an object falling from $r=2r_s$ at $t=0$. The light geodesic in red is the one calculated so that it just intercepts the object as $r \rightarrow r_{s}$ and has $(\Delta t)_{\rm EH} = 3.834 r_s/c$. But we see that this geodesic "overtakes" the falling object before it reaches the singularity at $r=0$. However, the green light geodesic, with $(\Delta t)_{\rm singularity} = 4.283 r_s/c$ intercepts the object geodesic at exactly $r=0$. • I had been working on this as well, but you beat me to it. I guess you're the photon and I'm the slow-moving object :) I just posted my own version. – Pulsar Mar 31 '18 at 13:22
• I think you missed an extra $-r_0$ in your last equation. Curiously, it seems that fig 31.4 in MTW contains a mistake: the $u$-coordinate of the path $FF''$ should be increasing rather than decreasing. At the singularity, $du/dv = u/v$. – Pulsar Apr 4 '18 at 4:20
• @Pulsar It is partly this uncertainty that stopped me trying to check the (added) results above in K-S coords. Are you sure? Doesn't the trajectory need to have decreasing $r$ and decreasing $t$ below the horizon, with a timelike geodesic cutting the lines of constant $t$ at right angles as it reaches the singularity? – Rob Jeffries Apr 4 '18 at 9:22
• I'm pretty sure. I found this article, which is consistent with my calculations. Look at fig 2, and eq 23. – Pulsar Apr 4 '18 at 10:00
• @Pulsar I agree (and got confused) a timelike geodesic should end up asymptotic to a line of constant $t$? Both AA'' and FF'' look wrong in that MTW diagram. – Rob Jeffries Apr 4 '18 at 10:21

Inspired by a similar question, I've been working on this topic at the same time as Rob Jeffries. Irritatingly, he beat me to it; but since I use a slightly different approach and since I don't want my efforts to be in vain, I'll post my own derivation. If nothing else, it serves as a confirmation of his fantastic answer :)

Let us start by stating the (region I) Kruskal–Szekeres coordinates

$$u = f(r)\cosh\left(\frac{ct}{2r_\text{s}}\right), \qquad v = f(r)\sinh\left(\frac{ct}{2r_\text{s}}\right),\\ f(r) = \left(\frac{r}{r_\text{s}}-1\right)^{\!1/2}\,\text{e}^{r/2r_\text{s}}.$$

As is well known, in these coordinates the geodesics of radially infalling light rays are straight lines at $-45^\circ$ angles. Indeed, if we plug $u + v= k$ into the equations, with $k$ a constant, then from $u^2 - v^2 = f(r)^2$ we find $k(u -v) = f(r)^2$, so that $$k\exp\left(-\frac{ct}{2r_\text{s}}\right) = f(r) = \left(\frac{r}{r_\text{s}}-1\right)^{\!1/2}\,\exp\left(\frac{r}{2r_\text{s}}\right),$$ or $$\frac{ct}{r_\text{s}} = \ln(k^2)-\frac{r}{r_\text{s}} - \ln\left(\frac{r}{r_\text{s}}-1\right),$$ which are indeed the geodesics of a radially infalling photon, with $k = f(r_{0,\gamma})$ and $r_{0,\gamma}$ the initial position of the photon at $t=0$.

Now, let us assume that we have a radially infalling object, which starts at rest at a position $r_0$ at $t=0$. Which radially infalling photons will reach the object before it crosses the event horizon? To answer this, we will try to derive the geodesic of a radially infalling photon such that it catches up with the object right at the event horizon.

The geodesic of a radially infalling object can be written in the form (Misner, Thorne & Wheeler Eq. (31.10), Pag. 824) \begin{align} r &= \frac{r_0}{2}(1+\cos\eta) = r_0\cos^2\eta/2,\\ \frac{c\tau}{r_\text{s}} &= \frac{1}{2}\left(\frac{r_0}{r_\text{s}}\right)^{\!3/2}(\eta + \sin\eta),\\ \frac{ct}{r_\text{s}} &= \ln\left(\frac{\sqrt{r_0/r_\text{s} -1} + \tan\eta/2}{\sqrt{r_0/r_\text{s} -1} - \tan\eta/2}\right) + \left(\frac{r_0}{r_\text{s}}-1\right)^{\!1/2}\left(\eta + \frac{r_0}{2r_\text{s}}(\eta + \sin\eta)\right). \end{align} It's also instructive to introduce the (dimensionless) total energy of the object $$E = \frac{\mathcal{E}}{mc^2} = \left(1- \frac{r_\text{s}}{r}\right)\frac{\text{d}t}{\text{d}\tau}.$$ Radial orbits satisfy the equation $$\left(\frac{\text{d}r}{c\,\text{d}\tau}\right)^2 = E^2 - \left(1- \frac{r_\text{s}}{r}\right),$$ so if the object is at rest at position $r_0$ at $t = \tau = 0$, then $$E = \sqrt{1- \frac{r_\text{s}}{r_0}}.$$ The equation for $t(\eta)$ can thus be rewritten as $$\frac{ct}{r_\text{s}} = \ln\left(\frac{E + \sqrt{1-E^2}\tan\eta/2}{E - \sqrt{1-E^2}\tan\eta/2}\right) + \frac{E}{\sqrt{1-E^2}}\left(\eta + \frac{\eta + \sin\eta}{2(1-E^2)}\right).$$ Next, I will follow this article (which contains a few errors) to derive how this equation behaves as $r$ approaches the event horizon. We write $$r = r_\text{s}(1+\varepsilon),\qquad \varepsilon\rightarrow 0.$$ Near the event horizon we can ignore higher-order terms in $\varepsilon$, so that \begin{align} \cos^2\eta/2 &= (1+\varepsilon)\frac{r_\text{s}}{r_0} = (1+\varepsilon)(1-E^2) = 1 - E^2 + \varepsilon(1-E^2),\\ \sin^2\eta/2 &= E^2 - \varepsilon(1-E^2), \end{align} and \begin{align} (1-E^2)\tan^2\eta/2 &= \frac{E^2 - \varepsilon(1-E^2)}{(1+\varepsilon)} \\ &\approx \left[E^2 - \varepsilon(1-E^2)\right](1-\varepsilon)\\ &\approx E^2 - \varepsilon(1-E^2) - \varepsilon)E^2 = E^2 - \varepsilon. \end{align} Therefore, \begin{align} E + \sqrt{1-E^2}\tan\eta/2&\approx E\left(1 + \sqrt{1 - \varepsilon/E^2}\right)\\ &\approx 2E - \frac{\varepsilon}{2E}\approx 2E, \end{align} and \begin{align} E - \sqrt{1-E^2}\tan\eta/2&\approx E\left(1 - \sqrt{1 - \varepsilon/E^2}\right)\\ &\approx \frac{\varepsilon}{2E}, \end{align} So that finally, as $r\rightarrow r_\text{s}$, $$\frac{ct_\text{s}}{r_\text{s}} \approx \ln\left(\frac{4E^2}{\varepsilon}\right) + \frac{E}{\sqrt{1-E^2}}\left(\eta_\text{s} + \frac{\eta_\text{s} + \sin\eta_\text{s}}{2(1-E^2)}\right),$$ with $\eta_\text{s}$ the value of $\eta$ at the event horizon. Since $\cosh(x) = \sinh(x) \rightarrow \text{e}^x/2$ as $x\rightarrow\infty$, the Kruskal–Szekeres coordinates of the object at the event horizon become (since $t\rightarrow\infty$) \begin{align} u_\text{s}^2 = v_\text{s}^2 &= \frac{1}{4}f(r_\text{s})^2\,\exp\left(\frac{ct_\text{s}}{r_\text{s}}\right) = \frac{\varepsilon\text{e}}{4}\exp\left(\frac{ct_\text{s}}{r_\text{s}}\right)\\ &= \text{e}E^2 \exp\left[\frac{E}{\sqrt{1-E^2}}\left(\eta_\text{s} + \frac{\eta_\text{s} + \sin\eta_\text{s}}{2(1-E^2)}\right)\right], \end{align} or \begin{align} u_\text{s} = v_\text{s} &= \sqrt{\text{e}}E \exp\left[\frac{E}{2\sqrt{1-E^2}}\left(\eta_\text{s} + \frac{\eta_\text{s} + \sin\eta_\text{s}}{2(1-E^2)}\right)\right]\\ &= \sqrt{\text{e}}\sqrt{\frac{r_\text{s}}{r_0}}\left(\frac{r_0}{r_\text{s}}- 1\right)^{\!1/2} \exp\left[\frac{1}{2}\left(\frac{r_0}{r_\text{s}}-1\right)^{\!1/2}\left(\eta_\text{s} + \frac{r_0}{2r_\text{s}}(\eta_\text{s} + \sin\eta_\text{s})\right)\right]. \end{align} The corresponding coordinates for a radially infalling photon satisfy $u_\text{s} + v_\text{s} = k_\text{b}$, for some boundary value $k_\text{b}$. Therefore $k_\text{b} = 2u_\text{s}$ and we find the corresponding null geodesic $$\frac{ct}{r_\text{s}} = \ln(k_\text{b}^2)-\frac{r}{r_\text{s}} - \ln\left(\frac{r}{r_\text{s}}-1\right).$$ We could solve this for $r$ at $t=0$, which yields a boundary radius $r_\text{b}$ beyond which radially infalling photons cannot catch up with the object before it crosses the event horizon. Alternatively, we could plug in $r=r_0$ and ask what the maximum time $\Delta t$ is such that photons sent out at $r_0$ before $t=\Delta t$ can still catch up with the object. We find \begin{align} \frac{c\Delta t}{r_\text{s}} &= \ln(k_\text{b}^2)-\frac{r_0}{r_\text{s}} - \ln\left(\frac{r_0}{r_\text{s}}-1\right)\\ &= 1 + \ln\left(\frac{4r_\text{s}}{r_0}\right) + \left[\left(\frac{r_0}{r_\text{s}}-1\right)^{\!1/2}\left(\eta_\text{s} + \frac{r_0}{2r_\text{s}}(\eta_\text{s} + \sin\eta_\text{s})\right)\right]-\frac{r_0}{r_\text{s}}, \end{align} which is exactly the same result as given by Rob Jeffries.

I made a plot to visualize the results, in Schwarzschild and Kruskal–Szekeres coordinates: The blue curve is the geodesic of an object, at rest at $t=0$ (here, $r_0 = 2r_\text{s}$). The orange curve is the geodesic of a photon which is at position $r_0$ at $t=0$. The red curve is the geodesic that I derived in this post. It starts at position $r_\text{b}$ at $t=0$ and catches up with the object right at the event horizon. Geodesics of photons that lie between the orange and red curves (I've plotted two, the dashed-lined curves) will catch up with the object, geodesics beyond the red curve will not.

• Puzzled by the right hand diagram. Is that the path of a freefalling object in K-S coordinates? – Rob Jeffries Mar 31 '18 at 13:36
• @RobJeffries Yes, it's the same object as the left hand diagram. It does cross the EH (at $u_s = 9.25$) albeit at a very shallow angle. I'm not sure why that is. Everything else checks out though. – Pulsar Mar 31 '18 at 14:00
• See my important edit. I have now accounted for (I think) the additional time available whilst the object falls from $r_s$ to the singularity. This should be included and is actually what is important to answer the question. It is much easier to calculate! – Rob Jeffries Mar 31 '18 at 17:06

To add to the excellent answers above, here is a spacetime diagram in Gullstrand-Painleve or "rain" coordinates. This is from the superb and accessible book Exploring Black Holes (2000) by Taylor & Wheeler, $\S B.6$. Their metaphor "rain" means a test particle with mass, which initially fell from rest far away from the black hole. Think of them as astronauts / observers, for this problem.

$t_\textrm{rain}$ is the proper time of a raindrop, which is used as a coordinate. $r$ is the usual curvature coordinate as in Schwarzschild[-Droste] coordinates, and $M$ is the mass of the black hole. The diagram shows that most "light pulses" never catch up to a given "rain plunger"; in particular they will not see the end of the universe. No. The black hole will completely evaporate in finite time, so by the end of the universe it will no longer exist.

• I wasn't the downvoter, but this isn't really an answer. The question was posed in the context of classical gravity. – Ben Crowell Nov 11 '14 at 3:17
• @Ben Crowell then this theory is not applicable to such range of time. – Anixx Nov 11 '14 at 3:51

Your question is due to some confusion with the spacetime concept of a black hole. You must distinguish between your coordinate system and what you see. Both are different concepts: One simple example is a Minkowski space: If a Minkowski diagram represents your coordinates, you get a fourdimensional view of the whole spacetime. In contrast, what you s e e are elements which are located on your lightcone which is showing into the past.

Near a black hole we must apply the same distinction of this twofold concept which can be shown in the following Kruskal diagram, with one infalling particle A and one particle remaining outside B: The time coordinates of a far-away observer are indicated by the lines going through the center: t = 0, t = 1, t = 2, limited by the event horizon where t = $\infty$. According to these time coordinates, the infalling particle will never reach the horizon. And conversely, when A is approaching the horizon, the clock of an outside observer will approach the end of time.

Perhaps this is the reason why you asked your question. But your question is not asking: what is the position with respect to the coordinates of an outside observer, but: what does the infalling particle see, and for this question you must refer (as shown in other answers) to the small 45°-arrows between the communicating particles A and B. The 3 diagonal arrows from the bottom to the left are showing that B is at a certain point when A touches the event horizon.

## protected by Qmechanic♦Oct 29 '13 at 22:26

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