Measuring Light Reflection in a Black Hole

[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.

 

[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 ...