General Relativity rocket puzzle

[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.