General Relativity rocket puzzle

[PeterDonis]

So the crew could say that the buoy moves approximately like light, except for just a very short time at the end of the journey.

I strongly suggest that you run the numbers, using the equations in the page you linked to, before making this claim.

 

[PAllen]

This is true, but "degree of flatness" is simply "range over which the Riemann tensor components are acceptably small". That's not the same as "change in proper acceleration of a stationary observer". See below.

Ok, also see below.

This can't be a correct criterion of flatness; it's much too strong, since even the flat spacetime scenario does not satisfy it. (Just compute the change in proper acceleration between two Rindler observers, one in the rocket as given in the scenario and one a million light-years above him, who is co-located with buoy #2 at the end of the scenario.)

Well, for the Rindler case, the difference in proper acceleration over any length approaches zero as distance from horizon approaches infinite. So my criterion could trivially be met in Rindler coordinates ... but not with the proper acceleration at one end of the length having 1.5g. Yet, you would still have exact flatness, with one end at 1.5g.

However, I don't see any way to propose physically motivated criterion on Riemann components. So what I propose to investigate when I get a chance is the notion of sectional curvature. Picking a criterion motivated by a sphere, we would want this to be related to the size of the r-t region of interest, e.g. 1/ (100 times largest 'distance' of region) squared. This would be analogous to requiring a radius of curvature of 100 times the largest distance within a section of a 2-sphere. The distances of the region might be taken to be proper length and proper time of curves within the rt region of interest.

Also, remember that the coordinate $r$ is not the same as radial distance, and since we must allow for the possibility that the LIF in question is close to the black hole's horizon, it might not even be approximately the same.

Yes, I realized I made 'a couple of approximations too many' in my prior analysis, and need to account that however large a horizon is, you have infinite distance outside of it, and proper acceleration becomes infinite on approach to the horizon, not on approach to the nonexistent 'center'. I used some approximations valid in scenarios when this didn't matter, - but they may matter very much in this case.

Also, I over simiplified the definition of the region of interest. It can't literally be defined in terms of the r coordinate values.

 

[PeterDonis]

So my criterion could trivially be met in Rindler coordinates ... but not with the proper acceleration at one end of the length having 1.5g. Yet, you would still have exact flatness, with one end at 1.5g.

Exactly; that's why your criterion won't work.

I don't see any way to propose physically motivated criterion on Riemann components.

Sure there is. Suppose the extent of the local inertial frame is a distance $L$. Then an upper bound for the tidal acceleration between any two points in the LIF is $RL$, where $R$ is the largest Riemann tensor component anywhere in the LIF. For Schwarzschild spacetime, the radial Riemann tensor component is $2M / r^3$, so $R$ will be that with the smallest value of $r$ anywhere in the LIF plugged in. If we take the smallest value of $r$ to be $2M$ (i.e., the LIF is just above the horizon), then we have an upper bound for the tidal acceleration anywhere in the LIF of

$$
a_t = \frac{L}{4M^2}
$$

The flatness criterion is then just that this acceleration must be much smaller than the smallest proper acceleration of a Rindler observer anywhere in the LIF. To put it another way, we want to ensure that geodesic deviation is much smaller than the smallest path curvature of any Rindler worldline within the LIF; that is what allows us to treat the geodesic worldlines as straight to a good enough approximation to realize the scenario.

 

[PeterDonis]

what I propose to investigate when I get a chance is the notion of sectional curvature.

This should turn out to be the same as the criterion I proposed in my previous post. Basically it's just another way of looking at the requirement that we want geodesic worldlines to be straight to a good enough approximation compared to the smallest path curvature we have to deal with.

 

[PAllen]

This should turn out to be the same as the criterion I proposed in my previous post. Basically it's just another way of looking at the requirement that we want geodesic worldlines to be straight to a good enough approximation compared to the smallest path curvature we have to deal with.

Same concept, but some key differences. My criteria would involve all Riemann components that are combinations of r and t subscripts. I think the time components are significant because of the long 'time' distance involved in the scenario.

[edit: I see that the expression you give is the only non-zero r/t component expressed in a local orthonormal basis. So, yes, I would think it is a logically equivalent criterion. On a coordinate basis, there are two relevant nonzero components, and neither is what you gave.]

 

[jartsa]

I strongly suggest that you run the numbers, using the equations in the page you linked to, before making this claim.

So the deceleration of the rocket takes ten proper years. Two of those years are spent to decelerate from the speed where gamma is 10, to stop. So 8 years are spent at speed where gamma is over 10.

I calculated that like this:

From Baez's page: "it's easier to use units of years for time and light years ("ly") for distance. Then c=1 ly/yr and g≃1.03 ly/yr²"

If proper deceleration is 1.5 light years/year² then rapidity goes from 3 light years / year to zero in 2 years, so gamma goes from 10 to zero in 2 years, because gamma=cosh(rapidity) and cosh(3)=10.0Well okay I made an error before, 2 years out of ten is not "just a very short time"

 

[1977ub]

I was describing what the ship has to do to switch from accelerating to decelerating: it has to turn around from its nose pointing towards buoy #3 to its tail pointing towards buoy #3. But I can see how the phrase could be ambiguous. I have seen this usage in plenty of sci-fi stories (for example, Larry Niven's stories involving ramscoop ships), but I admit I have not seen it in the scientific literature.

It came to me a bit later. Thank you for pointing it out. Normally when a vehicle "turns around" it has changed the direction it is *moving*. A rocket might have thrusters on both ends.

 

[PAllen]

Just a thought - the critical thing to find for a flatness criterion is the innermost r (in its standard SC definition), corresponding to the 'lowest relevant position' of the lowest buoy (the one that ends up momentarily at rest with the hovering rocket at the end of the process). To me at least, this is not obvious; I would need some real thought and calculation to figure it. The reason this is critical is, pretty obviously, this is where curvature would be maximum over the whole spacetime region of interest. It is here that a curvature constraint should be applied - my proposal being sectional r-t curvature < 10^-16 ly^-2.

 

[PeterDonis]

On a coordinate basis, there are two relevant nonzero components, and neither is what you gave.

Actually, in Schwarzschild coordinates, the r/t component is what I gave. (But you're right that this particular quantity is also the same in an orthonormal basis. That's one of the counterintuitive features of Schwarzschild coordinates.)

 

[PeterDonis]

Well okay I made an error before, 2 years out of ten is not "just a very short time"

Exactly.

 

[PeterDonis]

Normally when a vehicle "turns around" it has changed the direction it is *moving*.

I've already agreed that the expression was ambiguous, yes. It can also mean a change of orientation, not direction.

A rocket might have thrusters on both ends.

In principle this could be true, yes, but I don't think such a configuration is anything close to common enough to qualify as a default assumption (since it's never existed at all AFAIK).

 

[PAllen]

Actually, in Schwarzschild coordinates, the r/t component is what I gave. (But you're right that this particular quantity is also the same in an orthonormal basis. That's one of the counterintuitive features of Schwarzschild coordinates.)

Not as far as I know. For example:

https://physics.stackexchange.com/q...he-riemann-tensor-of-the-schwarzschild-metric

Also, consistent with this is:
https://en.m.wikipedia.org/wiki/Schwarzschild_metric

See the curvatures section

 

[PeterDonis]

the critical thing to find for a flatness criterion is the innermost r (in its standard SC definition), corresponding to the 'lowest relevant position' of the lowest buoy (the one that ends up momentarily at rest with the hovering rocket at the end of the process).

If we want to include this buoy in the scenario, yes. As at least a partial solution, you could also just find the $r$ of the rocket (assuming it was stationary) and treat that as the lowest $r$ (so the only buoy in the scenario is buoy #2). But a solution that includes buoy #3 raises some interesting issues that a partial solution only including buoy #2 does not.

o me at least, this is not obvious; I would need some real thought and calculation to figure it.

You don't need to know the lowest $r$ (or the highest one, for that matter) to find a solution; you can compute the range of $r$ after you've found the solution and then check the tidal gravity in order to verify that the solution has acceptable flatness.

Or, alternatively, you can make a reasonable assumption about the lowest $r$, following from the assumption that the massive body is a very, very, very large black hole. Can you see what that assumption would be? (Note that the lowest $r$ on this assumption is lower than the $r$ of the rocket.)

It is here that a curvature constraint should be applied

Yes, that's correct: the lowest $r$ will have the highest curvature, so Riemann tensor components at that $r$ will provide an upper bound on tidal gravity throughout the LIF.

my proposal being sectional r-t curvature < 10-16 ly-2.

This is reasonable (it's more or less the constraint I came up with).

 

[PeterDonis]

Not as far as I know.

Hm, yes, I forgot that raising and lowering indexes does make a difference in Schwarzschild coordinates (whereas in an orthonormal basis it doesn't, except possibly for a sign change). Strictly speaking, the Riemann component $R_{1212}$ in Schwarzschild coordinates, with all lower indexes, is the one that matches the orthonormal basis component. For this problem, the orthonormal basis component is the relevant one.

 

[PAllen]

If we want to include this buoy in the scenario, yes. As at least a partial solution, you could also just find the $r$ of the rocket (assuming it was stationary) and treat that as the lowest $r$ (so the only buoy in the scenario is buoy #2). But a solution that includes buoy #3 raises some interesting issues that a partial solution only including buoy #2 does not.

It seems to me that #3 is important. Consider the description from the LIF of these two buoys. Rocket is initially moving near c from #2 to #3, but decelerating such as to momentarily at rest at #3 (after a bit over a million years, in these LIF coordinates). Then, the spacetime region of interest is the rectangle of a million light years across, and a bit over a million years in the t direction. The lines of constant r go generally from the lower right to the upper left (assuming #2 is placed to the right of #3, with time going up). Then, the maximum curvature is at the lower left corner, corresponding to some unknown r below the rocket's position. This is where I would claim it is most valid to apply the curvature criterion. This is the only thing that guarantees a desired degree of resemblance of the this whole LIF description of the rocket's 'trip' from #2 to #3.

 

[PeterDonis]

the maximum curvature is at the lower left corner, corresponding to some unknown r below the rocket's position. This is where I would claim it is most valid to apply the curvature criterion.

Yes, this is a valid point. But note that, as I said in post #52, including buoy #3 raises some interesting issues (one of which is that you cannot adopt the reasonable assumption for the lowest $r$ that I hinted at in post #52, while still maintaining another reasonable assumption that we both made much earlier in this thread).

 

[PAllen]

Yes, this is a valid point. But note that, as I said in post #52, including buoy #3 raises some interesting issues (one of which is that you cannot adopt the reasonable assumption for the lowest $r$ that I hinted at in post #52, while still maintaining another reasonable assumption that we both made much earlier in this thread).

Well, I can think of a well defined recipe to fully include #3, and meet all requirements, but I am not enthusiastic about executing it.

1) Compute relative speed between rocket and buoy #2 initially. This is just application of the appropriate relativistic rocket equation. Using pure SR for this is valid by problem requirements.This is very easy.

2) Write out Fermi-Normal coordinates based on general radial geodesic in SC metric, including only first order terms (since by construction, these will eventually suffice). Match this to the requirements of buoy #2 (based on relative speed to a stationary path). Assuming t=0 at the start, x position of minus 1 million ly will correspond to some SC r value (presumably, in terms of M and lower right corner r). This is, for me, quite cumbersome.

3) Apply the sectional curvature condition to the r found in (2). For an r that satisfies, we are guaranteed that the approximations in (2) are ok. This should give us a minimum r as function of M that is flat enough over a large enough spacetime region. This is moderately cumbersome for me.

4) Now require that the proper acceleration at the r of (buoy #2, t=0) is 1.5 g. This should determine M. This should be very easy.

 

[PeterDonis]

I can think of a well defined recipe to fully include #3, and meet all requirements, but I am not enthusiastic about executing it.

You seem to be missing a much simpler procedure. Here is a hint at it: for a stationary observer sufficiently close to a black hole's horizon, the hole's horizon is also that stationary observer's Rindler horizon.

 

[PAllen]

You seem to be missing a much simpler procedure. Here is a hint at it: for a stationary observer sufficiently close to a black hole's horizon, the hole's horizon is also that stationary observer's Rindler horizon.

I am reluctant to assume that a minimum mass solution (that properly includes #3) is very close to the horizon.

 

[PeterDonis]

I am reluctant to assume that a minimum mass solution (that properly includes #3) is very close to the horizon.

You shouldn't be. The variation in "acceleration due to gravity" within the LIF needs to be indistinguishable from what it is for a family of Rindler observers. The only way to ensure that in Schwarzschild spacetime, if you want the rocket to be stationary with respect to the hole, is to be close enough to the hole's horizon so that the hole's horizon is also the rocket's Rindler horizon.

 

[PeterDonis]

I am reluctant to assume that a minimum mass solution (that properly includes #3) is very close to the horizon.

In addition to the reason I gave in post #59, there's another reason as well. Consider: buoy #2 is moving upward relative to the rocket with a gamma factor of 1.6 million (the value @jartsa posted earlier, which looks ok to me) but does not escape; it comes to rest a million light-years above the rocket. So we have an escape gamma factor larger than 1.6 million. That requires the radial coordinate $r$ of the rocket to be very close to $2M$ (because of the formula for escape velocity).

 

[PAllen]

In addition to the reason I gave in post #59, there's another reason as well. Consider: buoy #2 is moving upward relative to the rocket with a gamma factor of 1.6 million (the value @jartsa posted earlier, which looks ok to me) but does not escape; it comes to rest a million light-years above the rocket. So we have an escape gamma factor larger than 1.6 million. That requires the radial coordinate $r$ of the rocket to be very close to $2M$ (because of the formula for escape velocity).

All that is required is that SC radius (R) be close to r in ratio (in natural units) for escape velocity to approach c. The magnitude of difference (rather than ratio) can be made as large as you want.

I have a counter argument that placing the Rindler horizon for the rocket at the BH horizon is a false solution. For 1.5 g, this places the Rindler horizon about .65 ly away from the rocket in an LIF, which we are assuming is effectively flat. This means that buoy #3 is almost a million ly below the horizon. However, this part of the horizon is (in GR terms) the white hole horizon. This, per se, is not a problem, but it is so far below that curvature could hardly be minimized there, and I suspect you could not have #3 exist in #2 simultaneity at all (as of when the rocket and #2 coincide). Instead, #3 would have to come into existence later, per #2, emerging from the white hole singularity. I would call this an invalid realization of the problem.

Instead, the solution I have in mind has the rocket coincide with a very distant (in absolute terms) stationary observer. Other Rindler observers per the rocket then have nothing to do with stationary observers per the BH, and the Rindler horizon has nothing to do with the BH horizon (which is much further away, to allow buoy #3 to exist for the required span). It is conceivable that this then makes it impossible for the escape behavior of #2 to be realized, but I am not yet convinced of that. If so, then IMO opinion, there is no valid solution to problem using a BH, which is not a problem at all. There is no requirement per the POE that an arbitrarily large SR scenario be realizable in some class of metric.

 

[PAllen]

I now conclude that any escape velocity at all can be arranged to occur at stationary observer whose proper acceleration is e.g. 1.5g. However, this fully determines R and r. Then, either the other problem constraints can be met, or they can't. I haven't worked that yet. In any case, if this is possible, this is the only solution I would call correct. If not, then I would say no correct solution is possible.

 

[PAllen]

On further thought, I don't think escape velocity for #2 is relevant at all. What is strictly required is that #3 reach the stationary rocket at zero relative speed in its free fall trajectory. I see no specific requirement on #2, per se (there are some implicit requirements, but they should all fall out the flatness criterion).

 

[PeterDonis]

All that is required is that SR radius (R) be close to r in ratio (in natural units) for escape velocity to approach c.

Huh? The formula for escape velocity in Schwarzschild spacetime is $\sqrt{2M / r}$, where $r$ is the Schwarzschild $r$, i.e., the areal radius. There's no "SR radius" anywhere in there. The gamma factor corresponding to that escape velocity is $1 / \sqrt{1 - 2M / r}$; you can easily calculate the value of $r$ for which that gamma factor is 1.6 million.

I now conclude that any escape velocity at all can be arranged to occur at stationary observer whose proper acceleration is e.g. 1.5g.

This is true (just adjust the mass of the gravitating body appropriately), but irrelevant to the problem. In the problem as given, we know the velocity of buoy #2 relative to the rocket (gamma factor 1.6 million), and we know buoy #2 does not escape (it comes to rest relative to the rocket a million light-years above it). In a curved spacetime around a much smaller gravitating mass, for example a planet somewhat larger than the Earth (with a surface gravity of 1.5 g), an object flying upward in free-fall with a gamma factor of 1.6 million would escape; it would never come to rest relative to the planet, not even 10 million light-years away. (In fact the effect of the planet's gravity on the flying object's speed in that case would be negligible.)

Perhaps you are confused by the fact that in the flat spacetime case, there is no such thing as "escape velocity". That's true, but again, it's irrelevant to the problem, because we are not using any concept of escape velocity in the flat case. We are only using it in the curved case. And in the curved case, the fact that buoy #2 does not escape is obvious: it comes to rest relative to the rocket! No object on a free-fall escape trajectory will do that. So in order to satisfy the EP, we must be able to find a patch of a curved spacetime that satisfies the flatness criterion over the required range, and has strong enough gravity to make buoy #2 come to rest relative to the rocket, despite its huge initial gamma factor of 1.6 million.

On further thought, I don't think escape velocity for #2 is relevant at all.

I have no idea why you would think that. The fact that buoy #2 does not escape is required for the scenario. See above.

 

[PAllen]

Huh? The formula for escape velocity in Schwarzschild spacetime is $\sqrt{2M / r}$, where $r$ is the Schwarzschild $r$, i.e., the areal radius. There's no "SR radius" anywhere in there. The gamma factor corresponding to that escape velocity is $1 / \sqrt{1 - 2M / r}$; you can easily calculate the value of $r$ for which that gamma factor is 1.6 million.

Just a silly typo I corrected. I meant SC radius. As for the rest, M corresponds to some R (SC radius). For a chosen ratio of R and r (achieving an arbitrary escape velocity), r - R can be made as large as you want.

This is true (just adjust the mass of the gravitating body appropriately), but irrelevant to the problem. In the problem as given, we know the velocity of buoy #2 relative to the rocket (gamma factor 1.6 million), and we know buoy #2 does not escape (it comes to rest relative to the rocket a million light-years above it).

Perhaps you are confused by the fact that in the flat spacetime case, there is no such thing as "escape velocity". That's true, but again, it's irrelevant to the problem, because we are not using any concept of escape velocity in the flat case. We are only using it in the curved case. And in the curved case, the fact that buoy #2 does not escape is obvious: it comes to rest relative to the rocket! No object on a free-fall escape trajectory will do that. So in order to satisfy the EP, we must be able to find a patch of a curved spacetime that satisfies the flatness criterion over the required range, and has strong enough gravity to make buoy #2 come to rest relative to the rocket, despite its huge initial gamma factor of 1.6 million.
I have no idea why you would think that. The fact that buoy #2 does not escape is required for the scenario. See above.

But escape velocity per se, is not part of the problem, period. What is, that you ignore, is that #3, in free fall, achieve its peak altitude at the stationary observer, and the #2 is a million ly away at this time per #3 simultaneity.

 

[PeterDonis]

For 1.5 g, this places the Rindler horizon about .65 ly away from the rocket in an LIF, which we are assuming is effectively flat. This means that buoy #3 is almost a million ly below the horizon.

Bingo! You have hit on a key issue raised by including buoy #3 in the scenario.

However, this part of the horizon is (in GR terms) the white hole horizon.

Yes! In other words, to realize the scenario including buoy #3, we have to use maximally extended Schwarzschild spacetime. (Which, btw, should suggest an obvious coordinate chart to use to make it easier to show the correspondence with the flat spacetime case.)

it is so far below that curvature could hardly be minimized there

No. I strongly suggest that you do the calculation instead of guessing. Look at it this way: if we can find an LIF that is flat enough over a million light-years above the horizon (which just requires making the mass of the hole much greater than a million light-years--the flatness criterion you already posted allows us to estimate what mass would be required), a hole with a mass larger by a factor of order unity will allow you to extend the same LIF a million years below the horizon. (Note that it's a million years below, not a million light-years below, because we are extending the LIF in the timelike direction, i.e., into the past, not a spacelike direction.)

I suspect you could not have #3 exist in #2 simultaneity at all

You are wrong. You can. The events "buoy #3 meets the rocket" and "buoy #2 achieves maximum altitude above the rocket" are spacelike separated, and an LIF can be found in which they are simultaneous.

 

[PAllen]

You are wrong. You can. The events "buoy #3 meets the rocket" and "buoy #2 achieves maximum altitude above the rocket" are spacelike separated, and an LIF can be found in which they are simultaneous.

But that's not the problem case. It is buoy #2 meets the rocket, and buoy #3 is spacelike separated a million ly below the white hole horizon (and that this can be made flat enough such that this spacelike separation is mutually simultaneous in an approx. SR sense.

Note, of course, there is no issue with a free fall trajectory escaping a white hole horizon.

Perhaps this is achievable, I should calculate this, not guess.

 

[PeterDonis]

M corresponds to some R (SC radius). For a chosen ratio of R and r (achieving an arbitrary escape velocity), r - R can be made as large as you want.

It's true that, for a very large $M$, the size in light-years of $r - 2M$ will be large. But that doesn't mean that $r - 2M$ in mass of the hole units will be large.

In the case under discussion, yes, you will find that the altitude at which the escape velocity is a gamma of 1.6 million is much higher than the altitude of the rocket, in light-years. But that's just because the hole is so huge that even, say, a million light-years above the horizon is still only a miniscule increment in the radial coordinate $r$, i.e., in the areal radius.

 

[PeterDonis]

that's not the problem case. It is buoy #2 meets the rocket, and buoy #3 is spacelike separated a million ly below the white hole horizon (and that this can be made flat enough such that this spacelike separation is mutually simultaneous in an approx. SR sense.

This can be done as well. The coordinates of these two events are the same in the LIF as in the flat spacetime case: for buoy #2, when it is co-located with the rocket, t is minus a million years and x is a million light-years + 2/3 of a light-year (roughly), and for buoy #3 at the event that is simultaneous in the LIF/flat case, t is minus a million light-years and x is 2/3 of a light-year (roughly). The latter event, as you note, is below the Rindler horizon of the rocket, which in the LIF in the curved case is also below the hole's horizon (the white hole/past horizon).

Then, at the other pair of events, we have buoy #3 co-located with the rocket at t = 0, x = 2/3, and buoy #2 at its maximum altitude at t = 0, x = 1 million + 2/3. Everything in this LIF looks just like the flat spacetime case.

 

[PAllen]

Despite the above being a possible solution, I still see nothing clear preventing that the Rindler horizon have nothing to do with WH/BH horizon, and instead be far enough the true horizon such that the whole spacetime rectangle is outside the horizon. Only the rocket needs to coincide with a stationary world line. The other constraints (as I see it) are that #3, in free fall, reach its peak altitude at the the rocket, and that #2 is a million ly away at this point, per #3. Do you have any argument that this mandates that Rindler horizone coincide with the BH horizon? I don't see it in anything you've said.

Note, I don't see it as required for #2 to be at peak altidude when #3 reaches the rocket. It must be the right distance away, and not be moving very fast per #3 in the LIF, but what it does after #3 reaches the rocket is no longer part of the problem.

 

[PeterDonis]

I still see nothing clear preventing that the Rindler horizon have nothing to do with WH/BH horizon, and instead be far enough the true horizon such that the whole spacetime rectangle is outside the horizon

You can't do this and still have the rocket stationary relative to the hole.

Only the rocket needs to coincide with a stationary world line.

Of course. Obviously a free-falling worldline in Schwarzschild spacetime can't be stationary.

The other constraints (as I see it) are that #3, in free fall, reach its peak altitude at the the rocket, and that #2 is a million ly away at this point, per #3.

Buoys #2 and #3 are always at rest relative to each other, so if buoy #3 is at rest relative to the rocket at this point, buoy #2 must be as well (in the simultaneity convention of the LIF, which is just the simultaneity implied by the common rest frame of the buoys).

Do you have any argument that this mandates that Rindler horizone coincide with the BH horizon? I don't see it in anything you've said.

See posts #59 and #60.

I don't see it as required for #2 to be at peak altidude when #3 reaches the rocket.

If the rocket is stationary, it has to be. In addition to my comment above, in the flat spacetime case, this corresponds to the fact that, if the rocket keeps accelerating in the same direction (i.e,. toward buoy #2) after it meets buoy #3, it will start getting closer to buoy #2 again.

 

[PeterDonis]

what it does after #3 reaches the rocket is no longer part of the problem.

Not in terms of constructing the LIF, since the problem only requires the LIF to extend to the point where buoy #3 meets the rocket; but that doesn't mean what buoys #2 and #3 do after that point is not determined (assuming that they both remain in free fall).

 

[PAllen]

You can't do this and still have the rocket stationary relative to the hole.

This is just false, taken by itself. I can trivially make a stationary world line with 1.5 g be 10 million ly from the SC radius.

 

[PAllen]

I fail to see how #59 and #60 require that the rocket Rindler horizon correspond to the BH/WH horizon. In fact, #59 is simply a statement, without justification, of which I am utterly unconvinced and am looking for a rationale for. #60 I largely irrelevant, and I have explained, and you have admitted that this can be achieved as far from the SC radius as desired.

Specifically, neither, in any way explains what is wrong with a scenario where the only Rindler observer in the LIF that corresponds exactly to a BH stationary observer is the rocket.

I'll be concrete. Let's make r (of rocket) - R (SC radius) be e.g. > 3 million ly. We can chose R such that acceleration of stationary observer at r is 1.5 g. There is a free fall geodesic starting 1 million ly below r whose peak altitude is r. We can set this up so R is very large compared to 1 million ly (thus achieving desired flatness over the region). I do not see why this cannot work, especially not with anything said in #59 or #60.

I'm not claiming for sure it is possible, I just don't find any sufficient arguments in anything you've said to preclude it.