I Measuring Light Reflection in a Black Hole

[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.

 

[Mike S.]

@PeroK - I have to admire anyone who feels comfortable going to a Taylor expansion, especially when they're answering my math question ... I don't even remember about the tangent. But the figure you end up with for F(u) is even less than t0 - isn't t0 the time when the pulses started?

 

[PeroK]

@PeroK - I have to admire anyone who feels comfortable going to a Taylor expansion, especially when they're answering my math question ... I don't even remember about the tangent. But the figure you end up with for F(u) is even less than t0 - isn't t0 the time when the pulses started?

That post refers only to the equation for an infalling massive object starting at $u_0$ at $t=0$. It was to check that as $u_0 \to \infty$ we recover the standard formula for an infalling object. Which we do. That gives me a bit more confidence that the equation for $t(r)$ or $t(u)$ that I got is correct.

 

[Mike S.]

@PeroK - you seem to have solved this problem, not just for R = 4M but for any R, at least to a two-term approximation. But I'm still not clear how to go from R to some number of literal seconds of timestamps received at the probe, which seems essential for trying to apply intuition to this environment.

The other people talking about Kruskal coordinates are being helpful, but speaking as a by-now recognized authority on ignorance of these matters, I should say that coordinate system is still clear as mud to me. I see from the graphic that objects at a constant distance move as hyperbolas, so a sublight object can avoid a 45-degree angle without ever being *at* 45 degrees. It still seems strange though - a 45-degree angle represents the path of light, but at t = -3GM it doesn't cross from R = 2.4 to 2.6 GM within a GM clock tick, while at t=4GM it makes it all the way to 2.8 GM away in the same time. At the same gravity? The Earth starts out moving, I suppose, straight toward the horizon of Cygnus X-1, but must end up going at a nearly 45-degree angle right eventually? According to the drawing, the probe released in this case doesn't move straight up toward the horizon, but starts at a tangent to the hyperbola and keeps moving right, just not 45 degrees right. I mean, I recognize people are trying to make recognized arguments with these graphs but the rules about how to apply them are not easily apparent!

 

[PeterDonis]

I should say that coordinate system is still clear as mud to me.

The key thing to remember with Kruskal coordinates is that they mean nothing useful physically (although 45 degree lines do have a useful physical meaning--see below). Unfortunately, the similarity of a Kruskal diagram to a Minkowksi spacetime diagram in SR makes it easy to think that the Kruskal $X$ and $T$ must be something like the Minkowski $X$ and $T$. They're not. They're mathematical conveniences that make a diagram that is very useful for some things--in particular seeing causal structure, since radial light rays always move on 45 degree lines, just as in an SR spacetime diagram (diagrams that have this property are called "conformal" in the literature and are often used for this purpose)--but should not be taken too literally.

 

[PeroK]

Continuing with the calculations. We have the time for a massive object dropped from $u_0$ to reach $u$:
$$t = 2M\ln\bigg (\frac{\sqrt{1 - \frac{1}{u_0}} + \sqrt{\frac{1}{u} - \frac{1}{u_0}}}{\sqrt{1 -\frac{1}{u_0}} - \sqrt{\frac{1}{u} - \frac{1}{u_0}}} \bigg ) + F(u) \ \ \ (1)$$Where:
$$F(u) = 2M \bigg (1 - \frac 1 {u_0}\bigg )^{1/2} \bigg (1 - \frac{u}{u_0} \bigg )^{1/2}u_0u^{1/2}$$$$ + \ 2M\bigg (1 - \frac 1 {u_0}\bigg )^{1/2}(2 + u_0)u_0^{1/2}\tan^{-1}\bigg [\big (\frac{u_0}{u} \big )^{1/2} \big (1 - \frac{u}{u_0} \big )^{1/2}\bigg ]$$We can re-express the log term as before:
$$t = 2M\ln\bigg (\frac{u\big (\sqrt{1 - \frac{1}{u_0}} + \sqrt{\frac{1}{u} - \frac{1}{u_0}} \big )^2}{u - 1} \bigg ) + F(u)$$Now, we have the time for a inbound radial light pulse starting from $u_0$ to reach $u$:
$$T = 2M \bigg [u_0 - u + \ln \bigg (\frac{u_0 - 1}{u - 1} \bigg ) \bigg ]$$The difference is:
$$t - T = 2M\ln\bigg (\frac{u\big (\sqrt{1 - \frac{1}{u_0}} + \sqrt{\frac{1}{u} - \frac{1}{u_0}} \big )^2}{u_0 - 1} \bigg ) + F(u) - 2M(u_0 - u)$$ Having problems posting more ...

 

[PeterDonis]

I'm still not clear how to go from R to some number of literal seconds of timestamps received at the probe

It helps in these problems to have a sense of what the units of $2M$ mean in ordinary terms. A useful unit for $M$, at least for studying stellar mass black holes, is one solar mass. For an $M$ of one solar mass, $2M$ equates to 3 kilometers, or 10 microseconds (the time it takes light to travel 3 kilometers).

So if we apply the equations @PeroK or I posted to a one solar mass black hole, we are measuring time in units of 10 microseconds each, and distance in units of 3 kilometers each. Which means, for example, that an object free-falling from rest at infinity will take $2/3 \times 10$ microseconds, or $6.67$ microseconds, to fall from the horizon to the singularity by its own clock.

Or, to give a distance example, if an object starts free-falling towards a one solar mass black hole from a radial coordinate of $R = 10^6$ (1 million times $2M$), then they are starting from 3 million kilometers above the hole. (I know we cautioned earlier about the radial coordinate not being the same as physical distance, but for radial coordinates this large compared to $2M$, the correction involved is small enough to be ignored for most purposes.) Using my equation as an approximation, such an object would take the $3/2$ power of $10^6$, or $10^9$, time units of $2M$ to fall to the singularity (or the horizon, since the difference between the two in terms of fall time is negligible in this case), which equates to 10,000 seconds or somewhat less than 3 hours.

 

[PeroK]

... we have$$(t - T)_{max} = 2M\ln\big (\frac{4}{u_0} \big ) + 2M(2 + u_0)\sqrt{u_0}\sqrt{1 - \frac{1}{u_0}} \tan^{-1}\big ({\sqrt{u_0 -1}}\big )$$ That's the maximum coordinate time between the object and the last light pulse at $u_0$. We really want the local time at $u_0$, which has an additional factor of $\sqrt{1 - \frac 1 {u_0}}$:
$$\Delta \tau_0 = 2M \bigg [\sqrt{1 - \frac 1 {u_0}} \ln\big (\frac{4}{u_0} \big ) + (2 + u_0)\sqrt{u_0}\big (1 - \frac{1}{u_0} \big ) \tan^{-1}\big ({\sqrt{u_0 -1}} \big ) \bigg ]$$E.g. if $R = 4M$, then $u_0 = 2$ and:
$$\Delta \tau_0 = 2M\big [\frac 1 {\sqrt 2} \ln 2 + \frac {\pi} {\sqrt 2} \big ] = \sqrt 2 M(\ln 2 + \pi) \approx 5.42 M$$And if we take $M = 1.5 km$, i.e. one solar mass, then:
$$\Delta \tau_0 = 8135 km = 27\mu s$$

 

[PAllen]

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.

Continuing along a different path than @PeroK , I've been intermittently following up on the approach described above, using results from the indicated papers. So, using Lemaitre style coordinates, adapted to a "free fall from platform" congruence, discussed at p.8 of the first paper referenced, and changing notation from the paper as follows (all in units where c=G=1, with the idea that mass is expressed in terms of SC radius in light seconds, and spatial units - including SC radial coordinate - are in light seconds, direct computations yield seconds):

- I use P for the platform SC radial coordinate, R for SC radius of the BH
- I express e as used in the paper using a definition given earlier in the paper:
$$P=R/(1-e^2)$$

Then the metric for platform based Lemaitre style coordinates is:

$$ds^2=dT^2-R(1-R/P)^{-1}(1/r-1/P)d\rho^2$$

suppressing angular coordinates since we are treating a purely radial problem. Using results given in the paper for dT and $d\rho$ in terms of SC differentials, one can derive that $$\rho-T=\int (R/r - R/P)^{-1/2} dr $$

The f(r) defined by the integral can be evaluated to: $$-\sqrt{\frac {rP} R (P-r)} -\frac {P^{3/2}} {\sqrt R} \arctan(\sqrt{P/r-1}) $$ which can be seen to be equivalent to the formula given in the second paper (eq. 3.10) referenced above (taking $\rho=0$ and expressing as T).

Putting in r=P you verify get zero as desired, and then r=R and r=0 are readily computed as probe proper times for free fall from P to SC radius, and then to singularity.

Note, that in these coordinates, the chosen free fall world line from the platform takes the trivial form $\rho=0$, T varying. Note that generally, $$\rho-T = costant$$ gives a hovering world line (constant computed as f(r)). In our set up, the constant zero means the hovering platform. This shows the useful fact that proper time along the platform world line matches T coordinate time (just plug r=P into the metric above).

Thus, referring to my diagram, the coordinates of event A (the drop event) are simply $(\rho,T)=(0,0)$, event of probe reaching horizon are $(0,-f(R))$, and the limiting coordinate for reaching the singularity are $(0,-f(0))$.

So far, straight forward. But what I wanted to do next was derive the light path from the platform world line ending on the event of probe at horizon, and also the path from platform world line to probe at singularity. If these were derived, the T coordinates of the start events on the platform world line would directly be proper times for the platform from drop event to corresponding signal times, the second one described being the last signal that can reach the probe.

Unfortunately, here I have hit a major snag. It seems that light paths are very complicated to express in these coordinates, and I am not sure when I will find a way around this. As I can currently express this, I would need to numerically solve a truly complicated differential equation.

 

[ergospherical]

I have to admire anyone who feels comfortable going to a Taylor expansion

The study of Taylor expansions is also known by the alternative name "Physics"!

 

[PeroK]

I think I have a full solution for this. First, I'm going to take the slightly easier case where we have an object falling radially from infinity and passing a given radius $r_0$ at some time. We want to calculate the maximum time that we can wait at $r_0$ before sending a light signal that will reach the object a) before the event horizon and b) before the singularity.

I'll use $w = \sqrt u = \sqrt{\frac{r}{2M}}$.

First, we have the solution to the radially infalling object:
$$t = t_0 + 2M\bigg [\ln\bigg | \frac{w +1}{w-1}\bigg | - \frac 2 3 w^3 - 2w \bigg ]$$Where, if we take $t = 0$ at $r_0$, then $$t_0 = -2M\bigg [\ln\bigg | \frac{w_0 +1}{w_0-1}\bigg | - \frac 2 3 w_0^3 - 2w_0 \bigg ]$$Next, we have the solution for a radially inbound light pulse:
$$T = T_0 - 2M \big [w^2 + \ln|w^2 - 1| \big ]$$Where, if we take $T = 0$ at $r_0$, then:
$$T_0 = 2M \big [w_0^2 + \ln|w_0^2 - 1| \big ]$$We can do what we did before and look at the difference (Note: corrected an error in this line: I had $2w^2$):
$$t - T = t_0 - T_0 + 2M\big [2\ln(w + 1) -2w - \frac 2 3 w^3 + w^2 \big ]$$That gives us the maximum time we can wait for a light pulse to reach the object before the event horizon at $w = 1$: (corrected same error as above)
$$(t - T)_{max} = t_0 - T_0 + 2M\big [\ln 4 - \frac 5 3 \big ]$$Now, we have:
$$t_0 - T_0 = 2M \big [\frac 2 3 w_0^3 +2w_0 - w_0^2 - 2\ln(w_0 + 1) \big ]$$And, converting to local time at $w_0$, this gives the maximum time at that location if we want a light signal to reach the object before the event horizon:$$\Delta \tau_{max} = \sqrt{1 - \frac 1 {w_0^2}} \bigg (t_0 - T_0 + 2M\big [\ln 4 - \frac 5 3 \big ]\bigg )$$For example, if $r_0 = 4M$, then $w_0 = \sqrt 2$ and (another correction here, from above):
$$\Delta \tau_{max} = 2M(0.474) = 4.7 \mu s$$Taking $M$ to be 1 solar mass.

As a sanity check, we can compare this with the proper time of the object between $w_0$ and $w = 1$, which is:
$$\Delta \tau_{obj} = \frac{4M}{3}\big ( w_0^3 - 1 \big ) = 12.2 \mu s$$And, in fact, we can see that the dominant factor in the above equation for $t_0 - T_0$ is indeed $\dfrac{4M}{3}w_0^3$, which is, as we might expect, the proper time for the object to fall from $w_0$ to the singularity.

I'll post this now. More to come ...

 

[Mike S.]

@PeroK - very interesting work so far, and quite a lot of it! This suggests that the object, in this case falling past the station, sees just a tiny bit of redshift in the light it receives from the station. If I go by the factor-of-2 redshift at the horizon someone mentioned before, then this approximates to 11.4 microseconds of unchanged frequency received, plus .4 microseconds stretched to .8 microseconds. I have to say, that's less colorful than what I would have expected from such an extreme environment, but so far the hole is resisting my intuition quite effectively.

 

[PeroK]

@PeroK - very interesting work so far, and quite a lot of it! This suggests that the object, in this case falling past the station, sees just a tiny bit of redshift in the light it receives from the station. If I go by the factor-of-2 redshift at the horizon someone mentioned before, then this approximates to 11.4 microseconds of unchanged frequency received, plus .4 microseconds stretched to .8 microseconds. I have to say, that's less colorful than what I would have expected from such an extreme environment, but so far the hole is resisting my intuition quite effectively.

I had to correct a minor error to the above. Perhaps it looks better now.

 

[PeroK]

I'm not sure if this is totally valid, but anyway.

Now, we can do the same analysis for $0 < w < 1$ using the same constant of motion and find a finite difference between the coordinate times for the object and light in that region. The inbound light pulse has the same equation as before:
$$T = T_0 -2M\big [w^2 + \ln(1 - w^2) \big ]$$And, a solution for the motion of the object in this region is:
$$t = T_0 +2M \bigg[-\frac 2 3 w^3 - 2w + \ln\bigg (\frac{1 + w}{1-w} \bigg ) \bigg ]$$Note that the $t$ (and $T$) coordinate in this region is spacelike and runs from $+\infty$ at $w = 1$ to $T_0$ at $w = 0$.

As before, we have a finite difference for the two worldlines in this region:
$$t - T = 2M\big [-\frac 2 3 w^3 - 2w + w^2 + 2\ln(1 + w) \big ]$$And we get the same factor as before of $$2M[-\frac 5 3 + \ln 4]$$Note that $2M[\frac 5 3 - \ln 4] \approx 2M(0.28)$, which compares with $2M(\frac 2 3)$, which is the proper time for the object to fall from $w = 1$ to $w = 0$.

That suggests that if we allow this additional coordinate time at $w_0$ then that is the maximum that allows a light signal to reach the object before the singularity. That would allow us simply to plug $w = 0$ into the previous equation and get:
$$\Delta \tau_{max} = \sqrt{1 - \frac 1 {w_0^2}} \big (t_0 - T_0 \big) = 2M\sqrt{1 - \frac 1 {w_0^2}} \big [\frac 2 3 w_0^3 +2w_0 - w_0^2 - 2\ln(w_0 + 1) \big ]$$As the maximum time we can wait at $w_0$ to send a light pulse that will reach the object before the singularity at $w = 0$.

 

[PAllen]

Personally, I would have much more confidence in computing computing a triangle of 3 world lines: free fall, platform, and light with vertices being intersection of ingoing light and free fall at the singularity, platform and free fall at some drop point, and platform and light at transmission event. Then you can directly compute proper time along the platform world line between probe drop and signal transmission that arrives at the probe as the probe reaches the singularity. This avoids any coordinate dependent assumptions. This requires coordinates that cover the whole region, preferably chosen to make the invariant computations easier. The generalized Lemaitre I used were great for two of the 3 world lines, but terrible for the third (light paths). I think I will try generalized Gullestrand-Panlieve next, or maybe Eddington-Finkelstein (because they are great for light paths). While Kruskal is great to show the whole causal structure, it is a real pain to compute anything in.

 

[PeroK]

or maybe Eddington-Finkelstein (because they are great for light paths).

I was using Eddington-Finkelstein, but I didn't immediately see how to tie it all together. The solution is:

The inbound light pulse has $v = v_0$, which is constant. That means for the light path (using capitals for the Schwarzschild coordinates of the light pulse) we have:
$$v_0 = T + 2M\big (W^2 + \ln\big |W^2 - 1 \big| \big ) = T_0 + 2M\big (w_0^2 + \ln(w_0^2 - 1) \big )$$While for the object we have:
$$v = t + 2M \big (w^2 + \ln \big |w^2 - 1 \big | \big ) = t_0 + 2M\bigg [-\frac 2 3 w^3 -2w + \ln\bigg |\frac{w+1}{w-1}\bigg | \bigg ] + 2M \big (w^2 + \ln \big |w^2 - 1 \big | \big )$$And, when $w = 0$ we have:
$$v = t_0$$We want the object and light to coincide at $w = 0$, $v = v_0$, hence $t_0 = v_0$. And, if now we set $t = 0$ at $w_0$ for the object, then we have:
$$0 = v_0 + 2M\bigg [-\frac 2 3 w_0^3 -2w_0 + \ln\bigg |\frac{w_0+1}{w_0-1}\bigg | \bigg ]$$And, finally, that gives us:
$$T_0 = v_0 - 2M\big (w_0^2 + \ln(w_0^2 - 1) \big ) = 2M\bigg [\frac 2 3 w_0^3 + 2w_0 - \ln\bigg |\frac{w_0+1}{w_0-1}\bigg | \bigg ] - 2M\big (w_0^2 + \ln(w_0^2 - 1) \big )$$And:
$$T_0 = 2M\bigg [\frac 2 3 w_0^3 + 2w_0 - w_0^2 - 2\ln (w_0+1 ) \bigg ]$$

 

[PeroK]

So easy once you see it!

 

[PeroK]

Here's the solution for the problem starting at a finite coordinate $r_0$, with $u = \frac{r}{2M}$.

The equation for the particle in Schwarzschild coordinates is (with $t = 0$ at $u_0$):
$$t = 2M\ln\bigg (\frac{u\big (\sqrt{1 - \frac{1}{u_0}} + \sqrt{\frac{1}{u} - \frac{1}{u_0}} \big )^2}{u - 1} \bigg ) + F(u)$$Where $$F(u) = 2M \bigg (1 - \frac 1 {u_0}\bigg )^{1/2} \bigg (1 - \frac{u}{u_0} \bigg )^{1/2}u_0u^{1/2}$$$$ + \ 2M\bigg (1 - \frac 1 {u_0}\bigg )^{1/2}(2 + u_0)u_0^{1/2}\tan^{-1}\bigg [\big (\frac{u_0}{u} \big )^{1/2} \big (1 - \frac{u}{u_0} \big )^{1/2}\bigg ]$$We transform to Eddington-Finkelstein coordinates:
$$v = t + 2M \big [ u + \ln|u -1| \big ] = 2M \bigg [u + \ln\bigg (u \big (\sqrt{1 - \frac{1}{u_0}} + \sqrt{\frac{1}{u} - \frac{1}{u_0}} \big )^2 \bigg ) \bigg ] + F(u)$$And, the equation for the light pulse is ($t = T_0$ at $u_0$):
$$v = v_0 = T_0 +2M \big [ u_0 + \ln(u_0 - 1) \big ]$$If the light pulse catches the object at $u = 0$, then:
$$v_0 = F(0)$$Note that the above log tends to zero as $u \to 0$. Hence:
$$T_0 = F(0) - 2M \big [ u_0 + \ln(u_0 - 1) \big ]$$And:
$$F(0) = \pi M \big (\sqrt {u_0 - 1})(2 + u_0)$$And, if we look for the light signal to reach the object at the event horizon ($u = 1$), we get the same expression as before:
$$T_0 = 2M\ln\big (\frac{4}{u_0} \big ) + 2M(2 + u_0)\sqrt{u_0}\sqrt{1 - \frac{1}{u_0}} \tan^{-1}\big ({\sqrt{u_0 -1}}\big )$$

 

[PeterDonis]

While Kruskal is great to show the whole causal structure, it is a real pain to compute anything in.

Being a glutton for punishment, I went and did the beginnings of this anyway. Kruskal does have the advantage that all but one of the worldlines involved are either 45 degree lines (the light rays) or a hyperbola (the statite's worldline), and the computation can be reduced to two unknowns, one of which is just a choice of how close to the horizon the last light signal to be reflected back is emitted.

One other advantage of Kruskal coordinates is that the coordinates of the initial event, where the infalling object detaches from the statite and starts free-falling inward, can be given the convenient coordinates of $(X, T) = (X_0, 0)$, i.e., it can be placed on the $T = 0$ line. This can always be done because of the "boost" symmetry of the spacetime: if our chosen event were anywhere else on the statite's worldline (the hyperbola), we could just "rotate" everything using the Killing flow of the spacetime and put the event on the $T = 0$ line (because everything just flows along the hyperbolas).

Also, $X_0$ is just a known function of whatever $r$ coordinate we choose for the statite; I'll call this value $R$. Then we have

$$
X_0^2 = \left( \frac{R}{2M} - 1 \right) e^\frac{R}{2M}
$$

The statite's worldline is then just $X^2 - T^2 = X_0^2$.

We now label the events of interest. Event 0 is the one described above, where the infalling object starts falling inward. Event 1 is where the last light signal to be reflected back reaches the infalling object, just above the horizon. The coordinates of this event are $(X_1, T_1)$, where $X_1 = T_1 + \varepsilon$.

Event L (for "last") is where the last light signal to be reflected back is emitted from the statite. Its coordinates are $(X_L, T_L)$. Since this event is both on the statite's hyperbola and on the past light cone of event 1, we can compute its coordinates from the intersection of those two curves. The past light cone gives $X_L - X_1 = X_L - T_1 - \varepsilon = T_1 - T_L$, or $X_L = 2 T_1 + \varepsilon - T_L$, and the hyperbola gives $X_L^2 - T_L^2 = X_0^2$, which, after some computation and discarding terms quadratic in $\varepsilon$, gives

$$
T_L = \frac{X_0^2}{4 T_1} - \left( T_1 + \varepsilon \right)
$$

Event R (for "return") is where the last light signal reflected back is received by the statite. Its coordinates are $(X_R, T_R)$. This event is on both the statite's hyperbola and the future light cone of event 1, so we can compute its coordinates similarly to what we did above for event L. However, this computation turns out to be simpler because the light ray is now outgoing instead of ingoing; we have $X_R - X_1 = X_R - T_1 - \varepsilon = T_R - T_1$, which, after substituting into the hyperbola equation and some algebra, gives

$$
T_R = \frac{X_0^2}{2 \varepsilon}
$$

We now need to translate the coordinates of events L and R into proper time values for the statite. (The proper time of event 0 is $0$ by construction, so the proper times we are calculating are the elapsed times on the statite's clock from the time the infaller departs until the last light signal to be reflected back is emitted, and then received.) This is straightforward because we know the relationship between the Kruskal coordinates and the Schwarzschild $t$ coordinate, and between the Schwarzschild $t$ coordinate and the statite's proper time $\tau$. This gives us:

$$
\tau = \sqrt{1 - \frac{2M}{R}} 4M \tanh^{-1} \frac{T}{X} = \sqrt{1 - \frac{2M}{R}} 2M \ln \left( \frac{X + T}{X - T} \right)
$$

We then simply plug in the values from above; using the light ray equations to relate $X_L, T_L$ and $X_R, T_R$ is easier than using the statite hyperbola equation to do it. We obtain:

$$
\tau_L = \sqrt{1 - \frac{2M}{R}} 2M \ln \left( \frac{4 T_1^2 + 2 T_1 \varepsilon}{8 T_1^2 + 6 T_1 \varepsilon - X_0^2} \right)
$$

$$
\tau_R = \sqrt{1 - \frac{2M}{R}} 2M \ln \left( 1 + \frac{X_0^2}{\varepsilon^2} \right)
$$

The tedious part, of course, would be actually computing $T_1$, since that requires computing the infalling object's worldline using the geodesic equation in Kruskal coordinates, which is complicated because of the number of nonzero Christoffel symbols. But from the above we can already see the qualitative behavior we expect. First, for any finite $R$, we will have a finite $\tau_L$, so there will be a finite time on the statite's clock after which no ingoing light signal emitted will be received back. And second, as we decrease $\varepsilon$, i.e., as we move the reflection point of the last signal to be reflected closer to the horizon, $\tau_R$ increases without bound; the statite waits longer and longer to receive the last signal reflected back.

 

[pervect]

I'd like to give a quick overview of how I'd approach the problem conceptually, which might or might not be helpful to the original poster.

The general procedure is first to adopt a set of coordinates. Adopting a set of coordinates defines a metric tensor / line element, and vica versa. For the purposes of this exercise, one simply looks up the associated metric once one chooses the best set of coordinates. The metric tensor could in theory be found by solving Einstein's Field equations with the appropriate background condtions, but I'll assume that one is satisfied to use the known existing solutions. The EFE are partial differential equations, like Maxwell's equations for electromagnetism, but they're hard because they're nonlinear. But it's easiest to assume one just looks up the solution.

There are many choices for coordinates, because the OP is interested in the time light pulses are received, ingoing Eddington-Finklestein {EF} coordinates are a good choice. Kruskal is also good, but ingoing Eddingtion-Finklestein has as one coordinates a number which is equal to the number of radial ingoing flashes of light received from infinity by the infalling test particle (plus some offset) as one coordinates, which is very convenient. This coordinate is usually called v. See for instance https://en.wikipedia.org/wiki/Eddington–Finkelstein_coordinates. Kruskal has the same v coordinate, but it's paired with the "u" coordinate, which is an outgoing flash coordinate. To my mind it is easier to grasp the problem if one uses the v coordinate and the schwarzschild r cooordinate, which is what ingoing EF uses.

Upon writing and solving the geodesic equations (more below) one will have an expression for v(tau), v being the "flash number" EF coordinate, and r(tau), r being the schwarzschild r coordinate. Tau is the proper time, the time elapsed on a clock on the infalling clock, possibly plus some offset value. v is somewhat playing the role of "time" here, but it's a null coordinate $\partial_v$ is null, not timelike. Basically one can plot flash number vs proper time tau on one graph, and the r coordinate vs proper time tau on another graph, given the solution of the differential equations.

Given the metric, wiki describes how to write the 2nd order ordinary differential equations that describe the particles trajectory of a test particle in https://en.wikipedia.org/wiki/Solving_the_geodesic_equations. The procedure is different than the Newtonian procedure, but mathematically it all boils down to a set of second order differential equations, tghe same type of system of equations as the Newtonian problem.

While these geodesic equations could in principle be solved directly, it's much easier to take advantage of the existence of conserved quantities. https://www.fourmilab.ch/gravitation/orbits/ goes into the procedure for the Schwarzschild metric. The conserved quantity in this problem would generally be called "energy" or possibly "energy at infinity".

So all in all it's a fairly involved procedure, that requires a significant amount of background knowledge to carry out. But given the necessary background knowledge, it's "just" a matter of writing and solving a pair of second order ordinary differential equations. The details are messy, and time consuming. A program like SageMath that can do symbolic manipulations would be highly recommended. SageMath is free, it's not quite as convenient as some non-free programs. The manifolds package in SageMath is particularly applicable. There are some online papers and/or books that do go into detail with worked examples using SageMath for General Relativity, but I don't recall exactly where.

Non-free programs similar to SageMath would be Mathematica.

 

[PAllen]

Continuing along a different path than @PeroK , I've been intermittently following up on the approach described above, using results from the indicated papers. So, using Lemaitre style coordinates, adapted to a "free fall from platform" congruence, discussed at p.8 of the first paper referenced, and changing notation from the paper as follows (all in units where c=G=1, with the idea that mass is expressed in terms of SC radius in light seconds, and spatial units - including SC radial coordinate - are in light seconds, direct computations yield seconds):

- I use P for the platform SC radial coordinate, R for SC radius of the BH
- I express e as used in the paper using a definition given earlier in the paper:
$$P=R/(1-e^2)$$

Then the metric for platform based Lemaitre style coordinates is:

$$ds^2=dT^2-R(1-R/P)^{-1}(1/r-1/P)d\rho^2$$

suppressing angular coordinates since we are treating a purely radial problem. Using results given in the paper for dT and $d\rho$ in terms of SC differentials, one can derive that $$\rho-T=\int (R/r - R/P)^{-1/2} dr \tag {1.1}$$

The f(r) defined by the integral can be evaluated to: $$-\sqrt{\frac {rP} R (P-r)} -\frac {P^{3/2}} {\sqrt R} \arctan(\sqrt{P/r-1}) $$ which can be seen to be equivalent to the formula given in the second paper (eq. 3.10) referenced above (taking $\rho=0$ and expressing as T).

Putting in r=P you verify get zero as desired, and then r=R and r=0 are readily computed as probe proper times for free fall from P to SC radius, and then to singularity.

Note, that in these coordinates, the chosen free fall world line from the platform takes the trivial form $\rho=0$, T varying. Note that generally, $$\rho-T = costant$$ gives a hovering world line (constant computed as f(r)). In our set up, the constant zero means the hovering platform. This shows the useful fact that proper time along the platform world line matches T coordinate time (just plug r=P into the metric above).

Thus, referring to my diagram, the coordinates of event A (the drop event) are simply $(\rho,T)=(0,0)$, event of probe reaching horizon are $(0,-f(R))$, and the limiting coordinate for reaching the singularity are $(0,-f(0))$.

So far, straight forward. But what I wanted to do next was derive the light path from the platform world line ending on the event of probe at horizon, and also the path from platform world line to probe at singularity. If these were derived, the T coordinates of the start events on the platform world line would directly be proper times for the platform from drop event to corresponding signal times, the second one described being the last signal that can reach the probe.

Unfortunately, here I have hit a major snag. It seems that light paths are very complicated to express in these coordinates, and I am not sure when I will find a way around this. As I can currently express this, I would need to numerically solve a truly complicated differential equation.

I found a way to handle the light paths and can now give a complete alternative solution. Ideally, it should agree with @PeroK , as to numbers calculated, even if equivalence might be hard to show. But I suspect we might not agree and might not know which is right.

The key to handling light within the framework I started is to combine the platform-Lemaitre T coordinate as used above, with the SC r coordinate, rather than trying to use $\rho$. We have, from eqn (1.1) above:
$$d\rho-dT=(R/r-R/P)^{-1/2}dr$$ and thus $$d\rho=dT+(R/r-R/P)^{-1/2}dr$$
From the metric given in the quoted post (for light we must have $ds^2=0$), we get:
$$dT^2=R(1-R/P)^{-1}(1/r-1/P)d\rho^2$$
Substituting and rearranging a bit, gives quadratic:
$$(dr/dT)^2+2\sqrt{R(1/r-1/P)}(dr/dT)+(R/r-1)=0$$
Solving this gives:
$$dr/dT=-\sqrt {R/r-R/P}\pm \sqrt {1-R/P}$$
It turns out that taking the minus sign gives you ingoing light paths, while the plus gives outgoing light. Thus, for r=R, the outgoing light is frozen (derivative is zero). Further, for r<R, both derivatives are negative, implying both light directions are heading for r=0, as expected. Now, what we really want, to combine with earlier results, is T as a function of r, which we'll call g(r). So, we just need to take the reciprocal of the last equation and integrate. This is a surprisingly messy integral (I sought help from an online integrator). After a lot of algebra and simplification, I finally get:
$g(r)=\pm2R \sqrt{1-\frac R P} \ln(\left|\sqrt{1-\frac r P} \mp \sqrt {\frac r R - \frac r P}\right|)$
$~~~~~~~-\sqrt{RP}(1-\frac {2R} P) \arctan \left( \sqrt{\frac P r -1} \right) \pm r\sqrt {1- \frac R P}$
$~~~~~~~+\sqrt{rR(1-\frac r P)}+C$
Again, using the lower sign choices gives an ingoing light path, upper signs an outgoing path.
The constant is found by matching the light path to the world line events we previously computed. Thus, for the ingoing light to reach $(0,-f(0))$, we require $g(0)=-f(0)$. Then simply $g(P)$ gives the proper time since probe drop to send a light signal that will reach the probe as the probe reaches the singularity. Similarly, to get the platform proper time since probe drop for a signal to reach the probe at the horizon, we pick C to achieve $g(R)=-f(R)$, and then compute $g(P)$.
Later, I might compute for realistic numbers for a stellar BH, but for an easily computed case, I take the SC radius, R, as 1 light second, the platform r coordinate as 2 light seconds. Then the formulas will give answers in seconds.
For $-f(0)$, the probe proper time to reach the singularity, I get simply $\pi \sqrt 2$. This actually becomes the value of C, and then $g(P)=\sqrt 2(\pi-1)$. This is the time the platform must wait, per its own clock, after dropping the probe, to send a signal reaching the probe at its "singular death".
For $-f(R)$, the probe proper time to reach the horizon, I get $\sqrt 2 + (\pi/\sqrt 2)$. Then C to get $g(R)=-f(R)$ becomes $\sqrt 2 + \pi / \sqrt 2 + \ln 2 / \sqrt 2$. Finally, we have simply
$g(P) = (\pi + \ln 2)/ \sqrt 2)$ as the time the platform should wait (per its own clock) to send a signal to reach probe at horizon crossing.

Thus, you can see how fast everything happens even for a 1 light second BH. You send the signal to catch the probe at the horizon only about 2.7 seconds after dropping the probe, and then wait only another .3 seconds to signal it as it reaches the singularity.

 

[Mike S.]

I haven't kept up with this the past few days, but ... wow! Once you folks started doing math, oh *my* you did the math. Agreeing by different methods; I have no idea how long it'll take me to catch up. Many of the older threads linked below can now be updated based on your work - for example, https://www.physicsforums.com/threads/will-photons-fry-an-object-falling-into-black-hole.704578/ now has your answer that they light will blueshift by a little more than a factor of 2 overall, at least the part sent from 4GM to land on a freefalling probe.

Now there's one last straw I can think of to clutch at here (and you tend to do that if you're at risk of falling into a black hole). Hubble's law? If the time units of the Hubble constant corresponded to the Schwarzschild t, then I'm thinking the infalling object might move at an angle of 45 degrees or even more on the Kruskal diagram as the *space* it was in expanded faster than the speed of light. I'm very probably wrong, but I just had to toss it out there. :)

 

[PAllen]

I will now calculate this problem for a typical stellar mass BH. Eight solar masses seems to be a typical stellar BH. The SC radius of the sun is about 3 km, so such a BH would have an SC radius of 24 km. We take the platform to be at twice the SC radius, as in the OP (which works out nicely - a number of terms drop out for exactly this case). Thus, in light seconds, using my variables, we have: $R=8 \times 10^{-5}$ light-seconds, and $P= 1.6 \times 10^{-4}$ light-seconds.

First, we calculate the probe proper times to reach the horizon and singularity, using the f(r) I defined in my post #68. These are simply -f(R) and -f(0), respectively. Thus, 291 microseconds probe proper time to reach horizon, and 355 microseconds from drop to reach singularity.

Then using g(r) as for the simple case, we find the platform must send a signal to reach probe at horizon 217 microseconds after probe drop. The signal to reach probe at singularity must be sent 242 microseconds after probe drop. These are proper times for the platform - using its own local clock.

 

[PAllen]

Now there's one last straw I can think of to clutch at here (and you tend to do that if you're at risk of falling into a black hole). Hubble's law? If the time units of the Hubble constant corresponded to the Schwarzschild t, then I'm thinking the infalling object might move at an angle of 45 degrees or even more on the Kruskal diagram as the *space* it was in expanded faster than the speed of light. I'm very probably wrong, but I just had to toss it out there. :)

Expansion is irrelevant here. There are two contributions to the overall cosmological spacetime - the matter and energy (including dark matter) of the universe, and the cosmological constant (dark energy). The former contribution only applies when averaging over a region total matter/energy density similar to the universal average, and is approximately homogeneous. It doesn't apply at all within a galaxy, let alone a stellar region. The cosmological constant applies everywhere, but it is so tiny it could never have a measurable effect at less than galaxy cluster scale.

 

[PeterDonis]

If the time units of the Hubble constant corresponded to the Schwarzschild t

They don't. Hubble's law is not even applicable to a black hole spacetime.

 

[PAllen]

Since I used the rather unusual "platform adapted Lemaitre style" coordinates, unknown to me until I found this paper:

https://arxiv.org/abs/1911.05988 , page 8 etc.

I thought it would be worth discussing them. They only cover the region between a hovering platform and the singularity, They achieve a pure diagonal metric, in which free fall drops from the platform, as well as hovering world lines, are both straight lines. To clarify their character, I attach a picture showing the platform world line, horizon, and singularity, free fall paths, ingoing light signals, interior and exterior outgoing light paths, and several light cones. It is to scale for the case of a platform with twice the SC radial coordinate of the horizon.

 

[PeroK]

Then using g(r) as for the simple case, we find the platform must send a signal to reach probe at horizon 217 microseconds after probe drop. The signal to reach probe at singularity must be sent 242 microseconds after probe drop. These are proper times for the platform - using its own local clock.

This is what I get. It's $27\mu \ s$ and $30 \mu \ s$ per solar mass respectively.

 

[PAllen]

Can't seem to be completely done with this topic.

In my post #80, I define two functions, f(r) [defined in the quoted section], and g(r). Note that if taking differences of the g(r) for different r values, the constant of integration can be ignored. Systematizing the last part of that post, we can say:

The probe proper time for a probe dropped from a platform at SC r coordinate P, to reach some $r_0$ is $-f(r_0)$. This $r_0$ can be outside, at, or inside the horizon.

The platform proper time after probe drop to send a signal to catch the probe at $r_0$ is: $g(P)-g(r_0)-f(r_0)$. Noting that g(r) is strictly increasing as r decreases [assuming we pick the signs for ingoing light], it follows that the time the platform waits to send a signal to catch the probe at $r_0$ is always less than the probe time to reach $r_0$. This is in contrast to false arguments/intuitions that the platform would have to wait a long time or even infinite time. This is true for all values of $r_0$.

The above statements are completely general. The case of a platform at twice the SC radius, and $r_0$ values for the horizon and singularity, produce nice simplifications that allow the following simple exact expressions to hold:

1) probe proper time for probe to reach horizon: $R(\sqrt 2 + \pi/\sqrt 2)$
2) probe proper time for probe to reach singularity: $R\pi\sqrt 2$
3) platform proper time to wait to send signal to catch probe at horizon: $R(\pi + \ln (2))/\sqrt 2$
4) platform proper time to wait to send signal to catch probe singularity: $R \sqrt 2(\pi - 1)$

 

[pervect]

It turned out to be fairly easy to do a numerical solution of the infall problem for an initial r=2 r_s in SageMath. I am posting an image of the unevaluated worksheet to document the process of what commands need to be input. There are a fair number of comments and some non-essential displays. However, the program output isn't shown. I can try to post an image with the output present if there is interest.

Fair warning - I haven't used the program before, and I haven't compared my results to the various others posted.

Below is the .png of the unevaluated worksheet (sorry it got so squished in the conversion processes, viewing the image standalone with a right click should allow it to be readable).

And the plot output (the value of r as a function of the v coordinate - the v coordinate again represents the events at which regular ingoing light flashes from infinity are received, by the infalling test particle, the first flash being received at v=0.

 

[pervect]

The image of the worksheet was so bad, I decided to copy the commands with a few of the comments as that's the only way to make it legible.

version()
%display latex
M = Manifold(2, r"\mathcal{M}", structure='Lorentzian')
M
// EF is the ingoing Eddington Finklestein chart
EF.<v,r>=M.chart(r"v r:(0,+oo)")
EF
rs = var('r_s')
assume(rs > 0)
// Metric coefficients
g=M.metric()
g[0,0]=-1+rs/r
g[0,1]=1
g[1,0]=1
// define the start point p0
p0=M.point((0,2*rs),name='p_0')
// define the starting 4-velocity
v0=M.tangent_space(p0)((1/sqrt(1/2),0.0),name='v_0')
tau=var('tau',latex_name=r"\tau")
geod = M.integrated_geodesic(g, (tau, 0, 4), v0); geod
// display info about the geodesic system
geod.system(verbose=True)
// Solve and plot
sol = geod.solve(parameters_values={rs: 1}) # numerical integration
interp = geod.interpolate() # interpolation of the solution for the plot
graph = geod.plot_integrated()
graphI'll also append the system of differential equations generated:

Geodesic in the 2-dimensional Lorentzian manifold \mathcal{M} equipped with Lorentzian metric g on the 2-dimensional Lorentzian manifold \mathcal{M}, and integrated over the Real interval (0, 4) as a solution to the following geodesic equations, written with respect to Chart (\mathcal{M}, (v, r)):

Initial point: Point p_0 on the 2-dimensional Lorentzian manifold$\mathcal{M}$ with coordinates [0, 2*r_s] with respect to Chart $(\mathcal{M}, (v, r))$
Initial tangent vector: Tangent vector v_0 at Point p_0 on the 2-dimensional Lorentzian manifold $\mathcal{M}$ with components [2*sqrt(1/2), 0] with respect to Chart $(\mathcal{M}, (v, r))$

d(v)/dtau = Dv
d(r)/dtau = Dr
d(Dv)/dtau = -1/2*Dv^2*r_s/r^2
d(Dr)/dtau = 1/2*(Dv^2*r_s^2 + (2*Dr*Dv - Dv^2)*r*r_s)/r^3

 

[PAllen]

And the plot output (the value of r as a function of the v coordinate - the v coordinate again represents the events at which regular ingoing light flashes from infinity are received, by the infalling test particle, the first flash being received at v=0.
View attachment 297310

In terms of the functions I derived, r as a function of v is not expressible analytically (I doubt this is possible at all). However, v as a function of r is expressible analytically:
$$v(r)=\gamma(P) (g(P)-g(r)-f(r))$$
where P = 2 R, and R is SC radius, and $\gamma(P)$ is the time dilation factor between a stationary observer at P versus at infinity.

added:

In closed form, this becomes:

$$v(r) = \sqrt {r(2R-r)} + r -2R +2R \ln \left(\sqrt {1-r/2R} + \sqrt {r/2R}\right) + 4R \arctan \sqrt {2R/r-1}$$