Measuring Light Reflection in a Black Hole

[PeterDonis]

This should depend on the simultaneity convention used.

In this case there is a foliation that is picked out by an invariant symmetry of the spacetime: the family of spacelike hypersurfaces that is everywhere orthogonal to its timelike Killing vector field. The distance I referred to is distance in one of these hypersurfaces (it doesn't matter which one since the hovering observer's worldline is an integral curve of the timelike KVF).

 

[PeterDonis]

The coordinate time itself is not equal to the proper time of the statite

Yes, agreed. But since in the particular case under consideration there is a fixed proportionality between coordinate time and the statite's proper time, and since the events at each endpoint of the relevant interval are on the statite's worldline, calculating in terms of coordinate time can be used to obtain a physically relevant answer in terms of the statite's proper time.

 

[Orodruin]

calculating in terms of coordinate time can be used to obtain a physically relevant answer in terms of the statite's proper time.

That is true for any coordinate system. It is just that the Schwarzschild coordinates give a particularly simple relation because of using the hypersurfaces orthogonal to the timelike KVF as coordinate surfaces.

In this case there is a foliation that is picked out by an invariant symmetry of the spacetime: the family of spacelike hypersurfaces that is everywhere orthogonal to its timelike Killing vector field. The distance I referred to is distance in one of these hypersurfaces (it doesn't matter which one since the hovering observer's worldline is an integral curve of the timelike KVF).

True, but it could be discussed how meaningful this is. For example, the event that you are computing the distance to is the same for all of those hypersurfaces (the origin in a Kruskal diagram, which is the r = R event in of all of them) similar to the distance to the origin for a Rindler observer in SR.

 

[cianfa72]

In this case there is a foliation that is picked out by an invariant symmetry of the spacetime: the family of spacelike hypersurfaces that is everywhere orthogonal to its timelike Killing vector field. The distance I referred to is distance in one of these hypersurfaces (it doesn't matter which one since the hovering observer's worldline is an integral curve of the timelike KVF).

Just to check my understanding: consider two hovering observers. One is the given observer hovering at fixed $(R_1, \theta, \phi)$ Schwarzschild coordinates and the other is hovering at the event horizon with the same angular coordinates $(R_0, \theta, \phi)$.

Now since the worldlines of both hovering observers are integral curves of the timelike KVF if we pick the foliation above (namely the family of spacelike hypersurfaces that is everywhere orthogonal to the timelike KVF) the distance between them in any of those spacelike hypersurfaces is always the same.

 

[Orodruin]

the other is hovering at the event horizon with the same angular coordinates (R0,θ,ϕ).

It is impossible to hover at the event horizon. It is a lightlike surface.

The statement is however true for any stationary observer above the event horizon since the KVF by definition is a symmetry of the spacetime.

 

[cianfa72]

It is impossible to hover at the event horizon. It is a lightlike surface.

Ok, the point is that along the hovering observer worldline the spacelike hypersurfaces of constant Schwarzschild coordinate time $t$ have always the same 3D geometry. So on each of these spacelike hypersurfaces the distance between points at given coordinate $(R_1, \theta, \phi)$ and $(R_0, \theta, \phi)$ is always the same (the event horizon radial coordiante is always $R_0$).

 

[Orodruin]

Ok, the point is that along the hovering observer worldline the spacelike hypersurfaces of constant Schwarzschild coordinate time $t$ have always the same 3D geometry. So on each of these spacelike hypersurfaces the distance between points at given coordinate $(R_1, \theta, \phi)$ and $(R_0, \theta, \phi)$ is always the same (the event horizon radial coordiante is always $R_0$).

You have to be much more careful here. The point on the horizon that you are referring to is the same event for all of those hypersurfaces. This means that it is also not in any way or form a distance to ”where the dropped object crosses the horizon”. That event is in the causal future (actually on the light cone) of the event on the hypersurfaces.

 

[cianfa72]

You have to be much more careful here. The point on the horizon that you are referring to is the same event for all of those hypersurfaces.

Sorry I've some difficulty to grasp it. The exterior of BH is foliated by spacelike hypersurfaces of constant Schwarzschild coordinate time $t$.

Are you saying that all the spacelike hypersurfaces in the above family basically "share" a same event (i.e. they map a same event with different coordinate values) ?

 

[Orodruin]

Sorry I've some difficulty to grasp it. The exterior of BH is foliated by spacelike hypersurfaces of constant Schwarzschild coordinate time $t$.

Are you saying that all the spacelike hypersurfaces in the above family basically "share" a same event (i.e. they map a same event with different coordinate values) ?

The exterior does not include the event horizon. All of those hypersurfaces, when extended to the event horizon, include the same event on the horizon, much like all values of $\theta$ would map to the origin when $r=0$ in polar coordinates. This event is not the same event as where any observer would pass the event horizon.

Edit: This is of course related to the fact that no observer can pass the horizon in finite coordinate time in Schwarzschild coordinates.

 

[Ibix]

Are you saying that all the spacelike hypersurfaces in the above family basically "share" a same event (i.e. they map a same event with different coordinate values) ?

Have a look at a Kruskal diagram. Lines of constant Schwarzschild $t$ (the hypersurfaces you are talking about) are straight lines through the origin. So is the event horizon. This is the point @Orodruin is making - that all those surfaces meet at one event, the origin of the Kruskal diagram, which is in the causal past of any massive object's horizon crossing.

 

[cianfa72]

Have a look at a Kruskal diagram. Lines of constant Schwarzschild $t$ (the hypersurfaces you are talking about) are straight lines through the origin. So is the event horizon.

Ah ok, so the event horizon in Kruskal diagram is actually a straight line of 45 degree through the origin of the diagram. For instance in region I all the events that belong to the event horizon have Schwarzschild coordinates $(r=1,t=\infty)$ and varying $\theta, \phi$. In this sense - as @Orodruin pointed out with the example of polar coordinates - for a given couple $\theta, \phi$ there is an entire straight line of events all mapped to the same $(r=1,t=\infty)$.

that all those surfaces meet at one event, the origin of the Kruskal diagram, which is in the causal past of any massive object's horizon crossing.

That means that on all spacelike hypersurfaces of constant Schwarzschild coordinate time $t$ the event with coordinate $r=1$ is actually the same event (i.e. the event mapped in the origin of the Kruskal diagram). It is an event that belong to the event horizon.

 

[PeterDonis]

the event that you are computing the distance to is the same for all of those hypersurfaces

Hm, yes, this is an interesting point, because in a more realistic spacetime containing a black hole that forms by gravitational collapse (the Oppenheimer-Snyder model would be the idealized case of this), the event you describe is not there (because that portion of the Kruskal diagram is not present, it's replaced by the region occupied by the collapsing matter). I have seen the "distance to the horizon" formula (and the integral that is used to derive it) discussed in plenty of GR textbooks, but I have never seen any discussion of how it is to be interpreted in a more realistic model.

 

[Orodruin]

Hm, yes, this is an interesting point, because in a more realistic spacetime containing a black hole that forms by gravitational collapse (the Oppenheimer-Snyder model would be the idealized case of this), the event you describe is not there (because that portion of the Kruskal diagram is not present, it's replaced by the region occupied by the collapsing matter). I have seen the "distance to the horizon" formula (and the integral that is used to derive it) discussed in plenty of GR textbooks, but I have never seen any discussion of how it is to be interpreted in a more realistic model.

Indeed, I think this just goes to show that ”distances” are generally based on conventions and are not really physical. In a stationary spacetime there is a natural definition of distances due to the timelike KVF, but Schwarzschild spacetime is only stationary outside the event horizon.

 

[PeterDonis]

Schwarzschild spacetime is only stationary outside the event horizon.

And also, in the Oppenheimer-Snyder model, the region containing the collapsing matter (a portion of which is outside the horizon) is not stationary.

 

[DrGreg]

Here is an image I created and posted (at https://www.physicsforums.com/threads/oppenheimer-snyder-model-of-star-collapse.651362/post-4164435) nearly 10 years ago.

It's a Kruskal diagram of a black hole formed by collapse of matter. The white area in the bottom left is where the matter is collapsing, and not covered by these coordinates. In an eternal black+white hole, the pattern would continue in that area.

The straight pink lines represent the foliation into spacelike hypersurfaces of constant Schwarzschild coordinate time. If extrapolated (in an eternal black+white hole) they would all meet at the same event, but each one hits the surface of the collapsing matter (the grey line).

(The coloured gridlines were calculated and plotted using Matlab, but the grey line was just sketched and is an intelligent guess.)

So this reveals something I'd never consciously realized before: that, using the Schwartzschild definition of simultaneity, not only does no infalling person ever cross the horizon in finite Schwartzschild time, the horizon never even comes into existence in finite Schwartzschild time. (The Schwartzchild model does not take account of outward Hawking radiation, or any inward radiation or matter after the initial collapse.)

The "distance to the horizon" being discussed in previous posts is measured along the straight pink lines (via the metric of course), but the none of these lines actually reaches the horizon (unless it's an eternal black+white hole).

 

[PAllen]

There is one geometric definition of distance from an event to a body that I’ve seen used in GR manifolds. It is the geodesic interval along a spacelike geodesic 4-orthogonal to the world tube boundary that reaches the given event. There can possibly be more that one such geodesic, but then they are all physically plausible distances to different surface features.

In the case of the horizon, there are no such geodesics at all. This is consistent with the observation that spacelike geodesics between some event and a lightlike geodesic in SR can range from (0,infinity) in length. (Whereas, between an event a timelike geodesic, there is a maximum such geodesic length, which is also the orthgonal one).

Thus I claim that no meaningful definition of distance to a horizon is possible at all.

 

[PeterDonis]

the horizon never even comes into existence in finite Schwartzschild time

Yes, that's correct. Schwarzschild coordinates cannot cover the horizon at all. Even the "interior" Schwarzschild coordinate patch, which covers the region inside the horizon, does not cover the horizon itself.

 

[PeroK]

I don't have the full calculation at hand, but if it's meaningless to say that an object hovers at a distance D from an event horizon, and wrong to assume that just because horizon passes an object at the speed of light, the object passes the horizon at the speed of light, I should check to see if this math is also irrelevant.

According to https://cosmo.nyu.edu/yacine/teaching/GR_2019/lectures/lecture22.pdf , for light
dr/dt = -(1-2M/r)
t = ± [r + 2M ln(r/2M − 1)] + constant

And according to https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/ (Peter Donis), for matter at rest
dr/dt = -(1-2M/r) sqrt(2M/r)

Now, I ought to remember how to get that second differential equation sorted out quickly, but I'm afraid it's been a while. Is this going to yield an accepted result where the calculated difference in t actually has something to do with the elapsed time at the statite or anywhere else in the universe?

Using Schwarzschild coordinates, it's possible to prove that we can wait only a maximum finite time to send light after an infalling object in order for the light to overtake the object before the event horizon.

Starting from rest at initial radius $r = R$, we can generate the equation for the infalling object:
$$\frac{dr}{dt} = -\frac{1}{\sqrt{1 - \frac{2M}{R}}}\big ( 1 - \frac{2M}{r} \big ) \sqrt{\frac{2M}{r} - \frac{2M}{R}}$$This can be integrated to give:
$$t = 2M\ln\bigg (\frac{\sqrt{\frac{r}{2M} -\frac{r}{R}} + \sqrt{1 - \frac{r}{R}}}{\sqrt{\frac{r}{2M} -\frac{r}{R}} - \sqrt{1 - \frac{r}{R}}} \bigg ) + F(r)$$Where $F(r)$ is some additional finite function of $r$. I.e. finite as $r \to 2M$.

Then, we have the relatioship between $r$ and $T$, which is the coordinate time of a inbound light pulse stating from $r = R$:
$$T = -2M\ln\big (\frac{r}{2M} - 1\big ) - r + T_0$$Now, we for a given $r$, we can calculate the difference between the time, $t$, it takes the infalling object to reach $r$ and the time it takes a light pulse. This is:
$$t - T = 2M\ln\bigg (\frac{\sqrt{\frac{r}{2M} -\frac{r}{R}} + \sqrt{1 - \frac{r}{R}}}{\sqrt{\frac{r}{2M} -\frac{r}{R}} - \sqrt{1 - \frac{r}{R}}} \bigg ) + F(r) + 2M\ln\big (\frac{r}{2M} - 1\big ) +r - T_0$$What we need to show is that tends to some finite limit as $r \to 2M$. Everything is finite, except possibly the combined log term:
$$\ln\bigg (\frac{\sqrt{\frac{r}{2M} -\frac{r}{R}} + \sqrt{1 - \frac{r}{R}}}{\sqrt{\frac{r}{2M} -\frac{r}{R}} - \sqrt{1 - \frac{r}{R}}} \bigg ) + \ln\big (\frac{r}{2M} - 1\big )$$To see that this is finite, we re-express the term inside the first log as:
$$\frac{\sqrt{\frac{r}{2M} -\frac{r}{R}} + \sqrt{1 - \frac{r}{R}}}{\sqrt{\frac{r}{2M} -\frac{r}{R}} - \sqrt{1 - \frac{r}{R}}} = \frac{\frac{r}{2M} -\frac{2r}{R} + 1 + 2\sqrt{\frac{r}{2M} -\frac{r}{R}}\sqrt{1 - \frac{r}{R}}}{\frac{r}{2M} -1} $$And, when we combine the logs the denominator will cancel with the second log term, leaving a finite limit as $r \rightarrow 2M$.

In other words, the difference in time between a light pulse and an infalling object tends to some finite limit as $r \to 2M$, Which means that we have a finite time we can wait before sending a light signal that will reach the object before it gets to the event horizon.

 

[PeterDonis]

In the case of the horizon, there are no such geodesics at all.

I would say that a horizon does not have the kind of "world tube boundary" that is required. In particular, if the horizon is an integral curve of a KVF, since the KVF is null on the horizon, it is orthogonal to itself there, so the concept of "spacelike surface orthogonal to the KVF" is no longer valid.

In the case of the distance formula in Schwarzschild spacetime that I was referring to, that is standardly interpreted as the limit as $r \to 2M$ of the formula for the distance between stationary observers at two different values of $r$ that is valid outside the horizon. That limit is valid mathematically, but I'm no longer sure it has the physical meaning that is usually attributed to it. See further comments below.

no meaningful definition of distance to a horizon is possible at all.

In the light of the discussion we have been having about what portion of spacetime the spacelike hypersurfaces orthogonal to the timelike KVF in Schwarzschild spacetime actually cover, I now agree with this. There is a mathematical formula that will give a "distance to the horizon" in one of these hypersurfaces, but the spacelike geodesic segment it describes either doesn't exist as a complete segment in the spacetime (because a portion of it would be inside the collapsing matter region and the formula is not valid there), or goes to an event that is the intersection of multiple spacelike geodesics from different points on the timelike worldline of the hovering observer (the same issue that arises with the Rindler horizon in SR).

 

[PAllen]

@PAllen - your argument *is* valid if there is in fact a last received signal. That's why the question is about how you calculate what the last signal received by the probe is; if someone can show there is (or isn't) a specific cutoff, the qualitative arguments ought to be settled also.

I am going to take this up in two different posts. First, pictorially, using Kruskal coordinates, then quantitatively, with the computation made much easier by using the generalized Lemaitre coordinates derived in this paper:

https://arxiv.org/abs/1911.05988

around page 8. (This and another paper: https://arxiv.org/abs/1211.4337, are very interesting to see many generalizations of Gullestrand-Panlieve coordinates and their relationships to Lemaitre type coordinates).

So pictorially, we have:

https://www.physicsforums.com/attac...296920/?hash=0ada85ddb1326ac3dba60e793c42457e

In a later post, I will calculate the proper time elapsed for the statite between A and B and A and C. I will also calculate the proper time for the probe between A and its receipt of signal B, and between A and its receipt of signal C.

To address Hawking radiation, once the computations are done, you will see that events B and C, as well as their whole signal trajectories are in an era when the BH is growing due to absorption of CMB radiation. Only many trillions of years later, when expansion has reduced the CMB temperature to below the Hawking temperature for a BH, will the BH start losing mass. This is kilometers to the upper right in the attached diagram.

On a technical note, the true mathematical model applicable would be some fusion of an ingoing vaidya metric (for the BH growth phase), and outgoing vaidya metric for the shrinking phase, both modified to join with an FLRW metric. This has probably never been done exactly, but the differences in consequences for the region under consideration would only affect computations well beyond 10 significant digits.

 

[Mike S.]

@PAllen - Thanks; this looks like a powerful argument ... but I don't understand Kruskal coordinates. It looks like the statite needs a world line at a 45-degree angle, parallel to the event horizon, in order to not fall toward the hole, but doesn't that mean it's moving away from the hole at the speed of light?

 

[Orodruin]

@PAllen - Thanks; this looks like a powerful argument ... but I don't understand Kruskal coordinates. It looks like the statite needs a world line at a 45-degree angle, parallel to the event horizon, in order to not fall toward the hole, but doesn't that mean it's moving away from the hole at the speed of light?

No, it does not need that. In fact, it cannot have that as that would correspond to r growing without bound. A statite would have constant r, which corresponds to hyperbolae in a Kruskal diagram.

 

[PeterDonis]

It looks like the statite needs a world line at a 45-degree angle, parallel to the event horizon, in order to not fall toward the hole

No. The statite's worldline in Kruskal coordinates is a hyperbola in the right "wedge" of the diagram. The Insights article series I linked to discusses this.

 

[Mike S.]

@PeroK - your post seems to be the answer to the question. I'm terribly out of practice, but your solution seemed to check out (even if I ended up with the wrong overall sign somehow, and don't recall why a F(x) is needed).

I could still use much help understanding what this means in physical terms. According to your formula (but ignoring F(x)), I'm getting that it gives an object falling in from infinite distance "2M ln 4" additional timestamp signals before it reaches the event horizon, and an object dropped from r = 4M (twice the Schwarzschild radius) a total of "2M ln 2" worth of signals to reach the horizon. I know this used c=G=1, so for the hypothetical M = 1E30 kg (a half solar mass black hole, rather unusual I know), I can take 1E30 kg x 6.674E−11 m3⋅kg−1⋅s−2 / (299792458 m2 s-2) = 743 s, meaning it receives 515 s of timestamps for my example, and 989 s of timestamps for a fall from "infinity".

But I don't understand why the delay for an arbitrarily long fall (large R) doesn't work out to be arbitrarily large. Also, I'm not seeing how this fits versus a story like https://www.centauri-dreams.org/2021/09/16/predicting-a-supernova-in-2037, where light seems to be able to spend many years trapped in a gravity well before continuing on to Earth.

 

[PeterDonis]

I don't understand why the delay for an arbitrarily long fall (large R) doesn't work out to be arbitrarily large.

It does. I haven't fully checked the math that @PeroK posted, but the solution if we start by assuming a fall from rest at infinity is known and simple, and predicts an infinite time to fall. The easiest way to obtain it is to look at $dr / d\tau$, where $\tau$ is the proper time of the infalling object. This is:

$$
\frac{dr}{d\tau} = - \sqrt{\frac{2M}{r}}
$$

where I am using units in which $G = c = 1$. It is useful to rescale $r$ by defining $R = r / 2M$, so $dr = 2M dR$, and we then have for the time to fall to the horizon:

$$
\frac{\tau}{2M} = - \int \sqrt{R} dR = \left[ - \frac{2}{3} R^\frac{3}{2} \right]_{\infty}^{1} = \infty - \frac{2}{3}
$$

For the time to fall to the singularity, the upper limit of the integral would be $0$ and we would just get $\infty$. The formulas @PeroK posted should approach this one as a limit as $r \to \infty$.

Where this formula is often useful in giving good approximate finite answers is for cases of fall from some finite value of $r$ that is very large compared to $2M$, so the starting velocity at that $r$, while not zero, is very small, and the above formula gives a good approximation if we change the lower limit of the integral from $\infty$ to the large finite value. Then we get a finite answer that, as the result above shows, scales like the 3/2 power of $r$.

 

[PeterDonis]

I'm not seeing how this fits versus a story like https://www.centauri-dreams.org/2021/09/16/predicting-a-supernova-in-2037, where light seems to be able to spend many years trapped in a gravity well before continuing on to Earth.

The "many years" here translates to a couple of decades as compared to 10 billion years, i.e., about 1 part in a billion. That's a pretty small effect.

 

[PeroK]

But I don't understand why the delay for an arbitrarily long fall (large R) doesn't work out to be arbitrarily large.

Mathematically, the functions of $t(r)$ and $T(r)$ tend to $\infty$ as $r \to 2M$. The difference $r(t) - T(t)$ can do three things: it can tend to $\pm \infty$ or to some finite number.

A simple example of the latter would be $\sqrt{x +1} - \sqrt{x}$. Both functions individually tend to infinity, but their difference is bounded.

The functions in this problem are more complicated log functions, but the same conclusion applies.

Physically there are two outcomes: Either the light pulse reaches the probe at some point $r \ge 2M$; or, it doesn't. The mathematics shows that if you wait beyond a certain finite time, then the light pulse cannot catch the probe at any point $R \ge 2M$.

Here is a similar problem you may be familiar with: You fire a rocket off into space with a constant proper acceleration. Then fire a light pulse after it. Again, there is a finite time limit you can wait if you want to light pulse ever to reach the rocket. Beyond a certain time, the light pulse never reaches the rocket.

 

[Mike S.]

@PeroK - If I drop the probe from 100 AU away, it should receive more the 900 s of timing signals. Unless I'm missing something big in the F(x) or something?

 

[PeroK]

@PeroK - If I drop the probe from 100 AU away, it should receive more the 900 s of timing signals. Unless I'm missing something big in the F(x) or something?

I'll work on getting an answer when I have the chance. I want to check that my $F(r)$ is correct first. I'll hopefully post something tomorrow.

 

[PeroK]

The formulas @PeroK posted should approach this one as a limit as $r \to \infty$.

They do! The solution I get, using $u = \frac{r}{2M}$ and $u_0 = \frac{R}{2M}$, where $R$ is the starting radial coordinate, is:
$$t = 2M\ln\bigg (\frac{\sqrt{1 - \frac{1}{u_0}} + \sqrt{\frac{1}{u} - \frac{1}{u_0}}}{\sqrt{1 -\frac{1}{u_0}} - \sqrt{\frac{1}{u} - \frac{1}{u_0}}} \bigg ) + F(u) \ \ \ (1)$$Where:
$$F(u) = 2M \bigg (1 - \frac 1 {u_0}\bigg )^{1/2} \bigg (1 - \frac{u}{u_0} \bigg )^{1/2}u_0u^{1/2}$$$$ + \ 2M\bigg (1 - \frac 1 {u_0}\bigg )^{1/2}(2 + u_0)u_0^{1/2}\tan^{-1}\bigg [\big (\frac{u_0}{u} \big )^{1/2} \big (1 - \frac{u}{u_0} \big )^{1/2}\bigg ]$$The standard solution for $u_0 \to \infty$ is:
$$t = t_0 + 2M\bigg [\ln \bigg ( \frac{u^{1/2} +1}{u^{1/2} - 1} \bigg ) - 2u^{1/2} - \frac 2 3 u^{3/2} \bigg ]$$We need to show that equation (1) reduces to this in the limit. The log term in equation (1) can be evaluated in the limit simply:
$$\ln \bigg (\frac{1 + \sqrt{\frac{1}{u}}}{1 - \sqrt{\frac{1}{u}}} \bigg ) = \ln \bigg (\frac{u^{1/2} +1}{u^{1/2} - 1} \bigg )$$For the other two terms we need a Taylor expansion, neglecting terms in negative powers of $u_0$. The Taylor expansion of the first term (using $w = u^{1/2}$ for simplicity):
$$\bigg (1 - \frac 1 {u_0}\bigg )^{1/2} \bigg (1 - \frac{u}{u_0} \bigg )^{1/2}u_0 w = \bigg (1 - \frac 1 {2u_0}\bigg ) \bigg (1 - \frac{u}{2u_0} \bigg )u_0 w = u_0w - \frac 1 2 w -\frac 1 2 w^3$$The second term is:
$$\bigg (1 - \frac 1 {u_0}\bigg )^{1/2}(2 + u_0)u_0^{1/2}\tan^{-1}\bigg [\big (\frac{u_0}{u} \big )^{1/2} \big (1 - \frac{u}{u_0} \big )^{1/2}\bigg ]$$$$ = \bigg (\frac 3 2 u_0^{1/2} + u_0^{3/2} \bigg )\tan^{-1}\bigg [\big (\frac{u_0}{u} \big )^{1/2} \big (1 - \frac{u}{u_0} \big )^{1/2}\bigg ]$$Then, using
$$\tan^{-1} x = \frac \pi 2 - \frac 1 x + \frac 1{3x^3} + \dots$$gives:
$$\tan^{-1}\bigg [\big (\frac{u_0}{u} \big )^{1/2} \big (1 - \frac{u}{u_0} \big )^{1/2}\bigg ] = \frac \pi 2 - \frac w {u_0^{1/2}}\bigg (1 - \frac u {u_0} \bigg )^{-1/2} + \frac {w^3} {3u_0^{3/2}} \bigg (1 - \frac u {u_0} \bigg )^{-3/2}$$$$ = \frac \pi 2 - \frac w {u_0} - \frac {w^3} {6u_0^{3/2}}$$Putting these together gives us the second term:
$$t_0 - \frac 3 2 w - u_0w - \frac {w^3} 6$$where $t_0$ is the term in $u_0$ that is independent of $u$. Finally, we can put the two terms together to get the required expression for $F(u)$ in the limit $u_0 \to \infty$:
$$F(u) = t_0 + 2M \big (-2w - \frac 2 3 w^3 \big )$$And, it looks like we have a valid generalisation of the radial plunge.