I General Relativity rocket puzzle

[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.

 

[PeterDonis]

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.

Since all Rindler observers in the LIF are stationary relative to each other, if one is stationary relative to the hole, all of them are stationary relative to the hole. If you think about how the correspondence between the Killing fields works, this should be obvious.

 

[PAllen]

Since all Rindler observers in the LIF are stationary relative to each other, if one is stationary relative to the hole, all of them are stationary relative to the hole. If you think about how the correspondence between the Killing fields works, this should be obvious.

However, note that if the Rindler kvf is exact against the BH, the buoy kvf is only approximate. I don’t see a reason we can’t make the Rindler kvf approximate as well.

 

[PeterDonis]

if the Rindler kvf is exact against the BH, the buoy kvf is only approximate

The buoy kvf is not "approximate" in the LIF; it's exact. It just doesn't correspond to any kvf in the global Schwarzschild spacetime (not even approximately; it can't since there is only one timelike kvf, and that only above the horizon, and that's the one the Rindler kvf in the LIF corresponds to).

 

[PeterDonis]

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.

Please show an explicit set of numbers. I am unable to find a solution with $r - R$ 3 million ly and proper acceleration at $r$ of 1.5g. For that difference in radial coordinates $r - R$, I find a maximum of proper acceleration as $R$ is varied of about $6.4 \times 10^{-8}$ g, at an $R$ of around 6 million ly.

 

[PAllen]

Using units where c=G=1, and using R/2 for M, the proper acceleration formula for stationary observers in SC metric only involves R and r. It can be arranged into a quadratic R. Then substituting R+d for r gives you a quartic polynomial in R, for any value of proper accelaration and d. I did not verify it has a solution, I just assumed it did. I will try to play with this on my own, converting numbers to natural units to use the quartic in simple form.

 

[PeterDonis]

I am unable to find a solution with $r - R$ 3 million ly and proper acceleration at $r$ of 1.5g.

I now know why. See below.

I did not verify it has a solution, I just assumed it did.

You shouldn't have assumed. Whether there is a solution for $a = 1.5$ depends on the value of $r - R$. See below.

After quite a bit of spreadsheet tinkering, it occurred to me that this can be investigated analytically. Let $x = r - R$. Then we can write the formula for proper acceleration

$$
a = \frac{R}{2 \left(R + x\right)^2} \sqrt{\frac{R + x}{x}} = \frac{1}{2 \sqrt{x}} R \left( R + x \right)^{-3/2}
$$

Taking the derivative with respect to $R$ is straightforward:

$$
\frac{da}{dR} = \frac{1}{4 \sqrt{x}} \left( R + x \right)^{-5/2} \left( 2 x - R \right)
$$

This obviously goes to zero at $R = 2x$. Taking the second derivative and finding that it is negative at this zero point shows us that it is a maximum. So for a given value of $x$, there is a maximum possible proper acceleration; since we have $R = 2x$, we have $R + x = 3x$, and the formula for $a$ becomes

$$
a = \frac{1}{3 \sqrt{3} x} = \frac{0.19}{x}
$$

So for $x = 3 \times 10^6$, we get roughly $a = 6.3 \times 10^{-8}$ for the maximum proper acceleration, which is close to what I got from spreadsheet tinkering (I posted that in post #78).

 

[PAllen]

Ok, I finally agree that the unique solution is match the Rindler horizon with the BH horizon, making R very large compared to a million ly, and using the white hole region for buoy #3. The choice of R both satisfies the curvature condition, and ensures that buoy #3 is not far below the white hole horizon (relatively speaking), and near flatness extends here.

This also suggests that even my first over simplified approximations were telling me something - the whole solution can’t be built outside the horizon.

 

[PeterDonis]

I finally agree

Ok, good! Just to expand on the solution some: I chose $R = 2M = 10^8$ (all units are years/light-years, and acceleration in g; it's a nice coincidence that 1 g acceleration is close enough to 1 light-year per year squared). This equates to roughly $10^{20}$ solar masses.

For the radial coordinate of the rocket, at a physical distance of $2/3$ above the horizon, in the terminology of post #80 and the posts leading up to it, we have $r - R = 10^{-9}$.

For the radial coordinate of buoy #2 at maximum altitude, which is a physical distance of $10^6$ above the horizon, I get $r - R = 2.5 \times 10^3$. This quantity is also $R - r$, to a good approximation, for buoy #3 at minimum altitude; note the switch in signs, indicating that now we are looking at years below the horizon instead of light-years above the horizon--but these are radial coordinate years/light-years, not physical years/light-years.

As a check on flatness, we can estimate tidal acceleration as $RL / r^3$, where $L$ is the physical distance/time above/below the horizon at the edges of the LIF and $r$ is the corresponding radial coordinate. However, as can be seen from the above, $r - R$ is some 5 orders of magnitude smaller than $R$, so to a good approximation we can just plug in $r = R$ to obtain $L / R^2$. For $L = 10^6$, the maximum tidal acceleration is thus about $10^{-10}$.

 

[PeterDonis]

And just to follow up on the hint I gave in a previous post: since we are working in maximally extended Schwarzschild spacetime, we can view our LIF as a small patch of the Kruskal spacetime diagram which is very close to the origin (where the past and future horizons meet): in fact it will be a small rectangle a little bit to the right of the origin and extending down below the past horizon.

The line element in Kruskal coordinates is

$$
ds^2 = \frac{4 R^3}{r} e^{-r / R} \left( - dT^2 + dX^2 \right)
$$

If we plug in $r = R$, we obtain

$$
ds^2 = \frac{4 R^2}{e}\left( - dT^2 + dX^2 \right)
$$

So if we are sufficiently close to $r = 2M$ (which, as we can see from the previous post, we are if we are within the LIF), we can simply rescale the $T$ and $X$ coordinates to obtain the Minkowski line element; we just have $t = (2R / \sqrt{e}) T$ and $x = (2R / \sqrt{e}) X$.

The general formula for $r$ as a function of the Kruskal coordinates can be put in this convenient form:

$$
\frac{r}{R} - 1 = W_0 \left( \frac{X^2 - T^2}{e} \right)
$$

where $W_0$ is the Lambert $W$ function. For small values of its argument, the $W$ function can be approximated by its argument, so we have

$$
r - R = \frac{R}{e} \left( X^2 - T^2 \right)
$$

Rewriting this in terms of the LIF $t$ and $x$ coordinates gives

$$
r - R = \frac{1}{4R} \left( x^2 - t^2 \right)
$$

Plugging in the various $x$ and $t$ values known from how things look in the LIF should give the $r - R$ values I gave in my previous post.

 

[sweet springs]

Your drawings are too incomprehensible to me for me to comment on them.

The line buoy#1-buoy#2-buoy#3 has the same direction with gravity force or perpendicular to it was not sure for me.
I draw a perpendicular case. If perpendicular your statement

(1) We have three buoys, all free-falling upward in the body's gravitational field, at rest relative to each other, and one million light years spacing between them.

is surely satisfied.

 

[PeterDonis]

The line buoy#1-buoy#2-buoy#3 has the same direction with gravity force or perpendicular to it was not sure for me.

The only relevant coordinates in the problem are time and radial distance; there are no angular coordinates. The buoys are separated radially.

 

[PeterDonis]

match the Rindler horizon with the BH horizon

Just to belabor this a little more...

The construction using Kruskal coordinates that I gave in post #83 is one way to get to the Rindler viewpoint (just transform the LIF coordinates I obtained to Rindler coordinates). However, there is another way to approach it as well. Let $\varepsilon = r / R - 1$. We then obtain formulas for the proper acceleration $a$ and the distance above the horizon $d$ in terms of $\varepsilon$. We have

$$
a = \frac{1}{2 R \left( 1 + \varepsilon \right)^2} \sqrt{\frac{1 + \varepsilon}{\varepsilon}}
$$

For small $\varepsilon$, we can set $1 + \varepsilon = 1$ to obtain

$$
a = \frac{1}{2 R \sqrt{\varepsilon}}
$$

Now for the distance $d$; the general formula for that is

$$
d = \int_{R}^{r} \sqrt{\frac{r}{r - R}} dr = \sqrt{r (r - R)} + R \ln \left( \sqrt{\frac{r}{R}} + \sqrt{\frac{r}{R} - 1} \right)
$$

Substituting gives

$$
d = R \left[ \sqrt{\varepsilon (1 + \varepsilon)} + \ln \left( \sqrt{1 + \varepsilon} + \sqrt{\varepsilon} \right) \right]
$$

Making the same small $\varepsilon$ approximation as above, and using the fact that, for small $x$, $\ln(1 + x) = x$, we obtain

$$
d = 2 R \sqrt{\varepsilon} = 1 / a
$$

In other words, we have shown that, for stationary observers sufficiently close to the BH horizon, their proper acceleration is the reciprocal of their proper distance above the BH horizon--in other words, the BH horizon is their Rindler horizon.

For reference, the values of $\varepsilon$ for the solution I gave a few posts ago are $\varepsilon = 10^{-17}$ for the rocket, and $\varepsilon = 2.5 \times 10^{-5}$ for buoy #2 at maximum altitude and buoy #3 at minimum altitude. Plugging those values into the above formulas should give the appropriate results.

 

[sweet springs]

there are no angular coordinates. The buoys are separated radially.

(1) We have three buoys, all free-falling upward in the body's gravitational field, at rest relative to each other, and one million light years spacing between them.

So I am not sure the two statements go along with. Is the latter just an initial condition or surely kept during their free-falling motion in some frame of reference ?

Your very first case of SR in post #1 tells that rocket passengers observe by Lorentz contraciton the buoys are thin film shape and the distance between the buoys are much shorter than 1 milliion light year. So above latter statement does not refer to Rocket frame of reference (if your first SR case and your next GR case is equivalent, I am not sure about it) but which frame of reference ? As a possible way do local frames of buoy at rest coincide to form a global frame of reference under "real" (R is not zero somewhere) gravity of some celestrial body as well as they do in Rindler metric space ?

 

[PeterDonis]

Is the latter just an initial condition or surely kept during their free-falling motion in some frame of reference ?

It's true during the entire scenario in the LIF described in previous posts.

Your very first case of SR in post #1 tells that rocket passengers observe by Lorentz contraciton the buoys are thin film shape and the distance between the buoys are much shorter than 1 milliion light year.

In the inertial frame in which the rocket is momentarily at rest at the start of the scenario, this is true.

do local frames of buoy at rest coincide to form a global frame of reference

Within the LIF described in previous posts, there is only one "local frame of buoys at rest". But this frame is not the same as any familiar global frame in Schwarzschild spacetime (although, as my previous posts make clear, it can be viewed as a small patch of one).

under "real" (R is not zero somewhere) gravity of some celestrial body

Previous posts have already described the global spacetime (maximally extended Schwarzschild spacetime). Whether that qualifies as a "celestial body" is beyond the scope of this thread.

 

[sweet springs]

Within the LIF described in previous posts, there is only one "local frame of buoys at rest".

The LIF (LOCAL Inertial Frame, I take) ranging at least a few million light-year is an amazing thing. "For each time of the free fall process there exists an LIF where all the buoys are at rest and the distance between them is a million light-year" is also. Thanks.