How Much Thrust is Needed to Hover Above a Black Hole?

[greswd]

Let's say that I'm hovering using a jetpack at a fixed distance above the event horizon of a black hole. And that I'm perfectly well shielded from any radiation and whatnot.

Its a supermassive black hole with a super large Schwarzschild radius, so tidal forces are minimal.

Approximately how much thrust does the jetpack need to provide in order for me to hover?

 

[Orodruin]

Depends on how close you are to the event horizon. You can find out by computing the proper acceleration of a worldline with fixed spatial coordinates in the Schwarzschild metric.

 

[greswd]

Depends on how close you are to the event horizon. You can find out by computing the proper acceleration of a worldline with fixed spatial coordinates in the Schwarzschild metric.

Is there a simple, approximate formula for the thrust needed? In terms of the mass of the black hole and the height above the event horizon.

 

[Ibix]

There's an exact solution in terms of the Schwarzschild r coordinate. But that's not really distance from the horizon, which is a fairly arbitrary concept. It depends what coordinate system you use (similar to how distance depends on your reference frame in special relativity), and there's no particular significance to the choice. You can't even bounce a radar signal off the horizon in order to get some kind of frame-independent measure.

What is the context of your question? Why are you asking?

 

[greswd]

There's an exact solution in terms of the Schwarzschild r coordinate. But that's not really distance from the horizon, which is a fairly arbitrary concept. It depends what coordinate system you use (similar to how distance depends on your reference frame in special relativity), and there's no particular significance to the choice. You can't even bounce a radar signal off the horizon in order to get some kind of frame-independent measure.

What is the context of your question? Why are you asking?

let's say everything's at rest with respect to each other.

just curious as to whether one could hover above an event horizon.

 

[Dale]

just curious as to whether one could hover above an event horizon

Yes

 

[Ibix]

let's say everything's at rest with respect to each other.

just curious as to whether one could hover above an event horizon.

As Dale says, yes, at least in theory. The thrust requirements are fairly splatter-inducing. The relevant formula for acceleration needed to hover is $$a=\frac 1{\sqrt{1-R_S/r}}\frac{GM}{r^2}$$where $r$ is the Schwarzschild radial coordinate and $R_S=2GM/c^2$ is the Schwarzschild radius. Careful about $r$! Lower $r$ means closer to the horizon, but it isn't a straightforward measure of distance as the $r$ in Newtonian gravity is.

That said, at large $r$ the square root tends to one and the result is just Newton. This is where you can ignore the little bit in the middle where non-Euclidean effects are significant. Note also that the acceleration needed to hover goes to infinity at the Schwarzschild radius. You're stuck at that point.

 

[greswd]

Careful about $r$! Lower $r$ means closer to the horizon, but it isn't a straightforward measure of distance as the $r$ in Newtonian gravity is.

In this simplified scenario, are they equivalent?

 

[Dale]

No. It isn't a matter of simplification. They are defined differently.

 

[greswd]

No. It isn't a matter of simplification. They are defined differently.

I mean, in the case of Ibix's formula, can we take r to be the sum of the height at which one hovers above the event horizon and the Schwarzschild radius?

 

[Dale]

No. In that equation r is defined such that the surface area of the sphere at that r is $A=4\pi r^2$

 

[Ibix]

No. If you build a (massless) sphere around the black hole and its total area is A, then an observer standing on the sphere would use $r=\sqrt {A/4\pi}$ to calculate the gravitational acceleration they feel. If spacetime were flat then $r$ would be the radius, yes. But spacetime isn't flat and there isn't a unique way to define the distance to the event horizon.

 

[pervect]

Let's say that I'm hovering using a jetpack at a fixed distance above the event horizon of a black hole. And that I'm perfectly well shielded from any radiation and whatnot.

Its a supermassive black hole with a super large Schwarzschild radius, so tidal forces are minimal.

Approximately how much thrust does the jetpack need to provide in order for me to hover?

It depends on how you define distance, really. You can't use radar methods, for obvious reasons.

The formula for the proper acceleration required to hover at a Schwarzschild r coordinate r is

$$a = \frac{GM}{r^2\sqrt{1-\frac{r_s}{r}}} = \frac {c^2 r_s}{2 r^2 \sqrt{1-\frac{r_s}{r}}}$$

The later expression comes from utilizing the realtionship $r_s = 2GM/c^2$ to express GM in terms of $r_s$.

If we measure distances by rulers held by static observers (rulers held by observers maintaining a constant Schwarzschild r-coordinate), the distance from $r = r_s + \epsilon$ to the horizon by this definition would be:

$$d = \int_{r_s}^{r_s+\epsilon} \frac{dr}{\sqrt{ 1-r_s/r} }$$

If we let $r_s = 1$, so that my computer algebra package doesn't throw fits, I get

$$a \approx \frac{c^2}{2 \sqrt{\epsilon}}$$
$$ d \approx 2 \sqrt{\epsilon}$$

which means that the product a*d $\approx c^2$. Thus we conclude that $a \approx c^2/d$.

This is the same answer I got by more intuitive method, in which we approximate the black hole metric by the Rindler metric, and the black hole horizon by the Rindler horizon, and assume that the difference in Rindler coordinates serves as a reasonable definition of distance.

If you don't like the intuitive way, by all means use the more formal way, though it's much messier. The biggest issue is evaluating the messy integral for distance, (and taking the limit to remove the infinities), that's the most likely spot for an error.

 

[timmdeeg]

If of interest, the radial proper distance is given here in equation (9.4.21).

 

[pervect]

If of interest, the radial proper distance is given here in equation (9.4.21).

Thanks, that is helpful. Using this textbook formula, we can drop the assumption that $r_s=1$. With a fair amount of computer algebra, I get:

$$a \approx \frac{c^2}{2\,r_s\, \sqrt{\frac{\epsilon}{r_s}}} \quad d \approx 2 r_s \sqrt{\frac{\epsilon}{r_s}}$$

Here $r_s$ is the Schwarzschild radius of the black hole, and $r_s + \epsilon$ is the Schwarzschild coordinate of the static (hoovering) observer. We see that the product of the proper acceleration a, and the proper distance in the static frame, d, is approximately equal to c^2, independent of the value of $r_s$.

It's still messy, but most of the equations have a textbook reference now. I tracked down the reference to an expression for the proper acceleration in Wald's "General Realtivity", it's on pg 152. Wald's expression is in geometric units, so G and c are assumed to be equal to 1.

It is easy to check that static observers in the Schwarzschild spacetime must undergo a peroper acceleration (in order to "stand still" in the "gravitational field") given by a = $(1-2M/r)^{-1/2}\,M/r^2$, which diverges as r -> 2M.

Google books can poentitally find this, a search for "wald general realtivity acceleration of static observer" worked for me.

 

[pervect]

I thought of a way to mathematically illustrate the close tie between the Scwarzschild metric and the Rindler metric near the event horizoin based on the above.

Consider the transformation

$$r = r_s + \frac{R^2}{4r_s} $$

or, equivalently

$$R = 2 \sqrt{r_s \left(r-r_s \right) } $$

Then we can write

$$dr = \frac{R}{2 \, r_s}\,dR$$

In geometric units, with c=1, the line element for the Schwarzschild metric in the r-t plane is:

$$ds^2 = -\left(1-\frac{r_s}{r} \right) dt^2 + \frac{dr^2}{1-\frac{r_s}{r}}$$

Using the above, we can write the line element in terms of dt and dR as:

$$ds^2 = -\left(1-\frac{r_s}{r} \right) dt^2 + \frac{R^2}{4\,r_s^2 \left(1-\frac{r_s}{r} \right) } \, dR^2 $$

Substituting the expression for r in terms of R and simplifying, we get

$$ds^2 = -\frac{R^2}{R^2 + 4\,r_s^2}\,dt^2 + \left(1 + \frac{R^2}{4\,r_s^2} \right) dR^2$$

When $R << r_s$ this is approximately equal to

$$ds^2 = -\frac{1}{4\,r_s^2} \,R^2 \, dt^2 + dR^2$$

which is just one form of the Rindler metric $-a^2 R^2 dt^2 + dR^2$ with $a = 1 / 2\,r_s$. Note that a is just a parameter of the Rindler metric, it's not equal to the proper acceleration.

 

[timmdeeg]

I thought of a way to mathematically illustrate the close tie between the Scwarzschild metric and the Rindler metric near the event horizoin based on the above.

I wonder if "the close tie" means that the Equivalence Principle holds if we consider two observer in a uniformly accelerating rocket. Could they (one at the bottom, the other at the top) distinguish experimentally whether they are in a rocket or are shell observers close to a black hole?
The acceleration at the bottom is larger than that at the top. Having these values together with the proper distance between them, could they calculate pairs of r-coordinates and values for M which fit to their observation (acceleration, proper distance) or is there only one possibility for r and M?

 

[Nugatory]

I wonder if "the close tie" means that the Equivalence Principle holds if we consider two observer in a uniformly accelerating rocket. Could they (one at the bottom, the other at the top) distinguish experimentally whether they are in a rocket or are shell observers close to a black hole?

As an aside, that could read "close to any spherically symmetric mass". The Schwarzschild solution describes the vacuum region outside of things that aren't black holes as well as outside the event horizon of black holes.

 

[timmdeeg]

The Schwarzschild solution describes the vacuum region outside of things that aren't black holes as well as outside the event horizon of black holes.

My question is related to the Schwarzschild solutuion.

 

[Nugatory]

My question is related to the Schwarzschild solutuion.

Right - so whether there's a black hole involved or not is irrelevant. We have our experimenters in a windowless compartment, and your question is whether they can tell whether they're in an accelerating spaceship or sitting at a constant Schwarzschild $r$ coordinate near some gravitating mass (perhaps even parked on the surface of the earth). And the answer is that it depends on whether the compartment is large enough for tidal effects to be detectable - no matter how sensitive their instruments, we can always make the compartment small enough that they cannot tell the difference.

 

[timmdeeg]

We have our experimenters in a windowless compartment, and your question is whether they can tell whether they're in an accelerating spaceship or sitting at a constant Schwarzschild $r$ coordinate near some gravitating mass.

No, I will try to reformulate my question.
The rocket is uniformly accelerating in flat space. The astronauts measure the acceleration at the top and at the bottom (which is bigger) and the ruler length (which is their proper distance) of the rocket. This situation should correspond indistinguishable to two shell observers in Schwarzschild spacetime who are separated by the same proper distance and are measuring the same acceleration each.

Now, how would the astronauts calculate the corresponding $r-$coordinates and $M$? The radial acceleration to be stationary depends on $r$ and $M$. They don't know $M$, but instead their proper distance. My guess is they use the spacelike form of the metric to obtain $ds^2=(1-2M/r)^{-1/2}dr$ and the proper distance after integration. And that there is only one set of $r-$coordinates and $M$ which fits to what they have measured in the rocket.

 

[PeterDonis]

The astronauts measure the acceleration at the top and at the bottom (which is bigger) and the ruler length (which is their proper distance) of the rocket. This situation should correspond indistinguishable to two shell observers in Schwarzschild spacetime who are separated by the same proper distance and are measuring the same acceleration each.

In general, no, these two situations are not the same, because spacetime is flat in one and curved in the other. That means the relationship between the ruler length between the two observers and the accelerations they measure is different for the two cases.

Let's look at the math. We have three known quantities: two proper accelerations $a_1$ and $a_2$, with $a_1 > a_2$, and a proper distance $d$. Now, for the two cases, how are these quantities related?

First the flat spacetime case. Here we have, in Rindler coordinates, and in units where $c = 1$, $a_1 = 1 / x_1$ and $a_2 = 1 / x_2$. Since space is flat in Rindler coordinates, we have $d = x_2 - x_1$. So we have the following relationship between the known quantities:

$$
d = \frac{1}{a_2} - \frac{1}{a_1}
$$

Now the curved spacetime case. Here we have , in Schwarzschild coordinates:

$$
a_1 = \frac{M}{r_1^2 \sqrt{1 - 2M / r_1}}
$$
$$
a_2 = \frac{M}{r_2^2 \sqrt{1 - 2M / r_2}}
$$
$$
d = \int_{r_1}^{r_2} \frac{1}{\sqrt{1 - 2M / r}} dr
$$

The integral looks messy, but it actually has an exact solution:

$$
d = \sqrt{r_2 (r_2 - 2M)} - \sqrt{r_1 (r_1 - 2M)} + 2M \ln \frac{(\sqrt{r_2} + \sqrt{r_2 - 2M})}{(\sqrt{r_1} + \sqrt{r_1 - 2M})}
$$

We can rewrite the accelerations in a way that will help:

$$
a_1 = \frac{M}{r_1 \sqrt{r_1 (r_1 - 2M)}}
$$
$$
a_2 = \frac{M}{r_2 \sqrt{r_2 (r_2 - 2M)}}
$$

This let's us rewrite the formula for $d$ as follows:

$$
d = \frac{M}{r_2 a_2} - \frac{M}{r_1 a_1} + 2M \ln \left[ \frac{r_1 a_1}{r_2 a_2} \frac{ \frac{1}{\sqrt{r_2 - 2M}} + \frac{1}{\sqrt{r_2}} }{ \frac{1}{\sqrt{r_1 - 2M}} + \frac{1}{\sqrt{r_1}} } \right]
$$

As you can see, the two relationships, flat spacetime vs. curved spacetime, are quite different. It should also be evident that there is no way to extract a value for $M$ or $r_1$ or $r_2$ from the flat spacetime formulas.

 

[PAllen]

No, I will try to reformulate my question.
The rocket is uniformly accelerating in flat space. The astronauts measure the acceleration at the top and at the bottom (which is bigger) and the ruler length (which is their proper distance) of the rocket. This situation should correspond indistinguishable to two shell observers in Schwarzschild spacetime who are separated by the same proper distance and are measuring the same acceleration each.

Now, how would the astronauts calculate the corresponding $r-$coordinates and $M$? The radial acceleration to be stationary depends on $r$ and $M$. They don't know $M$, but instead their proper distance. My guess is they use the spacelike form of the metric to obtain $ds^2=(1-2M/r)^{-1/2}dr$ and the proper distance after integration. And that there is only one set of $r-$coordinates and $M$ which fits to what they have measured in the rocket.

I think you misunderstand the impact of M in the Schwarzschild (hereafter SC) case. It is responsible for two tidal effects: the difference in measured acceleration between top and bottom of ship will differ to second order from the Rindler expected result; and there will be second order non uniformity in direction of acceleration. Note that for any M, proper acceleration needed to hover can vary from zero at infinity to infinite on approach to horizon. All you can tell about M is a minimum value such that the second order tidal effects are undetectable to your available precision at the measured base acceleration and measured distance. Any M larger than this minimum would have a hover radius matching your observations.

[edit: I see I cross posted with Peter]

 

[timmdeeg]

In general, no, these two situations are not the same, because spacetime is flat in one and curved in the other. That means the relationship between the ruler length between the two observers and the accelerations they measure is different for the two cases.

Let's look at the math. We have three known quantities: two proper accelerations $a_1$ and $a_2$, with $a_1 > a_2$, and a proper distance $d$. Now, for the two cases, how are these quantities related?

First the flat spacetime case. Here we have, in Rindler coordinates, and in units where $c = 1$, $a_1 = 1 / x_1$ and $a_2 = 1 / x_2$. Since space is flat in Rindler coordinates, we have $d = x_2 - x_1$. So we have the following relationship between the known quantities:

$$
d = \frac{1}{a_2} - \frac{1}{a_1}
$$

Now the curved spacetime case. Here we have , in Schwarzschild coordinates:

$$
a_1 = \frac{M}{r_1^2 \sqrt{1 - 2M / r_1}}
$$
$$
a_2 = \frac{M}{r_2^2 \sqrt{1 - 2M / r_2}}
$$
$$
d = \int_{r_1}^{r_2} \frac{1}{\sqrt{1 - 2M / r}} dr
$$

The integral looks messy, but it actually has an exact solution:

$$
d = \sqrt{r_2 (r_2 - 2M)} - \sqrt{r_1 (r_1 - 2M)} + 2M \ln \frac{(\sqrt{r_2} + \sqrt{r_2 - 2M})}{(\sqrt{r_1} + \sqrt{r_1 - 2M})}
$$

We can rewrite the accelerations in a way that will help:

$$
a_1 = \frac{M}{r_1 \sqrt{r_1 (r_1 - 2M)}}
$$
$$
a_2 = \frac{M}{r_2 \sqrt{r_2 (r_2 - 2M)}}
$$

This let's us rewrite the formula for $d$ as follows:

$$
d = \frac{M}{r_2 a_2} - \frac{M}{r_1 a_1} + 2M \ln \left[ \frac{r_1 a_1}{r_2 a_2} \frac{ \frac{1}{\sqrt{r_2 - 2M}} + \frac{1}{\sqrt{r_2}} }{ \frac{1}{\sqrt{r_1 - 2M}} + \frac{1}{\sqrt{r_1}} } \right]
$$

As you can see, the two relationships, flat spacetime vs. curved spacetime, are quite different. It should also be evident that there is no way to extract a value for $M$ or $r_1$ or $r_2$ from the flat spacetime formulas.

Thanks for this clear explanation and for showing the math. So the two situations are not more than just similar. Choosing the same proper distance wouldn't yield the same accelerations.

 

[pervect]

Thanks for this clear explanation and for showing the math. So the two situations are not more than just similar. Choosing the same proper distance wouldn't yield the same accelerations.

In general, no. If you take $r_1 = r_s + \epsilon_1$ and $r_2 = r_s + \epsilon_2$, with $\epsilon_1 << r_s$ and $\epsilon_2 <<r_s$, you can do a series expansion in terms of $\epsilon / r_s$.

Add: the square roots, which appear in both expressions for a and d, complicate things slightly. It might be clearer to view it as a taylor series expansion in terms of $\sqrt{\epsilon / r_s}$.

Here $r_s$ is the Scwarzschild radius, which is 2M in geometric units with G=c=1. In that particular case, you should find , as I did in #13, that for the Schwarzschild metric, to first order in $\epsilon_i / r_s$ we have $a_i \approx \frac{1}{\sqrt{8 \,\epsilon_i \, M}}$ and $d_i \approx \sqrt{8 \epsilon_i \, M}$ , making the product $a_i \, d_i \approx 1$ (in geometric units) or $c^2$ (in non-geometric units), which is exactly true in the Rindler metric and approximately true in the Schwarzschild metric.

 

[timmdeeg]

Here $r_s$ is the Scwarzschild radius, which is 2M in geometric units with G=c=1. In that particular case, you should find , as I did in #13, that for the Schwarzschild metric, to first order in $\epsilon_i / r_s$ we have $a_i \approx \frac{1}{\sqrt{8 \,\epsilon_i \, M}}$ and $d_i \approx \sqrt{8 \epsilon_i \, M}$ , making the product $a_i \, d_i \approx 1$ (in geometric units) or $c^2$ (in non-geometric units), which is exactly true in the Rindler metric and approximately true in the Schwarzschild metric.

Thanks, this is interesting. If I understand you correctly Rindler- and Schwarzschild-metric describe almost the same geometry very close to the the horizon.

 

[pervect]

Thanks, this is interesting. If I understand you correctly Rindler- and Schwarzschild-metric describe almost the same geometry very close to the the horizon.

Yes. Post #16 gives explicitly a diffeomorphism (a coordinate transformation) that relates the two geometrical descriptions. To summarize, the diffeomorphism

$$r = r_s + \frac{R^2}{4\,r_s}$$

yields a new line element in terms of dt and dR, which is diffeomorphic to the Schwarzschild metric (and hence a description of the "same geometry"). This new line element is:

$$ds^2 = -\frac{R^2}{R^2 + 4\,r_s^2}\,dt^2 + \left(1 + \frac{R^2}{4\,r_s^2} \right) dR^2$$

This description is just a re-labelling of the Schwarzschild geometry with coordinates t and R instead of Schwarzschild coordinates t and r. The relation to the RIndler metric involves ignoring terms proportional to $R^2 / 4\,r_s^2$ which are consider to be "small", as we are assuming R << $r_s$. Ignoring these small terms, (which can be attributed to the presence of tidal forces in the Schwarzschld metric that are not present in the Rindler metric) we wind up with the Rindler metric.

The coefficient of dR^2 obviously differs from unity only by a "small" additive term. The coefficeint of dt^2 can be series expanded as:

$$\frac{R^2}{R^2 + 4\,r_s^2} \approx \frac{R^2}{4\,r_s^2} \left(1 - \frac{R^2}{4\,r_s^2} + O(R/r_s)^4 \right)$$

Dropping all the "small" terms, we have for our line element:

$$ds^2 \approx -k^2 dt^2 + dR^2$$

which is just the Rindler metric. The value of the parameter k does not affect the proper distance, nor the magnitude of the proper acceleration, so it's of little importance to the analysis. The magnitude of the proper acceleration of a static observer in the Rindler metric is proportional to 1/R in geometric units, or c^2/R in standard units and does not depend on k.

 

[timmdeeg]

$$ds^2 = -\frac{R^2}{R^2 + 4\,r_s^2}\,dt^2 + \left(1 + \frac{R^2}{4\,r_s^2} \right) dR^2$$

This description is just a re-labelling of the Schwarzschild geometry with coordinates t and R instead of Schwarzschild coordinates t and r. The relation to the RIndler metric involves ignoring terms proportional to $R^2 / 4\,r_s^2$ which are consider to be "small", as we are assuming R << $r_s$. Ignoring these small terms, (which can be attributed to the presence of tidal forces in the Schwarzschld metric that are not present in the Rindler metric) we wind up with the Rindler metric.

Which makes clear that $M$ can't be extracted anymore (as PeterDonis has already mentioned).

The coefficient of dR^2 obviously differs from unity only by a "small" additive term. The coefficeint of dt^2 can be series expanded as:

$$\frac{R^2}{R^2 + 4\,r_s^2} \approx \frac{R^2}{4\,r_s^2} \left(1 - \frac{R^2}{4\,r_s^2} + O(R/r_s)^4 \right)$$

Dropping all the "small" terms, we have for our line element:

$$ds^2 \approx -k^2 dt^2 + dR^2$$

which is just the Rindler metric. The value of the parameter k does not affect the proper distance, nor the magnitude of the proper acceleration, so it's of little importance to the analysis. The magnitude of the proper acceleration of a static observer in the Rindler metric is proportional to 1/R in geometric units, or c^2/R in standard units and does not depend on k.

Excellent, I have never seen this transition between Schwarzschild- and Rindler-metric. Thanks for valuable insights.

 

[pervect]

Which makes clear that $M$ can't be extracted anymore (as PeterDonis has already mentioned).


Excellent, I have never seen this transition between Schwarzschild- and Rindler-metric. Thanks for valuable insights.

As far as I know, it's something I came up with. The motivation was to use a radial coordinate that has a more straightforwards relationship to the proper distance.

Another result that I wanted to add to my previous calculations If we calculate the exact value of proper acceleration for the transformed metric, $u^a \nabla_a u^a$, I get

$$a = \frac{16\,r_s^4}{R\left( 4\,r_s^2 + R^2\right)^2}$$

This confirms that the proper acceleration $a \approx 1/R$ when $R<<r_s$